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Trigonometric Sum of a series

  • Thread starter asif zaidi
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I am not sure of the way in which I am solving. I am showing my approach and my questions are at end.

Plz advise why I maybe wrong?

Thanks

Asif

Problem Statement:

Use the formula for sum of a geometric progression to compute
exp(i[tex]\theta[/tex]) + exp(i2[tex]\theta[/tex]) +....+exp(in[tex]\theta[/tex])

and find formulas for trigonometric sums for

cos([tex]\theta[/tex]) + cos(2[tex]\theta[/tex])+....+cos(n[tex]\theta[/tex])

and

sin([tex]\theta[/tex]) + sin(2[tex]\theta[/tex])+....+sin(n[tex]\theta[/tex])


Solution

A geometric progression sum: 1/1-r (assuming sequence is r, r^2,...r^n

Therefore for this problem, the sum will be
1/(1-exp(i[tex]\theta[/tex])) = 1/(1-cos [tex]\theta[/tex]) - isin([tex]\theta[/tex])

Taking conjugate of denominator above equation reduces to

1/2 + i sin([tex]\theta[/tex])/(2(1-cos([tex]\theta[/tex]))

Therefore

cos([tex]\theta[/tex]) + cos(2[tex]\theta[/tex])+....+cos(n[tex]\theta[/tex]) = 1/2
and
sin([tex]\theta[/tex]) + sin(2[tex]\theta[/tex])+....+sin(n[tex]\theta[/tex]) = sin([tex]\theta[/tex])/(2(1-cos([tex]\theta[/tex]))

Problem:

1- Have I approached this problem in the right way
2- Does sum of cos = 1/2. Is this a property of cos? If so what is it called?
 

Answers and Replies

1,631
4
well, you have to consider the fact that a geometric progression sum converges only when [tex] |q|<1[/tex] and diverges for [tex] |q|>1[/tex] however in your problem you do not seem to have to calculate an infinite geometric progresion so, the formula for calculating the sum of n terms of a geometric progression is

[tex] S_n=b_1\frac{1-q^{n}}{1-q}[/tex] where q is the quotien, that is if this is a geom. sequence

[tex]b_1,b_2,...,b_n[/tex] then [tex]q=\frac{b_n}{b_n_ -_1}[/tex]
 
Last edited:
1,631
4
A geometric progression sum: 1/1-r (assuming sequence is r, r^2,...r^n

?
this is only valid when you take the limit of the sum of a geometric sequence as [tex] {n\rightarrow\infty}[/tex], and you need to have the first term of the sequence on the numerator, unless it is exactly 1.

that is

[tex]\lim_{n\rightarrow\infty}S_n=\lim_{n\rightarrow\infty}b_1\frac{1-q^{n}}{1-q}=\frac{b_1}{1-q}[/tex] for [tex] |q|<1[/tex]
and

[tex]\lim_{n\rightarrow\infty}S_n=\infty[/tex] when [tex] |q|>1[/tex]
 
Last edited:
MathematicalPhysicist
Gold Member
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it diverges also for |q|=1.
 
1,631
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229
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your answer is wrong because you didn't calculate the sum correctly, in general, when considering z + z^2 + z^3 + ... + z^N
let S = z + ... + Z^N, so zS = z^2 + ... + z^(N+1), so (1 - z)S = z - z^(N + 1)
so S = (z - z^(N + 1))/(1-z)
in your problem: z = e^(itheta)



note if your sums were infinite they would diverge
 
1,631
4
note if your sums were infinite they would diverge
NOT necessarly!!!

Then we would have to consider two cases:
1.

[tex]|e^{ni\theta}|<1[/tex] which would be true for any [tex]\theta<0[/tex] and

2.

[tex]|e^{ni\theta}|>1[/tex] which would be for any [tex]\theta>0[/tex] and also for

[tex]e^{ni\theta}=1[/tex] for [tex]\theta=0[/tex]

where n is from naturals.
 
Last edited:
229
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NOT necessarly!!!

Then we would have to consider two cases:
1.

[tex]|e^{ni\theta}|<1[/tex] which would be true for any [tex]\theta<0[/tex] and

2.

[tex]|e^{ni\theta}|>1[/tex] which would be for any [tex]\theta>0[/tex] and also for

[tex]e^{ni\theta}=1[/tex] for [tex]\theta=0[/tex]

where n is from naturals.
|e^(int)| = |cos(nt) + isin(nt)| = sqrt(cos(nt)^2 + sin(nt)^2) = sqrt(1) = 1
 
1,631
4
|e^(int)| = |cos(nt) + isin(nt)| = sqrt(cos(nt)^2 + sin(nt)^2) = sqrt(1) = 1
how do you prove that
|cos(nt) + isin(nt)| = sqrt(cos(nt)^2 + sin(nt)^2) ,this does not make any sens to me, please enlighten me!!!!!!!!

