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

Here is a toned down example of what I'm looking for:

Integration by solving for the unknown integral of $f(x)=x$:

$$\int x \, dx=x^2-\int x \, dx$$

$$2\int x \, dx=x^2$$

$$\int x \, dx=\frac{x^2}{2}$$

Can anyone think of any more examples?

P.S. This question was inspired by a question on MathOverflow that I found out about here. This question is meant to be more general, accepting things like solving integrals and using complex numbers to evaluate simple problems.

• I computed the integral $\int \frac{\ln(x)}{x}\,\text{d}x$ by integration by parts to show students in my class the danger of assuming that every integral with $\ln(x)$ must be done using integration by parts... Personally, my favorite is: Let $f\, : \, [0,1] \to \mathbb{R}$ be bounded measurable. Then by Carleson's Theorem, the partial Fourier Series gives a sequence of $C^\infty$ functions which converge to $f$ almost everywhere. (credit: Nate Eldredge at MO) May 16, 2014 at 17:54
• I feel like this answer from another thread (not mine) is relevant: With Fermat's Last Theorem, we can prove $\sqrt[n]{2}$ is irrational: math.stackexchange.com/questions/796236/… May 16, 2014 at 18:07
• I like this nuking. May 16, 2014 at 18:08
• @user2357112 Consider $\, D(1) = D(\color{#c00}1\cdot 1) = D(\color{#c00}1)\cdot 1 +\color{#c00}1\cdot D(1)\,\Rightarrow\, D(1) = 0.\,$ Do you object to that too? May 16, 2014 at 19:11
• He's using the product rule, $\int u \, dv = uv - \int v \, du$, with $u=x, dv=dx$ May 16, 2014 at 19:57

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.

• You have a sick and twisted sense of humor ;0 May 17, 2014 at 0:57
• +1 for an awesome overcomplication of an already complicated integral. May 17, 2014 at 2:03
• al-Khwarizmi would be mightily impressed! May 17, 2014 at 18:59
• Your teacher probably hated you, however. May 17, 2014 at 20:58
• This is certainly an impressive show of algebra, which I can see is the point of this whole question, however for people like me with much less intuitional capabilities, I should point out that since $\left(\sqrt{a^2+b^2}+a\right)\left(\sqrt{a^2+b^2}-a\right)=b^2$, by rationalising the denominator and difference of two squares we have $$\frac{\left(\sqrt{a^2+b^2}+a\right)^4-b^4}{4\left(\sqrt{a^2+b^2}+a\right)^2}=\frac{1}{4}\left(\left(\sqrt{a^2+b^2}+a\right)^2-\left(\sqrt{a^2+b^2}-a\right)^2\right)=a\sqrt{a^2+b^2}$$ Jan 6, 2020 at 18:57

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 $$\int_0^1 x^2\,dx = \int_0^\infty \frac{t^2}{(1+t)^4}\,dt = -\operatorname{Res}\limits_{z=-1} \frac{z^2\log z}{(1+z)^4} = \frac13.$$

• I like this example !! Nice ! May 19, 2014 at 8:13

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?"

• ahem those are imaginary numbers, not complex numbers! ;) May 16, 2014 at 20:57
• @Arkamis: Filthundred and neeb? THAT'S NUMBERWANG! youtube.com/watch?v=zJDu5D_IXbc May 16, 2014 at 21:10
• Seen the Calvin and Hobbes, but the rest are new. +1 May 16, 2014 at 22:14
• +1 for the foxtrot. Never seen that one before. I read most of the books in fifth grade before I knew how to take logarithms or integrate, so I didn't get most of the math jokes. May 16, 2014 at 22:23
• Whenever somebody pulls out that C&H comic, I always tell them that, coincidentally, eleventeen is exactly half thirty-twelve. May 17, 2014 at 22:58

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.

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

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

I've always had the feeling that using the calculus of variations to prove that a line is the shortest path between two points is a bit overkill...

$$I(f)=\int_a^b \sqrt{1+y'^2}\,\mathrm{d}x$$ $$\frac{\partial f}{\partial y}=\frac{d}{dx}\frac{\partial f}{\partial y'}$$ $$0=\frac{d}{dx}\frac{y'}{\sqrt{1+y'^2}}$$ $$y'=C\sqrt{1+y'^2}\implies y'^2(C^2-1)=-C^2\implies y'=c$$ so that we finally get $$y=cx+d.$$

Who knew that proving such an simple statement could be so hard!

