Приклади рядів Тейлора поширених функцій

Trigonometric functions: $\sin x$ and $\cos x$

We will begin by finding the series for $\sin x$

The first question we must tackle, which is of the utmost importance, is:

Question 2:

This means that $f^{(n)}(0)$ will cycle after every fourth derivative, so we really only need to find the value of the first four!

$f^{(0)}(0)=\sin 0=0$

$f^{(1)}(0)=\cos 0=1$

$f^{(2)}(0)=-\sin 0=0$

$f^{(3)}(0)=-\cos 0=-1$

As $\sin x$ is infinitely differentiable, we may write: $$\sin x=\sum_{n=0}^\infty f^{(n)}(0)\frac{x^n}{n!}=(0)\frac{x^0}{0!}+(1)\frac{x^1}{1!}+(0)\frac{x^2}{2!}+(-1)\frac{x^3}{3!}+\cdots$$ $$\sin x=x-\frac1{3!}x^3+\frac{1}{5!}x^5-\frac{1}{7!}x^7+\cdots=\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}$$

But what about $\cos x$?

Well, as we said earlier, $\frac{\text d\sin x}{\text dx}=\cos x$, right?

It follows that $$\cos x=\frac{\text d}{\text dx}\left(\sum_{n=0}^\infty \frac{(-1)^nx^{2n+1}}{(2n+1)!}\right)\\=\sum_{n=0}^\infty\frac{\text d}{\text dx}\left(\frac{(-1)^nx^{2n+1}}{(2n+1)!}\right)=\sum_{n=0}^\infty \frac{(-1)^n(2n+1)x^{2n}}{(2n+1)!}$$

Therefore, $$\cos x=\sum_{n=0}^\infty\frac{(-1)^nx^{2n}}{(2n)!}$$

Exponential function: $e^x$

Simple, right?

Still keeping up? Good. Now for the important part:

That wasn't a trick, it's still one. Since the $n^{th}$ derivative of $e^x$ is $e^x$ for all $n$, we observe the $n^{th}$ derivative to be $1$ at $x=0$ for all $n$.

Since $f(x)=e^x$ is infinitely differentiable, we may write (see Taylor Series):

$$e^x=\sum_{n=0}^\infty f^{(n)}(0)\frac{x^n}{n!}=\sum_{n=0}^\infty \frac{x^n}{n!}$$

DIY: Show this series is convergent for all finite $x$ using the ratio test.

Hyperbolic Function: $\frac1{1-x}$

There are several ways to find the Taylor series for this function. One way is to try to get rid of the denominator:

Note that, if we multiply $(1-x)$ by $(1+x)$, we get $1-x^2$ by difference of squares.

Similarly, by multiplying $(1-x)$ by $\sum_{n=0}^k x^n$, we obtain a telescoping series of $\sum_{n=0}^k(x^n-x^{n+1})$, which results in $1-x^{n+1}$.

Thus we have $$\frac{1}{1-x}=\frac{\sum_{n=0}^kx^n}{(1-x)\left(\sum_{n=0}^kx^n\right)}=\frac{\sum_{n=0}^kx^n}{1-x^{k+1}}$$

Now, if $|x|>1$, as $k$ grows, the denominator simply gets bigger and bigger, which does not help us in "removing" the denominator. The function is not defined where $x=1$, so we will only consider $|x|<1$ (we will address $x=-1$ later).

For $|x|<1$, $\lim_{k\to\infty} x^{k+1}=0$, so $$\frac1{1-x}=\lim_{k\to\infty}\frac{\sum_{n=0}^kx^n}{1-x^{k+1}}=\frac{\sum_{n=0}^\infty x^n}{1-0}=\sum_{n=0}^\infty x^n\;\text{for }|x|<1$$

Now, consider $x=-1$. The series at $x=-1$ is $$\sum_{n=0}^\infty(-1)^n=1-1+1-1+1-1+1-1\ldots$$ The partial sums of this series are $\{1,0,1,0,1,0\ldots\}$. Since the series does not converge to a point, it is not a convergent series and is neither $0$ nor $1$ (nor is it $\frac{1}{1-(-1)}=\frac12$). Therefore we consider the series only to converge for $|x|<1$, and not for $x=-1$.

Natural Logarithm: $\ln x$

We obviously cannot have a Taylor series centered at $x=0$ for $f(x) = \ln x,$ so we will instead pick $x=1$ as an obvious convenient location.

All we need to do is to start differentiating $\ln x$ and hope we find a pattern. Fortunately, we will: $$f(x) = \ln x$$ $$f'(x) = \frac{1}{x}$$ $$f''(x) = -\frac{1}{x^2}$$ $$f^{(3)}(x) = \frac{2}{x^3}$$ $$...$$ One can easily see from the Power Rule and a straightforward proof by induction that: $$f^{(n)}(x) = \frac{(-1)^{n-1}(n-1)!}{x^n}$$ for all positive integers $n.$ It is now clear to see that $f(1) = 0$ and $f^{(n)}(1) = (-1)^{n-1}(n-1)!$ for all positive integers $n.$ Now, all that remains is to write out the Taylor series: $$f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(1)}{n!}(x-1)^n$$ $$\ln x = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}(x-1)^n\\ = \frac{x-1}{1} - \frac{(x-1)^2}{2} + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + ...$$ You will sometimes see this written as: $$\ln(x+1) = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n}x^n = \frac{x}{1} - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + ...$$

e^x is 1+ x + x^2/2! + x^3/3! + .....