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Question Number 216830 by MrGaster last updated on 22/Feb/25
Prove:∀x∈R,∣cos x∣+∣cos 2x∣+…+∣cos nx∣≥((n−1)/2)(n∈Z_(>0) )
Prove:xR,cosx+cos2x++cosnx∣≥n12(nZ>0)
Commented by MathematicalUser2357 last updated on 25/Feb/25
Or you could do {n∣n∈Z∧n>0} (or {n∣n∈N})
Oryoucoulddo{nnZn>0}(or{nnN})
Answered by MrGaster last updated on 23/Feb/25
Prove:Σ_(k=1) ^n ∣cos(kπ)∣≥((n−1)/2) (n ∈ Z_(>0) )  Let S_n (x)=Σ_(k=1) ^n ∣cos(kx)∣  ⇒S_n (x)=∣cos(x)∣+∣cos(2x)∣+…+∣cos(nx)∣  Consider T_n (x)=Σ_(k=1) ^n cos(kπ)  =((sin(((nx)/2))cos((((n+1)x)/2)))/(sin((x/2))))  ⇒∣T_n (x)∣≤(1/(∣sin((x/2))∣))  Let U_n (x)=Σ_(k=1) ^n cos^2 (kπ)  =(n/2)+(1/2)Σ_(k=1) ^n cos(2kx)  =(n/2)+(1/2)(((sin(nx)cos((n+1)x))/(sin(x))))  ⇒U_n (x)≤(n/2)+(1/2)  ∵ ∣cos(kx)∣≥cos^2 (kx)  ⇒S_n (x)≥U_n (x)  ⇒S_n (x)≥(n/2)−(1/2)  ∴S_n (x)≥((n−1)/2)  [Q.E.D]
Prove:nk=1cos(kπ)∣≥n12(nZ>0)LetSn(x)=nk=1cos(kx)Sn(x)=∣cos(x)+cos(2x)++cos(nx)ConsiderTn(x)=nk=1cos(kπ)=sin(nx2)cos((n+1)x2)sin(x2)⇒∣Tn(x)∣≤1sin(x2)LetUn(x)=nk=1cos2(kπ)=n2+12nk=1cos(2kx)=n2+12(sin(nx)cos((n+1)x)sin(x))Un(x)n2+12cos(kx)∣≥cos2(kx)Sn(x)Un(x)Sn(x)n212Sn(x)n12[Q.E.D]

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