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## Taylor Series

A function $$f$$ may or may not have a power series representation. But if it has one, we can find it explicitly.

Theorem. Suppose that a function $$f$$ is infinitely differentiable at $$c$$ and $$f$$ has a power series representation about $$c$$: $f(x)=\sum_{n=0}^{\infty} a_n (x-c)^n,\; |x-c| < R.$ Then $$a_n=\displaystyle\frac{f^{(n)}(c)}{n!},\; n\geq 0$$, i.e., $f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!} (x-c)^n =f(c)+\frac{f'(c)}{1!}(x-c)+\frac{f''(c)}{2!}(x-c)^2+\frac{f'''(c)}{3!}(x-c)^3+\cdots.$

Plugging $$x=c$$ in $$f(x)=\displaystyle\sum_{n=0}^{\infty} a_n (x-c)^n$$, we get $$a_0=f(c)$$. Taking derivative of $$f$$ at $$c$$, we have $f'(c)=\sum_{n=1}^{\infty} a_n n(x-c)^{n-1} \Big\vert_{x=c} \implies a_1=f'(c).$ Similarly taking $$k$$th derivative of $$f$$ at $$c$$, we have $f^{(k)}(c)=\sum_{n=k}^{\infty} a_n n!(x-c)^{n-k} \Big\vert_{x=c} \implies f^{(k)}(c)=k!a_k \implies a_k=\frac{f^{(k)}(c)}{k!}.$

Definition. The Taylor Series of $$f$$ about $$c$$ is $\sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!} (x-c)^n =f(c)+\frac{f'(c)}{1!}(x-c)+\frac{f''(c)}{2!}(x-c)^2+\frac{f'''(c)}{3!}(x-c)^3+\cdots.$ The Maclaurin Series of $$f$$ is the Taylor Series of $$f$$ about $$0$$: $\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n =f(0)+\frac{f'(0)}{1!}x+\frac{f''(0)}{2!}x^2+\frac{f'''(0)}{3!}x^3+\cdots.$

Example.

1. Find the Taylor Series of $$f(x)=e^x$$ about $$2$$.

2. Find the Maclaurin Series of $$f(x)=e^x$$.

Solution.

1. Since $$f^{(n)}(x)=e^x$$ for all integers $$n\geq 0$$, $$f^{(n)}(2)=e^2$$ for all integers $$n\geq 0$$. Then the Taylor Series of $$f(x)=e^x$$ about $$2$$ is $\sum_{n=0}^{\infty} \frac{f^{(n)}(2)}{n!} (x-2)^n =\sum_{n=0}^{\infty} \frac{e^2}{n!} (x-2)^n =e^2+\frac{e^2}{1!}(x-2)+\frac{e^2}{2!}(x-2)^2+\frac{e^2}{3!}(x-2)^3+\cdots.$
2. Since $$f^{(n)}(0)=e^0=1$$ for all integers $$n\geq 0$$, the Maclaurin Series of $$f(x)=e^x$$ is $\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n =\sum_{n=0}^{\infty} \frac{1}{n!} x^n =1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots.$ Note that if $$f(x)=e^x$$ has a power series representation about $$0$$, then $e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!} =1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots.$

Question. When does a function $$f$$ have a power series representation? When is $$f(x)$$ equal to its Taylor series?

To answer these questions, we define the $$n$$th degree Taylor polynomial of $$f$$ about $$c$$, denoted by $$T_n$$, as $T_n(x)=\sum_{i=0}^{n} \frac{f^{(i)}(c)}{i!} (x-c)^i =f(c)+\frac{f'(c)}{1!}(x-c)+\frac{f''(c)}{2!}(x-c)^2+\cdots+\frac{f^{(n)}(c)}{n!}(x-c)^n.$ We define the corresponding remainder, denoted by $$R_n$$, as $$R_n(x)=f(x)-T_n(x)$$. Then $f(x)=T_n(x)+R_n(x).$ Taking limit as $$n\to \infty$$, we have $f(x)=\lim_{n\to \infty} T_n(x)+ \lim_{n\to \infty}R_n(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!} (x-c)^n +\lim_{n\to \infty}R_n(x) =\sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!} (x-c)^n,$ when $$\displaystyle\lim_{n\to \infty}R_n(x)=0$$.

