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A recurrence relation $$\frac{b_{j+2}}{b_{j}} = \frac{j-\xi}{(j+2)(j+1)}$$ defines the general term for the power series where $\xi$ is a constant. $$g(x)=\sum^{\infty}_{j=0} b_{j}x^{j}.$$ I want to show this series then terminates when $\xi=m\in\mathbb{Z}$. So evidently if $m$ is even then the even part of the power series terminates but the odd part doesn't, similarly if $m$ is odd then the odd part terminates, but there is no number for which both odd and even parts terminate. What am I doing wrong?

user2850514
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1 Answers1

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If $m$ is even, there is a solution with non-zero even terms that terminates.
We can let the odd terms all equal zero.
The general solution for $m=0$ is $b_0+A(b_1x+b_3x^3+b_5x^5+...)$. The one with $A=0$ is a polynomial.
If $m$ is odd, there is a solution, with zero even terms, and non-zero odd terms that terminate.
The general solution for $m=1$ is $b_1x+A(b_0+b_2x^2+b_4x^4+...)$. Again, to get a polynomial, let $A=0$.

Empy2
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  • So for $m=0$ then $b_2=b_4=b_6=...=0$ but $b_1=b_3=b_5=...\not=0$ so $g(x) = b_1x+b_3x^3+b_5x^5+...$ and for $m=1$ then $b_3=b_5=b_7=...=0$ but $b_0=b_2=b_4=b_6=...\not=0$ so $g(x)=b_0+b_2x^2+b_4x^4+b_6x^6+...$, so clearly neither of them terminate into a polynomial.. – user2850514 Nov 15 '14 at 15:27
  • If $m=0$ then $b_2=0b_0=0$. For a polynomial, we need to choose $b_1=0$, as that is the only way to make any of the other $b_{2k+1}=0$ as well. and still have a non-zero polynomial because $b_0\ne0$. When $m=1$, if we want a polynomial we need $b_0=0$ for the same reason. – Empy2 Nov 15 '14 at 15:37
  • Surely if we let $A=0$ then this is no longer equal to $g(x)$? – user2850514 Nov 15 '14 at 16:12
  • If $m$ is not an integer, then both odd and even terns continue forever, so the only one that terminates is $g(x)=0$. When $m$ is a positive integer, at least we get one nonzero polynomial. The series may terminate. – Empy2 Nov 15 '14 at 16:24
  • But you've just shown that if $m$ is a positive integer then $g(x)$ continues forever unless you truncate it which is therefore no longer equal to $g(x)$. – user2850514 Nov 15 '14 at 16:30
  • For any $m$, the set of solutions is two dimensional. Don't think of a single $g(x)$, but the full set of functions $Ag_1(x)+Bg_2(x)$, where $A$ and $B$ are any real numbers. It is possible that the power series terminates when $m\in\mathbb{Z_+}$, by setting either $A$ or $B$ to zero. – Empy2 Nov 15 '14 at 17:24
  • Suppose we split this series up into even and odd terms, ie. $g(x) = a_{0}(1+\frac{a_{2}}{a_{0}}x^{2}+\frac{a_{4}}{a_{0}}x^{4}+...)+a_{1}(x+\frac{a_{3}}{a_{1}}x^{3}+\frac{a_{5}}{a_{1}}x^{5}+...)$ then you say we can just let $a_{1}$ or $a_{0}$ equal to zero out of no where? There doesn't seem to be any logic behind that and say if $g$ was a solution to a differential equation, setting part of it to zero would therefore mean it is no longer a solution. – user2850514 Nov 15 '14 at 22:01