how can solve this " sin(X)+sin(2x)+sin(3x)+....+sin(nx) " ?

Question

how can solve this " sin(X)+sin(2x)+sin(3x)+....+sin(nx) " ?

this Q is in "Complex number part" and i think it solve with "demo over" can you solve this? in this way

Answer 1

sin(X)+sin(2x)+sin(3x)+....+sin(nx)
= img[exp(ix)+exp(i2x)+exp(i3x)+...+exp(inx)]
=img[(exp(i(n+1)x) -1) / (exp(ix) - 1)]
=img[(exp(i(n+1)(x/2)) -1) /exp(i(n+1)(x/2)) * (exp(i(n+1)(x/2)) -exp(-i(n+1)(x/2))) / (exp(i(x/2)) - exp(-i(x/2)))
=[sin((1/2)(n+1)x)/sin((1/2)x)]* img(exp(inx/2))
=[sin((1/2)nx)/sin((1/2)x)]*[sin((1/2)(n+1)x)/sin((1/2)x)]/[sin((1/2)x]

[exp is exponential, img is imaginary part, 'i' is usual imaginary i in complex numbers] I also suggest you write down this so that you will be more clear with the equations. Sorry but without equation editorits bit difficult to type math.

If you want an infinite sum, please take the limit as x-->infinity of the last term obtained.

I hope this answeres your question. All the sest.

Answer 2

3 cos(x) + sin(2x) = 0 3 cos(x) + 2 sin(x)cos(x) = 0 cos(x) (3 + 2 sin(x)) = 0 ? x = ?/2 (4k ± a million) { ok ? ? } .......... cos(x) = 0 ? x = 2?k - asin(3/2) or ?(2k + a million) + asin(3/2) { ok ? ? } notice that asin(3/2) is a complicated fee. the only actual root is ?/2 (4k ± a million) answer: x = ?/2 (4k ± a million) { ok ? ? }

Answer 3

It is proven by an intuitive application of Taylor's theorem that,
e^iz = cos z + isin z , for real or complex z
then consider,
(e^iz0).(e^iz1).(e^iz2).....(e^izN) = e^i(z0+z1+...zN)
= cos [z0+z1+....+zN] + i.sin [z0+z1+....+zN]

when z0 = z1 =....=zN = z
(e^iz)^n = e^i(nz) = cos nz + i.sin nz

Now consider,
e^iz + e^2iz + e^3iz +......+ e^niz
= [cos z+ cos 2z+....+ cos nz] + i[sin z+ sin 2z+....+ sin nz]

Note now that Im {Σe^irz} is the required sum, now Σe^irz is a geometric progression with first term e^iz and ratio e^iz
so let the sum Σe^irz =C + iS
then C is the series of cosines and S is that of sines.
summing up
C + iS = (e^iz) (e^niz -1) / (e^iz - 1)
= (e^iz) [(e^niz -1)(e^-iz -1)] / [(e^iz-1)(e^-iz -1)]
= (e^iz) [e^(n-1)iz - e^niz - e^-iz +1] / [1 - (e^iz +e^-iz) +1]
= [e^niz - e^(n+1)iz + e^iz -1] / [2(1 - cos z)]
= [(cos nz - cos (n+1)z +cos z -1) + i(sin nz - sin (n+1)z
+sin z) ] / [2sin^2 (z/2)]
Now equate real and imaginary parts and S the imaginary part (equal to S) is the required series.

Answer 4

e^iθ = cosθ + i sinθ
e^2iθ = cos2θ + i sin2θ
e^3iθ = cos3θ + i sin3θ
..............................
e^niθ = cosnθ + i sinnθ

Sum these equations:
S = e^iθ + e^2iθ + e^3iθ + ... + e^niθ
= cosθ + cos2θ + cos3θ + ... + cosnθ + i(sinθ + sin2θ + sin3θ + ... + sinnθ)
= e^iθ (e^niθ - 1)/(e^iθ - 1) on summing a GP with a = e^iθ and r = e^iθ
= (e^niθ - 1)/(1 - e^-iθ) = (cosnθ - 1 + isinnθ)/(1 - cosθ - isinθ) * (1 - cosθ + isinθ)/(1 - cosθ + isinθ)
= [(cosnθ - 1)(1 - cosθ) + isinnθ*isinθ +i(sinnθ(1 - cosθ) - cosnθsinθ)/[(1 - cosθ)² - i²sin²θ)]
= [cosnθ - 1 - cosnθcosθ + cosθ - sinnθsinθ + i(sinnθ - sinnθcosθ - cosnθsinθ)]/[1 - 2 cosθ + cos²θ + sin²θ]
= [cosnθ + cosθ - 1 - cos(nθ - θ) + i (sinnθ - sin(nθ + θ)]/(2 - 2 cosθ)

Equating imaginery parts

sinθ + sin2θ + sin3θ + ... + sinnθ
= (sinnθ - sin(nθ + θ)]/(2 - 2 cosθ)
= (sin(nθ + θ) - sinnθ]/(2cos - 2)
= [2cos((2n+1)θ/2)sinθ/2]/[2(1 - 2 sin²(θ/2)) - 2]
= - cos((2n+1)θ/2)/2sin(θ/2)