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In this section we present numerous examples that provide a number of useful procedures to find new Taylor series from Taylor series that we already know. However, we are only worried about ``computing'' and we don't worry (for now) about the convergence of the series we find.
We know ``by heart'' the following series:
\begin{eqnarray*} e^x &=& 1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\ldots = \sumno \frac{x^n}{n!}\\ \cos x &=& 1- \frac{x^2}{2!} +\frac{x^4}{4!}- \frac{x^6}{6!}+\ldots = \sumno (-1)^n \frac{x^{2n}}{(2n)!}\\ \sin x &=& x- \frac{x^3}{3!} +\frac{x^5}{5!}- \frac{x^7}{7!}+\ldots = \sumno (-1)^n \frac{x^{2n+1}}{(2n+1)!}\\ (1+x)^p &=& 1+px + \frac{p(p-1)}{2!}x^2+\frac{p(p-1)(p-2)}{3!}x^3+\ldots + \frac{p(p-1)\cdot\ldots\cdot(p-(n-1))}{n!}x^n+\ldots\\ \ln (1+x) &=& x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots = \sumn (-1)^{n+1}\frac{x^n}{n} \end{eqnarray*}
Example 1 Find the Taylor series about $x=0$ for $\sin(x^2)$ . If we try to take derivatives then we soon realize that consecutive derivatives get extremely hard to compute. However, one can do a simple trick. Since we know the Taylor series for $\sin(x)$ we can evaluate it at $x^2$ : $$\sin x = x- \frac{x^3}{3!} +\frac{x^5}{5!}- \frac{x^7}{7!}+\ldots = \sumno (-1)^n \frac{x^{2n+1}}{(2n+1)!}$$ $$\sin(x^2) = (x^2) -\frac{(x^2)^3}{3!} + \frac{(x^2)^5}{5!} - \ldots =
x^2-\frac{x^6}{3!}+\frac{x^{10}}{5!}-\ldots$$ One can also use the $\Sigma$ notation: $$\sin(x^2)=\sumno (-1)^n \frac{(x^2)^{2n+1}}{(2n+1)!}=\sumno (-1)^n \frac{x^{2(2n+1)}}{(2n+1)!}.$$
Example 2 Find the Taylor series about $x=0$ for $e^{-x^2}$ (this is a very important function, for example in probability theory). Again, we use the simple Taylor series of $e^x$ : $$e^x = 1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\ldots = \sumno \frac{x^n}{n!}$$ $$e^{-x^2} = 1+(-x^2)+\frac{(-x^2)^2}{2}+\frac{(-x^2)^3}{3!}+\ldots= 1-x^2+\frac{x^4}{2}-\frac{x^6}{3!}+\ldots$$ Using the sigma notation we obtain: $$e^{-x^2} = \sumno \frac{(-x^2)^n}{n!}=\sumno \frac{(-1)^nx^{2n}}{n!}.$$
Example 3 Series can also be multiplied by $x$ . For example, we find the Taylor series for $xe^x$ : $$xe^x= x(1+x+\frac{x^2}{2}+\frac{x^3}{3!}+\ldots)= x+x^2+\frac{x^3}{2}+\frac{x^4}{3!}+\ldots)$$ or $$xe^x= x \left( \sumno \frac{x^n}{n!} \right)=\sumno \frac{x^{n+1}}{n!}$$
Example 4 Series can also be divided by $x$ provided that the result has only non-negative exponents. For example, we find the Taylor series for $\frac{\ln(1+x)}{x}$ : $$\frac{\ln(1+x)}{x}=\frac{x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\ldots}{x}= 1-\frac{x}{2}+\frac{x^2}{3}-\frac{x^3}{4}+\ldots $$ or $$\frac{\ln(1+x)}{x}= \frac{\sumn (-1)^{n+1}\frac{x^n}{n}}{x}=\sumn (-1)^{n+1}\frac{x^{n-1}}{n}$$
Example 5 As well, we can multiply two Taylor series ( term by term). Suppose we want to find the Taylor polynomial of degree $3$ about $x=0$ of $e^x\cos x$ . Then we can multiply the respective Taylor polynomials of degree $3$ of $e^x$ and $\cos x$ and disregard any term higher than $3$ : $$\left(1+x+\frac{x^2}{2}+\frac{x^3}{6}\right)\left(1-\frac{x^2}{2}\right)$$ $$= \left(1-\frac{x^2}{2}\right) +