• In some introductions to Euclidean geometry, where the Euclidean plane is simply defined as $\mathbb{R}^2$ and the lines as sets of the form $\{(x,y):ax+by+c=0\}$, this apparent overkill is necessary to show that segments are length-minimizing curves (geodesics) and reconcile the metric definition of distance through the Pythagrean theorem and the definition of distance as length of the shortest path. Nov 19, 2017 at 21:07
• Echoing Jack's comment, the fact that the solution sets of linear equations (the algebraically simplest ones) are length-minimizing strikes me as something rather deep, linking algebra (polynomials) to geometric analysis (length minimization). Jan 16, 2020 at 20:41

Let $x,y\in\mathbb{R}^n$. Let $A$ be the $n\times 2$ matrix whose columns are $x$ and $y$. A standard argument shows that any eigenvalue of $A^TA$ is nonnegative (consider $\langle v,A^TAv\rangle$); since $A^TA$ is symmetric, it's diagonalizable, so it follows that $\det(A^TA)\ge 0$. Thus: $$0 \le \det(A^TA) = \left|\begin{matrix} \|x\|^2 & \langle x,y\rangle \\ \langle x,y\rangle & \|y\|^2 \end{matrix}\right| = \|x\|^2\|y\|^2 - \langle x,y\rangle^2$$ and we have proved the Cauchy-Schwarz inequality.

• Why is this an example of what the question is asking for? May 17, 2014 at 19:59
• I would say that because one can prove CS without the notions of eigenvalue, nor matrix, nor transpose, nor semipositiveness. May 18, 2014 at 23:58

If you estimate the area of a circle by circumscribing a hexagon, you get the inequality $\pi < 2\sqrt3$. An alternative proof is: $$\frac{\pi^2}{6} = \sum_{n=1}^\infty \frac1{n^2} < 1 + \sum_{n=2}^\infty \frac1{n(n-1)} = 1 + \sum_{n=2}^\infty \left(\frac1{n-1} - \frac1n\right) = 2$$

Lemma: Let $X \sim \mathcal{N}(0,1)$ be normally distributed. Then $Y = \sigma X + \mu$ is normally distributed as $Y \sim \mathcal{N}(\mu,\sigma^2)$.

Proof:

Let $Y$ be represented by its polynomial chaos expansion: $$Y = \sum_{i=0}^\infty y_i\Phi_i(\zeta).$$ Choose $\zeta$ to to have a zero mean normal distribution inducing a choice of $\Phi_i = H_i$, where $H_i$ is the $i$th Hermite Polynomial appropriately scaled so that $\left\langle H_i H_j\right\rangle = \delta_{ij}$.

We compute the expansion coefficients using the Galerkin method by projecting each orthogonal polynomial basis function onto both sides of the expansion:

$$\left\langle Y H_j(\zeta)\right\rangle = \left\langle \sum_{i=0}^\infty y_iH_i(\zeta) H_j(\zeta)\right\rangle \\ y_j = \frac{1}{\left\langle H_j^2 (\zeta)\right\rangle} \int_{\mathbb{R}} YH_j(\zeta) w(\zeta)\ d\zeta,$$ where $w(\zeta)$ is the weighting function of the Hermite polynomials, appropriately scaled.

Since $Y$ and $\zeta$ are fully correlated, perform an inverse transform of their distribution functions to the same uniformly-distributed random variable $u$:

$$F(Y) = u = G(\zeta) \implies h(u) \equiv F^{-1}(u) = Y, l(u) \equiv G^{-1}(u) = \zeta.$$

Note that the CDF of the standard normal distribution can be written in terms of the error function: $$G(\zeta) = \frac12\left(1+\textrm{erf}\left(\frac{\zeta}{\sqrt{2}}\right)\right),$$ so we can write $$l(u) = \sqrt{2} \textrm{erf}^{-1}(2u-1).$$

Similarly, it is easy to show that $$h(u) = \sqrt{2\sigma^2}\textrm{erf}^{-1}(2u-1)+\mu.$$

