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Question Number 171484 by infinityaction last updated on 16/Jun/22
let f(x) = x+(2/(1.3))x^3 +((2.4)/(1.3.5))x^5 +((2.4.6)/(1.3.5.7))x^7 +......... ∀x∈(0,1) the value of f((1/( (√2)))) = ?
$$ \\ $$$$\:\:\:\:{let}\:{f}\left({x}\right)\:=\:{x}+\frac{\mathrm{2}}{\mathrm{1}.\mathrm{3}}{x}^{\mathrm{3}} +\frac{\mathrm{2}.\mathrm{4}}{\mathrm{1}.\mathrm{3}.\mathrm{5}}{x}^{\mathrm{5}} +\frac{\mathrm{2}.\mathrm{4}.\mathrm{6}}{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}}{x}^{\mathrm{7}} +……… \\ $$$$\:\:\:\:\forall{x}\in\left(\mathrm{0},\mathrm{1}\right)\:\:{the}\:{value}\:{of}\:\:{f}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)\:=\:? \\ $$
Commented by mr W last updated on 16/Jun/22
i got f(x)=((sin^(−1) x)/( (√(1−x^2 )))) f((1/( (√2))))=(((√2)π)/4)
$${i}\:{got} \\ $$$${f}\left({x}\right)=\frac{\mathrm{sin}^{−\mathrm{1}} {x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$${f}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)=\frac{\sqrt{\mathrm{2}}\pi}{\mathrm{4}} \\ $$
Answered by mr W last updated on 16/Jun/22
f(x)=x+(2/(1.3))x^3 +((2.4)/(1.3.5))x^5 +((2.4.6)/(1.3.5.7))x^7 +... xf(x)=x^2 +(2/(1.3))x^4 +((2.4)/(1.3.5))x^6 +((2.4.6)/(1.3.5.7))x^8 +... ∫_0 ^x xf(x)dx=(1/3)x^3 +(2/(1.3.5))x^5 +((2.4)/(1.3.5.7))x^7 +((2.4.6)/(1.3.5.7.9))x^9 +... (1/x)∫_0 ^x xf(x)dx=(1/(1.3))x^2 +(2/(1.3.5))x^4 +((2.4)/(1.3.5.7))x^6 +((2.4.6)/(1.3.5.7.9))x^8 +... [(1/x)∫_0 ^x xf(x)dx]′=(2/(1.3))x+((2.4)/(1.3.5))x^3 +((2.4.6)/(1.3.5.7))x^5 +((2.4.6.8)/(1.3.5.7.9))x^7 +... x^2 [(1/x)∫_0 ^x xf(x)dx]′=(2/(1.3))x^3 +((2.4)/(1.3.5))x^5 +((2.4.6)/(1.3.5.7))x^7 +((2.4.6.8)/(1.3.5.7.9))x^9 +... x^2 [(1/x)∫_0 ^x xf(x)dx]′=−x+x+(2/(1.3))x^3 +((2.4)/(1.3.5))x^5 +((2.4.6)/(1.3.5.7))x^7 +((2.4.6.8)/(1.3.5.7.9))x^9 +... x^2 [(1/x)∫_0 ^x xf(x)dx]′=−x+f(x) x^2 [−(1/x^2 )∫_0 ^x xf(x)dx+f(x)]=−x+f(x) ∫_0 ^x xf(x)dx=(x^2 −1)f(x)+x xf(x)=2xf(x)+(x^2 −1)f′(x)+1 (x^2 −1)f′(x)+xf(x)+1=0 ⇒f′(x)+(x/(x^2 −1))f(x)=(1/(1−x^2 )) I=e^(∫((x dx)/(x^2 −1))) =e^((ln ∣1−x^2 ∣)/2) =(√(1−x^2 )) ⇒f(x)=(1/( (√(1−x^2 ))))∫((√(1−x^2 ))/( 1−x^2 ))dx=((sin^(−1) x)/( (√(1−x^2 ))))+C f(0)=0 ⇒C=0 ⇒f(x)=((sin^(−1) x)/( (√(1−x^2 )))) ⇒f((1/( (√2))))=(((√2)π)/4)
$${f}\left({x}\right)={x}+\frac{\mathrm{2}}{\mathrm{1}.\mathrm{3}}{x}^{\mathrm{3}} +\frac{\mathrm{2}.\mathrm{4}}{\mathrm{1}.\mathrm{3}.\mathrm{5}}{x}^{\mathrm{5}} +\frac{\mathrm{2}.\mathrm{4}.\mathrm{6}}{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}}{x}^{\mathrm{7}} +… \\ $$$${xf}\left({x}\right)={x}^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{1}.\mathrm{3}}{x}^{\mathrm{4}} +\frac{\mathrm{2}.\mathrm{4}}{\mathrm{1}.\mathrm{3}.\mathrm{5}}{x}^{\mathrm{6}} +\frac{\mathrm{2}.\mathrm{4}.\mathrm{6}}{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}}{x}^{\mathrm{8}} +… \\ $$$$\int_{\mathrm{0}} ^{{x}} {xf}\left({x}\right){dx}=\frac{\mathrm{1}}{\mathrm{3}}{x}^{\mathrm{3}} +\frac{\mathrm{2}}{\mathrm{1}.