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I originally planned to continue the last post by discussing the math battle problem in Episode 1, but I think it’s mostly sufficiently explained in the episode already, though there are some logical holes: Sayuri’s first “proof” isn’t really a proof, and not really an argument either. It just sets up part of the geometry, which Nina extends to actually derive something. The whole thing also relies a bit too much on assumed accuracy of measurements, I think, which I don’t really buy, but once you get past the assumption, things are sound. If you’d like more explanation, however, I can write up something on it.

I also refer you to Prettycatchy’s video explanation:

Instead, in this post, I’ll discuss (and solve) the seven problems posed to Kazuki in Episode 2. In context, they’re presented in reverse order of difficulty, from university-level to lower-middle-school-level, and only the last problem is explained. I’ll go through all of them in order.

1. Prove that when 0 < x < π/4, the integral of cosine t from 0 to x is greater than twice the integral of sine t from 0 to x.

In mathematical notation, the inequality to be proven is

$\int_0^x \cos t\, dt > 2\int_0^x \sin t\, dt$

Each side is a definite integral, which gives you the area of the region underneath the curve given inside the integral (or the negative area when the curve is below the horizontal axis). In this case, we have t as the independent variable of the function inside the integral, instead of x as is usually the case (since x is the independent variable of the entire expression). The left-hand side is the area of the region bounded by the t-axis, the curve $y = \cos t$, the vertical line $t = 0$ and the vertical line $t = x$. The right-hand side is the same except the curve is $y = \sin t$ and you multiply the resulting area by 2. So the inequality states that the gray area is strictly larger than twice the orange area, as shown in the two graphs below:

You can evaluate a definite integral by finding the antiderivative and taking the difference of the antiderivative evaluated at the upper point and the antiderivative evaluated at the lower point. The antiderivative of $\cos t$ is $\sin t$, and the antiderivative of $\sin t$ is $-\cos t$. Doing this and simplifying results in an equivalent inequality:

$\left.\sin t\right|_0^x > 2\left.(-\cos t)\right|_0^x$
$(\sin x - \sin 0) > 2(-\cos x + \cos 0)$
$\sin x > 2(1-\cos x)$
$\frac{1}{2}\sin x + \cos x > 1$

So if we prove this is true for $0 < x < \pi/4$, we prove the original inequality.

Let $f(x) = \frac12\sin x + \cos x$.

You can plot $f(x)$ over that interval and see that it’s always greater than 1:

But arguing by graph isn’t a foolproof method (and isn’t feasible to do by hand), so let’s do this analytically. We can do a proof by contradiction: we suppose that there is a point q in $(0, \pi/4)$ such that the inequality does not hold (that is, $f(q) = \frac12\sin q + \cos q \leq 1$), and show that this leads to a contradiction.

We can now use the mean value and intermediate value theorems to prove that if such a point q exists, then there must be two distinct points in $(0, \pi/4)$ where the derivative (the slope of the line tangent to the curve at a given point) is 0, which we will show is a contradiction.

We start by checking the coordinates of the endpoints of the interval:

$f(0) = \frac12{\sin 0} + \cos 0 = 0 + 1 = 1$
$f(\pi/4) = \frac{1}{2}\sin\frac{\pi}{4} + \cos\frac{\pi}{4} = \frac{\sqrt{2}}{4} + \frac{\sqrt{2}}{2} = \frac{3\sqrt{2}}{4} > 1$

For the first endpoint, $f(0)$ is 1, but since 0 is not itself part of the open interval $(0, \pi/4)$, this is not a problem. The right endpoint is greater than one (this can be easily verified by hand by squaring both sides), so it is also consistent with the inequality. Next, we compute the derivative and evaluate it at the endpoints:

$f'(x) = \frac12{\cos x} - \sin x$
$f'(0) = \frac12{\cos 0} - \sin 0 = \frac12 - 0 = \frac12$
$f'(\pi/4) = \frac{1}{2}\cos \frac{\pi}{4} - \sin \frac{\pi}{4} = \frac{\sqrt{2}}{4} - \frac{\sqrt{2}}{2} = -\frac{\sqrt{2}}{4}$

So the derivative at 0 is positive, and the derivative at π/4 is negative.

If we connect the left endpoint $(0, 1)$ and $(q, f(q))$ by a straight line, the slope of the line is

$\frac{f(q)-1}{q}$

Since $q > 0$ and $f(q) \leq 1$, the slope is either zero (if $f(q) = 1$) or negative. By the mean value theorem, since f is continuous and differentiable (a property of all sinusoids), there must exist a point p in $(0, q)$ such that the derivative at p is equal to this slope. So $f'(p) \leq 0$.

Next, we connect $(q, f(q))$ to the right endpoint $(\pi/4, 3\sqrt{2}/4)$ by a straight line. The slope here is

$\frac{3\sqrt{2}/4 - f(q)}{\pi/4 -q}$

Since $3\sqrt{2}/4 > 1$ and $f(q) \leq 1$, their difference must be positive. $\pi/4 - q$ is also positive, so the slope here is positive. Again, by the mean value theorem, this means that there is a point r in $(q, \pi/4)$ such that $f'(r) > 0$.

We now have constraints on the derivative at the two endpoints as well as some points p and r within the interval.

$f'(0) = 1/2 > 0$
$f'(p) \leq 0$
$f'(r) > 0$
$f'(\pi/4) = -\frac{\sqrt{2}}{4} < 0$

Since $0 < p < r < \pi/4$, this means the value of $f'$ goes from positive to zero/negative to positive to negative.

The intermediate value theorem tells us that since $f'$ is continuous, for any closed interval $[a, b]$, for every d in $[f(a), f(b)]$ (assuming $f(a) \leq f(b)$; switch the two otherwise), there exists a point c in $[a, b]$ where $f'(c) = d$. This means that any interval $[a, b]$ where a is positive and b is zero/negative, or vice versa, must contain a point c where $f'(c) = 0$.

So there must be a point u in $[0, p]$ such that $f'(u) = 0$, a point v in $[p, r]$ such that $f'(v) = 0$, and a point w in $[r, \pi/4]$ such that $f'(w) = 0$. Since these are closed intervals with overlapping endpoints, it is possible for u and v to be identical. It is not possible for u and w to be identical, however, since p is strictly less than r.

This shows that there are two distinct points u and w in $[0, \pi/4]$ whose derivative is 0.

We will now show that this contradicts the properties of the particular f we have. Let us solve for any points c in the interval $(0, \pi/4)$ with a derivative of 0:

$f'(c) = \frac12{\cos c} - \sin c = 0$
$\frac12\cos c = \sin c$
$\frac12 = \frac{\sin c}{\cos c} = \tan c$

The tangent function is one-to-one in the interval $(-\pi/2, \pi/2)$, so there is at most one c in $(0, \pi/4)$ such that $f'(c) = 0$. (We can can quickly prove the one-to-one property analytically. First, the tangent function defined for all points in this interval, since the cosine of all points in this range is nonzero. Suppose it’s not one-to-one; then there must be two distinct points m and n in $(-\pi/2, \pi/2)$ such that $\tan m = \tan n$. This means the slope of the line connecting the two points is 0, so by the mean value theorem, there is a k in $(m, n)$ such that the derivative of the tangent function is 0. The derivative is $\sec^2 x$, which is always positive, so there is no such point.)

Our assumption that there is a point q in $(0, \pi/4)$ where $f(q) \leq 1$ leads to the conclusion that there are two distinct points meeting this condition, so there is a contradiction. Therefore, the assumption is not valid, so there is no such q.

So for all points x in $(0, \pi/4)$, $f(x) > 1$. This proves the desired inequality.

2. Find the volume V of a regular octahedron with side length a.

A regular octahedron is a polyhedron with 8 faces (each one an equilateral triangle) and 12 sides of equal length. We can compute the volume by splitting the octahedron into two pyramids of equal volume, finding the volume of one of the pyramids, and multiplying by 2.

Let’s look at one of the pyramids. We can compute the volume $V = \frac{1}{3}Bh$, where B is the area of the base, and h is the height. In the figure below, each of the edges is of length a, the height is h, and the diagonal of the square base is d:

The area of the square base is $B = a^2$. To determine the height h, we can take a cross-section along one of the diagonals. This yields a triangle formed from two of the sides of the octahedron and the diagonal of the square:

The diagonal has length $d = a\sqrt{2}$. Since two of the sides have the same length, we can divide the isosceles triangle into two right triangles. This yields a triangle with one of the octahedron’s sides as the hypotenuse and two legs: one is half the diagonal of the base square, and the other is the height of the pyramid.

We can use the Pythagorean theorem to relate these:

$a^2 = h^2 + \left(a\frac{\sqrt{2}}{2}\right)^2$
$a^2 = h^2 + a^2/2$
$a^2/2 = h^2$
$h = \frac{a}{\sqrt{2}}$

Multiplying to get the volume of the entire octahedron, we have:

$V = 2[\frac{1}{3}Bh] = \frac{2}{3}(a^2)(\frac{a}{\sqrt{2}})$
$V = \frac{a^3\sqrt{2}}{3}$

3. Find a vector c of length 1 orthogonal to both vector a = (5, 7, 3) and vector b = (7, 6, 5).

One can compute the cross product of two linearly independent vectors to get a vector orthogonal to both.

There are several ways to compute the cross product. One is to make a 3×3 matrix out of the standard basis vectors i, j, and k in the first row and the two vectors in the second and third rows, and then compute the determinant of this matrix. In this case we have

$\mathbf{a} \times \mathbf{b} = \left|\begin{array}{ccc}\mathbf{i}&\mathbf{j}&\mathbf{k} \\ 5 & 7 & 3 \\ 7 & 6 & 5\end{array}\right|$

We can break this into three smaller determinants via Laplace expansion:

$\mathbf{a} \times \mathbf{b} = \left|\begin{array}{ccc}\mathbf{i}&\mathbf{j}&\mathbf{k} \\ 5 & 7 & 3 \\ 7 & 6 & 5\end{array}\right| = \left|\begin{array}{cc}7 & 3 \\ 6 & 5\end{array}\right|\mathbf{i} - \left|\begin{array}{cc}5 & 3 \\ 7 & 5\end{array}\right|\mathbf{j} + \left|\begin{array}{cc}5 & 7 \\ 7 & 6\end{array}\right|\mathbf{k}$

Then we can easily compute each of the three determinants by multiplying the main diagonal elements and subtracting the product of the other two elements:

$\mathbf{a} \times \mathbf{b} = (7\cdot 5 - 3\cdot 6)\mathbf{i} - (5\cdot 5 - 3\cdot 7)\mathbf{j} + (5\cdot 6 - 7\cdot 7)\mathbf{k}$
$\mathbf{a} \times \mathbf{b} = 17\mathbf{i} - 4\mathbf{j} - 19\mathbf{k}$

So the cross product is $\mathbf{a} \times \mathbf{b} = (17, -4, -19)$.

