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siddharth

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Discuss the uniform convergence of the following sequence in the interval indicated

[tex] {x^n} , 0< x <1 [/tex]

Now,

[tex] f(x) = \lim_{n\rightarrow \infty} f_{n}(x) = 0[/tex]

Therefore given any small [tex] \epsilon > 0 [/tex], if there exists [tex] N [/tex] such that [tex] |f_n(x)-f(x)| < \epsilon [/tex] for all [tex] n \geq N [/tex] for all x in the given interval, then f_n(x) is uniformly convergent.

That gives

[tex] x^n < \epsilon [/tex]

[tex]n > \frac{\log \epsilon }{\log x} [/tex]

So, it is not possible to fix an [tex] N [/tex] 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 [tex] x^n [/tex] is not uniformly convergent in the given interval.

Is my above reasoning correct?

[tex] {x^n} , 0< x <1 [/tex]

Now,

[tex] f(x) = \lim_{n\rightarrow \infty} f_{n}(x) = 0[/tex]

Therefore given any small [tex] \epsilon > 0 [/tex], if there exists [tex] N [/tex] such that [tex] |f_n(x)-f(x)| < \epsilon [/tex] for all [tex] n \geq N [/tex] for all x in the given interval, then f_n(x) is uniformly convergent.

That gives

[tex] x^n < \epsilon [/tex]

[tex]n > \frac{\log \epsilon }{\log x} [/tex]

So, it is not possible to fix an [tex] N [/tex] 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 [tex] x^n [/tex] is not uniformly convergent in the given interval.

Is my above reasoning correct?

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