Substituting all this into the integral for $y_j$, we find

\begin{align*} y_j & = \int_0^1 h(u)H_j(l(u))\ du \\ &= \int_0^1 \sqrt{2\sigma^2}\textrm{erf}^{-1}(2u-1)H_j(\sqrt{2} \textrm{erf}^{-1}(2u-1))\ du + \int_0^1 \mu H_j(\sqrt{2} \textrm{erf}^{-1}(2u-1))\ du \\ &= \underbrace{\int_{\mathbb{R}} \sqrt{2\sigma^2}\zeta H_j(\zeta) w(\zeta)\ d\zeta}_{\sqrt{\sigma^2}\left\langle H_1 H_j\right\rangle} + \underbrace{\int_\mathbb{R} \mu H_j(\zeta)w(\zeta)\ d\zeta}_{\left\langle H_0 , 1\right\rangle} \end{align*}

Hence, the first integral is non-zero only for $j=1$, and the second is non-zero for only $j=0$. By the appropriate choice of scaling of the Hermite polynomials, we have

$$Y = \mu + \sigma \zeta$$

and we finally note that $\zeta$ is identically distributed to $X$.

$(1)$ By far the most complicated way of doing something I've seen came from my real analysis course. We were instructed to manipulate the power series of $\cos(x),\sin(x)$ namely, $$\cos(x) = 1 - \frac{x^2}{2!} +\frac{x^4}{4!} - \cdots , \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$$ to derive addition formulas for $\cos,\sin$ versus this or using rotation matrices to derive the result.

$(2)$ Other interesting results can be seen when proving the "big theorems" in analysis such as mean value, extreme value and intermediate value theorem. In topology although very general, the results are reached fairly quickly with the knowledge of connectedness, continuity. \

Check out analysis proof of IVT versus this topological proof: Let $a,b \in X$, and let $r \in Y$ lie between $f(a)$ and $f(b)$. Define sets $A=f(X)\cap(−\infty,r)$ and $B=f(X)\cap(r,\infty)$. These sets are clearly disjoint, and they are clearly nonempty since one contains $f(a)$ and the other contains $f(b)$. We can also see that they are both open by definition as the intersection of open sets. Assume there is no point c such that $f(c)=r$. Then $f(X)=A \cup B$, so $A$ and $B$ constitute a separation of $X$. But this contradicts the fact that the image of a connected space under a continuous mapping is connected.

• This argument is just an application of the binomial theorem. As far as generating functions go, this particular proof is relatively short and simple.
– blue
May 16, 2014 at 18:14
• @seaturtles: Yes we all know that, but proving the addition formulas of $\sin,\cos$ via manipulating power series is not trivial at all and involved very clever manipulations of double sums. May 16, 2014 at 18:17
• It's true that some Analysis theorems appear rather "trivial" once topology and its terminology are introduced, but the simplification is only apparent because the difficulty is shifted to prove that $[a,b]$ is compact and $\Bbb R$ is connected. May 16, 2014 at 18:45
• This derivation of trig identities is on the contrary very simple: you get quickly all you need, including 2pi periodicity, from very little material (only the definition of $e^z$), and without having to resort too much to geometric intuition (and much "fun" with many cases according to quadrant) nor to inegrating $1/(1+t^2)$, taking inverse then completing by periodicity (and that would only give you the real case, not complex trig functions). So, if you want a rigorous derivation of trig functions and identities, it's by far the simplest. May 19, 2014 at 14:11

You can also say when you do an integral by parts, that you are using the Stoke's theorem...

• Isn't this true for all integrals? (not just integration by parts) May 17, 2014 at 5:14
• Stokes theorem is a generalization of the fundamental theorem of calculus, not specifically integration by parts. May 18, 2014 at 16:52

Once during my math studies I encountered a task of creating an infinite sequence of numbers, which sums up to some arbitrary value. Since there were no other restrictions on the sequence, the simplest and most obvious solution was:

$$a, 0, 0, 0, ...$$

But instead I created a function, which, when given real $$x$$ and positive non-zero integer $$n$$ returned $$n$$th digit of that real value in appropriate 10th power. So for $$\pi$$ we would get $$3, 0.1, 0.04, 0.001, 0.0005$$ and so on.