Theorem. Suppose that a function $$f$$ is infinitely differentiable at $$c$$ with Taylor polynomial $$T_n$$ about $$c$$ and the remainder $$R_n$$. If $$\displaystyle\lim_{n\to \infty}R_n(x)=0$$ for all $$x$$ satisfying $$|x-c| < R$$, then $f(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(c)}{n!} (x-c)^n,\; |x-c| < R.$

It is difficult to show $$\displaystyle\lim_{n\to \infty}R_n(x)=0$$. We use the following result:
Theorem. If $$M$$ is a constant such that $$|f^{(n)}(x)|\leq M$$ for all $$x$$ satisfying $$|x-c| < R$$, then $|R_n(x)|\leq \frac{M}{(n+1)!} |x-c|^{n+1},\;|x-c| < R.$

Example. Prove that $$f(x)=e^x$$ is equal to its Maclaurin Series for all $$x$$: $e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!} =1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots.$
Solution. Let $$R$$ be an arbitrary positive real number. It suffices to show that $$f(x)=e^x$$ is equal to its Maclaurin Series for all $$x$$ satisfying $$|x| < R$$.
Since $$f^{(n)}(x)=e^x$$ for all integers $$n\geq 0$$, $$|f^{(n)}(x)|\leq e^R$$ for all $$x$$ satisfying $$|x| < R$$. Then by the preceding theorem, $|R_n(x)|\leq \frac{e^R}{(n+1)!} |x|^{n+1},\;|x| < R.$ Taking limit as $$n\to \infty$$, we have $0\leq \lim_{n\to \infty}|R_n(x)|\leq \lim_{n\to \infty}\frac{e^R}{(n+1)!} |x|^{n+1}=0 \implies \lim_{n\to \infty}|R_n(x)|=0,\;|x| < R.$ $f(x)=\lim_{n\to \infty} T_n(x)+ \lim_{n\to \infty}R_n(x)=\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n!} x^n=\sum_{n=0}^{\infty} \frac{x^n}{n!},\;|x| < R.$ Since $$R$$ is an arbitrary positive real number, for all $$x$$, $e^x=\sum_{n=0}^{\infty} \frac{x^n}{n!} =1+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots.$

By arguments similar to that above, we have the following results: \begin{align*} \cos x =\sum_{n=0}^{\infty} (-1)^n\frac{x^{2n}}{(2n)!} &=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots, \text{ for all } x \;(R=\infty)\\ \sin x =\sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{(2n+1)!} &=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots, \text{ for all } x \;(R=\infty)\\ \tan^{-1} x =\sum_{n=0}^{\infty} (-1)^n\frac{x^{2n+1}}{2n+1} &=x-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots, \text{ for } |x|<1 \;(R=1)\\ \ln(1+x) =\sum_{n=0}^{\infty} (-1)^{(n-1)}\frac{x^n}{n} &=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots, \text{ for } |x|<1 \;(R=1) \end{align*}

Example. Write $$\displaystyle\int e^{-x^2}\;dx$$ and $$\displaystyle\int_0^1 e^{-x^2}\;dx$$ as infinite series.

Solution. We know that $$e^x=\displaystyle\sum_{n=0}^{\infty} \frac{x^n}{n!}$$ for all $$x$$. Substituting $$x$$ by $$-x^2$$, we get $e^{-x^2}=\sum_{n=0}^{\infty} \frac{(-x^2)^n}{n!} =\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{n!} =1-\frac{x^2}{1!}+\frac{x^4}{2!}-\frac{x^6}{3!}+\cdots.$ By term by term integration, we have $\int e^{-x^2} \;dx=\int\left[\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n}}{n!} \right]dx =\sum_{n=0}^{\infty} \left[ \int (-1)^n\frac{x^{2n}}{n!}\;dx \right] =C+\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{n!(2n+1)}.$ $\int_0^1 e^{-x^2} \;dx =\left. C+\sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{n!(2n+1)} \right\vert_0^1 =\sum_{n=0}^{\infty} \frac{(-1)^n}{n!(2n+1)}.$ Note that we can approximate $$\displaystyle\int_0^1 e^{-x^2}\;dx$$ by truncating the above series after a few terms. $\int_0^1 e^{-x^2} \;dx \approx\sum_{n=0}^{2} \frac{(-1)^n}{n!(2n+1)} =\frac{1}{1}-\frac{1}{3}+\frac{1}{10}=\frac{23}{30}.$

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