$$f(x)= \frac{x}{1+5x^2}$$ I got the power series: $$\sum_{n=0}^\infty (-1)^n (5^n)(x^{2n+1})$$ Assuming this is correct I would think the domain would be $$(-5^{1/3}, 5^{1/3})$$ because the absolute value for convergence would be $$|5x^3|<1$$However, my classmate thinks the answer is 1, because the series is derived from the basic geometric series. Any ideas?
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Power series...about what number? – DonAntonio Apr 21 '16 at 16:32
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The MacLaurin series about 0 – user333535 Apr 21 '16 at 16:33
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I assume you meant $\left(-5^{-\frac 13},5^{-\frac13}\right)$... – abiessu Apr 21 '16 at 16:33
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$f$ is defined over $\mathbb{R}$, and a MacLaurin series converges over any compact (closed interval) of $\mathbb{R}$. What is exactly the domain issue? – Vincent Apr 21 '16 at 16:36
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Yes, is that the correct answer? – user333535 Apr 21 '16 at 16:36
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I'm not sure, but that is just the correction of your interval statement. – abiessu Apr 21 '16 at 16:37
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@Vincent I had to find the domain for the given power series. I assumed that the domain was equivalent to the interval of convergence. – user333535 Apr 21 '16 at 16:39
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I think you almost nailed it. Using the $\;n\,-$ th root test (for the Cauchy- Hadamard formula):
$$\sqrt[n]{\left|(-1)^n5^nx^{2n+1}\right|}=5|x|^2\sqrt[n]{|x|}\xrightarrow[n\to\infty]{}5|x|^2<1\implies |x|<5^{-1/2}$$
Don't forget now to check the interval's end points.
DonAntonio
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Your power series is correct.
The binomial expansion gives $$\frac{1}{1+5x^2} \equiv (1+5x^2)^{-1} = 1- 5x^2 + (5x^2)^2 - (5x^2)^3 + \cdots$$ and so we need $|5x^2| < 1$ for the infinite series to converge.
Multiplying by $x$ does not change the domain of convergence. When we substitute a value for $x$, $x$ will just be a finite number. It's only infinite series we need to worry about.
$$\frac{x}{1+5x^2} \equiv x(1+5x^2)^{-1} = x(1- 5x^2 + (5x^2)^2 - (5x^2)^3 + \cdots)$$ will converge for all $|5x^2| < 1$, i.e. $|x| < \frac{1}{5}\sqrt 5$.
Fly by Night
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