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Thread: De Moivres series problem.

  1. #1
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    De Moivres series problem.

    I managed the first parts, its the last part im having trouble with. (let x be theta)
    An infinite series is given by z - z^2 + z^3 - z^4
    i.) Assuming the series converges, find an expression for the sum.
    ii.) Given that z = 1/4(cos x + isin x) explain why the series converges for all values of x.
    This part was fine and the sum of the convergence series was
    $\displaystyle
    \frac{z}{1 + z}
    $
    So i assume we use this for part b.)
    b.) By using de moivres theorm or otherwise, prove that the sum of the infinite series
    $\displaystyle
    \frac{sin x }{4} - \frac{sin 2x}{4^2} + \frac{sin 3x}{4^3}...(-1)\frac{sin nx}{4^n}
    $
    Converges into:
    $\displaystyle
    \frac{4sinx}{17 + 8cos x}
    $

    Well im not sure how to do this by de moivres theorm...but by series:
    Sum of infinite series:
    $\displaystyle \frac{a}{1 - r}$
    So for a being the first in the series, this must be the imaginary part of
    $\displaystyle \frac{e^ix}{4}$
    And r, the common ration must be imaginary part of
    $\displaystyle \frac{- e^ix}{4}$

    So the convergence is imaginary part of:
    $\displaystyle
    \frac{\frac{e^ix}{4}}{1 + \frac{e^ix}{4}}
    $

    Which can be re-arranged to get:
    $\displaystyle
    \frac{4sin x}{16 + 4sin x}
    $
    Well the numerator is the same as required in the question however im not sure how to get the denominator...or in fact if i have even done this right.
    Any help would be greatley appriciated
    Thankyou
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  2. #2
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    From the way you have broken up this question, I am not entirely sure of its meaning.
    Are you saying that the series sums to $\displaystyle \frac{z}{{1 + z}}$? Which is correct.
    And it converges for all x if $\displaystyle z = \frac{{\cos (x) + i\sin (x)}}{4}$. Correct.
    Using that z, express the sum in terms of x?

    If so, here is what I would do.
    $\displaystyle \frac{z}{{1 + z}} = \frac{{z\overline {\left( {1 + z} \right)} }}
    {{\left| {1 + z} \right|^2 }} = \frac{{z\left( {1 + \overline z } \right)}}
    {{\left| {1 + z} \right|^2 }} = \frac{{z + \left| z \right|^2 }}
    {{\left| {1 + z} \right|^2 }}$.
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    Quote Originally Posted by Plato View Post
    From the way you have broken up this question, I am not entirely sure of its meaning.
    Are you saying that the series sums to $\displaystyle \frac{z}{{1 + z}}$? Which is correct.
    And it converges for all x if $\displaystyle z = \frac{{\cos (x) + i\sin (x)}}{4}$. Correct.
    Using that z, express the sum in terms of x?

    If so, here is what I would do.
    $\displaystyle \frac{z}{{1 + z}} = \frac{{z\overline {\left( {1 + z} \right)} }}
    {{\left| {1 + z} \right|^2 }} = \frac{{z\left( {1 + \overline z } \right)}}
    {{\left| {1 + z} \right|^2 }} = \frac{{z + \left| z \right|^2 }}
    {{\left| {1 + z} \right|^2 }}$.
    Thankyou for your reply .

    The question was broken into bits but i thought i'd provide the earlier parts of the question incase it was of some use because i assume from looking at the z series, part b of the question was related to it in some way:

    This was part b, the part i am stuck on
    b.) By using de moivres theorm or otherwise, prove that the sum of the infinite series
    $\displaystyle
    \frac{sin x }{4} - \frac{sin 2x}{4^2} + \frac{sin 3x}{4^3}...(-1)\frac{sin nx}{4^n}
    $
    Converges into:
    $\displaystyle
    \frac{4sinx}{17 + 8cos x}
    $

    But im not quite sure of your working?

