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Potential series method

 
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Dec19-12, 12:44 PM   #1
 

Potential series method


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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Dec20-12, 05:44 PM   #2
 
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hi matematikuvol!
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)
 
Dec21-12, 01:22 AM   #3
 
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.
 
Dec21-12, 03:11 AM   #4
 
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Potential series method


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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