Properties of Power Series

Addition and subtraction

When two functions f and g are decomposed into power series around the same center c, the power series of the sum or difference of the functions can be obtained by termwise addition and subtraction. That is, if

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x) = \sum_{n=0}^\infty a_n (x - c)^n} and $g(x)=\sum _{n=0}^{\infty }b_{n}(x-c)^{n}$

then

$f(x)\pm g(x)=\sum _{n=0}^{\infty }(a_{n}\pm b_{n})(x-c)^{n}.$

It is not true that if two power series ${\textstyle \sum _{n=0}^{\infty }a_{n}x^{n}}$ and ${\textstyle \sum _{n=0}^{\infty }b_{n}x^{n}}$ have the same radius of convergence, then ${\textstyle \sum _{n=0}^{\infty }\left(a_{n}+b_{n}\right)x^{n}}$ also has this radius of convergence. If ${\textstyle a_{n}=(-1)^{n}}$ and ${\textstyle b_{n}=(-1)^{n+1}\left(1-{\frac {1}{3^{n}}}\right)}$, then both series have the same radius of convergence of 1, but the series ${\textstyle \sum _{n=0}^{\infty }\left(a_{n}+b_{n}\right)x^{n}=\sum _{n=0}^{\infty }{\frac {(-1)^{n}}{3^{n}}}x^{n}}$ has a radius of convergence of 3.

Multiplication and division

With the same definitions for $f(x)$ and $g(x)$, the power series of the product and quotient of the functions can be obtained as follows:

${\begin{aligned}f(x)g(x)&=\left(\sum _{n=0}^{\infty }a_{n}(x-c)^{n}\right)\left(\sum _{n=0}^{\infty }b_{n}(x-c)^{n}\right)\\&=\sum _{i=0}^{\infty }\sum _{j=0}^{\infty }a_{i}b_{j}(x-c)^{i+j}\\&=\sum _{n=0}^{\infty }\left(\sum _{i=0}^{n}a_{i}b_{n-i}\right)(x-c)^{n}.\end{aligned}}$

The sequence ${\textstyle m_{n}=\sum _{i=0}^{n}a_{i}b_{n-i}}$ is known as the convolution of the sequences $a_{n}$ and $b_{n}$.

For division, if one defines the sequence $d_{n}$ by

${\frac {f(x)}{g(x)}}={\frac {\sum _{n=0}^{\infty }a_{n}(x-c)^{n}}{\sum _{n=0}^{\infty }b_{n}(x-c)^{n}}}=\sum _{n=0}^{\infty }d_{n}(x-c)^{n}$

then

$f(x)=\left(\sum _{n=0}^{\infty }b_{n}(x-c)^{n}\right)\left(\sum _{n=0}^{\infty }d_{n}(x-c)^{n}\right)$

and one can solve recursively for the terms $d_{n}$ by comparing coefficients.

Solving the corresponding equations yields the formulae based on determinants of certain matrices of the coefficients of $f(x)$ and $g(x)$

$d_{0}={\frac {a_{0}}{b_{0}}}$

$d_{n}={\frac {1}{b_{0}^{n+1}}}{\begin{vmatrix}a_{n}&b_{1}&b_{2}&\cdots &b_{n}\\a_{n-1}&b_{0}&b_{1}&\cdots &b_{n-1}\\a_{n-2}&0&b_{0}&\cdots &b_{n-2}\\\vdots &\vdots &\vdots &\ddots &\vdots \\a_{0}&0&0&\cdots &b_{0}\end{vmatrix}}$

Differentiation and integration

Once a function Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle f(x)} is given as a power series as above, it is differentiable on the interior of the domain of convergence. It can be differentiated and integrated quite easily, by treating every term separately:

Failed to parse (MathML with SVG or PNG fallback (recommended for modern browsers and accessibility tools): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \begin{align} f'(x) &= \sum_{n=1}^\infty a_n n (x - c)^{n-1} = \sum_{n=0}^\infty a_{n+1} (n + 1) (x - c)^n, \\ \int f(x)\,dx &= \sum_{n=0}^\infty \frac{a_n (x - c)^{n+1}}{n + 1} + k = \sum_{n=1}^\infty \frac{a_{n-1} (x - c)^n}{n} + k. \end{align}}

Both of these series have the same radius of convergence as the original one.

Resources

Differentiation and Integration of Power Series

• Differentiation and Integration Using Power Series Video by James Sousa, Math is Power 4U
• Estimate a Definite Integral Using Power Series Video by James Sousa, Math is Power 4U
• Differentiating and Integrating Power Series Video by Patrick JMT
• Integrating a Power Series Video by Patrick JMT
• Integrating a Power Series, Example 2 Video by Patrick JMT
• Power Series Differentiation Video by Krista King
• Expressing the Integral as a Power Series Video by Krista King
• Using Power Series to Estimate a Definite Integral Video by Krista King
• Power Series - Differentiation and Integration Video by The Organic Chemistry Tutor

Power Series Representation of Functions Using Differentiation and Integration

• Find a Power Series to Represent arctan(x) Using Integration Video by Math is Power 4U
• Find a Power Series to Represent a Rational Function Using Differentiation Video by Math is Power 4U
• Finding a Power Series by Differentiation Video by Patrick JMT
• Finding Power Series by Differentiation - Three Examples Video by Patrick JMT
• Power Series Representation Using Integration Video by Patrick JMT
• Power Series Representation by Differentiation Video by The Organic Chemistry Tutor
• Power Series Representation by Integration Video by The Organic Chemistry Tutor

Licensing

Content obtained and/or adapted from:

• Power series, Wikipedia under a CC BY-SA license