Menu Close

Find-the-Talor-series-of-ln-1-x-1-x-2-at-x-0-

Tinku Tara 0 Comments
Pin




Question Number 148720 by qaz last updated on 30/Jul/21
Find the Talor series of ((ln(1−x))/((1−x)^2 )) at x=0.
FindtheTalorseriesofln(1−x)(1−x)2atx=0.
Answered by mathmax by abdo last updated on 30/Jul/21
f(x)=((ln(1−x))/((1−x)^2 )) we have Σ_(n=0) ^∞ x^n =(1/(1−x)) if ∣x∣<1 ⇒ Σ_(n=1) ^∞ nx^(n−1) =(1/((1−x)^2 )) ⇒(1/((1−x)^2 ))=Σ_(n=0) ^∞ (n+1)x^n (d/dx)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) =−Σ_(n=1) ^∞ (x^n /n) ⇒f(x)=−(Σ_(n=0) ^∞ (n+1)x^n )(Σ_(n=1) ^∞ (1/n)x^n ) =−Σ_(n=1) ^∞ (x^n /n)−(Σ_(n=1) ^∞ (n+1)x^n )(Σ_(n=1) ^∞ (1/n)x^n ) =−Σ_(n=1) ^∞ (x^n /n)−Σ_(n=1) ^∞ C_n x^n C_n =Σ_(i+j=n) a_i b_j =Σ_(i=1) ^(n−1) (i+1)×(1/(n−i)) ⇒ f(x)=−Σ_(n=1) ^∞ (x^n /n)−Σ_(n=1) ^∞ (Σ_(i=1) ^n ((i+1)/(n−i)))x^n
f(x)=ln(1−x)(1−x)2wehave∑n=0∞xn=11−xif∣x∣<1⇒∑n=1∞nxn−1=1(1−x)2⇒1(1−x)2=∑n=0∞(n+1)xnddxln(1−x)=−11−x=−∑n=0∞xn⇒ln(1−x)=−∑n=0∞xn+1n+1+c(c=0)=−∑n=1∞xnn⇒f(x)=−(∑n=0∞(n+1)xn)(∑n=1∞1nxn)=−∑n=1∞xnn−(∑n=1∞(n+1)xn)(∑n=1∞1nxn)=−∑n=1∞xnn−∑n=1∞CnxnCn=∑i+j=naibj=∑i=1n−1(i+1)×1n−i⇒f(x)=−∑n=1∞xnn−∑n=1∞(∑i=1ni+1n−i)xn
Commented by mathmax by abdo last updated on 30/Jul/21
f(x)=−Σ_(n=1) ^∞ (x^n /n)−Σ_(n=1) ^∞ (Σ_(i=1) ^(n−1) ((i+1)/(n−i)))x^n
f(x)=−∑n=1∞xnn−∑n=1∞(∑i=1n−1i+1n−i)xn
Answered by Kamel last updated on 30/Jul/21
Find the Talor series of ((ln(1−x))/((1−x)^2 )) at x=0. f(x)=((Ln(1−x))/((1−x)^2 )) (1/((1−ax)(1−bx)))=(1/(a−b))((a/(1−ax))−(b/(1−bx))) =(1/(a−b))Σ_(n=0) ^(+∞) (a^(n+1) −b^(n+1) )x^n ∴ f(x)=−Σ_(n=0) ^(+∞) ∫_0 ^1 (((n+1)(1−b)−1+b^(n+1) )/((1−b)^2 ))dbx^n =−Σ_(n=0) ^(+∞) (−n+(n+1)∫_0 ^1 ((1−b^n )/(1−b)))x^n ∴ ((Ln(1−x))/((1−x)^2 )) =Σ_(n=0) ^(+∞) (n−(n+1)H_n )x^n
FindtheTalorseriesofln(1−x)(1−x)2atx=0.f(x)=Ln(1−x)(1−x)21(1−ax)(1−bx)=1a−b(a1−ax−b1−bx)=1a−b∑+∞n=0(an+1−bn+1)xn∴f(x)=−∑+∞n=0∫01(n+1)(1−b)−1+bn+1(1−b)2dbxn=−∑+∞n=0(−n+(n+1)∫011−bn1−b)xn∴Ln(1−x)(1−x)2=∑+∞n=0(n−(n+1)Hn)xn
