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Discuss the uniform convergence of the following sequence in the interval indicated
{x^n} , 0< x <1
Now,
f(x) = \lim_{n\rightarrow \infty} f_{n}(x) = 0
Therefore given any small \epsilon > 0, if there exists N such that |f_n(x)-f(x)| < \epsilon for all n \geq N for all x in the given interval, then f_n(x) is uniformly convergent.
That gives
x^n < \epsilon
n > \frac{\log \epsilon }{\log x}
So, it is not possible to fix an N such that the above condition is satisfied for all values of n>N because for a given value of N, I can always find a value of x close to 1 such that the above condition is not valid.
Hence x^n is not uniformly convergent in the given interval.
Is my above reasoning correct?
{x^n} , 0< x <1
Now,
f(x) = \lim_{n\rightarrow \infty} f_{n}(x) = 0
Therefore given any small \epsilon > 0, if there exists N such that |f_n(x)-f(x)| < \epsilon for all n \geq N for all x in the given interval, then f_n(x) is uniformly convergent.
That gives
x^n < \epsilon
n > \frac{\log \epsilon }{\log x}
So, it is not possible to fix an N such that the above condition is satisfied for all values of n>N because for a given value of N, I can always find a value of x close to 1 such that the above condition is not valid.
Hence x^n is not uniformly convergent in the given interval.
Is my above reasoning correct?
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