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Question Number 83074 by ~blr237~ last updated on 27/Feb/20
find all function satisfying ∀ x∈R\{kπ , k∈Z} f(x)+∫_0 ^1 f^2 (x)dx=(x/(sin(πx)))
$${find}\:{all}\:{function}\:\:{satisfying}\:\:\forall\:{x}\in\mathbb{R}\backslash\left\{{k}\pi\:,\:\:{k}\in\mathbb{Z}\right\} \\ $$$${f}\left({x}\right)+\int_{\mathrm{0}} ^{\mathrm{1}} {f}^{\mathrm{2}} \left({x}\right){dx}=\frac{{x}}{{sin}\left(\pi{x}\right)} \\ $$
Commented by mr W last updated on 28/Feb/20
∫_0 ^1 f^2 (x)dx=k=constant f(x)=(x/(sin(πx)))−k f^2 (x)=(x^2 /(sin^2 (πx)))−2k(x/(sin πx))+k^2 ∫_0 ^1 f^2 (x)dx=∫_0 ^1 ((x^2 dx)/(sin^2 (πx)))−2k∫_0 ^1 ((xdx)/(sin πx))+k^2 =k k^2 −(1+2∫_0 ^1 ((xdx)/(sin πx)))k+∫_0 ^1 ((x^2 dx)/(sin^2 (πx)))=0 k^2 −(1+2A)k+B=0 ⇒k=((1+2A±(√((1+2A)^2 −4B)))/2) ⇒f(x)=(x/(sin(πx)))−((1+2A±(√((1+2A)^2 −4B)))/2) with A=∫_0 ^1 ((xdx)/(sin πx))=(1/π^2 )∫_0 ^π ((tdt)/(sin t)) B=∫_0 ^1 ((x^2 dx)/(sin^2 (πx)))=(1/π^3 )∫_0 ^π ((t^2 dt)/(sin^2 t)) but A and B may not exist, so the question may be wrong.
$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}^{\mathrm{2}} \left({x}\right){dx}={k}={constant} \\ $$$${f}\left({x}\right)=\frac{{x}}{{sin}\left(\pi{x}\right)}−{k} \\ $$$${f}^{\mathrm{2}} \left({x}\right)=\frac{{x}^{\mathrm{2}} }{{sin}^{\mathrm{2}} \left(\pi{x}\right)}−\mathrm{2}{k}\frac{{x}}{\mathrm{sin}\:\pi{x}}+{k}^{\mathrm{2}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} {f}^{\mathrm{2}} \left({x}\right){dx}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} {dx}}{{sin}^{\mathrm{2}} \left(\pi{x}\right)}−\mathrm{2}{k}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{xdx}}{\mathrm{sin}\:\pi{x}}+{k}^{\mathrm{2}} ={k} \\ $$$${k}^{\mathrm{2}} −\left(\mathrm{1}+\mathrm{2}\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{xdx}}{\mathrm{sin}\:\pi{x}}\right){k}+\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} {dx}}{{sin}^{\mathrm{2}} \left(\pi{x}\right)}=\mathrm{0} \\ $$$${k}^{\mathrm{2}} −\left(\mathrm{1}+\mathrm{2}{A}\right){k}+{B}=\mathrm{0} \\ $$$$\Rightarrow{k}=\frac{\mathrm{1}+\mathrm{2}{A}\pm\sqrt{\left(\mathrm{1}+\mathrm{2}{A}\right)^{\mathrm{2}} −\mathrm{4}{B}}}{\mathrm{2}} \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{{x}}{{sin}\left(\pi{x}\right)}−\frac{\mathrm{1}+\mathrm{2}{A}\pm\sqrt{\left(\mathrm{1}+\mathrm{2}{A}\right)^{\mathrm{2}} −\mathrm{4}{B}}}{\mathrm{2}} \\ $$$${with} \\ $$$${A}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{xdx}}{\mathrm{sin}\:\pi{x}}=\frac{\mathrm{1}}{\pi^{\mathrm{2}} }\int_{\mathrm{0}} ^{\pi} \frac{{tdt}}{\mathrm{sin}\:{t}} \\ $$$${B}=\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{x}^{\mathrm{2}} {dx}}{{sin}^{\mathrm{2}} \left(\pi{x}\right)}=\frac{\mathrm{1}}{\pi^{\mathrm{3}} }\int_{\mathrm{0}} ^{\pi} \frac{{t}^{\mathrm{2}} {dt}}{\mathrm{sin}^{\mathrm{2}} \:{t}} \\ $$$${but}\:{A}\:{and}\:{B}\:{may}\:{not}\:{exist},\:{so}\:{the} \\ $$$${question}\:{may}\:{be}\:{wrong}. \\ $$
Commented by ~blr237~ last updated on 28/Feb/20
nice sir ! the both don′t exist :it′s feel as sometimes it′s not easy to prove that the solution set of an equation is empty
$${nice}\:{sir}\:!\:\:{the}\:{both}\:{don}'{t}\:{exist}\:\::{it}'{s}\:\:{feel}\:{as}\:{sometimes}\:{it}'{s}\:{not}\:{easy}\:{to}\:{prove}\:{that}\:{the}\:{solution}\:{set}\:{of}\:{an}\: \\ $$$${equation}\:{is}\:{empty} \\ $$

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