Question

If the power series {summation} a_n xⁿ converges uniformly in ] α ,β [ then so does {summation} a_n (-x)ⁿ . true or false ? Justify

Answer 1

ANSWER. The statement is true in the following cases :

1) the radius of convergence of the series $\sum_{n=0}^{\infty}a_{n}x^{n}$ is $\infty$

2) the convergence set of the series is the segment $[-R,R]$

3) Points $\alpha, \beta$ are not end points of the set $]-R,R[$ (the convergence set of the series is the $]-R,R]$ or $[-R,R[, or ]-R,R[$ )

If at least point one of the points $\alpha, \beta$ is the end point of the set indicated in 3) , then the statement is not true.

EXPLANATION

Note , that $|a_{n}x^{n}|=|a_{n}(-x)^{n}|$ . Therefore, if one of the series

$\sum_{n=0}^{\infty}a_{n}x^{n}$ (1)

$\sum_{n=0}^{\infty}a_{n}(-x)^{n}$ (2)

in a point $c$ absolutely converges , then the other converges absolutely. Since the power series in the interval of converges $]-R,R[$ converge absolutely , then the series (1) and (2) have the same radius of converges. Since the power series converges uniformly on any segment $[\alpha, \beta]\subset ]-R,R[$ , then both series (1), (2) converge uniformly on $]\alpha,\beta[\subset[\alpha, \beta]$ . This proves 1),2),3).

Let series (1) diverge at one of the point $-R,R$ (or both)

Suppose , that $\alpha=-R$ and both series (1), (2) converge uniformly on $]-R,\beta[$ . There are limits at the point $(-R)$ for all $n\in \N$

$\lim_{x\rightarrow(-R)^{-}} {a_{n}}x^{n}=a_{n}(-R)^{n}\, \, , \, \lim_{x\rightarrow(-R)^{-}} {a_{n}}(-x) ^{n}=a_{n}( R)^{n}\, .$ By the theorem on the

limit of the sum of the power series we have the equalities

  $\lim_{x\rightarrow (-R)^{-}}\sum_{n=0}^{\infty} a_{n}x^{n}=\sum_{n=0}^{\infty} a_{n}(-R )^{n} \\\lim_{x\rightarrow (-R)^{-}}\sum_{n=0}^{\infty} a_{n}(-x) ^{n}=\sum_{n=0}^{\infty} a_{n}R^{n}$

and the series on the right converge.Therefore, the series (1) converges on the $[-R,R]$ in contrast to the assumption.