Question Number 30752 by abdo imad last updated on 25/Feb/18
$${let}\:{f}\left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{1}+{x}^{\mathrm{2}} }\:\:{calculate}\:{f}^{\left({n}\right)} \left({x}\right). \\ $$
Commented by abdo imad last updated on 28/Feb/18
![we have f(x)= (1/((x+i)(x−i)))=(1/(2i))( (1/(x−i)) −(1/(x+i))) ⇒ f^((n)) (x)= (1/(2i))( ((1/(x−i)))^((n)) −((1/(x+i)))^((n)) ) =(1/(2i))( (((−1)^n n!)/((x−i)^(n+1) )) −(((−1)^n n!)/((x+i)^(n+1) ))) =((n!(−1)^n )/(2i))( (((x+i)^(n+1) −(x−i)^(n+1) )/((1+x^2 )^(n+1) ))) but (x+i)^(n+1) −(x−i)^(n+1) = Σ_(k=0) ^(n+1) C_(n+1) ^k i^k x^(n+1−k) −Σ_(k=0) ^(n+1) (−i)^k x^(n+1−k) =Σ_(k=0) ^(n+1) C_(n+1) ^k ( i^k −(−i)^k )x^(n+1−k) =Σ_(p=0) ^([(n/2)]) C_(n+1) ^(2p+1) (2i) x^(n+1−2p−1) ⇒ f^((n)) (x)=n! (−1)^n ((Σ_(p=0) ^([(n/2)]) C_(n+1) ^(2p+1) x^(n−2p) )/((1+x^2 )^(n+1) )) . also wecan give f^((n)) at form of arctan....](https://www.tinkutara.com/question/Q30908.png)
$${we}\:{have}\:{f}\left({x}\right)=\:\frac{\mathrm{1}}{\left({x}+{i}\right)\left({x}−{i}\right)}=\frac{\mathrm{1}}{\mathrm{2}{i}}\left(\:\frac{\mathrm{1}}{{x}−{i}}\:−\frac{\mathrm{1}}{{x}+{i}}\right)\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\:\frac{\mathrm{1}}{\mathrm{2}{i}}\left(\:\left(\frac{\mathrm{1}}{{x}−{i}}\right)^{\left({n}\right)} −\left(\frac{\mathrm{1}}{{x}+{i}}\right)^{\left({n}\right)} \right) \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{i}}\left(\:\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}−{i}\right)^{{n}+\mathrm{1}} }\:−\frac{\left(−\mathrm{1}\right)^{{n}} {n}!}{\left({x}+{i}\right)^{{n}+\mathrm{1}} }\right) \\ $$$$=\frac{{n}!\left(−\mathrm{1}\right)^{{n}} }{\mathrm{2}{i}}\left(\:\frac{\left({x}+{i}\right)^{{n}+\mathrm{1}} \:−\left({x}−{i}\right)^{{n}+\mathrm{1}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}+\mathrm{1}} }\right)\:{but} \\ $$$$\left({x}+{i}\right)^{{n}+\mathrm{1}} \:−\left({x}−{i}\right)^{{n}+\mathrm{1}} =\:\sum_{{k}=\mathrm{0}} ^{{n}+\mathrm{1}} \:\:{C}_{{n}+\mathrm{1}} ^{{k}} \:{i}^{{k}} \:{x}^{{n}+\mathrm{1}−{k}} \\ $$$$−\sum_{{k}=\mathrm{0}} ^{{n}+\mathrm{1}} \:\left(−{i}\right)^{{k}} \:{x}^{{n}+\mathrm{1}−{k}} =\sum_{{k}=\mathrm{0}} ^{{n}+\mathrm{1}} \:{C}_{{n}+\mathrm{1}} ^{{k}} \left(\:{i}^{{k}} \:−\left(−{i}\right)^{{k}} \right){x}^{{n}+\mathrm{1}−{k}} \\ $$$$=\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \:\:{C}_{{n}+\mathrm{1}} ^{\mathrm{2}{p}+\mathrm{1}} \:\left(\mathrm{2}{i}\right)\:{x}^{{n}+\mathrm{1}−\mathrm{2}{p}−\mathrm{1}} \:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)={n}!\:\left(−\mathrm{1}\right)^{{n}} \:\:\frac{\sum_{{p}=\mathrm{0}} ^{\left[\frac{{n}}{\mathrm{2}}\right]} \:\:{C}_{{n}+\mathrm{1}} ^{\mathrm{2}{p}+\mathrm{1}} \:\:{x}^{{n}−\mathrm{2}{p}} }{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{{n}+\mathrm{1}} }\:.\:{also}\:{wecan}\:{give} \\ $$$${f}^{\left({n}\right)} {at}\:{form}\:{of}\:{arctan}…. \\ $$