Use the Geometric Series Test to help you find a power series representation of f(x)=x(2+x3) centered at 0. Find the interval and radius of convergence.

Given a function f(x)=x(2+x3) To Find: power of series Note: The geometric power series is a series of the form a1−r=∑n=0∞arn,|r|&lt;1 Where a is the first term and r is the common ratio. This implies f(x)=x2+x3 =x2⋅(11+x32) Now, the power series for f(x) is 11+x32=∑n=0∞an ∑n=0∞(−(x213)3)n ∑n=0∞(−1)nx3n2n Converging for |x32&lt;1|, i.e. −213&lt;x&lt;213 Now, x2+x3=x2(11+x32)=x2∑n=0∞(−1)nx3n2n=∑n=0∞(−1)nx3n+12n+1 With radius of radius of convergence R=213 Therefore, the power series of the function is x2+x3=∑n=0∞(−1)nx3n+12n+1

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Yulia

Answered question

2021-02-12

Use the Geometric Series Test to help you find a power series representation of

f (x) = \frac{x}{(2 + x^{3})}

centered at 0. Find the interval and radius of convergence.

Answer & Explanation

un4t5o4v

Skilled2021-02-13Added 105 answers

Given a function
$f (x) = \frac{x}{(2 + x^{3})}$
To Find: power of series
Note: The geometric power series is a series of the form
$\frac{a}{1 - r} = \sum_{n = 0}^{\infty} a r^{n}, | r | < 1$
Where a is the first term and r is the common ratio.
This implies $f (x) = \frac{x}{2 + x^{3}}$
$= \frac{x}{2} \cdot (\frac{1}{1 + \frac{x^{3}}{2}})$
Now, the power series for f(x) is
$\frac{1}{1 + \frac{x^{3}}{2}} = \sum_{n = 0}^{\infty} a_{n}$
$\sum_{n = 0}^{\infty} (- (\frac{x}{2^{\frac{1}{3}}})^{3})^{n}$
$\sum_{n = 0}^{\infty} (- 1)^{n} \frac{x^{3 n}}{2^{n}}$
Converging for
$| \frac{x^{3}}{2} < 1 |,$ i.e. $- 2^{\frac{1}{3}} < x < 2^{\frac{1}{3}}$
Now,
$\frac{x}{2 + x^{3}} = \frac{x}{2} (\frac{1}{1 + \frac{x^{3}}{2}}) = \frac{x}{2} \sum_{n = 0}^{\infty} (- 1)^{n} \frac{x^{3 n}}{2^{n}} = \sum_{n = 0}^{\infty} (- 1)^{n} \frac{x^{3 n + 1}}{2^{n + 1}}$
With radius of radius of convergence $R = 2^{\frac{1}{3}}$
Therefore, the power series of the function is
$\frac{x}{2 + x^{3}} = \sum_{n = 0}^{\infty} (- 1)^{n} \frac{x^{3 n + 1}}{2^{n + 1}}$

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