Also: you need not put e^(int) in abs values since it is always positive.

Moreover, as far as i am concerned you have made tons of mistakes here, because the following also is not true:
sqrt(cos(nt)^2 + sin(nt)^2) = sqrt(1)

.....lol......
 
Last edited:
morphism
Science Advisor
Homework Helper
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The absolute value signs |.| ircdan is using denote the norm of a complex number, i.e. |a+ib|=sqrt(a^2+b^2). Be careful to note that e^(int) isn't necessary positive - it may not even be a real number.

His penultimate equality follows from the trig identity sin^2+cos^2=1.
 
tiny-tim
Science Advisor
Homework Helper
25,789
249
1- Have I approached this problem in the right way?
Hi asif! :smile:

No. You've treated this as an infinite sum - that's not what the question asked for!

So your solution has no n in it …

And you've proceeded as if the sum starts with 1 +; it doesn't, and you'll have to put the 1 + in yourself, apply the forumula, and then take the 1 + out again (or just divide by the first term: see below)! :smile:

(Ignore what some people have been saying about sums not converging for |q| = 1: that only applies to infinite series, and this isn't infinite!)

Your [tex]e^{i\theta}\,+\,e^{2i\theta}\,+\,.\,.\,.\,+\,e^{ni\theta}[/tex]

can be written: [tex]e^{i\theta}\,+\,(e^{i\theta})^2\,+\,.\,.\,.\,+\,(e^{i\theta})^n[/tex]

which I think you can see is: [tex]e^{i\theta}\,\frac{1\,-\,e^{i\theta}^n}{1\,-\,e^{i\theta}}\,.[/tex]

Then, of course, to get the cos and sin sums, you take the conjugate of the denominator, and then separate out the real and imaginary parts (as you've correctly shown us).
 
1,631
4
The absolute value signs |.| ircdan is using denote the norm of a complex number, i.e. |a+ib|=sqrt(a^2+b^2). Be careful to note that e^(int) isn't necessary positive - it may not even be a real number.

His penultimate equality follows from the trig identity sin^2+cos^2=1.
here at the end he suared the angles rather than the functions itself. i see now what he meant to do.
 
56
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To everyone: thanks for the reply.

Hi Tim:

You are right that a geometric has 1 + r + r^2 + .... I saw that but didn't think too much about it.

So for my specific h/w here are the steps I am doing - I will explain what I didn't understand in your reply


Step1:
e[tex]^{i\theta}[/tex] + e[tex]^{i2\theta}[/tex] +... + e[tex]^{in\theta}[/tex]

Step2:
e[tex]^{i\theta}[/tex] (1 + e[tex]^{i\theta}[/tex] +... + e[tex]^{i(n-1)\theta}[/tex])
This is where I have the difference with you. You are saying it should be e[tex]^{in\theta}[/tex]

Step3:
The terms in the bracket converge to
(1-e[tex]^{i(n-1)\theta}[/tex])/(1-e[tex]^{i\theta}[/tex])

Step4:
Multiply Step3 by e[tex]^{i\theta}[/tex]
e[tex]^{i\theta}[/tex] * ((1-e[tex]^{i(n-1)\theta}[/tex])/(1-e[tex]^{i\theta}[/tex])

Step5:
Solve by taking conjugate of denominator and separating real and imaginary terms.

My question:

Is Step2 right?

Thanks

Asif
 
tiny-tim
Science Advisor
Homework Helper
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Step2:
e[tex]^{i\theta}[/tex] (1 + e[tex]^{i\theta}[/tex] +... + e[tex]^{i(n-1)\theta}[/tex])
This is where I have the difference with you. You are saying it should be e[tex]^{in\theta}[/tex]

Step3:
The terms in the bracket converge to
(1-e[tex]^{i(n-1)\theta}[/tex])/(1-e[tex]^{i\theta}[/tex])

Is Step2 right?
Hi asif! :smile:

(btw, don't say "converge" - that's only for infinite sums - say "are equal to"!)

Your step 2 is correct, but in step 3, you should have [tex](1-e^{in\theta})[/tex], not [tex](1-e^{i(n-1)\theta})[/tex], on the top!

Your step 2 and Step 3 (as corrected) together are the same as my third step.

And steps 4 and 5, of course, are correct.

Very good (except for the easy mistake)! :smile:
 

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