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Question Number 170800 by thfchristopher last updated on 31/May/22
Prove that, for any real number x and odd positive integer n, cos^n x=(1/2^(n+1) )Σ_(k=0) ^((n−1)/2) C_k ^n cos (n−2k)x
$$\mathrm{Prove}\:\mathrm{that},\:\mathrm{for}\:\mathrm{any}\:\mathrm{real}\:\mathrm{number}\:{x}\:\mathrm{and}\:\mathrm{odd}\:\mathrm{positive}\:\mathrm{integer}\:{n}, \\ $$$$\:\:\:\:\:\:\:\mathrm{cos}^{{n}} {x}=\frac{\mathrm{1}}{\mathrm{2}^{{n}+\mathrm{1}} }\underset{{k}=\mathrm{0}} {\overset{\left({n}−\mathrm{1}\right)/\mathrm{2}} {\sum}}{C}_{{k}} ^{{n}} \mathrm{cos}\:\left({n}−\mathrm{2}{k}\right){x} \\ $$
Answered by aleks041103 last updated on 04/Jun/22
cosx=((e^(ix) +e^(−ix) )/2) cos^n x=(1/2^n )(e^(ix) +e^(−ix) )^n = =(1/2^n )(Σ_(k=0) ^n C_k ^n e^(ikx) e^(−i(n−k)x) )= =(1/2^n )(Σ_(k=0) ^n C_k ^n e^(ix(2k−n)) ) Σ_(k=0) ^n C_k ^n e^(ix(2k−n)) = =Σ_(k=0) ^((n−1)/2) C_k ^n e^(ix(2k−n)) +Σ_(k=(n+1)/2) ^n C_k ^n e^(ix(2k−n)) Σ_(k=(n+1)/2) ^n C_k ^n e^(ix(2k−n)) let n−2j=2k−n ⇒2j=2n−2k⇒j=n−k, k=n−j j_1 =n−((n+1)/2)=((n−1)/2) j_2 =n−n=0 ⇒Σ_(k=(n+1)/2) ^n C_k ^n e^(ix(2k−n)) =Σ_(j=0) ^((n−1)/2) C_(n−j) ^n e^(−ix(2j−n)) but C_(n−j) ^n =C_j ^n changing j to k ⇒Σ_(k=(n+1)/2) ^n C_k ^n e^(ix(2k−n)) =Σ_(k=0) ^((n−1)/2) C_k ^n e^(−ix(2k−n)) ⇒Σ_(k=0) ^n C_k ^n e^(ix(2k−n)) = =Σ_(k=0) ^((n−1)/2) C_k ^n e^(ix(2k−n)) +Σ_(k=0) ^((n−1)/2) C_k ^n e^(−ix(2k−n)) = =Σ_(k=0) ^((n−1)/2) C_k ^n (e^(ix(2k−n)) +e^(−ix(2k−n)) )= =2Σ_(k=0) ^((n−1)/2) C_k ^n cos((2k−n)x) ⇒cos^n x=(1/2^(n−1) )Σ_(k=0) ^((n−1)/2) C_k ^n cos((2k−n)x) cos is even function cos^n x=(1/2^(n−1) )Σ_(k=0) ^((n−1)/2) C_k ^n cos((n−2k)x)
$${cosx}=\frac{{e}^{{ix}} +{e}^{−{ix}} }{\mathrm{2}} \\ $$$${cos}^{{n}} {x}=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\left({e}^{{ix}} +{e}^{−{ix}} \right)^{{n}} = \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\left(\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} {e}^{{ikx}} {e}^{−{i}\left({n}−{k}\right){x}} \right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{n}} }\left(\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} {e}^{{ix}\left(\mathrm{2}{k}−{n}\right)} \right) \\ $$$$\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} {e}^{{ix}\left(\mathrm{2}{k}−{n}\right)} = \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\left({n}−\mathrm{1}\right)/\mathrm{2}} {\sum}}{C}_{{k}} ^{{n}} {e}^{{ix}\left(\mathrm{2}{k}−{n}\right)} +\underset{{k}=\left({n}+\mathrm{1}\right)/\mathrm{2}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} {e}^{{ix}\left(\mathrm{2}{k}−{n}\right)} \\ $$$$\underset{{k}=\left({n}+\mathrm{1}\right)/\mathrm{2}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} {e}^{{ix}\left(\mathrm{2}{k}−{n}\right)} \\ $$$${let}\:{n}−\mathrm{2}{j}=\mathrm{2}{k}−{n} \\ $$$$\Rightarrow\mathrm{2}{j}=\mathrm{2}{n}−\mathrm{2}{k}\Rightarrow{j}={n}−{k},\:{k}={n}−{j} \\ $$$${j}_{\mathrm{1}} ={n}−\frac{{n}+\mathrm{1}}{\mathrm{2}}=\frac{{n}−\mathrm{1}}{\mathrm{2}} \\ $$$${j}_{\mathrm{2}} ={n}−{n}=\mathrm{0} \\ $$$$\Rightarrow\underset{{k}=\left({n}+\mathrm{1}\right)/\mathrm{2}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} {e}^{{ix}\left(\mathrm{2}{k}−{n}\right)} =\underset{{j}=\mathrm{0}} {\overset{\left({n}−\mathrm{1}\right)/\mathrm{2}} {\sum}}{C}_{{n}−{j}} ^{{n}} {e}^{−{ix}\left(\mathrm{2}{j}−{n}\right)} \\ $$$${but}\:{C}_{{n}−{j}} ^{{n}} ={C}_{{j}} ^{{n}} \\ $$$${changing}\:{j}\:{to}\:{k} \\ $$$$\Rightarrow\underset{{k}=\left({n}+\mathrm{1}\right)/\mathrm{2}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} {e}^{{ix}\left(\mathrm{2}{k}−{n}\right)} =\underset{{k}=\mathrm{0}} {\overset{\left({n}−\mathrm{1}\right)/\mathrm{2}} {\sum}}{C}_{{k}} ^{{n}} {e}^{−{ix}\left(\mathrm{2}{k}−{n}\right)} \\ $$$$\Rightarrow\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}{C}_{{k}} ^{{n}} {e}^{{ix}\left(\mathrm{2}{k}−{n}\right)} = \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\left({n}−\mathrm{1}\right)/\mathrm{2}} {\sum}}{C}_{{k}} ^{{n}} {e}^{{ix}\left(\mathrm{2}{k}−{n}\right)} +\underset{{k}=\mathrm{0}} {\overset{\left({n}−\mathrm{1}\right)/\mathrm{2}} {\sum}}{C}_{{k}} ^{{n}} {e}^{−{ix}\left(\mathrm{2}{k}−{n}\right)} = \\ $$$$=\underset{{k}=\mathrm{0}} {\overset{\left({n}−\mathrm{1}\right)/\mathrm{2}} {\sum}}{C}_{{k}} ^{{n}} \left({e}^{{ix}\left(\mathrm{2}{k}−{n}\right)} +{e}^{−{ix}\left(\mathrm{2}{k}−{n}\right)} \right)= \\ $$$$=\mathrm{2}\underset{{k}=\mathrm{0}} {\overset{\left({n}−\mathrm{1}\right)/\mathrm{2}} {\sum}}{C}_{{k}} ^{{n}} {cos}\left(\left(\mathrm{2}{k}−{n}\right){x}\right) \\ $$$$\Rightarrow{cos}^{{n}} {x}=\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }\underset{{k}=\mathrm{0}} {\overset{\left({n}−\mathrm{1}\right)/\mathrm{2}} {\sum}}{C}_{{k}} ^{{n}} {cos}\left(\left(\mathrm{2}{k}−{n}\right){x}\right) \\ $$$${cos}\:{is}\:{even}\:{function} \\ $$$${cos}^{{n}} {x}=\frac{\mathrm{1}}{\mathrm{2}^{{n}−\mathrm{1}} }\underset{{k}=\mathrm{0}} {\overset{\left({n}−\mathrm{1}\right)/\mathrm{2}} {\sum}}{C}_{{k}} ^{{n}} {cos}\left(\left({n}−\mathrm{2}{k}\right){x}\right) \\ $$

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