Nuking the Mosquito — ridiculously complicated ways to achieve very simple results

Solution 1:

Below is an excerpt of integrating a helical staircase I submitted as a solution in a vector calculus class years ago on April Fools when I was still an undergraduate. I got full marks and no regrets.

…therefore our integral is \begin{align*} &a\int^{2\pi}_{0}\int^{1}_{0} \sqrt{a^{2}u^{2}+b^{2}} \, dudv\\ &= 2\pi a \int_{0}^{1} \sqrt{a^{2}u^{2}+b^{2}} \, du. \qquad (1) \end{align*}

Now clearly, we should use the substitution $au = b\tan{\theta}$, but we have already demonstrated how to integrate $\sec^{3}\theta$ in Assignment 8, $\S 15.3$ Question 6. So instead, suppose we restrict ourselves to non-trigonometric substitution. Then we should instead let $\sqrt{a^{2}u^{2}+ b^{2}} = t - ua$. Then squaring both sides gives $$a^{2}u^{2}+b^{2} = t^{2} - 2uat + u^{2}a^{2}$$ and solving for u gives $$\frac{t^{2}-b^{2}}{2at} = u.$$ Differentiating both sides gives $$du = \frac{4at^{2} - 2at^{2} + 2ab^{2}}{4a^{2}t^{2}} dt.$$ Substituting this in we have $(1)$ equal to \begin{align*} &2\pi a\int_{t(0)}^{t(1)} \left(t - \frac{a(t^{2}-b^{2})}{2at}\right) \left(\frac{4at^{2} - 2at^{2} + 2ab^{2}}{4a^{2}t^{2}}\right) dt\\ &=2\pi a \int_{t(0)}^{t(1)} \left(t - \frac{t^{2}-b^{2}}{2t}\right) \left(\frac{1}{a} -\left(\frac{1}{2a}\right) +\frac{b^{2}}{2at^{2}}\right)dt\\ &= 2\pi \int_{t(0)}^{t(1)} \left(\frac{2t^{2} - t^{2}+b^{2}}{2t}\right)\left(\left(\frac{1}{2}\right) +\frac{b^{2}}{2t^{2}}\right)dt\\ &= 2\pi\int_{t(0)}^{t(1)} \left(\frac{t^{2} + b^{2}}{2t}\right) \left(\frac{t^{2}+b^{2}}{2t^{2}}\right)\, dt\\ &= 2\pi\int_{t(0)}^{t(1)} \frac{t^{4} + t^{2}b^{2} + b^{2}t^{2} + b^{4}}{4t^{3}} dt\\ &= \pi \int_{t(0)}^{t(1)} \frac{t^{4} + t^{2}b^{2} + b^{2}t^{2} + b^{4}}{2t^{3}} dt\\ &= \pi \int_{t(0)}^{t(1)} \frac{t}{2} + \frac{b^{2}}{2t} + \frac{b^{2}}{2t} + \frac{b^{4}}{2t^{3}} dt\\ &= \pi \int_{t(0)}^{t(1)} \frac{t}{2} + \frac{b^{2}}{t} + \frac{b^{4}}{2t^{3}} \, dt\\ &= \pi \left( \frac{t^{2}}{4} + b^{2}\ln |t| - \frac{b^{4}}{4t^{2}}\right) \bigg|_{t(0)}^{t(1)}. \qquad (2) \end{align*} Now when $u = 1$, we have $t = \sqrt{a^{2}+b^{2}}+a$ and when $u = 0$ we have $t = b$. Plugging these in, we have $(2)$ equal to \begin{align*} &\pi \left(\frac{(\sqrt{a^{2}+b^{2}}+a)^{2}}{4} + b^{2}\ln \left| \sqrt{a^{2}+b^{2}}+a\right| - \left(\frac{b^{4}}{4(\sqrt{a^{2}+b^{2}}+a)^{2}}\right) - \frac{b^{2}}{4} - b^{2}\ln\left|b\right| + \frac{b^{4}}{4b^{2}}\right)\\ &= \pi \left( \frac{(\sqrt{a^{2}+b^{2}}+a)^{2}}{4} - \left(\frac{b^{4}}{4(\sqrt{a^{2}+b^{2}}+a)^{2}}\right) + b^{2}\ln\left(\frac{\sqrt{a^{2}+b^{2}}+a}{b}\right) \right). \qquad (3) \end{align*} Now noticing the similarity