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What is the coefficient of $x^{11}$ in the power series expansion of $\frac 1{1-x-x^4}$?

How do I do power series expansions?

user228320
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  • Assuming your power series is centered at $0$, As a hint: try expanding it out as a Taylor series expansion, the coefficients have a very nice formula. – DaveNine Jun 04 '15 at 23:50
  • Look at the series expansion of $\frac{1}{1-t}=1+t+t^2+\cdots$. and replace $t$ by $x+x^4=x(1+x^3)$. Then thin a bit about the ways to get an $x^{11}$. Not much work if one keeps one's eyes open. Awful by hand for $x^{1111}$. – André Nicolas Jun 04 '15 at 23:51
  • @user228320: more easy way here http://math.stackexchange.com/questions/1312734 is it gets answered :) – Alexey Burdin Jun 05 '15 at 00:30

2 Answers2

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suppose $$\frac{1}{1-x-x^4} = 1 + x + a_2x^2 + a_3x^3 + \cdots. $$ then we have $$1 =(1-x-x^4)(1 + x + a_2x^2 + a_3x^3 +\cdots)=1+(a_2-1)x^2 + (a_3-a_2)x^3+\\(a_4-a_3-1)x^4+\cdots+(a_n-a_{n-1} -a_{n-4})x^n+\cdots $$ equating the coefficients of $x^2, x^3, \cdots$, we find that $$ a_0 = 1, a_1 = 1, a_2 = 1, a_3 =1, a_4=2 \text{ and } a_n=a_{n-1} +a_{n-4} $$
$\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|} \hline n&0&1&2&3&4&5&6&7&8&9&10&11&12 \\ \hline a_n&1&1&1&1&2&3&4&5&7&10&14&19&26 \\ \hline \end{array}$

$$ \text{ i found }19 \text{ to be the coefficient of } x^{11}.$$

abel
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Hint: $$\frac{1}{1-x(1+x^3)}=1+x(1+x^3)+x^2(1+x^3)^2+x^3(1+x^3)^3+.....x^{11}(1+x^3)^{11}+.........$$ Cancel the terms after the twelfth term because the power of $x$ becomes more than $11$

E.H.E
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