But we’re not done yet. The problem wants an orthogonal vector of length one, so we have to normalize the cross product. We can compute the magnitude of the cross product by squaring each component, adding, and taking the square root:

$|\mathbf{a} \times \mathbf{b}| = \sqrt{17^2 + (-4)^2 + (-19)^2} = \sqrt{666} = 3\sqrt{74}$

We can divide each of components of the cross product by the magnitude to produce a vector pointing in the same direction with length 1:

$\frac{1}{3\sqrt{74}}(17, -4, -19)$

There are two possible answers, actually. The vector can point in the same direction as the cross product, as given above, or it can point the opposite way. Either of these is acceptable:

$\mathbf{c} = \pm\frac{1}{3\sqrt{74}}(17, -4, -19)$

4. Find the radius of the inscribed circle in triangle ABC, where AB = 5, BC = 7, and CA = 3.

The circle inscribed in a triangle is the circle that is tangent to all three sides of the triangle:

Here we have triangle $\triangle ABC$ with the inscribed circle at center O with radius r. The triangle can be partitioned into three triangles $\triangle ABO$, $\triangle BCO$, and $\triangle ACO$.

The area of each of the triangles is equal to half of the radius r times the length of the edge tangent to the circle.

The area of the entire circle can be found by using Heron’s formula:

$T = \sqrt{s(s-a)(s-b)(s-c)}$

where s is the semiperimeter:

$s = (a + b + c)/2$

For the triangle given in the problem, we have

$s = (5 + 7 + 3)/2 = 15/2$
$T = \sqrt{\frac{15}{2}(\frac{15}{2}-5)(\frac{15}{2}-7)(\frac{15}{2}-3)} = \sqrt{(\frac{15}{2})(\frac{5}{2})(\frac12)(\frac92)} = \frac{15\sqrt{3}}{4}$

The total area is equal to the sum of the areas of the three smaller triangles:

$T = T_{\triangle ABO} + T_{\triangle BCO} + T_{\triangle ACO} = \frac12(5r) + \frac12(7r) + \frac12(3r) = \frac{15r}{2}$

Set these equal to each other and solve for r:

$T = \frac{15\sqrt{3}}{4} = \frac{15r}{2}$
$15\sqrt{3} = 30r$
$r = \frac{\sqrt{3}}2$

5. Find the smallest positive integer n such that 765n is a square number.

A square number is an integer whose square root is also an integer. This requires that each prime factor of the number have an even multiplicity, so that you can divide the factors into two identical groups, each of them multiplying to form the square root.

So for 765n to be square, each of the factors in its prime factorization must appear an even number of times. Since we don’t know what n is yet, we can do it just for 765 (factorization in general is a hard problem, but this one can be done by hand easily, once you observe that it’s a multiple of both 3 and 5):

$765 = 3 \times 3 \times 5 \times 17$

3 already appears twice, but 5 and 17 each appear once. Therefore n needs to have an odd number of occurrences of 5 and 17 for 765n to be square. Since we want the smallest positive integer that works, we can just multiply 5 and 17: $n = 85$

6. Solve the system of equations 5x + 7y = 3, 7x – 6y = 5.

Sayumi’s favorite pastime, solving systems of equations!

This can be done using several methods, but some form of Gaussian elimination is probably the most fun.

To do so, we take the system of equations and put the coefficients in an augmented matrix, with the right-hand values in a column on the right:

$\left(\begin{array}{cc|c}5 & 7 & 3 \\ 7 & -6 & 5\end{array}\right)$

We divide the first row by the leftmost element, 5, to get a 1 in the upper left:

$\left(\begin{array}{cc|c}1 & 7/5 & 3/5 \\ 7 & -6 & 5\end{array}\right)$

Then we subtract 7 times the resulting first row from the second row, to put a 0 in the lower left corner:

$\left(\begin{array}{cc|c}1 & 7/5 & 3/5 \\ 0 & -79/5 & 4/5\end{array}\right)$

Now we divide the second row by -79/5 to get a 1 there as well:

$\left(\begin{array}{cc|c}1 & 7/5 & 3/5 \\ 0 & 1 & -4/79\end{array}\right)$

Finally we subtract 7/5 times the second row from the first row to get an identity matrix:

$\left(\begin{array}{cc|c}1 & 0 & 53/79 \\ 0 & 1 & -4/79\end{array}\right)$

This translates back into $x = 53/79, y = -4/79$, which is our solution.

7. The time is 3:00. Between 3:00 and 4:00, what hour, minute, and second will the minute hand and the hour hand overlap exactly?

This was sufficiently explained in the episode, I think, so I won’t discuss it in detail. The idea is to determine the angular velocity of the minute hand and the hour hand (6°/min and 0.5°/min, respectively), find their relative difference (5.5°/min), and divide the initial angular displacement (90°) by this relative velocity to get the time it takes for the hands to overlap, which is 16 and 4/11 minutes. 4/11 of 60 seconds is 21.8181… seconds, or approximately 22 seconds. So the time would be 3:16:22.

Prettycatchy has another video explanation here:

With those 7 problems solved, I look forward to seeing what awaits us in the next episode!! (^_^)

I’ve been a bit slow in responding to the new math drama Suugaku♥Joshi Gakuen starring a whole bunch of H!P members, but now that the first episode is out, here’s a post.

H!P math drama??!!! I’ve been waiting for this my whole life! (Never mind that my life predates H!P.) And of course I’m happy to see Sayu starring as a math geek, given the interest she’s expressed in the subject.

I don’t have time right now to do a full review of the first episode, but I’ve found it quite awesome so far, with respect to the geekiness, though the number of typos I’ve spotted have been quite a distraction. The series has apparently been developed in cooperation with the Mathematical Olympiad Foundation of Japan, so the math is essentially correct, as far as I can tell so far (I haven’t dissected most of it yet), though the editing work seems to have been rather sloppy.

Also, I’m not sure who’s planning to fansub this drama, but I would like to volunteer as a math consultant. So if you are working on this, please sign me up. I think the kinds of mistakes that have slipped through the editing can also pop up in translation, and I can help look over the translations and correct any possible math errors.

Here are some typos (and non-typos) I’ve found:

The first equation is given as $\lim_{x\to a} f(x)=a$. This says that the limit of f(x) as x approaches a is equal to a. As written, this is only correct for all values of a if f is the identity function. This should probably be $\lim_{x\to a} f(x)=f(a)$, which is true for all continuous functions.

The second equation says $\frac{d}{dx} = a^x = a^x \log(a)$. There is an extra equal sign here. It should be $\frac{d}{dx} a^x = a^x \log(a)$. This is the derivative of ax respect to x.

The third equation is fine as is, and expresses linearity of integration.

This clock is just awesome. It has expressions evaluating to or otherwise indicating each of the integers from 1 to 12, which we would see on a normal clock.

1. $\tan(45^{\circ})$ — the tangent of 45°, equal to 1.
2. $\sum_{i=0}^{\infty} 1/2$ — this is a typo; should be $\sum_{i=0}^{\infty} 1/2^i$, an infinite series converging to 2.
3. &#x33i — this is another typo; should be &#x33;, an HTML code for the character 3.
4. $2^{-1}\mbox{ }(\mbox{mod }7)$ — the modular multiplicative inverse of 2 (mod 7); 4 and 2 multiply to 8, which is congruent to 1 (mod 7), so 4 is an inverse of 2 (though not the only inverse; 11 is the other integer on the clock that satisfies this property, which makes this a flawed clock).
5. $X^2=3^2+4^2$ — solve for X, which can be either 5 or -5.
6. $3!$ — the factorial of 3, which is 6.
7. $6.\overline{9}$ — this repeating decimal is equal to 7.
8. $\sqrt{64}$ — the square root of 64 is 8.
9. $3(\pi -.14)$ — since π is irrational, this is actually 9.004777960769…, but it’s approximately 9.
10. $-8 = 2-X$ — solve for X, which is 10.
11. 0x0Bhexadecimal for 11, with the standard hexadecimal prefix of “0x” used in Unix-like shells and C.
12. $\sqrt[3]{1728}$ — the cube root of 1728 is 12.

This is given as $v = \sqrt{2}gh$, but the radical should extend over the entire right-hand side: $v = \sqrt{2gh}$.

In context, Nina (Reina’s character) swings on a rope into the classroom. This expresses her velocity at the lowest point of her trajectory.

Nina starts out with a gravitational potential energy of mgh (m is mass, g is gravitational acceleration, and h is height, measured relative to the lowest point of her trajectory) and no kinetic energy prior to swinging on the rope. At the lowest point, she has a kinetic energy of $\frac{1}{2} mv^2$ (m is mass and v is velocity) and no potential energy. Due to conservation of energy, these two are equal:

$\frac{1}{2}mv^2 = mgh$
$\frac{1}{2}v^2 = gh$
$v^2 = 2gh$
$v = \sqrt{2gh}$

It seems Tsunku♂ has joined the eminent ranks of Napier, Fermat, Gauss, Cauchy, Lagrange, et al…. :o

#### Introduction: Hello! Project’s Central Mystery

For ages, it has defied explanation, piggybacking on one group name after another. It waits, announcing its presence in a subtle yet forceful fashion, catching the eye with its unexpected existence yet never explaining its purpose. This elusive typographical flourish is none other than the ideographic full stop (。) at the end of several Hello! Project group names: モーニング娘。 (Morning Musume。), ココナッツ娘。 (Coconuts Musume。), カントリー娘。 (Country Musume。), アイスクリー娘。 (Ice Creamusume。), ミニモニ。 (Mini Moni。), エコモニ。 (Eco Moni。)—but interestingly, not プッチモニ (Pucchi Moni), a puzzle that we may be able to solve shortly. Unlike the names that precede it, this hardy symbol has proven immune to transformations like romanization: while “モーニング娘” can be written “MORNING MUSUME” in official documents (often, but not always, in all caps), the stop itself is never replaced by its Latin counterpart (.). There is something intrinsic to the ideographic full stop, then, that would be lost in translation. But what is it? As a distinguishing feature of the flagship group’s name since the very beginning, this little circular symbol lies at the heart of the Hello! Project enterprise, and understanding its true meaning may very well be the key to unlocking H!P’s greatest mysteries.

