Question and Answers Forum

All Questions      Topic List

Differentiation Questions

Previous in All Question      Next in All Question      

Previous in Differentiation      Next in Differentiation      

Question Number 168348 by mnjuly1970 last updated on 08/Apr/22

Answered by mathman1234 last updated on 09/Apr/22

plz help if 0<a<b and x>0 proove that (2/π)(1−(a/b))≤sup∣((sin(ax))/(ax)) − ((sin(ax))/(ax))∣≤4(1−(a/b))

$$\mathrm{plz}\:\mathrm{help} \\ $$$$\mathrm{if}\:\mathrm{0}<\mathrm{a}<\mathrm{b}\:\mathrm{and}\:\mathrm{x}>\mathrm{0}\:\mathrm{proove}\:\mathrm{that} \\ $$$$\frac{\mathrm{2}}{\pi}\left(\mathrm{1}−\frac{\mathrm{a}}{\mathrm{b}}\right)\leqslant\mathrm{sup}\mid\frac{\mathrm{sin}\left(\mathrm{ax}\right)}{\mathrm{ax}}\:−\:\frac{\mathrm{sin}\left(\mathrm{ax}\right)}{\mathrm{ax}}\mid\leqslant\mathrm{4}\left(\mathrm{1}−\frac{\mathrm{a}}{\mathrm{b}}\right) \\ $$

Commented by mr W last updated on 09/Apr/22

please don′t put your new question in existing threads of other people′s questions! please open a new post for your own question!

$${please}\:{don}'{t}\:{put}\:{your}\:{new}\:{question}\:{in} \\ $$$${existing}\:{threads}\:{of}\:{other}\:{people}'{s} \\ $$$${questions}!\:{please}\:{open}\:{a}\:{new}\:{post} \\ $$$${for}\:{your}\:{own}\:{question}! \\ $$

Answered by mindispower last updated on 11/Apr/22

Ψ(1+x)=−γ−Σ_(n≥) ζ(n+1)(−x)^n Ψ(1−x)=−γ−Σ_(n≥1) ζ(n+1)x^n Ψ(1−x)−Ψ(1+x)=Σ_(n≥1) ζ(n+1)(−(x)^n +(−x)^n ) =−2Σ_(n≥0) ζ(2n+2)x^(2n+1) =Ψ(1−x)−Ψ(1+x) ⇒Σ_(n≥0) ((ζ(2n+2))/(2n+3))x^(2n+3) =−(1/2)∫_0 ^x t(Ψ(1−t)−Ψ(1+t))dt =−(1/2)∫_0 ^x t(πcot(πt)−(1/t))dt =−(1/2)∫_0 ^x πtcot(πt)+(x/2) x=(1/2)⇔(1/4)−(1/2)∫_0 ^(π/2) ucot(u)(du/π) =(1/4)+(1/(2π))∫_0 ^(π/2) ln(sin(u))du(1/4)−(1/(2π)).(π/2)ln(2) =(1/4)−(1/4)ln(2)=Σ_(n≥1) ((ζ(2n))/(2n+1)).(1/2^(2n+1) ) ⇒Σ_(n≥1) ((ζ(2n))/(2^(2n−1) (2n+1)))=1−ln(2)=ln((e/2))

$$\Psi\left(\mathrm{1}+{x}\right)=−\gamma−\underset{{n}\geqslant} {\sum}\zeta\left({n}+\mathrm{1}\right)\left(−{x}\right)^{{n}} \\ $$$$\Psi\left(\mathrm{1}−{x}\right)=−\gamma−\underset{{n}\geqslant\mathrm{1}} {\sum}\zeta\left({n}+\mathrm{1}\right){x}^{{n}} \\ $$$$\Psi\left(\mathrm{1}−{x}\right)−\Psi\left(\mathrm{1}+{x}\right)=\underset{{n}\geqslant\mathrm{1}} {\sum}\zeta\left({n}+\mathrm{1}\right)\left(−\left({x}\right)^{{n}} +\left(−{x}\right)^{{n}} \right) \\ $$$$=−\mathrm{2}\underset{{n}\geqslant\mathrm{0}} {\sum}\zeta\left(\mathrm{2}{n}+\mathrm{2}\right){x}^{\mathrm{2}{n}+\mathrm{1}} =\Psi\left(\mathrm{1}−{x}\right)−\Psi\left(\mathrm{1}+{x}\right) \\ $$$$\Rightarrow\underset{{n}\geqslant\mathrm{0}} {\sum}\frac{\zeta\left(\mathrm{2}{n}+\mathrm{2}\right)}{\mathrm{2}{n}+\mathrm{3}}{x}^{\mathrm{2}{n}+\mathrm{3}} =−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{{x}} {t}\left(\Psi\left(\mathrm{1}−{t}\right)−\Psi\left(\mathrm{1}+{t}\right)\right){dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{{x}} {t}\left(\pi{cot}\left(\pi{t}\right)−\frac{\mathrm{1}}{{t}}\right){dt} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{{x}} \pi{tcot}\left(\pi{t}\right)+\frac{{x}}{\mathrm{2}} \\ $$$${x}=\frac{\mathrm{1}}{\mathrm{2}}\Leftrightarrow\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ucot}\left({u}\right)\frac{{du}}{\pi} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}\pi}\int_{\mathrm{0}} ^{\frac{\pi}{\mathrm{2}}} {ln}\left({sin}\left({u}\right)\right){du}\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{2}\pi}.\frac{\pi}{\mathrm{2}}{ln}\left(\mathrm{2}\right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}}{ln}\left(\mathrm{2}\right)=\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\zeta\left(\mathrm{2}{n}\right)}{\mathrm{2}{n}+\mathrm{1}}.\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}{n}+\mathrm{1}} } \\ $$$$ \\ $$$$\Rightarrow\underset{{n}\geqslant\mathrm{1}} {\sum}\frac{\zeta\left(\mathrm{2}{n}\right)}{\mathrm{2}^{\mathrm{2}{n}−\mathrm{1}} \left(\mathrm{2}{n}+\mathrm{1}\right)}=\mathrm{1}−{ln}\left(\mathrm{2}\right)={ln}\left(\frac{{e}}{\mathrm{2}}\right) \\ $$$$ \\ $$

Commented by mnjuly1970 last updated on 13/Apr/22

grateful sir power...

$$\:\:\:\:{grateful}\:{sir}\:\:{power}... \\ $$

Commented by mindispower last updated on 13/Apr/22

withe Pleasur sir Have nice Day Ramadan Mobarak

$${withe}\:{Pleasur}\:{sir}\:{Have}\:{nice}\:{Day} \\ $$$${Ramadan}\:{Mobarak} \\ $$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com