# Trigonometric Sum of a series

1. Mar 13, 2008

### asif zaidi

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$$\theta$$) + exp(i2$$\theta$$) +....+exp(in$$\theta$$)

and find formulas for trigonometric sums for

cos($$\theta$$) + cos(2$$\theta$$)+....+cos(n$$\theta$$)

and

sin($$\theta$$) + sin(2$$\theta$$)+....+sin(n$$\theta$$)

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$$\theta$$)) = 1/(1-cos $$\theta$$) - isin($$\theta$$)

Taking conjugate of denominator above equation reduces to

1/2 + i sin($$\theta$$)/(2(1-cos($$\theta$$))

Therefore

cos($$\theta$$) + cos(2$$\theta$$)+....+cos(n$$\theta$$) = 1/2
and
sin($$\theta$$) + sin(2$$\theta$$)+....+sin(n$$\theta$$) = sin($$\theta$$)/(2(1-cos($$\theta$$))

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?

2. Mar 13, 2008

### sutupidmath

well, you have to consider the fact that a geometric progression sum converges only when $$|q|<1$$ and diverges for $$|q|>1$$ 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

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

$$b_1,b_2,...,b_n$$ then $$q=\frac{b_n}{b_n_ -_1}$$

Last edited: Mar 13, 2008
3. Mar 13, 2008

### sutupidmath

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

that is

$$\lim_{n\rightarrow\infty}S_n=\lim_{n\rightarrow\infty}b_1\frac{1-q^{n}}{1-q}=\frac{b_1}{1-q}$$ for $$|q|<1$$
and

$$\lim_{n\rightarrow\infty}S_n=\infty$$ when $$|q|>1$$

Last edited: Mar 13, 2008
4. Mar 13, 2008

### MathematicalPhysicist

it diverges also for |q|=1.

5. Mar 13, 2008

### sutupidmath

yeah, sure.

how does one write in latex, smaller or equal, or : greater or equal??

6. Mar 13, 2008

### ircdan

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

7. Mar 13, 2008

### sutupidmath

NOT necessarly!!!

Then we would have to consider two cases:
1.

$$|e^{ni\theta}|<1$$ which would be true for any $$\theta<0$$ and

2.

$$|e^{ni\theta}|>1$$ which would be for any $$\theta>0$$ and also for

$$e^{ni\theta}=1$$ for $$\theta=0$$

where n is from naturals.

Last edited: Mar 13, 2008
8. Mar 13, 2008

### ircdan

|e^(int)| = |cos(nt) + isin(nt)| = sqrt(cos(nt)^2 + sin(nt)^2) = sqrt(1) = 1

9. Mar 14, 2008

### sutupidmath

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: Mar 14, 2008
10. Mar 14, 2008

### morphism

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.

11. Mar 14, 2008

### tiny-tim

Hi asif!

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)!

(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 $$e^{i\theta}\,+\,e^{2i\theta}\,+\,.\,.\,.\,+\,e^{ni\theta}$$

can be written: $$e^{i\theta}\,+\,(e^{i\theta})^2\,+\,.\,.\,.\,+\,(e^{i\theta})^n$$

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

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).

12. Mar 14, 2008

### sutupidmath

here at the end he suared the angles rather than the functions itself. i see now what he meant to do.

13. Mar 14, 2008

### asif zaidi

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$$^{i\theta}$$ + e$$^{i2\theta}$$ +... + e$$^{in\theta}$$

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

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

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

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

My question:

Is Step2 right?

Thanks

Asif

14. Mar 14, 2008

### tiny-tim

Hi asif!

(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 $$(1-e^{in\theta})$$, not $$(1-e^{i(n-1)\theta})$$, 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)!