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Question Number 4714 by 123456 last updated on 28/Feb/16
lets f:[0,T]→R such that  ∫_0 ^T [f(t)]^2 dt<+∞  ωT=2π  if a(n)=(2/T)∫_0 ^T f(t)cos(ωnt)dt  and b(n)=(2/T)∫_0 ^T f(t)sin (ωnt)dt  does?  lim_(n→0)  a(n)=a(0)  lim_(n→0)  b(n)=0
letsf:[0,T]RsuchthatT0[f(t)]2dt<+ωT=2πifa(n)=2TT0f(t)cos(ωnt)dtandb(n)=2TT0f(t)sin(ωnt)dtdoes?limn0a(n)=a(0)limn0b(n)=0
Commented by prakash jain last updated on 29/Feb/16
I think in this case we can directly substitute  the value of n before evaluating the integral.  So lim_(n→0)  a(n)=a(0)  lim_(n→0)  b(n)=0  The given condition∫_0 ^T [f(t)]^2 dt<+∞ implies that limit  exists.
Ithinkinthiscasewecandirectlysubstitutethevalueofnbeforeevaluatingtheintegral.Solimn0a(n)=a(0)limn0b(n)=0ThegivenconditionT0[f(t)]2dt<+impliesthatlimitexists.

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