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

The Frobenius automorphism

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

Answer 4

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

Answer 5

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 6

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

It comes from the zeta function.

Answer 7

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

Answer 8

\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 9

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

Answer 10

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 can produce at most 14 distinct sets. It's called as the Kuratowski's Closure complement problem.

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

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

and

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

Answer 14

\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 15

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 16

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

Answer 17

Facts about $\pi$ are always fun!

\begin{equation} \frac{\pi}{2} = \frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot\frac{6}{5}\cdot\frac{6}{7}\cdot\frac{8}{7}\cdot\ldots\\ \end{equation} \begin{equation} \frac{\pi}{4} = 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\ldots\\ \end{equation} \begin{equation} \frac{\pi^2}{6} = 1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\ldots\\ \end{equation} \begin{equation} \frac{\pi^3}{32} = 1-\frac{1}{3^3}+\frac{1}{5^3}-\frac{1}{7^3}+\frac{1}{9^3}+\ldots\\ \end{equation} \begin{equation} \frac{\pi^4}{90} = 1+\frac{1}{2^4}+\frac{1}{3^4}+\frac{1}{4^4}+\frac{1}{5^4}+\ldots\\ \end{equation} \begin{equation} \frac{2}{\pi} = \frac{\sqrt{2}}{2}\cdot\frac{\sqrt{2+\sqrt{2}}}{2}\cdot\frac{\sqrt{2+\sqrt{2+\sqrt{2}}}}{2}\cdot\ldots\\ \end{equation} \begin{equation} \pi = \cfrac{4}{1+\cfrac{1^2}{3+\cfrac{2^2}{5+\cfrac{3^2}{7+\cfrac{4^2}{9+\ldots}}}}}\\ \end{equation}

Answer 18

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

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

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 21

$$ \frac{e}{2} = \left(\frac{2}{1}\right)^{1/2}\left(\frac{2\cdot 4}{3\cdot 3}\right)^{1/4}\left(\frac{4\cdot 6\cdot 6\cdot 8}{5\cdot 5\cdot 7\cdot 7}\right)^{1/8}\left(\frac{8\cdot 10\cdot 10\cdot 12\cdot 12\cdot 14\cdot 14\cdot 16}{9\cdot 9\cdot 11\cdot 11\cdot 13\cdot 13\cdot 15\cdot 15}\right)^{1/16}\cdots $$ [Nick Pippenger, Amer. Math. Monthly, 87 (1980)]. Set all the exponents to 1 and you get the Wallis formula for $\pi/2$.

Answer 22

The following number is prime

$p = 785963102379428822376694789446897396207498568951$

and $p$ in base 16 is

$89ABCDEF012345672718281831415926141424F7$

which includes counting in hexadecimal, and digits of $e$, $\pi$, and $\sqrt{2}$.

Do you think this's surprising or not?

$$11 \times 11 = 121$$ $$111 \times 111 = 12321$$ $$1111 \times 1111 = 1234321$$ $$11111 \times 11111 = 123454321$$ $$\vdots$$

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

By excluding the first two primes, Euler's Prime Product becomes a square:

$$\prod _{n=3}^{\infty } \frac{1}{1-\frac{1}{(p_n)^{2}}}=\frac{\pi ^2}{9}$$

By using multiples of the product of the first two primes, we get the square root:

$$\prod _{n=1}^{\infty } \frac{1}{1-\frac{1}{(n p_1 p_2)^{2}}}=\frac{\pi }{3}$$

Answer 25

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 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.

Answer 27

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

Answer 28

What is 42?

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

Answer 29

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 $$n(n+1)(n+2)(n+3) = k^2 - 1.$$

Answer 30

\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}