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Infinite series of sin + cos both to the 2n power

  • #1
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Homework Statement


For the following series ∑an determine if they are convergent or divergent. If convergent find the sum.

(ii) ∑n=0 cos(θ)2n+sin(θ)2n[/B]


Homework Equations


geometric series, [/B]


The Attempt at a Solution



First I have to show that the equation is convergent.

Both cos(θ) and sin(θ) alternate between 1 and -1 depending on the value of θ. It was hinted to me by my instructor that I am to define a θ that the series is convergent.

Because cos(θ)2n is basically the same as as r2n, the same rule follows that it will be convergent if |r|<1. I need to find a θ that will let the cos and sin functions fit this rule. [/B]

Cos(0) = 1 and sin (pi/2) = 1. If i give the range 0<θ<(pi/2) then both terms in the series will be convergent because their magnitudes will be less than 1.

"find the sum"
here is where I am having trouble. I have a rule that Σrn = 1/(1-r) which does not seem to apply when you have r2n

Another problem I am having is that I cannot seem to define the sum in terms of θ. I believe that according to the question, the only appropriate way to find a sum is in terms of θ, but i do not have the tools to do this.


What would be the right way to attempt this problem? Right now in my course we are learning about ways of figuring out convergence or divergence of series and have actually done very little on the subject of actually finding the sum of these series. I have done some research and haven't found anything on my issue with r2n. Additionally if anyone could recommend me a good resource for learning this material, could you link it to me? My instructor doesn't use a book and seems to be diving into the deep end very quickly.

I greatly appreciate any help you guys can give me.
 

Answers and Replies

  • #2
34,070
9,957
If i give the range 0<θ<(pi/2) then both terms in the series will be convergent because their magnitudes will be less than 1.
That is true, but it is not the only range.

"find the sum"
here is where I am having trouble. I have a rule that Σrn = 1/(1-r) which does not seem to apply when you have r2n[/quote]
You can re-write r2n a bit to fix that.

Another problem I am having is that I cannot seem to define the sum in terms of θ. I believe that according to the question, the only appropriate way to find a sum is in terms of θ, but i do not have the tools to do this.
Well, r depends on θ, that does not matter for the calculation (θ is a constant).
 
  • #3
33
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That is true, but it is not the only range.

"find the sum"
here is where I am having trouble. I have a rule that Σrn = 1/(1-r) which does not seem to apply when you have r2n
You can re-write r2n a bit to fix that.

Well, r depends on θ, that does not matter for the calculation (θ is a constant).
I think i was using the term range incorrectly but I think you know what i was getting at.


I was playing around a bit with the values of θ with my calculator. I plugged in a few different θ's that obeyed my "range" for θ to make the series convergent and received different values each time.
When θ = 1 the sum is 4.83
when θ = 0.5 the sum is 5.65
when θ = 0.1 the sum is 101.34

as a sort of cop out after spending too much time on this problem, i decided to let θ = (pi/4) because it is in the middle of the "range" for θ I had defined and then that brought me back to my r2n problem.

Looking at it again, when I bring θ to pi/4 then the formula can be rewritten as 2*cos(pi/4)2n
Then I rewrite that as 2Σ (cos(pi/4)2)n
which simplifies to 2Σ 0.5n
the sum of this series is much easier to calculate and it sums to 4.

I am still not sure i answered appropriately since there is definitely a different sum with each θ. I have an idea though.

I will rewrite both the cos and sin terms in the sum
cos(θ)2n = (cos(θ)2)n
Then I will call cos(θ) A and rewrite it as (A2)n
I do the same to sin(θ) but call it B because it is a different value.

I know that A2 and B2 each are < 1 in magnitude.

Then I use the rule Σ rn = 1/(1-r)
Σ = 1/(1-A2) + 1/(1-B2)

I cannot simplify it more than this, but when i substitute sin and cos back in I get a sum in terms of θ

Σ = 1/(1-cos(θ)2) + 1/(1-sin(θ)2)


Thank you for getting my mind working, I think I might understand this a bit better, I had just forgotten that r2n is the same as (r2)n. I was also a little substitution shy i suppose.
 
  • #4
694
114
I know this might sound a little stupid, but in the question, each entire trigonometric function is raised to the power 2n, and not just theta, right?
 
  • #5
33
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I know this might sound a little stupid, but in the question, each entire trigonometric function is raised to the power 2n, and not just theta, right?
yeah you are correct. The question on my assignment is actually written out as cos2n(θ) which really gets on my nerves. Probably more correct that way.
 
  • #6
34,070
9,957
I think i was using the term range incorrectly but I think you know what i was getting at.
Yes, and you are missing values where the sum converges. The sum converges for values of θ you did not consider yet. Consider θ = 3/4 pi for example.
I cannot simplify it more than this, but when i substitute sin and cos back in I get a sum in terms of θ

Σ = 1/(1-cos(θ)2) + 1/(1-sin(θ)2)
Right.
 
  • #7
19,948
4,106
I cannot simplify it more than this, but when i substitute sin and cos back in I get a sum in terms of θ

Σ = 1/(1-cos(θ)2) + 1/(1-sin(θ)2)

Of course you can simplify it more than this.

[tex]1-\cos^2θ=?[/tex]
[tex]1-\sin^2θ=?[/tex]

Then reduce the two terms to a common denominator. Then make use of 2sinθcosθ=sin(2θ)

Chet
 

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