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I just started learning about formal power series and I'm trying to find the coefficients in the expansion of a formal power series, specifically $$[x^6](1-3x^2)^{-9}$$

Using the binomial theorem and power series form, I get that the coefficient should be $$\binom{3+8}{8} * 3^3 $$ but my friend insists it's $$\binom{6+8}{8} * 3^3 $$ If he's right, why is that so ? I would think that we would only need n to be 3 to get the $x^6$ from $x^{2n}$ in the power series.

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  • $\begingroup$ The binomial theorem gives neither expression. Are you expanding the reciprocal? $\endgroup$ – André Nicolas Jan 21 '16 at 21:12
  • $\begingroup$ I'm talking about the power series form of (1-x)^-k, $\endgroup$ – P-L Jan 21 '16 at 21:24
  • $\begingroup$ Then some editing is needed. $\endgroup$ – André Nicolas Jan 21 '16 at 21:32
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I think the coefficient you get is the correct one. The inverse of $(1-3x^2)$ in $\mathbb{C}[[x]]$ is $$ 1+3x^2+3^2x^4+3^3x^6+3^4x^8+... $$ So, $(1-3x^2)^{-9}$ is equal to $$ \left(1+3x^2(1+3x^2+3^2x^4+3^3x^6+...)\right)^9. $$ Using the Binomial Theorem we get that the coefficient of $x^6$ in this expansion is $$ 9.3^3+\frac{9.8}{2}.3^2.6+\frac{9.8.7}{6}.3^3=4455. $$

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I assume you mean that you want to find the coefficient of $x^6$ in $(1-3x^2)^{-9}$. Using the binomial theorem we have:$$(1-x)^{-n}=\sum_{k=0}^{\infty}\binom{n}{k}(-x)^k$$Note that this is only valid for $-1\lt x\lt1$. Hence we have:$$(1-3x^2)^9=\sum_{k=0}^{\infty}\binom{9}{k}(-3x^2)^k$$This is only valid for $-\frac{1}{3}\lt x^2\lt\frac{1}{3}$.

Now $x^6=(x^2)^3$ hence the coefficient you want is obtained when $k=3$. Hopefully you can complete from here.

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  • $\begingroup$ Sorry I made a mistake, I the exponent was -9 not 9, I'll fix it $\endgroup$ – P-L Jan 21 '16 at 21:43

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