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Question Number 144000 by bluberry508 last updated on 20/Jun/21

prove that ∀_m ∈N , a_k ,b_k ∈R cos^(2m) x =Σ_(k=1) ^m a_k cos 2kx cos^(2m−1) x=Σ_(k=1) ^m b_k cos (2k−1)x and find expr of a_k ,b_k in terms of k.

$$\mathrm{prove}\:\mathrm{that}\: \\ $$$$ \\ $$$$\forall_{{m}} \in\mathbb{N}\:,\:{a}_{{k}} ,{b}_{{k}} \in\mathbb{R} \\ $$$$\mathrm{cos}\:^{\mathrm{2}{m}} {x}\:=\underset{{k}=\mathrm{1}} {\overset{{m}} {\sum}}{a}_{{k}} \mathrm{cos}\:\mathrm{2}{kx} \\ $$$$\mathrm{cos}\:^{\mathrm{2}{m}−\mathrm{1}} {x}=\underset{{k}=\mathrm{1}} {\overset{{m}} {\sum}}{b}_{{k}} \mathrm{cos}\:\left(\mathrm{2}{k}−\mathrm{1}\right){x} \\ $$$$ \\ $$$$\mathrm{and}\:\:\mathrm{find}\:\mathrm{expr}\:\:\mathrm{of}\:\:{a}_{{k}} \:,{b}_{{k}} \:\mathrm{in}\:\mathrm{terms}\:\mathrm{of}\:\mathrm{k}. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Answered by mathmax by abdo last updated on 20/Jun/21

cos^(2m) x =(((e^(ix) +e^(−ix) )/2))^(2m) =(1/2^(2m) )Σ_(k=0) ^(2m) C_(2m) ^k (e^(ix) )^k (e^(−ix) )^(2m−k) =(1/2^(2m) )Σ_(k=0) ^(2m) e^(ikx) .e^(−i(2m−k)x) =(1/2^(2m) )Σ_(k=0) ^(2m) C_(2m) ^k e^(i(2k−2m)x) =(1/2^(2m) )Σ_(k=0) ^(2m) C_(2m) ^k e^(2i(k−m)x) =(1/2^(2m) )Σ_(k=0) ^(2m) C_(2m) ^k cos2(k−m)x +(i/2^(2m) )Σ_(k0) ^(2m) C_(2m) ^k sin2(k−m)x but cos^(2m) x is real ⇒cos^(2m) x=(1/2^(2m) )Σ_(k=0) ^(2m) C_(2m) ^k cos2(k−m)x =_(k−m=j) (1/2^(2m) )Σ_(j=−m) ^m C_(2m) ^(m+j) cos2jx =(1/2^(2m) )Σ_(j=−m) ^(−1) C_(2m) ^(m+j) cos(2jx) +(C_(2m) ^m /2^(2m) ) +Σ_(j=1) ^m C_(2m) ^(m+j) cos(2jx) =_(j=−k) (1/2^(2m) )Σ_(k=1) ^m C_(2m) ^(m−k) cos(2kx) +(C_(2m) ^m /2^(2m) ) +(1/2^(2m) )Σ_(k=1) ^m C_(2m) ^(m−k) cos(2kx) =1+(2/2^(2m) )Σ_(k=1) ^m C_(2m) ^(m−k) cos(2kx) =(C_(2m) ^m /2^(2m) )+(1/2^(2m−1) )Σ_(k=1) ^m C_(2m) ^(m−k) cos(2kx) ⇒a_k =Σ_(k=1) ^m (C_(2m) ^(m−k) /2^(2m−1) ) for k≥1 and a_0 =(C_(2m) ^m /2^(2m) ) cos^(2m) x=a_0 +Σ_(k=1) ^m a_k cos(2kx)

