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Question Number 72260 by mathmax by abdo last updated on 26/Oct/19

let f(x)=((2x+3)/(x^2 +1)) calculate f^((n)) (x) 2)find f^((10)) (x) and f^((15)) (x) 3)calculate f^((10)) (0) and f^((15)) (0) 4)developp f at integr serie 5)let g(x)=∫_0 ^x f(t)dt developp g at integr serie.

$${let}\:{f}\left({x}\right)=\frac{\mathrm{2}{x}+\mathrm{3}}{{x}^{\mathrm{2}} +\mathrm{1}} \\ $$$${calculate}\:\:{f}^{\left({n}\right)} \left({x}\right) \\ $$$$\left.\mathrm{2}\right){find}\:{f}^{\left(\mathrm{10}\right)} \left({x}\right)\:{and}\:{f}^{\left(\mathrm{15}\right)} \left({x}\right) \\ $$$$\left.\mathrm{3}\right){calculate}\:{f}^{\left(\mathrm{10}\right)} \left(\mathrm{0}\right)\:{and}\:{f}^{\left(\mathrm{15}\right)} \left(\mathrm{0}\right) \\ $$$$\left.\mathrm{4}\right){developp}\:{f}\:{at}\:{integr}\:{serie} \\ $$$$\left.\mathrm{5}\right){let}\:{g}\left({x}\right)=\int_{\mathrm{0}} ^{{x}} {f}\left({t}\right){dt}\:\:{developp}\:{g}\:{at}\:{integr}\:{serie}. \\ $$

Commented by mathmax by abdo last updated on 27/Oct/19

1) f(x)=((2x+3)/(x^2 +1)) =((2x+3)/((x−i)(x+i))) =(a/(x−i)) +(b/(x+i)) a=((2i+3)/(2i)) =1+(3/(2i)) =1−((3i)/2) =((2−3i)/2) b=((−2i+3)/((−2i))) =((2i−3)/(2i)) =1−(3/(2i)) =1+((3i)/2) =((2+3i)/2) ⇒ f(x)=((2−3i)/(2(x−i))) +((2+3i)/(2(x+i))) ⇒f^((n)) (x)=((2−3i)/2)((1/(x−i)))^((n)) +((2+3i)/2)((1/(x+i)))^((n)) =((2−3i)/2) (((−1)^n n!)/((x−i)^(n+1) )) +((2+3i)/2) (((−1)^n n!)/((x+i)^(n+1) )) =(((−1)^n n!)/2){ ((2+3i)/((x+i)^(n+1) )) +((2−3i)/((x−i)^(n+1) ))} =(((−1)^n n!)/2){(((2+3i)(x−i)^(n+1) +(2−3i)(x+i)^(n+1) )/((x^2 +1)^(n+1) ))} =(((−1)^n n!)/((x^2 +1)^(n+1) ))Re{(2+3i)(x−i)^(n+1) } but 2+3i =(√(13))e^(iarctan((3/2))) and x−i =(√(x^2 +1))e^(−iarctan((1/x))) ⇒ (x−i)^(n+1) =(x^2 +1)^((n+1)/2) e^(−i(n+1)arctan((1/x))) ⇒ (2+3i)(x−i)^(n+1) =(√(13))(x^2 +1)^((n+1)/2) e^(i(arctan((3/2))−(n+1)arctan((1/x)))) ⇒ f^((n)) (x)=(((−1)^n n!)/((x^2 +1)^(n+1) ))(√(13))(x^2 +1)^((n+1)/2) cos(arctan((3/2))−(n+1)arctan((1/x))) f^((n)) (x)=(√(13))(−1)^n n!(x^2 +1)^(−((n+1)/2)) cos(arctan((3/2))−(n+1)arctan((1/x)))

