I am trying to do an example in the textbook for series solutions to second order linear equations, and I am not quite understanding the recurrence relation as follows:

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I understand and follow the steps but where I am a bit confused is as to how equation (23) was arrived at, I know that it relates to the recurrence relation but I cant seem to piece together the series.

EDIT: I tried writing out the series as follows: $$\sum_{n=0}^\infty a_{n+2}(n+2)(n+1)(x-1)^n = a_2(2)(1)(x-1)^0+a_3(3)(2)(x-1)^1+a_4(4)(3)(x-1)^2 \ldots$$ $$\sum_{n=0}^{\infty}a_n(x-1)^n = a_0(x-1)^0 + a_1(x-1)^1 + a_2(x-1)^2 + a_3(x-1)^3 + a_4(x-1)^4$$ $$\sum_{n=1}^{\infty}a_{n-1}(x-1)^n = a_0(x-1)^1 + a_1(x-1)^2 + a_2(x-1)^3 + a_3(x-1)^4$$

So I can see that the recurrence relation holds but where did the $\frac{(x-1)^3}{6}$ come from ? If it is matched to the first series written above, then shouldnt it have 20 on the denominator ? However, it appears that it is matched according to the coefficients of the $$\sum_{n=0}^{\infty}a_n(x-1)^n$$ in which case it would lead me to ask why do we only look at the coefficients of the first series and not the second..

Could someone help me please ? Advice will be greatly appreciated.

  • $\begingroup$ Equation (23) simply writes out the terms in the series solution for $a_{0}$, $a_{1}$, $\ldots$, $a_{5}$, which were worked out in the previous line. The author hasn't given a general formula for $a_{n}$, although you might easily figure out the pattern. $\endgroup$ – Brian Borchers Jun 28 '16 at 21:49
  • $\begingroup$ I made an edit to give more details on where I am getting confused at $\endgroup$ – PutsandCalls Jun 28 '16 at 22:02
  • $\begingroup$ $a_{3}=a_{0}/6+a_{1}/6$. The $(x-1)^{3}$ term has coefficients of $a_{0}/6+a_{1}/6$. $\endgroup$ – Brian Borchers Jun 28 '16 at 22:20
  • $\begingroup$ Yes, and the $a_3(x-1)^3$ corresponds to the series $\sum_{n=0}^\infty a_n(x-1)^n$? $\endgroup$ – PutsandCalls Jun 28 '16 at 22:24
  • $\begingroup$ Yes. Keep in mind that (23) is written in terms of $a_{0}$ and $a_{1}$, not $a_{n}$, $a_{n-1}$, and $a_{n-2}$, as in (22). $\endgroup$ – Brian Borchers Jun 28 '16 at 22:47

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