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FP3 Trigonometric Series Help

Sum the following trigonometric series.

(a) sinθ - 1/3 sin2θ + 1/9 sin3θ - 1/27 sin4θ + ...

(b) 1 -
(n1)\displaystyle \binom{n}{1} cosθcosθ + (n2)\displaystyle \binom{n}{2} cos^2θcos2θ -(n3)\displaystyle \binom{n}{3} cos^3θcos3θ +...+ (-1)^n cos^nθcosnθ

I'm pretty much stumped on both and the textbook isn't being particularly helpful
.

The only starting point I can think of for part (a) is that the series is convergent, with a limit 1/(1-(1/3)e^​θi).

Any help would be really appreciated
:confused:.
Reply 1
Original post by tbhlouise
Sum the following trigonometric series.

(a) sinθ - 1/3 sin2θ + 1/9 sin3θ - 1/27 sin4θ + ...

(b) 1 -
(n1)\displaystyle \binom{n}{1} cosθcosθ + (n2)\displaystyle \binom{n}{2} cos^2θcos2θ -(n3)\displaystyle \binom{n}{3} cos^3θcos3θ +...+ (-1)^n cos^nθcosnθ

I'm pretty much stumped on both and the textbook isn't being particularly helpful
.

The only starting point I can think of for part (a) is that the series is convergent, with a limit 1/(1-(1/3)e^​θi).

Any help would be really appreciated
:confused:.


sorry to be a pain but
post a picture of the question

what level work is this so I can see what you may use
Reply 2
Original post by TeeEm
sorry to be a pain but
post a picture of the question

what level work is this so I can see what you may use


Here's the q, it's A Level Further Pure
Reply 3
Original post by tbhlouise
Here's the q, it's A Level Further Pure


these can be summed by demoivre by turning them into geometric progression of a complex exponential (and occasionally binomials) and then equate real and imaginary parts.


here is 2 examples

pdf.pdf

I hope you work your problems out.

If you have no luck I can run you though one of them tomorrow afternoon after 4 p.m.

.
Reply 4
Original post by TeeEm
these can be summed by demoivre by turning them into geometric progression of a complex exponential (and occasionally binomials) and then equate real and imaginary parts.


here is 2 examples

pdf.pdf

I hope you work your problems out.

If you have no luck I can run you though one of them tomorrow afternoon after 4 p.m.

.


Thanks :smile:. I'll work through them now and hope I get clarity on this topic
Reply 5
Original post by tbhlouise
Thanks :smile:. I'll work through them now and hope I get clarity on this topic


otherwise I will look at one of your questions tomorrow...
way too busy now.
Reply 6
Original post by tbhlouise
Sum the following trigonometric series.

(a) sinθ - 1/3 sin2θ + 1/9 sin3θ - 1/27 sin4θ + ...

(b) 1 -
(n1)\displaystyle \binom{n}{1} cosθcosθ + (n2)\displaystyle \binom{n}{2} cos^2θcos2θ -(n3)\displaystyle \binom{n}{3} cos^3θcos3θ +...+ (-1)^n cos^nθcosnθ

I'm pretty much stumped on both and the textbook isn't being particularly helpful
.

The only starting point I can think of for part (a) is that the series is convergent, with a limit 1/(1-(1/3)e^​θi).

Any help would be really appreciated
:confused:.


You've got the right idea for (a) but you can't have a real series that sums to a complex resuit - you need to multiply your series by i and add to the corresponding series with cosines and then take the imaginary part of the result.

N.B, you might need to multiply the resultant series by eiθe^{i\theta} to get a more recognizable GP but I haven't worked it through, so that might not be necessary!
Original post by tbhlouise
Sum the following trigonometric series.

(b) 1 (n1)\displaystyle \binom{n}{1} cosθcosθ + (n2)\displaystyle \binom{n}{2} cos^2θcos2θ -(n3)\displaystyle \binom{n}{3} cos^3θcos3θ +...+ (-1)^n cos^nθcosnθ

I'm pretty much stumped on both and the textbook isn't being particularly helpful
.


1. Handy hint: learn latex.
2. Consider the binomial theorem and "take the real part" trick for b)
Reply 8
Original post by davros
You've got the right idea for (a) but you can't have a real series that sums to a complex resuit - you need to multiply your series by i and add to the corresponding series with cosines and then take the imaginary part of the result.

N.B, you might need to multiply the resultant series by eiθe^{i\theta} to get a more recognizable GP but I haven't worked it through, so that might not be necessary!


Thanks, that worked! I got an expression for cosθ+isinθ and multiplied top and bottom by its conjugate and then equated its imaginary part to get the right answer :smile:.

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