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Question Number 34849 by math khazana by abdo last updated on 11/May/18
let f(x) = (e^(−x) /(2+x)) developp f at integr serie.
$${let}\:{f}\left({x}\right)\:=\:\:\:\:\frac{{e}^{−{x}} }{\mathrm{2}+{x}} \\ $$$${developp}\:{f}\:{at}\:{integr}\:{serie}. \\ $$
Commented by math khazana by abdo last updated on 13/May/18
we have f(x)= (e^(−x) /(2(1+(x/2)))) ⇒ for ∣x∣<2 2f(x)= (Σ_(n=0) ^∞ (−1)^n ((x/2))^n ) .(Σ_(n=0) ^∞ (((−1)^n x^n )/(n!))) = (Σ_(n=0) ^∞ (((−1)^n )/2^n ) x^n ).(Σ_(n=0) ^∞ (((−1)^n )/(n!)) x^n ) =Σ_(n=0) ^∞ c_n x^n with c_n = Σ_(i+j =n) a_i b_j = Σ_(i+j =n) (((−1)^i )/2^i ) (((−1)^j )/(j!)) =Σ_(i=0) ^n a_i b_(n−i) = Σ_(i=0) ^n (((−1)^i )/2^i ) (((−1)^(n−i) )/((n−i)!)) =Σ_(i=0) ^n (((−1)^n )/((n−i)!2^i )) ⇒ f(x)= (1/2) Σ_(n=0) ^∞ ( Σ_(i=0) ^n (((−1)^n )/(2^i (n−i)!)))x^n .
$${we}\:{have}\:{f}\left({x}\right)=\:\frac{{e}^{−{x}} }{\mathrm{2}\left(\mathrm{1}+\frac{{x}}{\mathrm{2}}\right)}\:\Rightarrow\:{for}\:\mid{x}\mid<\mathrm{2}\: \\ $$$$\mathrm{2}{f}\left({x}\right)=\:\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(−\mathrm{1}\right)^{{n}} \left(\frac{{x}}{\mathrm{2}}\right)^{{n}} \right)\:.\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} \:{x}^{{n}} }{{n}!}\right) \\ $$$$=\:\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{n}} }\:{x}^{{n}} \:\right).\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:{x}^{{n}} \right) \\ $$$$=\sum_{{n}=\mathrm{0}} ^{\infty} \:{c}_{{n}} \:{x}^{{n}} \:\:\:{with}\:{c}_{{n}} =\:\sum_{{i}+{j}\:={n}} \:{a}_{{i}} \:{b}_{{j}} \\ $$$$=\:\sum_{{i}+{j}\:={n}} \:\frac{\left(−\mathrm{1}\right)^{{i}} }{\mathrm{2}^{{i}} }\:\frac{\left(−\mathrm{1}\right)^{{j}} }{{j}!}\:=\sum_{{i}=\mathrm{0}} ^{{n}} \:{a}_{{i}} \:{b}_{{n}−{i}} \\ $$$$=\:\sum_{{i}=\mathrm{0}} ^{{n}} \:\:\:\frac{\left(−\mathrm{1}\right)^{{i}} }{\mathrm{2}^{{i}} }\:\frac{\left(−\mathrm{1}\right)^{{n}−{i}} }{\left({n}−{i}\right)!}\:=\sum_{{i}=\mathrm{0}} ^{{n}} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}−{i}\right)!\mathrm{2}^{{i}} }\:\Rightarrow \\ $$$${f}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}}\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(\:\sum_{{i}=\mathrm{0}} ^{{n}} \:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{i}} \left({n}−{i}\right)!}\right){x}^{{n}} \:. \\ $$$$ \\ $$
Commented by math khazana by abdo last updated on 13/May/18
another method f(x)= Σ_(n=0) ^∞ ((f^((n)) (0))/(n!)) x^n let calculate f^((n)) (0) leibnitz formula give f^((n)) (x)= Σ_(k=0) ^n C_n ^k ((1/(2+x)))^((k)) (e^(−x) )^((n−k)) =Σ_(k=0) ^n C_n ^k (((−1)^k k!)/((x+2)^(k+1) )) (−1)^(n−k) e^(−x) ⇒ f^((n)) (0) = Σ_(k=0) ^n C_n ^k ((k! (−1)^n )/2^(k+1) ) ⇒ f(x) = Σ_(n=0) ^∞ (Σ_(k=0) ^n C_n ^k ((k!)/(n!)) (((−1)^n )/2^(k+1) )) x^n but C_n ^k .((k!)/(n!)) = ((n!)/(k!(n−k)!)) ((k!)/(n!)) = (1/((n−k)!)) ⇒ f(x) = Σ_(n=0) ^∞ (Σ_(k=0) ^n (((−1)^n )/((n−k)! 2^(k+1) )))x^n .
$${another}\:{method} \\ $$$${f}\left({x}\right)=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \:\:\:{let}\:{calculate}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$${leibnitz}\:{formula}\:{give}\:{f}^{\left({n}\right)} \left({x}\right)=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left(\frac{\mathrm{1}}{\mathrm{2}+{x}}\right)^{\left({k}\right)} \:\left({e}^{−{x}} \right)^{\left({n}−{k}\right)} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \:\:\:\frac{\left(−\mathrm{1}\right)^{{k}} \:{k}!}{\left({x}+\mathrm{2}\right)^{{k}+\mathrm{1}} }\:\:\left(−\mathrm{1}\right)^{{n}−{k}} \:{e}^{−{x}} \:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \:\:\frac{{k}!\:\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{k}+\mathrm{1}} }\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \left(\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \:\:\:\:\:\:\frac{{k}!}{{n}!}\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}^{{k}+\mathrm{1}} }\right)\:{x}^{{n}} \:\:{but} \\ $$$${C}_{{n}} ^{{k}} \:\:.\frac{{k}!}{{n}!}\:\:=\:\frac{{n}!}{{k}!\left({n}−{k}\right)!}\:\frac{{k}!}{{n}!}\:=\:\frac{\mathrm{1}}{\left({n}−{k}\right)!}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=\:\sum_{{n}=\mathrm{0}} ^{\infty} \:\left(\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\:\frac{\left(−\mathrm{1}\right)^{{n}} }{\left({n}−{k}\right)!\:\mathrm{2}^{{k}+\mathrm{1}} }\right){x}^{{n}} \:\:. \\ $$

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