Previous in Relation and Functions Next in Relation and Functions
Question Number 33699 by math khazana by abdo last updated on 22/Apr/18
![let S(x)=Σ_(n=0) ^∞ f_n (x) with f_n (x)= (((−1)^n )/(n!(x+n))) x∈]0,+∞[ 1) prove that S id defined .calculate S(1) and prove that ∀x>0 xS(x) −S(x+1) =(1/e) 2) prove that S is C^∞ on R^(+∗) 3) prove that S(x) ∼ (1/x) (x→0^+ ) .](Q33699.png)
$${let}\:{S}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:{f}_{{n}} \left({x}\right)\:\:{with}\:{f}_{{n}} \left({x}\right)=\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!\left({x}+{n}\right)} \\ $$ $$\left.{x}\in\right]\mathrm{0},+\infty\left[\right. \\ $$ $$\left.\mathrm{1}\right)\:\:{prove}\:{that}\:{S}\:{id}\:{defined}\:.{calculate}\:{S}\left(\mathrm{1}\right)\:{and} \\ $$ $${prove}\:{that}\:\forall{x}>\mathrm{0}\:\:{xS}\left({x}\right)\:−{S}\left({x}+\mathrm{1}\right)\:=\frac{\mathrm{1}}{{e}} \\ $$ $$\left.\mathrm{2}\right)\:{prove}\:{that}\:{S}\:{is}\:{C}^{\infty} \:{on}\:{R}^{+\ast} \\ $$ $$\left.\mathrm{3}\right)\:{prove}\:{that}\:{S}\left({x}\right)\:\sim\:\frac{\mathrm{1}}{{x}}\:\left({x}\rightarrow\mathrm{0}^{+} \right)\:. \\ $$
Commented bymath khazana by abdo last updated on 29/Apr/18
![we have ∣f_n (x)∣= (1/(n!(x+n))) ≤ (1/(n(n!))) but the seSrie Σ (1/(n(n!))) is convergent so S is defined on]0,+∞[ S(1)=Σ_(n=0) ^∞ f_n (1) =Σ_(n=0) ^∞ (((−1)^n )/(n!(n+1))) let w(x) = Σ_(n=0) ^∞ (((−1)^n )/(n!(n+1))) x^(n+1) w^′ (x) = Σ_(n=0) ^∞ (((−1)^n x^n )/(n!)) = Σ_(n=0) ^∞ (((−x)^n )/(n!)) =e^(−x) ⇒ w(x)=−e^(−x) +λ we have w(0)=0 =−1 +λ ⇒ λ=1 ⇒w(x)=1−e^(−x) S(1)=w(1)= 1−(1/e) . xS(x) −S(x+1)=Σ_(n=0) ^∞ ((x(−1)^n )/(n!(x+n))) −Σ_(n=0) ^∞ (((−1)^n )/(n!(x+n+1))) =1+Σ_(n=1) ^∞ (((−1)^n )/(n!)) ( ((x+n −n)/(x+n))) −Σ_(n=0) ^∞ (((−1)^n )/(n!(x+n+1))) =Σ_(n=0) ^∞ (((−1)^n )/(n!)) −Σ_(n=1) ^∞ (((−1)^n )/((n−1)!(x+n))) −Σ_(n=0) ^∞ (((−1)^n )/(n!(x+n+1))) = (1/e) −Σ_(n=0) ^∞ (((−1)^(n+1) )/(n!(x+n +1))) −Σ_(n=0) ^∞ (((−1)^n )/(n!(x+n+1))) =(1/e) ⇒ xS(x) −S(x+1) =(1/e)](Q34015.png)
$${we}\:{have}\:\mid{f}_{{n}} \left({x}\right)\mid=\:\frac{\mathrm{1}}{{n}!\left({x}+{n}\right)}\:\leqslant\:\frac{\mathrm{1}}{{n}\left({n}!\right)}\:{but}\:{the}\:{seSrie} \\ $$ $$\left.\Sigma\:\:\frac{\mathrm{1}}{{n}\left({n}!\right)}\:{is}\:{convergent}\:{so}\:{S}\:{is}\:{defined}\:{on}\right]\mathrm{0},+\infty\left[\right. \\ $$ $${S}\left(\mathrm{1}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:{f}_{{n}} \left(\mathrm{1}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!\left({n}+\mathrm{1}\right)} \\ $$ $${let}\:{w}\left({x}\right)\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!\left({n}+\mathrm{1}\right)}\:{x}^{{n}+\mathrm{1}} \\ $$ $${w}^{'} \left({x}\right)\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{x}^{{n}} }{{n}!