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Question Number 204244 by pticantor last updated on 09/Feb/24
Answered by pticantor last updated on 09/Feb/24
please can some one help me to solve question 2−b for this exercise?
$${please}\:{can}\:{some}\:{one}\:{help}\:{me}\:{to}\:{solve}\:{question}\:\mathrm{2}−{b}\: \\ $$$${for}\:{this}\:{exercise}? \\ $$
Answered by witcher3 last updated on 10/Feb/24
2−b ∫_0 ^π f(x)dx=lim_(n→∞) Σ_(k=0) ^(n−1) (1/n)f(((kπ)/n)) =lim_(n→∞) (1/n)Σ_(k=0) ^∞ ln(x^2 −2xcos(((kπ)/n))+1) =lim_(n→∞) ln(Π_(k=0) ^(n−1) (x^2 −2xcos(((kπ)/n))+1))=I X^(2n) −1=Π_(k=0) ^(2n−1) (x−e^((ikπ)/n) )=Π_(k=0) ^(n−1) (x−e^(i((kπ)/n)) )Π_(k=n) ^(2n−1) (x−e^(i((kπ)/n)) ) k=n+k dans le 2 eme produits,k est muete =Π_(k=0) ^(n−1) (x−e^((ikπ)/n) )(x−e^(−((ikπ)/n)) )=Π(x^2 −2xcos(((kπ)/n))+1)=X^(2n) −1 ⇒I=lim_(n→∞) (1/n)ln(X^(2n) −1),∣x∣>1 =lim_(n→∞) .(1/2)(2nln(∣x∣)+ln(1−(1/x^(2n) )))=ln(x^2 ) −1<x<1;x≠0I=∫ln(x^2 )+ln(1−(2/x)cos(t)+(1/x^2 ))dt =πln(x^2 )+ln((1/x^2 ))=(π−1)ln(x^2 ) x=0 I=0
$$\mathrm{2}−\mathrm{b} \\ $$$$\int_{\mathrm{0}} ^{\pi} \mathrm{f}\left(\mathrm{x}\right)\mathrm{dx}=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\sum}}\frac{\mathrm{1}}{\mathrm{n}}\mathrm{f}\left(\frac{\mathrm{k}\pi}{\mathrm{n}}\right) \\ $$$$=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{n}}\underset{\mathrm{k}=\mathrm{0}} {\overset{\infty} {\sum}}\mathrm{ln}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2xcos}\left(\frac{\mathrm{k}\pi}{\mathrm{n}}\right)+\mathrm{1}\right) \\ $$$$=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}ln}\left(\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\prod}}\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2xcos}\left(\frac{\mathrm{k}\pi}{\mathrm{n}}\right)+\mathrm{1}\right)\right)=\mathrm{I} \\ $$$$\mathrm{X}^{\mathrm{2n}} −\mathrm{1}=\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{2n}−\mathrm{1}} {\prod}}\left(\mathrm{x}−\mathrm{e}^{\frac{\mathrm{ik}\pi}{\mathrm{n}}} \right)=\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\prod}}\left(\mathrm{x}−\mathrm{e}^{\mathrm{i}\frac{\mathrm{k}\pi}{\mathrm{n}}} \right)\underset{\mathrm{k}=\mathrm{n}} {\overset{\mathrm{2n}−\mathrm{1}} {\prod}}\left(\mathrm{x}−\mathrm{e}^{\mathrm{i}\frac{\mathrm{k}\pi}{\mathrm{n}}} \right) \\ $$$$\mathrm{k}=\mathrm{n}+\mathrm{k}\:\mathrm{dans}\:\mathrm{le}\:\mathrm{2}\:\mathrm{eme}\:\mathrm{produits},\mathrm{k}\:\mathrm{est}\:\mathrm{muete} \\ $$$$=\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{n}−\mathrm{1}} {\prod}}\left(\mathrm{x}−\mathrm{e}^{\frac{\mathrm{ik}\pi}{\mathrm{n}}} \right)\left(\mathrm{x}−\mathrm{e}^{−\frac{\mathrm{ik}\pi}{\mathrm{n}}} \right)=\Pi\left(\mathrm{x}^{\mathrm{2}} −\mathrm{2xcos}\left(\frac{\mathrm{k}\pi}{\mathrm{n}}\right)+\mathrm{1}\right)=\mathrm{X}^{\mathrm{2n}} −\mathrm{1} \\ $$$$\Rightarrow\mathrm{I}=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}\frac{\mathrm{1}}{\mathrm{n}}\mathrm{ln}\left(\mathrm{X}^{\mathrm{2n}} −\mathrm{1}\right),\mid\mathrm{x}\mid>\mathrm{1} \\ $$$$=\underset{\mathrm{n}\rightarrow\infty} {\mathrm{lim}}.\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{2nln}\left(\mid\mathrm{x}\mid\right)+\mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2n}} }\right)\right)=\mathrm{ln}\left(\mathrm{x}^{\mathrm{2}} \right) \\ $$$$−\mathrm{1}<\mathrm{x}<\mathrm{1};\mathrm{x}\neq\mathrm{0I}=\int\mathrm{ln}\left(\mathrm{x}^{\mathrm{2}} \right)+\mathrm{ln}\left(\mathrm{1}−\frac{\mathrm{2}}{\mathrm{x}}\mathrm{cos}\left(\mathrm{t}\right)+\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)\mathrm{dt} \\ $$$$=\pi\mathrm{ln}\left(\mathrm{x}^{\mathrm{2}} \right)+\mathrm{ln}\left(\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)=\left(\pi−\mathrm{1}\right)\mathrm{ln}\left(\mathrm{x}^{\mathrm{2}} \right) \\ $$$$\mathrm{x}=\mathrm{0}\:\mathrm{I}=\mathrm{0} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

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