Perhaps it is a simple matter of topology. The ideographic and Latin full stops can be viewed as essentially a zero-dimensional point and a one-dimensional circle, respectively. Since they are represented in two dimensions on the page, however, we can also consider them to be projections of two-dimensional objects in a three-dimensional space (the page itself is a two-dimensional Euclidean plane, so we must go to higher dimensions to have other types of two-dimensional objects). From this perspective, the ideographic full stop is homeomorphic to a 1-torus, while its Latin counterpart is homeomorphic to a sphere. These are distinct homeomorphism classes since there is no continuous deformation from one to the other, just as there is no way to continuously deform a donut into a ball.

But of course this does not explain the relevance of the full stop to Hello! Project. One important characteristic of the full stop is that it appears in the names of groups, so it is inextricably linked to H!P’s concept of a group. We may therefore be able to come to a fuller understanding of H!P by studying the structure of its constituent groups.

The term group also has a mathematical meaning, of course, and while on the surface the two meanings may seem disparate and unrelated, an in-depth analysis from a group-theoretic standpoint does yield a surprising number of coincidences, suggesting that H!P groups do have inherent mathematical structure worth taking a closer look at, whether or not this is the intention of producer Tsunku♂ et al. We may be able to gain insight into the nature of Hello! Project by viewing its groups in mathematical terms, and using group theory as a tool to aid our understanding.

#### The Basics of Groups

(See also Wikipedia’s article on elementary group theory, which may be a more useful intro to group theory.)

The mathematical discipline of group theory, a part of abstract algebra, examines the algebraic structures called groups. Groups are a central concept in algebra and other areas of mathematics, and have also found applications to physics, chemistry, computer science, and linguistics, among other fields.

Definition: A group consists of a set of objects along with a binary operator that can be applied to any two objects in the set. We can denote the set as G and the operator as ×, which for explanatory purposes we’ll call “multiplication” in a general sense, though numerical multiplication is only one example of a binary operator. Furthermore, the four group axioms must hold:

• Closure: For all a, b in G, a × b is also in G. (The result of multiplying any two elements in the set produces an element in the set.)
• Associativity: For all a, b, c in G, (a × b) × c = a × (b × c). (When there is more than one multiplication, it doesn’t matter which pair gets multiplied first—the result is the same.)
• Identity: There exists an element e in G such that for all a in G, a × e = e × a = a. (There is an element that can be multiplied with any element, in either order, without changing it.)
• Inverse: For each a in G, there exists an element b in G such that a × b = b × a = e. (Every element can be multiplied with some other element, in either order, to produce the identity.)

If a set and a binary operator defined on the set meet these conditions, then they form a group. We can specify further restrictions to define particular types of groups. For example, we can add commutativity to define the notion of an abelian group:

• Commutativity: For all a, b in G, a × b = b × a. (The order in which you multiply elements doesn’t matter.)

One example of a group (which also happens to be abelian) is the set of real numbers, R, under addition. (Note that here the group operator is addition, not multiplication). We can easily verify that the four group axioms, as well as commutativity, hold (this is not a proof, just an example):

• Closure: The sum of two real numbers is a real number.
• Associativity: For any real numbers a, b, c, (a + b) + c = a + (b + c).
• Identity: 0 is the additive identity. You can add it to any real number without changing it.
• Inverse: For any real number a, -a is its inverse. The sum of the two is 0.
• Commutativity: For any real numbers a, b, a + b = b + a.

The set of real numbers under multiplication is not a group, however, since 0 does not have a multiplicative inverse (there’s no real number you can multiply with 0 to make 1, the multiplicative identity). But you can make this a group by excluding 0.

As we will see shortly, H!P groups consist of both abelian and non-abelian groups.

#### Hello! Project Groups as Mathematical Groups

##### Set Composition

There are a number of ways in which to regard H!P groups as groups in the group-theoretic sense. A naïve approach would be to define a set consisting of the members of the group and define some appropriate binary operator over the members. Though this makes some intuitive sense and is relatively simple to start out with, we can easily show that it fails to capture the nature of H!P groups.

A counterexample would be Coconuts Musume。 after Ayaka’s graduation on 2008-04-30. According to the official graduation announcement on Hello! Project’s site, Ayaka, the last remaining member of the group, graduated from Hello! Project as well as Coconuts Musume。 (“２００８年４月３０日（水）をもちまして『ココナッツ娘。』及び『ハロー！プロジェクト』から卒業することになりました。”) This implies that Coconuts Musume。 as a group did not graduate from Hello! Project. After losing its last member, it must remain as a zero-member group.

Unfortunately, there is no such thing as a zero-element group. That would violate the identity property, which states that there exists an identity element. The smallest group is the trivial group, which contains just the identity and nothing else.

And that is the key to understanding how H!P groups work in a group-theoretic sense. All groups must have an identity, and H!P groups are no exception. What Coconuts Musume。 retained after Ayaka’s graduation was its identity: the name “Coconuts Musume。”. The name did not graduate along with Ayaka but remained as the single persisting feature of the group. Here, the term identity is applicable in both its mathematical sense and its sense of being a name or label, an identifier.

So all true H!P groups have an identity, a name that is distinct from the members who constitute the group but is itself an element of the group. Due to the potential for confusion in terminology, I’ll clarify that I’m using member to refer to a human constituent of a group and element to refer to a constituent of the group in the theoretical sense. In this study, I assume that a group’s elements are solely its identity and its members, but that assumption need not hold in possible alternative group-theoretic conceptions of H!P groups, which I will not consider here. Thus the current lineup of Morning Musume。 has ten elements (that is, it’s a group of order 10): nine members and an identity.

One could argue that the identity is just an insignificant auxiliary feature of a group, but I think treating it as an element in its own right does better to characterize H!P’s conception of its groups. The only element of Morning Musume。 that has remained throughout its changing generations is the identity, and it is solely due to the inclusion of the identity that the current lineup can still claim older songs like “LOVE Machine” or “Renai Revolution 21″ as its own, even though none of the current members were around at the time of those releases. The Hello! Project web site acknowledges the importance of the group identity: the official profile of the group has a section devoted to each of its ten elements, and each of the ten elements has its name highlighted in blue. The identity has its own section at the top, where it belongs, since it is the longest lasting and arguably most important element of the group.

Interestingly, H!P uses the term unit (ユニット) to refer to its groups. This term appears, for example, in reference to H!P’s special and shuffle units, and Tsunku♂ uses it frequently. In a mathematical sense, unit can refer to some form of the number 1 (for example, the unit circle, the set of points with a distance of 1 from the origin), which is often the identity element in groups (especially when the group operator is regarded as multiplication). In a way, H!P’s use of the term associates a group with its identity.

##### Group Operator

Having considered the question of set composition, we now turn to what kinds of binary operators exist in H!P groups. This is a crucial aspect of a group, as the group operator determines the structure of a group, and without it, a group is simply a set, an unordered, unstructured collection of distinct elements. It can also be particularly tricky to discern the existence of a group operator, since it is not always clear what the inherent structure of an H!P group is. That said, we can find examples of group operators in some H!P groups.

Let us turn to the tracklist for Morning Musume。’s album Platinum 9 DISC. Several of the songs are credited to a single member (Mitsui, Michishige, Kamei) or to a subset of the members ({Takahashi, Niigaki, Tanaka}, {Kusumi, Junjun, Linlin}). Especially noteworthy is the use of an interpunct (・) to separate the names. The interpunct is commonly used in mathematics to denote multiplication (a · b), alongside other representations such as a × b, a * b, or simply ab. (My physical copy of the album uses a space to separate the names, which is another common way to denote multiplication.) We can assume that this indicates the application of the group operator to the members in question. As the group operator is associative, we need not specify an order in which pairs should be operated on, but since groups are not in general commutative, the order in which the members are listed may be relevant. (In the next section, I argue that Morning Musume。 is an abelian group, in which case the order of the members is actually not relevant.)

Given the prominence of the “Produced by Tsunku♂” credit on the cover of most of Tsunku♂’s productions, it seems sensible to call the group operator the “Tsunku♂ product” (or maybe “Tsunku♂ produce”) for those groups produced by Tsunku♂.

##### Groups vs. Non-Groups

We have been discussing H!P groups without really defining what counts as an H!P group and what doesn’t. Here is a good place to distinguish between the two. A common taxonomy of H!P divides it into two categories: “groups” and “soloists”. This classification is largely consistent with our group-theoretic formulation of H!P groups, but there are a few exceptions, which I will now examine more closely. In particular, there are acts commonly classified as “soloists” that are actually groups, and there are acts consisting of more than one member that are not actually groups.

One example of the former is Tsukishima Kirari starring Kusumi Koharu (Morning Musume。). Commonly regarded as a soloist, this is actually a two-element group consisting of the identity “Tsukishima Kirari starring Kusumi Koharu (Morning Musume。)” and the single member Kusumi Koharu. The two elements are distinct from each other: Kusumi is also a member of Morning Musume。, but Tsukishima Kirari isn’t.

Most soloists are not groups. The key criterion is whether or not the soloist’s performing identity is distinct from their own. If so, then the soloist and the identity together form a group; if not, then the soloist is not a group.

On the other hand, we have collections of members that do not constitute groups. One example is the aforementioned “Takahashi Ai · Niigaki Risa · Tanaka Reina”, who are credited for two songs on Platinum 9 DISC. In this case, there is no distinct identity associated with this trio. They are simply individual members appearing together, credited as such. If one applies the group operator to the trio, the result is an element of Morning Musume。, but which one it is depends on the definition of the group operator, a topic we will address in the next section.

In practice, there is not always a clear distinction between groups and non-groups in H!P. Some acts, like Abe Natsumi & Yajima Maimi (°C-ute), have names that are composed of individual members’ names, so it may be equally valid to regard them as individual members appearing together, or as a group with a distinct identity that just happens to resemble the individual members’ names. H!P acts vary a bit in this regard, and some composite names seem more like distinct group identities than others.

