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Question

Find a power series representation for the function and determine the interval of convergence. \

f(x)=11+9x2f(x)=\dfrac{1}{1+9 x^2}

Solution

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Answered 2 years ago
Answered 2 years ago
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Let's rewrite the function as a more familiar expression:

11+9x2=11(9x2)=n=0(9x2)n=n=0(9)nx2n\begin{align*} \frac{1}{1+9x^2} = \frac{1}{1-(-9x^2)} = \sum_{n=0}^\infty \left(-9x^2\right)^n = \sum_{n=0}^\infty (-9)^n \, x^{2n} \end{align*}

The series converges for:

9x2lt;1    9x2lt;1    x2lt;19    xlt;13\begin{align*} \left|-9x^2\right| < 1 \iff 9|x|^2 < 1 \iff |x|^2 < \frac{1}{9} \iff |x| < \frac{1}{3} \end{align*}

Therefore, the interval of convergence is (13,13).\left(-\frac{1}{3},\frac{1}{3}\right).

Note:\textit{Note:} We do not need to check the endpoints because we know the series diverges for x13|x| \geq \frac{1}{3}.

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