}\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−{x}\right)^{{n}} }{{n}!}\:={e}^{−{x}} \:\Rightarrow \\ $$ $${w}\left({x}\right)=−{e}^{−{x}} \:+\lambda\:\:{we}\:{have}\:{w}\left(\mathrm{0}\right)=\mathrm{0}\:=−\mathrm{1}\:+\lambda\:\Rightarrow \\ $$ $$\lambda=\mathrm{1}\:\Rightarrow{w}\left({x}\right)=\mathrm{1}−{e}^{−{x}} \\ $$ $${S}\left(\mathrm{1}\right)={w}\left(\mathrm{1}\right)=\:\mathrm{1}−\frac{\mathrm{1}}{{e}}\:. \\ $$ $${xS}\left({x}\right)\:−{S}\left({x}+\mathrm{1}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{x}\left(−\mathrm{1}\right)^{{n}} }{{n}!\left({x}+{n}\right)}\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!\left({x}+{n}+\mathrm{1}\right)} \\ $$ $$=\mathrm{1}+\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:\left(\:\frac{{x}+{n}\:−{n}}{{x}+{n}}\right)\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!\left({x}+{n}+\mathrm{1}\right)} \\ $$ $$=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:−\sum_{{n}=\mathrm{1}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}−\mathrm{1}\right)!\left({x}+{n}\right)}\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!\left({x}+{n}+\mathrm{1}\right)} \\ $$ $$=\:\frac{\mathrm{1}}{{e}}\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}+\mathrm{1}} }{{n}!\left({x}+{n}\:+\mathrm{1}\right)}\:−\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!\left({x}+{n}+\mathrm{1}\right)} \\ $$ $$=\frac{\mathrm{1}}{{e}}\:\Rightarrow\:{xS}\left({x}\right)\:−{S}\left({x}+\mathrm{1}\right)\:=\frac{\mathrm{1}}{{e}} \\ $$
Commented bymath khazana by abdo last updated on 29/Apr/18
![2) due to uniform convergence and f_n are C^∞ S will be C^ on ]0,+∞[ and S^((p)) (x) =Σ_(n=0) ^∞ f_n ^((p)) (x)= Σ_(n=0) ^∞ (((−1)^n )/(n!)) (((−1)^p p!)/((x+n)^(p+1) )) .](Q34016.png)
$$\left.\mathrm{2}\right)\:{due}\:{to}\:{uniform}\:{convergence}\:{and}\:{f}_{{n}} \:{are}\:{C}^{\infty} \\ $$ $$\left.{S}\:{will}\:{be}\:{C}^{} \:{on}\:\right]\mathrm{0},+\infty\left[\:{and}\right. \\ $$ $${S}^{\left({p}\right)} \left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:{f}_{{n}} ^{\left({p}\right)} \left({x}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:\:\frac{\left(−\mathrm{1}\right)^{{p}} {p}!}{\left({x}+{n}\right)^{{p}+\mathrm{1}} }\:. \\ $$ $$ \\ $$
Commented bymath khazana by abdo last updated on 29/Apr/18
$$\left.\mathrm{3}\right)\:{we}\:{have}\:{proved}\:{that}\:\:{xS}\left({x}\right)−{S}\left({x}+\mathrm{1}\right)=\frac{\mathrm{1}}{{e}}\:\Rightarrow \\ $$ $${x}\:{S}\left({x}\right)\:−{S}\left(\mathrm{1}\right)\sim\:\frac{\mathrm{1}}{{e}}\:\:\left({x}\rightarrow\mathrm{0}^{+} \right)\:\Rightarrow \\ $$ $${xS}\left({x}\right)\:−\mathrm{1}+\frac{\mathrm{1}}{{e}}\:\sim\:\frac{\mathrm{1}}{{e}}\:\Rightarrow\:{S}\left({x}\right)\sim\:\frac{\mathrm{1}}{{x}}\:\left({x}\rightarrow\mathrm{0}^{+} \right)\:. \\ $$