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](Q4714.png)
$$\mathrm{lets}\:{f}:\left[\mathrm{0},\mathrm{T}\right]\rightarrow\mathbb{R}\:\mathrm{such}\:\mathrm{that} \\ $$ $$\underset{\mathrm{0}} {\overset{\mathrm{T}} {\int}}\left[{f}\left({t}\right)\right]^{\mathrm{2}} {dt}<+\infty \\ $$ $$\omega\mathrm{T}=\mathrm{2}\pi \\ $$ $$\mathrm{if}\:{a}\left({n}\right)=\frac{\mathrm{2}}{\mathrm{T}}\underset{\mathrm{0}} {\overset{\mathrm{T}} {\int}}{f}\left({t}\right)\mathrm{cos}\left(\omega{nt}\right){dt} \\ $$ $$\mathrm{and}\:{b}\left({n}\right)=\frac{\mathrm{2}}{\mathrm{T}}\underset{\mathrm{0}} {\overset{\mathrm{T}} {\int}}{f}\left({t}\right)\mathrm{sin}\:\left(\omega{nt}\right){dt} \\ $$ $$\mathrm{does}? \\ $$ $$\underset{{n}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{a}\left({n}\right)={a}\left(\mathrm{0}\right) \\ $$ $$\underset{{n}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{b}\left({n}\right)=\mathrm{0} \\ $$
Commented byprakash 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.](Q4718.png)
$$\mathrm{I}\:\mathrm{think}\:\mathrm{in}\:\mathrm{this}\:\mathrm{case}\:\mathrm{we}\:\mathrm{can}\:\mathrm{directly}\:\mathrm{substitute} \\ $$ $$\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:{n}\:\mathrm{before}\:\mathrm{evaluating}\:\mathrm{the}\:\mathrm{integral}. \\ $$ $$\mathrm{So}\:\underset{{n}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{a}\left({n}\right)={a}\left(\mathrm{0}\right) \\ $$ $$\underset{{n}\rightarrow\mathrm{0}} {\mathrm{lim}}\:{b}\left({n}\right)=\mathrm{0} \\ $$ $$\mathrm{The}\:\mathrm{given}\:\mathrm{condition}\underset{\mathrm{0}} {\overset{\mathrm{T}} {\int}}\left[{f}\left({t}\right)\right]^{\mathrm{2}} {dt}<+\infty\:\mathrm{implies}\:\mathrm{that}\:\mathrm{limit} \\ $$ $$\mathrm{exists}. \\ $$