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Question Number 27790 by abdo imad last updated on 14/Jan/18
let give f(x)= x^(n−1) ln(1+x) with n fromN^∗ find f^((n)) (x)
letgivef(x)=xn−1ln(1+x)withnfromN∗findf(n)(x)
Commented by abdo imad last updated on 15/Jan/18
by leibnitz formule f^((n)) (x)= Σ_(k=0) ^n (x^(n−1) )^((k)) (ln(1+x))^((n−k)) and for p≤n (x^n )^((p)) =n(n−1)...(n−p+1)x^(n−p) so for k∈[[0,n−1]] (x^(n−1) )^k =(n−1)(n−2)....(n−1−k +1)x^(n−1−k) =(n−1)(n−2)...(n−k) x^(n−1−k) and we have ln(1+x)^((p)) =(((−1)^(p−1) (p−1)!)/((1+x)^p )) for p≥1 ⇒ f^((n)) (x)= Σ_(k=0) ^(n−1) (n−1)(n−2)...(n−k) x^(n−1−k) (((−1)^(n−k−1) (n−k−1)!)/((1+x)^(n−k) )) ( (x^(n−1) )^((n)) =0)
byleibnitzformulef(n)(x)=∑k=0n(xn−1)(k)(ln(1+x))(n−k)andforp⩽n(xn)(p)=n(n−1)…(n−p+1)xn−psofork∈[[0,n−1]](xn−1)k=(n−1)(n−2)….(n−1−k+1)xn−1−k=(n−1)(n−2)…(n−k)xn−1−kandwehaveln(1+x)(p)=(−1)p−1(p−1)!(1+x)pforp⩾1⇒f(n)(x)=∑k=0n−1(n−1)(n−2)…(n−k)xn−1−k(−1)n−k−1(n−k−1)!(1+x)n−k((xn−1)(n)=0)
Commented by abdo imad last updated on 15/Jan/18
f^((n)) (x)= Σ_(k=0) ^n C_n ^k (x^(n−1) )^((k)) (ln(1+x)^((n−k)) and f^((n)) (x)= Σ_(k=0) ^(n−1) C_n ^k (n−1)(n−2)...(n−k)x^(n−1−k) (((−1)^(n−k−1) (n−k−1)!)/((1+x)^(n−k) ))
f(n)(x)=∑k=0nCnk(xn−1)(k)(ln(1+x)(n−k)andf(n)(x)=∑k=0n−1Cnk(n−1)(n−2)…(n−k)xn−1−k(−1)n−k−1(n−k−1)!(1+x)n−k

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