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Question Number 145634 by mathmax by abdo last updated on 06/Jul/21
let s(x)=Σ_(n=1) ^∞ (((−1)^n )/((2x^2 +2x(√(1+x^2 ))+1)^n )) 1) explicite s(x) 2) calculate ∫_0 ^1 s(x)dx
$$\mathrm{let}\:\mathrm{s}\left(\mathrm{x}\right)=\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{2x}^{\mathrm{2}} +\mathrm{2x}\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }+\mathrm{1}\right)^{\mathrm{n}} } \\ $$$$\left.\mathrm{1}\right)\:\mathrm{explicite}\:\mathrm{s}\left(\mathrm{x}\right) \\ $$$$\left.\mathrm{2}\right)\:\mathrm{calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{s}\left(\mathrm{x}\right)\mathrm{dx} \\ $$
Answered by Olaf_Thorendsen last updated on 06/Jul/21
S(x) = Σ_(n=1) ^∞ (((−1)^n )/((2x^2 +2x(√(1+x^2 ))+1)^n )) S(x) = Σ_(n=1) ^∞ (((−1)^n )/((x^2 +2x(√(1+x^2 ))+(1+x^2 ))^n )) S(x) = Σ_(n=1) ^∞ (((−1)^n )/((x+(√(1+x^2 )))^(2n) )) S(x) = Σ_(n=0) ^∞ (((−1)^n )/((x+(√(1+x^2 )))^(2n) ))−1 S(x) = (1/(1+(1/((x+(√(1+x^2 )))^2 ))))−1 S(x) = −(1/((x+(√(1+x^2 )))^2 +1)) ∫_0 ^1 S(x)dx = −∫_0 ^1 (dx/((x+(√(1+x^2 )))^2 +1)) = −∫_0 ^(sh^(−1) (1)) ((chu)/((shu+(√(1+sh^2 u)))^2 +1))du = −∫_0 ^(sh^(−1) (1)) ((chu)/((shu+chu)^2 +1))du = −∫_0 ^(sh^(−1) (1)) ((chu)/(2ch^2 u+2shuchu))du = −(1/2)∫_0 ^(sh^(−1) (1)) (du/(chu+shu)) = −(1/2)∫_0 ^(sh^(−1) (1)) (du/e^u ) = (1/2)[e^(−u) ]_0 ^(sh^(−1) (1)) = (1/2)(e^(−sh^(−1) (1)) −1) = (1/2)((1/(1+(√(1+1^2 ))))−1) = (1/2)((1/(1+(√2)))−1) = (1/2)((√2)−2) = (1/( (√2)))−1
$$\mathrm{S}\left({x}\right)\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }+\mathrm{1}\right)^{{n}} } \\ $$$$\mathrm{S}\left({x}\right)\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({x}^{\mathrm{2}} +\mathrm{2}{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }+\left(\mathrm{1}+{x}^{\mathrm{2}} \right)\right)^{{n}} } \\ $$$$\mathrm{S}\left({x}\right)\:=\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}{n}} } \\ $$$$\mathrm{S}\left({x}\right)\:=\:\underset{{n}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}{n}} }−\mathrm{1} \\ $$$$\mathrm{S}\left({x}\right)\:=\:\frac{\mathrm{1}}{\mathrm{1}+\frac{\mathrm{1}}{\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} }}−\mathrm{1} \\ $$$$\mathrm{S}\left({x}\right)\:=\:−\frac{\mathrm{1}}{\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$$\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{S}\left({x}\right){dx}\:=\:−\int_{\mathrm{0}} ^{\mathrm{1}} \frac{{dx}}{\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)^{\mathrm{2}} +\mathrm{1}} \\ $$$$=\:−\int_{\mathrm{0}} ^{\mathrm{sh}^{−\mathrm{1}} \left(\mathrm{1}\right)} \frac{\mathrm{ch}{u}}{\left(\mathrm{sh}{u}+\sqrt{\mathrm{1}+\mathrm{sh}^{\mathrm{2}} {u}}\right)^{\mathrm{2}} +\mathrm{1}}{du} \\ $$$$=\:−\int_{\mathrm{0}} ^{\mathrm{sh}^{−\mathrm{1}} \left(\mathrm{1}\right)} \frac{\mathrm{ch}{u}}{\left(\mathrm{sh}{u}+\mathrm{ch}{u}\right)^{\mathrm{2}} +\mathrm{1}}{du} \\ $$$$=\:−\int_{\mathrm{0}} ^{\mathrm{sh}^{−\mathrm{1}} \left(\mathrm{1}\right)} \frac{\mathrm{ch}{u}}{\mathrm{2ch}^{\mathrm{2}} {u}+\mathrm{2sh}{u}\mathrm{ch}{u}}{du} \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{sh}^{−\mathrm{1}} \left(\mathrm{1}\right)} \frac{{du}}{\mathrm{ch}{u}+\mathrm{sh}{u}} \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{sh}^{−\mathrm{1}} \left(\mathrm{1}\right)} \frac{{du}}{{e}^{{u}} } \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left[{e}^{−{u}} \right]_{\mathrm{0}} ^{\mathrm{sh}^{−\mathrm{1}} \left(\mathrm{1}\right)} \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left({e}^{−\mathrm{sh}^{−\mathrm{1}} \left(\mathrm{1}\right)} −\mathrm{1}\right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{1}^{\mathrm{2}} }}−\mathrm{1}\right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{1}+\sqrt{\mathrm{2}}}−\mathrm{1}\right) \\ $$$$=\:\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{2}}−\mathrm{2}\right)\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}−\mathrm{1} \\ $$
