Difference between revisions of "Properties of Power Series"

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<strong>Differentiation and Integration of Power Series</strong>
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== Addition and subtraction ==
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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
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: <math>f(x) = \sum_{n=0}^\infty a_n (x - c)^n</math> and <math>g(x) = \sum_{n=0}^\infty b_n (x - c)^n</math>
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then
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: <math>f(x) \pm g(x) = \sum_{n=0}^\infty (a_n \pm b_n) (x - c)^n.</math>
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It is not true that if two power series <math display="inline">\sum_{n=0}^\infty a_n x^n</math> and <math display="inline">\sum_{n=0}^\infty b_n x^n</math> have the same radius of convergence, then <math display="inline">\sum_{n=0}^\infty \left(a_n + b_n\right) x^n</math> also has this radius of convergence. If <math display="inline">a_n = (-1)^n</math> and <math display="inline">b_n = (-1)^{n+1} \left(1 - \frac{1}{3^n}\right)</math>, then both series have the same radius of convergence of 1, but the series <math display="inline">\sum_{n=0}^\infty \left(a_n + b_n\right) x^n = \sum_{n=0}^\infty \frac{(-1)^n}{3^n} x^n</math> has a radius of convergence of 3.
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== Multiplication and division ==
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With the same definitions for <math>f(x)</math> and <math>g(x)</math>, the power series of the product and quotient of the functions can be obtained as follows:
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: <math>\begin{align}
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  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) \\
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          &= \sum_{i=0}^\infty \sum_{j=0}^\infty  a_i b_j (x - c)^{i+j} \\
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          &= \sum_{n=0}^\infty \left(\sum_{i=0}^n a_i b_{n-i}\right) (x - c)^n.
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\end{align}</math>
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The sequence <math display="inline">m_n = \sum_{i=0}^n a_i b_{n-i}</math> is known as the convolution of the sequences <math>a_n</math> and <math>b_n</math>.
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For division, if one defines the sequence <math>d_n</math> by
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: <math>\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</math>
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then
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: <math>f(x) = \left(\sum_{n=0}^\infty b_n (x - c)^n\right)\left(\sum_{n=0}^\infty d_n (x - c)^n\right)</math>
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and one can solve recursively for the terms <math>d_n</math> by comparing coefficients.
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Solving the corresponding equations yields the formulae based on [[determinant]]s of certain matrices of the coefficients of <math>f(x)</math> and <math>g(x)</math>
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:<math>d_0=\frac{a_0}{b_0}</math>
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:<math>d_n=\frac{1}{b_0^{n+1}} \begin{vmatrix}
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a_n    &b_1  &b_2  &\cdots&b_n    \\
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a_{n-1}&b_0  &b_1  &\cdots&b_{n-1}\\
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a_{n-2}&0    &b_0  &\cdots&b_{n-2}\\
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\vdots &\vdots&\vdots&\ddots&\vdots \\
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a_0    &0    &0    &\cdots&b_0\end{vmatrix}</math>
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== Differentiation and integration==
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Once a function <math>f(x)</math> 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:
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: <math>\begin{align}
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    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, \\
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  \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.
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\end{align}</math>
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Both of these series have the same radius of convergence as the original one.
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==Resources==
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===Differentiation and Integration of Power Series===
  
 
* [https://www.youtube.com/watch?v=zCQuEN7TFNE, Differentiation and Integration Using Power Series] Video by James Sousa, Math is Power 4U
 
* [https://www.youtube.com/watch?v=zCQuEN7TFNE, Differentiation and Integration Using Power Series] Video by James Sousa, Math is Power 4U
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* [https://www.youtube.com/watch?v=nD6hai32ykQ, Power Series - Differentiation and Integration] Video by The Organic Chemistry Tutor
 
* [https://www.youtube.com/watch?v=nD6hai32ykQ, Power Series - Differentiation and Integration] Video by The Organic Chemistry Tutor
  
 
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===Power Series Representation of Functions Using Differentiation and Integration===
<strong>Power Series Representation of Functions Using Differentiation and Integration</strong>
 
 
* [https://www.youtube.com/watch?v=bvNPmQRY8Ts, Find a Power Series to Represent arctan(x) Using Integration] Video by Math is Power 4U
 
* [https://www.youtube.com/watch?v=bvNPmQRY8Ts, Find a Power Series to Represent arctan(x) Using Integration] Video by Math is Power 4U
 
* [https://www.youtube.com/watch?v=ujB-5zQWZAY, Find a Power Series to Represent a Rational Function Using Differentiation] Video by Math is Power 4U
 
* [https://www.youtube.com/watch?v=ujB-5zQWZAY, Find a Power Series to Represent a Rational Function Using Differentiation] Video by Math is Power 4U

Revision as of 16:06, 10 October 2021

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

and

then

It is not true that if two power series and have the same radius of convergence, then also has this radius of convergence. If and , then both series have the same radius of convergence of 1, but the series has a radius of convergence of 3.

Multiplication and division

With the same definitions for and , the power series of the product and quotient of the functions can be obtained as follows:

The sequence is known as the convolution of the sequences and .

For division, if one defines the sequence by

then

and one can solve recursively for the terms by comparing coefficients.

Solving the corresponding equations yields the formulae based on determinants of certain matrices of the coefficients of and

Differentiation and integration

Once a function 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:

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

Resources

Differentiation and Integration of Power Series

Power Series Representation of Functions Using Differentiation and Integration