Uniform convergence

Aug 2008
172
1
Discuss the Uniform convergence of the following series of functions :
1- \(\displaystyle \sum \frac{x^n}{1+ x^n} \ on \ [0,a] , a<1 \ and \ on \ [0,1)\)
2- \(\displaystyle \sum (-1)^n x^n (1-x) \ on \ [0,1]\)
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1- on [0,a] :
BY Weierstrass M- Test :
\(\displaystyle \left | \frac{x^n}{1+x^n} \right |\leq \frac{a^n}{1} \)
\(\displaystyle \sum a^n \) is conv {geometric series }
\(\displaystyle \rightarrow \) \(\displaystyle \sum \frac{x^n}{1+ x^n} \) is Uniformly convergent on [0,a]
on [0,1):
?????
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2-
 

chisigma

MHF Hall of Honor
Mar 2009
2,162
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near Piacenza (Italy)
The series...

\(\displaystyle S = \sum_{n=0}^{\infty} (-1)^{n} x^{n} (1-x)\) (1)

... converges is \(\displaystyle [0,1]\). For \(\displaystyle x=1\) all terms os (1) vanish and is \(\displaystyle S=0\) and for \(\displaystyle 0 \le x <1\) is...

\(\displaystyle S=\sum_{n=0}^{\infty} (-1)^{n} x^{n} (1-x)= \frac{1-x}{1+x}\) (2)

Now if we compute the remainder...

\(\displaystyle R_{k} = \sum_{n=k}^{\infty} (-1)^{n} x^{n} (1-x) = (-1)^{k} x^{k} \frac{1-x}{1+x}\) (3)

... for any \(\displaystyle \varepsilon >0\) the inequality \(\displaystyle |R_{k}|< \varepsilon\) is verified only if is ...

\(\displaystyle k \ln x < \ln \varepsilon + \ln (1+x) - \ln (1-x) \rightarrow k> \frac{\ln \varepsilon + \ln (1+x) - \ln (1-x)}{\ln x}\) (4)

But the limit for \(\displaystyle x \rightarrow 1-0\) of the (4) is unbounded so that the (1) doesn't converge uniformely in \(\displaystyle [0,1)\) ...

Kind regards

\(\displaystyle \chi\) \(\displaystyle \sigma\)
 
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