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Question Number 65285 by mathmax by abdo last updated on 27/Jul/19
let f(x) =e^(−x^2 ) ln(1−x) developp f at integr serie.
$${let}\:{f}\left({x}\right)\:={e}^{−{x}^{\mathrm{2}} } {ln}\left(\mathrm{1}−{x}\right) \\ $$$${developp}\:{f}\:{at}\:{integr}\:{serie}. \\ $$
Commented by mathmax by abdo last updated on 30/Jul/19
we have e^(−x^2 ) =Σ_(n=0) ^∞ (((−x^2 )^n )/(n!)) =Σ_(n=0) ^∞ (((−1)^n x^(2n) )/(n!)) ln^′ (1−x) =−(1/(1−x)) =−Σ_(n=0) ^∞ x^n ⇒ln(1−x) =−Σ_(n=0) ^∞ (x^(n+1) /(n+1)) +c (c=0)⇒ln(1−x) =−Σ_(n=1) ^∞ (x^n /n) ⇒ f(x) =−(Σ_(n=0) ^∞ (((−1)^n x^(2n) )/(n!)))(Σ_(n=1) ^∞ (x^n /n)) =−(1+Σ_(n=1) ^∞ (((−1)^n x^(2n) )/(n!)))(Σ_(n=1) ^∞ (x^n /n)) −Σ_(n=1) ^∞ (x^n /n) −(Σ_(n=1) ^∞ (((−1)^n x^(2n) )/(n!)))(Σ_(n=1) ^∞ (x^n /n)) (Σ_(n=1) ^∞ a_n )(Σ_(n=1) ^∞ b_n ) =Σ c_n with c_n =Σ_(i+j=n) a_i b_j =Σ_(i=1) ^(n−1) a_i b_(n−i) =Σ_(i=1) ^(n−1) (((−1)^i x^(2i) )/(i!)) (x^(n−i) /((n−i))) =Σ_(i=1) ^(n−1) (((−1)^i )/((n−i)i!)) x^(n+i) ⇒ f(x) =Σ_(n=1) ^∞ {Σ_(i=1) ^(n−1) (((−1)^i )/((n−i)i!))x^(n+i) }−Σ_(n=1) ^∞ (x^n /n) .
$${we}\:{have}\:{e}^{−{x}^{\mathrm{2}} } =\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{\left(−{x}^{\mathrm{2}} \right)^{{n}} }{{n}!}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}} }{{n}!} \\ $$$${ln}^{'} \left(\mathrm{1}−{x}\right)\:=−\frac{\mathrm{1}}{\mathrm{1}−{x}}\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \:{x}^{{n}} \:\Rightarrow{ln}\left(\mathrm{1}−{x}\right)\:=−\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{x}^{{n}+\mathrm{1}} }{{n}+\mathrm{1}}\:+{c} \\ $$$$\left({c}=\mathrm{0}\right)\Rightarrow{ln}\left(\mathrm{1}−{x}\right)\:=−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}\:\Rightarrow \\ $$$${f}\left({x}\right)\:=−\left(\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}} }{{n}!}\right)\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}\right) \\ $$$$=−\left(\mathrm{1}+\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} \:{x}^{\mathrm{2}{n}} }{{n}!}\right)\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}\right) \\ $$$$−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}\:−\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{\left(−\mathrm{1}\right)^{{n}} {x}^{\mathrm{2}{n}} }{{n}!}\right)\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}\right) \\ $$$$\left(\sum_{{n}=\mathrm{1}} ^{\infty} \:{a}_{{n}} \right)\left(\sum_{{n}=\mathrm{1}} ^{\infty} {b}_{{n}} \right)\:=\Sigma\:{c}_{{n}} \:\:\:{with}\:{c}_{{n}} =\sum_{{i}+{j}={n}} \:{a}_{{i}} {b}_{{j}} \:\:\:=\sum_{{i}=\mathrm{1}} ^{{n}−\mathrm{1}} {a}_{{i}} {b}_{{n}−{i}} \\ $$$$=\sum_{{i}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\frac{\left(−\mathrm{1}\right)^{{i}} {x}^{\mathrm{2}{i}} }{{i}!}\:\frac{{x}^{{n}−{i}} }{\left({n}−{i}\right)}\:=\sum_{{i}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\frac{\left(−\mathrm{1}\right)^{{i}} }{\left({n}−{i}\right){i}!}\:{x}^{{n}+{i}} \:\Rightarrow \\ $$$${f}\left({x}\right)\:=\sum_{{n}=\mathrm{1}} ^{\infty} \left\{\sum_{{i}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\frac{\left(−\mathrm{1}\right)^{{i}} }{\left({n}−{i}\right){i}!}{x}^{{n}+{i}} \right\}−\sum_{{n}=\mathrm{1}} ^{\infty} \:\frac{{x}^{{n}} }{{n}}\:. \\ $$$$ \\ $$

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