$$\left.\mathrm{1}\right)\:{f}\left({x}\right)=\frac{\mathrm{2}{x}+\mathrm{3}}{{x}^{\mathrm{2}} +\mathrm{1}}\:=\frac{\mathrm{2}{x}+\mathrm{3}}{\left({x}−{i}\right)\left({x}+{i}\right)}\:=\frac{{a}}{{x}−{i}}\:+\frac{{b}}{{x}+{i}} \\ $$$${a}=\frac{\mathrm{2}{i}+\mathrm{3}}{\mathrm{2}{i}}\:=\mathrm{1}+\frac{\mathrm{3}}{\mathrm{2}{i}}\:=\mathrm{1}−\frac{\mathrm{3}{i}}{\mathrm{2}}\:=\frac{\mathrm{2}−\mathrm{3}{i}}{\mathrm{2}} \\ $$$${b}=\frac{−\mathrm{2}{i}+\mathrm{3}}{\left(−\mathrm{2}{i}\right)}\:=\frac{\mathrm{2}{i}−\mathrm{3}}{\mathrm{2}{i}}\:=\mathrm{1}−\frac{\mathrm{3}}{\mathrm{2}{i}}\:=\mathrm{1}+\frac{\mathrm{3}{i}}{\mathrm{2}}\:=\frac{\mathrm{2}+\mathrm{3}{i}}{\mathrm{2}}\:\Rightarrow \\ $$$${f}\left({x}\right)=\frac{\mathrm{2}−\mathrm{3}{i}}{\mathrm{2}\left({x}−{i}\right)}\:+\frac{\mathrm{2}+\mathrm{3}{i}}{\mathrm{2}\left({x}+{i}\right)}\:\Rightarrow{f}^{\left({n}\right)} \left({x}\right)=\frac{\mathrm{2}−\mathrm{3}{i}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{x}−{i}}\right)^{\left({n}\right)} +\frac{\mathrm{2}+\mathrm{3}{i}}{\mathrm{2}}\left(\frac{\mathrm{1}}{{x}+{i}}\right)^{\left({n}\right)} \\ $$$$=\frac{\mathrm{2}−\mathrm{3}{i}}{\mathrm{2}}\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}−{i}\right)^{{n}+\mathrm{1}} }\:+\frac{\mathrm{2}+\mathrm{3}{i}}{\mathrm{2}}\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}+{i}\right)^{{n}+\mathrm{1}} } \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\mathrm{2}}\left\{\:\frac{\mathrm{2}+\mathrm{3}{i}}{\left({x}+{i}\right)^{{n}+\mathrm{1}} }\:+\frac{\mathrm{2}−\mathrm{3}{i}}{\left({x}−{i}\right)^{{n}+\mathrm{1}} }\right\} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\mathrm{2}}\left\{\frac{\left(\mathrm{2}+\mathrm{3}{i}\right)\left({x}−{i}\right)^{{n}+\mathrm{1}} +\left(\mathrm{2}−\mathrm{3}{i}\right)\left({x}+{i}\right)^{{n}+\mathrm{1}} }{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{{n}+\mathrm{1}} }\right\} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{{n}+\mathrm{1}} }{Re}\left\{\left(\mathrm{2}+\mathrm{3}{i}\right)\left({x}−{i}\right)^{{n}+\mathrm{1}} \right\}\:{but} \\ $$$$\mathrm{2}+\mathrm{3}{i}\:=\sqrt{\mathrm{13}}{e}^{{iarctan}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)} \:\:\:{and}\:{x}−{i}\:=\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}{e}^{−{iarctan}\left(\frac{\mathrm{1}}{{x}}\right)} \:\Rightarrow \\ $$$$\left({x}−{i}\right)^{{n}+\mathrm{1}} =\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{{n}+\mathrm{1}}{\mathrm{2}}} \:{e}^{−{i}\left({n}+\mathrm{1}\right){arctan}\left(\frac{\mathrm{1}}{{x}}\right)} \:\Rightarrow \\ $$$$\left(\mathrm{2}+\mathrm{3}{i}\right)\left({x}−{i}\right)^{{n}+\mathrm{1}} =\sqrt{\mathrm{13}}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\frac{{n}+\mathrm{1}}{\mathrm{2}}} \:\:{e}^{{i}\left({arctan}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)−\left({n}+\mathrm{1}\right){arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right)} \:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{{n}+\mathrm{1}} }\sqrt{\mathrm{13}}\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{\frac{{n}+\mathrm{1}}{\mathrm{2}}} \:{cos}\left({arctan}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)−\left({n}+\mathrm{1}\right){arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right) \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\sqrt{\mathrm{13}}\left(−\mathrm{1}\right)^{{n}} {n}!\left({x}^{\mathrm{2}} \:+\mathrm{1}\right)^{−\frac{{n}+\mathrm{1}}{\mathrm{2}}} {cos}\left({arctan}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)−\left({n}+\mathrm{1}\right){arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right) \\ $$

