Question Number 46421 by maxmathsup by imad last updated on 25/Oct/18
$${let}\:{f}\left({x}\right)=\left({x}+\frac{\mathrm{1}}{{x}}\right)^{{n}} −\left({x}−\frac{\mathrm{1}}{{x}}\right)^{{n}} \:{with}\:{n}\:{integr}\:{natural}\:{and}\:{x}\:{from}\:{R}\:\left({x}\neq\mathrm{0}\right) \\ $$$$\left.\mathrm{1}\right)\:{simplify}\:{f}\left({x}\right) \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:{lim}_{{x}\rightarrow+\infty} {f}\left({x}\right) \\ $$$$\left.\mathrm{3}\right)\:{calculate}\:\:\int_{\mathrm{1}} ^{\mathrm{3}} \:{f}\left({x}\right){dx} \\ $$
Commented by maxmathsup by imad last updated on 27/Oct/18
![1) the binome formulae give f(x)=Σ_(k=0) ^n C_n ^k x^k (1/x^(n−k) ) −Σ_(k=0) ^n C_n ^k x^k (((−1)^(n−k) )/x^(n−k) ) = Σ_(k=0) ^n C_n ^k x^(2k−n) −Σ_(k.=0) ^n C_n ^k (−1)^(n−k) x^(2k−n) =Σ_(k=0) ^n C_n ^k (1−(−1)^(n−k) )x^(2k−n) =Σ_(k=0) ^n (1−(−1)^(n−k) )C_n ^k x^(2k−n) =_(n−k =p) Σ_(p=0) ^n (1−(−1)^p ) C_n ^(n−p) x^(2(n−p)−n) =Σ_(p=0) ^n (1−(−1)^p ) C_n ^p x^(n−2p) =Σ_(k=0) ^([((n−1)/2)]) 2 C_n ^(2k+1) x^(n−2(2k+1)) = 2 Σ_(k=0) ^([((n−1)/2)]) C_n ^(2k+1) x^(n−4k−2) .](https://www.tinkutara.com/question/Q46518.png)
$$\left.\mathrm{1}\right)\:{the}\:{binome}\:{formulae}\:{give}\:{f}\left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{x}^{{k}} \:\frac{\mathrm{1}}{{x}^{{n}−{k}} }\:−\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{x}^{{k}} \:\frac{\left(−\mathrm{1}\right)^{{n}−{k}} }{{x}^{{n}−{k}} } \\ $$$$=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:{x}^{\mathrm{2}{k}−{n}} \:−\sum_{{k}.=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left(−\mathrm{1}\right)^{{n}−{k}} \:{x}^{\mathrm{2}{k}−{n}} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \left(\mathrm{1}−\left(−\mathrm{1}\right)^{{n}−{k}} \right){x}^{\mathrm{2}{k}−{n}} \:=\sum_{{k}=\mathrm{0}} ^{{n}} \left(\mathrm{1}−\left(−\mathrm{1}\right)^{{n}−{k}} \right){C}_{{n}} ^{{k}} \:{x}^{\mathrm{2}{k}−{n}} \\ $$$$=_{{n}−{k}\:={p}} \:\:\:\sum_{{p}=\mathrm{0}} ^{{n}} \:\left(\mathrm{1}−\left(−\mathrm{1}\right)^{{p}} \right)\:{C}_{{n}} ^{{n}−{p}} \:\:{x}^{\mathrm{2}\left({n}−{p}\right)−{n}} \\ $$$$=\sum_{{p}=\mathrm{0}} ^{{n}} \:\left(\mathrm{1}−\left(−\mathrm{1}\right)^{{p}} \right)\:{C}_{{n}} ^{{p}} \:\:{x}^{{n}−\mathrm{2}{p}} \:\:=\sum_{{k}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\mathrm{2}\:{C}_{{n}} ^{\mathrm{2}{k}+\mathrm{1}} \:\:{x}^{{n}−\mathrm{2}\left(\mathrm{2}{k}+\mathrm{1}\right)} \\ $$$$=\:\mathrm{2}\:\sum_{{k}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:{C}_{{n}} ^{\mathrm{2}{k}+\mathrm{1}} \:\:{x}^{{n}−\mathrm{4}{k}−\mathrm{2}} \:. \\ $$$$ \\ $$
