Question Number 593 by jitendrarathod2556@gmail.com last updated on 06/Feb/15
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Commented by 123456 last updated on 17/Feb/15
$$\mathrm{1}\leqslant\digamma_{\mathrm{264}} \leqslant\mathrm{1}+\frac{\pi}{\mathrm{4}}\lesssim\mathrm{1}+\mathrm{0}.\mathrm{79}=\mathrm{1}.\mathrm{79} \\ $$
Answered by 123456 last updated on 08/Feb/15
![f∈C^0 ,∀x∈[0,T] f(x+T)=f(x) f_n (t)=Σ_(i=0) ^n a_i cos (ωit)+b_i sin (ωit) a_0 =(1/T)∫_0 ^T f(t)dt a_n =(2/T)∫_0 ^T f(t)cos (ωnt)dt,n>0 b_0 =0 b_n =(2/T)∫_0 ^T f(t)sin (ωnt)dt,n>0 ∀ε>0,∃N,∀n>N⇒∣f_n (t)−f(t)∣<ε](Q595.png)
$${f}\in{C}^{\mathrm{0}} ,\forall{x}\in\left[\mathrm{0},{T}\right] \\ $$$${f}\left({x}+{T}\right)={f}\left({x}\right) \\ $$$${f}_{{n}} \left({t}\right)=\underset{{i}=\mathrm{0}} {\overset{{n}} {\sum}}{a}_{{i}} \mathrm{cos}\:\left(\omega{it}\right)+{b}_{{i}} \mathrm{sin}\:\left(\omega{it}\right) \\ $$$${a}_{\mathrm{0}} =\frac{\mathrm{1}}{{T}}\underset{\mathrm{0}} {\overset{{T}} {\int}}{f}\left({t}\right){dt} \\ $$$${a}_{{n}} =\frac{\mathrm{2}}{{T}}\underset{\mathrm{0}} {\overset{{T}} {\int}}{f}\left({t}\right)\mathrm{cos}\:\left(\omega{nt}\right){dt},{n}>\mathrm{0} \\ $$$${b}_{\mathrm{0}} =\mathrm{0} \\ $$$${b}_{{n}} =\frac{\mathrm{2}}{{T}}\underset{\mathrm{0}} {\overset{{T}} {\int}}{f}\left({t}\right)\mathrm{sin}\:\left(\omega{nt}\right){dt},{n}>\mathrm{0} \\ $$$$\forall\epsilon>\mathrm{0},\exists{N},\forall{n}>{N}\Rightarrow\mid{f}_{{n}} \left({t}\right)−{f}\left({t}\right)\mid<\epsilon \\ $$
Commented by 123456 last updated on 08/Feb/15
$$\underset{{n}\rightarrow+\infty} {\mathrm{lim}}{f}_{{n}} \left({x}\right)={f}\left({x}\right) \\ $$
Commented by 123456 last updated on 17/Feb/15
![1.∫_0 ^T f(x)<+∞ 2.∫_0 ^T ∣f(x)∣dx<+∞ 3.N={x∈[0,T],f′(x)=0},∣N∣<+∞](Q642.png)
$$\mathrm{1}.\underset{\mathrm{0}} {\overset{{T}} {\int}}{f}\left({x}\right)<+\infty \\ $$$$\mathrm{2}.\underset{\mathrm{0}} {\overset{{T}} {\int}}\mid{f}\left({x}\right)\mid{dx}<+\infty \\ $$$$\mathrm{3}.{N}=\left\{{x}\in\left[\mathrm{0},{T}\right],{f}'\left({x}\right)=\mathrm{0}\right\},\mid{N}\mid<+\infty \\ $$