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Question Number 74501 by mathmax by abdo last updated on 25/Nov/19 | ||
$$\:{let}\:{P}\left({x}\right)=\frac{\mathrm{1}}{{n}!}\left({x}^{\mathrm{2}} −\mathrm{1}\right)^{{n}} \\ $$$${calculate}\:{P}^{\left({n}\right)} \left({x}\right)\:\:{and}\:{P}^{\:\left({n}\right)} \left(\mathrm{0}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\: \\ $$ | ||
Commented by mathmax by abdo last updated on 28/Nov/19 | ||
$${we}\:{have}\:{P}\left({x}\right)=\frac{\mathrm{1}}{{n}!}\sum_{{k}=\mathrm{0}} ^{{n}} \:{C}_{{n}} ^{{k}} \:{x}^{\mathrm{2}{k}} \left(−\mathrm{1}\right)^{{n}+{k}} \\ $$$$=\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:\sum_{{k}=\mathrm{0}} ^{{n}} \:\left(−\mathrm{1}\right)^{{k}} \:{C}_{{n}} ^{{k}} \:{x}^{\mathrm{2}{k}} \:\Rightarrow{P}^{\left({n}\right)} \left({x}\right)=\:\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}+\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\sum_{{k}=\mathrm{1}} ^{{n}} \:\left(−\mathrm{1}\right)^{{k}} \:{C}_{{n}} ^{{k}} \:\left({x}^{\mathrm{2}{k}} \right)^{\left.\right)\left.{n}\right)} \\ $$$$\left({x}^{\mathrm{2}{k}} \right)^{\left(\mathrm{1}\right)} =\left(\mathrm{2}{k}\right){x}^{\mathrm{2}{k}−\mathrm{1}} \:,\left({x}^{\mathrm{2}{k}} \right)^{\left(\mathrm{2}\right)} =\left(\mathrm{2}{k}\right)\left(\mathrm{2}{k}−\mathrm{1}\right)\:{x}^{\mathrm{2}{k}−\mathrm{2}} \:.... \\ $$$$\left({x}^{\mathrm{2}{k}} \right)^{\left({n}\right)} =\left(\mathrm{2}{k}\right)\left(\mathrm{2}{k}−\mathrm{1}\right)....\left(\mathrm{2}{k}−{n}+\mathrm{1}\right){x}^{\mathrm{2}{k}−\mathrm{1}} \\ $$$$=\frac{\left(\mathrm{2}{k}\right)!}{\left(\mathrm{2}{k}−{n}\right)!}\:{x}^{\mathrm{2}{k}−\mathrm{1}} \:\Rightarrow \\ $$$${P}^{\left({n}\right)} \left({x}\right)=\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}+\frac{\left(−\mathrm{1}\right)^{{n}} }{{n}!}\:\sum_{{k}=\mathrm{1}} ^{{n}} \left(−\mathrm{1}\right)^{{k}} \:{C}_{{n}} ^{{k}} \:\:\frac{\left(\mathrm{2}{k}\right)!}{\left(\mathrm{2}{k}−{n}\right)!}\:{x}^{\mathrm{2}{k}−\mathrm{1}} \\ $$ | ||