One important feature of an identity, however, is its immutability. If you change the name of an identity, it becomes a different identity. So this could be used to determine what counts as a distinct identity. In the case of “Takahashi Ai · Niigaki Risa · Tanaka Reina”, different official representations have used interpuncts and spaces to separate the members’ names, so there is no single correct representation of the name, and hence no distinct identity. On the other hand, Shigepink, Kohappink is an actual group, even if it looks as though it could be a composite of smaller groups Shigepink and Kohappink. The half-stop or comma between the names is invariable, as can be seen in the track listing for 7.5 Fuyu Fuyu Morning Musume。 Mini!. While an interpunct separates Yoshizawa, Niigaki, and Kamei in the track credit above, Shigepink, Kohappink’s credit retains a half-stop.

##### Group Isomorphisms

Once we can characterize the structure of a group, it is useful to relate it to a more abstract notion of a group. We can say that an H!P group is isomorphic to some abstract group—that the two exhibit the same group structure. In particular, this means that there is a one-to-one correspondence between H!P group elements and elements of the abstract group such that the relationships expressed by the group operator are preserved: if Takahashi, Michishige, and Kusumi correspond to the elements a, b, and c, respectively, in another group, and multiplying Takahashi and Michishige produces Kusumi, then applying the group operator to a and b in the other group should yield c.

One can then apply what is already known about such an abstract group to the H!P group in question, and also relate different H!P groups that are isomorphic to the same abstract group (and hence to each other).

Up to this point, we have not yet conclusively defined the group operator for any group, but for groups with three or fewer elements, we do not need any additional information to determine an isomorphism, as long as we know which element is the identity.

##### Field Guide???

If this were, say, a field guide to H!P groups in their natural habitats (or should I say group guide, since a field is another kind of algebraic structure), each group could have its own entry with relevant info, statistics, and taxonomic classification. In this post, I will attempt to provide a brief entry for several of the groups, and if any of you are interested, you can try adding entries for the other groups.

The first three entries will include one representative group each of orders 1, 2, and 3; the groups have already been mentioned above.

(Note that the concepts of subgroup and center don’t exactly correspond to their conventional H!P meanings. I’ll discuss these in Part 2.)

Coconuts Musume。 (post-Ayaka era, May 2008 — ???)
Order 1
Identity ココナッツ娘。
Elements and inverses
 Element Inverse ココナッツ娘。 (e) e
Cayley table
 × e e e
Nontrivial proper subgroups and cosets
 Subgroup (* = normal) Cosets None –
Minimal generating sets {}
Center {e}
Cycle graph

Isomorphic to Trivial group
Finite? Yes
Simple? No
Abelian? Yes
Cyclic? Yes
Perfect? Yes
Nilpotent? Yes
Solvable? Yes

Tsukishima Kirari starring Kusumi Koharu (Morning Musume。) (2006 — 2009)
Order 2
Identity 月島きらり starring 久住小春(モーニング娘。)
Elements and inverses
 Element Inverse 月島きらり starring 久住小春(モーニング娘。) (e) e Kusumi Koharu (K) K
Cayley table
 × e K e e K K K e
Nontrivial proper subgroups and cosets
 Subgroup (* = normal) Cosets None –
Minimal generating sets {K}
Center {e,K}
Cycle graph

Isomorphic to Cyclic group C2
Finite? Yes
Simple? Yes
Abelian? Yes
Cyclic? Yes
Perfect? No
Nilpotent? Yes
Solvable? Yes

Shigepink, Kohappink (2006 — ???)
Order 3
Identity 重ピンク、こはっピンク
Elements and inverses
 Element Inverse 重ピンク、こはっピンク (e) e Michishige Sayumi (M) K Kusumi Koharu (K) M
Cayley table
 × e M K e e M K M M K e K K e M
Nontrivial proper subgroups and cosets
 Subgroup (* = normal) Cosets None –
Minimal generating sets {M},{K}
Center {e,M,K}
Cycle graph

Isomorphic to Cyclic group C3
Finite? Yes
Simple? Yes
Abelian? Yes
Cyclic? Yes
Perfect? No
Nilpotent? Yes
Solvable? Yes

#### The Cyclic Groups of H!P

It is interesting that Morning Musume。 was founded on a notion of “cycling” through members to establish successive generations, by inducting new members and graduating existing members. Both words in the name also reflect a notion of periodic cycles: mornings are part of the daily cycle, and musume (daughters) are part of a cycle of familial generations. The full stop also adds a notion of periodicity—both in its more common North American name, the period, and its circular shape. All of this suggests that a notion of cyclicity is integral to the concept of Morning Musume。 as a group, and perhaps to others as well, particularly those with the full stop in their names.

Indeed, groups in the mathematical sense can also be cyclic. A cyclic group is one whose elements can be generated by a single element of the group: every element can be produced by successively applying the group operator to the generator. In a multiplicative sense, every element is a power of the generator.

We can then conclude that every H!P group with a circular full stop in its name is a cyclic group. If a group has n members, then it is isomorphic to the cyclic group of order n + 1, which is isomorphic to both Zn+1, the group of integers under addition modulo n + 1, and the complex (n + 1)st roots of unity, $\{e^{2i\pi k/(n+1)} : 0 \leq k \leq n\}$. For example, here are the 6th complex roots of unity, a cyclic group of order 6:

(It is worth noting that the circle group, being an uncountably infinite group, is not a cyclic group, so the pictorial representation of the full stop may not be accurate if perceived as a circle. However, it is possible to generate a dense cyclic subgroup of the circle group by using any irrational number as a generator, so there is a valid cyclic group interpretation.)

The three groups detailed above are all cyclic groups, but as we go higher in order, it is possible for groups not to be cyclic, so it is useful to be able to recognize cyclic structure where it exists.

The presence of the full stop establishes Morning Musume。 as a cyclic group of order 10. However, from just this information alone, we cannot conclusively determine Morning Musume。’s group structure. We know that it is isomorphic to the cyclic group of order 10, but we don’t know which isomorphism is correct. Depending on our choice of isomorphism, we could identify different subsets of the members as generators. (In the previous three groups, all isomorphisms are equivalent: any permutation of the members yields the same group structure; but in Morning Musume。, different permutations of the members can yield different group structures. There are in fact 90720 possible distinct group structures you can produce depending on which of the possible permutations of the nine members you pick—I leave the proof of this fact as an exercise for the reader.)

We therefore look to the standard order of H!P’s groups. Hello! Project generally orders its groups by seniority, and the standard order features prominently in musical releases, concerts, and public appearances. It is not an arbitrary order but an integral feature of each group. Accordingly, we use the standard order to enumerate the elements of an H!P cyclic group: the kth member in a group with n members is mapped to k in Zn+1 and to $e^{2i\pi k/(n+1)}$ in the group of complex (n + 1)st roots of unity.

This yields the following entry for the current lineup of Morning Musume。:

Morning Musume。 (Takahashi era, June 2007 — present)
Order 10
Identity モーニング娘。
Elements and inverses
 Element Inverse モーニング娘。 (e) e Takahashi Ai (TA) L Niigaki Risa (N) J Kamei Eri (KE) MA Michishige Sayumi (MS) KK Tanaka Reina (TR) TR Kusumi Koharu (KK) MS Mitsui Aika (MA) KE Junjun (J) N Linlin (L) TA
Cayley table
 × e TA N KE MS TR KK MA J L e e TA N KE MS TR KK MA J L TA TA N KE MS TR KK MA J L e N N KE MS TR KK MA J L e TA KE KE MS TR KK MA J L e TA N MS MS TR KK MA J L e TA N KE TR TR KK MA J L e TA N KE MS KK KK MA J L e TA N KE MS TR MA MA J L e TA N KE MS TR KK J J L e TA N KE MS TR KK MA L L e TA N KE MS TR KK MA J
Nontrivial proper subgroups and cosets
 Subgroup (* = normal) Cosets *{e,N,MS,KK,J} {e,N,MS,KK,J}{TA,KE,TR,MA,L} *{e,TR} {e,TR}{TA,KK}{N,MA}{KE,J}{MS,L}
Minimal generating sets {TA}, {KE}, {MA}, {L}
Center {e,TA,N,KE,MS,TR,KK,MA,J,L}
Cycle graph

Isomorphic to Cyclic group C10
Finite? Yes
Simple? No
Abelian? Yes
Cyclic? Yes
Perfect? No
Nilpotent? Yes
Solvable? Yes

With this group structure, we can evaluate the combinations of group members mentioned earlier (and since all cyclic groups are abelian, the order in which we apply the group operator does not matter):

• Takahashi · Niigaki · Tanaka = 1 + 2 + 5 = 8 = Junjun
• Kusumi · Junjun · Linlin = 6 + 8 + 9 = 23 ≡ 3 (mod 10) = Kamei

Or, equivalently, treating the operation as a product (the “Tsunku♂ product”) of complex numbers:

• Takahashi · Niigaki · Tanaka = $e^{2i\pi (1)/10} e^{2i\pi (2)/10} e^{2i\pi (5)/10} =e^{i\pi /5} e^{2i\pi/5} e^{i\pi} = e^{8i\pi/5} = e^{2i\pi(8)/10}$ = Junjun
• Kusumi · Junjun · Linlin = $e^{2i\pi (6)/10} e^{2i\pi (8)/10} e^{2i\pi (9)/10} =e^{6 i\pi /5} e^{8i\pi/5} e^{9i\pi/5} = e^{23i\pi/5}$ $= e^{4i\pi}e^{3i\pi/5} = 1*e^{3i\pi/5} = e^{2i\pi(3)/10}$ = Kamei

Of course, this raises newer questions that we have yet to answer, but it seems Kamei is somehow responsible for “Guruguru JUMP”.

