Table of Content

### Taylor polynomial

##### Example

Compare the graph of $$\sin x$$ (red) with the graphs of the polynomials $x-\frac{x^3}{3!}+\frac{x^5}{5!}-\dots + \frac{(-1)^nx^{2n+1}}{(2n+1)!}$ (blue) for $$n=1,2,3,\dots,12$$.

##### Definition: Taylor polynomial

Let $$f$$ be $$k$$ times differentiable at the point $$x_{0}$$. Then the Taylor polynomial \begin{align} P_n(x)&=P_n(x;x_0)\\\ &=f(x_0)+f'(x_0)(x-x_0)+\frac{f''(x_0)}{2!}(x-x_0)^2+ \\ & \dots +\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n\\ &=\sum_{k=0}^n\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k\\ \end{align} is the best polynomial approximation of degree $$n$$ (with respect to the derivative) for a function $$f$$, close to the point $$x_0$$.

Note. The special case $$x_0=0$$ is often called the Maclaurin polynomial.

If $$f$$ is $$n$$ times differentiable at $$x_0$$, then the Taylor polynomial has the same derivatives at $$x_0$$ as the function $$f$$, up to the order $$n$$ (of the derivative).

The reason (case $$x_0=0$$): Let $P_n(x)=c_0+c_1x+c_2x^2+c_3x^3+\dots +c_nx^n,$ so that \begin{align} P_n'(x)&=c_1+2c_2x+3c_3x^2+\dots +nc_nx^{n-1}, \\ P_n''(x)&=2c_2+3\cdot 2 c_3x\dots +n(n-1)c_nx^{n-2} \\ P_n'''(x)&=3\cdot 2 c_3\dots +n(n-1)(n-2)c_nx^{n-3} \\ \dots && \\ P^{(k)}(x)&=k!c_k + x\text{ terms} \\ \dots & \\ P^{(n)}(x)&=n!c_n \\ P^{(n+1)}(x)&=0. \end{align}

From these way we obtain the coefficients one by one: \begin{align} c_0= P_n(0)=f(0) &\Rightarrow c_0=f(0) \\ c_1=P_n'(0)=f'(0) &\Rightarrow c_1=f'(0) \\ 2c_2=P_n''(0)=f''(0) &\Rightarrow c_2=\frac{1}{2}f''(0) \\ \vdots & \\ k!c_k=P_n^{(k)}(0)=f^{(k)}(0) &\Rightarrow c_k=\frac{1}{k!}f^{(k)}(0). \\ \vdots &\\ n!c_n=P_n^{(n)}(0)=f^{(n)}(0) &\Rightarrow c_k=\frac{1}{n!}f^{(n)}(0). \end{align} Starting from index $$k=n+1$$ we cannot pose any new conditions, since $$P^{(n+1)}(x)=0$$.

##### Taylor's Formula

If the derivative $$f^{(n+1)}$$ exists and is continuous on some interval $$I=\, ]x_0-\delta,x_0+\delta[$$, then $$f(x)=P_n(x;x_0)+E_n(x)$$ and the error term $$E_n(x)$$ satisfies $E_n(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-x_0)^{n+1}$ at some point $$c\in [x_0,x]\subset I$$. If there is a constant $$M$$ (independent of $$n$$) such that $$|f^{(n+1)}(x)|\le M$$ for all $$x\in I$$, then $|E_n(x)|\le \frac{M}{(n+1)!}|x-x_0|^{n+1} \to 0$ as $$n\to\infty$$.

\neq omitted here (mathematical induction or integral).

Examples of Maclaurin polynomial approximations: \begin{align} \frac{1}{1-x} &\approx 1+x+x^2+\dots +x^n =\sum_{k=0}^{n}x^k\\ e^x&\approx 1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\dots + \frac{1}{n!}x^n =\sum_{k=0}^{n}\frac{x^k}{k!}\\ \ln (1+x)&\approx x-\frac{1}{2}x^2+\frac{1}{3}x^3-\dots + \frac{(-1)^{n-1}}{n}x^n =\sum_{k=1}^{n}\frac{(-1)^{k-1}}{k}x^k\\ \sin x &\approx x-\frac{1}{3!}x^3+\frac{1}{5!}x^5-\dots +\frac{(-1)^n}{(2n+1)!}x^{2n+1} =\sum_{k=0}^{n}\frac{(-1)^k}{(2k+1)!}x^{2k+1}\\ \cos x &\approx 1-\frac{1}{2!}x^2+\frac{1}{4!}x^4-\dots +\frac{(-1)^n}{(2n)!}x^{2n} =\sum_{k=0}^{n}\frac{(-1)^k}{(2k)!}x^{2k} \end{align}

##### Example

Which polynomial $$P_n(x)$$ approximates the function $$\sin x$$ in the interval $$[-\pi,\pi]$$ so that the absolute value of the error is less than $$10^{-6}$$?