The function looked more less like the following:

$$f(x,n) = signum(x) \cdot (\lfloor \lvert x \rvert \cdot 10^{\lfloor -log_{10}(\lvert x \rvert) \rfloor} \cdot 10^n \rfloor - (\lfloor \lvert x \rvert \cdot 10^{\lfloor-log_{10}(\lvert x \rvert \rfloor} \rfloor \cdot 10^{n-1}) \cdot 10) \cdot 10^{\lfloor log_{10}(\lvert x \rvert)) \rfloor - (n-1)}$$

$$\text{where} ~ n = 1, 2, ...$$

Then I declared the sequence to be:

$$f(a, 1), f(a, 2), f(a, 3), ...$$

Output from Maxima:

(%i1) log10(x) := log(x) / log(10);

(%o1) $${log10}\left( x\right) :=\frac{\mathrm{log}\left( x\right) }{\mathrm{log}\left( 10\right) }$$

(%i2) f(x,n):=signum(x)*(floor(abs(x)*10^(floor(-log10(abs(x))))*10^n) - floor(abs(x)*10^(floor(-log10(abs(x))))*10^(n-1))*10)*10^(floor(log10(abs(x))) - (n-1));

(%o2) $$\mathrm{f}\left( x,n\right) :=\mathrm{signum}\left( x\right) \,\left( \mathrm{floor}\left( \left| x\right| \,{10}^{\mathrm{floor}\left( -\mathrm{log10}\left( \left| x\right| \right) \right) }\,{10}^{n}\right) -\mathrm{floor}\left( \left| x\right| \,{10}^{\mathrm{floor}\left( -\mathrm{log10}\left( \left| x\right| \right) \right) }\,{10}^{n-1}\right) \,10\right) \,{10}^{\mathrm{floor}\left( \mathrm{log10}\left( \left| x\right| \right) \right) -\left( n-1\right) }$$

(%i3) f(123.456, 2);

(%o3) $$20.0$$

(%i4) f(-%pi, 3);

(%o4) $$-0.04$$

• What is the mosquito and where is the nuking?
– Did
May 18, 2014 at 10:17
• @Did Mosquito: A sequence summing to an arbitrary value, with no other restrictions. Nuke: Anything involving floors or logs. May 19, 2014 at 1:31

Here is a major number-theoretical nuking. By a result of Gronwall (1913) the Generalized Riemann Hypothesis (GRH) implies that the only quadratic number fields $\,K$ whose integers have unique factorization are $\,\Bbb Q[\sqrt {-d}],\,$ for $\,d\in \{1,2,3,7,11,19,43,67,163\}.\,$ Therefore, if $\,K$ is not in this list then it has an integer with a nonunique factorization into irreducibles.

But that can be proved much more simply in any particular case, e.g. the classic elementary proof that $\,2\cdot 3 = (1-\sqrt{-5})(1+\sqrt{-5})\,$ is a nonunique factorization into irreducibles in $\,\Bbb Z[\sqrt{-5}],\,$ which can easily be comprehended by a bright high-school student.

Similarly, other sledgehammers arise by applying general classification theorems to elementary problems, e.g. classifications of (finite) (abelian) (simple) groups. Examples of such sledgehammers can be found here and on MathOverflow by keyword searches.

Do you also accept algebraic propositions? If so:

Finite domains are fields (the standard proof is a one liner: note that multiplication by a nonzero element is injective hence surjective):

Let $A$ be a finite domain. Let $K$ its field of fractions. Certainly $K$ is a finite $A$-module. Consider $K/A \otimes_A K/A$. This is obviously $0$. But for finitely generated $A$ modules $M, N$ it is $\operatorname{supp}(M \otimes_A N) = \operatorname{supp}(M) \cap \operatorname{supp}(N)$. In other words $M \otimes M = 0$ implies $M=0$. Hence in our situation $K=A$.

Here's one I experienced on this site not too long ago:

Question: Assume $K/F$ is a field extension of finite degree. Prove that the extension is algebraic.

My "atomic bomb":

To show that it is algebraic, select any $a \in K$ \ $F$. Then consider the evaluation homomorphism $ev_a:F[x] \rightarrow K$ defined as follows: $g(x) \mapsto g(\alpha)$. Certainly, this map cannot be injective because $L$ is finitely generated, so its kernel must be a nontrivial ideal. By the isomorphism theorems, we know that $F[x]/\ker(ev_a) \cong K$.

Next, $K$ is a field, so it is definitely an integral domain. We know that $F[x]/\ker(ev_a)$ is an integral domain $\iff$ $\ker(ev_a)$ is a prime ideal. This is only possible if $\ker(ev_a)$ is generated by an irreducible polynomial in $F[x]$.

