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What is the favorite equation of your life? I know this might be a subjective question, and may be not-so-on-topic here, so if anyone decides to close this, could you link me somewhere I can ask this?

It doesn't have to be simple, it doesn't have to be complicated. It could be something you came across in grad school, it could be something you came across in the fifth grade. It just needs to elegant to you.

For some context, the inspiration for this question is another question I came across here earlier today, asking for,

$$1 - \frac13 + \frac15 - \frac17 +\cdots = $$

I didn't actually answer the question, since it was "homework" with no effort whatsoever. From what I can tell, he was given this question in sigma-notation, he didn't even bother expanding to obtain the individual terms.

Anyway, I didn't have to calculate, mainly because I remember this result from the first time I derived it, simply because I was so impressed by the fact that a sum involving the reciprocals of odd numbers can come out to be $\pi/4$, the sum had nothing to do with circles! This was admittedly one of the turning points on my outlooks towards mathematics. I never found math hard or terrible, intellectually challenging and engaing yes, but never unbearable. There was a satisfaction with "cracking a puzzle". But I would never consider math "beautiful" before this, this one revelation changed my life(at least the bit that had to do with math, maybe more).

So the question is, what was your revelation, if any? It could be more than one too. You could say this question is in part out of a curiosity to find more such awe-inspiring revelations.

Here for any readers who are not familiar with above expression, here is one short, and in my opinion elegant, proof.

$$1 - x^2 + x^4 - x^6 + \cdots = \frac1{1+x^2}$$

A simple geometric series, which on integrating both sides, from $x = 0$ to $1$, gives the result.

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closed as primarily opinion-based by Najib Idrissi, vonbrand, Davide Giraudo, Your Ad Here, Andrew D. Hwang Mar 5 '14 at 21:53

Many good questions generate some degree of opinion based on expert experience, but answers to this question will tend to be almost entirely based on opinions, rather than facts, references, or specific expertise.If this question can be reworded to fit the rules in the help center, please edit the question.

You can try, they love this kind of question. On MSE it's IMO off-topic. – Najib Idrissi Mar 5 '14 at 19:48
To whatever extent this isn't off-topic, it should be CW. – MJD Mar 5 '14 at 19:49
up vote 2 down vote accepted

Eulers Reflection Formula for the Gamma function is hard to beat for elegance , simplicity and beauty , $$ \Gamma(z) \cdot \Gamma(1 - z) = \frac{ \pi}{ \sin( \pi z)}$$

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The factorial of a positive number is the Gaussian of its reciprocal: $$n!=\mathcal G\bigg(\frac1n\bigg),\qquad\text{where}\qquad\mathcal G(n)=\int_0^\infty e^{-x^n}dx$$

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whoa! I have never ever seen this. wow, thanks man. – Sabyasachi Mar 5 '14 at 19:33
I will upvote this tommorow, vote limit for the day. – Sabyasachi Mar 5 '14 at 19:35
See $\Gamma$ function. The two expressions are equivalent, by a simple change in variable. – Lucian Mar 5 '14 at 19:39
ah yes. although this raises one question, how does this integral have anything to do with the factorial function. – Sabyasachi Mar 5 '14 at 19:46
It can easily be proven, using a simple integration by parts, that $\Gamma(n+1)=n\cdot\Gamma(n)$. Then, since $\displaystyle\Gamma(1)=\int_0^\infty x^0e^{-x}dx=1$, the desired identity follows immediately. – Lucian Mar 5 '14 at 19:56


I was playing with a calculator with an exponent function. I was doing compound interest calculations and found that this converged to e as N went to infinity. I found this to be pretty amazing, as I'd never seen this in my schooling, only e as the area under a curve, as if e was defined to create an answer to an integral of 1/x.

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Shame that no one told you this, since this is the definition of $e$. But might have been a blessing in disguise, since I can only imagine your elation at discovering it yourself. – Sabyasachi Mar 5 '14 at 19:28
you forgot to include e in the equation though. – Sabyasachi Mar 5 '14 at 19:29
Here's a little something you might like – Sabyasachi Mar 5 '14 at 19:30
added e. Thanks. Yes, it really was a delight to discover. – JoeTaxpayer Mar 5 '14 at 19:34
@Sabayasachi There are quite a few equivalent definitions of Euler's constant. – PVAL Mar 5 '14 at 19:38

I was playing around some time with a sequence recursively defined as following:

$$a_0 = 1,\quad a_{n+1}=a_n + \frac 1{a_n}.$$ $$1, 2, 2.5, 2.9, \dotsc$$

Using spreadsheets I suspected that this sequence gets close to $\sqrt{2n}$ for big $n$, i.e. $$\lim_{n\to\infty} \frac{a_n}{\sqrt{2n}} = 1.$$

But I had no clue how to prove it. But then I had an idea:

Note that $a_{n+1}^2 = a_n^2+\frac1{a_n^2}+2$. Since $a_n^2$ is positive we have $a_n^2 \ge 2n$. By induction

$$a_{n+1}^2 = 2n + a_0^2 +\sum_{i=0}^n \frac1{a_n^2} \le 2n+1+\sum_{i=0}^n\frac1{2i}$$

The harmonic series grows logarithmically, so it gets small if we divide this by $2n$. So we have

$$1 \le \frac{a_n^2}{2n} \le 1+\text{some terms converging to zero}$$

Basically, the big step for me in proving this was looking at the squares and finding the above formula with the sum. It was kind of an accomplishment that I could finally prove this.

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wow, nice observation you got there. I wouldn't have even seen the $\sqrt{2n}$. Good catch. – Sabyasachi Mar 5 '14 at 19:53
I saw it in a spreadsheet by looking at values for large n, for example $a_{5000}$ is roughly $100$. That gave me the conjecture. – ljfa Mar 5 '14 at 19:55
AND after seeing it, even the similarity between 2.9 and 2.828 is apparent. :D math! – Sabyasachi Mar 5 '14 at 19:56

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