Маніпулювання рядами Тейлора

There are several manipulations that you can perform on Taylor series, all of which are fairly easy to understand.

First, we will add two Taylor series. To do this, simply add like terms, as you would in any other case of adding two algebraic expressions. Here is an example: $$xe^x=x+x^2+\frac{1}{2}x^3+...+\frac{1}{(n-1)!}x^n+...$$ $$x^2e^x=x^2+x^3+\frac{1}{2}x^4+...+\frac{1}{(n-2)!}x^n+...$$ $$(x+x^2)e^x=x+2x^2+\frac{3}{2}x^3+...+\left(\frac{1}{(n-1)!}+\frac{1}{(n-2)!}\right)x^n+...$$ Notice that we have simply added the $x$ terms, added the $x^2$ terms, and so on.

Now, we will apply a composition to a Taylor series. This is also very simple. Suppose that we have a Taylor series for $f(x)$, and we would like the composition $f\circ g(x)=f(g(x))$. To find a Taylor series for the composition, simply replace every instance of $x$ with $g(x)$. Here is an example: $$\cos x=1-\frac{1}{2}x^2+\frac{1}{24}x^4-\frac{1}{720}x^6+...+\frac{(-1)^n}{(2n)!}x^{2n}+...$$ $$\cos\sqrt x=1-\frac{1}{2}(\sqrt x)^2+\frac{1}{24}(\sqrt x)^4-\frac{1}{720}(\sqrt x)^6+...+\frac{(-1)^n}{(2n)!}(\sqrt x)^{2n}+...$$ $$\cos\sqrt x=1-\frac{1}{2}x+\frac{1}{24}x^2-\frac{1}{720}x^3+...+\frac{(-1)^n}{(2n)!}x^{n}+...$$

Last, we will multiply two Taylor series. This is a bit more challenging. Since Taylor series are simply really, really long polynomials, we may still multiply them as if they were polynomials. Although sometimes finding the coefficient of a certain power of $x$ may be tough, it is often quite manageable. Here is an example: $$\sin x=x-\frac{1}{6}x^3+\frac{1}{120}x^5-\frac{1}{5040}x^7+...+\frac{(-1)^n}{(2n+1)!}x^{2n+1}+...$$ $$\cos x=1-\frac{1}{2}x^2+\frac{1}{24}x^4-\frac{1}{720}x^6+...+\frac{(-1)^n}{(2n)!}x^{2n}+...$$ $$\sin x\cos x=\left(\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}\right)\left(\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}\right)$$ $$\small{\sin x\cos x=\sum_{n=0}^{\infty}(-1)^nx^{2n+1}\left(\frac{1}{(0)!(2n+1)!}+\frac{1}{(2)!(2n-1)!}+...+\frac{1}{(2n)!(1)!}\right)}$$ $$\small{\sin x\cos x=}\tiny{\sum_{n=0}^{\infty}(-1)^nx^{2n+1}\frac{1}{(2n+1)!}\left[{{2n+1}\choose{0}}+{{2n+1}\choose{2}}+...+{{2n+1}\choose{2n}}\right]}$$ Now, we may use the identity ${k\choose 0}+{k\choose 2}+{k\choose 4}+...=2^{k-1}$ to obtain our result: $$\sin x\cos x=\sum_{n=0}^{\infty}(-1)^nx^{2n+1}\frac{2^{2n}}{(2n+1)!}$$ $$\sin x\cos x=\sum_{n=0}^{\infty}\frac{(-1)^n2^{2n}}{(2n+1)!}x^{2n+1}$$ Note that this equals $\frac{1}{2}\sin(2x)$; you can confirm this by plugging $2x$ into the Taylor series for $\sin x$ above. This should not be surprising, but the fact that we have satisfied the double-angle identity gives us confidence in our answer.

On this particular page, I have actually only used Maclaurin series for the examples, but that was just for simplicity. Of course, the same rules will work for other Taylor series as well.

So there you have it: addition, substitution, and multiplication for Taylor series! The core principle to remember is that the Taylor series' infinite extent should not intimidate you; you should deal with them as the docile and reasonably well-behaved polynomials that they really are.