Funny identities [closed]

Here is a funny exercise $$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$ (If you prove it don't publish it here please). Do you have similar examples?

• This is related. Commented Nov 3, 2010 at 22:35
• I noticed that your expression can be also written as $\sin(x - y) \sin(x + y) = (\cos y + \cos x) (\cos y - \cos x)$ Commented Nov 4, 2010 at 11:09
• I have tripped up many calculus students with this one: $log(1+2+3)=log1+log2+log3$. I am evil...
– user641
Commented Dec 8, 2012 at 1:23
• @SteveD If only we could find an odd example... Commented Jan 13, 2013 at 0:31
• Almost an identity: $$\sqrt{123456790}\approx 11111.11111\,.$$
– user158047
Commented Jul 12, 2014 at 18:47

Best near miss

$$\int_{0}^{\infty }\cos\left ( 2x \right )\prod_{n=0}^{\infty}\cos\left ( \frac{x}{n} \right )~\mathrm dx\approx \frac{\pi}{8}-7.41\times 10^{-43}$$

One can easily be fooled into thinking that it is exactly $\dfrac{\pi}{8}$.

References:

$$\int_0^\infty\frac1{1+x^2}\cdot\frac1{1+x^\pi}dx=\int_0^\infty\frac1{1+x^2}\cdot\frac1{1+x^e}dx$$

• Is this true for any exponent, or just $\pi$ and $e$? Commented Aug 17, 2015 at 19:43
• It's a simple exercise, try it. Commented Aug 17, 2015 at 21:41
• @Vladimir: Would you mind giving a hint? Commented Nov 28, 2015 at 19:51
• The identity also holds if you replace $\pi$ with Euler's $\gamma$ :) Commented Nov 29, 2015 at 1:19
• Obviously... :) Commented Feb 21, 2016 at 14:01

Let $f$ be a symbol with the property that $f^n = n!$. Consider $d_n$, the number of ways of putting $n$ letters in $n$ envelopes so that no letter gets to the right person (aka derangements). Many people initially think that $d_n = (n-1)! = f^{n-1}$ (the first object has $n-1$ legal locations, the second $n-2$, ...). The correct answer isn't that different actually:

$d_n = (f-1)^n$.

• Hooray for Umbral calculus! Commented Jun 18, 2012 at 22:41
• To make this more rigorous, in a sense: We can define a linear operator $L$ acting on $\mathbb{C}[f]$ such that $L(f^n)=n!$ and $L(1)=1$. Thus, we can write $d_n=L\left((f-1)^n\right)$. (Am I doing this right?) Commented Aug 28, 2014 at 4:00

$(x-a)(x-b)(x-c)\ldots(x-z) = 0$

• I don't get it? Commented Aug 17, 2015 at 19:52
• @columbus8myhw $(x-a)(x-b)(x-c)\ldots (x-w){\color{red}{(x-x)}}(x-y)(x-z)=0$
– Surb
Commented Aug 23, 2015 at 16:00

$$71 = \sqrt{7! + 1}.$$

Besides the amusement of reusing the decimal digits $7$ and $1$, this is conjectured to be the last solution of $n!+1 = x^2$ in integers. ($n=4$ and $n=5$ also work.) Even finiteness of the set of solutions is not known except using the ABC conjecture.

Heres a interesting one again
$3435=3^3+4^4+3^3+5^5%$

I actually think currying is really cool:

$$(A \times B) \to C \; \simeq \; A \to (B \to C)$$

Though not strictly an identity, but an isomorphism.

When I met it for the first time it seemed to be a bit odd but it is so convenient and neat. At least in programming.

We have by block partition rule for determinant $$\det \left[ \begin{array}{cc} U & R \\ L & D \end{array} \right] = \det U\cdot \det ( D-LU^{-1}R)$$ But if $U,R,L$ and $D$ commute we have that $$\det \left[ \begin{array}{cc} U & R \\ L & D \end{array} \right] = \det (UD-LR)$$

• Actually, only U and L need to commute.
– Dirk
Commented Aug 22, 2018 at 14:08

The Cayley-Hamilton theorem:

