# A Question On Euler's Proof Of the Basel Problem

I've studied the proof that Euler gave for the famous Basel Problem, and it would seem that while it is technically correct, he does not justify all of his steps properly. Namely, he assumes that

$$\frac{\sin(x)}{x}=\left(1-\frac{x}{\pi}\right)\left(1+\frac{x}{\pi}\right)\left(1-\frac{x}{2\pi}\right)\left(1+\frac{x}{2\pi}\right)\dots$$

simply because they have the same roots, which is really not a strong enough condition. How do you really show that the equality holds?

He then notices that if you use the above equality, and consider it against the Taylor expansion for $\frac{\sin(x)}{x}$, then you can equate the coefficients of the two infinite expansions at each order, and the result of the Basel problem follows. But how do you know that if you have two different expansions for a function, then their coefficients at each order must be equal?

I would really appreciate if someone could show me how to make these two intuitive, yet informal, steps rigorous.

• I think I can guess from your post what the Basel problem is, but it would be helpful if you edited your post to include an explanation. – TonyK Mar 25 '13 at 21:35
• Often this proof is referred to as "non-rigorous" exactly because of your first complaint. We know the result is true now, as mentioned in Mike's solution, but at the time this was not so. – Relsiark Mar 25 '13 at 23:06

Two power series for a function, $f$, at $x_0$ that converge in some neighborhood of $x_0$ to $f$ must have the same coefficients. Euler computed the formal power series for the infinite product given for $\frac{\sin(x)}{x}$. Proving that this is actually the convergent power series requires a bit more work, but the idea is good.
• @Coffee_Table: suppose that $$\sum\limits_{n=0}^\infty a_n(x-x_0)^n=\sum\limits_{n=0}^\infty b_n(x-x_0)^n$$ in some neighborhood of $x_0$. Subtract and plug in $x=x_0$ to get that $a_0=b_0$. Since they converge in a neighborhood of $x_0$, we can differentiate the series term by term. Again, subtracting and plugging in $x=x_0$ yields $a_1=b_1$. Repeat as often as needed. – robjohn Mar 28 '13 at 17:50