If we consider °C-ute a cyclic group as well, interpreting the degree symbol (°) to be analogous to the full stop, we have the following entry (assuming standard order to resolve among the 120 possible distinct group structures):

°C-ute (post-Arihara era, July 2009 — present)
Order 7
Identity °C-ute
Elements and inverses
 Element Inverse °C-ute (e) e Umeda Erika (U) H Yajima Maimi (Y) O Nakajima Saki (N) S Suzuki Airi (S) N Okai Chisato (O) Y Hagiwara Mai (H) U
Cayley table
 × e U Y N S O H e e U Y N S O H U U Y N S O H e Y Y N S O H e U N N S O H e U Y S S O H e U Y N O O H e U Y N S H H e U Y N S O
Nontrivial proper subgroups and cosets
 Subgroup (* = normal) Cosets None –
Minimal generating sets {U}, {Y}, {N}, {S}, {O}, {H}
Center {e,U,Y,N,S,O,H}
Cycle graph

Isomorphic to Cyclic group C7
Finite? Yes
Simple? Yes
Abelian? Yes
Cyclic? Yes
Perfect? No
Nilpotent? Yes
Solvable? Yes

Being a cyclic group of prime order, °C-ute is a simple group, as opposed to Morning Musume。, which is not currently a simple group but has been so several times in the past. Prior to losing Arihara, however, °C-ute was also not a simple group, and in addition, had the curious distinction of being a group whose leader (Yajima) was not a generator, since Yajima’s index, 2, is not relatively prime to °C-ute’s group order at the time, 8. In most groups, the leader has the highest seniority and accordingly has an index of 1, which is relatively prime to everything, so it is usually the case that the leader is also a generator.

With that, we end our discussion of cyclic groups in H!P. Not all groups are cyclic, however, and while one valid approach would be to categorize all H!P groups as cyclic (since there is a cyclic group for every positive integer order), the characteristics of some H!P groups suggest that they exhibit different group structures not isomorphic to cyclic groups. And this is where H!P groups start to get very interesting, but as I’ve already written way too much for this post, I’ll end it here and promise to resume next time with a survey of a variety of non-cyclic Hello! Project groups, as well as a possible explanation for the lack of a full stop in Pucchi Moni’s name and the meaning behind the V in Pucchi Moni V.

to be continued…

This is my account of going to see Morning Musume。 at Anime Expo in Los Angeles last week, which is a bit delayed due to technical and other difficulties. It was quite an adventure, and there is much more to say than I have room for here, but this should be a nice recap of my experience. I’ve left out parts that I think have been covered comprehensively elsewhere, like details of the MM Q&A panel and the concert. So, without further ado, LET’S STARTING!!! (o_0)

Preparation

When it was announced that Morning Musume。 would be signing autographs, I wanted to make the experience super awesome, so I got some additional merchandise that hopefully I’d be able to get signed while at AX. This included a full-size Papancake poster (Taiwan version):

(The Hello! Hello Kitty toaster is my roomie’s, not mine. ^_^)

This poster hung above our breakfast/dinner table for a while, and it was definitely enjoyable to eat breakfast with Koha (or eat breakfast while being stared at by Koha…).

I also got a Best☆Kirari poster and a Koha mini-poster, and my roomie got a Kirari to Fuyu poster.

After it was announced that we could get a different item signed by each member (for up to nine separate items), I wanted to do just that, since having all of their autographs on one item is kinda boring. Things didn’t exactly turn out as planned, but I was going to have these items signed:

• Mini Moni ja Movie: Okashi na Daibouken! DVD cover — Takahashi
• Souda! We’re ALIVE CD cover — Niigaki
• Hare Ame Nochi Suki ♡ CD cover — Kamei
• Ai no Sono ~Touch My Heart!~ CD cover — Michishige
• FIRST KISS CD cover — Tanaka
• Papancake poster — Kusumi
• Onna ni Sachi Are CD cover — Mitsui
• Shouganai Yume Oibito CD cover — Junjun
• Platinum 9 DISC CD cover — Linlin
• Osaka Koi no Uta CD cover — HANGRY

If it turned out I’d be able to attend more than one autograph session, I’d get something else signed by everyone. Or maybe everyone except Koha, because I had a ton of Koha stuff. ^_^

In addition, since giving gifts was allowed, I decided to give each member one. Being broke, I decided to get MIT t-shirts for Aichan and Sayumi and a plush lobster for Koha. On top of that, each member would get an individual personalized card.

I had a lot to say to Koha, Aichan, and Sayu (who also happen to be my roomie’s favorite members—we like a lot of the same things; it’s like we’re the same person ^_^), so I filled up their cards, but I didn’t know what to say to the others. So rather than putting something generic on all the remaining cards, I came up with some pretty random messages for a few of the other members.

Eri got a pretty terrible (and self-referential) haiku:

 亀が好き kame ga suki I like turtles 亀井 大好き Kamei daisuki I like Kamei a lot 俳句です haiku desu this is a haiku

It was so bad it deserved another haiku to comment on its atrociousness:

 詩は酷い shi wa hidoi poetry so bad 墓に芭蕉の haka ni Bashou no Bashou in his grave 寝返りね negaeri ne is rolling, isn’t he?

Continuing with the haiku theme, I decided to write Reina some nonsensical Dadaist ones, all having a seasonal theme but otherwise not making a whole lot of sense (though they all have some extratextual resonance of one kind or another) … these actually took a while to write; nonsense isn’t as easy to write as it might seem.

 夏休み natsuyasumi summer vacation サスカチュワンに Sasukachuwan ni to Saskatchewan 果実 飛ぶ kajitsu tobu fruit flies 紅葉の絵 momiji no e picture of autumn colors 手押し車は teoshi kuruma wa the wheelbarrow 青いです aoi desu is blue 冬の道 fuyu no michi winter road あいにくケーキ ainiku keeki unfortunately the cake うそじゃない uso ja nai is not a lie 春か猫 haru ka neko spring or cat? 量子力学 ryuushirikigaku quantum mechanics 知らないよ shiranai yo I don’t understand

Having run out of creative haikujuices, I offered Aika the Y combinator in PostScript:

Apologies to Risa, Junjun, and Linlin … yours weren’t as spectacular, alas, other than a banana drawing for JunJun. :-o

As for the other three, Aichan got a green MIT t-shirt with a design like this but in black and white:

Her card had this on the back (“MIT ♡ Aichan”):

Inside, I told her how I really liked Q.E.D. Shomei Shuryo especially due to its MIT connections and that I knew of at least five current MIT students coming to AX just to see Morning Musume。 (which is a considerable number given that the school only has about 10,000 students), and several more who are H!P fans.

Due to her professed interest in math, Sayu also got an MIT t-shirt, but one that was more disguised. It had this design (not my photo, and not the shirt I bought):

In her card, I commented that the three letters clued by the formulae also happened to be the first three letters of her family name in Kunreishiki romanization (Mitisige). It’s like Sayu and MIT were made for each other… XD SayuMIT!

The back of Sayu’s card (featuring Euler’s identity):

Koha’s was by far the longest, as I had plenty to say. I told her about how her music was largely responsible for getting me into the H!P fandom, and that the “sine cosine tangent” line from “Hana wo Pu~n” was too awesome. I drew diagrams corresponding to her hand movements in the PV (the angle is in the bottom left, and she outlines the two sides whose ratio is the sine, cosine, or tangent of the angle; note that in Japanese convention, you put the denominator in a ratio first—1/3 is “sanbun no ichi”):

(The above was actually my first ever YouTube video, posted under my old account, back before this blog even existed.)

Also, I told Koha I shared this clip in my electronic music class and passed around my copy of Kirarin☆Land, and that my classmates and the instructor were very impressed (which is true … one classmate even wanted a rip of the album tracks).

In addition to this, I wrote down all eight of the double dactyls I had written to celebrate Koha’s birthday last year on a larger card and rolled it up.

That sums up my preparation, I think. Now to attend my first ever non-classical concert and my first ever anime convention! XD

Day 0 – Airport Awesomeness and Line-waiting Lameness

I flew into Orange County on June 29, and stayed there with my relatives for a week, other than the days of the convention. As this is irrelevant, I’ll jump to Day 0, July 1, when Morning Musume。 arrived at the airport.

I wasn’t originally planning on going to meet them at the airport, due to my lack of transportation and not wanting to inconvenience my relatives. But as June 30 rolled by without a sign of the group, it became clear that they’d definitely arrive on July 1, which was eventually confirmed by a fellow passenger on their flight.

I decided to get myself to the airport by public transportation only, as part of the charm of visiting a new city for me is getting lost in its public transportation, which inevitably happens regardless of how much planning I do. This has happened to me with Boston, New York, San Francisco, and Chicago, so I was eager to add Los Angeles to the list.

It took me 3.5 hours to get to LAX, using two local Orange County buses, an Orange County express bus, two Metra trains, and an airport shuttle. I did in fact get lost, as I failed to request the Harbor Freeway station stop to switch to the Metra Green Line. The bus didn’t announce its stops, and since no one requested that stop, it drove right on by. :-o Eventually, we got to downtown LA, driving right by the convention center, so that was my first glimpse of the place I’d be spending most of the next few days. I got off at 7th street and took the Metra from there, adding about forty-five minutes to my planned trip.

Arriving at the airport with plenty of time to spare, I saw that a small group had already gathered at the waiting area in the Tom Bradley International Terminal:

It was here that I met Bryan and Jen, hosts of the truly epic Hello! Party to come, and Claudia and Ivet, creators of the 3,2,1 BREAKIN’ OUT! OPV that won the Morning Musume。 prize. XDDD

After much waiting, during which several passengers from another flight from Japan did a double take after seeing the huge banner we had up (“Ehh~~??? Morning Musume。‽‽‽”), the members themselves appeared, to much fanfare and chaotic screaming:

We chased after them (from a respectable distance) and saw them get into a white van. Seeing them up close (and getting waves from them ^_^) was so awesome! One fellow passenger on their flight, after asking what all the commotion was and being told they were a popular J-pop group coming to America to perform for the first time, had this to say (regarding Aichan, I think): “She had nice legs.” :-o

I next went to the LA Convention Center with Bryan, Jen, Claudia, and Ivet, and got in line for badges. I was melting and hungry, so I parted ways and went to grab food, which in retrospect was probably not the best decision. Returning to the line a bit later, I saw that it had grown to a ridiculous size. Oh well. Unfortunately they had computer trouble, so I ended up waiting 90 minutes in the blazing sun. Ugh. And after that line came the concert ticket line, which was fortunately mostly in the shade. After four hours of waiting, I finally had badges and concert tickets for myself, my roomie Kyttyee, and our dear friend Aaron the Apathetic Wota (whom I will affectionately call A~ron in this post), who earlier this year had managed to accomplish the amazing feat of unintentionally attending a Johnny’s concert in Japan.

I also met up with CatchFiveBats, maiZe, johpan, Greg, and Paul Thomas.