Commented by mathmax by abdo last updated on 27/Oct/19

2)f^((10)) (x) =(√(13))(10)!(x^2 +1)^(−((11)/2)) cos(arctan((3/2))−11arctan((1/x)) f^((15)) (x)=−(√(13))(15)! (x^2 +1)^(−8) cos(arctan((3/2))−16 arctan((1/x)))

$$\left.\mathrm{2}\right){f}^{\left(\mathrm{10}\right)} \left({x}\right)\:=\sqrt{\mathrm{13}}\left(\mathrm{10}\right)!\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{−\frac{\mathrm{11}}{\mathrm{2}}} \:{cos}\left({arctan}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)−\mathrm{11}{arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right. \\ $$$${f}^{\left(\mathrm{15}\right)} \left({x}\right)=−\sqrt{\mathrm{13}}\left(\mathrm{15}\right)!\:\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{−\mathrm{8}} {cos}\left({arctan}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)−\mathrm{16}\:{arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right) \\ $$

Commented by mathmax by abdo last updated on 27/Oct/19

4) we have f^((n)) (x)= (√(13))(−1)^n n!(x^2 +1)^(−((n+1)/2)) cos(arctan((3/2))−(n+1)arctan((1/x))) ⇒ f^((n)) (0) =(√(13))(−1)^n n!cos(arctan((3/2))+^− (n+1)(π/2)) f(x)=Σ_(n=0) ^∞ ((f^((n)) (x))/(n!)) =Σ_(n=0) ^∞ (√(13))(−1)^n cos(arctan((3/2))+^− (n+1)(π/2))x^n

$$\left.\mathrm{4}\right)\:{we}\:{have}\:{f}^{\left({n}\right)} \left({x}\right)= \\ $$$$\sqrt{\mathrm{13}}\left(−\mathrm{1}\right)^{{n}} {n}!\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{−\frac{{n}+\mathrm{1}}{\mathrm{2}}} \:{cos}\left({arctan}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)−\left({n}+\mathrm{1}\right){arctan}\left(\frac{\mathrm{1}}{{x}}\right)\right)\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\sqrt{\mathrm{13}}\left(−\mathrm{1}\right)^{{n}} {n}!{cos}\left({arctan}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\overset{−} {+}\left({n}+\mathrm{1}\right)\frac{\pi}{\mathrm{2}}\right) \\ $$$${f}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{f}^{\left({n}\right)} \left({x}\right)}{{n}!}\:=\sum_{{n}=\mathrm{0}} ^{\infty} \sqrt{\mathrm{13}}\left(−\mathrm{1}\right)^{{n}} {cos}\left({arctan}\left(\frac{\mathrm{3}}{\mathrm{2}}\right)\overset{−} {+}\left({n}+\mathrm{1}\right)\frac{\pi}{\mathrm{2}}\right){x}^{{n}} \\ $$

