Taylor Series

In Section 2.6 we introduced the preliminaries of the Taylor Series. Its core idea is to find a good approximation of a given function using a power series. And that is why we introduced the infinite series and power series in the before sections.

Uniqueness Theorem

This is the basic theorem of the Taylor Series, which described the uniqueness of the Taylor expansion. That is:

*If a function can be expanded in the form of a power series, this expansion must be the Taylor Expansion of the original function.

For a function ${\displaystyle f(x)=c_0+c_1(x-a)+c_2(x-a{)}^2+\cdots +c_n(x-a{)}^n}$, to make it the approximation of ${\displaystyle P(x)}$, we let ${\displaystyle \frac{dP}{dx}=\frac{df}{dx}}$, ${\displaystyle \frac{d^{2}P}{d{P}^2}=\frac{d^{2}f}{d{x}^2}}$, and ${\displaystyle \frac{d^{2}P}{d{P}^n}=\frac{d^{n}f}{d{x}^n}}$, while finding the ${\displaystyle n}$th derivative step by step, we will naturally find ${\displaystyle c_n=\frac{f^{(n)}(a)}{n!}}$. This process is biconditional, so the existence and uniqueness of the Taylor Expansion ${\displaystyle f(x)}$ of the original function ${\displaystyle P(x)}$ at ${\displaystyle x=a}$ is determined.

So now we have the Taylor Series. But since it is the approximation of a function, it can not be calculate while it is infinite, and it can not be ${\displaystyle 100\mathrm{\%}}$ accurate while it is finite. So we next introduce the remainder theorem.

Remainder Theorem

While we use Taylor polynomials to approach a function, we need to estimate how accurate the approximation is. For a ${\displaystyle n}$-order Taylor polynomial ${\displaystyle f(x)}$, we can only say that ${\displaystyle f(x)\approx P(x)}$, but not ${\displaystyle f(x)=P(x)}$, since the finite polynomial can not make accurate approximation.

So we introduce the remainder ${\displaystyle R_n(x)}$ for ${\displaystyle n}$-order Taylor polynomial. We use it to ensure that the approximation fits the original function well in some interval, so that we can write ${\displaystyle f(x)+R_n(x)=P(x)}$.

In the Mean Value Theorem of Derivative, we said in a interval ${\displaystyle [a,b]}$, it must have a point ${\displaystyle x=c}$ makes that ${\displaystyle f^{\prime }(c)=\frac{f(b)-f(a)}{b-a}}$. We expand this theorem into higher order, for example in ${\displaystyle n}$-order, so that it could be used in the remainder theorem.

The remainder:

${\displaystyle R_n(x)=\frac{f^{n+1}(c)}{(n+1)!}\cdot (x-a{)}^{n+1}}$

This method works well when we try to approach ${\displaystyle P(x)}$ near ${\displaystyle x=a}$. The Taylor polynomial itself is ${\displaystyle 100\mathrm{\%}}$ accurate at the expand point. But it means that we need to know the value of ${\displaystyle P(x)}$ at the target point if we do the expansion at there, then it is meaningless to do this since we have known the information we want to find. That is why we introduce the remainder here. A example is given below:

e.g.

Try to find the value of ${\displaystyle \ln 1.1}$.

We do the Taylor expansion at ${\displaystyle a=1}$:
${\displaystyle (\ln x{)}^{\prime }=\frac{1}{x}}$
${\displaystyle (\ln x{)}^{″}=-\frac{1}{x^2}}$
${\displaystyle (\ln x{)}^{‴}=\frac{2}{x^3}}$
Since we are just giving an example here, we simply use the there-order Taylor expansion. But in practical cases we can make polynomial longer for a more accurate approximation.

${\displaystyle f(x)=f(a)+f^{\prime }(a)(x-a)+\frac{f^{″}(a)}{2!}(x-a{)}^2+\frac{f^{‴}(a)}{3!}(x-a{)}^3+R_3}$

${\displaystyle f(x)=\ln (a)+\frac{1}{a}(x-a)-\frac{1}{2a^2}(x-a{)}^2+\frac{1}{3a^3}(x-a{)}^3+R_3}$

So, when ${\displaystyle a=1}$, the approximation ${\displaystyle f(x)=0+(x-1)-\frac{1}{2}(x-1{)}^2+\frac{1}{3}(x-1{)}^3+R_3}$ can be used to approach the value of ${\displaystyle \ln x}$ near ${\displaystyle x=1}$. We will further explain why this approximation could be used.

For now we can calculate ${\displaystyle f(x)}$ using ${\displaystyle x=1.1}$. The value is ${\displaystyle 0.095333}$, while ${\displaystyle \ln 1.1\approx 0.095310}$. It is actually a good approximation.

We can calculate the remainder using the formula to verify our result:

${\displaystyle R_3(x)=\frac{f^{⁗}(c)}{4!}(x-1{)}^4}$, where ${\displaystyle 1<c<1.1}$.

${\displaystyle f^{⁗}(c)=-\frac{6}{c^4}}$

When ${\displaystyle 1<c<1.1}$, ${\displaystyle R_3(x)<0.000025}$. And our error is ${\displaystyle 0.000023}$, the remainder works.

The example above shows how to use Taylor expansion to approach a function near the expanding point, and then use the remainder theorem to estimate the error.

But still we need to know whether we can use this method to make the approximation and when we could say it is a good approximation.

Existence Theorem

To prove the Taylor expansion in finite terms of a function exists, we need to prove that:

${\displaystyle \underset{n\to \mathrm{\infty }}{lim}{R}_n(x)=0}$

Where ${\displaystyle R_n(x)}$ is the remainder we introduced in the Remainder Theorem.

When a Taylor Series fit this requirement, it means that when the expansion become longer and longer, the error between the Taylor Series and the original function becomes ${\displaystyle 0}$, and that is a good approximation we can use.