Today we will look at a very fascinating proof due to Euler for the following identity: \frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \frac{1}{5^2} + \dots = \frac{\pi^2}{6}
The mathematician Euler formulated this proof in 1734 when he was 28 year old.
We will present Euler's proof step by step.
Step 1. Using Taylor series expansion for f(x) = \sin(x)
We have
f(x) = \sin(x) ~\Rightarrow ~ f(0) = 0 f'(x) = \cos(x) ~\Rightarrow ~ f'(0) = 1 f''(x) = -\sin(x) ~\Rightarrow ~ f''(0) = 0 f^{(3)}(x) = -\cos(x) ~\Rightarrow ~ f^{(3)}(0) = -1 f^{(4)}(x) = \sin(x) ~\Rightarrow ~ f^{(4)}(0) = 0 f^{(5)}(x) = \cos(x) ~\Rightarrow ~ f^{(5)}(0) = 1 f^{(6)}(x) = -\sin(x) ~\Rightarrow ~ f^{(6)}(0) = 0 f^{(7)}(x) = -\cos(x) ~\Rightarrow ~ f^{(7)}(0) = -1
So the Taylor series for the function f(x) = \sin(x) is as follows:
f(x) = f(0) + f'(0) x + \frac{f''(0)}{2!} x^2 + \frac{f^{(3)}(0)}{3!} x^3 + \frac{f^{(4)}(0)}{4!} x^4 + \dots + \frac{f^{(n)}(0)}{n!} x^n + \dots
\sin(x) = x - \frac{1}{3!} x^3 + \frac{1}{5!} x^5 - \frac{1}{7!} x^7 + \dots
\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots
Step 2. Express \sin(x) as product of linear factors
Since the equation f(x) = \sin(x) = 0 has the following roots x=0, ~x = \pm \pi, ~x = \pm 2 \pi, ~x = \pm 3 \pi, \dots
we have
\sin(x) = C x (x - \pi)(x + \pi)(x - 2\pi)(x + 2 \pi)(x - 3\pi)(x + 3\pi) \dots
\sin(x) = C x (x^2 - \pi^2)(x^2 - 2^2 \pi^2)(x^2 - 3^2 \pi^2)(x^2 - 4^2 \pi^2) \dots
Step 3. Combining step 1 and step 2
Comparing the two representations of \sin(x) in step 1 and step 2, we have
\sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \dots = C x (x^2 - \pi^2)(x^2 - 2^2 \pi^2)(x^2 - 3^2 \pi^2)(x^2 - 4^2 \pi^2) \dots
So
1 - \frac{x^2}{3!} + \frac{x^4}{5!} - \frac{x^6}{7!} + \dots = C (x^2 - \pi^2)(x^2 - 2^2 \pi^2)(x^2 - 3^2 \pi^2)(x^2 - 4^2 \pi^2) \dots
Substitute x^2 by x, we have
1 - \frac{x}{3!} + \frac{x^2}{5!} - \frac{x^3}{7!} + \dots = C (x - \pi^2)(x - 2^2 \pi^2)(x - 3^2 \pi^2)(x - 4^2 \pi^2) \dots
Step 4. Normalization to get rid of the constant factor C
If we have an equation of the form
f(x) = C g_1(x) g_2(x) g_3(x) \dots
we can get rid of the constant factor C by normalization as follows
\frac{f(x)}{f(0)} = \frac{g_1(x)}{g_1(0)} \frac{g_2(x)}{g_2(0)} \frac{g_3(x)}{g_3(0)} \dots
With the equation that we have in step 3
1 - \frac{x}{3!} + \frac{x^2}{5!} - \frac{x^3}{7!} + \dots = C (x - \pi^2)(x - 2^2 \pi^2)(x - 3^2 \pi^2)(x - 4^2 \pi^2) \dots
by normalization, we obtain
1 - \frac{x}{3!} + \frac{x^2}{5!} - \frac{x^3}{7!} + \dots = \frac{(x - \pi^2)}{-\pi^2} \frac{(x - 2^2 \pi^2)}{-2^2 \pi^2} \frac{(x - 3^2 \pi^2)}{-3^2 \pi^2} \frac{(x - 4^2 \pi^2)}{-4^2 \pi^2} \dots = (1 - \frac{x}{\pi^2}) (1 - \frac{x}{2^2 \pi^2}) (1 - \frac{x}{3^2 \pi^2}) (1 - \frac{x}{4^2 \pi^2})\dots
Step 5. Use Vieta formula
Comparing the coefficients of x in both sides of the equation
1 - \frac{x}{3!} + \frac{x^2}{5!} - \frac{x^3}{7!} + \dots = (1 - \frac{x}{\pi^2}) (1 - \frac{x}{2^2 \pi^2}) (1 - \frac{x}{3^2 \pi^2}) (1 - \frac{x}{4^2 \pi^2})\dots
we have - \frac{x}{3!} = - \frac{x}{\pi^2} - \frac{x}{2^2 \pi^2} - \frac{x}{3^2 \pi^2} - \frac{x}{4^2 \pi^2} - \dots
Therefore, \frac{1}{3!} = \frac{1}{\pi^2} + \frac{1}{2^2 \pi^2} + \frac{1}{3^2 \pi^2} + \frac{1}{4^2 \pi^2} + \dots
Thus, we obtain
\frac{1}{1^2} + \frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \dots =\frac{\pi^2}{3!}= \frac{\pi^2}{6}
Let us stop here for now. In the homework section, you can try Euler's method with the function f(x) = \cos(x) to derive another proof of the above identity. Hope to see you again next time.
Homework.
1. Show that the Taylor series for f(x) = \cos(x) is as follows \cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots
2. Using the fact that the equation \cos(x) = 0 has the following roots x = \pm \frac{\pi}{2}, ~x = \pm \frac{3 \pi}{2}, ~x = \pm \frac{5 \pi}{2}, \dots
show that
\cos(x) = 1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \frac{x^6}{6!} + \dots = (1 - \frac{4 x^2}{\pi^2})(1 - \frac{4 x^2}{3^2 \pi^2})(1 - \frac{4 x^2}{5^2 \pi^2}) \dots
3. Prove that
\frac{1 }{1^2} + \frac{1 }{3^2} + \frac{1 }{5^2} + \dots = \frac{\pi^2}{8}
4. Prove that
\frac{1 }{1^2} + \frac{1 }{3^2} + \frac{1 }{5^2} + \dots = \frac{3}{4} \left( \frac{1 }{1^2} + \frac{1 }{2^2} + \frac{1 }{3^2} + \frac{1 }{4^2} + \dots \right)
thus deriving
\frac{1 }{1^2} + \frac{1 }{2^2} + \frac{1 }{3^2} + \frac{1 }{4^2} + \dots = \frac{\pi^2}{6}
5. Go to google.com and search about the Basel problem and the Riemann zeta function.