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Question Number 28428 by abdo imad last updated on 25/Jan/18

let give f_n (x)= ((x^2 −1)^n )^((n))    find  f_n  .

$${let}\:{give}\:{f}_{{n}} \left({x}\right)=\:\left(\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{{n}} \right)^{\left({n}\right)} \:\:\:{find}\:\:{f}_{{n}} \:. \\ $$

Commented by abdo imad last updated on 27/Jan/18

 we have  f_n (x)= (p(x))^((n))   with p(x)=(x^2 −1)^n   p(x)= Σ_(k=0) ^n  C_n ^k   x^(2k) (−1)^(n−k) =(−1)^n  Σ_(k=0) ^n (−1)^k  C_n ^k  x^(2k)   (p(x))^((n))   =(−1)^n  Σ_(k=0) ^n (−1)^k  C_n ^k  (x^(2k) )^((n))   but if  2k<n     (x^(2k) )^((n)) =0    so   f_n (x)= (−1)^n  Σ_(k=[((n−1)/2)] +1) ^n   C_n ^k  (x^(2k) )^((n)) .let find  (x^p )^((n))   for p≥n  wehave (x^p )^((1)) =p x^(p−1)   ,(x^p )^((2)) =p(p−1)x^(p−2)   so  (x^p )^((n)) = p(p−1)....(p−n+1)x^(p−n)   =((p(p−1)....(p−n+1)(p−n)!)/((p−n)!)) x^(p−n)  =((p!)/((p−n)!)) x^(p−n)    ⇒  (x^(2k) )^((n)) =  (((2k)!)/((2k−n)!)) x^(2k−n)  so  f_n (x)= (−1)^n  Σ_(k=[((n−1)/2)]+1) ^n  C_n ^k    (((2k)!)/((2k−n)!)) x^(2k−n)   .

$$\:{we}\:{have}\:\:{f}_{{n}} \left({x}\right)=\:\left({p}\left({x}\right)\right)^{\left({n}\right)} \:\:{with}\:{p}\left({x}\right)=\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{{n}} \\ $$$${p}\left({x}\right)=\:\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:{x}^{\mathrm{2}{k}} \left(−\mathrm{1}\right)^{{n}−{k}} =\left(−\mathrm{1}\right)^{{n}} \:\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \:{C}_{{n}} ^{{k}} \:{x}^{\mathrm{2}{k}} \\ $$$$\left({p}\left({x}\right)\right)^{\left({n}\right)} \:\:=\left(−\mathrm{1}\right)^{{n}} \:\sum_{{k}=\mathrm{0}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \:{C}_{{n}} ^{{k}} \:\left({x}^{\mathrm{2}{k}} \right)^{\left({n}\right)} \:\:{but}\:{if} \\ $$$$\mathrm{2}{k}<{n}\:\:\:\:\:\left({x}^{\mathrm{2}{k}} \right)^{\left({n}\right)} =\mathrm{0}\:\:\:\:{so}\: \\ $$$${f}_{{n}} \left({x}\right)=\:\left(−\mathrm{1}\right)^{{n}} \:\sum_{{k}=\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]\:+\mathrm{1}} ^{{n}} \:\:{C}_{{n}} ^{{k}} \:\left({x}^{\mathrm{2}{k}} \right)^{\left({n}\right)} .{let}\:{find} \\ $$$$\left({x}^{{p}} \right)^{\left({n}\right)} \:\:{for}\:{p}\geqslant{n}\:\:{wehave}\:\left({x}^{{p}} \right)^{\left(\mathrm{1}\right)} ={p}\:{x}^{{p}−\mathrm{1}} \:\:,\left({x}^{{p}} \right)^{\left(\mathrm{2}\right)} ={p}\left({p}−\mathrm{1}\right){x}^{{p}−\mathrm{2}} \\ $$$${so}\:\:\left({x}^{{p}} \right)^{\left({n}\right)} =\:{p}\left({p}−\mathrm{1}\right)....\left({p}−{n}+\mathrm{1}\right){x}^{{p}−{n}} \\ $$$$=\frac{{p}\left({p}−\mathrm{1}\right)....\left({p}−{n}+\mathrm{1}\right)\left({p}−{n}\right)!}{\left({p}−{n}\right)!}\:{x}^{{p}−{n}} \:=\frac{{p}!}{\left({p}−{n}\right)!}\:{x}^{{p}−{n}} \: \\ $$$$\Rightarrow\:\:\left({x}^{\mathrm{2}{k}} \right)^{\left({n}\right)} =\:\:\frac{\left(\mathrm{2}{k}\right)!}{\left(\mathrm{2}{k}−{n}\right)!}\:{x}^{\mathrm{2}{k}−{n}} \:{so} \\ $$$${f}_{{n}} \left({x}\right)=\:\left(−\mathrm{1}\right)^{{n}} \:\sum_{{k}=\left[\frac{{n}−\mathrm{1}}{\mathrm{2}}\right]+\mathrm{1}} ^{{n}} \:{C}_{{n}} ^{{k}} \:\:\:\frac{\left(\mathrm{2}{k}\right)!}{\left(\mathrm{2}{k}−{n}\right)!}\:{x}^{\mathrm{2}{k}−{n}} \:\:. \\ $$$$ \\ $$

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