$$\mathrm{cos}^{\mathrm{2m}} \mathrm{x}\:=\left(\frac{\mathrm{e}^{\mathrm{ix}} +\mathrm{e}^{−\mathrm{ix}} }{\mathrm{2}}\right)^{\mathrm{2m}} \:=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2m}} }\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{2m}} \:\mathrm{C}_{\mathrm{2m}} ^{\mathrm{k}} \:\left(\mathrm{e}^{\mathrm{ix}} \right)^{\mathrm{k}} \left(\mathrm{e}^{−\mathrm{ix}} \right)^{\mathrm{2m}−\mathrm{k}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2m}} }\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{2m}} \:\mathrm{e}^{\mathrm{ikx}} .\mathrm{e}^{−\mathrm{i}\left(\mathrm{2m}−\mathrm{k}\right)\mathrm{x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2m}} }\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{2m}} \mathrm{C}_{\mathrm{2m}} ^{\mathrm{k}} \:\mathrm{e}^{\mathrm{i}\left(\mathrm{2k}−\mathrm{2m}\right)\mathrm{x}} =\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2m}} }\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{2m}} \mathrm{C}_{\mathrm{2m}} ^{\mathrm{k}} \:\mathrm{e}^{\mathrm{2i}\left(\mathrm{k}−\mathrm{m}\right)\mathrm{x}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2m}} }\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{2m}} \:\mathrm{C}_{\mathrm{2m}} ^{\mathrm{k}} \:\mathrm{cos2}\left(\mathrm{k}−\mathrm{m}\right)\mathrm{x}\:+\frac{\mathrm{i}}{\mathrm{2}^{\mathrm{2m}} }\sum_{\mathrm{k0}} ^{\mathrm{2m}} \:\mathrm{C}_{\mathrm{2m}} ^{\mathrm{k}} \:\mathrm{sin2}\left(\mathrm{k}−\mathrm{m}\right)\mathrm{x} \\ $$$$\mathrm{but}\:\mathrm{cos}^{\mathrm{2m}} \mathrm{x}\:\mathrm{is}\:\mathrm{real}\:\Rightarrow\mathrm{cos}^{\mathrm{2m}} \mathrm{x}=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2m}} }\sum_{\mathrm{k}=\mathrm{0}} ^{\mathrm{2m}} \:\mathrm{C}_{\mathrm{2m}} ^{\mathrm{k}} \:\mathrm{cos2}\left(\mathrm{k}−\mathrm{m}\right)\mathrm{x} \\ $$$$=_{\mathrm{k}−\mathrm{m}=\mathrm{j}} \:\:\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2m}} }\sum_{\mathrm{j}=−\mathrm{m}} ^{\mathrm{m}} \:\mathrm{C}_{\mathrm{2m}} ^{\mathrm{m}+\mathrm{j}} \:\mathrm{cos2jx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2m}} }\sum_{\mathrm{j}=−\mathrm{m}} ^{−\mathrm{1}} \mathrm{C}_{\mathrm{2m}} ^{\mathrm{m}+\mathrm{j}} \mathrm{cos}\left(\mathrm{2jx}\right)\:+\frac{\mathrm{C}_{\mathrm{2m}} ^{\mathrm{m}} }{\mathrm{2}^{\mathrm{2m}} }\:+\sum_{\mathrm{j}=\mathrm{1}} ^{\mathrm{m}} \:\mathrm{C}_{\mathrm{2m}} ^{\mathrm{m}+\mathrm{j}} \:\mathrm{cos}\left(\mathrm{2jx}\right) \\ $$$$=_{\mathrm{j}=−\mathrm{k}} \:\:\:\:\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2m}} }\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{m}} \:\mathrm{C}_{\mathrm{2m}} ^{\mathrm{m}−\mathrm{k}} \mathrm{cos}\left(\mathrm{2kx}\right)\:+\frac{\mathrm{C}_{\mathrm{2m}} ^{\mathrm{m}} }{\mathrm{2}^{\mathrm{2m}} }\:+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2m}} }\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{m}} \:\mathrm{C}_{\mathrm{2m}} ^{\mathrm{m}−\mathrm{k}} \:\mathrm{cos}\left(\mathrm{2kx}\right) \\ $$$$=\mathrm{1}+\frac{\mathrm{2}}{\mathrm{2}^{\mathrm{2m}} }\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{m}} \:\mathrm{C}_{\mathrm{2m}} ^{\mathrm{m}−\mathrm{k}} \:\mathrm{cos}\left(\mathrm{2kx}\right)\:=\frac{\mathrm{C}_{\mathrm{2m}} ^{\mathrm{m}} }{\mathrm{2}^{\mathrm{2m}} }+\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2m}−\mathrm{1}} }\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{m}} \:\mathrm{C}_{\mathrm{2m}} ^{\mathrm{m}−\mathrm{k}} \:\mathrm{cos}\left(\mathrm{2kx}\right) \\ $$$$\Rightarrow\mathrm{a}_{\mathrm{k}} =\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{m}} \:\frac{\mathrm{C}_{\mathrm{2m}} ^{\mathrm{m}−\mathrm{k}} }{\mathrm{2}^{\mathrm{2m}−\mathrm{1}} }\:\:\mathrm{for}\:\mathrm{k}\geqslant\mathrm{1}\:\mathrm{and}\:\mathrm{a}_{\mathrm{0}} =\frac{\mathrm{C}_{\mathrm{2m}} ^{\mathrm{m}} }{\mathrm{2}^{\mathrm{2m}} } \\ $$$$\mathrm{cos}^{\mathrm{2m}} \mathrm{x}=\mathrm{a}_{\mathrm{0}} +\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{m}} \:\mathrm{a}_{\mathrm{k}} \mathrm{cos}\left(\mathrm{2kx}\right) \\ $$

Commented by bluberry508 last updated on 20/Jun/21

wow... I haven′t ever expected this view thank you sir

$$\mathrm{wow}...\:\:\mathrm{I}\:\mathrm{haven}'\mathrm{t}\:\mathrm{ever}\:\mathrm{expected}\:\mathrm{this}\:\mathrm{view} \\ $$$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$

Commented by Dwaipayan Shikari last updated on 20/Jun/21

Sorry i have marked your comment as inappropiate by mistake .

$${Sorry}\:{i}\:{have}\:{marked}\:{your}\:{comment}\:{as}\:{inappropiate} \\ $$$${by}\:{mistake}\:.\: \\ $$

Commented by Rasheed.Sindhi last updated on 20/Jun/21

And I′ve cleared the red mark by liking the post!

$${And}\:{I}'{ve}\:{cleared}\:{the}\:{red}\:{mark}\:{by} \\ $$$${liking}\:{the}\:{post}! \\ $$

Commented by Dwaipayan Shikari last updated on 20/Jun/21

Thanks sir

$${Thanks}\:{sir} \\ $$

Commented by mathmax by abdo last updated on 20/Jun/21

you are welcome

$$\mathrm{you}\:\mathrm{are}\:\mathrm{welcome} \\ $$

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