\mathrm{3}.\mathrm{5}}{x}^{\mathrm{5}} +\frac{\mathrm{2}.\mathrm{4}}{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}}{x}^{\mathrm{7}} +\frac{\mathrm{2}.\mathrm{4}.\mathrm{6}}{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}.\mathrm{9}}{x}^{\mathrm{9}} +… \\ $$$$\frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{{x}} {xf}\left({x}\right){dx}=\frac{\mathrm{1}}{\mathrm{1}.\mathrm{3}}{x}^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{1}.\mathrm{3}.\mathrm{5}}{x}^{\mathrm{4}} +\frac{\mathrm{2}.\mathrm{4}}{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}}{x}^{\mathrm{6}} +\frac{\mathrm{2}.\mathrm{4}.\mathrm{6}}{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}.\mathrm{9}}{x}^{\mathrm{8}} +… \\ $$$$\left[\frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{{x}} {xf}\left({x}\right){dx}\right]'=\frac{\mathrm{2}}{\mathrm{1}.\mathrm{3}}{x}+\frac{\mathrm{2}.\mathrm{4}}{\mathrm{1}.\mathrm{3}.\mathrm{5}}{x}^{\mathrm{3}} +\frac{\mathrm{2}.\mathrm{4}.\mathrm{6}}{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}}{x}^{\mathrm{5}} +\frac{\mathrm{2}.\mathrm{4}.\mathrm{6}.\mathrm{8}}{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}.\mathrm{9}}{x}^{\mathrm{7}} +… \\ $$$${x}^{\mathrm{2}} \left[\frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{{x}} {xf}\left({x}\right){dx}\right]'=\frac{\mathrm{2}}{\mathrm{1}.\mathrm{3}}{x}^{\mathrm{3}} +\frac{\mathrm{2}.\mathrm{4}}{\mathrm{1}.\mathrm{3}.\mathrm{5}}{x}^{\mathrm{5}} +\frac{\mathrm{2}.\mathrm{4}.\mathrm{6}}{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}}{x}^{\mathrm{7}} +\frac{\mathrm{2}.\mathrm{4}.\mathrm{6}.\mathrm{8}}{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}.\mathrm{9}}{x}^{\mathrm{9}} +… \\ $$$${x}^{\mathrm{2}} \left[\frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{{x}} {xf}\left({x}\right){dx}\right]'=−{x}+{x}+\frac{\mathrm{2}}{\mathrm{1}.\mathrm{3}}{x}^{\mathrm{3}} +\frac{\mathrm{2}.\mathrm{4}}{\mathrm{1}.\mathrm{3}.\mathrm{5}}{x}^{\mathrm{5}} +\frac{\mathrm{2}.\mathrm{4}.\mathrm{6}}{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}}{x}^{\mathrm{7}} +\frac{\mathrm{2}.\mathrm{4}.\mathrm{6}.\mathrm{8}}{\mathrm{1}.\mathrm{3}.\mathrm{5}.\mathrm{7}.\mathrm{9}}{x}^{\mathrm{9}} +… \\ $$$${x}^{\mathrm{2}} \left[\frac{\mathrm{1}}{{x}}\int_{\mathrm{0}} ^{{x}} {xf}\left({x}\right){dx}\right]'=−{x}+{f}\left({x}\right) \\ $$$${x}^{\mathrm{2}} \left[−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\int_{\mathrm{0}} ^{{x}} {xf}\left({x}\right){dx}+{f}\left({x}\right)\right]=−{x}+{f}\left({x}\right) \\ $$$$\int_{\mathrm{0}} ^{{x}} {xf}\left({x}\right){dx}=\left({x}^{\mathrm{2}} −\mathrm{1}\right){f}\left({x}\right)+{x} \\ $$$${xf}\left({x}\right)=\mathrm{2}{xf}\left({x}\right)+\left({x}^{\mathrm{2}} −\mathrm{1}\right){f}'\left({x}\right)+\mathrm{1} \\ $$$$\left({x}^{\mathrm{2}} −\mathrm{1}\right){f}'\left({x}\right)+{xf}\left({x}\right)+\mathrm{1}=\mathrm{0} \\ $$$$\Rightarrow\boldsymbol{{f}}'\left(\boldsymbol{{x}}\right)+\frac{\boldsymbol{{x}}}{\boldsymbol{{x}}^{\mathrm{2}} −\mathrm{1}}\boldsymbol{{f}}\left(\boldsymbol{{x}}\right)=\frac{\mathrm{1}}{\mathrm{1}−\boldsymbol{{x}}^{\mathrm{2}} } \\ $$$${I}={e}^{\int\frac{{x}\:{dx}}{{x}^{\mathrm{2}} −\mathrm{1}}} ={e}^{\frac{\mathrm{ln}\:\mid\mathrm{1}−{x}^{\mathrm{2}} \mid}{\mathrm{2}}} =\sqrt{\mathrm{1}−{x}^{\mathrm{2}} } \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}\int\frac{\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}{\:\mathrm{1}−{x}^{\mathrm{2}} }{dx}=\frac{\mathrm{sin}^{−\mathrm{1}} {x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }}+{C} \\ $$$${f}\left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow{C}=\mathrm{0} \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{\mathrm{sin}^{−\mathrm{1}} {x}}{\:\sqrt{\mathrm{1}−{x}^{\mathrm{2}} }} \\ $$$$\Rightarrow{f}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)=\frac{\sqrt{\mathrm{2}}\pi}{\mathrm{4}} \\ $$
Commented by infinityaction last updated on 16/Jun/22
thank you sir nice solution
$${thank}\:{you}\:{sir} \\ $$$${nice}\:{solution} \\ $$
Commented by Tawa11 last updated on 16/Jun/22
Great sir.
$$\mathrm{Great}\:\mathrm{sir}. \\ $$
Answered by qaz last updated on 16/Jun/22
f(x)=Σ_(k=0) ^∞ (((2k)!!)/((2k+1)!!))x^(2k+1) =Σ_(k=0) ^∞ x^(2k+1) ∫_0 ^(π/2) sin^(2k+1) tdt=∫_0 ^(π/2) ((xsin t)/(1−(xsin t)^2 ))dt ⇒f((1/( (√2))))=(1/( (√2)))∫_0 ^(π/2) ((sin t)/(1−(1/2)sin^2 t))dt =(√2)∫_0 ^(π/2) ((sin t)/(1+cos^2 t))dt =(√2)∙arctan (cos t)∣_(π/2) ^0 =(((√2)π)/4)
$${f}\left({x}\right)=\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{2}{k}\right)!!}{\left(\mathrm{2}{k}+\mathrm{1}\right)!!}{x}^{\mathrm{2}{k}+\mathrm{1}} =\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}{x}^{\mathrm{2}{k}+\mathrm{1}} \int_{\mathrm{0}} ^{\pi/\mathrm{2}} \mathrm{s}{in}^{\mathrm{2}{k}+\mathrm{1}} {tdt}=\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{{x}\mathrm{sin}\:{t}}{\mathrm{1}−\left({x}\mathrm{sin}\:{t}\right)^{\mathrm{2}} }{dt} \\ $$$$\Rightarrow{f}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\right)=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{\mathrm{sin}\:{t}}{\mathrm{1}−\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sin}\:^{\mathrm{2}} {t}}{dt} \\ $$$$=\sqrt{\mathrm{2}}\int_{\mathrm{0}} ^{\pi/\mathrm{2}} \frac{\mathrm{sin}\:{t}}{\mathrm{1}+\mathrm{cos}\:^{\mathrm{2}} {t}}{dt} \\ $$$$=\sqrt{\mathrm{2}}\centerdot\mathrm{arctan}\:\left(\mathrm{cos}\:{t}\right)\mid_{\pi/\mathrm{2}} ^{\mathrm{0}} \\ $$$$=\frac{\sqrt{\mathrm{2}}\pi}{\mathrm{4}} \\ $$
Commented by infinityaction last updated on 16/Jun/22
amazing solution sir
$${amazing}\:{solution}\:{sir} \\ $$
Answered by infinityaction last updated on 16/Jun/22

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