Find both the Maclaurin and Taylor series expansion for the given function
Sketch truncated Taylor polynomial of a Taylor series expansion
Work with limits when evaluating a given series
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Power Series : Extra video by Sam recapping power series from the lecture.
Problem sheet
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:
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:
$$f(x)=\frac{1}{1+x}=1+(-x)+x^2-x^3+x^4+...+(-1)^nx^n= \boxed{\sum_{n=0}^{\infty}(-1)^nx^n}$$
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:
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:
$$\boxed{\lim\limits_{x\rightarrow0}\frac{x^2}{x^2}=1.}$$
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.
(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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