1
$\begingroup$

Consider the sequence of functions $$f_n:[0,1]\longrightarrow \mathbb{R},\qquad f_n:=(-1)^n(1-x)x^n,\qquad n\geq 0.$$

Show that the series $f(x):=\sum_{n=0}^{\infty}f_n(x)$ converges to $f$ uniformly on $[0,1]$.

I'm having trouble with the proof and looking for help, thanks :)

$\endgroup$
  • 1
    $\begingroup$ The series doesn't have to be absolutely convergent. $\endgroup$ – copper.hat Apr 11 '13 at 16:36
2
$\begingroup$

Using the summation formula of geometric sums, the partial sums are $$ \sum_{k=0}^nf_k(x)=\frac{1-x}{1+x}(1-(-x)^{n+1}). $$ Considering the cases $0\leq x<1$ and $x=1$ separately, we see that this converges pointwise to

$$ f(x)=\sum_{k=0}^{+\infty}f_k(x)=\frac{1-x}{1+x}\qquad\forall x\in[0,1]. $$

So the remainder is $$ R_n(x)=f(x)-\sum_{k=0}^nf_k(x)=\frac{1-x}{1+x}\left(1-(1-(-x)^{n+1})\right)=\frac{(1-x)(-x)^{n+1}}{1+x}. $$ An easy study of the derivative of the nonnegative function $g_n(x):=(1-x)x^{n+1}$ on $[0,1]$ shows that its maximum is attained for $x=\frac{n+1}{n+2}=1-\frac{1}{n+2}$. Hence $$ \sup_{[0,1]}\;|R_n(x)|\leq g_n\left(1-\frac{1}{n+2}\right)=\frac{1}{n+2}\left(1-\frac{1}{n+2}\right)^{n+1}\leq \frac{1}{n+2}. $$ It follows that $$ \lim_{n\rightarrow +\infty}\;\; \sup_{[0,1]}\;|R_n(x)|=0 $$ that is, the series converges uniformly to $f$ on $[0,1]$.

$\endgroup$
1
$\begingroup$

First, since $f_n(x)$ is alternating, and $f_n(x) \to 0$, the series $\sum_n f_n(x)$ converges, so $f(x)$ is well defined.

If $x \in[0,1)$ we have $\sum_{n=0}^\infty (-1)^n x^n = \frac{1}{1+x}$, and so we have $(1-x)\sum_{n=0}^\infty (-1)^n x^n = \frac{1-x}{1+x}$ for all $x \in [0,1]$. In particular, $|(1-x)\sum_{n=0}^\infty (-1)^n x^n| \le 1-x $ for all $x \in [0,1]$.

Now consider \begin{eqnarray} |f(x)-\sum_{n=0}^N (-1)^n (1-x)x^n| &=& |\sum_{n=N+1}^\infty (-1)^n (1-x)x^n |\\ &=& |\sum_{n=0}^\infty (-1)^{n+N+1} (1-x)x^{n+N+1} | \\ &=& x^{N+1} |\sum_{n=1}^\infty (-1)^{n} (1-x)x^{n} | \\ &\le& x^{N+1}(1-x) \\ & \le & \frac{1}{N+1}(\frac{N+1}{N+2})^{N+1} \\ & \le & \frac{1}{N} \end{eqnarray} It follows that the convergence is uniform.

$\endgroup$
  • $\begingroup$ Thanks for making me realize that $x^n\leq 1$ when $|x|\leq 1$...(+1). $\endgroup$ – Julien Apr 11 '13 at 18:47
0
$\begingroup$

Assume $\:\:\frac{\text d }{\text dx}\sum_j f_j(x)=\sum_j \frac{\text d }{\text dx} f_j(x)\:$ and try to show that the latter converges uniformly over $\:\bigcup_{\delta>0}[\delta,1-\delta]\:$ for any such fixed $\:\delta\:$ and since $\:f_n\:$ converges uniformly trivially at $\:x_n\in\{0,1\}\:$ this will show that the series of functions is continuous over the compact $\:\mathcal K=[0,1],\:$ which is a necessary condition for it to converge uniformly.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.