Let f(x)=\sin(2x). Prove or disprove there are a sequnce of polynomials P_n(x) which convenes to f(x) uniformly on (0,\infty).

vazelinahS 2021-09-04 Answered
Let f(x)=sin(2x)
Prove or disprove there are a sequnce of polynomials Pn(x) which convenes to f(x) uniformly on (0,).
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Expert Answer

StrycharzT
Answered 2021-09-05 Author has 102 answers

We have to prove or disprove there are a sequence of polynomials Pn(x) which converges to f(x) uniformly on (0,)
We have to prove by contradiction.
Suppose there exists a sequence of polynomial fn(x) which converges uniformly to sin(2x)
So Sup|fn(x)sin(2x)|<ϵnn0
In particular for ϵ=12
We have
Sup|fn(x)sin(2x)|<12
As sin(2x) is bounded function by 1(By triangle inequality)
So, |fn(x)|<32
This means, sequence of polynomial fn is bounded by 32
That is almost all fn are constant polynomials
But fn(0)sin(20)=0 as n that is fn(0)0
And fn(π4)sin(2p4)=1 as n that is fn(π4)1
Which leads to contradiction to fact that almost all fn are constant polynomial.
Hence, there doesn't exists any sequence of polynomial converges uniformly to sin(2x)

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