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Sum of reciprocal squares



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.






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