How to solve this? The first three terms of a different geometric progression are (4/3)cos^2(3theta), (16/9)cos^4(3theta), and (64/27)cos^6(3theta)

Kulla_9289 said:
i dont understand. pls could you break it down?
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From you posts I can't figure out what you understand and what you don't. Which means that short of posting a whole lecture or duplicating Wikipedia pages I don't know how to help you. Have you looked at the Wikipedia page I linked in an earlier post? Do you have questions about it? Also:
  1. Can you write a formula for a geometric progression with the first member [imath]a[/imath] and ratio [imath]r[/imath] ?
  2. Do you understand what a partial sum is?
  3. Can you see how partial sums look for [imath]r=\pm 1[/imath] ?
Please break down your own answer so that we can see how to help you.

Thanks.
 
Kulla_9289 said:
r = 4(cos^2(3theta))/3. I found the angles as pi/18, 11pi/18, 5pi/18, 7pi/18. I am stuck here.
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Kulla_9289 said:
the solution given is this: I dont understand how they got lesser than or equal to sign and thet
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Kulla_9289 said:
pi/18 < r < 5pi/18
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For the record, I think their answer is wrong, and @Kulla_9289 is right (except that the inequality is about theta, not r!); but then, I am not sure of the interpretation of the problem (what does "has a sum to infinity" mean?).

I've avoided joining in here while we wait for some substantive information from the OP about exactly what he does and does not understand; but on realizing this, I think it may be the underlying source of confusion.
 
Dr.Peterson said:
I am not sure of the interpretation of the problem (what does "has a sum to infinity" mean?).
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My interpretation is that the partial sums' limit is infinity, i.e. they are not limited by any finite value. For [imath]r=-1[/imath] the partial sums do not converge, but their absolute value remains limited.
 
blamocur said:
My interpretation is that the partial sums' limit is infinity, i.e. they are not limited by any finite value. For [imath]r=-1[/imath] the partial sums do not converge, but their absolute value remains limited.
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But if their answer is as shown, the reality is that |r| < 1 for pi/18 < theta < 5pi/18, so in the interior of their interval, the series will converge, not go to infinity! And if you take the problem as you do, at both endpoints, r is 1, not -1 (since the cosine is squared), so the partial sum still go to infinity, and there is no reason for < vs <=. Their answer seems to be wrong no matter how we interpret it.

Here is a graph of r vs theta:

1710711351081.png​

My initial understanding of the problem agreed with this: "the progression has a sum [i.e. converges, as the terms are taken] to infinity". Neither interpretation strikes me as normal wording.

I'm just confused.
 
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