We use Taylor's Formula for $$f(x)=\sin x$$ at $$x_0=0$$. Then $$|f^{(n+1)}(c)|\le 1$$ independently of $$n$$ and the point $$c$$. Also, in the interval in question, we have $$|x-x_0|=|x|\le \pi$$. The requirement will be satisfied (at least) if $|E_n(x)|\le \frac{1}{(n+1)!}\pi^{n+1} < 10^{-6}.$ This inequality must be solved by trying different values of $$n$$; it is true for $$n\ge 16$$.

The required approximation is achieved with $$P_{16}(x)$$, which fo sine is the same as $$P_{15}(x)$$.

Check from graphs: $$P_{13}(x)$$ is not enough, so the theoretical bound is sharp!

### Taylor polynomial and extreme values

If $$f'(x_0)=0$$, then also some higher derivatives may be zero: $f'(x_0)=f''(x_0)= \dots = f^{(n)}(x_0) =0,\ f^{(n+1)}(x_0) \neq 0.$ Then the behaviour of $$f$$ near $$x=x_0$$ is determined by the leading term (after the constant term $$f(x_0)$$) $\frac{f^{(n+1)}(x_0)}{(n+1)!}(x-x_0)^{n+1}.$ of the Taylor polynomial.

This leads to the following result:

##### Extreme values
• If $$n$$ is even, then $$x_0$$ is not an extreme point of $$f$$.
• If $$n$$ is odd and $$f^{(n+1)}(x_0)>0$$, then $$f$$ has a local minimum at $$x_0$$.
• If $$n$$ is odd and $$f^{(n+1)}(x_0)<0$$, then $$f$$ has a local maximum at $$x_0$$.

### Newton's method

The first Taylor polynomial $$P_1(x)=f(x_0)+f'(x_0)(x-x_0)$$ is the same as the linearization of $$f$$ at the point $$x_0$$. This can be used in some simple approximations and numerical methods.

##### Newton's method

The equation $$f(x)=0$$ can be solved approximately by choosing a starting point $$x_0$$ (e.g. by looking at the graph) and defining $x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$ for $$n=0,1,2,\dots$$ This leads to a sequence $$(x_0,x_1,x_2,\dots )$$, whose terms usually give better and better approximations for a zero of $$f$$.

The recursion formula is based on the geometric idea of finding an approximative zero of $$f$$ by using its linearization (i.e. the tangent line).

##### Example

Find an approximate value of $$\sqrt{2}$$ by using Newton's method.

We use Newton's method for the function $$f(x)=x^2-2$$ and initial value $$x_0=2$$. The recursion formula becomes $x_{n+1}= x_n-\frac{x_n^2-2}{2x_n} = \frac{1}{2}\left(x_n+\frac{2}{x_n}\right),$ from which we obtain $$x_1=1{,}5$$, $$x_2\approx 1{,}41667$$, $$x_3\approx 1{,}4142157$$ and so on.

By experimenting with these values, we find that the number of correct decimal places doubles at each step, and $$x_7$$ gives already 100 correct decimal places, if intermediate steps are calculated with enough precision.

### Taylor series

##### Taylor series

If the error term $$E_n(x)$$ in Taylor's Formula goes to zero as $$n$$ increases, then the limit of the Taylor polynomial is the Taylor series of $$f$$ (= Maclaurin series for $$x_0=0$$).

The Taylor series of $$f$$ is of the form $\sum_{k=0}^{\infty}\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k = \lim_{n\to\infty} \sum_{k=0}^{n}\frac{f^{(k)}(x_0)}{k!}(x-x_0)^k .$ This is an example of a power series.

The Taylor series can be formed as soon as $$f$$ has derivatives of all orders at $$x_0$$ and they are substituted into this formula. There are two problems related to this: Does the Taylor series converge for all values of $$x$$?

Answer: Not always; for example, the function $f(x)=\frac{1}{1-x}$ has a Maclaurin series (= geometric series) converging only for $$-1 < x < 1$$, although the function is differentiable for all $$x\neq 1$$: $f(x)=\frac{1}{1-x} = 1+x+x^2+x^3+x^4+\dots$

If the series converges for some $$x$$, then does its sum equal $$f(x)$$? Answer: Not always; for example, the function $f(x)=\begin{cases} e^{-1/x^2}, & x\neq 0,\\ 0, & x=0,\\ \end{cases}$ satisfies $$f^{(k)}(0)=0$$ for all $$k\in \mathbf{N}$$ (elementary but difficult calculation). Thus its Maclaurin series is identically zero and converges to $$f(x)$$ only at $$x=0$$.

Conclusion: Taylor series should be studied carefully using the error terms. In practice, the series are formed by using some well known basic series.