Answered by mind is power last updated on 26/Oct/19

((2x+3)/(x^2 +1))=(a/(x+i))+(c/(x−i)) a+c=2,i(c−a)=3⇒c=((2−3i)/2),a=((2+3i)/2) f(x)=((2+3i)/(2(x+i)))+((2−3i)/(2(x−i))) f^n (x)=((2+3i)/2)(d^n x/dx^n ) (1/(x+i))+((2−3i)/2)(d^n x/dx^n ) (1/(x−i)) =((2+3i)/2) (((−1)^n n!)/((x+i)^(n+1) ))+((2−3i)/2).(((−1)^n n!)/((x−i)^(n+1) )) =(((−1)^n n!)/2)((((2+3i)(x−i)^(n+1) +(2−3i)(x+i)^(n+1) )/((x^2 +1)^(n+1) ))) =(((−1)^n n! (2.Re{ (2+3i)(x+i)^(n+1) }))/((x^2 +1)^(n+1) )) Re(2+3i)(x+i)^(n+1) (x+i)^(n+1) =ΣC_(n+1) ^k .i^k .(x)^(n+1−k) =Σ_(k=0) ^(E(((n+1)/2))) C_(n+1) ^(2k) .(−1)^k .X^(n+1−2k) +Σ_(k=0) ^(E((n/2))) C_(n+1) ^(2k+1) (−1)^k i.X^(n−2k) Re (2+3i).(x+i)^(n+1) = 2Σ_(k=0) ^(E(((n+1)/2))) C_(n+1) ^(2k) .(−1)^k .X^(n+1−2k) +3Σ_(k=0) ^(E((n/2))) C_(n+1) ^(2k+1) (−1)^(k+1) .X^(n−2k) f^n (x)=(((−1)^n n!)/((x^2 +1)^(n+1) )).2(2Σ_(k=0) ^(E(((n+1)/2))) C_(n+1) ^(2k) .(−1)^k .X^(n+1−2k) +3Σ_(k=0) ^(E((n/2))) C_(n+1) ^(2k+1) (−1)^(k+1) .X^(n−2k) ) f^((10)) (x)=2.((10!)/((x^2 +1)^(11) )).(2Σ_(k=0) ^5 C_(11) ^(2k) .(−1)^k X^(11−2k) +3Σ_(k=0) ^5 C_(11) ^(2k+1) (−1)^(k+1) X^(10−2k) ) f^(15) (x) same idee 3) f^(10) (x)=2.10!.(3.)=6.10! f^(15) (0)=2(−1)^(15) .15!.(2.)=−4.15! 4) f^n (0) if n=2s f^n (0)=(−1)^(2s) .(2s)!.2.(3.(−1)^(s+1) )=(((−1)^(s+1) .6.(2s)!)/) n=2s+1⇒f^n (0)=(((−1)^(2s+1) .(2s+1)!.2(2.(−1)^(s+1) ))/)=(−1)^s .4(2s+1)! f(x)=Σ_(n≥0) ((f^n (0).x^n )/(n!))=Σ((f^(2s) (0)x^(2s) )/((2s)!))+Σ((f^(2s+1) (0)x^(2s+1) )/((2s+1)!)) =Σ_(s=0) ^(+∞) (((f^(2s) (0)x^(2s) )/((2s)!))+((f^(2s) (0)x^(2s+1) )/((2s+1)!)))=Σ_(s=0) ^(+∞) (6(−1)^(s+1) x^(2s) +4(−1)^s x^(2s+1) ) =Σ_(s≥0) (−1)^s x^(2s) (−6+4x) 5)gx=∫_0 ^x f g(x)=Σ_(s≥0) (∫_0 ^x 6(−1)^(s+1) t^(2s) +4(−1)^s t^(2s+1) ) =Σ_(s≥0) .(6(−1)^(s+1) .(x^(2s+1) /(2s+1))+((2(−1)^s x^(2s+2) )/(s+1)))