Unfortunately, this meant I missed the last express bus out of LA, so I had to get a ride from my relatives. :-/

Day 1 – Opening Ceremony, Q&A Panel, and More Waiting in Line

The next day, A~ron and Kyttyee picked me up, and we arrived at the convention center in time for the opening ceremony, details of which you can find elsewhere. Afterward, A~ron and I got in line for the exhibit hall, which was ridiculously long and extended the entire length of the South Hall and a bit further. Once we got inside, shortly after noon, we rushed to the JapanFiles/JPopHouse booths to get autograph tickets. Unfortunately, the lines for the two booths had merged, and the staff were trying to untangle them. As it turned out, I was in the JPopHouse line when I thought I was in the JapanFiles line, and the rearranging resulted in the section of the line I was in getting shuffled to the back of the JapanFiles line. So after thirty minutes of waiting, I was now back at the end of the line. >_<

A~ron, having gotten into what looked like the JPopHouse line, ended up further ahead of me in the JapanFiles line, and was able to get a HANGRY ticket, but no MM ticket. So we managed to get something after all. Later on, we were able to get a second HANGRY ticket from JPopHouse after the panel, as they still had them for a while longer.

We headed to the panel and got there as the one before was letting out. We were able to get seats at the right of the fourth row, and after the panel was over (details of which you can find elsewhere), we got our autograph tickets and followed the directions to the back of Hall K in the Exhibit Hall. There we found hundreds of people running around in circles following each other, quite an amazing thing to experience. After overhearing some more sensible directions from the info booth, we headed upstairs to Room 301 and found the actual autograph line. Unfortunately, there were two lines on opposite sides of the building, and we were told to go to the other line, despite being right by one line of people who had already gathered. I decided to go to the other line while A~ron secured a spot for Kyttyee in the first line. The other line, when I got to it, wasn't much longer, so I stayed there. As it turned out, that was a mistake, as Kyttyee was able to get autographs in like 30 minutes (and a 3-minute conversation with Aika! :-o), while I ended up having the door shut in front of me with about twenty people ahead of me (I was #182).

Oh wells. Try again on Saturday?

We (CFB, Greg, maiZe, joh, and Paul, I think) headed down to the JPopHouse booth and got in line (yet again!) for some H!P merchandise. I bought a photoset and a Platinum 9 Disco poster, since the autograph session had specified one item for all members to sign, and I didn’t have anything I wanted to have signed that had all nine members on it. I also purchased some glowsticks and glow bracelets from JapanFiles. They had run out of all the colored glowsticks, but luckily found they still had a box of the red ones (Koha!!) as I got there. So yayz.

Having not eaten anything since dinner the day before, I was ready for some dinner, which I had at IHOP with CFB. Afterward I chilled for a bit with CFB and friends. We heard that Morning Musume。 was staying at the Westin, and discussed going there, but alas, I never actually went there the whole time I was in LA. :(

Returning to the convention center, exhausted, I napped in the lobby (should have brought a save point to set up next to me XD), and eventually got to the Econolodge motel in Hollywood or somewhere, where I would stay the next two nights with Kyttyee and Sekai no Melody friends Vivi and Kimi. Before catching some Z’s, we sang, danced, and listened to some of my mashups, and fangirled over the Koha posters Kyttyee and I had brought. ^_^ Kohaaaaa~~~!! *squee!*

Day 2 – Concert, HANGRY, and Hello! Party

In the morning, Kyttyee, A~ron, and I found ourselves back at the convention center for some more awesomeness. We met up with our friend “Edwin », who introduced me to Morning Musume。 in the first place, back when I was just getting into J-pop way back in 2007. Unfortunately, his choice of “LOVE Machine” left me disappointed with the group (I stopped watching the PV a third of the way in because I couldn’t stand it), and it wasn’t until several months later that I found that MM had awesome songs I liked.

Anyway, “Edwin » (being “Edwin ») didn’t stick around for the concert (!) and parted ways with us. We made our way over to the main hall, and somewhere along the way, I lost the poster that Kyttyee, Vivi, and I had made for Aichan (saying “MIT ♥ Aichan! We’ll always support you! Ganbatte ne!”) and took pictures of us with in front of MIT’s Great Dome, which we were planning to give to Aichan as a present (sorry! :( ). In addition, I had forgotten my glowsticks and glow bracelets at the hotel, so all I had were the eleven glow bracelets that had broken and which I was wearing. Kyttyee got us some more glowsticks (no red, alas…) and glow necklaces, and while waiting for the concert to start, I hooked them together into an awesome glowstarball that I waved during the concert (though it would have been even more awesome with non-dead glow bracelets forming the star, and with red glowsticks).

I must say, I was disappointed with “LOVE Machine” being performed, but the rest of the concert was pretty cool, other than having “3, 2, 1 BREAKIN’ OUT!” performed way too many times and being too short overall. But the medley and “Sono Bamen de Bibiccha Ikenai jan!” were definitely awesome (I use that word too much…).

Afterward, Kyttyee and I got in line for the HANGRY autograph signing, and since they wouldn’t let us form a line past a certain point, we walked around in circles and grabbed some food, which we ended up eating in line and actually brought into the autograph booth with us (“Hi HANGRY! Have some nachos!”).

ZOMG, HANGRY in person! It was surreal. I gave her the card I had written for her, and asked her to sign Osaka Koi no Uta, which she did, as “HANGRY → Yossie” (in hiragana):

Due to the excitement (which for the second time I just typed as “exciteness” … ???? O_o), I think I forgot to shake her hand. Blargz.

I grabbed dinner at IHOP (the third time I ate there while in LA), and headed to the awesome Hello! Party, where I encountered lots of fellow H!P fans, including Amy, Rocky, Lysa, CK (who I had met earlier in the day), mp122984, and delrey28.

The party was super awesome, and I stayed until the end. Kyttyee and I were planning to karaoke Renai♥Chance!, but we were too far back in the queue and didn’t get to. But I’m still eager to see what mashup karaoke sounds like and what kind of reaction it would get from a large gathering of H!P fans. (Next Hello! Party? XDDD)

Day 3 – Ch-ch-chance!

Waking up at 8:00 was FAIL. Checking online, I found that the line for the morning autograph session was already more than a hundred people long. Nevertheless, we rushed to the convention center and found the line (thereby missing the second half of the TSUNKU♂ panel >_<). Kyttyee and Kimi chilled in the standby line, while I joined the ticketed line with my ticket from Thursday. And to my surprise, the line was shorter than I expected, with me being only #183 in line (one spot higher than on Thursday). Since Thursday's session had started late and ended early and was rather unorganized, I was hoping it'd be speedier this time and get through more than 160 people.

Unfortunately, after waiting 90 minutes and having had my hopes up, I had the door shut in front of me again, with only twelve people in front of me. >_<

Thrice thwarted!

Dejected, I made my way down to the exhibit hall and joined Kyttyee and Vivi for some lunch before going over to the afternoon autograph session in hopes of finding someone who was getting individual items signed and didn't care much for Koha (so I could get my Papancake poster signed instead). As it turned out, they had changed their policy at the last minute to be only one item for all the members (like the other sessions), so this wouldn’t have worked anyway. But at the time I didn’t know that, which worked out in my favor, since after unsuccessfully asking a few people in line, I ran into Claudia and Ivet, who had an extra ticket (thank you so much! you are awesome! XDDDD). Unbelievable joy!

So I got in line (in front of Amy and Mozenator) and had all my individual items out, along with everyone’s gifts. It was hard to hold everything at the same time and I accidentally dropped some of my stuff on the floor (>_<). The line was moving really quickly, compared to the previous lines I was in, and in no time, we were right in front of the booth MM were in, which I had walked past before getting in line, seeing them through the curtains.

At this point, they told us that we could only have one item for all the members, so I was desperately trying to stuff all my individual items into my bag, pull out and unwrap my group poster, and keep from dropping everything on the floor at the same time while walking into the booth—where I found MM in not their expected order (!), so I had to re-sort my gift cards as well. I handed my poster to a staff member (who promptly unrolled it in front of Reina, sitting at the front) and asked if it was ok to give gifts, then handed over my pile of gifts as well. As a result of this chaos, I didn’t get to say anything to Reina other than a quick “arigatou gozaimasu” since she had already finished signing. >__<

Having failed to shake HANGRY’s hand on Friday, I wanted to shake hands with Morning Musume。 this time. Apparently this session was a “no handshakes, no touching, no chitchat” session, unlike the previous two (probably because they needed to rush through 200 autographs—I think only 200 of the announced 300 tickets were actually given out) in a short amount of time. But I didn’t know that—I assumed handshakes were ok because they were fine in the previous sessions. Either they didn’t announce the policy for this session, or I was too busy trying to do too many things at once and didn’t register what they were saying. So I came in expecting to be able to shake hands and say a few reasonably short things to the MM members.

I remembered to ask for a handshake only after Reina was already done signing, so I attempted to shake Linlin’s hand. She didn’t appear to notice (and my brain was too frozen to say anything), so I retracted my hand and moved on. >_< I don't think any of the staff noticed either, because I didn't get any response from them.

Next was Aichan. I managed to say hello to her and tell her I'm an MIT student (showing her my MIT ID card), and that I loved her drama Q.E.D. Shomei Shuryo. She seemed pretty excited and said “arigatou”. ^_^ I was cut off by a staff member saying something (presumably telling me to hurry up), so I thanked her and moved on. XDDD

After Aichan were Risa, Aika, and Eri. Unfortunately, I don’t remember what happened at this point because I got an unexpected “HI!!!!” and enthusiastic waves from both Eri and Aika (at different points in time) while I was still in front of another member. I hope I responded appropriately to their greetings and remembered to thank all three of them for their autographs… :o

Next was Sayu, and I was going to say something mathy, but everything was rushed and confused, so I wasn’t able to say much. I do remember that my poster rolled itself up, and I got a “sorry, sorry!” from Sayu and maybe Koha while I helped them unroll the poster.

Seeing Koha unoccupied (and looking pretty tired, I think), I greeted her with a “KONNICHI PA!!!” and got a “KONNICHI PA!!!” in return, along with an ever-delightful Koha-esque 8D of utter glee (XDDDD). Since I had missed my chance to shake hands with all of the previous members, I offered my hand, and Koha, after a brief moment of hesitation (well, it was a “no handshakes” autograph session after all …), shook it (♥ ^_^ ♥). KOHANDSHAKE!!! XDDD *squee~!*

Unfortunately, I didn’t get to tell her how awesome she was, but hopefully she’ll have a chance to read the card I wrote for her.