##### Examples

\begin{align} \frac{1}{1-x} &= \sum_{k=0}^{\infty} x^k,\ \ |x|< 1 \\ e^x &= \sum_{k=0}^{\infty} \frac{1}{k!}x^k, \ \ x\in \mathbb{R} \\ \sin x &= \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k+1)!} x^{2k+1}, \ \ x\in \mathbb{R} \\ \cos x &= \sum_{k=0}^{\infty} \frac{(-1)^{k}}{(2k)!} x^{2k},\ \ x\in \mathbb{R} \\ (1+x)^r &= 1+\sum_{k=1}^{\infty} \frac{r(r-1)(r-2)\dots (r-k+1)}{k!}x^k, |x|<1 \end{align} The last is called the Binomial Series and is valid for all $$r\in \mathbb{R}$$. If $$r=n \in \mathbb{N}$$, then starting from $$k=n+1$$, all the coefficients are zero and in the beginning $\binom{n}{k} =\frac{n!}{k!(n-k)!} = \frac{n(n-1)(n-2)\dots (n-k+1)}{k!}.$

Compare this to the Binomial Theorem: $(a+b)^n=\sum_{k=0}^n\binom{n}{k} a^{n-k}b^k =a^n +na^{n-1}b+\dots +b^n$ for $$n\in\mathbb{N}$$.

### Power series

##### Definition: Power series

A power series is of the form $\sum_{k=0}^{\infty} c_k(x-x_0)^k = \lim_{n\to\infty} \sum_{k=0}^{n}c_k(x-x_0)^k.$ The point $$x_0$$ is the centre and the $$c_k$$ are the coefficients of the series.

The series converges at $$x$$ if the above limit is defined.

There are only three essentially different cases:

##### Abel's Theorem.
• The power series converges only for $$x=x_0$$ (and then it consists of the constant $$c_0$$ only)
• The power series converges for all $$x\in \mathbb{R}$$
• The power series converges on some interval $$]x_0-R,x_0+R[$$ (and possibly in one or both of the end points), and diverges for other values of $$x$$.

The number $$R$$ is the radius of convergence of the series. In the first two cases we say that $$R=0$$ or $$R=\infty$$ respectively.

##### Example

For which values of the variable $$x$$ does the power series $\sum_{k=1}^{\infty} \frac{k}{2^k}x^k$ converge?

We use the ratio test with $$a_k=kx^k/2^k$$. Then $\left| \frac{a_{k+1}}{a_k} \right| = \left| \frac{(k+1)x^{k+1}/2^{k+1}}{kx^k/2^k} \right| = \frac{k+1}{2k}|x| \to \frac{|x|}{2}$ as $$k\to\infty$$. By the ratio test, the series converges for $$|x|/2<1$$, and diverges for $$|x|/2>1$$. In the border-line cases $$|x|/2= 1\Leftrightarrow x=\pm 2$$ the general term of the series does not tend to zero, so the series diverges.

Result: The series converges for $$-2< x< 2$$, and diverges otherwise.

##### Definition: Sum function

In the interval $$I$$ where the series converges, we can define a function $$f\colon I\to \mathbb{R}$$ by setting $$\label{summafunktio} f(x) = \sum_{k=0}^{\infty} c_k(x-x_0)^k, \tag{1}$$ which is called the sum function of the power series.

The sum function $$f$$ is continuous and differentiable on $$]x_0-R,x_0+R[$$. Moreover, the derivative $$f'(x)$$ can be calculated by differentiating the sum function term by term: $f'(x)=\sum_{k=1}^{\infty}kc_k(x-x_0)^{k-1}.$ Note. The constant term $$c_0$$ disappears and the series starts with $$k=1$$. The differentiated series converges in the same interval $$x\in \, ]x_0-R,x_0+R[$$; this may sound a bit surprising because of the extra coefficient $$k$$.

##### Example

Find the sum function of the power series $$1+2x+3x^2+4x^3+\dots$$

This series is obtained by differentiating termwise the geometric series (with $$q=x$$). Therefore, \begin{align} 1+2x+3x^2+4x^3+\dots &= D(1+x+x^2+x^3+x^4+\dots ) \\ &= \frac{d}{dx}\left( \frac{1}{1-x}\right) = \frac{1}{(1-x)^2}. \end{align} Multiplying with $$x$$ we obtain $\sum_{k=1}^{\infty}kx^{k} = x+2x^2+3x^3+4x^4+\dots = \frac{x}{(1-x)^2},$ which is valid for $$|x|<1$$.

In the case $$[a,b]\subset\ ]x_0-R,x_0+R[$$ we can also integrate the sum function termwise: $\int_a^b f(x)\, dx = \sum_{k=0}^{\infty}c_k\int_a^b (x-x_0)^k\, dx.$ Often the definite integral can be extended up to the end points of the interval of convergence, but this is not always the case.

##### Example

Calculate the sum of the alternating harmonic series.

Let us first substitute $$q=-x$$ to the geometric series. This yields $1-x+x^2-x^3+x^4-\dots =\frac{1}{1-(-x)} = \frac{1}{1+x}.$ By integrating both sides from $$x=0$$ to $$x=1$$ we obtain $1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\dots =\int_0^1\frac{1}{1+x} =\ln 2.$ Note. Extending the limit of integration all the way up to $$x=1$$ should be justified more rigorously here. We shall return to integration later on the course.

Last modified: Thursday, 21 January 2021, 6:29 PM
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