# 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
$
\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
$
\frac{sin x }{4} - \frac{sin 2x}{4^2} + \frac{sin 3x}{4^3}...(-1)\frac{sin nx}{4^n}
$

Converges into:
$
\frac{4sinx}{17 + 8cos x}
$

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

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

Which can be re-arranged to get:
$
\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 $\frac{z}{{1 + z}}$? Which is correct.
And it converges for all x if $z = \frac{{\cos (x) + i\sin (x)}}{4}$. Correct.
Using that z, express the sum in terms of x?

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

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 $\frac{z}{{1 + z}}$? Which is correct.
And it converges for all x if $z = \frac{{\cos (x) + i\sin (x)}}{4}$. Correct.
Using that z, express the sum in terms of x?

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

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
$
\frac{sin x }{4} - \frac{sin 2x}{4^2} + \frac{sin 3x}{4^3}...(-1)\frac{sin nx}{4^n}
$

Converges into:
$
\frac{4sinx}{17 + 8cos x}
$

But im not quite sure of your working?

Thanks

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

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

$\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:

$

$
$\frac{4sin x}{16 + 4sin x}
" alt="\frac{4sin x}{16 + 4sin x}
" />

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

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

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

7. Originally Posted by Plato
We know that $\sum\limits_{k = 1}^\infty {\left( { - 1} \right)^{k + 1} z^k } = \frac{z}{{1 + z}} = \frac{{z + \left| z \right|^2 }}{{\left| {1 + z} \right|^2 }}$.

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

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