beautiful proofs

Beautiful proofs (#4) – When Gauss was a young child…

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The legend goes something like this:

Gauss’s teacher wanted to occupy his students by making them add large sets of numbers and told everyone in class to find the sum of 1+2+3+ …. + 100.

And Gauss, who was a young child (age ~ 10) quickly found the sum by just pairing up numbers:

Using this ingenious method used by Gauss allows us to write a generic formula for the sum of first n positive integers as follows:

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Beautiful Proofs(#3): Area under a sine curve !

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So, I read this post on the the area of the sine curve some time ago and in the bottom was this equally amazing comment :

Screenshot from 2017-06-07 00:19:11

\sum sin(\theta)d\theta =   Diameter of the circle/ The distance covered along the x axis starting from 0 and ending up at \pi.

And therefore by the same logic, it is extremely intuitive to see why:

\int\limits_{0}^{2\pi} sin/cos(x) dx = 0

Because if a dude starts at 0 and ends at 0/ 2\pi/ 4\pi \hdots, the effective distance that he covers is 0.

Circle_cos_sin.gif

If you still have trouble understanding, follow the blue point in the above gif and hopefully things become clearer.

 

Beautiful proofs(#2): Euler’s Sum

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1 + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \hdots = \frac{\pi^2}{6}

Say what? This one blew my mind when I first encountered it. But it turns out Euler was the one who came up with it and it’s proof is just beautiful!

Prerequisite
Say you have a quadratic equation f(x) whose roots are r_1,r_2 , then you can write f(x) as follows:

f(x) = (x-r_1)(x-r_2) =  0   (or)

f(x) = (r_1-x)(r_2-x) =  0   (or)

f(x) =  (1- \frac{x}{r_1})(1- \frac{x}{r_2}) =  0

f(x) = 1 - (\frac{1}{r_1} + \frac{1}{r_2}) + \frac{x^2}{r_1 r_2} = 0

As for as this proof is concerned we are only worried about the coefficient of x , which you can prove that for a n-degree polynomial is:

a_1 = - (\frac{1}{r_1} + \frac{1}{r_2} + \hdots + + \frac{1}{r_n})

where r_1,r_2 \hdots r_n are the n-roots of the polynomial.

 

Now begins the proof

It was known to Euler that

f(y) = \frac{sin(\sqrt{y})}{\sqrt{y}} = 1 - \frac{1}{3!}y + \hdots

But this could also be written in terms of the roots of the equation as:

f(y) = \frac{sin(\sqrt{y})}{\sqrt{y}} = 1 - (\frac{1}{r_1} + \frac{1}{r_2} + \hdots + + \frac{1}{r_n})y + \hdots

Now what are the roots of f(y) ?. Well, f(y) = 0 when \sqrt{y} = n \pi i.e y = n^2 \pi^2 *

The roots of the equation are y = \pi^2, 4 \pi^2, 9 \pi^2, \hdots

Therefore,

f(y) = \frac{sin(\sqrt{y})}{\sqrt{y}} = 1 - \frac{1}{3!}y + \hdots = 1 -( \frac{1}{\pi^2} + \frac{1}{4 \pi^2} + \hdots )y + \hdots

Comparing the coefficient of y on both sides of the equation we get that:

\frac{1}{6} = \frac{1}{\pi^2} + \frac{1}{4 \pi^2} + \frac{1}{ 9 \pi^2} + \hdots

\zeta(2) = \frac{\pi^2}{6} = 1 + \frac{1}{4} + \frac{1}{9} + \hdots 

Q.E.D

* n=0 is not a root since
\frac{sin(\sqrt{y})}{\sqrt{y}} = 1 at y = 0

** Now if all that made sense but you are still thinking : Why on earth did Euler use this particular form of the polynomial for this problem, read the first three pages of this article. (It has to do with convergence)

Beautiful proofs (#1) : Divergence of the harmonic series

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The harmonic series are as follows:

image

And it has been known since as early as 1350 that this series diverges. Oresme’s proof to it is just so beautiful.

image
image

Now replace ever term in the bracket with the lowest term that is present in it. This will give a lower bound on S1.

image
image

Clearly the lower bound of S1 diverges and therefore S1 also diverges.
But it interesting to note that of divergence is incredibly small: 10 billion terms in the series only adds up to around 23.6 !

Beautiful proofs (#1) : Divergence of the harmonic series

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The harmonic series are as follows:

\sum\limits_{n=1}^{\infty} \frac{1}{n} = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \hdots

And it has been known since as early as 1350 that this series diverges. Oresme’s proof to it is just so beautiful.

S_1 = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \hdots

S_1 = 1 + \left(\frac{1}{2}\right) + \left(\frac{1}{3} + \frac{1}{4}\right) + \left(\frac{1}{5} + \frac{1}{6} + \frac{1}{7} + \frac{1}{8} \right) \hdots

Now replace ever term in the bracket with the lowest term that is present in it. This will give a lower bound on S_1 .

S_1 > 1 + \left(\frac{1}{2}\right) + \left(\frac{1}{4} + \frac{1}{4}\right) + \left(\frac{1}{8} + \frac{1}{8} + \frac{1}{8} + \frac{1}{8} \right) \hdots

S_1 > 1 + \left(\frac{1}{2}\right) + \left(\frac{1}{2}\right) + \left(\frac{1}{2}\right) + \left(\frac{1}{2}\right) + \hdots

Clearly the lower bound of S_1 diverges and therefore S_1 also diverges. 😀
But it interesting to note that of divergence is incredibly small: 10 billion terms in the series only adds up to around 23.6 !