We conclude that, for every $a \in K$ \ $F$, there exists an irreducible polynomial in $F[x]$ with $a$ as a root. Therefore, $K$ is an algebraic extension of $F$.

A much more elegant response by the user Fretty:

To show $K/F$ is algebraic if finite we must show that every element of $K$ satisfies a polynomial over $F$.

Suppose $[K : F] = n$ and choose $\alpha\in K$. Then consider the elements $1,\alpha,\alpha^2,...,\alpha^n$.

This is a list of $n+1$ elements in an $n$ dimensional $F$-vector space so must be linearly dependent. Thus there exists $a_0,a_1,...,a_n\in F$ not all zero such that $a_n \alpha^n + ... + a_2\alpha^2 + a_1\alpha + a_0 = 0$.

But then $\alpha$ is a root of the polynomial $a_nx^n + ... + a_2x^2 + a_1x + a_0$ over $F$.

• You were done when you noted the kernel is nontrivial. May 17, 2014 at 19:55
• Well I'll be! You're right. I'm glad I contributed to this thread because now I'm walking away with a little more experience under my belt. May 18, 2014 at 5:55
• This is probably a naive question but how does a non-trivial kernel of an evaluation map of $a$ imply that $a$ is algebraic, without the rest of the proof? May 18, 2014 at 14:18
• Well, the kernel of a homomorphism $\phi:R \rightarrow S$ is the set of elements in $R$ that map to $0$ in $S$. In this case, $ev_a$ sends $g(x)$ to $g(a)$, and since there's a nontrivial kernel, then that means there is some polynomial other than the zero polynomial such that $g(a) = 0$. In other words, some irreducible $g(x) \in F[x]$ has $a$ as a root, and therefore $a \in K$ is algebraic. May 18, 2014 at 19:02

Here's one!

Prove that $2^\frac13$ is irrational.

Assume the contrary, that $2^\frac13$ is rational.

Then: $$2^\frac13 = \frac{a}{b}$$ $$2=\frac{a^3}{b^3}$$ $$2b^3=a^3$$ $$b^3+b^3=a^3$$ Which contradicts Fermat's Last Theorem!

Hence, $2^\frac13$ is irrational.

Very elegant in itself, but nonetheless, quite overpowered.

• Wiles's proof of FLT assumes the $n = 3$ case, and doesn't prove it. Here's a deleted answer to this question that makes the same claim, essentially.
– user296602
Apr 2, 2018 at 2:35
• I can't see the deleted answer for some reason... but I certainly see your point. Can we assume FLT to be true ;) ? Apr 2, 2018 at 2:40

$$n!$$ is not a power of a prime for $$n\geq 3$$.

Boring proof: $$n!$$ is divisible by two distinct primes $$2$$ and $$3$$.

Cool proof: For $$n\geq 5$$, the only normal subgroups of $$S_n$$ are $$\{1\},A_n$$ and $$S_n$$. Since $$A_n,S_n$$ are not commutative, it follows that the center $$Z(S_n)$$ is trivial. On the other hand, if $$|S_n|=n!$$ was a power of a prime, then $$S_n$$ would be a (nontrivial) $$p$$-group, and such always has a nontrivial center, which is a contradiction.

To prove that there are infinitely many primes, you can use the prime number theorem, since if there were finitely many the asymptotics of the number of primes less than or equal to $x$ would be ${\displaystyle {k \over x}}$, where $k$ is the number of primes.

Or you can just invoke the Green-Tao theorem....

(And don't give me any stuff about circular reasoning ok? ;))

This is the first example that comes to my mind:

Let $$f(x)=\sin^2(x)+\cos^2(x)$$ then $$f(0)=1$$ and

$$f'(x)=2\sin(x)\cos(x)-2\cos(x)\sin(x)=0$$

so $$f$$ is a constant function and since $$f(0)=1 \implies f(x)=1=\sin^2(x)+\cos^2(x)$$

Explanation: The explanation mostly depends on the discussion in the below comments;

When we think the usual definition of $$\sin,\cos$$ function by unit circle it is obvious fact that $$\sin^2(x)+\cos^2(x)=1$$ as they are coordinate of points of the circle.

