Question Number 39214 by math khazana by abdo last updated on 03/Jul/18 | ||
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$${let}\:{f}\left({x}\right)=\:\frac{{e}^{−\mathrm{3}{x}} }{{x}^{\mathrm{2}} \:+\mathrm{4}} \\ $$$${developp}\:{f}\:\:{at}\:{integr}\:{serie}. \\ $$ | ||
Commented by math khazana by abdo last updated on 04/Jul/18 | ||
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$${f}\left({x}\right)=\sum_{{n}=\mathrm{0}} ^{\infty} \:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \:\:\:{let}\:{find}\:{f}^{\left({n}\right)} \left(\mathrm{0}\right) \\ $$$${f}\left({x}\right)=\frac{{e}^{−\mathrm{3}{x}} }{\left({x}−\mathrm{2}{i}\right)\left({x}+\mathrm{2}{i}\right)}\:=\frac{{e}^{−\mathrm{3}{x}} }{\mathrm{4}{i}}\left\{\:\:\frac{\mathrm{1}}{{x}−\mathrm{2}{i}}\:−\frac{\mathrm{1}}{{x}+\mathrm{2}{i}}\right\} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}{i}}\left\{\:\:\frac{{e}^{−\mathrm{3}{x}} }{{x}−\mathrm{2}{i}}\:−\frac{{e}^{−\mathrm{3}{x}} }{{x}+\mathrm{2}{i}}\right\}\:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}{i}}\left\{\:\left(\:\frac{{e}^{−\mathrm{3}{x}} }{{x}−\mathrm{2}{i}}\right)^{\left({n}\right)} \:−\left(\frac{{e}^{−\mathrm{3}{x}} }{{x}+\mathrm{3}{i}}\right)^{\left({n}\right)} \right\}\:{but} \\ $$$${leibniz}\:{formula}\:{give} \\ $$$$\left(\frac{{e}^{−\mathrm{3}{x}} }{{x}−\mathrm{2}{i}}\right)^{\left({n}\right)} =\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \:\:\left(\frac{\mathrm{1}}{{x}−\mathrm{2}{i}}\right)^{\left({k}\right)} \:\left({e}^{−\mathrm{3}{x}} \right)^{\left.{n}−{k}\right)} \\ $$$$=\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \:\:\frac{\left(−\mathrm{1}\right)^{{k}} \:{k}!}{\left({x}−\mathrm{2}{i}\right)^{{k}+\mathrm{1}} }\:\left(−\mathrm{3}\right)^{{n}−{k}} \:{e}^{−\mathrm{3}{x}\:} \:\:{also} \\ $$$$\left(\:\frac{{e}^{−\mathrm{3}{x}} }{{x}+\mathrm{2}{i}}\right)^{\left({n}\right)} =\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \:\:\:\:\frac{\left(−\mathrm{1}\right)^{{k}} \:{k}!}{\left({x}+\mathrm{2}{i}\right)^{{k}+\mathrm{1}} }\:\left(−\mathrm{3}\right)^{{n}−{k}} \:{e}^{−\mathrm{3}{x}} \\ $$$${f}^{\left({n}\right)} \left({x}\right)=\frac{\mathrm{1}}{\mathrm{4}{i}}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\left(−\mathrm{1}\right)^{{k}} {k}!\left(−\mathrm{3}\right)^{{n}−{k}} \:{C}_{{n}} ^{{k}} \left\{\:\frac{\mathrm{1}}{\left({x}−\mathrm{2}{i}\right)^{{k}+\mathrm{1}} }−\frac{\mathrm{1}}{\left({x}+\mathrm{2}{i}\right)^{{k}+\mathrm{1}} }\right\} \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)=\frac{\mathrm{1}}{\mathrm{4}{i}}\:\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \:{k}!\left(−\mathrm{3}\right)^{{n}−{k}} \:{C}_{{n}} ^{{n}} \:\left\{\:\:\frac{\mathrm{1}}{\left(−\mathrm{2}{i}\right)^{{k}+\mathrm{1}} }\:−\frac{\mathrm{1}}{\left(\mathrm{2}{i}\right)^{{k}+\mathrm{1}} }\right\} \\ $$$${but}\:\:\frac{\mathrm{1}}{\left(−\mathrm{2}{i}\right)^{{k}+\mathrm{1}} }\:+\:\frac{\mathrm{1}}{\left(\mathrm{2}{i}\right)^{{k}+\mathrm{1}} }\:=\frac{\left(\mathrm{2}{i}\right)^{{k}+\mathrm{1}} \:−\left(−\mathrm{2}{i}\right)^{{k}+\mathrm{1}} }{\mathrm{4}^{{k}+\mathrm{1}} } \\ $$$$=\frac{\mathrm{2}{iIm}\left(\:\left(\mathrm{2}{i}\right)^{{k}+\mathrm{1}} \right)}{\mathrm{4}^{{k}+\mathrm{1}} }\:=\:\frac{\mathrm{2}{i}\:\mathrm{2}^{{k}+\mathrm{1}} \:{e}^{\frac{{i}\left({k}+\mathrm{1}\right)\pi}{\mathrm{2}}} }{\mathrm{2}^{{k}+\mathrm{1}} \:\:\mathrm{2}^{{k}+\mathrm{1}} } \\ $$$$=\frac{{i}}{\mathrm{2}^{{k}} }\:{e}^{\frac{{i}\left({k}+\mathrm{1}\right)\pi}{\mathrm{2}}} \:\Rightarrow \\ $$$${f}^{\left({n}\right)} \left(\mathrm{0}\right)\:=\frac{\mathrm{1}}{\mathrm{4}}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\:\left(−\mathrm{1}\right)^{{k}} \:{k}!\left(−\mathrm{3}\right)^{{n}−{k}} \:{C}_{{n}} ^{{k}} \:\:\frac{\mathrm{1}}{\mathrm{2}^{{k}} }\:{e}^{{i}\frac{\left({k}+\mathrm{1}\right)\pi}{\mathrm{2}}} \\ $$$${and}\:{f}\left({x}\right)\:=\sum_{{n}=\mathrm{0}} ^{\infty} \:\:\frac{{f}^{\left({n}\right)} \left(\mathrm{0}\right)}{{n}!}\:{x}^{{n}} \\ $$ | ||