Commented by maxmathsup by imad last updated on 27/Oct/18
$$\left.\mathrm{2}\right)\:{we}\:{have}\:{f}\left({x}\right)={x}^{{n}} \left\{\:\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{{n}} \:−\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{{n}} \right\}\:{but}\: \\ $$$$\left(\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{{n}} \:\:\sim\:\mathrm{1}+\frac{{n}}{{x}^{\mathrm{2}} }\:\:{and}\:\left(\mathrm{1}−\frac{\mathrm{1}}{{x}^{\mathrm{2}} }\right)^{{n}} \sim\:\mathrm{1}−\frac{{n}}{{x}^{\mathrm{2}} }\:\left({x}\rightarrow\infty\right)\:\Rightarrow \\ $$$${f}\left({x}\right)\:\sim\:{x}^{{n}} \:\left\{\:\frac{\mathrm{2}{n}}{{x}^{\mathrm{2}} }\right\}\:\:\Rightarrow{f}\left({x}\right)\sim\:\mathrm{2}{n}\:{x}^{{n}−\mathrm{2}} \: \\ $$$${n}>\mathrm{2}\:\Rightarrow{lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)=+\infty \\ $$$${n}=\mathrm{2}\:\:\Rightarrow{lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)=\mathrm{2}{n}\:=\mathrm{4} \\ $$$${n}=\mathrm{0}\:\Rightarrow{f}\left({x}\right)=\mathrm{0} \\ $$$${n}=\mathrm{1}\:\Rightarrow\:{lim}_{{x}\rightarrow+\infty} {f}\left({x}\right)=\mathrm{0} \\ $$
Commented by maxmathsup by imad last updated on 27/Oct/18
![3) we have f(x)=2 Σ_(k=0) ^([((n−1)/2)]) C_n ^(2k+1) x^(n−4k−2) ⇒ ∫_1 ^3 f(x)dx =2 Σ_(k=0) ^([((n−1)/2)]) C_n ^(2k+1) [(1/(n−4k−1)) x^(n−4k−1) ]_1 ^3 =2 Σ_(k=0) ^([((n−1)/2)]) (C_n ^(2k+1) /(n−4k−1)){3^(n−4k−1) −1} .](https://www.tinkutara.com/question/Q46521.png)
$$\left.\mathrm{3}\right)\:{we}\:{have}\:{f}\left({x}\right)=\mathrm{2}\:\sum_{{k}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:{C}_{{n}} ^{\mathrm{2}{k}+\mathrm{1}} \:{x}^{{n}−\mathrm{4}{k}−\mathrm{2}} \:\Rightarrow \\ $$$$\int_{\mathrm{1}} ^{\mathrm{3}} {f}\left({x}\right){dx}\:=\mathrm{2}\:\sum_{{k}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:{C}_{{n}} ^{\mathrm{2}{k}+\mathrm{1}} \:\left[\frac{\mathrm{1}}{{n}−\mathrm{4}{k}−\mathrm{1}}\:{x}^{{n}−\mathrm{4}{k}−\mathrm{1}} \right]_{\mathrm{1}} ^{\mathrm{3}} \\ $$$$=\mathrm{2}\:\sum_{{k}=\mathrm{0}} ^{\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]} \:\:\:\frac{{C}_{{n}} ^{\mathrm{2}{k}+\mathrm{1}} }{{n}−\mathrm{4}{k}−\mathrm{1}}\left\{\mathrm{3}^{{n}−\mathrm{4}{k}−\mathrm{1}} \:−\mathrm{1}\right\}\:. \\ $$