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.

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

Commented bymathmax 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

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

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

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

Commented byqaz last updated on 30/Jul/21

I found... ∫_a ^b f(x)dx=∫_a ^b f(a+b−x)dx Σ_a ^b f(n)=Σ_a ^b f(a+b−n)