I hope I didn’t completely skip over Sayu while doing this (if so, I’m sorry!). D-:

And I don’t even remember what happened with Junjun. O_o

And … curses! I was going to wish Sayu and Koha a happy birthday, and I forgot! >_< gahhhhhhhh … well, I put it in their cards, so I hope they saw them… ^_^

Overall, getting to see Morning Musume。 in person and interact with them was a blast, even if the session was extremely rushed, and the members didn't seem to be enjoying it too much (from other people's reports, apparently several members were told to hurry up and not to interact with the fans, and at least one fan had their hand slapped by a staff member for wanting to get a handshake >_<). I don't recall making eye contact with Reina, Linlin, Risa, Sayu, or Junjun (T_T), but being greeted by Eri and Aika and getting to talk a bit with Aichan and Koha (with a Kohandshake!) was definitely too awesome for words. XDDD

Afterward, there was much ecstatic squeeing as we met up with CFB, CK, joh, maiZe, Rocky, and Paul and looked at what we’d had signed. ^_^

And that is where I’ll end my report, because I left the con later in the evening and didn’t do anything on Sunday. Maybe I should have gone to the airport to see MM off (Koha was wearing glasses and Sayu, Aichan, and Junjun were wearing what appear to be snowboots! I totally approve, having worn glasses and snowboots at the con myself… XDDD), but it was nice just to relax after several days of rushing around, waiting around, and not getting enough sleep.

I’d like to thank Morning Musume。, HANGRY, and TSUNKU♂ for taking the time and effort to come to LA and meet us fans, and the AX/UFA/JapanFiles/JPopHouse staff who made all this possible, even if things could have been done a bit differently. And a bucketful of huge thanks to Bryan and Jen for organizing Hello! Party, and thank you to all H!P fans for your support, whether or not you attended AX; without you, this event wouldn’t have happened. Until next time! ^_^

I’ve been anticipating the new detective drama Q.E.D. starring Takahashi Ai and Nakamura Aoi since it looked like it would involve a lot of awesome math, and now that the first episode is out, we can see that there is indeed a lot of awesome math! And correct, non-trivial math to boot!

Hello! Project member + stereotypical MIT nerd = WIN!

Meet Touma Sou (Nakamura), stereotypical MIT math nerd! Who, having graduated from MIT at a very early age, is for some reason reattending high school … I’m sure this will be explained in future episodes …

Mizuhara Kana (Takahashi) does an impression of MIT! D-:

Of course I am highly amused that “Massachusetts Institute of Technology” appears on the screen at the same time as Takahashi, who should definitely drop by for a visit sometime. I can show you around this place, Aichan, if you’re ever in the area.

Since “Q.E.D.” is traditionally written at the end of a math proof, this first episode of course features a math proof. As the show’s primary purpose is entertainment and not educational, it doesn’t focus on the details, but the details nevertheless are there and they’re intellectually stimulating, so I’ll cover that aspect here.

To begin with, the teacher presents the following proof in class:

[If this is hard to read, it and the below proofs are available as a PDF.]

Let’s go through the details. The statement being proved is given in line (1). This is giving a closed-form expression for the sum of the product of three consecutive integers over a range. In line (2), the summed expression is expanded as a cubic polynomial: $k^3+3k^2+2k$. This lets us use linearity of summation to split the sum into three parts (3). Then using summation identities, each of the three sums is expressed in closed form (4). In line (5), we factor out $\frac{n(n+1)}{4}$, and in line (6), we group like terms of the remaining expression to produce a quadratic polynomial: $n^2+5n+6$. This polynomial can be factored into $(n+2)(n+3)$, leaving the desired expression (7).

Of course, for Sou this is not nearly intellectually challenging enough. (The subtitles say that the teacher’s derivation is erroneous, but it’s correct as far as I can tell; if this is what Sou actually says, maybe the producers just wanted to avoid getting into details about the distinction between generalizing a theorem and offering a better proof of the same theorem.) Sou realizes that this fact can be generalized, establishing a more powerful theorem. Instead of having a product of just three consecutive integers $k(k+1)(k+2)$, why not have arbitrarily many? Sou generalizes the theorem to apply to a product of $r+1$ consecutive integers: $k(k+1)\cdots(k+r)$. And indeed, there is a closed-form expression for this as well:

Note that this reduces to the previous theorem if you substitute $r=2$.

It’s a fun exercise to try to prove this theorem, which I encourage you to do if you’re interested. I did so, and I’ll present my own proof later on. As it turns out, Sou’s proof is simpler than mine, but due to camera angles and low resolution, it’s pretty hard to actually see what the proof is.

After much squinting and deducing, I finally figured it out:

This is pretty short and makes some jumps that may not be intuitive, so I’ll go through the details:

In line (9), each term of the summation is multiplied by a ratio equal to 1, which does not change the value of the expression. You can easily do the math and see that the numerator is equal to $r+2$. In line (10), the denominator of this ratio is pulled out of the summation, since it’s a constant factor, and the two terms in the numerator are multiplied with the product $k(k+1)\cdots(k+r)$, using the distributive law. Line (11) is probably the biggest jump in this proof; it’s using a telescoping sum, reducing a sum of terms by canceling out intermediate pairs. If you plug in $n$ for $k$, you get $n(n+1)\cdots(n+r+1) - (n-1)n(n+1)\cdots(n+r)$, and likewise, for $k=n-1$, you get $(n-1)n(n+1)\cdots(n+r) - (n-2)(n-1)n(n+1)\cdots(n+r-1)$ Notice that the $(n-1)n(n+1)\cdots(n+r)$ terms cancel when you add these together. If you keep doing this all the way to $k=1$, all the intermediate pairs cancel out, and you’re left with just two terms: $n(n+1)\cdots(n+r+1) - 0(1)(2)\cdots(r+1)$. The second term vanishes, so the overall expression is reduced to (11), which completes the proof.

The following is my own proof, which is a bit lengthier and uses binomial coefficients, along with Pascal’s second identity. Let’s first prove Pascal’s second identity, stated here as a lemma:

We use the principle of induction, which is showing that a statement holds for some base case (here $n=0$) and showing that if the statement holds for some $n$, then it has to hold for $n+1$ as well. Putting these parts together, you prove that the statement holds for all integers greater than or equal to the base case.

Lines (13) through (17) prove the base case. This makes use of the definition of the binomial coefficient ${n+k \choose k} = \frac{(n+k)!}{n!k!}$, which for $k=0$ is equal to 1.

The inductive step is proved in lines (18) through (25). We assume the statement holds for $n$ and show that it also holds for $n+1$. In line (18), we split the sum of $n+1$ terms into two parts. The first part is identical to the sum of $n$ terms, which we can replace with ${n+m+1 \choose n}$ by the inductive assumption (19). In line (20), we replace the binomial coefficients with their equivalent factorial ratios, and in line (21), we factor $m+1$ and $n+1$ out of the two denominators, respectively. In lines (22) and (23), we factor out common factors of both terms, and add together the remaining fractions. Combining the two parts, we get (24), which can be written as (25), thus proving the lemma.

My alternate proof is as follows:

In line (26), we rewrite the product of $r+1$ consecutive integers as a ratio of factorials. We then reindex the summation (27) for convenience, so that $k$ starts at 0. Next we multiply the entire expression by $(r+1)!$ and divide it by the same amount (28). Each term of the summation can be expressed in binomial form (29). Using Pascal’s second identity, we can rewrite the sum as (30). Turning this back into a ratio of factorials, we get (31). We can cancel out a $(r+1)!$ factor, leaving (32). Finally, expressing this ratio of factorials as a product of consecutive integers, we have (33), as desired.

(I’d stick a “Q.E.D.” at the end of that proof, but no one uses that anymore. The square box is sufficient.)

I’m definitely looking forward to more math and awesomeness in the upcoming episodes….

[The above proofs are available as a PDF.]

EDIT: Upon rewatching the episode, I discovered that Sou has excellent taste in reading material:

Douglas R. Hofstadter’s Gödel, Escher, Bach: an Eternal Golden Braid. I approve.

EDIT 2: Agggh!! What was I thinking? There’s a really easy inductive proof that I would have seen had I inducted on the right variable the first time. >_<

It’s, uh, self-explanatory.

CUZ I DONT DO REVIEWS YO

this is a DIALOGUE on the nature of computation brought to you by °C-ute or actually just THE AIRI AND MAIMI SHOW cuz everybody else is like wut im a computer n00b i dont even know how to type zomgwtf but AIRI AND MAIMI are like SUPER FREAKING GENIUSES or something and they totally know everything about computers and programming and data structures and algorithms and complexity theory and all that

because contrary to popular belief °C-ute is actually short for

°COMPUTE

thats right THATS A DASH your supposed to fill in the missing letters duh

(betcha didnt know that huh?)

WUT

oh hay look its °C-ute in the middle of their STRUCTURE AND INTERPRETATION OF COMPUTER PROGRAMS class and today their going over HIGHER ORDER PROCEDURES and how to call a procedure using another procedure as an argument or even reflexively using the procedure itself

so erika and mai are like WUUUHHHHHH???? but airi and maimi totally know wuts going on cuz their like SUPER FREAKING GENIUSES and their already jumping ahead to next weeks topic…..

and airi goes hey check this out zomglolololololol

… one level of embedding …

….. TWO levels of embedding …..

……. FOUR LEVELS OF EMBEDDING ……..

……… EIGHT LEVELS OF EMBEDDING IS THAT AMAZING OR WUT OMG???!!! ………

XDDDDDDDDDDDDDDDDDDDDDD

and maimis like WUT

and airis like RECURSION B****ES!!!

but maimis like STFU N00B that terminates after only eight iterations

betcha cant do an infinite loop wut

so airi sez OH YEAH WELL CHECK THIS OUT

(define (FOREVER x)
(display x) (FOREVER x))
(FOREVER "LOVE ")


SEE IT PRINTS OUT LOVE AND RECURSIVELY CALLS ITSELF SO IT CAN PRINT OUT LOVE AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN AND AGAIN

its like LOVE …….. FOREVER!!!!!

wut

FOREVER LOVE

get it? GET IT??!!!

but maimis like SRSLY zomgrofl THAT IS SO TRIVIAL IT MAKES THE TRIVIAL GROUP LOOK NONCOMMUTATIVE lol

NONCOMMUTATIVE?????!!!!!!!

