Power Series Tutorial Sheet, Sheet #5A

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Skill Building Questions

Problem 1.

Compute the coefficients of the following power series:

(a) $f(x)=\sin{x}\quad\quad\quad(x=0,n=6)$

Maclaurin series (x=0) in the form: $$\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}x^n$$ Table of $f^{(n)}(0)$ coefficients for this series: $$\begin{align*} & n& f&^{(n)}(x)& f&^{(n)}(0)& \\ \hline & 0& &\sin{x}& &0& \\ & 1& & \cos{x}& &1&\\ & 2& -&\sin{x}& &0& \\ & 3& -&\cos{x}& -&1& \\ & 4& &\sin{x}& &0& \\ & 5& &\cos{x}& &1& \\ & 6& -&\sin{x}& &0& \\ \hline \end{align*}$$ Coefficients only exist for odd values of n. In addition, the sign of the coefficients will alternate. Thus, the resulting power series is: $$f(x)=f(0)+f^{'}(0)x+f^{''}(0)\frac{x^2}{2!}+f^{'''}(0)\frac{x^3}{3!}+f^{''''}(0)\frac{x^4}{4!}+...$$ $$\boxed{f(x)=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+...}$$








(b) $f(x)=e^{-x^2}\quad\quad\quad(x=0,n=3)$

Table of $f^{(n)}(0)$ coefficients for this series (Maclaurin, x=0): $$\begin{align*} & n& f&^{(n)}(x)& f&^{(n)}(0)& \\ \hline & 0& e&^{-x^2}& &1& \\ & 1& -&2xe^{-x^2}& &0& \\ & 2& (4&x^2-2)e^{-x^2}& -&2& \\ & 3& -&4(2x^3-3x)e^{-x^2}& &0& \\ \hline \end{align*}$$ Thus, the resulting power series is: $$\boxed{f(x)=1-x^2+...}$$








(c) $f(x)=\frac{1}{2+x}\quad\quad\quad (x=0,n=3)$

Table of $f^{(n)}(0)$ coefficients for this series (Maclaurin, x=0): $$\begin{align*} & n& f&^{(n)}(x)& f&^{(n)}(0)& \\ \hline & 0& &\frac{1}{x+2}& &\frac{1}{2}& \\ & 1& -&\frac{1}{(x+2)^2}& -&\frac{1}{4}& \\ & 2& &\frac{2}{(x+2)^3}& &\frac{1}{4}& \\ & 3& -&\frac{6}{(x+2)^4}& -&\frac{3}{8}& \\ \hline \end{align*}$$ Thus, the resulting power series is: $$\boxed{f(x)=\frac{1}{2}-\frac{x}{4}+\frac{x^2}{8}-\frac{x^3}{16}+...}$$








Problem 2.

Write the general equation of the $n^{th}$ order term for function $f(x)$ from $n=0$ to infinity and re-express function f(x) in power series form:

(a) $f(x) =e^x= e+e(x-1)+\frac{e}{2}(x-1)^2+\frac{e}{3!}(x-1)^3+\frac{e}{4!}(x-1)^4+…$

Determine the $n^{th}$ term of the sequence $$n^{th}=e+e(x-1)+\frac{e}{2}(x-1)^2+\frac{e}{3!}(x-1)^3+\frac{e}{4!}(x-1)^4+...+\frac{e(x-1)^n}{n!}$$ Re-express function f(x) in power series form: $$f(x)=e+e(x-1)+\frac{e}{2}(x-1)^2+\frac{e}{3!}(x-1)^3+\frac{e}{4!}(x-1)^4+...+\frac{e(x-1)^n}{n!}=\boxed{\sum_{n=0}^{\infty}\frac{e(x-1)^n}{n!}}$$








(b) $f(x)=\frac{1}{1+x}=1-x+x^2-x^3+x^4+…$

Determine the $n^{th}$ term of the sequence $$n^{th}=\frac{1}{1+x}=1-x+x^2-x^3+x^4+...+(-1)^nx^n$$ Re-express function f(x) in the form of power series: $$f(x)=\frac{1}{1+x}=1+(-x)+x^2-x^3+x^4+...+(-1)^nx^n= \boxed{\sum_{n=0}^{\infty}(-1)^nx^n}$$








