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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.

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closed as too broad by Grigory M, Magdiragdag, Hakim, Sujaan Kunalan, Pedro Tamaroff May 19 '14 at 0:27

There are either too many possible answers, or good answers would be too long for this format. Please add details to narrow the answer set or to isolate an issue that can be answered in a few paragraphs.If this question can be reworded to fit the rules in the help center, please edit the question.

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) – Nicholas Stull May 16 '14 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:… – Kaj Hansen May 16 '14 at 18:07
@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? – Bill Dubuque May 16 '14 at 19:11
He's using the product rule, $\int u \, dv = uv - \int v \, du$, with $u=x, dv=dx$ – Snowbody May 16 '14 at 19:57
I hate such questions to be put on hold or closed. Sure, this is not about a specific mathematical problem, but is fun to read and actually sometimes raises some interesting issues. I believe, that the hermetic SE rules should have exceptions for such posts. I mean, doesn't it look like ridiculous? +32 question with 18 answers (including +33 and +25) and closed - the same as -5 question with 0 answers. Something is not right here. – Spook May 19 '14 at 10:03

17 Answers 17

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.

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You have a sick and twisted sense of humor ;0 – PA6OTA May 17 '14 at 0:57
+1 for an awesome overcomplication of an already complicated integral. – Nicholas Stull May 17 '14 at 2:03
al-Khwarizmi would be mightily impressed! – Cor_Blimey May 17 '14 at 18:59
Your teacher probably hated you, however. – Zarrax May 17 '14 at 20:58

solving integrals and using complex numbers to evaluate simple problems



Complex numbers:

Calvin and Hobbes

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

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ahem those are imaginary numbers, not complex numbers! ;) – Emily May 16 '14 at 20:57
@Arkamis: Filthundred and neeb? THAT'S NUMBERWANG! – Eric Lippert May 16 '14 at 21:10
Seen the Calvin and Hobbes, but the rest are new. +1 – Aidan F. Pierce May 16 '14 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. – recursive recursion May 16 '14 at 22:23
Whenever somebody pulls out that C&H comic, I always tell them that, coincidentally, eleventeen is exactly half thirty-twelve. – Joe Z. May 17 '14 at 22:58

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: enter image description here

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

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I like this example !! Nice ! – jibounet May 19 '14 at 8:13

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.

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I wonder if any of them tried imagining the square. The official solution is fairly amusing :) – Navin May 17 '14 at 5:18

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

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

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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!

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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.

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Why is this an example of what the question is asking for? – Ragib Zaman May 17 '14 at 19:59
I would say that because one can prove CS without the notions of eigenvalue, nor matrix, nor transpose, nor semipositiveness. – Martin Argerami May 18 '14 at 23:58
What @MartinArgerami said. – user21467 May 19 '14 at 23:56
(Also, the theorem that symmetric matrices are diagonalizable is rather a big gun for the target.) – user21467 May 20 '14 at 0:04

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


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 $X$.

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

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

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

$(1)$ 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.

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I've seen that proof. It wasn't a particularly fun proof to work through either... – Nicholas Stull May 16 '14 at 18:05
Not at all. I would replicate the result, but by the time I finished it would be irrelevant. – Mr.Fry May 16 '14 at 18:07
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 '14 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. – Mr.Fry May 16 '14 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. – Andrea Mori May 16 '14 at 18:45

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.

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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) $-0.04$

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What is the mosquito and where is the nuking? – Did May 18 '14 at 10:17
@Did It turns out, that you do not always have to use integrals and complex numbers to make something simple a lot more complicated. – Spook May 18 '14 at 10:32
@Did Mosquito: A sequence summing to an arbitrary value, with no other restrictions. Nuke: Anything involving floors or logs. – michaelb958 May 19 '14 at 1:31

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 $F$.

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You were done when you noted the kernel is nontrivial. – Ragib Zaman May 17 '14 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. – Kaj Hansen May 18 '14 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? – DanZimm May 18 '14 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. – Kaj Hansen May 18 '14 at 19:02

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 $K=A$.

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This is the first example that comes to my mind:

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


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 :))

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This is a very natural, simple proof, not a "ridicuously complicated" one. – Bill Dubuque May 16 '14 at 17:58
So anything that was not known 2000 years ago is "ridiculously complicated"? – Bill Dubuque May 16 '14 at 18:02
this is wrong logic. – mesel May 16 '14 at 18:03
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. – Alex Zorn May 16 '14 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. – mesel May 16 '14 at 18:41

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? ;))

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