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Find the Maclaurin series of $f(x)=\dfrac {1}{(1+x+x^2)} $and the radius of convergence of the series.

$f(x) = \dfrac {(1-x)}{(1-x^3)} = (1-x) \sum _{0} ^\infty x^{(3n)}$

and so?

How this can be the shape of $\sum_0 ^\infty \dfrac {f^n}{n!}x^n$ ?

And I don't know how to get a radius.

jimjim
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Gagomse
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2 Answers2

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Ok, taking it from where you leave it:

$$f(x)=(1-x)\sum_{n=0}^\infty x^{3n}=\sum_{n=0}^\infty x^{3n}(1-x)=$$

For the radius of convergence - quotient test:

$$\left|\frac{\;x^{3n+3}(1-x)}{x^{3n}(1-x)}\;\right|=|x|^3<1\implies |x|<1\ldots$$

DonAntonio
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  • Now I understood for the radius. But isn't 1-x a form of (f^n}/(n!) ? – Gagomse Jun 11 '13 at 06:41
  • I'm not sure I understand your question, @Gagomse: the coefficients of the series we obtained indeed have the form $$\frac{f^{(n)}(0)}{n!}$$ as you can check at least for the first few cases, but as you continue differentiating things get really ugly, so it's not easy to see that. – DonAntonio Jun 11 '13 at 06:49
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Calculate $$(1-x)(1+x^3+x^6+x^9+x^{12}+\cdots).$$ We get $$1-x+x^3-x^4+x^6-x^7+x^9-x^{10}+x^{12}-x^{13}+\cdots.$$ For the radius of convergence, note that the terms of the series do not approach $0$ if $|x|\ge 1$. So the series diverges if $|x|\ge 1$.

We will prove (absolute) convergence if $|x|\lt 1$.
Take the absolute values of the terms. This gives us the series $$1+|x|+|x|^3+|x|^4+|x|^6+|x|^7+|x|^9+\cdots.\tag{1}$$ By comparison with the geometric series $$1+|x|+|x|^2 +|x|^3+|x|^4+|x|^5+|x|^6+\cdots,$$ which converges if $|x|\lt 1$, we can see that (1) converges if $|x|\lt 1$. It follows that the radius of convergence of our series is $1$.

Remark: Let $f(x)=\frac{1}{1+x+x^2}$. The series we obtained does indeed have the shape $\sum_{n=0}^\infty \frac{f^{(n)}(0)}{n!}x^n$. However, computing all but the first few derivatives of $f$ is fairly painful, which is one reason that the series for $\frac{1}{1+x+x^2}$ was obtained in a somewhat indirect way.

André Nicolas
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