(c) $f(x)=\cos{x}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+…$

Determine $n^{th}$ term of the sequence $$n^{th}=\cos{x}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...+(-1)^n\frac{x^{2n}}{(2n)!}$$ Re-express function f(x) in the form of power series: $$f(x)=\cos{x}=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+...+(-1)^n\frac{x^{2n}}{(2n)!}=\boxed{\sum_{n=0}^{\infty}(-1)^n\frac{x^{2n}}{(2n!)}}$$








Problem 3.

(a) Find the Maclaurin series for : \(\int_0^x\cos(t^3)dt\)

Consider the Maclaurin series for $\mathrm{\cos}(x)$ is $$1-\frac{x^2}{2!}+\frac{x^4}{4!}-...=\sum^{\infty}_{n=0}(-1)^n\frac{x^{2n}}{(2n)!},$$ But for $t^3$: $x$ can be replaced with $t^3$ to get the Maclaurin series for $\mathrm{cos}(t^3)$: $$1-\frac{t^6}{2!}+\frac{t^{12}}{4!}-...=\sum^{\infty}_{n=0}(-1)^n\frac{t^{6n}}{(2n)!},$$ Therefore, by integrating the sum: $$\int^{x}_{0}\cos(t^3)dt=\int_{0}^{x}\sum^{\infty}_{n=0}(-1)^n\frac{t^{6n}}{(2n)!}dt=\Bigg[\sum^{\infty}_{n=0}(-1)^n\frac{t^{6n+1}}{(6n+1)(2n)!}\Bigg]^{x}_{0}=$$ $$=\sum^{\infty}_{n=0}(-1)^n\frac{x^{6n+1}}{(6n+1)(2n)!}$$ Where the integral can be evalutated term-by-term. The first four terms of this series are: $$\boxed{x-\frac{x^{7}}{14}+\frac{x^{13}}{312}-\frac{x^{19}}{13680}+...}$$








Problem 4.

Find the Maclaurin Series expansion (up to and including the 4th term) near $x=0$ for the following equation:

$f(x)=\mathrm{sin}(x)$

Find the Maclaurin Series expansion near $x =0$ for $\displaystyle f(x)=\mathrm{sin}x$, by finding the first, second, third, etc derivatives and evaluating them at $x = 0$. Table of $f^{(n)}(0)$ coefficients for this series (Maclaurin, x=0): \begin{align*} & n& f&^{(n)}(x)& f&^{(n)}(0)& \\ \hline & 0& &\mathrm{sin}x& &0&\\ & 1& &\mathrm{cos}x& &1&\\ & 2& -&\mathrm{sin}x& &0& \\ & 3& -&\mathrm{cos}x& -&1& \\ & 4& &\mathrm{sin}x& &0& \\ \hline \end{align*} As this function can be differentiated infinitely, $n$ will continue forever.
Substituting the values of the derivatives into the Maclaurin series: $$f(x)\approx{f(0)+f{'}(0)x+\frac{f{''}(0)}{2!}x^2+\frac{f{'''}(0)}{3!}x^3+\frac{f{''''}(0)}{4!}x^4+...}$$ Resulting in: $$\mathrm{sin}x=0+(1)(x)+0+\frac{-1}{3!}x^3+0+\frac{1}{5!}x^5+0+\frac{-1}{7!}x^7+...$$ Giving: $$\boxed{\mathrm{sin}x=x-\frac{1}{6}x^3+\frac{1}{120}x^5-\frac{1}{5040}x^7+...}$$ Plotting the polynomial to check if it is a good approximation to $f(x)=\mathrm{sin}x$: Once again, it can be observed that our polynomial (in blue) is a good approximation to $\displaystyle f(x)=\mathrm{sin}x$ (in red) between $[-\pi,+\pi]$.








Problem 5.