Above proof used two strong tools one of them is derivatives second of them is a corollary of mean value theorem.

$$f'(x)=0 \implies f$$ is constant function. (most books prove this by the mean value theorem)

Of course it is a valid proof but I thought it as "killing a fly with atomic bomb".

By the way as @Alex zorn point out if we just start from $$\sin'=\cos$$ and $$\cos'=-\sin$$ it is a natural proof which was not my intention and @Bill thought that it is completely natural proof which I respect his opinion. (He also thought that my opinion about this proof will change when I get enough experience, but I do not think so:) )

Please notice that there is no exact definition of being "ridiculously complicated" and it is kind of subjective topic but it is natural since it is a soft question.

I feel need to make this explanation to show my thought behind it, thanks for all who thought about this example and give some reaction. (including the ones who downvote :))

• This is a very natural, simple proof, not a "ridicuously complicated" one. May 16, 2014 at 17:58
• This is actually a really nice proof if your starting point for sin and cos are the assumptions that sin' = cos, cos' = -sin, sin(0) = 0, cos(0) = 1. May 16, 2014 at 18:13
• @BillDubuque: I did not say calculus proof is ridiculously complicated. I think it is again wrong logic. If you think usual defination of $sin,cos$ by the unit circle, it is obvious fact that $sin^2+cos^2=1$ and to show this, using srtong tools(weapon) like derivative is not needed. Even the case, $f'=0$ then $f$ is a constant function is a strong result. That is why I thought it as an good example. If you do not agree with me, it is okey but please do not reach the conclusion I did not say and mean. May 16, 2014 at 18:41
• I think it is quite logical to attempt to infer your definition of "unreasonably complicated" from what you wrote above. If those comments do not accurately reflect the definition that you have in mind then please do elaborate. Precisely what is it about the calculus proof that makes you deem it "ridiculously complicated" if it is not the fact that it uses calculus? May 16, 2014 at 18:56
• Your last comment does help to disambiguate your earlier ones. However, I think we have different ideas of ridiculously complicated. To me that signifies proofs that would never be used in practice (except as a joke), which is not the case for the proof in your answer. – Bill Dubuque 5 mins ago May 16, 2014 at 19:29

$$\frac{1}{n(n + 1)} < \frac{1}{n^2} \hspace{2.0em} \text{for } n \geq 1$$ Therefore: \begin{align*}\sum_{n = 1}^{\infty}{\frac{1}{n(n + 1)}} &< \sum_{n = 1}^{\infty}{\frac{1}{n^2}} = \zeta(2) = \frac{\pi^2}{6} \\ \sum_{n = 1}^{\infty}{\left(\frac{1}{n} - \frac{1}{n + 1}\right)} &< \frac{\pi^2}{6} \\ \sum_{n = 1}^{\infty}{\frac{1}{n}} - \sum_{n = 1}^{\infty}{\frac{1}{n + 1}} &< \frac{\pi^2}{6} \end{align*} And then telescoping: \begin{align*}\sum_{n = 1}^{\infty}{\frac{1}{n}} - \sum_{n = 2}^{\infty}{\frac{1}{n}} &< \frac{\pi^2}{6} \\ \frac{1}{1} &< \frac{\pi^2}{6} \\ \pi^2 &> 6\end{align*} And so: $$\pi > \sqrt{6}$$

You can also use this method to get an upper bound of $\sqrt{12}$ (use $\frac{1}{n(n - 1)}$). If you modify this to use $\zeta(4), \zeta(6)...$ etc, it actually gives worse bounds. E.g, for $\zeta(4)$ a similar method gives $\pi > 1.86$.

Computing sums of powers of roots using the argument principle.

Example:

Let $$\alpha,\beta,\gamma,\delta$$ be the distinct roots of $$x^4+x^2+1$$. Find $$\alpha^6+\beta^6+\gamma^6+\delta^6$$.

Simple algebraic solution:

Note that $$(x^2-1)(x^4+x^2+1)=x^6-1=0\Rightarrow x^6=1$$, so each the answer is just $$1+1+1+1=4$$.

Using the argument principle:

Let $$\sigma_n=\alpha^n+\beta^n+\gamma^n+\delta^n$$.