of the first two terms, our intuition suggests this is easily simplified, so bringing this under common denominator we have $(3)$ equal to \begin{align*} &\pi\left( \frac{(\sqrt{a^{2}+b^{2}} + a)^{4} - b^{4}}{4(\sqrt{a^{2}+b^{2}} + a)^{2}} + b^{2}\ln\left(\frac{\sqrt{a^{2}+b^{2}}+a}{b}\right) \right)\\ &= \pi \left(\frac{\left(2a^{2} + 2a\sqrt{a^{2}+b^{2}} + b^{2}\right)^{2} -b^{4}}{4(\sqrt{a^{2}+b^{2}} + a)^{2}} + b^{2}\ln\left(\frac{\sqrt{a^{2}+b^{2}}+a}{b}\right) \right). \end{align*} Expanding again we have \begin{align*} &\pi \left(\frac{4a^{4} + 4a^{3}\sqrt{a^{2}+b^{2}} + 2a^{2}b^{2} + 4a^{3}\sqrt{a^{2}+b^{2}}+4a^{2}(a^{2}+b^{2}) + 2ab^{2}\sqrt{a^{2}+b^{2}}}{4\left(\sqrt{a^{2}+b^{2}}+a\right)^{2}}\right. \dots\\ &\left.\dots +\frac{2a^{2}b^{2} + 2ab^{2}\sqrt{a^{2}+b^{2}} + b^{4} - b^{4}}{4\left(\sqrt{a^{2}+b^{2}}+a\right)^{2}} +b^{2}\ln\left(\frac{\sqrt{a^{2}+b^{2}}+a}{b}\right)\right)\\ &= \pi \left( \frac{8a^{4} + 8a^{2}b^{2} + 8a^{3}\sqrt{a^{2}+b^{2}}+4ab^{2}\sqrt{a^{2}+b^{2}}}{4(\sqrt{a^{2}+b^{2}}+a)^{2}} + b^{2}\ln\left(\frac{\sqrt{a^{2}+b^{2}}+a}{b}\right)\right)\\ &= \pi \left(\frac{2a^{4}+2a^{2}b^{2}+2a^{3}\sqrt{a^{2}+b^{2}}+ab^{2}\sqrt{a^{2}+b^{2}}}{(\sqrt{a^{2}+b^{2}}+a)^{2}} + b^{2}\ln \left(\frac{\sqrt{a^{2}+b^{2}}+a}{b}\right)\right). \qquad (4) \end{align*} Using our intuition we know that the only term that was canceled after expansion was $b^{4}$ so we should examine powers of $(\sqrt{a^{2}+b^{2}}+a$ before using more complicated methods. We know from earlier that $(\sqrt{a^{2}+b^{2}} + a)^{2} = 2a^{2} + 2a\sqrt{a^{2}+b^{2}} + b^{2}$ and by examining the numerator of the first term of $(4)$, we can see that $2a^{2}$, $2a\sqrt{a^{2}+b^{2}}$ and $b^{2}$ all share the common factor of $a\sqrt{a^{2}+b^{2}}$. Therefore, $a\sqrt{a^{2}+b^{2}}(\sqrt{a^{2}+b^{2}}+a)^{2}$ is a reasonable candidate for the correct factorization of the numerator. A quick check to confirm the cross-terms match shows that \begin{align*} a\sqrt{a^{2}+b^{2}}(\sqrt{a^{2}+b^{2}}+a)^{2} &= a\sqrt{a^{2}+b^{2}} (2a^{2} + 2a\sqrt{a^{2}+b^{2}} + b^{2})\\ &= 2a^{3}\sqrt{a^{2}+b^{2}} + 2a^{2}(a^{2}+b^{2}) + ab^{2}\sqrt{a^{2}+b^{2}}\\ &= 2a^{3}\sqrt{a^{2}+b^{2}} + 2a^{4} + 2a^{2}b^{2} + ab^{2}\sqrt{a^{2}+b^{2}}\\ &= 2a^{4} + 2a^{2}b^{2} + 2a^{3}\sqrt{a^{2}+b^{2}} +ab^{2}\sqrt{a^{2}+b^{2}}. \qquad (5) \end{align*} So, now that we have verified that the cross-terms match we can use $(5)$ and thus have $(4)$ equal to \begin{align*} \pi\left(\frac{a\sqrt{a^{2}+b^{2}}(\sqrt{a^{2}+b^{2}}+a)^{2}}{(\sqrt{a^{2}+b^{2}}+a)^{2}} + b^{2}\ln(\phi)\right) &= a\pi\sqrt{a^{2}+b^{2}} + \pi b^{2}\ln\left(\frac{\sqrt{a^{2}+b^{2}}+a}{b}\right) \end{align*} which was what was to be shown.

