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f-n-x-x-1-x-2x-2-x-nx-n-x-n-N-f-n-x-f-n-1-x-nx-n-x-f-1-x-x-1-x-f-n-1-x-f-n-x-nx-n-x-n-Z-f-0-x-f-1-x-x-1-x




Question Number 5585 by 123456 last updated on 21/May/16
f_n (x)=(x/(1−x))+((2x)/(2−x))+...+((nx)/(n−x))  n∈N^∗   −−−−−−−−−−−−−−−−  f_n (x)=f_(n−1) (x)+((nx)/(n−x))  f_1 (x)=(x/(1−x))  f_(n−1) (x)=f_n (x)−((nx)/(n−x))  n∈Z  −−−−−−−−−−−−−−−−  f_0 (x)=f_1 (x)−(x/(1−x))=0  f_(−1) (x)=f_0 (x)=0  f_(−2) (x)=f_(−1) (x)−(x/(1+x))=−(x/(1+x))  −−−−−−−−−−−−−−−−  f_(−n) (x)=−(x/(x+1))−((2x)/(x+2))−...−(((n−1)x)/(x+(n−1)))  n∈Z_− ^∗ ,n<−1
$${f}_{{n}} \left({x}\right)=\frac{{x}}{\mathrm{1}−{x}}+\frac{\mathrm{2}{x}}{\mathrm{2}−{x}}+…+\frac{{nx}}{{n}−{x}} \\ $$$${n}\in\mathbb{N}^{\ast} \\ $$$$−−−−−−−−−−−−−−−− \\ $$$${f}_{{n}} \left({x}\right)={f}_{{n}−\mathrm{1}} \left({x}\right)+\frac{{nx}}{{n}−{x}} \\ $$$${f}_{\mathrm{1}} \left({x}\right)=\frac{{x}}{\mathrm{1}−{x}} \\ $$$${f}_{{n}−\mathrm{1}} \left({x}\right)={f}_{{n}} \left({x}\right)−\frac{{nx}}{{n}−{x}} \\ $$$${n}\in\mathbb{Z} \\ $$$$−−−−−−−−−−−−−−−− \\ $$$${f}_{\mathrm{0}} \left({x}\right)={f}_{\mathrm{1}} \left({x}\right)−\frac{{x}}{\mathrm{1}−{x}}=\mathrm{0} \\ $$$${f}_{−\mathrm{1}} \left({x}\right)={f}_{\mathrm{0}} \left({x}\right)=\mathrm{0} \\ $$$${f}_{−\mathrm{2}} \left({x}\right)={f}_{−\mathrm{1}} \left({x}\right)−\frac{{x}}{\mathrm{1}+{x}}=−\frac{{x}}{\mathrm{1}+{x}} \\ $$$$−−−−−−−−−−−−−−−− \\ $$$${f}_{−{n}} \left({x}\right)=−\frac{{x}}{{x}+\mathrm{1}}−\frac{\mathrm{2}{x}}{{x}+\mathrm{2}}−…−\frac{\left({n}−\mathrm{1}\right){x}}{{x}+\left({n}−\mathrm{1}\right)} \\ $$$${n}\in\mathbb{Z}_{−} ^{\ast} ,{n}<−\mathrm{1} \\ $$
Commented by Rasheed Soomro last updated on 21/May/16
What is required?
$$\mathrm{What}\:\mathrm{is}\:\mathrm{required}? \\ $$
Commented by 123456 last updated on 21/May/16
(df_n /dx)
$$\frac{{df}_{{n}} }{{dx}} \\ $$
Commented by FilupSmith last updated on 22/May/16
f_n (x)=(x/(1−x))+((2x)/(2−x))+...+((nx)/(n−x))  f_n (x)=Σ_(i=1) ^n ((ix)/(i−x))  (df_n /dx)=(d/dx)Σ_(i=1) ^n ((ix)/(i−x))  (df_n /dx)=Σ_(i=1) ^n ((i(i−x)+ix)/((i−x)^2 ))     quotant rule  (df_n /dx)=Σ_(i=1) ^n (i^2 /((i−x)^2 ))  ????
$${f}_{{n}} \left({x}\right)=\frac{{x}}{\mathrm{1}−{x}}+\frac{\mathrm{2}{x}}{\mathrm{2}−{x}}+…+\frac{{nx}}{{n}−{x}} \\ $$$${f}_{{n}} \left({x}\right)=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{ix}}{{i}−{x}} \\ $$$$\frac{{df}_{{n}} }{{dx}}=\frac{{d}}{{dx}}\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{ix}}{{i}−{x}} \\ $$$$\frac{{df}_{{n}} }{{dx}}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{i}\left({i}−{x}\right)+{ix}}{\left({i}−{x}\right)^{\mathrm{2}} }\:\:\:\:\:{quotant}\:{rule} \\ $$$$\frac{{df}_{{n}} }{{dx}}=\underset{{i}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{i}^{\mathrm{2}} }{\left({i}−{x}\right)^{\mathrm{2}} } \\ $$$$???? \\ $$

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