Calculus – Sequences and series – Power series representations

We start with the power series \(\displaystyle\sum_{n=0}^{\infty}x^n=1+x+x^2+\cdots\), which is a geometric series with common ratio \(x\). Hence: the series converges for \(|x| < 1\) and diverges for \(|x|\geq1\). Moreover, for \(|x| < 1\) we have that \(\displaystyle\sum_{n=0}^{\infty}x^n=\frac{1}{1-x}\).

If we replace, for instance, \(x\) by \(-x^2\) we obtain that

\[\frac{1}{1+x^2}=\sum_{n=0}^{\infty}(-x^2)^n=\sum_{n=0}^{\infty}(-1)^nx^{2n}\quad\text{for}\quad|-x^2| < 1\quad\Longleftrightarrow\quad|x| < 1.\]

Theorem: If the power series \(\displaystyle\sum_{n=0}^{\infty}c_n(x-a)^n\) has radius of convergence \(R>0\), then the function \(f\) defined by

\[f(x)=c_0+c_1(x-a)+c_2(x-a)^2+\cdots=\sum_{n=0}^{\infty}c_n(x-a)^n,\quad|x-a| < R\]

is differentiable (and therefore continuous) on the interval \((a-R,a+R)\) and

\[f'(x)=c_1+2c_2(x-a)+3c_3(x-a)^2+\cdots=\sum_{n=1}^{\infty}nc_n(x-a)^{n-1},\quad|x-a| < R\]

and

\[\int f(x)\,dx=C+c_0(x-a)+\frac{1}{2}c_1(x-a)^2+\frac{1}{3}c_2(x-a)^3+\cdots=C+\sum_{n=0}^{\infty}\frac{c_n}{n+1}(x-a)^{n+1},\quad|x-a| < R\]

with \(C\) an arbitrary constant.

Remark:The central point and the radius of convergence stay the same; the interval of convergence may be different.

Hence we have

\[\arctan(x)=\int\frac{dx}{1+x^2}=C+\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}x^{2n+1},\quad|x| < 1.\]

Since \(\arctan(0)=0\) we conclude that \(C=0\). This implies that \(\displaystyle\arctan(x)=\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+1}x^{2n+1}\) for \(|x| < 1\).

We also have:

\[\frac{2x}{1+x^2}=2\sum_{n=0}^{\infty}(-1)^nx^{2n+1}\quad\Longrightarrow\quad\ln(1+x^2)=C+2\sum_{n=0}^{\infty}\frac{(-1)^n}{2n+2}x^{2n+2},\quad|x| < 1.\]

Since \(\ln(1)=0\) we conclude that \(\ln(1+x^2)=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}x^{2n+2}\) for \(|x| < 1\). Hence we have: \(\ln(1+x)=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}x^{n+1}\) for \(|x| < 1\).

Similarly, we have:

\[\frac{1}{1+x}=\sum_{n=0}^{\infty}(-1)^nx^n\quad\Longrightarrow\quad\ln(1+x)=C+\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}x^{n+1}, \quad|x| < 1.\]

Since \(\ln(1)=0\) we conclude that \(\ln(1+x)=\displaystyle\sum_{n=0}^{\infty}\frac{(-1)^n}{n+1}x^{n+1}\) for \(|x| < 1\).

Another example is:

\[\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n\quad\Longrightarrow\quad\frac{1}{(1-x)^2}=\sum_{n=1}^{\infty}nx^{n-1} \quad\Longrightarrow\quad\frac{x}{(1-x)^2}=\sum_{n=1}^{\infty}nx^n,\quad|x| < 1.\]

For instance, this implies that \(\displaystyle\sum_{n=1}^{\infty}\frac{n}{2^n}=\sum_{n=1}^{\infty}n\left(\tfrac{1}{2}\right)^n =\frac{\frac{1}{2}}{(1-\frac{1}{2})^2}=2\). Differentiating once more we obtain:

\[\frac{1}{(1-x)^2}=\sum_{n=1}^{\infty}nx^{n-1}\quad\Longrightarrow\quad\frac{2}{(1-x)^3}=\sum_{n=2}^{\infty}n(n-1)x^{n-2} \quad\Longrightarrow\quad\frac{2x^2}{(1-x)^3}=\sum_{n=2}^{\infty}n(n-1)x^n,\quad|x| < 1.\]

For instance, this implies that \(\displaystyle\sum_{n=2}^{\infty}\frac{n(n-1)}{2^n}=\sum_{n=2}^{\infty}n(n-1)\left(\tfrac{1}{2}\right)^n =\frac{2\left(\frac{1}{2}\right)^2}{(1-\frac{1}{2})^3}=4\).

Application: Consider the integral \(\displaystyle\int_0^{0.5}\frac{x^2}{1+x^8}\,dx\). It is difficult to evaluate this integral exactly. However, we may use a power series representation of the integrand to find an approximation. Start with \(\displaystyle\frac{1}{1-x}=\sum_{n=0}^{\infty}x^n\) for \(|x| < 1\). This implies that

\[\frac{1}{1+x^8}=\sum_{n=0}^{\infty}(-1)^nx^{8n}\quad\Longrightarrow\quad\frac{x^2}{1+x^8}=\sum_{n=0}^{\infty}(-1)^nx^{8n+2} \quad\Longrightarrow\quad\int\frac{x^2}{1+x^8}\,dx=C+\sum_{n=0}^{\infty}\frac{(-1)^n}{8n+3}x^{8n+3},\quad |x| < 1.\]

This implies that

\[\int_0^{0.5}\frac{x^2}{1+x^8}\,dx=\sum_{n=0}^{\infty}\frac{(-1)^n(0.5)^{8n+3}}{8n+3}=\frac{(0.5)^3}{3}-\frac{(0.5)^{11}}{11} +\frac{(0.5)^{19}}{19}-\frac{(0.5)^{27}}{27}+\cdots\approx\frac{(0.5)^3}{3}=0.041666\ldots.\]

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Last modified on March 15, 2021