# How do you find a power series representation for #x^2 arctan(x^3# and what is the radius of convergence?

The radius of convergence is

Start with the well known Maclaurin series:

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To find a power series representation for (x^2 \arctan(x^3)), we can use the formula for the power series expansion of (\arctan(x)):

[ \arctan(x) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{2n+1}}{2n+1} ]

First, replace (x) with (x^3) in the series formula:

[ \arctan(x^3) = \sum_{n=0}^{\infty} (-1)^n \frac{(x^3)^{2n+1}}{2n+1} ]

[ = \sum_{n=0}^{\infty} (-1)^n \frac{x^{6n+3}}{2n+1} ]

Then, multiply by (x^2):

[ x^2 \arctan(x^3) = x^2 \sum_{n=0}^{\infty} (-1)^n \frac{x^{6n+3}}{2n+1} ]

[ = \sum_{n=0}^{\infty} (-1)^n \frac{x^{6n+5}}{2n+1} ]

So, the power series representation for (x^2 \arctan(x^3)) is:

[ x^2 \arctan(x^3) = \sum_{n=0}^{\infty} (-1)^n \frac{x^{6n+5}}{2n+1} ]

The radius of convergence of this series can be found using the ratio test. Let (a_n = (-1)^n \frac{1}{2n+1} x^{6n+5}). Applying the ratio test:

[ \lim_{n \to \infty} \left| \frac{a_{n+1}}{a_n} \right| = \lim_{n \to \infty} \left| \frac{(-1)^{n+1} \frac{1}{2(n+1)+1} x^{6(n+1)+5}}{(-1)^n \frac{1}{2n+1} x^{6n+5}} \right| ]

[ = \lim_{n \to \infty} \left| \frac{-1}{2n+3} x^{6n+11} \cdot \frac{2n+1}{1} x^{-6n-5} \right| ]

[ = \lim_{n \to \infty} \left| -\frac{x^6}{2n+3} \right| ]

Since (\lim_{n \to \infty} \frac{x^6}{2n+3} = 0) for all (x), the radius of convergence is infinite, which means the series converges for all real numbers (x).

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When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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