Does <munderover> &#x2211;<!-- ∑ --> <mrow class="MJX-TeXAtom-ORD"> k =

ownerweneuf

ownerweneuf

Answered question

2022-05-30

Does k = 1 2 k 1 x 2 k 1 1 1 + x 2 k 1 converges uniformly.

Answer & Explanation

minnegodks

minnegodks

Beginner2022-05-31Added 10 answers

The series does not converge uniformly on ( 0 , 1 ) and hence also not on ( 1 , 1 )
To see this, let f n ( x ) = 2 n 1 x 2 n 1 1 1 + x 2 n 1 . Then each f n is continuous and (hence) bounded on [ 0 , 1 ]
Now, let f ( x ) := n f n ( x ). If n = 1 N f n f would hold with uniform convergence on ( 0 , 1 ), it would follow that f is bounded on ( 0 , 1 )
But since each term of the sum is nonnegative for x ( 0 , 1 ), we have
f sup f ( x ) f n ( x ) 2 n 1 x 2 n 1 1 1 + x 2 n 1 2 n 1 2  for  x 1.
Since this holds for arbitrary n N , we conclude f sup =
Hence, f is not bounded on ( 0 , 1 ), so that the series is not uniformly convergent on ( 0 , 1 )

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?