Compute the following limits using Taylor and Maclaurin expansions:

(a) $\lim\limits_{x\rightarrow0}\big(\frac{e^x -\mathrm{sin}x-\mathrm{cos}x}{e^{x^2}-e^{x^3}}\big)$

Note: $O(x^n)$ denotes all terms with powers greater or equal to $n$, which can be considered negligible):
Express each term separately using Maclaurin series equation $$e^x-\mathrm{sin}x-\mathrm{cos}x= $$ $$=\Big(1+x+\frac{1}{2}x^2+O(x^3)\Big)-\Big(x+O(x^3)\Big)-\Big(1-\frac{1}{2}x^2+O(x^3)\Big)=x^2+O(x^3),$$ Similarly, the function in the denominator can be written as: $$e^{x^{2}}-e^{x^{3}}=\Big(1+x^2+O(x^3)\Big)-\Big(1+x^3+O(x^4)\Big)=x^2+O(x^3)$$ Note: Given $e^x \approx{} (1+x+\frac{1}{2}x^2+O(x^3))$ Replace $x$ with $x^n$ in the Maclaurin series above to get: $$e^{x^n} \approx{} (1+x^{n}+\frac{1}{2}(x^n)^2+O((x^n)^3))$$ Hence: $$\lim\limits_{x\rightarrow0}\frac{e^x-\mathrm{sin}x-\mathrm{cos}x}{e^{x^{2}}-e^{x^{3}}}=\lim\limits_{x\rightarrow0}\frac{x^2+O(x^3)}{x^2+O(x^3)}=$$ Ignoring all negligible summands: $$\boxed{\lim\limits_{x\rightarrow0}\frac{x^2}{x^2}=1.}$$








(b) $\lim\limits_{x\rightarrow0}\big(\frac{(\sqrt[5]{1-5x^2+x^4}-1+x^2)}{x^4}\big)$

Find the Maclaurin series for $\sqrt[5]{1+x}$, $x$ will be substituted for $(-5x^2+x^4)$ later: \begin{align*} & n& f&^{(n)}(x)& f&^{(n)}(0)& \newline \hline & 0& &(1+x)^\frac{1}{5}& &1& \\\ & 1& \frac{1}{5}&(1+x)^{-\frac{4}{5}}& &\frac{1}{5}& \\\ & 2& -\frac{4}{25}&(1+x)^{-\frac{9}{5}}& -&\frac{4}{25}& \newline \hline \end{align*} As a result: $$\sqrt[5]{1+x}\approx1+\frac{1}{5}x-\frac{2}{25}x^2$$ Therefore: $$\sqrt[5]{1-5x^2+x^4}=\sqrt[5]{1+(-5x^2+x^4)}\approx1+\frac{1}{5}(-5x^2+x^4)-\frac{2}{25}(-5x^2+x^4)^2+...$$ Thus the numerator can be rewritten as: $$\sqrt[5]{1-5x^2+x^4}-1+x^2=$$ $$\big(1+\frac{1}{5}(-5x^2+x^4)-\frac{2}{25}(-5x^2+x^4)^2+O(x^5))-1+x^2=$$ $$=1-x^2+\frac{1}{5}x^4+-2x^4+O(x^5)-1+x^2=-\frac{9}{5}x^4+O(x^5)$$ Hence: $$ \lim\limits_{x\rightarrow0}\frac{\sqrt[5]{1-5x^2+x^4}-1+x^2}{x^4}=\lim\limits_{x\rightarrow0}\frac{-\frac{9}{5}x^4+O(x^5)}{x^4}= $$ Ignoring all negligible summands: $$=\boxed{\lim\limits_{x\rightarrow0}\frac{-\frac{9}{5}x^4}{x^4}=-\frac{9}{5}}$$








Problem 6.

Study the local behaviour of function: \(f(x) = 1-(x-2)^2+\frac{1}{4}(x-2)^3+O((x-2)^4),\quad x\rightarrow 2\) in a neighbourhood of $x_0=2$, discuss whether it is a stationary point, and, if yes, of which type.

