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# Power series method

In mathematics, the power series method is used to seek a power series solution to certain differential equations.

Method

Consider the second-order linear differential equation: $a_2\left(z\right)f"\left(z\right)+a_1\left(z\right)f\text{'}\left(z\right)+a_0\left(z\right)f\left(z\right)=0;!$Suppose "a"2 is nonzero for all "z". Then we can divide throughout to obtain: $f"+\left\{a_1\left(z\right)over a_2\left(z\right)\right\}f\text{'}+\left\{a_0\left(z\right)over a_2\left(z\right)\right\}f=0$Suppose further that "a"1/"a"2 and "a"0/"a"2 are analytic functions.

The power series method calls for the construction of a power series solution:$f=sum_\left\{k=0\right\}^infty A_kz^k$

If "a"2 is zero for some "z", then the Frobenius method, a variation on this method, is suited to deal with so called "singular points".

Example usage

Let us look at the Hermite differential equation,: $f"-2zf\text{'}+lambda f=0;;lambda=1$

We can try and construct a series solution: $f=sum_\left\{k=0\right\}^infty A_kz^k$: $f\text{'}=sum_\left\{k=0\right\}^infty kA_kz^\left\{k-1\right\}$: $f"=sum_\left\{k=0\right\}^infty k\left(k-1\right)A_kz^\left\{k-2\right\}$

Substituting these in the differential equation: $sum_\left\{k=0\right\}^infty k\left(k-1\right)A_kz^\left\{k-2\right\}-2zsum_\left\{k=0\right\}^infty kA_kz^\left\{k-1\right\}+sum_\left\{k=0\right\}^infty A_kz^k=0$: $=sum_\left\{k=0\right\}^infty k\left(k-1\right)A_kz^\left\{k-2\right\}-sum_\left\{k=0\right\}^infty 2kA_kz^k+sum_\left\{k=0\right\}^infty A_kz^k$Making a shift on the first sum: $=sum_\left\{k+2=0\right\}^infty \left(k+2\right)\left(\left(k+2\right)-1\right)A_\left\{k+2\right\}z^\left\{\left(k+2\right)-2\right\}-sum_\left\{k=0\right\}^infty 2kA_kz^k+sum_\left\{k=0\right\}^infty A_kz^k$: $=sum_\left\{k=-2\right\}^infty \left(k+2\right)\left(k+1\right)A_\left\{k+2\right\}z^k-sum_\left\{k=0\right\}^infty 2kA_kz^k+sum_\left\{k=0\right\}^infty A_kz^k$: $=\left(0\right)\left(-1\right)A_\left\{0\right\}z^\left\{-2\right\} + \left(-1\right)\left(0\right)A_\left\{1\right\}z^\left\{-1\right\}+sum_\left\{k=0\right\}^infty \left(k+2\right)\left(k+1\right)A_\left\{k+2\right\}z^k-sum_\left\{k=0\right\}^infty 2kA_kz^k+sum_\left\{k=0\right\}^infty A_kz^k$: $=sum_\left\{k=0\right\}^infty \left(k+2\right)\left(k+1\right)A_\left\{k+2\right\}z^k-sum_\left\{k=0\right\}^infty 2kA_kz^k+sum_\left\{k=0\right\}^infty A_kz^k$: $=sum_\left\{k=0\right\}^infty left\left(\left(k+2\right)\left(k+1\right)A_\left\{k+2\right\}+\left(-2k+1\right)A_k ight\right)z^k$Now, if this series is a solution, all these coefficients must be zero, so:: $\left(k+2\right)\left(k+1\right)A_\left\{k+2\right\}+\left(-2k+1\right)A_k=0;!$We can rearrange this to get a recurrence relation for "A""k"+2.: $\left(k+2\right)\left(k+1\right)A_\left\{k+2\right\}=-\left(-2k+1\right)A_k;!$: $A_\left\{k+2\right\}=\left\{\left(2k-1\right)over \left(k+2\right)\left(k+1\right)\right\}A_k;!$

Now, we have: $A_2 = \left\{-1 over \left(2\right)\left(1\right)\right\}A_0=\left\{-1over 2\right\}A_0,, A_3 = \left\{1 over \left(3\right)\left(2\right)\right\} A_1=\left\{1over 6\right\}A_1$We can determine "A"0 and "A"1 if there are initial conditions, ie., if we have an initial value problem.

So, we have: $A_4=\left\{1over 4\right\}A_2 = left\left(\left\{1over 4\right\} ight\right)left\left(\left\{-1 over 2\right\} ight\right)A_0 = \left\{-1 over 8\right\}A_0$

: $A_5=\left\{1over 4\right\}A_3 = left\left(\left\{1over 4\right\} ight\right)left\left(\left\{1 over 6\right\} ight\right)A_1 = \left\{1 over 24\right\}A_1$

: $A_6=\left\{7over 30\right\}A_4 = left\left(\left\{7over 30\right\} ight\right)left\left(\left\{-1 over 8\right\} ight\right)A_0 = \left\{-7 over 240\right\}A_0$

: $A_7=\left\{3over 14\right\}A_5 = left\left(\left\{3over 14\right\} ight\right)left\left(\left\{1 over 24\right\} ight\right)A_1 = \left\{1 over 112\right\}A_1$

and the series solution is

: $f=A_0x^0+A_1x^1+A_2x^2+A_3x^3+A_4x^4+A_5x^5+A_6x^6+A_7x^7+cdots$

: $=A_0x^0+A_1x^1+\left\{-1over 2\right\}A_0x^2+\left\{1over 6\right\}A_1x^3+\left\{-1 over 8\right\}A_0x^4+\left\{1 over 24\right\}A_1x^5+\left\{-7 over 240\right\}A_0x^6+\left\{1 over 112\right\}A_1x^7+cdots$

: $=A_0x^0+\left\{-1over 2\right\}A_0x^2+\left\{-1 over 8\right\}A_0x^4+\left\{-7 over 240\right\}A_0x^6+A_1x+\left\{1over 6\right\}A_1x^3+\left\{1 over 24\right\}A_1x^5+\left\{1 over 112\right\}A_1x^7+cdots$

which we can break up into the sum of two linearly independent series solutions:

: $f=A_0\left(1+\left\{-1over 2\right\}x^2+\left\{-1 over 8\right\}x^4+\left\{-7 over 240\right\}x^6+cdots\right)+A_1\left(x+\left\{1over 6\right\}x^3+\left\{1 over 24\right\}x^5+\left\{1 over 112\right\}x^7+cdots\right)$

which can be further simplified by the use of hypergeometric series (which goes beyond the scope of this article).

*
* [http://math.fullerton.edu/mathews/n2003/FrobeniusSeriesMod.html Module for Frobenius Series Solution]
* [http://www.ntu.edu.sg/home/mwtang/odesite.htm A Concise Introductory Course in Ordinary Differential Equations (with a chapter on series solutions)]

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