Solution 2:

Inspired by your (far too simple!) example in the question.

Let us compute $\int_0^1 x^2\,dx$. We start with the obvious change of variables: $$ t = \frac{x}{1-x} \quad\Leftrightarrow\quad x = \frac{t}{1+t} $$ from which we get $$ dx =\frac{dt}{(1+t)^2}. $$ and the integral transforms to $$ \int_0^1 x^2\,dx = \int_0^\infty \frac{t^2}{(1+t)^4}\,dt $$ which can be attacked using residue calculus. Define $$ f(z) = \frac{z^2\log z}{(1+z)^4} $$ where $\log$ is chosen as the natural branch of the complex logarithm and integrate over a keyhole contour:

Standard estimates on the various parts of the contour shows that on $C_R$: $$ \left| \frac{z^2\log z}{(1+z)^4} \right| \le \frac{R^2(\ln R + 2\pi)}{R^4-1} $$ so $$ \left| \int_{C_R} f(z)\,dz \right| \le 2\pi R \cdot \frac{R^2(\ln R + 2\pi)}{R^4-1} $$ which tends to $0$ as $R \to \infty$. Similarly, on $C_\varepsilon$: $$ \left| \frac{z^2\log z}{(1+z)^4} \right| \le \frac{\varepsilon^2(\ln \varepsilon + 2\pi)}{(1/2)^4} $$ (if $\varepsilon < 1/2$) so $$ \left| \int_{C_R} f(z)\,dz \right| \le 2\pi R \cdot 16 \varepsilon^2(\ln \varepsilon + 2\pi), $$ which tends to $0$ as $\varepsilon \to 0^+$. It remains to investigate what happens on $I^+$ and $I^-$. On $I^+$ we get $$ \int_{I^+} f(z)\,dz = \int_{\varepsilon}^R \frac{x^2\ln x}{(1+x)^4}\,dx $$ and on $I^-$: $$ \int_{I^+} f(z)\,dz = \int_R^{\varepsilon} \frac{x^2(\ln x+2\pi i}{(1+x)^4}\,dx. $$

Putting everything together, using the residue theorem and letting $R\to\infty$, $\varepsilon\to0^+$ (note that the integrals containing $\ln x$ cancel) we get $$ -2\pi i \int_0^\infty \frac{x^2}{(1+x)^4}\,dx = 2\pi i\operatorname{Res}\limits_{z=-1} \frac{z^2\log z}{(1+z)^4}. $$ Finally, $$ \operatorname{Res}\limits_{z=-1} \frac{z^2\log z}{(1+z)^4} = \frac1{3!} (z^2\log z)^{'''}\big|_{z=-1} = -\frac13 $$ (omitting tedious algebra), and we reach the amazing result $$ \int_0^1 x^2\,dx = \int_0^\infty \frac{t^2}{(1+t)^4}\,dt = -\operatorname{Res}\limits_{z=-1} \frac{z^2\log z}{(1+z)^4} = \frac13. $$

Solution 3:

solving integrals and using complex numbers to evaluate simple problems

Integrals:

Complex numbers:

And of course you can't help but use complex arithmetic to solve the problem of "is it numberwang?"

• https://www.youtube.com/watch?v=ZH-cXBhkl-E
• https://www.youtube.com/watch?v=zJDu5D_IXbc
• https://www.youtube.com/watch?v=AIxz6BDmTNU

Solution 4:

I recall this from an actual math exam:

$ABCD$ is a square in a three-dimensional space.
The vectors $\overrightarrow{A}$, $\overrightarrow{AB}$, and $\overrightarrow{AD}$ are given. Find $\overrightarrow{C}$!

Half of the students were not able to solve it. The official solution was to solve a set of equations, and using the dot product. It was something like $\overrightarrow{AB} + \overrightarrow{BC} = \overrightarrow{AD} + \overrightarrow{DC}$; $\overrightarrow{AB} \cdot \overrightarrow{BC} = 0$; and $\overrightarrow{AD} \cdot \overrightarrow{DC} = 0$.

The topic of the exam was equation systems, so it made sense that such a solution was expected and all students, me included, tried to solve it that way. The task was worth 5 points, which means that the average student was expected to solve it in 5 minutes.

But of course, since we are talking about a square, we know that $\overrightarrow{BC} = \overrightarrow{AD}$. Thus, all it took was $\overrightarrow{C} = \overrightarrow{A} + \overrightarrow{AB} + \overrightarrow{AD}$, i.e., just add the three vectors that were given.

Solution 5:

Here is a proof of a basic combinatorial identity, using algebraic topology.

For $n \ge 0$, the $n$-simplex $\Delta^n$ can be constructed using $\binom{n+1}{k+1}$ cells of dimension $k$, for $k=0,1,\dots,n$. (There are $n+1$ vertices present, and a $k$-cell $c$ is determined uniquely by a choice of $k+1$ of these vertices to be incident on $c$.)

Therefore the euler characteristic $\chi(\Delta^n)$ is given by the alternating sum

$$\binom{n+1}{1} - \binom{n+1}{2} + \dots = \sum_{i=1}^n (-1)^{i+1}\binom{n+1}{i}.$$

On the other hand, $\Delta^n$ is contractible, so $\chi(\Delta^n) = 1 = \binom{n+1}{0}$. Rearranging the equation

$$\binom{n+1}{0} = \sum_{i=1}^{n+1} (-1)^{i+1}\binom{n+1}{i}$$

yields the familiar

$$\sum_{i=0}^{n+1} (-1)^i \binom{n+1}{i} = 0.$$