If $A \in \mathbb{R}^{n \times n}$ and $I_{n} \in \mathbb{R}^{n \times n}$ is the identity matrix, then the characteristic polynomial of $A$ is $p(\lambda) = \det(\lambda I_n - A)$. Then the Cayley Hamilton theorem can be obtained by "substituting" $\lambda = A$, since $$p(A) = \det(AI_n-A) = \det(0-0) = 0$$

$$\frac{1}{998901}=0.000001002003004005006...997999000001...$$

• Similar results for $1/((9...9)^2)$, and any further analysis will explain why :-) Commented Jul 12, 2012 at 4:03
• Indeed, I saw it explained here initially :-) Commented Jul 14, 2012 at 20:19
• 1/98.99 = 0.010102030508132134... Commented Sep 26, 2013 at 14:41

\begin{eqnarray} \zeta(0) = \sum_{n \geq 1} 1 = -\frac{1}{2} \end{eqnarray}

• Actually, this can be made rigorous by noting that $$\zeta(z)=\lim_{n\to\infty}\left(\sum_{k=1}^nk^{-z}-\frac{1}{1-z}n^{1-z}-\frac12n^{-z}\right)$$ for $\mathrm{Re}(z)>-1$.
– robjohn
Commented Jun 21, 2012 at 0:52

$$\dfrac{1}{2}=\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\dfrac{\dfrac{1}{2}}{\dfrac{1}{2}+\cdots}}}}}}$$

and more generally we have $$\dfrac{1}{n+1}=\frac{\dfrac{1}{n(n+1)}}{\dfrac{1}{n(n+1)}+\dfrac{\dfrac{1}{n(n+1)}}{\dfrac{1}{n(n+1)}+\dfrac{\dfrac{1}{n(n+1)}}{\dfrac{1}{n(n+1)}+\dfrac{\dfrac{1}{n(n+1)}}{\dfrac{1}{n(n+1)}+\dfrac{\dfrac{1}{n(n+1)}}{\dfrac{1}{n(n+1)}+\dfrac{\frac{1}{n(n+1)}}{\dfrac{1}{n(n+1)}+\ddots}}}}}}$$

\begin{eqnarray} \sum_{k = 0}^{\lfloor q - q/p) \rfloor} \left \lfloor \frac{p(q - k)}{q} \right \rfloor = \sum_{k = 1}^{q} \left \lfloor \frac{kp}{q} \right \rfloor \end{eqnarray}

• I don't see the 'punch' here. Isn't that just reversing the order of summation and truncating some zeros?
– Ofir
Commented Jan 13, 2013 at 0:10

\begin{align}\frac{64}{16}&=\frac{6\!\!/\,4}{16\!\!/}\\&=\frac41\\&=4\end{align}

For more examples of these weird fractions, see "How Weird Are Weird Fractions?", Ryan Stuffelbeam, The College Mathematics Journal, Vol. 44, No. 3 (May 2013), pp. 202-209.

• Commented Nov 15, 2013 at 16:59
• Commented Oct 24, 2016 at 21:12

$$\frac{\pi}{4}=\sum_{n=1}^{\infty}\arctan\frac{1}{f_{2n+1}},$$ where $f_{2n+1}$ there are fibonacci numbers, $n=1,2,...$

$$2592=2^59^2$$ Found this in one of Dudeney's puzzle books

$$\sin \theta \cdot \sin \bigl(60^\circ - \theta \bigr) \cdot \sin \bigl(60^\circ + \theta \bigr) = \frac{1}{4} \sin 3\theta$$

$$\cos \theta \cdot \cos \bigl(60^\circ - \theta \bigr) \cdot \cos \bigl(60^\circ + \theta \bigr) = \frac{1}{4} \cos 3\theta$$

$$\tan \theta \cdot \tan \bigl(60^\circ - \theta \bigr) \cdot \tan \bigl(60^\circ + \theta \bigr) = \tan 3\theta$$

• Not much idea what you want me to understand ? :) Commented Nov 4, 2010 at 9:38
• I just wanted to mention that your first identity is equivalent to the case $n=3$ of the formula for $\sin nx$ given there. (Just replace $\sin(60^{\circ}-\theta)$ by $\sin(\theta+120^{\circ})$.) Commented Nov 4, 2010 at 9:56
• considering your first two identities the thirth should be $$\tan \theta \cdot \tan \bigl(60 - \theta \bigr) \cdot \tan \bigl(60 + \theta \bigr) = \tan 3\theta$$ Commented Mar 6, 2011 at 16:08

$$27\cdot56=2\cdot756,$$ $$277\cdot756=27\cdot7756,$$ $$2777\cdot7756=277\cdot77756,$$ and so on.

