Question Number 31499 by abdo imad last updated on 09/Mar/18
$${find}\:{the}\:{polynial}\:{p}\:{wich}\:{verify}\:{p}\left({x}\right)−{p}^{'} \left({x}\right)={x}^{{n}} \:{then} \\ $$$${calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} {p}\left({x}\right){dx}. \\ $$
Commented by abdo imad last updated on 14/Mar/18
![let put p(x)=Σ_(k=0) ^n a_k x^k we have p^′ (x)=Σ_(k=1) ^n k a_k x^(k−1) p(x)−p^′ (x)=x^n ⇔Σ_(k=0) ^n a_k x^k −Σ_(k=1) ^n k a_k x^(k−1) =x^n ⇔ Σ_(k=0) ^n a_k x^k −Σ_(k=0) ^(n−1) (k+1)a_(k+1) x^k =x^n ⇔ Σ_(k=0) ^(n−1) (a_k −(k+1)a_(k+1) )x^k +a_n x^n =x^n ⇔ a_k −(k+1)a_(k+1) =0 ∀k∈[[0,n−1]] and a_n =1 ⇔a_n =1 and a_(k+1) = (a_k /(k+1)) ⇒ Π_(k=0) ^(n−1) a_(k+1) = ((Π_(k=0) ^(n−1) a_k )/(Π_(k=0) ^(n−1) (k+1))) ⇒ a_1 .a_2 ...a_n =(1/(n!)) a_0 .a_1 .a_2 ...a_(n−1) ⇒a_n = (a_0 /(n!)) ⇒ p(x)= Σ_(k=0) ^(n−1) (a_0 /(k!)) x^k +x^n p(o)−p^′ (0)=0 ⇒a_0 =0 ⇒ p(x)= Σ_(k=1) ^(n−1) (x^k /(k!)) +x^n 2) ∫_0 ^1 p(x)dx=Σ_(k=1) ^(n−1) (1/(k!)) ∫_0 ^1 x^k dx +(1/(n+1)) =Σ_(k=1) ^(n−1) (1/((k+1)!)) +(1/(n+1))](https://www.tinkutara.com/question/Q31782.png)
$${let}\:{put}\:{p}\left({x}\right)=\sum_{{k}=\mathrm{0}} ^{{n}} {a}_{{k}} {x}^{{k}} \:{we}\:{have}\:{p}^{'} \left({x}\right)=\sum_{{k}=\mathrm{1}} ^{{n}} {k}\:{a}_{{k}} {x}^{{k}−\mathrm{1}} \\ $$$${p}\left({x}\right)−{p}^{'} \left({x}\right)={x}^{{n}} \:\Leftrightarrow\sum_{{k}=\mathrm{0}} ^{{n}} {a}_{{k}} {x}^{{k}} \:−\sum_{{k}=\mathrm{1}} ^{{n}} {k}\:{a}_{{k}} \:{x}^{{k}−\mathrm{1}} \:={x}^{{n}} \\ $$$$\Leftrightarrow\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{a}_{{k}} {x}^{{k}} \:\:−\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left({k}+\mathrm{1}\right){a}_{{k}+\mathrm{1}} \:{x}^{{k}} \:={x}^{{n}} \:\Leftrightarrow \\ $$$$\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\left({a}_{{k}} \:−\left({k}+\mathrm{1}\right){a}_{{k}+\mathrm{1}} \right){x}^{{k}} \:\:+{a}_{{n}} {x}^{{n}} ={x}^{{n}} \:\Leftrightarrow \\ $$$$\:{a}_{{k}} \:−\left({k}+\mathrm{1}\right){a}_{{k}+\mathrm{1}} =\mathrm{0}\:\forall{k}\in\left[\left[\mathrm{0},{n}−\mathrm{1}\right]\right]\:{and}\:{a}_{{n}} =\mathrm{1}\:\Leftrightarrow{a}_{{n}} =\mathrm{1}\:{and} \\ $$$${a}_{{k}+\mathrm{1}} =\:\frac{{a}_{{k}} }{{k}+\mathrm{1}}\:\Rightarrow\:\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{a}_{{k}+\mathrm{1}} =\:\frac{\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:{a}_{{k}} }{\prod_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \left({k}+\mathrm{1}\right)}\:\Rightarrow \\ $$$${a}_{\mathrm{1}} .{a}_{\mathrm{2}} …{a}_{{n}} =\frac{\mathrm{1}}{{n}!}\:{a}_{\mathrm{0}} .{a}_{\mathrm{1}} .{a}_{\mathrm{2}} …{a}_{{n}−\mathrm{1}} \:\Rightarrow{a}_{{n}} =\:\frac{{a}_{\mathrm{0}} }{{n}!}\:\Rightarrow \\ $$$${p}\left({x}\right)=\:\sum_{{k}=\mathrm{0}} ^{{n}−\mathrm{1}} \:\frac{{a}_{\mathrm{0}} }{{k}!}\:{x}^{{k}} \:\:\:+{x}^{{n}} \\ $$$${p}\left({o}\right)−{p}^{'} \left(\mathrm{0}\right)=\mathrm{0}\:\Rightarrow{a}_{\mathrm{0}} \:=\mathrm{0}\:\Rightarrow\:{p}\left({x}\right)=\:\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\frac{{x}^{{k}} }{{k}!}\:+{x}^{{n}} \\ $$$$\left.\mathrm{2}\right)\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{p}\left({x}\right){dx}=\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \frac{\mathrm{1}}{{k}!}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:{x}^{{k}} {dx}\:\:+\frac{\mathrm{1}}{{n}+\mathrm{1}} \\ $$$$=\sum_{{k}=\mathrm{1}} ^{{n}−\mathrm{1}} \:\:\frac{\mathrm{1}}{\left({k}+\mathrm{1}\right)!}\:+\frac{\mathrm{1}}{{n}+\mathrm{1}} \\ $$$$ \\ $$