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Hi everyone :) We learnt what a power series is in class, but that coefficient thing is new. How do we find coefficients of power series using that equation? What do we do?

Power Series

If someone can help me with any hints on (a), that would be great. I'll do my best for the rest. Thanks lots.

UPDATE:Thanks to @Macavity, I did get to this point:

$$\frac{\left(a_{k-2}+Axa_{k-2}+a_k+Aa_{k-1}\right)}{a_{k-2}}\sum _{k=0}^{infinity}a_kx^k$$

But what do I do next?

soupynoodles
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1 Answers1

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$$\eqalign{(1+Ax+Bx^2)\sum_{k=0}^\infty a_kx^k&=\sum_{k=0}^\infty\bigl(a_k x^k+Aa_k x^{k+1}+Ba_k x^{k+2}\bigr)\cr&=\sum_{k=0}a_k x^k+\sum_{k'=1}^\infty Aa_{k'-1}x^{k'}+\sum_{k'=2}^\infty Ba_{k'-2}x^{k'}\cr &=\ldots+\sum_{k=2}^\infty\bigl(a_k+Aa_{k-1}+Ba_{k-2}\bigr)x^k\ .\cr}$$ Now take care of the terms incorporated in the $\ \ldots\ $.

Christian Blatter
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  • Thank you, Professor Blatter. May I please ask - for the last two terms in the second step, did you take the derivative? And how to expand the whole thing to obtain the "..." part? – soupynoodles Apr 01 '16 at 18:11
  • No derivatives, just arranging, re-indexing, and re-collecting terms. Some terms of the first two sums in the second line do not appear in the sum beginning with $k=2$ in the third line. They should be taken care of separately where the $\ldots\ $ are. – Christian Blatter Apr 01 '16 at 18:18
  • That helped! Professor, could you have a look at my working below? I don't know what I need to do here. $$\sum {k=0}^{infinity}a_kx^k+\sum _{k=0}^{infinity}Aa_kx^{\left(k+1\right)}+\sum _{k=0}^{infinity}Ba_kx^{\left(k+2\right)}$$ $$Aa_0x+Ba_0x^2+Ba_1x^3+\sum _{k=0}^{infinity}a_kx^k+\sum _{k=1}^{infinity}Aa{k-1}x^{\left(k\right)}+\sum {k=2}^{infinity}Ba{k-2}x^{\left(k\right)}$$ Apologies for the repeated edits, I'm clearly bad with MathJax. – soupynoodles Apr 01 '16 at 18:34