$$\frac{\mathrm{2x}+\mathrm{3}}{\mathrm{x}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{a}}{\mathrm{x}+\mathrm{i}}+\frac{\mathrm{c}}{\mathrm{x}−\mathrm{i}} \\ $$$$\mathrm{a}+\mathrm{c}=\mathrm{2},\mathrm{i}\left(\mathrm{c}−\mathrm{a}\right)=\mathrm{3}\Rightarrow\mathrm{c}=\frac{\mathrm{2}−\mathrm{3i}}{\mathrm{2}},\mathrm{a}=\frac{\mathrm{2}+\mathrm{3i}}{\mathrm{2}} \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\frac{\mathrm{2}+\mathrm{3i}}{\mathrm{2}\left(\mathrm{x}+\mathrm{i}\right)}+\frac{\mathrm{2}−\mathrm{3i}}{\mathrm{2}\left(\mathrm{x}−\mathrm{i}\right)} \\ $$$$\mathrm{f}^{\mathrm{n}} \left(\mathrm{x}\right)=\frac{\mathrm{2}+\mathrm{3i}}{\mathrm{2}}\frac{\mathrm{d}^{\mathrm{n}} \mathrm{x}}{\mathrm{dx}^{\mathrm{n}} }\:\frac{\mathrm{1}}{\mathrm{x}+\mathrm{i}}+\frac{\mathrm{2}−\mathrm{3i}}{\mathrm{2}}\frac{\mathrm{d}^{\mathrm{n}} \mathrm{x}}{\mathrm{dx}^{\mathrm{n}} }\:\frac{\mathrm{1}}{\mathrm{x}−\mathrm{i}} \\ $$$$=\frac{\mathrm{2}+\mathrm{3i}}{\mathrm{2}}\:\:\:\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \:\mathrm{n}!}{\left(\mathrm{x}+\mathrm{i}\right)^{\mathrm{n}+\mathrm{1}} }+\frac{\mathrm{2}−\mathrm{3i}}{\mathrm{2}}.\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{n}!}{\left(\mathrm{x}−\mathrm{i}\right)^{\mathrm{n}+\mathrm{1}} } \\ $$$$=\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{n}!}{\mathrm{2}}\left(\frac{\left(\mathrm{2}+\mathrm{3i}\right)\left(\mathrm{x}−\mathrm{i}\right)^{\mathrm{n}+\mathrm{1}} +\left(\mathrm{2}−\mathrm{3i}\right)\left(\mathrm{x}+\mathrm{i}\right)^{\mathrm{n}+\mathrm{1}} }{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }\right) \\ $$$$=\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{n}!\:\left(\mathrm{2}.\mathrm{Re}\left\{\:\left(\mathrm{2}+\mathrm{3i}\right)\left(\mathrm{x}+\mathrm{i}\right)^{\mathrm{n}+\mathrm{1}} \right\}\right)}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} } \\ $$$$\mathrm{Re}\left(\mathrm{2}+\mathrm{3i}\right)\left(\mathrm{x}+\mathrm{i}\right)^{\mathrm{n}+\mathrm{1}} \\ $$$$\left(\mathrm{x}+\mathrm{i}\right)^{\mathrm{n}+\mathrm{1}} =\Sigma\mathrm{C}_{\mathrm{n}+\mathrm{1}} ^{\mathrm{k}} .\mathrm{i}^{\mathrm{k}} .\left(\mathrm{x}\right)^{\mathrm{n}+\mathrm{1}−\mathrm{k}} \\ $$$$=\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{E}\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}}\right)} {\sum}}\mathrm{C}_{\mathrm{n}+\mathrm{1}} ^{\mathrm{2k}} .\left(−\mathrm{1}\right)^{\mathrm{k}} .\mathrm{X}^{\mathrm{n}+\mathrm{1}−\mathrm{2k}} +\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{E}\left(\frac{\mathrm{n}}{\mathrm{2}}\right)} {\sum}}\mathrm{C}_{\mathrm{n}+\mathrm{1}} ^{\mathrm{2k}+\mathrm{1}} \left(−\mathrm{1}\right)^{\mathrm{k}} \mathrm{i}.