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: $a\_2(z)f"(z)+a\_1(z)f\text{'}(z)+a\_0(z)f(z)=0;!$Suppose "a"₂ is nonzero for all "z". Then we can divide throughout to obtain: $f"+\{a\_1(z)over\; a\_2(z)\}f\text{'}+\{a\_0(z)over\; a\_2(z)\}f=0$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:$f=sum\_\{k=0\}^infty\; A\_kz^k$

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: $f=sum\_\{k=0\}^infty\; A\_kz^k$: $f\text{'}=sum\_\{k=0\}^infty\; kA\_kz^\{k-1\}$: $f"=sum\_\{k=0\}^infty\; k(k-1)A\_kz^\{k-2\}$

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

Now, we have: $A\_2\; =\; \{-1\; over\; (2)(1)\}A\_0=\{-1over\; 2\}A\_0,,\; A\_3\; =\; \{1\; over\; (3)(2)\}\; A\_1=\{1over\; 6\}A\_1$We can determine "A"₀ and "A"₁ if there are initial conditions, ie., if we have an initial value problem.

So, we have: $A\_4=\{1over\; 4\}A\_2\; =\; left(\{1over\; 4\}\; ight)left(\{-1\; over\; 2\}\; ight)A\_0\; =\; \{-1\; over\; 8\}A\_0$

: $A\_5=\{1over\; 4\}A\_3\; =\; left(\{1over\; 4\}\; ight)left(\{1\; over\; 6\}\; ight)A\_1\; =\; \{1\; over\; 24\}A\_1$

: $A\_6=\{7over\; 30\}A\_4\; =\; left(\{7over\; 30\}\; ight)left(\{-1\; over\; 8\}\; ight)A\_0\; =\; \{-7\; over\; 240\}A\_0$

: $A\_7=\{3over\; 14\}A\_5\; =\; left(\{3over\; 14\}\; ight)left(\{1\; over\; 24\}\; ight)A\_1\; =\; \{1\; over\; 112\}A\_1$

and the series solution is

: $f=A\_0x^0+A\_1x^1+A\_2x^2+A\_3x^3+A\_4x^4+A\_5x^5+A\_6x^6+A\_7x^7+cdots$

: $=A\_0x^0+A\_1x^1+\{-1over\; 2\}A\_0x^2+\{1over\; 6\}A\_1x^3+\{-1\; over\; 8\}A\_0x^4+\{1\; over\; 24\}A\_1x^5+\{-7\; over\; 240\}A\_0x^6+\{1\; over\; 112\}A\_1x^7+cdots$

: $=A\_0x^0+\{-1over\; 2\}A\_0x^2+\{-1\; over\; 8\}A\_0x^4+\{-7\; over\; 240\}A\_0x^6+A\_1x+\{1over\; 6\}A\_1x^3+\{1\; over\; 24\}A\_1x^5+\{1\; over\; 112\}A\_1x^7+cdots$

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

: $f=A\_0(1+\{-1over\; 2\}x^2+\{-1\; over\; 8\}x^4+\{-7\; over\; 240\}x^6+cdots)+A\_1(x+\{1over\; 6\}x^3+\{1\; over\; 24\}x^5+\{1\; over\; 112\}x^7+cdots)$

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

External links

*
* [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)]