Rewriting Taylor expansion for $f(x)$ of order $3$ centered at $x_0=2$ as: \begin{equation*} f(x)=f(2)+f'(2)(x-2)+\frac{1}{2!}f''(2)(x-2)^2+\frac{1}{3!}f'''(2)(x-2)^3+O\big((x-2)^4\big),\quad x\rightarrow2 \end{equation*} Then by comparing the coefficients of the rewritten form for f(x) to the f(x) in the question: \begin{align*} &\textrm{Coefficient for}& &\textrm{Question}& &\textrm{Rewritten}& \newline \hline & f(x-2)& &1& &f(2)& \\\ & f'(x-2)& &0& &f'(2)& \\\ & f''(x-2)& -&1& &\frac{1}{2!}f''(2)& \\\ & f'''(x-2)& &\frac{1}{4}& &\frac{1}{3!}f'''(2)& \newline \hline \end{align*} Comparing the coefficients of the various degrees of $(x-2)$ in $f(x)$ and solving for each $f^{n}(2)$: \begin{equation*} f(2)=1,\quad f'(2)=0,\quad f''(2)=-1\times2!=-2,\quad f'''(2)=\frac{1}{4}\times3!=\frac{3}{2}. \end{equation*} After finding the values of f'(2) and f''(2), it can be concluded that \ans{$x_0=2$ is a stationary point and, in particular, it is a local maximum of $f$.








Exam Style Questions

Problem 7.

(a) Evaluate the first five truncated Taylor polynomials (i.e. $0^{th}$, $1^{st}$…$5^{th}$) for: $f(x)=\mathrm{\ln}(x),$ around ${}x=10$

Find the Taylor Expansion of $f(x)=lnx$ near $x=10$ by finding the first, second, third, etc derivatives and evaluate them at $x = 10$. Table of $f^{(n)}(0)$ coefficients for this series (Taylor, x=10): \begin{align*} & n& f&^{(n)}(x)& f&^{(n)}(10)& \newline \hline & 0& &\mathrm{ln}x& &2.302585093& \\\ & 1& &\frac{1}{x}& &\frac{1}{10}& \\\ & 2& -&\frac{1}{x^2}& -&\frac{1}{100}& \\\ & 3& &\frac{2}{x^3}& &\frac{2}{1000}& \\\ & 4& &\frac{6}{x^4}& -&\frac{6}{10000}& \newline \hline \end{align*} This function can be differentiated infinitely so the pattern in the table will continue forever. Substituting these values into the Taylor Series: \begin{equation*} f(x)\approx{f(a)+f{'}(a)(x-a)+\frac{f{''}(a)}{2!}(x-a)^2+\frac{f{'''}(a)}{3!}(x-a)^3+\frac{f{''''}(a)}{4!}(x-a)^4+...} \end{equation*} Results in: \begin{equation*} f(x)\approx{f(10)+f{'}(10)(x-10)+\frac{f{''}(10)}{2!}(x-10)^2+\frac{f{'''}(10)}{3!}(x-10)^3+\frac{f{''''}(10)}{4!}(x-10)^4+...} \end{equation*} $$f(x)\approx{2.302585093+0.1(x-10)-\frac{0.01}{2!}(x-10)^2+\frac{0.002}{3!}(x-10)^3}$$ $$-\frac{0.0006}{4!}(x-10)^4+...$$ Expanding this all out and collecting like terms, we obtain the polynomial which approximates $lnx$: \begin{equation*} f(x)\approx0.21925+0.4x-0.03x^2+0.00133x^3-0.000025x^4+.. \end{equation*} We see from the graph that our polynomial (in blue) is a good approximation for the graph of the natural logarithm function $f(x)=\mathrm{ln}x$ (in red) in the region near $x=10.$








(b) Sketch $f(x)$ and the first two truncated Taylor polynomials for the function expanded around the point $c=10$









Problem 8.

The first five terms of the Maclaurin series of function$f(x)=\frac{sin(\pi x)}{x}$ are
$g_{4}(x)=\pi -\frac{\pi ^{3}x^{2}}{6}+\frac{\pi ^{5}x^{4}}{120}-\frac{\pi ^{7}x^{6}}{5040}+\frac{\pi ^{9}x^{8}}{362880}$

(a) Find an expression for the term containing $x^{n}$ of the above series.