• How does it work? Commented Oct 2, 2015 at 10:45

$$\sum_{n=1}^{+\infty}\frac{\mu(n)}{n}=1-\frac12-\frac13-\frac15+\frac16-\frac17+\frac1{10}-\frac1{11}-\frac1{13}+\frac1{14}+\frac1{15}-\cdots=0$$ This relation was discovered by Euler in 1748 (before Riemann's studies on the $\zeta$ function as a complex variable function, from which this relation becomes much more easier!).

Then one of the most impressive formulas is the functional equation for the $\zeta$ function, in its asimmetric form: it highlights a very very deep and smart connection between the $\Gamma$ and the $\zeta$: $$\pi^{\frac s2}\Gamma\left(\frac s2\right)\zeta(s)= \pi^{\frac{1-s}2}\Gamma\left(\frac{1-s}2\right)\zeta(1-s)\;\;\;\forall s\in\mathbb C\;.$$

Moreover no one seems to have wrote the Basel problem (Euler, 1735): $$\sum_{n=1}^{+\infty}\frac1{n^2}=\frac{\pi^2}{6}\;\;.$$

$\lnot$(A$\land$B)=($\lnot$A$\lor$$\lnotB) and \lnot(A\lorB)=(\lnotA\land$$\lnot$B), because they mean that negation is an "equal form".

Here's one clever trigonometric identity that impressed me in high-school days. Add $\sin \alpha$, to both the numerator and the denominator of $\sqrt{\frac{1-\cos \alpha}{1 + \cos \alpha}}$ and get rid of the square root and nothing changes. In other words:

$$\frac{1 - \cos \alpha + \sin \alpha}{1 + \cos \alpha + \sin \alpha} = \sqrt{\frac{1-\cos \alpha}{1 + \cos \alpha}}$$

If you take a closer look you'll notice that the RHS is the formula for tangent of a half-angle. Actually if you want to prove those, nothing but the addition formulas are required.

$\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3) = \pi$ (using the principal value), but if you blindly use the addition formula $\tan^{-1}(x) + \tan^{-1}(y) = \tan^{-1}\dfrac{x+y}{1-x y}$ twice, you get zero:

$\tan^{-1}(1) + \tan^{-1}(2) = \tan^{-1}\dfrac{1+2}{1-1*2} =\tan^{-1}(-3)$; $\tan^{-1}(1) + \tan^{-1}(2) + \tan^{-1}(3) =\tan^{-1}(-3) + \tan^{-1}(3) =\tan^{-1}\dfrac{-3+3}{1-(-3)(3)} = 0$.

$$\begin{array}{rcrcl} \vdots & \vdots & \vdots & \vdots & \vdots \\[1mm] \int{1 \over x^{3}}\,{\rm d}x & = & -\,{1 \over 2}\,{1 \over x^{2}} & \sim & x^{\color{#ff0000}{\large\bf -2}} \\[1mm] \int{1 \over x^{2}}\,{\rm d}x & = & -\,{1 \over x} & \sim & x^{\color{#ff0000}{\large\bf -1}} \\[1mm] \int{1 \over x}\,{\rm d}x & = & \ln\left(x\right) & \sim & x^{\color{#0000ff}{\LARGE\bf 0}} \color{#0000ff}{\LARGE\quad ?} \\[1mm] \int x^{0}\,{\rm d}x & = & x^{1} & \sim & x^{\color{#ff0000}{\large\bf 1}} \\[1mm] \int x\,{\rm d}x & = & {1 \over 2}\,x^{2} & \sim & x^{\color{#ff0000}{\large\bf 2}} \\[1mm] \vdots & \vdots & \vdots & \vdots & \vdots \end{array}$$