Answered by mathmax by abdo last updated on 30/Jul/21
another way f(x)=((ln(1−x))/((1−x)^2 )) f(x)=Σ_(n=0) ^∞ ((f^((n)) (0))/(n!))x^n =f(0) +Σ_(n=1) ^∞ ((f^((n)) (0))/(n!))x^n =Σ_(n=1) ^∞ ((f^((n)) (0))/(n!))x^n but f^((n)) (0)? f^((n)) (x)=Σ_(k=0) ^n C_n ^k (ln(1−x))^((k)) ×((1/((x−1)^2 )))^((n−k)) we have (ln(1−x))^((1)) =−(1/(1−x))=(1/(x−1)) ⇒ (ln(1−x))^((k)) =(((−1)^(k−1) (k−1)!)/((x−1)^k )) also ((1/((x−1)^2 )))^((n−k)) =(((−1)^(n−k) (n−k)!)/((x−1)^(n−k+1) )) ⇒ f^((n)) (x)=(((−1)^n n!)/((x−1)^(n+1) ))ln(1−x)+Σ_(k=1) ^n C_n ^k (((−1)^(k−1) (k−1)!)/((x−1)^k ))×(((−1)^(n−k) (n−k)!)/((x−1)^(n−k+1) )) ⇒f^((n)) (0)=Σ_(k=1) ^n (((−1)^(k−1) (k−1)!)/((−1)^k ))×(((−1)^(n−k) (n−k)!)/((−1)^(n−k+1) )) =Σ_(k=1) ^n (((k−1)!(n−k)!)/1) ⇒ f(x)=Σ_(n=1) ^∞ (Σ_(k=1) ^n (((k−1)!(n−k)!)/(n!)))x^n
anotherwayf(x)=ln(1−x)(1−x)2f(x)=∑n=0∞f(n)(0)n!xn=f(0)+∑n=1∞f(n)(0)n!xn=∑n=1∞f(n)(0)n!xnbutf(n)(0)?f(n)(x)=∑k=0nCnk(ln(1−x))(k)×(1(x−1)2)(n−k)wehave(ln(1−x))(1)=−11−x=1x−1⇒(ln(1−x))(k)=(−1)k−1(k−1)!(x−1)kalso(1(x−1)2)(n−k)=(−1)n−k(n−k)!(x−1)n−k+1⇒f(n)(x)=(−1)nn!(x−1)n+1ln(1−x)+∑k=1nCnk(−1)k−1(k−1)!(x−1)k×(−1)n−k(n−k)!(x−1)n−k+1⇒f(n)(0)=∑k=1n(−1)k−1(k−1)!(−1)k×(−1)n−k(n−k)!(−1)n−k+1=∑k=1n(k−1)!(n−k)!1⇒f(x)=∑n=1∞(∑k=1n(k−1)!(n−k)!n!)xn
Answered by qaz last updated on 30/Jul/21
((ln(1−x))/((1−x)^2 )) =−(Σ_(k=1) ^∞ (x^k /k))(Σ_(k=1) ^∞ x^(k−1) )(Σ_(k=1) ^∞ x^(k−1) ) =−(Σ_(k=1) ^∞ (x^k /k))(Σ_(n=1) ^∞ Σ_(k=1) ^n x^(k−1) ∙x^(n+1−k−1) ) =−(Σ_(k=1) ^∞ (x^k /k))(Σ_(n=1) ^∞ nx^(n−1) ) =−Σ_(n=1) ^∞ Σ_(k=1) ^n (x^k /k)∙(n+1−k)∙x^(n+1−k−1) =−Σ_(n=1) ^∞ Σ_(k=1) ^n (((n+1)/k)−1)∙x^n =Σ_(n=1) ^∞ [n−(n+1)H_n ]x^n
ln(1−x)(1−x)2=−(∑∞k=1xkk)(∑∞k=1xk−1)(∑∞k=1xk−1)=−(∑∞k=1xkk)(∑∞n=1∑nk=1xk−1⋅xn+1−k−1)=−(∑∞k=1xkk)(∑∞n=1nxn−1)=−∑∞n=1∑nk=1xkk⋅(n+1−k)⋅xn+1−k−1=−∑∞n=1∑nk=1(n+1k−1)⋅xn=∑∞n=1[n−(n+1)Hn]xn

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *