Taylor series

December 25, 2022
Útulek Series, 5 | CF1761D Chapter, 5

I have been reviewing the Taylor series, which are defined for $f(x)$ about $a$ as

$$f_i(x)=\sum_{i=0}^j a_i(x-a)^i$$

with the property that $f_i(a)=f(a)$, $f_i'(a)=f'(a)$, and so on, as long as $f$ is differentiable at $a$. This forces the $a_i$ to be the values of the derivatives themselves,

$$a_i=f^{(i)}(a)/i!,$$

and intuitively, the Taylor series is simply a deriviative-based approximation around $a$ and its neighborhood. We assume, perhaps with overconfidence, that any function is the limit of a series of polynomials, and by that definition, these polynomials necessarily satisfy the form of a sum of $a_i(x-a)^i$. Tony provides an intuitive derivation of Taylor series via the Fundamental Theorem of Calculus, which supports this intuition.

Indeed, the limit of the Taylor series, $f_\infty(x)$, is $f(x)$, while the error of the series in general can be bounded by what’s known as Taylor’s remainder theorem.

1 Taylor Series of $f(x)=\ln x$ at $a=1$

We have

$$\begin{aligned} a_0&=f(a)/0!=0,\\ a_1&=f'(a)/1!=1,\\ a_2&=f''(a)/2!=-a^{-2}/2=-1/2,\\ a_3&=f'''(a)/3!=a^{-3}/3=1/3, \end{aligned}$$

and indeed as we claimed last time,

$$f_{\infty,1}(x)=\ln x=(x-1)-(x-1)^2/2+(x-1)^3/3-\ldots.$$

2 Proof for $\zeta(2)=\pi^2/6$

Take $f(x)=\sin(x)$. We have for $a=0$

$$\begin{aligned} a_0&=f(a)/0!=0,\\ a_1&=f'(a)/1!=1,\\ a_2&=f''(a)/2!=0,\\ a_3&=f'''(a)/3!=-1, \end{aligned}$$

and hence

$$\begin{aligned} f_{\infty,0}&=\sin x=x-x^3/3!+x^5/5!-\ldots\\ &=x(1-x^2/3!+x^4/5!-\ldots). \end{aligned}$$

Recalling that $\sin x$ has roots at $k\pi$, we must be able to factor $f_{\infty,0}$ similarly with these roots:

$$\begin{aligned} f_{\infty,0}/x&=(1-x^2/3!+x^4/5!-\ldots)\\ &=C(x-\pi)(x+\pi)(x-2\pi)(x+2\pi)\ldots\\ &=(1-x/\pi)(1+x/\pi)(1-x/2\pi)(1+x/2\pi)\ldots\\ &=(1-x^2/\pi^2)(1-x^2/4\pi^2)\ldots \end{aligned}$$

where we see the first inklings of $\zeta(2)$ on the RHS. In fact, the coefficient of $x^2$ in this representation of $f_{\infty,0}$ is exactly $-\zeta(2)/\pi^2$. Since we know this coefficient is $-1/6$, we have $\zeta(2)=\pi^2/6$.

3 Proof of the Wallis Product

In The Recurrence Detanglement Conjecture and the Harmonic Series, I introduced the Wallis product

$$\prod_{i=1}^\infty\frac{4i^2}{4i^2-1}=(\frac{2}{1}\cdot\frac{2}{3})(\frac{4}{3}\cdot\frac{4}{5})(\frac{6}{5}\cdot\frac{6}{7})\cdots.$$

Again, we have for $f(x)=\frac{\sin x}{x}$,

$$\begin{aligned} f_{\infty,0}/x&=2/\pi=(1-1/4)(1-1/16)(1-1/36)\ldots\\ &=\prod_{i=1}^\infty\frac{4i^2}{4i^2-1}. \end{aligned}$$