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Find the interval of convergence for f(x) = x^{0.6} as a power series.

Question:

Find the interval of convergence for

{eq}f(x) = x^{0.6} {/eq} as a power series.

Power Series; Interval of Convergence:

The center of expansion for the power series will affect the interval of convergence. To consider a general situation we'll pick a center {eq}x=c>0 {/eq} so that the function {eq}f(x) = x^{0.6} {/eq} is defined in a neighborhood of {eq}c. {/eq} We'll write power series {eq}f(x)=x^{0.6}=\sum_{n=1}^{\infty}a_{n}(x-c)^n {/eq} for some Taylor coefficients {eq}a_n {/eq}.

Answer and Explanation:

The problem statement does not specify the center for the power series, so we'll work on the general case of the center at {eq}x=c>0 {/eq} so that the function {eq}f {/eq} is defined on a neighborhood of {eq}c. {/eq}

We'll rewrite {eq}f(x) {/eq} in terms of a new variable {eq}u=x-c {/eq} as follows:

{eq}\begin{align*} f(x)&=x^{0.6}\\ &=(c+(x-c))^{0.6}\\ &=(c+u)^{0.6}\\ &=c^{0.6}\cdot\left (1+\frac{u}{c}\right )^{0.6}\\ &=c^{\frac{3}{5}}\cdot\left (1+\frac{u}{c}\right )^{\frac{3}{5}}. \end{align*} {/eq}

We'll use the binomial expansion:

{eq}\begin{align*} (1+x)^p&=\sum_{n=0}^{\infty}\binom{p}{n}x^n\\ &\text{recalling that }\binom{p}{0}=1\text{ for any }p\ne 0,\\ &=1+\sum_{n=1}^{\infty}\frac{p(p-1)(p-2)\cdots (p-n+1)}{n!}x^n\\ &=1+px+\frac{p(p-1)}{2!}x^2 +\frac{p(p-1)(p-2)}{3!}x^3+\\ &\qquad\cdots+\frac{p(p-1)(p-2)\cdots (p-n+1)}{n!}x^n+\cdots\\ &\text{which in general converges for }|x|<1. \end{align*} {/eq}

In the particular case of a positive exponent {eq}p=0.6=\frac{3}{5}>0 {/eq} the corresponding binomial series also converges at the endpoints, so it has interval of convergence {eq}[-1,1] {/eq}. Thus we have:

{eq}\begin{align*} (1+x)^{\frac{3}{5}}&=\sum_{n=0}^{\infty}\binom{\frac{3}{5}}{n}x^n,\\ &\text{or expanding the binomials,}\\ &=1+\frac{3}{5}x+\frac{\frac{3}{5}(\frac{3}{5}-1)}{2!}x^2 +\frac{\frac{3}{5}(\frac{3}{5}-1)(\frac{3}{5}-2)}{3!}x^3+\\ &\qquad\cdots+\frac{\frac{3}{5}(\frac{3}{5}-1)(\frac{3}{5}-2)\cdots \left (\frac{3}{5}-n+1\right )}{n!}x^n+\cdots.\\ &\text{In abreviated form we write3:}\\ &\boxed{{(1+x)^{\frac{3}{5}}}=\sum_{n=0}^{\infty}\binom{\frac{3}{5}}{n}x^n}, \text{ which converges for }|x|\le 1. \end{align*} {/eq}

We can use the binomial series above, substituting {eq}\frac{u}{c} {/eq} in place of {eq}x {/eq}, to write:

{eq}\begin{align*} f(x)&=x^{0.6}\\ &=c^{\frac{3}{5}}\cdot\left (1+\frac{u}{c}\right )^{\frac{3}{5}}\\ &=c^{\frac{3}{5}}\cdot\sum_{n=0}^{\infty}\binom{\frac{3}{5}}{n}\left(\frac{u}{c}\right)^n,\text{ for }\left|\frac{u}{c}\right|\le1 \\ &=\sum_{n=0}^{\infty}\frac{c^{\frac{3}{5}}\cdot\binom{\frac{3}{5}}{n}}{c^n}\,u^n,\text{ for }\left|u\right|<c,\\ &\text{we return to the original variable making }u=x-c,\\ &\text{and we get,}\\ &\boxed{f(x)=x^{0.6}=\sum_{n=0}^{\infty}\frac{c^{\frac{3}{5}}\cdot\binom{\frac{3}{5}}{n}}{c^n}\,(x-c)^n},\text{ for }\left|x-c\right|\le c. \end{align*} {/eq}

This power series converges for {eq}\left|x-c\right|\le c {/eq} but diverges for {eq}\left|x-c\right|>c, {/eq} thus we can write that the interval of convergence is {eq}\boxed{[0,2c].} {/eq}

Observation: The particular case of the power series with center at {eq}x=1 {/eq} {eq}(c=1) {/eq} we get:

{eq}f(x)=x^{0.6}=\sum_{n=0}^{\infty} \binom{\frac{3}{5}}{n}\,(x-1)^n,\text{ for }\left|x-1\right|\le 1, {/eq}

and the interval of convergence is {eq}\boxed{[0,2].} {/eq}


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