$\quad \Rightarrow{} \boxed{(-1)^{\frac{n}{2}}\frac{\pi ^{n+1}x^{n}}{(n+1)!}}$








(b) Use the series above to help you sketch this function in the region $-3\leq x\leq 3$.










(c)Add sketches of the zeroth, second, and fourth order truncated series approximations to your graph.









Extension Questions

Problem 9.

Find power series of $\frac{1}{x^{2}+4x+3}$ in term of $(x-1)$.

$\frac{1}{x^{2}+4x+3}=\frac{1}{\left ( x+1 \right )\left ( x+3 \right )}= \frac{1}{2\left ( 1+x \right )}-\frac{1}{2\left ( 3+x \right )}$

write $x$ in form of $(x-1)$, knowing that $\left | x-1 \right |< 2$

$\Rightarrow \quad \frac{1}{4\left ( 1+\frac{x-1}{2} \right )}-\frac{1}{8\left ( 1+\frac{x-1}{4} \right )}$

$\Rightarrow \quad \frac{1}{4}\left [ 1-\frac{x-1}{2}+\frac{\left ( x-1 \right )^{2}}{2^{2}} +...+\left ( -1 \right )^{n}\frac{\left ( x-1 \right )^{n}}{2^{n}}+...\right ]-\frac{1}{8}\left [ 1-\frac{x-1}{4}+\frac{\left ( x-1 \right )^{2}}{4^{2}} +...+\left ( -1 \right )^{n}\frac{\left ( x-1 \right )^{n}}{4^{n}}+...\right ] $

$\Rightarrow \quad \boxed{\sum_{n=0}^{\infty }\left ( -1 \right )^{n}\left ( \frac{1}{2^{n+2}} - \frac{1}{2^{2n+3}}\right )\left ( x-1 \right )^{n}} \quad $ for$ \quad \boxed{ -1< x < 3}$








Problem 10.

(a) Write the Maclaurin series expansion of the function up to the $4^{th}$ power:

\[f(x) = x \cos \left(\frac{x}{\sqrt{3}}\right)-(\alpha-x^3)\sin x \text{ for all } \alpha\in\mathbb{R}\]

Using this expansion, find for which $\alpha$ values of the point $x = 0$ is stationary for $f$ and specify of which type.

Using the Maclaurin expansion of functions $\cos x$ and $\sin x$ we get that as $x\rightarrow0$ \begin{equation*} f(x)=x \cos \left( \frac{x}{\sqrt{3}}-(\alpha-x^3) \right) \sin x= \end{equation*} \begin{equation*} x\left(1-\frac{1}{6}x^2+O(x^4)\right)-(\alpha-x^3)\big(x-\frac{1}{6}x^3+O(x^5)\big)= \end{equation*} \begin{equation*} x-\frac{1}{6}x^3+O(x^5)-\alpha{x}+\frac{1}{6}\alpha{x^3}+x^4+O(x^5)=(1-\alpha)x-\frac{1}{6}(1-\alpha)x^3+x^4+O(x^5). \end{equation*} Hence the Maclaurin expansion of order $4$ of $f(x)$ is: \begin{equation*} f(x)=(1-\alpha)x-\frac{1}{6}(1-\alpha)x^3+x^4+O(x^5). \end{equation*} Since the Maclaurin expansion of order $4$ of $f$ is \begin{equation*} f(x)=f(0)+f'(0)x+\frac{1}{2}f''(0)x^2+\frac{1}{6}f'''(0)x^3+\frac{1}{24}f^{(4)}(0)x^4+O(x^5), \end{equation*} comparing the coefficients of each of the same order: \begin{equation*} f(0)=0,\quad f'(0)=1-\alpha,\quad f''(0)=0,\quad f'''(0)=\alpha-1,\quad f^{(4)}(0)=24. \end{equation*} For $x=0$ to be a stationary point, $\alpha=1$ so that $f'(0)=0$: $$f''(0) = f'''(0) = 0 \text{ and } f^{(4)}(0) = 24$$ It follows that $\boxed{\text{if } \alpha = 1 \text{ then } x = 0 \text{ is a local minimum of } f}$.








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