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    Hello

    I am well and truly stuck on a question:

    Use cos n0 = 0.5((z^n)+(z^-n)) to express cos0+cos30+cos50+...+cos(2n-1)0 (where I have used 0 to represent theta) as a geometric series in terms of z. Hence find this sum in terms of 0.

    I have written the series as 0.5(((z^1)+(z^-1))+((z^3)+(z^-3)+((z^5)+(z^-5))+...+((z^(2n-1))+(z^(1-2n))). However, I cannot find a common ratio for this sequence and so do not know whether I have gone about it the wrong way.

    Please could someone help me?

    (P.S. Please quote me)
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    (Original post by Magu1re)
    Hello

    I am well and truly stuck on a question:

    Use cos n0 = 0.5((z^n)+(z^-n)) to express cos0+cos30+cos50+...+cos(2n-1)0 (where I have used 0 to represent theta) as a geometric series in terms of z. Hence find this sum in terms of 0.

    I have written the series as 0.5(((z^1)+(z^-1))+((z^3)+(z^-3)+((z^5)+(z^-5))+...+((z^(2n-1))+(z^(1-2n))). However, I cannot find a common ratio for this sequence and so do not know whether I have gone about it the wrong way.

    Please could someone help me?

    (P.S. Please quote me)
    Write it as a series with the first term very negative:

    \frac{1}{2}(z^{1-2n}+z^{3-2n}+...+z^{-3}+z^{-1}+z^1+z^3+z^5+...+z^{2n-1})
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    (Original post by ttoby)
    Write it as a series with the first term very negative:

    \frac{1}{2}(z^{1-2n}+z^{3-2n}+...+z^{-3}+z^{-1}+z^1+z^3+z^5+...+z^{2n-1})
    Thank you. I just could not work that out! haha
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    (Original post by ttoby)
    Write it as a series with the first term very negative:

    \frac{1}{2}(z^{1-2n}+z^{3-2n}+...+z^{-3}+z^{-1}+z^1+z^3+z^5+...+z^{2n-1})
    Sorry for bringing this up again but my last post was a bit premature. I did not follow the problem through to the end. I have
    0.5((z^(1-2n))+(z^(3-2n))+...+(z^(2n-3))+(z^(2n-1))
    which is equal to
    0.5(((z^(1-2n))((z^(4n))-1))/((z^(2))-1))

    How do I then write this in its simplest form? (in terms of theta)
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    (Original post by Magu1re)
    Sorry for bringing this up again but my last post was a bit premature. I did not follow the problem through to the end. I have
    0.5((z^(1-2n))+(z^(3-2n))+...+(z^(2n-3))+(z^(2n-1))
    which is equal to
    0.5(((z^(1-2n))((z^(4n))-1))/((z^(2))-1))

    How do I then write this in its simplest form? (in terms of theta)
    So you have so far:
    \dfrac{z^{1-2n}}{2}\left(\dfrac{z^{4n}-1}{z^2-1}\right)

    Using a similar method to how you would have established \cos n\theta = 0.5(z^n+z^{-n}), you can work out that i\sin n\theta = 0.5 (z^n-z^{-n}). This is useful because  0.5 (z^n-z^{-n})=0.5(z^n-\frac{1}{z^n})=0.5\frac{z^{2n}-1}{z^n} so 2i\sin n\theta = \frac{z^{2n}-1}{z^n}.

    Your answer so far can be re-written as \dfrac{1}{2}\left(\dfrac{(z^{4n}-1)/z^{2n}}{(z^2-1)/z}\right) and you can put this in terms of theta using the above formula.

    I'll give you an idea of the thought process that I went through in answering this question. My first instinct was to try and use the formula \cos n\theta = 0.5(z^n+z^{-n}) and perhaps replacing n with some other number as appropriate. I noticed that \cos n\theta = 0.5(z^n+z^{-n})=0.5(z^{2n}+1)/z^n which looked very similar to the terms at the top and bottom of the fraction in the sum of the g.p. but the problem here was the minus sign so I wouldn't have been able to use these workings to make a substitution. So then I thought, if the only difference here is a minus sign then perhaps I might have more luck getting a formula for the g.p. in terms of sin? So I put together a similar formula in terms of sin n theta and it turns out that I can use it to replace the terms involving z.
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    (Original post by ttoby)
    So you have so far:
    \dfrac{z^{1-2n}}{2}\left(\dfrac{z^{4n}-1}{z^2-1}\right)

    Using a similar method to how you would have established \cos n\theta = 0.5(z^n+z^{-n}), you can work out that i\sin n\theta = 0.5 (z^n-z^{-n}). This is useful because  0.5 (z^n-z^{-n})=0.5(z^n-\frac{1}{z^n})=0.5\frac{z^{2n}-1}{z^n} so 2i\sin n\theta = \frac{z^{2n}-1}{z^n}.

    Your answer so far can be re-written as \dfrac{1}{2}\left(\dfrac{(z^{4n}-1)/z^{2n}}{(z^2-1)/z}\right) and you can put this in terms of theta using the above formula.

    I'll give you an idea of the thought process that I went through in answering this question. My first instinct was to try and use the formula \cos n\theta = 0.5(z^n+z^{-n}) and perhaps replacing n with some other number as appropriate. I noticed that \cos n\theta = 0.5(z^n+z^{-n})=0.5(z^{2n}+1)/z^n which looked very similar to the terms at the top and bottom of the fraction in the sum of the g.p. but the problem here was the minus sign so I wouldn't have been able to use these workings to make a substitution. So then I thought, if the only difference here is a minus sign then perhaps I might have more luck getting a formula for the g.p. in terms of sin? So I put together a similar formula in terms of sin n theta and it turns out that I can use it to replace the terms involving z.

    Thank you so much for your help with this question. This time I have followed it through right up to the final answer and am pleased to say it is the same as that at the back of the book.

    That was an excellent explaination by the way!
 
 
 
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