Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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?

share|improve this question
3  
Maybe a moderator should put the zeta ones together since there are three already? –  anon Nov 3 '10 at 22:29
3  
Perhaps this should be a community wiki question. –  Nuno Nov 3 '10 at 22:31
7  
This is related. –  J. M. Nov 3 '10 at 22:35
85  
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
5  
@SteveD If only we could find an odd example... –  peoplepower Jan 13 '13 at 0:31
show 8 more comments

61 Answers

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

share|improve this answer
1  
1/98.99 = 0.010102030508132134... –  Michael Sep 26 '13 at 14:41
show 2 more comments

\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}

share|improve this answer
4  
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
add comment

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

share|improve this answer
show 2 more comments

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

share|improve this answer
add comment

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

share|improve this answer
add comment

\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}

share|improve this answer
3  
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
add comment

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

share|improve this answer
2  
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
add comment

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

share|improve this answer
add comment

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

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

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

share|improve this answer
1  
See also the comments here: math.stackexchange.com/q/8385/1242 –  Hans Lundmark Nov 4 '10 at 8:53
1  
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
1  
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
show 2 more comments

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

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

share|improve this answer
add comment

\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.

share|improve this answer
show 1 more comment

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

share|improve this answer
1  
Remind me of this: http://xkcd.com/184/ –  alex.jordan Nov 3 '13 at 17:31
show 1 more comment

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

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

share|improve this answer
add comment

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:

share|improve this answer
add comment

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

share|improve this answer
1  
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. –  Travis Bemrose Sep 28 '13 at 10:11
show 2 more comments

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

share|improve this answer
add comment

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

share|improve this answer
add comment

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

share|improve this answer
add comment

\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}

share|improve this answer
add comment

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

share|improve this answer
add comment

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

share|improve this answer
add comment

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

share|improve this answer
add comment

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.

share|improve this answer
add comment

$\textbf{Claim:}\quad$$$\frac{\sin x}{n}=6$$ for all $n,x$ ($n\neq 0$).

$\textit{Proof:}\quad$$$\frac{\sin x}{n}=\frac{\dfrac{1}{n}\cdot\sin x}{\dfrac{1}{n}\cdot n}=\frac{\operatorname{si}x}{1}=\text{six}.\quad\blacksquare$$

share|improve this answer
add comment

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 $p\not\mid k$.

share|improve this answer
add comment

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.

Link to proof: Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$

share|improve this answer
add comment

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

share|improve this answer
add comment

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

share|improve this answer
add comment

$$ \sum_{n=1}^{+\infty}\frac{\mu(n)}{n}=1-\frac12-\frac13-\frac15+\frac16-\frac17+\frac1{10}-\frac1{11}-\frac1{13}+\frac1{14}+\frac1{15}-\cdots=0 $$ 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): $$ \sum_{n=1}^{+\infty}\frac1{n^2}=\frac{\pi^2}{6}\;\;. $$

share|improve this answer
add comment

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

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.