Funny identities

Question

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?

Answer 1

$$\int_0^1\frac{\mathrm{d}x}{x^x}=\sum_{k=1}^\infty \frac1{k^k}$$

Answer 2

$$\left(\sum\limits_{k=1}^n k\right)^2=\sum\limits_{k=1}^nk^3 .$$

Answer 3

$$\frac{1}{\sin(2\pi/7)} + \frac{1}{\sin(3\pi/7)} = \frac{1}{\sin(\pi/7)}$$

Answer 4

Machin's Formula: \begin{eqnarray} \frac{\pi}{4} = 4 \arctan \frac{1}{5} - \arctan \frac{1}{239}. \end{eqnarray}

Answer 5

\begin{eqnarray} 1^{3} + 2^{3} + 2^{3} + 2^{3} + 4^{3} + 4^{3} + 4^{3} + 8^{3} = (1 + 2 + 2 + 2 + 4 + 4 + 4 + 8)^{2} \end{eqnarray} More generally, let $D_{k} = ${ $d$ } be the set of unitary divisors of a positive integer $k$, and let $\mathsf{d}^{*} \colon \mathbb{N} \to \mathbb{N}$ denote the number-of-unitary-divisors (arithmetic) function. Then \begin{eqnarray} \sum_{d \in D} \mathsf{d}^{*}(d)^{3} = \left( \sum_{d \in D} \mathsf{d}^{*}(d) \right)^{2} \end{eqnarray}

Note that $\mathsf{d}^{*}(k) = 2^{\omega(k)}$, where $\omega(k)$ is the number distinct prime divisors of $k$.

Answer 6

The Frobenius automorphism

$$(x + y)^p = x^p + y^p$$

Answer 7

Ah, this is one identity which comes into use for proving the Euler's Partition Theorem. The identity is as follows: $$ (1+x)(1+x^{2})(1+x^{3}) \cdots = \frac{1}{(1-x)(1-x^{3})(1-x^{5}) \cdots}$$

Answer 8

Well, i don't know whether to classify this as funny or surprising, but ok it's worth posting.

• Let $(X,\tau)$ be a topological space and let $A \subset X$ . By iteratively applying operations of closure and complemention, one cannot produce more than 14 disjoint sets. It's called as the Kuratowski's Closure complement problem.

Answer 9

$$ \infty! = \sqrt{2 \pi} $$

It comes from the zeta function.

Answer 10

$\displaystyle\big(a^2+b^2\big)\cdot\big(c^2+d^2\big)=\big(ac \mp bd\big)^2+\big(ad \pm bc\big)^2$

Answer 11

\[\sqrt{n}^{\log n}=n^{\log \sqrt{n}}\]

Answer 12

$$ \sum\limits_{n=1}^{\infty} n = 1 + 2 +3 + \cdots \text{ad inf.} = - \frac{1}{12}$$

Now doesn't this sound funny. You can also see many more on http://terrytao.wordpress.com/2010/04/10/the-euler-maclaurin-formula-bernoulli-numbers-the-zeta-function-and-real-variable-analytic-continuation/

Answer 13

\begin{eqnarray} \sum_{i_1 = 0}^{n-k} \, \sum_{i_2 = 0}^{n-k-i_1} \cdots \sum_{i_k = 0}^{n-k-i_1 - \cdots - i_{k-1}} 1 = \binom{n}{k} \end{eqnarray}

Answer 14

Two related integrals:

$$\int_0^\infty\sin\;x\quad\mathrm{d}x=1$$

$$\int_0^\infty\ln\;x\;\sin\;x\quad \mathrm{d}x=-\gamma$$

Answer 15

$$\sec^2(x)+\csc^2(x)=\sec^2(x)\csc^2(x)$$

Answer 16

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

Answer 17

\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}

Answer 18

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:

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

Answer 19

$$\left|z+z'\right|^{2}+\left|z-z'\right|^{2}=2\times\left(\left|z\right|^{2}+\left|z'\right|^{2}\right)$$

The sum of the squares of the sides equals the sum of the squares of the diagonals.

Answer 20

$$\large{1,741,725 = 1^7 + 7^7 + 4^7 + 1^7 + 7^7 + 2^7 + 5^7}$$

and

$$\large{111,111,111 \times 111,111,111 = 12,345,678,987,654,321}$$

Answer 21

M.V Subbarao's identity: an integer n>22 is a prime number iff it satisfies,

$$n\sigma(n)\equiv 2 \pmod {\phi(n)}$$

Answer 22

What is 42?

$$ 6 \times 9 = 42 \text{ base } 13 $$ I always knew that there is something wrong with this universe.

Answer 23

Considering the main branches

$$i^i = \exp\left(-\frac{\pi}{2}\right)$$

$$\root i \of i = \exp\left(\frac{\pi}{2}\right) $$

And $$ \frac{4}{\pi } = \displaystyle 1 + \frac{1}{{3 +\displaystyle \frac{{{2^2}}}{{5 + \displaystyle\frac{{{3^2}}}{{7 +\displaystyle \frac{{{4^2}}}{{9 +\displaystyle \frac{{{n^2}}}{{\left( {2n + 1} \right) + \cdots }}}}}}}}}} $$

Answer 24

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

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

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

Answer 25

$\lnot$(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".

Answer 26

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.