Find the Maclaurin series of $f(x)=\dfrac{1}{1+x+x^2}$ and the radius of convergence of the series.
I can't solve this problem.
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1what have you done so far? – WiseStrawberry Jun 10 '13 at 18:54
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1Start by doing a partial fractions decomposition – mrf Jun 10 '13 at 18:54
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I'm studying Calculus and I just learned about power series and Taylor (+ Maclaurin series). I want to know how to approach solution. – Gagomse Jun 10 '13 at 18:57
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1@mrf: Partial fractions is a very useful general method that OP should become acquainted with, but in this case there is a much easier way. – André Nicolas Jun 10 '13 at 19:00
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To find the radius of convergence, you only need to look for the zeros of $1 + x + x^2$ – Cocopuffs Jun 10 '13 at 19:02
2 Answers
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Hint: $$\frac{1}{1+x+x^2}=\frac{1-x}{1-x^3},\quad \frac{1}{1-y}=\sum_{n=0}^\infty y^n.$$
Hu Zhengtang
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@Gagomse: $\frac{1-x}{1-x^3}=(1-x)\cdot\sum_{n=0}^\infty x^{3n}$. Hope you can complete the solution from here. – Hu Zhengtang Jun 10 '13 at 19:15
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@sidht: To obtain the Maclaurin series of $f$ from your series, you still have to do the binomial expansion of $(x+x^2)^n$. – Hu Zhengtang Jun 11 '13 at 06:02
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@sidht: It is not a Maclaurin series. A Maclaurin series, say of $f$, must be of the form $\sum_{n=0}^\infty a_nx^n$, where $a_n=\frac{f^{(n)}(0)}{n!}$. – Hu Zhengtang Jun 12 '13 at 03:22
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A grneral method:
Maclaurin series: $f(x)=f(0)+f^{'}(0)x+\ldots+\frac{f^{(n)}(0)}{n!}x^n+\ldots $
Denote $a_n=\frac{f^{(n)}(0)}{n!}$, using Leibniz's rule, $$ (1+x+x^2)f(x)=1 \Rightarrow (1+x+x^2)f^{(n)}+{n\choose 1}(1+2x)f^{(n-1)}+{n\choose 2}\cdot2\cdot f^{(n-2)}=0 $$
Let $x=0$, we obtain $$ f^{(n)}(0)+nf^{(n-1)}(0)+n(n-1)f^{(n-2)}(0)=0 $$ which is equivalent to $$ \frac{f^{(n)}(0)}{n!}+\frac{f^{(n-1)}(0)}{(n-1)!}+\frac{f^{(n-2)}(0)}{(n-2)!}=0 $$ Hence $a_n=-(a_{n-1}+a_{n-2})$, and $a_0=1,a_1=-1$. Now you know the Maclaurin series.
Coiacy
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