\left(x-\frac{x^3}{2}\right) + \left(\frac{x^2}{2}\right) + \left(\frac{x^3}{6}\right)+\ldots $$ Since every other term in the product is of degree higher than $3$ we disregard them. Thus: $$T_3(x)=\left(1-\frac{x^2}{2}\right) + \left(x-\frac{x^3}{2}\right) + \left(\frac{x^2}{2}\right) + \left(\frac{x^3}{6}\right)=1+x-\frac{x^3}{2}+\frac{x^3}{6}=1+x-\frac{x^3}{3}.$$
Example 6 Find the first three terms of the Taylor series for $\sqrt{1+2\sin x}$ . Since $\sqrt{1+x}=(1+x)^{1/2}$ we know that: $$\sqrt{1+x}=1+\frac{x}{2}-\frac{x^2}{8}+\ldots$$ Thus: $$\sqrt{1+2\sin x}= 1+\frac{2\sin x}{2}-\frac{(2\sin x)^2}{8}+\ldots$$ Moreover, $\sin x = x - \frac{x^3}{3!}+\ldots$ : $$\sqrt{1+2\sin x}= 1+\frac{2(x - \frac{x^3}{3!}+\ldots)}{2}-\frac{(2(x - \frac{x^3}{3!}+\ldots))^2}{8}+\ldots$$ By disregarding other than the first term in $\sin x$ we obtain the first three terms of the series are: $$\sqrt{1+2\sin x}=1 + \frac{2x}{2} - \frac{(2x)^2}{8}+\ldots=1+x-\frac{x^2}{2}+\ldots$$
Example 7 ( Differentiation) Notice that we can deduce the series of $\sin x$ from the series for $\cos x$ by differentiating. Indeed $\frac{d}{dx} \cos x =- \sin x$ and we differentiate (term by term) the Taylor series of $\cos x$ we obtain the Taylor series of $\sin x$ (DO IT!).
Another example. Let us deduce the Taylor series of $\frac{1}{(1-x)^2}$ . Notice that $\frac{d}{dx} \frac{1}{(1-x)}= \frac{1}{(1-x)^2}$ . Since: $$\frac{1}{1-x}= 1+x+x^2+x^3+\ldots$$ by deriving both sides we obtain: $$\frac{1}{(1-x)^2}=0+1+2x+3x^2+\ldots=\sumno (n+1)x^n$$ and if we derive again we obtain: $$\frac{2}{(1-x)^3}=2+6x+12x^2\ldots$$ Thus, $$\frac{1}{(1-x)^3}=1+3x+6x^2+10x^3+\ldots=\sumno \frac{(n+2)(n+1)}{2}x^n.$$
Example 8 (Integration) Finally, we will deduce the Taylor series for $\arctan x$ using integration (term by term). Notice that: $$\int \frac{1}{1+x^2} dx = \arctan x + C$$ Moreover, since $1/(1-x)=1+x+x^2+x^3+\ldots$ by substituting $x$ by $-x^2$ we obtain: $$\frac{1}{1+x^2}=1-x^2+x^4-x^6+\ldots=\sumno (-1)^n x^{2n}$$ Thus, the Taylor series of $\arctan x$ can be constructed integrating the previous one: $$\arctan x = \int \frac{1}{1+x^2} dx = \int (1-x^2+x^4-x^6+\ldots )dx = x -\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\ldots$$ In $\Sigma$ notation: $$\arctan x = \int \frac{1}{1+x^2} dx = \int \sumno (-1)^n x^{2n} dx = \sumno (-1)^n \int x^{2n} dx = \sumno (-1)^n
\frac{x^{2n+1}}{2n+1}$$
Example 9 As an application of the previous example, we compute $\pi$ . Indeed: $$\arctan x = \sumno (-1)^n \frac{x^{2n+1}}{2n+1}$$ converges between $-1\leq x\leq 1$ and in particular $$\arctan 1 = \frac{\pi}{4}$$ by the definition of $\tan x$ and $\arctan x$ . Thus: $$\frac{\pi}{4}=\sumno (-1)^n \frac{1^{2n+1}}{2n+1}=\sumno \frac{(-1)^n}{2n+1}$$ and $$\pi = 4\left(\sumno \frac{(-1)^n}{2n+1}\right)=4\left(1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\frac{1}{9}-\ldots\right)$$ For example, if one adds up to the $1/9$ term,
one obtains the approximation $\pi\approx 3.33$ . Unfortunately, the convergence is very slow. If you want to have about $m$ correct digits then you have to add about $10^m/2$ terms. For example, if you add $10^3/2=500$ terms we get $3.143588\ldots$ .
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