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Potential series method

by matematikuvol
Tags: method, potential, series
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matematikuvol
#1
Dec19-12, 12:44 PM
P: 192
Why sometimes we search solution of power series in the way:
[tex]y(x)=\sum^{\infty}_{n=0}a_nx^n[/tex]
and sometimes
[tex]y(x)=\sum^{\infty}_{n=0}a_nx^{n+1}[/tex]???
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tiny-tim
#2
Dec20-12, 05:44 PM
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hi matematikuvol!
Quote Quote by matematikuvol View Post
Why sometimes we search solution of power series in the way:
[tex]y(x)=\sum^{\infty}_{n=0}a_nx^n[/tex]
and sometimes
[tex]y(x)=\sum^{\infty}_{n=0}a_nx^{n+1}[/tex]???
no particular reason

sometimes one gives neater equations than the other

they'll both work (provided, of course, that y(0) = 0)
matematikuvol
#3
Dec21-12, 01:22 AM
P: 192
I think that in the case when
[tex]\alpha(x)y''(x)+\beta(x)y'(x)+\gamma(x)y(x)=0[/tex]
if ##\alpha(0)=0## you must work with ##\sum^{\infty}_{n=0}a_nx^{n+k}##, but I'm not sure.

tiny-tim
#4
Dec21-12, 03:11 AM
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Potential series method

Quote Quote by matematikuvol View Post
I think that in the case when
[tex]\alpha(x)y''(x)+\beta(x)y'(x)+\gamma(x)y(x)=0[/tex]
if ##\alpha(0)=0## you must work with ##\sum^{\infty}_{n=0}a_nx^{n+k}##, but I'm not sure.
but that's the same as ##\sum^{\infty}_{n=0}b_nx^n## with ##b_n = a_{n-k}## for n ≥ k, and ##b_n = 0## otherwise


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