Answered by mathmax by abdo last updated on 07/Jul/21
1)we have (x/( (√(1+x^2 ))))=_(x=sht) ((sht)/(cht))=((e^t −e^(−t) )/(e^t +e^(−t) ))=((1−e^(−2t) )/(1+e^(−2t) )) =(1−e^(−2t) )Σ_(n=0) ^∞ (−1)^n e^(−2nt) =Σ_(n=0) ^(∞ ) (−1)^n e^(−2nt) −Σ_(n=0) ^∞ (−1)^n e^(−(2n+2)t) =Σ_(n=0) ^∞ (−1)^n e^(−2nt) +Σ_(n=1) ^∞ (−1)^n e^(−2nt) =1+2Σ_(n=0) ^∞ (((−1)^n )/e^(2nt) )=1+2Σ_(n=1) ^∞ (((−1)^n )/e^(2nlog(x+(√(1+x^2 )))) ) =1+2Σ_(n=1) ^∞ (((−1)^n )/((x+(√(1+x^2 )))^(2n) ))=1+2Σ_(n=1) ^∞ (((−1)^n )/((x^2 +2x(√(1+x^2 ))+1+x^2 )^n )) =1+2s(x) ⇒2s(x)=(x/( (√(1+x^2 ))))−1 ⇒s(x)=(1/2)((x/( (√(1+x^2 ))))−1)
$$\left.\mathrm{1}\right)\mathrm{we}\:\mathrm{have}\:\frac{\mathrm{x}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}=_{\mathrm{x}=\mathrm{sht}} \:\:\:\frac{\mathrm{sht}}{\mathrm{cht}}=\frac{\mathrm{e}^{\mathrm{t}} \:−\mathrm{e}^{−\mathrm{t}} }{\mathrm{e}^{\mathrm{t}} +\mathrm{e}^{−\mathrm{t}} }=\frac{\mathrm{1}−\mathrm{e}^{−\mathrm{2t}} }{\mathrm{1}+\mathrm{e}^{−\mathrm{2t}} } \\ $$$$=\left(\mathrm{1}−\mathrm{e}^{−\mathrm{2t}} \right)\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{2nt}} \: \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty\:} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{2nt}} −\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{e}^{−\left(\mathrm{2n}+\mathrm{2}\right)\mathrm{t}} \\ $$$$=\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{e}^{−\mathrm{2nt}} +\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{e}^{−\mathrm{2nt}} \\ $$$$=\mathrm{1}+\mathrm{2}\sum_{\mathrm{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{e}^{\mathrm{2nt}} }=\mathrm{1}+\mathrm{2}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\mathrm{e}^{\mathrm{2nlog}\left(\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right)} } \\ $$$$=\mathrm{1}+\mathrm{2}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{x}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right)^{\mathrm{2n}} }=\mathrm{1}+\mathrm{2}\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} }{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2x}\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }+\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{n}} } \\ $$$$=\mathrm{1}+\mathrm{2s}\left(\mathrm{x}\right)\:\Rightarrow\mathrm{2s}\left(\mathrm{x}\right)=\frac{\mathrm{x}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}−\mathrm{1}\:\Rightarrow\mathrm{s}\left(\mathrm{x}\right)=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{x}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}−\mathrm{1}\right) \\ $$$$ \\ $$
Commented by mathmax by abdo last updated on 07/Jul/21
2)∫_0 ^1 s(x)dx=(1/2)∫_0 ^1 ((xdx)/( (√(1+x^2 ))))−(1/2) =(1/2)[(√(1+x^2 ))]_0 ^1 −(1/2) =(1/2)((√2)−1)−(1/2)=((√2)/2)−1
$$\left.\mathrm{2}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \mathrm{s}\left(\mathrm{x}\right)\mathrm{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\mathrm{xdx}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\right]_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{2}}\:=\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}−\mathrm{1} \\ $$

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