\mathrm{X}^{\mathrm{n}−\mathrm{2k}} \\ $$$$\mathrm{Re}\:\left(\mathrm{2}+\mathrm{3i}\right).\left(\mathrm{x}+\mathrm{i}\right)^{\mathrm{n}+\mathrm{1}} = \\ $$$$\mathrm{2}\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{E}\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}}\right)} {\sum}}\mathrm{C}_{\mathrm{n}+\mathrm{1}} ^{\mathrm{2k}} .\left(−\mathrm{1}\right)^{\mathrm{k}} .\mathrm{X}^{\mathrm{n}+\mathrm{1}−\mathrm{2k}} +\mathrm{3}\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{E}\left(\frac{\mathrm{n}}{\mathrm{2}}\right)} {\sum}}\mathrm{C}_{\mathrm{n}+\mathrm{1}} ^{\mathrm{2k}+\mathrm{1}} \left(−\mathrm{1}\right)^{\mathrm{k}+\mathrm{1}} .\mathrm{X}^{\mathrm{n}−\mathrm{2k}} \\ $$$$\mathrm{f}^{\mathrm{n}} \left(\mathrm{x}\right)=\frac{\left(−\mathrm{1}\right)^{\mathrm{n}} \mathrm{n}!}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{n}+\mathrm{1}} }.\mathrm{2}\left(\mathrm{2}\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{E}\left(\frac{\mathrm{n}+\mathrm{1}}{\mathrm{2}}\right)} {\sum}}\mathrm{C}_{\mathrm{n}+\mathrm{1}} ^{\mathrm{2k}} .\left(−\mathrm{1}\right)^{\mathrm{k}} .\mathrm{X}^{\mathrm{n}+\mathrm{1}−\mathrm{2k}} +\mathrm{3}\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{E}\left(\frac{\mathrm{n}}{\mathrm{2}}\right)} {\sum}}\mathrm{C}_{\mathrm{n}+\mathrm{1}} ^{\mathrm{2k}+\mathrm{1}} \left(−\mathrm{1}\right)^{\mathrm{k}+\mathrm{1}} .\mathrm{X}^{\mathrm{n}−\mathrm{2k}} \right) \\ $$$$ \\ $$$$\mathrm{f}^{\left(\mathrm{10}\right)} \left(\mathrm{x}\right)=\mathrm{2}.\frac{\mathrm{10}!}{\left(\mathrm{x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{11}} }.\left(\mathrm{2}\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{5}} {\sum}}\mathrm{C}_{\mathrm{11}} ^{\mathrm{2k}} .\left(−\mathrm{1}\right)^{\mathrm{k}} \mathrm{X}^{\mathrm{11}−\mathrm{2k}} +\mathrm{3}\underset{\mathrm{k}=\mathrm{0}} {\overset{\mathrm{5}} {\sum}}\mathrm{C}_{\mathrm{11}} ^{\mathrm{2k}+\mathrm{1}} \left(−\mathrm{1}\right)^{\mathrm{k}+\mathrm{1}} \mathrm{X}^{\mathrm{10}−\mathrm{2k}} \right) \\ $$$$\mathrm{f}^{\mathrm{15}} \left(\mathrm{x}\right)\:\mathrm{same}\:\mathrm{idee} \\ $$$$\left.\mathrm{3}\right) \\ $$$$\mathrm{f}^{\mathrm{10}} \left(\mathrm{x}\right)=\mathrm{2}.\mathrm{10}!.\left(\mathrm{3}.\right)=\mathrm{6}.\mathrm{10}! \\ $$$$\mathrm{f}^{\mathrm{15}} \left(\mathrm{0}\right)=\mathrm{2}\left(−\mathrm{1}\right)^{\mathrm{15}} .\mathrm{15}!.\left(\mathrm{2}.\right)=−\mathrm{4}.\mathrm{15}! \\ $$$$\left.\mathrm{4}\right)\:\mathrm{f}^{\mathrm{n}} \left(\mathrm{0}\right)\:\mathrm{if}\:\mathrm{n}=\mathrm{2s} \\ $$$$\mathrm{f}^{\mathrm{n}} \left(\mathrm{0}\right)=\left(−\mathrm{1}\right)^{\mathrm{2s}} .\left(\mathrm{2s}\right)!.\mathrm{2}.\left(\mathrm{3}.