Question Number 37901 by math khazana by abdo last updated on 19/Jun/18
$${let}\:{f}\left({x}\right)=\:\left(\mathrm{1}+{e}^{−{x}} \right)^{{n}} \\ $$$$\left.\mathrm{1}\right)\:{calculate}\:{f}^{\left({p}\right)} \left({x}\right)\:\:{and}\:{f}^{\left({p}\right)} \left({o}\right) \\ $$$$\left.\mathrm{2}\right){calculate}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{3}\right){developp}\:{f}\:{at}\:{integr}\:{serie}\:. \\ $$
Commented by math khazana by abdo last updated on 21/Jun/18
$${we}\:{have}\:{f}\left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:{e}^{−{kx}} \:\Rightarrow \\ $$$${f}^{\left({p}\right)} \left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:\left({e}^{−{kx}} \right)^{\left({p}\right)} \:\:{but} \\ $$$$\left({e}^{−{kx}} \right)^{\left(\mathrm{1}\right)} =−{k}\:{e}^{−{kx}} \:{and}\:\left({e}^{−{kx}} \right)^{\left(\mathrm{2}\right)} =\left(−{k}^{} \right)^{\mathrm{2}} \:{e}^{−{kx}} \\ $$$$\left({e}^{−{kx}} \right)^{\left({p}\right)} =\left(−{k}\right)^{{p}} \:{e}^{−{kx}} \:\Rightarrow \\ $$$${f}^{\left({p}\right)} \left({x}\right)=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:\left(−{k}\right)^{{p}} \:{e}^{−{kx}} \:\Rightarrow \\ $$$${f}^{\left({p}\right)} \left(\mathrm{0}\right)\:=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\left(−{k}\right)^{{p}} \:{C}_{{n}} ^{{k}} \:\: \\ $$$$\left.\mathrm{2}\right)\:{f}^{\left({n}\right)} \left(\mathrm{0}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} \left(−{k}\right)^{{n}} \:{C}_{{n}} ^{{k}} \\ $$$$\left.\mathrm{3}\right){f}\left({x}\right)=\sum_{{p}=\mathrm{0}} ^{\infty} \:\frac{{f}^{\left({p}\right)} \left(\mathrm{0}\right)}{{p}!}\:{x}^{{p}} \\ $$$$=\sum_{{p}=\mathrm{0}} ^{\infty} \:\:\:\frac{\mathrm{1}}{{p}!}\left\{\:\sum_{{k}=\mathrm{0}} ^{{n}} \left(−{k}\right)^{{p}} \:{C}_{{n}} ^{{k}} \right\}\:{x}^{{p}\:} \:\:\:. \\ $$