    Thanks
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  4. #4
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    $\displaystyle z = \frac{{\cos (x) + i\sin (x)}}
    {4}\; \Rightarrow \;\sum\limits_{k = 1}^\infty {\left( { - 1} \right)^{k + 1} z^k } = \sum\limits_{k = 1}^\infty {\left( { - 1} \right)^{k + 1} \left[ {\frac{{\cos (kx)}}
    {{4^k }} + i\frac{{\sin (kx)}}
    {{4^k }}} \right]} $

    $\displaystyle \text{Im}\left( {\sum\limits_{k = 1}^\infty {\left( { - 1} \right)^{k + 1} \left[ {\frac{{\cos (kx)}}
    {{4^k }} + i\frac{{\sin (kx)}}
    {{4^k }}} \right]} } \right) = \sum\limits_{k = 1}^\infty {\left( { - 1} \right)^{k + 1} \frac{{\sin (kx)}}
    {{4^k }}}
    $
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  5. #5
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    Quote Originally Posted by Plato View Post
    $\displaystyle z = \frac{{\cos (x) + i\sin (x)}}
    {4}\; \Rightarrow \;\sum\limits_{k = 1}^\infty {\left( { - 1} \right)^{k + 1} z^k } = \sum\limits_{k = 1}^\infty {\left( { - 1} \right)^{k + 1} \left[ {\frac{{\cos (kx)}}
    {{4^k }} + i\frac{{\sin (kx)}}
    {{4^k }}} \right]} $

    $\displaystyle \text{Im}\left( {\sum\limits_{k = 1}^\infty {\left( { - 1} \right)^{k + 1} \left[ {\frac{{\cos (kx)}}
    {{4^k }} + i\frac{{\sin (kx)}}
    {{4^k }}} \right]} } \right) = \sum\limits_{k = 1}^\infty {\left( { - 1} \right)^{k + 1} \frac{{\sin (kx)}}
    {{4^k }}}
    $
    Sorry, i don't understand...How does the above prove the series converges to:

    $\displaystyle

    $
    $\displaystyle \frac{4sin x}{16 + 4sin x}
    $

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  6. #6
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    We know that $\displaystyle \sum\limits_{k = 1}^\infty {\left( { - 1} \right)^{k + 1} z^k } = \frac{z}{{1 + z}} = \frac{{z + \left| z \right|^2 }}{{\left| {1 + z} \right|^2 }}$.

    It follows from that $\displaystyle \text{Im}\left[ {\sum\limits_{k = 1}^\infty {\left( { - 1} \right)^{k + 1} z^k } } \right] = \left[ {\sum\limits_{k = 1}^\infty {\left( { - 1} \right)^{k + 1} \sin (kx)} } \right]$.
    AND $\displaystyle \text{Im}\left[ {\frac{{z + \left| z \right|^2 }}
    {{\left| {1 + z} \right|^2 }}} \right] = \frac{{\text{Im}(z)}}
    {{\left| {1 + z} \right|^2 }}$.

    Therefore, $\displaystyle z = \frac{{\cos (x) + i\sin (x)}}
    {4}\; \Rightarrow \;\frac{{\text{Im}(z)}}
    {{\left| {1 + z} \right|^2 }} = \frac{{\frac{{\sin (x)}}
    {4}}}{{\frac{{\left( {4 + \cos (x)} \right)^2 }}
    {{16}} + \frac{{\sin ^2 (x)}}
    {{16}}}} = \frac{{4\sin (x)}}
    {{17 + 8\cos (x)}}$
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  7. #7
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    Quote Originally Posted by Plato View Post
    We know that $\displaystyle \sum\limits_{k = 1}^\infty {\left( { - 1} \right)^{k + 1} z^k } = \frac{z}{{1 + z}} = \frac{{z + \left| z \right|^2 }}{{\left| {1 + z} \right|^2 }}$.

    It follows from that $\displaystyle \text{Im}\left[ {\sum\limits_{k = 1}^\infty {\left( { - 1} \right)^{k + 1} z^k } } \right] = \left[ {\sum\limits_{k = 1}^\infty {\left( { - 1} \right)^{k + 1} \sin (kx)} } \right]$.
    AND $\displaystyle \text{Im}\left[ {\frac{{z + \left| z \right|^2 }}
    {{\left| {1 + z} \right|^2 }}} \right] = \frac{{\text{Im}(z)}}
    {{\left| {1 + z} \right|^2 }}$.

    Therefore, $\displaystyle z = \frac{{\cos (x) + i\sin (x)}}
    {4}\; \Rightarrow \;\frac{{\text{Im}(z)}}
    {{\left| {1 + z} \right|^2 }} = \frac{{\frac{{\sin (x)}}
    {4}}}{{\frac{{\left( {4 + \cos (x)} \right)^2 }}
    {{16}} + \frac{{\sin ^2 (x)}}
    {{16}}}} = \frac{{4\sin (x)}}
    {{17 + 8\cos (x)}}$
    Oh i see. Thankyou very much!
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