# Thread: De Moivres series problem.

1. ## 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

2. 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.
$\displaystyle \frac{z}{{1 + z}} = \frac{{z\overline {\left( {1 + z} \right)} }} {{\left| {1 + z} \right|^2 }} = \frac{{z\left( {1 + \overline z } \right)}} {{\left| {1 + z} \right|^2 }} = \frac{{z + \left| z \right|^2 }} {{\left| {1 + z} \right|^2 }}$.

3. Originally Posted by Plato
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.
$\displaystyle \frac{z}{{1 + z}} = \frac{{z\overline {\left( {1 + z} \right)} }} {{\left| {1 + z} \right|^2 }} = \frac{{z\left( {1 + \overline z } \right)}} {{\left| {1 + z} \right|^2 }} = \frac{{z + \left| z \right|^2 }} {{\left| {1 + z} \right|^2 }}$.

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

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

$\displaystyle \text{Im}\left( {\sum\limits_{k = 1}^\infty {\left( { - 1} \right)^{k + 1} \left[ {\frac{{\cos (kx)}} {{4^k }} + i\frac{{\sin (kx)}} {{4^k }}} \right]} } \right) = \sum\limits_{k = 1}^\infty {\left( { - 1} \right)^{k + 1} \frac{{\sin (kx)}} {{4^k }}}$

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

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

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

6. 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, $\displaystyle z = \frac{{\cos (x) + i\sin (x)}} {4}\; \Rightarrow \;\frac{{\text{Im}(z)}} {{\left| {1 + z} \right|^2 }} = \frac{{\frac{{\sin (x)}} {4}}}{{\frac{{\left( {4 + \cos (x)} \right)^2 }} {{16}} + \frac{{\sin ^2 (x)}} {{16}}}} = \frac{{4\sin (x)}} {{17 + 8\cos (x)}}$

7. Originally Posted by Plato
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!