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

StrycharzT · Accepted Answer

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)|&amp;lt;ϵ∀n≥n0 In particular for ϵ=12 We have Sup|fn(x)−sin⁡(2x)|&amp;lt;12 As sin⁡(2x) is bounded function by 1(By triangle inequality) So, |fn(x)|&amp;lt;32 This means, sequence of polynomial fn is bounded by 32 That is almost all fn are constant polynomials But fn(0)→sin⁡(2⋅0)=0 as n→∞ that is fn(0)→0 And fn(π4)→sin⁡(2⋅p4)=1 as n→∞ that is fn(π4)→1 Which leads to contradiction to fact that almost all fn are constant polynomial. Hence, there doesn&#039;t exists any sequence of polynomial converges uniformly to sin⁡(2x)

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).

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