Is the sum of cos = 1/2 a property of cos?

[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?

 

[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}

 

[sutupidmath]

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 {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

 

[MathematicalPhysicist]

it diverges also for |q|=1.

 

[sutupidmath]

it diverges also for |q|=1.

yeah, sure.

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

 

[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

 

[sutupidmath]

note if your sums were infinite they would diverge

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.

 

[ircdan]

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.

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

 

[sutupidmath]

|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...

 

[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.

 

[tiny-tim]

1- Have I approached this problem in the right way?

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).

 

[sutupidmath]

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.

 

[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

 

[tiny-tim]

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})

Is Step2 right?

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)!