Question Number 27790 by abdo imad last updated on 14/Jan/18
$${let}\:{give}\:\:{f}\left({x}\right)=\:{x}^{{n}−\mathrm{1}} \:{ln}\left(\mathrm{1}+{x}\right)\:\:{with}\:{n}\:{fromN}^{\ast} \:\:\:{find}\:{f}^{\left({n}\right)} \left({x}\right) \\ $$
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)](https://www.tinkutara.com/question/Q27831.png)
$$\:{by}\:{leibnitz}\:{formule}\:{f}^{\left({n}\right)} \left({x}\right)=\:\sum_{{k}=\mathrm{0}} ^{{n}} \left({x}^{{n}−\mathrm{1}} \right)^{\left({k}\right)} \left({ln}\left(\mathrm{1}+{x}\right)\right)^{\left({n}−{k}\right)} \\ $$$$\:{and}\:{for}\:{p}\leqslant{n}\:\:\:\:\left({x}^{{n}} \right)^{\left({p}\right)} ={n}\left({n}−\mathrm{1}\right)…\left({n}−{p}+\mathrm{1}\right){x}^{{n}−{p}} \:{so}\:{for} \\ $$$${k}\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right]\:\:\:\:\left({x}^{{n}−\mathrm{1}} \right)^{{k}} =\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)….\left({n}−\mathrm{1}−{k}\:+\mathrm{1}\right){x}^{{n}−\mathrm{1}−{k}} \\ $$$$=\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)…\left({n}−{k}\right)\:{x}^{{n}−\mathrm{1}−{k}} \:\: \\ $$$${and}\:{we}\:{have}\:\:\:{ln}\left(\mathrm{1}+{x}\right)^{\left({p}\right)} =\frac{\left(−\mathrm{1}\right)^{{p}−\mathrm{1}} \left({p}−\mathrm{1}\right)!}{\left(\mathrm{1}+{x}\right)^{{p}} }\:{for}\:{p}\geqslant\mathrm{1} \\ $$$$\Rightarrow\:{f}^{\left({n}\right)} \left({x}\right)=\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)…\left({n}−{k}\right)\:{x}^{{n}−\mathrm{1}−{k}} \:\frac{\left(−\mathrm{1}\right)^{{n}−{k}−\mathrm{1}} \left({n}−{k}−\mathrm{1}\right)!}{\left(\mathrm{1}+{x}\right)^{{n}−{k}} } \\ $$$$\left(\:\:\:\:\left({x}^{{n}−\mathrm{1}} \right)^{\left({n}\right)} =\mathrm{0}\right) \\ $$$$ \\ $$
Commented by abdo imad last updated on 15/Jan/18
$${f}^{\left({n}\right)} \left({x}\right)=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\left({x}^{{n}−\mathrm{1}} \right)^{\left({k}\right)} \:\left({ln}\left(\mathrm{1}+{x}\right)^{\left({n}−{k}\right)} \right. \\ $$$${and}\:{f}^{\left({n}\right)} \left({x}\right)=\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{C}_{{n}} ^{{k}} \:\left({n}−\mathrm{1}\right)\left({n}−\mathrm{2}\right)…\left({n}−{k}\right){x}^{{n}−\mathrm{1}−{k}} \:\frac{\left(−\mathrm{1}\right)^{{n}−{k}−\mathrm{1}} \left({n}−{k}−\mathrm{1}\right)!}{\left(\mathrm{1}+{x}\right)^{{n}−{k}} } \\ $$