Difference between revisions of "Basel Problem"
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The answer has been proven to be <math>\frac{\pi^2}{6}.</math> | The answer has been proven to be <math>\frac{\pi^2}{6}.</math> | ||
| − | == | + | ==Proof== |
The derivative of a function <math>f(x)</math> is defined as <math>f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.</math> So the derivative of <math>f(x) = e^x</math> is | The derivative of a function <math>f(x)</math> is defined as <math>f'(x) = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}.</math> So the derivative of <math>f(x) = e^x</math> is | ||
Revision as of 17:32, 31 August 2025
The Basel Problem asks for the precise sum of the infinite sum of the reciprocal of squares or:
![\[\sum_{n=1}^{\infty} \frac{1}{n^2} = 1 + \frac{1}{4} + \frac{1}{9} + ... = ?\]](http://latex.artofproblemsolving.com/6/1/f/61fa499ac36a3a79cc2fd5d8fb1f2fde0f5da348.png)
The answer has been proven to be 
Proof
The derivative of a function 
is defined as 
So the derivative of 
is
![\[\lim_{h \to 0} \frac{e^{x+h} - e^x}{h} = e^x \lim_{h \to 0} \frac{e^h - 1}{h}.\]](http://latex.artofproblemsolving.com/c/2/8/c2851e621eec493cb34725cd6003416ed6880d30.png)
Let 
Then as 
, 
and 
![\[e^x \lim_{h \to 0} \frac{e^h - 1}{h} = e^x \lim_{y \to 0} \frac{y}{\ln(1+y)} = e^x \lim_{y \to 0} \frac{1}{\ln(1+y)^{1/y}} = e^x \cdot \frac{1}{\ln \left( \lim_{y \to 0} (1+y)^{1/y} \right)} = e^x \cdot \frac{1}{\ln e} = e^x.\]](http://latex.artofproblemsolving.com/9/d/6/9d609f2d09520d94d32a7ebf129cfaa00b755386.png)
Since the derivative of 
is itself, the 
th derivative of will be 
Now, consider the infinite sum 
The derivatives at 
are 
and so on, with 
Let this sum equal the function 
Then we must have 
which means 
so 
The series for is thus 
Now, from Euler's Formula, we have 
Using the infinite series expansion and substituting 
for 
we get:
![\[e^{ix} = \sum_{n=0}^{\infty} \frac{(ix)^n}{n!} = 1 + ix + \frac{(ix)^2}{2!} + \frac{(ix)^3}{3!} + \frac{(ix)^4}{4!} + \frac{(ix)^5}{5!} + \dots\]](http://latex.artofproblemsolving.com/9/a/6/9a6c8b55d4806710448dfb9f510e21ec992231c5.png)
![\[= 1 + ix - \frac{x^2}{2!} - i\frac{x^3}{3!} + \frac{x^4}{4!} + i\frac{x^5}{5!} + \dots\]](http://latex.artofproblemsolving.com/c/2/7/c27054aced612378daac3505eedb44df307c75f4.png)
![\[= \left( 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots \right) + i \left( x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots \right).\]](http://latex.artofproblemsolving.com/1/d/5/1d59275953a7e0494a9850e51709b786b0ecd998.png)
By equating the real and imaginary parts of the series with Euler's formula, we find the series expansions for sine and cosine:
![\[\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots\]](http://latex.artofproblemsolving.com/4/b/d/4bd3b054bfcec556dc77ead10ef00c31f0953bd3.png)
![\[\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots\]](http://latex.artofproblemsolving.com/9/5/8/95895b06c04d78cb6d7908885b34594cc14c0873.png)
Since 
for all integers 
the roots of the function 
are at 
We can factor the polynomial expansion in terms of its roots:
![\[\sin(x) = x( 1 - \frac{x^2}{\pi^2})( 1 - \frac{x^2}{(2\pi)^2})( 1 - \frac{x^2}{(3\pi)^2}) \dots\]](http://latex.artofproblemsolving.com/0/8/1/081bfdae4a8ec03b07e6d86368d6cbd86573baa4.png)
Now, consider the coefficient of the 
term in the expansion of 
From the infinite product, the coefficient of the term is found by multiplying the 
with each term with the 
![\[\text{Coefficient of } x^3 = -\left( \frac{1}{\pi^2} + \frac{1}{(2\pi)^2} + \frac{1}{(3\pi)^2} + \dots \right) = -\frac{1}{3!}\]](http://latex.artofproblemsolving.com/e/0/0/e00708b3ed6b9011e0e90597fd3249ee4920e88c.png)
Equating the two expressions for the coefficient, we get:
![\[-\frac{1}{6} = -\frac{1}{\pi^2} \sum_{n=1}^{\infty} \frac{1}{n^2}\]](http://latex.artofproblemsolving.com/9/1/3/913e9408c69b7696b0467c60c39f56523b669637.png)
![\[\implies \sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^2}{6}.\]](http://latex.artofproblemsolving.com/1/a/0/1a0d29a91111160fa368cfd8732d6510f29d1f0c.png)