By the generalization of the argument principle,

$$\sigma_6=\frac1{2\pi i}\oint_C\frac{4z^3+2z}{z^4+z^2+1}z^6~\mathrm dz$$

where $$C$$ is a circle of radius $$2$$. By long division this becomes

$$\frac{4z^3+2z}{z^4+z^2+1}z^6=4z^5-2z^3+2z+\frac{4z^3+2z}{z^4+z^2+1}$$

but by the argument principle, this simple becomes

$$\sigma_6=4$$

This can be used instead of things such as Newton's identities and Vieta's formulas, albeit very messily sometimes.

This just showed up linked from a duplicate, so here's one of mine, reproduced from AoPS:

Problem (original author Koukai):

Let $$E$$ be the set of all continuous functions $$u:[0,1]\longmapsto\mathbb R$$ that satisfy $$|u(x)-u(y)|\le|x-y|$$ for $$0\le x,y\le1$$ and $$u(0)=0.$$ Let $$f:E\longmapsto\mathbb R$$ be defined by $$f(u)=\int_0^1(u^2(x)-u(x))\,dx.$$ Show that $$f$$ attains its maximum value at some element of $$E.$$

The easy way (original author WWW):

We have $$|u(x)|\le x,x\in [0,1],$$ for all $$u \in E.$$ So for all $$u\in E,$$ $$\int_0^1(u^2-u) \le \int_0^1(u^2+|u|) \le \int_0^1(x^2+x)\,dx = \frac{5}{6}.$$ This value is attained when $$u(x)=-x,$$ which belongs to $$E,$$ so we're done.

The ridiculous way (original author jmerry):

First, note that given any compact topological space $$K$$, a continuous function from $$K$$ to $$\mathbb{R}$$ attains its maximum and minimum - this is a version of the Extreme Value Theorem, which follows from the fact that the continuous image of a compact set is compact. We wish to show that under some reasonable topology, $$E$$ is compact and $$f$$ is continuous.

To define the topology on $$E$$, let's try a natural choice - the uniform norm $$\|u\|_{\infty}=\sup_x |u(x)|$$ on bounded functions. The distance between two functions $$u$$ and $$v$$ is then $$\|u-v\|_{\infty}$$, and it is fairly easy to verify the axioms of a metric space. We do need to note that $$\|u(x)\|\le 1$$ for all $$u\in E$$ and all $$x$$, so everything in our space has a norm. Since our $$u$$ are continuous, the supremum in the definition is actually a maximum.

Next, we show that $$f$$ is continuous. We have $$(u^2(x)-u(x))-(v^2(x)-v(x))=(u(x)-v(x))(u(x)+v(x)+1)$$, so $$\left|(u^2(x)-u(x))-(v^2(x)-v(x))\right|\le 3(u(x)-v(x))\le 3\|u-v\|_{\infty}$$. Integrate that and $$|f(u)-f(v)|\le 3\|u-v\|_{\infty}$$. That's a Lipschitz condition, which is more than enough for continuity.

Now, what does that $$|u(x)-u(y)|\le |x-y|$$ get us? Since that applies everywhere in the interval, it's uniform (Lipschitz) continuity. Since it applies to all $$u$$ in $$E$$, it's also equicontinuity. $$E$$ is a uniformly bounded and equicontinuous family of functions on a compact interval... so we can apply the Arzelà-Ascoli Theorem. Any sequence in $$E$$ has a uniformly convergent subsequence; if we can show that the limit of such a subsequence is always in $$E$$, we have that $$E$$ is compact.

So, the final lemma: if $$\|u_n-u\|_{\infty}\to 0$$ with each $$u_n\in E$$, $$u$$ is also in $$E$$. For any $$\epsilon>0$$, there is some $$N$$ such that $$|u_n-u|_{\infty}<\epsilon$$ whenever $$n>N$$. Choose $$x$$ and $$y$$, as well as some particular $$n>N$$; we then have $$|u(x)-u(y)|\le |u(x)-u_n(x)|+|u_n(x)-u_n(y)|+|u_n(y)-u(y)|< 2\epsilon+|x-y|$$ Now, since $$\epsilon$$ was arbitrary, we let it go to zero; if $$|u(x)-u(y)|<2\epsilon+|x-y|$$ for all $$\epsilon>0$$, $$|u(x)-u(y)|\le |x-y|$$. That's the definition of $$E$$, and we're done.

A nonconstructive proof that the maximum exists by invoking an advanced theorem - when all we needed to do was maximize it pointwise with an explicit construction.