I OWN TSUUGAKU VECTOR☂

I AM THE LIVING EMBODIMENT OF THE INNER PRODUCT OF ANY TWO VECTORS IN A HILBERT SPACE

cuz thats like CONJUGATE SYMMETRY

which when restricted to a real scalar field means that under the inner product any two vectors COMMUTE

FOR REAL

oh yeah well thats nothing compared to the Y COMBINATOR

(lambda (f)
((lambda (x) (f (lambda (y) ((x x) y))))
(lambda (x) (f (lambda (y) ((x x) y))))))


your procedure is so weak it needs a name zomglol

WITH THE Y COMBINATOR YOU CAN DO ANONYMOUS RECURSION ZOMGAWESOME

(((lambda (f)
((lambda (x) (f (lambda (y) ((x x) y))))
(lambda (x) (f (lambda (y) ((x x) y))))))
(lambda (p) (lambda (s) (display s) (p s))))
"LOVE ")


LOOK NO NAMES WUT

………………… O__________________O ……………………

THATS RIGHT THE Y COMBINATOR IS SHORT FOR YAJIMA COMBINATOR!!!!!!

o rly?

YA RLY LAMBDA CALCULUS FTW

well lambda calculus is ok but….

KAPPA CALCULUS IS THE BEST!!!!!! =(^w^)=

and maimis like *facepalm*

but then saki butts in and is like

i like SKI calculus cuz its like SAKI but like without the A

and airi and maimi are like LOLWUT

NEXT TIME ON THE AIRI AND MAIMI SHOW FEATURING °C-UTE

WTFAWESOME BATTLE OF THE BRAINZ LIKE ZOMG

SUPER FREAKING GENIUS IDOLS DISCUSS COMBINATORY LOGIC AND FORMAL GRAMMARS AND THE CHOMSKY HIERARCHY AND THE COMPUTATIONAL EQUIVALENCE OF RECURSION AND TURING MACHINES!!!!!!

and °C-ute perform AN INTERPRETIVE DANCE interpreting THE RUNTIME EXECUTION OF A SCHEME INTERPRETER INTERPRETING ITS OWN SOURCE CODE

LOLWUT kthxbai <33333333 XD

… I’ve just been busy, sorry.

Well … I’ve been living it up (?) as a graduate teaching assistant for a course in mathematics for computer science, taken mostly by second-year undergrads, with a total enrollment of around ~180. As part of my duties, I get to teach two sections that meet twice a week and also contribute to writing problems for assignments and quizzes. What fun!

Actually, the making-up-awesome-problems part really is fun! I managed to whip together an entire problem set on the topic of sums and asymptotic relations in which all the problems are Hello! Project-themed.

Alas, after discussion with the other staff members, we decided that while the problems were awesome and hilarious (maybe more so for me than for them), they were a bit on the challenging side, not straightforward enough, and touched on a few topics we weren’t really covering (Problem 4d in particular “would kill the students”). So it got scratched, and a more boring replacement was released instead.

But all is not lost! We’ve decided to release this problem set as optional, not-for-credit “challenge problems”, and you can try them out here:

If you wish, you can send your solutions to kirarinsnow@mit.edu, and I’ll respond with comments.

Enjoy!

Countdown! The Top 100 Hello! Project PVs

At this rate, we’ll get to the #1 video in the year 2023 or so…

Happy birthday Koharu!

In celebration of Koha’s birthday (07.15), here are a few double dactyls I’ve composed:

Papancake

Pancakey pancakey,
Master chef Kirari
Proves she can pancake-sort
Faster than SHIPS.

Quite inexplicably,
This method takes but a
Subpolynomial
Number of flips.

SUGAO-flavor

Flavor Flav, baklava,
Koha-chan’s talk of a
Genuine flavor” is
Just a disguise—

Hard-to-find particle
Actually is quantum
Chromodynamics
‘s
Largest in size.

Hana wo Pu~n

Kirari pikari,
Koha and Mai, though
Experts at tangent and
Cosine and sine
,

Find themselves thwarted by
Trigonometrical
Hippohypoteni:
Transforms affine.

Konnichi pa

Konnichi pa-pa-pa!
Kirari’s tra-la-la
Seizes the heartstrings and
Moves one to tears.

Poignantly touched by her
Einführungsmelodie,
Passersby nonetheless
Cover their ears.

More double dactyls are in the works, so stay tuned!

(see Part 2)

Going over the last few installments of Pocket Morning Weekly Q&A, posted in translation at Hello!Online, one might notice a developing interest in mathematics by none other than Michishige Sayumi:

2008.03.30
Question: Is there something about which you’ve thought, “Certainly this year, I want to challenge myself with this!”?
Michishige: Math Problems ☆

2008.05.11
Question: Fill in the blank to the right with one word. “I’m surprisingly ___”
Michishige: I’m surprisingly intellectual.
Please try to understand that somehow. m(・-・)m

2008.06.15
Question: Among your fellow members, what about you makes you think, “At this, I definitely can’t lose!”
Michishige: Simultaneous equations!!

The evidence is indisputable. Sayumi is a math geek. XD

While her fellow MoMusu are busy with more mundane interests, our Sayumi is off challenging herself with math problems (here, Sayu, try Project Euler) and has apparently discovered the wonders of linear algebra (I’m assuming at least some of those simultaneous equations are linear). No doubt Sayumi has mastered the techniques of Gauss-Jordan elimination, Cramer’s rule, and LU decomposition and is well on her way to achieving world domination.

In addition to this, Sayumi has listed Tetris as a hobby and as a “special skill”. This is by far the geekiest interest I’ve seen in any H!P member. Because Tetris is not your average video game. It is a mind-stretching mathematical puzzle, and several of its subproblems are NP-complete. NP-complete, I tell you! This places it in the class of difficult problems that includes Boolean k-satisfiability, determining the existence of a Hamiltonian path, and Minesweeper.

Sayumi is hardcore.

For this, she gets an Excellence in Unabashed Geekitude Award.

And I still need to give Koharu one, don’t I?

Countdown! The Top 100 Hello! Project PVs

After a three-month hiatus, the Countdown continues! And you thought I’d abandoned it!

Countdown! The Top 100 Hello! Project PVs

Welcome to yet another installment of Kawaiirrhea Goes on a Rampage and Invokes Left and Right the Names of Topic After Topic That Should Never Be Mentioned in Connection with Hello! Project.

As I’ve mentioned elsewhere, while I love compiling rankings of works released by Hello! Project, I refuse to give a subjective ranking of individual people, for various reasons. But Paul Thomas’s new poll is just so shiny I couldn’t resist…

So here is my objective ranking of Morning Musume, Berryz Koubou, and ºC-ute, using criteria that (I assume) no fan actually uses, consciously or not, to determine their favorites, though the near-perfect (except Kamei) spectral order of MoMusu is rather interesting and makes me suspect someone on the inside might be using a similar criterion to assign costume colors…

Sure, it might be a coincidence, but really, the probability of at least n – 1 out of n elements randomly appearing in a specified order is $\frac{n^2-2n+2}{n!}$.* In MoMusu’s case, n = 9, so the odds are about 1 in 5582.77. Hmm…

* I leave the proof of this formula as an exercise for the reader.

Further thoughts regarding that great pi sequence from Buono!’s “Honto no Jibun” PV:

Countdown! The Top 100 Hello! Project PVs

Disclaimer: I’m sorry; I have a tendency to burst out into bizarre math (and compsci and physics and linguistics and … and [field of index n] and …) references every now and then. If you get confused, don’t worry about it. It’s not essential to the content of the post. If you’re intrigued, though, I would recommend researching these topics further. Wikipedia is usually a good place to start. A little extra knowledge is rarely a bad thing.

I. On Rankings

Assembling rankings is definitely a favorite pastime with many H!P fans. People seem to put up new lists of their favorite H!P members, songs, etc., every other day. There are even online tools to assist people in satisfying their cravings.

I must say, though, I feel uncomfortable ranking individual members, even though I do have my favorites and my not-so-favorites. It’s kind of like discussing with your friends the order in which they matter to you. Even though it’s a discussion of perfect strangers, they’re still people…. I don’t know; it just doesn’t feel quite right. Though I don’t mean to criticize anyone else at all; everyone has their own approach to enjoying what they like. As for me, I’d much rather judge a piece of artwork than judge the artist who produces it.

And indeed, I’m not averse to compiling a ranking of H!P releases. For the last month or so, I’ve endeavored to produce a ranking of my favorite H!P PVs. It seemed like a natural thing to do, given that the music video/promotional video (whatever you want to call it) is my sixth favorite art form.* So I set about compiling a list of my top 20 and realized that there were quite a lot of videos out there that I hadn’t even seen yet. So I started watching them. And so … the list gradually grew from 20 …

… to …

… 100.

That’s right. I now have a list of my favorite 100 Hello! Project PVs.

And I’m going to discuss them all. In order.

And it’s going to be awesome.

* After the novelette, the quine, the puzzle, the cartographic map, and the tool-assisted video game simulation, particularly its dominant genre of speedruns. (Please don’t ask me what else is in my top ten. It’s like comparing apples and orthogonal bases.) Though I would be quite pleasantly surprised to see Hello! Project novelettes, quines, (original) puzzles, maps, and speedruns…

To start things off, I am announcing a new award given to displays of geekiness and/or celebrations of academic topics in music releases in any format, with special attention given to J-Pop releases. Though the first couple of winners are math-oriented, this award is by no means restricted to math-related stuff. If you would like to nominate any recording, PV, concert clip, etc., for this award, please let me know.

2007 was a truly awesome year, not just because I discovered Hello! Project in July 2007, but also because within four months of my discovering H!P, H!P released two works of such overwhelmingly mathtastic awesomeness that (being a math major) my appreciation for H!P went through the roof.

So these are the first two recipients of the Excellence in Unabashed Geekitude Award. Since I have other stuff I need to be doing today, I’ll save the second for later and write up my reasons for awarding the first award now.

Without further ado, the First Excellence in Unabashed Geekitude Award goes to…

### Remixes

DJ Kirarin☆Snow ☃'s remixes are now appearing at K!☆Mixed.