\left(−\mathrm{1}\right)^{\mathrm{s}+\mathrm{1}} \right)=\frac{\left(−\mathrm{1}\right)^{\mathrm{s}+\mathrm{1}} .\mathrm{6}.\left(\mathrm{2s}\right)!}{} \\ $$$$\mathrm{n}=\mathrm{2s}+\mathrm{1}\Rightarrow\mathrm{f}^{\mathrm{n}} \left(\mathrm{0}\right)=\frac{\left(−\mathrm{1}\right)^{\mathrm{2s}+\mathrm{1}} .\left(\mathrm{2s}+\mathrm{1}\right)!.\mathrm{2}\left(\mathrm{2}.\left(−\mathrm{1}\right)^{\mathrm{s}+\mathrm{1}} \right)}{}=\left(−\mathrm{1}\right)^{\mathrm{s}} .\mathrm{4}\left(\mathrm{2s}+\mathrm{1}\right)! \\ $$$$\mathrm{f}\left(\mathrm{x}\right)=\underset{\mathrm{n}\geqslant\mathrm{0}} {\sum}\frac{\mathrm{f}^{\mathrm{n}} \left(\mathrm{0}\right).\mathrm{x}^{\mathrm{n}} }{\mathrm{n}!}=\Sigma\frac{\mathrm{f}^{\mathrm{2s}} \left(\mathrm{0}\right)\mathrm{x}^{\mathrm{2s}} }{\left(\mathrm{2s}\right)!}+\Sigma\frac{\mathrm{f}^{\mathrm{2s}+\mathrm{1}} \left(\mathrm{0}\right)\mathrm{x}^{\mathrm{2s}+\mathrm{1}} }{\left(\mathrm{2s}+\mathrm{1}\right)!} \\ $$$$=\underset{\mathrm{s}=\mathrm{0}} {\overset{+\infty} {\sum}}\left(\frac{\mathrm{f}^{\mathrm{2s}} \left(\mathrm{0}\right)\mathrm{x}^{\mathrm{2s}} }{\left(\mathrm{2s}\right)!}+\frac{\mathrm{f}^{\mathrm{2s}} \left(\mathrm{0}\right)\mathrm{x}^{\mathrm{2s}+\mathrm{1}} }{\left(\mathrm{2s}+\mathrm{1}\right)!}\right)=\underset{\mathrm{s}=\mathrm{0}} {\overset{+\infty} {\sum}}\left(\mathrm{6}\left(−\mathrm{1}\right)^{\mathrm{s}+\mathrm{1}} \mathrm{x}^{\mathrm{2s}} +\mathrm{4}\left(−\mathrm{1}\right)^{\mathrm{s}} \mathrm{x}^{\mathrm{2s}+\mathrm{1}} \right) \\ $$$$=\underset{\mathrm{s}\geqslant\mathrm{0}} {\sum}\left(−\mathrm{1}\right)^{\mathrm{s}} \mathrm{x}^{\mathrm{2s}} \left(−\mathrm{6}+\mathrm{4x}\right) \\ $$$$\left.\mathrm{5}\right)\mathrm{gx}=\int_{\mathrm{0}} ^{\mathrm{x}} \mathrm{f}\: \\ $$$$\mathrm{g}\left(\mathrm{x}\right)=\underset{\mathrm{s}\geqslant\mathrm{0}} {\sum}\left(\int_{\mathrm{0}} ^{\mathrm{x}} \mathrm{6}\left(−\mathrm{1}\right)^{\mathrm{s}+\mathrm{1}} \mathrm{t}^{\mathrm{2s}} +\mathrm{4}\left(−\mathrm{1}\right)^{\mathrm{s}} \mathrm{t}^{\mathrm{2s}+\mathrm{1}} \right) \\ $$$$=\underset{\mathrm{s}\geqslant\mathrm{0}} {\sum}.\left(\mathrm{6}\left(−\mathrm{1}\right)^{\mathrm{s}+\mathrm{1}} .\frac{\mathrm{x}^{\mathrm{2s}+\mathrm{1}} }{\mathrm{2s}+\mathrm{1}}+\frac{\mathrm{2}\left(−\mathrm{1}\right)^{\mathrm{s}} \mathrm{x}^{\mathrm{2s}+\mathrm{2}} }{\mathrm{s}+\mathrm{1}}\right) \\ $$

Commented by mathmax by abdo last updated on 27/Oct/19

thank you sir.

$${thank}\:{you}\:{sir}. \\ $$

Commented by mind is power last updated on 27/Oct/19

y′re welcom

$$\mathrm{y}'\mathrm{re}\:\mathrm{welcom} \\ $$

Commented by mind is power last updated on 27/Oct/19

y′re welcom

$$\mathrm{y}'\mathrm{re}\:\mathrm{welcom} \\ $$

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