# sin^4 θ as linear combo of e^ikθ

• Oct 23rd 2010, 09:25 AM
SyNtHeSiS
sin^4 θ as linear combo of e^ikθ
Use the Binomial Theorem to write $sin^4\theta$ as a linear combo of functions of the form $e^{ik\theta}$, where k E Z.

Attempt:

$sin^4 \theta$
$= (\frac{e^{i\theta} - e^{-i \theta}}{2i})^4$
$= \frac{1}{16}(e^{i\theta} - e^{-i\theta})^4$

$(e^{i\theta} - e^{-i\theta})^4
= \sum_{k=0}^{4}\binom{4}{k}$
$(e^{i\theta})^{4 - k}(-e^{-i\theta})^k$
$=\binom{4}{0}(e^{i\theta})^4 + \binom{4}{1}(e^{i\theta})^3(-e^{-i\theta}) + \binom{4}{2}(e^{i\theta})^2(-e^{-i\theta})^2 + \binom{4}{3}(e^{i\theta})(-e^{-i\theta})^3 + (-e^{-i\theta})^4$
$=e^{i4\theta} - 4e^{i3\theta - i\theta} + 6e^{i2\theta + i2\theta} - 4e^{i\theta - i3\theta} + e^{i4\theta}$
$=e^{i4\theta} - 4e^{i2\theta} + 6e^{i4\theta} - 4e^{-i2\theta} + e^{i4\theta}$
$=8e^{i4\theta} - 4e^{i2\theta} - 4e^{-i2\theta}$

$sin^4 \theta$
$= \frac{1}{16}(8e^{i4\theta} - 4e^{i2\theta} - 4e^{-i2\theta})$
$=\frac{1}{2}e^{4\theta} - \frac{1}{4}e^{i2\theta} - \frac{1}{4}e^{-{i2\theta}}$

but I think I made a mistake somewhere
• Oct 23rd 2010, 10:30 AM
Plato
It should be just $6$ not $6e^{4\theta i}$.
• Oct 23rd 2010, 10:30 AM
Jhevon
Quote:

Originally Posted by SyNtHeSiS
Use the Binomial Theorem to write $sin^4\theta$ as a linear combo of functions of the form $e^{ik\theta}$, where k E Z.

Attempt:

$sin^4 \theta$
$= (\frac{e^{i\theta} - e^{-i \theta}}{2i})^4$
$= \frac{1}{16}(e^{i\theta} - e^{-i\theta})^4$

$(e^{i\theta} - e^{-i\theta})^4
= \sum_{k=0}^{4}\binom{4}{k}$
$(e^{i\theta})^{4 - k}(-e^{-i\theta})^k$
$=\binom{4}{0}(e^{i\theta})^4 + \binom{4}{1}(e^{i\theta})^3(-e^{-i\theta}) + \binom{4}{2}(e^{i\theta})^2(-e^{-i\theta})^2 + \binom{4}{3}(e^{i\theta})(-e^{-i\theta})^3 + (-e^{-i\theta})^4$
$=e^{i4\theta} - 4e^{i3\theta - i\theta} + 6e^{i2\theta + i2\theta} - 4e^{i\theta - i3\theta} + e^{i4\theta}$
$=e^{i4\theta} - 4e^{i2\theta} + 6e^{i4\theta} - 4e^{-i2\theta} + e^{i4\theta}$
$=8e^{i4\theta} - 4e^{i2\theta} - 4e^{-i2\theta}$

$sin^4 \theta$
$= \frac{1}{16}(8e^{i4\theta} - 4e^{i2\theta} - 4e^{-i2\theta})$
$=\frac{1}{2}e^{4\theta} - \frac{1}{4}e^{i2\theta} - \frac{1}{4}e^{-{i2\theta}}$

but I think I made a mistake somewhere

you did. note that $(-e^{-x})^2 = e^{-2x}$ and similarly, $(-e^{-x})^4 = e^{-4x}$

EDIT: which explains Plato's comment
• Oct 24th 2010, 04:09 AM
SyNtHeSiS
Thanks. So would the correct answer be:

$sin^4\theta = \frac{1}{8}cos4\theta = \frac{1}{2}cos2\theta + \frac{3}{8}$?
• Oct 24th 2010, 09:18 AM
Jhevon
Quote:

Originally Posted by SyNtHeSiS
Thanks. So would the correct answer be:

$sin^4\theta = \frac{1}{8}cos4\theta = \frac{1}{2}cos2\theta + \frac{3}{8}$?

??

• Oct 25th 2010, 05:41 AM
SyNtHeSiS
Quote:

Originally Posted by Jhevon

Yeah I did. Here is my working out:

$sin^4 \theta$
$= (\frac{e^{i\theta} - e^{-i \theta}}{2i})^4$
$= \frac{1}{16}(e^{i\theta} - e^{-i\theta})^4$

$(e^{i\theta} - e^{-i\theta})^4
= \sum_{k=0}^{4}\binom{4}{k}$
$(e^{i\theta})^{4 - k}(-e^{-i\theta})^k$
$=\binom{4}{0}(e^{i\theta})^4 + \binom{4}{1}(e^{i\theta})^3(-e^{-i\theta}) + \binom{4}{2}(e^{i\theta})^2(-e^{-i\theta})^2 + \binom{4}{3}(e^{i\theta})(-e^{-i\theta})^3 + (-e^{-i\theta})^4$
$=e^{i4\theta} - 4e^{i3\theta - i\theta} + 6e^{i2\theta - i2\theta} - 4e^{i\theta - i3\theta} + e^{-i4\theta}$
$=e^{i4\theta} - 4e^{i2\theta} + 6 - 4e^{-i2\theta} + e^{-i4\theta}$

$sin^4 \theta$
$=\frac{1}{16}(e^{i4\theta} - 4e^{i2\theta} + 6 - 4e^{-i2\theta} + e^{-i4\theta})$
$=\frac{1}{8}(\frac{e^{i4\theta + e^{-i4\theta}}}{2}) - \frac{1}{2}(\frac{e^{i2\theta} + e^{-i2\theta}}{2}) + \frac{3}{8}$
$=\frac{1}{8}cos4\theta = \frac{1}{2}cos2\theta + \frac{3}{8}$