• $\ln(x)$ converges slowly enough, so indeed $\ln(x) \sim 1$ might not be completely stupid :p Commented Sep 28, 2013 at 8:31
• Hmm, considering that logarithms get at the exponent, and $x$ has a constant exponent ... Since $\ln\left(x^a\right)=a\ln\left(x\right)$ (the log of an expression equals the exponent times the log of the base), then $\ln\left(x^1\right)=1\ln\left(x\right)=x^0\ln\left(x\right)$ might be saying something to the effect that it's more important that your exponent is a constant, than the fact that the log of your base $\ln\left(x\right)$ is growing slowly. Commented Sep 28, 2013 at 10:11
• @Travis Wow, nice one! But I think you took my comment too seriously. Commented Sep 28, 2013 at 13:13
• The mystery, perhaps, lies in the constant of integration. All of the other integrals are evaluated from 0 to x, while the $\dfrac1x$ one is evaluated from 1 to x. (Note, by the way, that $\displaystyle\ln x=\lim_{t\to0}\frac{x^t-1}t$, which perhaps fits the pattern better.) Commented Aug 28, 2014 at 4:08

$$\lim_{\omega\to\infty}3=8$$ The "proof" is by rotation through $\pi/2$. More of a joke than an identity, I suppose.

• Remind me of this: http://xkcd.com/184/ Commented Nov 3, 2013 at 17:31
• Or maybe the "proof" is through taking the closure? Commented Dec 15, 2013 at 5:57

For all $n\in\mathbb{N}$ and $n\neq1$ $$\prod_{k=1}^{n-1}2\sin\frac{k \pi}{n} = n$$

For some reason, the proof involves complex numbers and polynomials.

\begin{align} E &= \sqrt{\left(pc\right)^{2} + \left(mc^{2}\right)^{2}} = mc^{2} + \left[\sqrt{\left(pc\right)^{2} + \left(mc^{2}\right)^{2}} - mc^{2}\right] \\[3mm]&= mc^{2} + {\left(pc\right)^{2} \over \sqrt{\left(pc\right)^{2} + \left(mc^{2}\right)^{2}} + mc^{2}} = mc^{2} + {p^{2}/2m \over 1 + {\sqrt{\left(pc\right)^{2} + \left(mc^{2}\right)^{2}} - mc^{2} \over 2mc^{2}}} \\[3mm]&= mc^{2} + {p^{2}/2m \over 1 + {p^{2}/2m \over \sqrt{\left(pc\right)^{2} + \left(mc^{2}\right)^{2}} + mc^{2}}} = mc^{2} + {p^{2}/2m \over 1 + {p^{2}/2m \over 1 + {p^{2}/2m \over \sqrt{\left(pc\right)^{2} + \left(mc^{2}\right)^{2}} - mc^{2}}}} \end{align}

$\textbf{Claim:}\quad$$\frac{\sin x}{n}=6$$ for all$n,x$($n\neq 0$).$\textit{Proof:}\quad$$\frac{\sin x}{n}=\frac{\dfrac{1}{n}\cdot\sin x}{\dfrac{1}{n}\cdot n}=\frac{\operatorname{si}x}{1}=\text{six}.\quad\blacksquare$$

• how do you obtain $\dfrac{six}{1}$ ?
– ASB
Commented Jan 29, 2015 at 13:49
• @Tgymasb Denominator: $1/n\cdot n=1$. Numerator: $1/\textbf{n}\cdot\operatorname{si}\!\textbf{n}\, x=\operatorname{si}x$. Commented Jan 29, 2015 at 14:00

$$\frac{\pi}{2}=1+2\sum_{k=1}^{\infty}\frac{\eta(2k)}{2^{2k}}$$ $$\frac{\pi}{3}=1+2\sum_{k=1}^{\infty}\frac{\eta(2k)}{6^{2k}}$$ where $\eta(n)=\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^{n}}$

If we define $P$ as the infinite lower triangular matrix where $P_{i,j} = \binom{i}{j}$ (we can call it the Pascal Matrix), then $$P^k_{i,j} = \binom{i}{j}k^{i-j}$$

where $P^k_{i,j}$ is the element of $P^k$ in the position $i,j.$

Voronoi summation formula:

$\sum \limits_{n=1}^{\infty}d(n)(\frac{x}{n})^{1/2}\{Y_1(4\pi \sqrt{nx})+\frac{2}{\pi}K_1(4\pi \sqrt{nx})\}+x \log x +(2 \gamma-1)x +\frac{1}{4}=\sum \limits _{n\leq x}'d(n)$