Funny identities

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?

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Maybe a moderator should put the zeta ones together since there are three already? –  anon Nov 3 '10 at 22:29
Perhaps this should be a community wiki question. –  Nuno Nov 3 '10 at 22:31
This is related. –  Ｊ. Ｍ. Nov 3 '10 at 22:35
I have tripped up many calculus students with this one: $log(1+2+3)=log1+log2+log3$. I am evil... –  user641 Dec 8 '12 at 1:23
@SteveD If only we could find an odd example... –  peoplepower Jan 13 '13 at 0:31

The product of any four consecutive integers is one less than a perfect square.

To phrase it more like an identity:

For every integer $n$, there exists an integer $k$ such that $$n(n+1)(n+2)(n+3) = k^2 - 1.$$

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You can also write that as $$n(n+1)(n+2)(n+3)=((n+1)^2+1)^2 - 1$$ –  AD. Jun 19 '12 at 6:53
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\begin{align} \frac{\mathrm d}{\mathrm dx}(x^x) &= x\cdot x^{x-1} &\text{Power Rule?}&\text{False}\\ \frac{\mathrm d}{\mathrm dx}(x^x) &= x^{x}\ln(x) &\text{Exponential Rule?}&\text{False}\\ \frac{\mathrm d}{\mathrm dx}(x^x) &= x\cdot x^{x-1}+x^{x}\ln(x) &\text{Sum of these?}&\text{True}\\ \end{align}

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This is a special case of $\frac{d}{dx} h(f(x),g(x)) = \partial_1 h f' + \partial_2h g'$ –  ronno Dec 20 '13 at 13:56

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 $$p(A) = \det(AI_n-A) = \det(0-0) = 0$$

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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 $$\det \left[ \begin{array}{cc} U & R \\ L & D \end{array} \right] = \det (UD-LR)$$

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\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}

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

$$\binom{n}{0}+\binom{n}{1}+\binom{n}{2}+\cdots+\binom{n}{n}=2^n$$

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

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I have one: In a $\Delta ABC$, $$\tan A+\tan B+\tan C=\tan A\tan B\tan C.$$

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Also $\cot(A/2)+\cot(B/2)+ \cot(C/2)=\cot(A/2)\cot(B/2)\cot(C/2)$. –  N. S. Apr 11 '13 at 22:39

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

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Remind me of this: http://xkcd.com/184/ –  alex.jordan Nov 3 '13 at 17:31
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$$\sin \theta \cdot \sin \bigl(60^\circ - \theta \bigr) \cdot \sin \bigl(60^\circ + \theta \bigr) = \frac{1}{4} \sin 3\theta$$

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

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

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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 $\sin(\theta+120^{\circ})$.) –  Hans Lundmark Nov 4 '10 at 9:56
considering your first two identities the thirth should be $$\tan \theta \cdot \tan \bigl(60 - \theta \bigr) \cdot \tan \bigl(60 + \theta \bigr) = \tan 3\theta$$ –  Neves Mar 6 '11 at 16:08

\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.

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$\tan^{-1}(1)+\tan^{-1}(2)+\tan^{-1}(3) = \pi$ (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:

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

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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:

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

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

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

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

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

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\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}

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$$\sqrt{\vphantom{\large A}2025\,} = 20 + 25 = 45$$

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

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

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

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

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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.

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$$\int_{-\infty}^{\infty}{\sin\left(x\right) \over x}\,{\rm d}x = \pi\int_{-1}^{1}\delta\left(k\right)\,{\rm d}k$$

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Let $\sigma(n)$ denote the sum of the divisors of $n$.

If $$p=1+\sigma(k),$$ then $$p^a=1+\sigma(kp^{a-1})$$ where $a,k$ are positive integers and $p$ is a prime such that $p\not\mid k$.

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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.

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

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$$\frac{\pi}{4}=\sum_{n=1}^{\infty}\arctan\frac{1}{f_{2n+1}},$$ where $f_{2n+1}$ there are fibonacci numbers, $n=1,2,...$
Here is an Asian kid paradox - perhaps they will understand (apologies for not being strictly mathematical). If $$Study=No \;Fail$$ and $$No \; Study=Fail$$ then $$Study+No\;Study=Fail+No\;Fail$$ $$\implies (1+No)Study=(1+No)Fail$$ Cancelling gives $$Study=Fail$$ Isn't that weird???
$1+No$ is 0 in both sides. You can't cancel zero... –  CODE Jun 4 '13 at 16:47