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Question Number 103314 by Ar Brandon last updated on 14/Jul/20

Linearise cos^(2(p−1)) (t)

Linearisecos2(p−1)(t)

Answered by OlafThorendsen last updated on 14/Jul/20

cos^(2(p−1)) (t) = [((e^(it) +e^(−it) )/2)]^(2(p−1)) cos^(2(p−1)) (t) = (1/4^(p−1) )Σ_(k=0) ^(2(p−1)) C_(2(p−1)) ^k e^(ikt) e^(−it(2p−2−k)) cos^(2(p−1)) (t) = (1/4^(p−1) )Σ_(k=0) ^(2(p−1)) C_(2(p−1)) ^k e^(−it(2p−2−2k)) cos^(2(p−1)) (t) = (1/4^(p−1) )Σ_(k=0) ^(2(p−1)) C_(2(p−1)) ^k e^(2it(k+1−p)) cos^(2(p−1)) (t) = (((2p−2)!)/4^(p−1) )Σ_(k=0) ^(2(p−1)) ((cos[2(k+1−p)t])/(k!(2p−2−k)!))

cos2(p−1)(t)=[eit+e−it2]2(p−1)cos2(p−1)(t)=14p−1∑2(p−1)k=0C2(p−1)keikte−it(2p−2−k)cos2(p−1)(t)=14p−1∑2(p−1)k=0C2(p−1)ke−it(2p−2−2k)cos2(p−1)(t)=14p−1∑2(p−1)k=0C2(p−1)ke2it(k+1−p)cos2(p−1)(t)=(2p−2)!4p−1∑2(p−1)k=0cos[2(k+1−p)t]k!(2p−2−k)!

Commented by Ar Brandon last updated on 14/Jul/20

Thanks

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