Note 12 - 14/04/2023
TABLE OF CONTENTS
I. Power Series
A power series is a series of the form:
\[\sum_{n=0}^\infty c_nx^n=c_0+c_1x+c_2x^2+c_3x^3+\ldots\]More generally, a series of the form
\[\sum_{n=0}^\infty c_n(x-a)^n=c_0+c_1(x-a)+c_2(x-a)^2+c_3(x-a)^3+\ldots\]is called power series in $(x-a)$ or a power series centered at $a$ or a power series about $a$.
Theorem
For a given power series $\sum_{n=0}^\infty c_n(x-a)^n$ there are only three possibilities:
- The series converges only when $x=a$.
- The series converges for all $x$.
- There is a positive number $R$ such that the series converges if $\vert x-a \vert < R$ and diverges if $\vert x-a \vert > R$
The number $R$ is called the radius of convergence of the power series.
In general, the Ratio Test (or sometimes the Root Test) should be used to determine the radius of convergence $R$. The Ratio Test and Root Test always fail when $x$ is an endpoint of the interval of convergence, so the endpoints must be checked with some other test.
Example
Find the radius of convergence and interval of convergence of the series
a. $\sum_{n=0}^\infty (-1)^nnx^n$
The given power series is centered at $a=0$. Let $a_n=(-1)^nnx^n$ then
\[\lim_{n \to \infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert=\lim_{n \to \infty} \left\vert \frac{(-1)^{n+1}(n+1)x^{n+1}}{(-1)^nnx^n} \right\vert = \lim_{n \to \infty} \frac{n+1}{n} \vert x \vert = \vert x \vert\]By the Ratio Test, we have:
- The series converges iff $\vert x \vert < 1$
- The series diverges iff $\vert x \vert > 1$
$\Rightarrow$ The radius of convergence is $R=1$
However, we should check if $\vert x \vert=1$ then the power series converges or diverges.
- $x=1$ then the power series $\sum_{n=0}^\infty a_n=\sum_{n=0}^\infty (-1)^nn$ is divergent by using the Alternating Series Test.
- $x=-1$ then the power series $\sum_{n=0}^\infty a_n=\sum_{n=0}^\infty n$ is divergent by using the Test for Divergence.
Hence, the interval of convergence is $I=(-1,1)$
b. $\sum_{n=0}^\infty \dfrac{(-3)^nx^n}{\sqrt{n+1}}$
The provided power series is centered at $a=0$. Let $a_n=\dfrac{(-3)^nx^n}{\sqrt{n+1}}$ then
\[\lim_{n\to\infty} \left\vert \frac{a_{n+1}}{a_n} \right\vert = \lim_{n\to\infty} \left\vert \frac{(-3)^{n+1}x^{n+1}}{\sqrt{n+2}} \cdot \frac{\sqrt{n+1}}{(-3)^nx^n} \right\vert = 3\lim_{n\to\infty} \sqrt{\frac{n+1}{n+2}} \vert x \vert = 3\vert x \vert\]By the Ratio Test, we have:
- The series converges iff $3\vert x \vert<1$
- The series converges iff $3\vert x \vert>1$
So $\vert x \vert < \dfrac{1}{3} \Rightarrow$ The radius of convergence is $R=\dfrac{1}{3}$
However, we should check if $\vert x \vert = \dfrac{1}{3}$ then the power series converges or diverges.
- $x=\dfrac{1}{3}$ then $\sum_{n=0}^\infty a_n = \sum_{n=0}^\infty \dfrac{(-1)^n}{\sqrt{n+1}}$ is convergent by using the Alternating Series Test.
- $x=-\dfrac{1}{3}$ then $\sum_{n=0}^\infty a_n = \sum_{n=0}^\infty \dfrac{1}{\sqrt{n+1}}$ is divergent by using the Integral Test.
Hence, the interval of convergence is $I=\left(-\dfrac{1}{3},\dfrac{1}{3}\right]$
II. Taylor and Maclaurin Series
Suppose that $f$ is any function that can be represented by a power series
$\displaystyle f(x)=c_0+c_1(x-a)+c_2(x-a)^2+c_3(x-a)^3+c_4(x-a)^4+\cdots \ \ \ \ \ \vert x-a \vert < R$
Subtitute $x=a$ into the above equation and we get
\[f(a)=c_0\]We can differentiate the series term by term:
$\displaystyle f’(x)=c_1+2c_2(x-a)+3c_3(x-a)^2+4c_4(x-a)^3+\cdots \ \ \ \ \ \vert x-a \vert < R$
Again we put $x=a$ in the above equation. The result is
\[f'(a)=c_1\]By now you can see the pattern. If we continue to differentiate and subtitute $x=a$, we obtain
\[f^{(n)}(a)=2\cdot3\cdot4\cdots nc_n=n!c_n\]So
\[c_n=\frac{f^{(n)}(a)}{n!}\]Thus we have proved the following theorem
Theorem
If $f$ has a power series representation (expansion) at $a$, that is, if
\[f(x)=\sum_{n=0}^\infty c_n(x-a)^n \ \ \ \ \ \ \ \vert x-a \vert < R\]then its coefficients are given by the formula
\[c_n=\frac{f^{(n)}(a)}{n!}\]Taylor series
\[\begin{align*} f(x)&=\sum_{n=0}^\infty \frac{f^{(n)}(a)}{n!} (x-a)^n \\ &=f(a)+\frac{f'(a)}{1!}(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+\cdots \end{align*}\]Taylor series of the function $f$ at $a$ (or about $a$ or centered at $a$).
For the special case $a=0$ the Taylor series becomes Maclaurin series.