Tinkutara Equation Editor: Math Forum Question #: 101595


Question Number 101595 by mathmax by abdo last updated on 03/Jul/20

	Answered by mathmax by abdo last updated on 05/Jul/20
	$1) we have f(x) =cos^n x ⇒f(x) =(((e^(ix) +e^(−ix) )/2))^n =(1/2^n )Σ_(k=0) ^n C_n ^k (e^(ix) )^k (e^(−ix) )^(n−k) =(1/2^n ) Σ_(k=0) ^n C_n ^k e^(ikx) e^(−i(n−k)x) =(1/2^n ) Σ_(k=0) ^n C_n ^k e^(i(k−n+k)x) =(1/2^n )Σ_(k=0) ^n C_n ^k e^(i(2k−n)x) ⇒ f^((p)) (x) =(1/2^n ) Σ_(k=0) ^n C_n ^k (e^(i(2k−n)x) )^((p)) =(1/2^n ) Σ_(k=0) ^n C_n ^k (i(2k−n))^p e^(i(2k−n)x) ⇒ f^((n)) (x) =(1/2^n ) Σ_(k=0) ^n C_n ^k (i(2k+n))^n e^(i(2k−n)x) and f^((n)) (0) =(1/2^n ) Σ_(k=0) ^n (2k+n)^n i^n C_n ^k 2)f(x) =Σ_(p=0) ^∞ ((f^((p)) (x))/(p!)) x^p ⇒f(x) =Σ_(p=0) ^∞ (1/(p!)){(1/2^n ) Σ_(k=0) ^n C_n ^k i^p (2k−n)^p }x^p =Σ_(p=0) ^∞ A_p x^p with A_p =(1/2^n )×(1/(p!)) Σ_(k=0) ^n C_n ^k i^p (2k−n)^p$
	Answered by mathmax by abdo last updated on 05/Jul/20
	$3) we have cos^n x =(((e^(ix) +e^(−ix) )/2))^n =(1/2^n ) Σ_(k=0) ^n C_n ^k (e^(ix) )^k (e^(−ix) )^(n−k) =(1/2^n ) Σ_(k=0) ^n C_n ^k e^(ikx) e^((k−n)ix) =(1/2^n ) Σ_(k=0) ^n C_n ^k e^(i(2k−n)x) ⇒ ∫f(x)dx =(1/2^n ) Σ_(k=0) ^n C_n ^k ×(1/(i(2k−n)))e^(i(2k−n)x) +C =((−i)/2^n ) Σ_(k=0) ^n C_n ^k (1/(2k−n)){cos(2k−n)x +isin(2k−n)x} +C =−(i/2^n ) Σ_(k=0) ^n (C_n ^k /(2k−n))cos(2k−n)x +(1/2^n ) Σ_(k=0) ^n (C_n ^k /(2k−n)) sin(2k−n)x +c but ∫ cos^n x dx ∈ R ⇒ ∫ cos^n x dx =(1/2^n ) Σ_(k=0) ^n (C_n ^k /(2k−n))sin(2k−n)x +c$