# proof by induction, trigonometric identity help:

• Aug 8th 2010, 12:25 AM
stalker123
proof by induction, trigonometric identity help:
first up, sorry if this is the wrong forum to post but i was unsure where to post it under.

i am trying to prove that:

$\displaystyle 1 + 2* sum( cos(ku), k=1 to n) = sin( (2n+1)u/2) / sin(u/2):$
(ment to read sum( cos(ku), k=1 to n))

my induction assumption is that:
$\displaystyle 1 + 2( cos(u) +.....+ cos(ku) ) = sin( ku + u/2)/sin(u/2)$

and so i am trying to reduce:
$\displaystyle sin(ku + u/2)/sin(u/2) + 2cos( ku + u)$
to become
$\displaystyle sin( ku + 3u/2) / sin(u/2)$

this is where i am stuck. any help would be appreciated greatly
• Aug 8th 2010, 01:10 AM
Also sprach Zarathustra
A solution without using induction!

$\displaystyle cos(u) + cos(2u) + cos(3u) + ... + cos(nu)$

Let$\displaystyle T = \Sigma_{k = 1}^ {n}[ cos(ku) ]$

Multiplying by $\displaystyle 2 sin(\frac{u}{ 2})$:

$\displaystyle 2T sin(\frac{u}{ 2}) = \Sigma_{k = 1}^ {n}[2 sin(\frac{u}{ 2})cos(ku) ]$

Now, using the identity $\displaystyle 2 sin(\frac{u}{ 2}) cos(ku) = sin( [ \frac{(2k + 1) }{ 2} ]u ) - sin( [ \frac{(2k - 1)}{ 2} ]u )$:

$\displaystyle 2T sin(\frac{u}{ 2}) = \Sigma_{k = 1}^ {n}[ sin( [ \frac{(2k + 1)}{ 2 }]u ) - sin( [\frac{(2k - 1)} { 2} ]u ) ]$

Splitting the R.H.S into two separate summations, we will get:

$\displaystyle 2T sin(\frac{u}{ 2}) = \Sigma_{k = 1}^ {n} [ sin( [\frac{ (2k + 1)} { 2} ]u ) ] - \Sigma_{k = 1}^ {n} sin( [ \frac{(2k - 1)} { 2} ]u ) ]$

Or:

$\displaystyle 2T sin(\frac{u}{ 2})) = \Sigma _{k = 2}^ {n } [ sin( [ \frac{(2k - 1)} { 2} ]u ) ] - \Sigma_{k = 1}^ {n} [ sin( [\frac{ (2k - 1)} { 2} ]u ) ]$

The above is telescopic series:

$\displaystyle 2T sin(\frac{u}{ 2})$
$\displaystyle = sin( [ \frac{(2n + 1)} { 2} ]u ) - sin(\frac{u}{ 2})$
$\displaystyle = sin [\frac{ (2n + 1)} { 2} ]u ) - sin(\frac{u}{ 2})$
$\displaystyle = 2 sin(\frac{nu }{ 2}) cos( \frac{[ n + 1 ]u} { 2} )$

Hence:
$\displaystyle T =\frac{ sin(\frac{nu}{ 2}) cos( \frac{[ n + 1 ]u} { 2} ) }{sin(\frac{u}{ 2})}$
• Aug 8th 2010, 01:45 AM
Prove It
First of all, let's write this in a form that is readable...

$\displaystyle 1 + 2\sum_{k = 1}^{n}\cos{(ku)} = \frac{\sin{\left[\frac{(2n + 1)u}{2}\right]}}{\sin{\left(\frac{u}{2}\right)}}$.

Base step $\displaystyle n = 1$:

$\displaystyle LHS = 1 + 2\cos{u}$.

$\displaystyle RHS = \frac{\sin{\left(\frac{3u}{2}\right)}}{\sin{\left( \frac{u}{2}\right)}}$

$\displaystyle = \frac{\sqrt{\frac{1 - \cos{3u}}{2}}}{\sqrt{\frac{1 - \cos{u}}{2}}}$

$\displaystyle = \sqrt{\frac{\frac{1 - \cos{3u}}{2}}{\frac{1 - \cos{u}}{2}}}$

$\displaystyle = \sqrt{\frac{1 - \cos{3u}}{1 - \cos{u}}}$

$\displaystyle = \sqrt{\frac{1 - (4\cos^3{u} - 3\cos{u})}{1 - \cos{u}}}$

$\displaystyle = \sqrt{\frac{1 + 3\cos{u} - 4\cos^3{u}}{1 - \cos{u}}}$

$\displaystyle = \sqrt{\frac{(1 - \cos{u})(2\cos{u} + 1)^2}{1 - \cos{u}}}$

$\displaystyle = \sqrt{(2\cos{u} + 1)^2}$

$\displaystyle = 2\cos{u} + 1$

$\displaystyle = LHS$.

Inductive step - Assume the statement is true for $\displaystyle n$.

So we assume $\displaystyle 1 + 2\sum_{k = 1}^{n}\cos{(ku)} = \frac{\sin{\left[\frac{(2n + 1)u}{2}\right]}}{\sin{\left(\frac{u}{2}\right)}}$.

Now we need to show that the statement is true for $\displaystyle n + 1$, i.e. that
$\displaystyle 1 + 2\sum_{k = 1}^{n+1}\cos{(ku)} = \frac{\sin{\left[\frac{(2n + 3)u}{2}\right]}}{\sin{\left(\frac{u}{2}\right)}}$.

$\displaystyle LHS = 1 + 2\sum_{k = 1}^{n+1}\cos{(ku)}$.

$\displaystyle RHS = \frac{\sin{\left[\frac{(2n + 3)u}{2}\right]}}{\sin{\left(\frac{u}{2}\right)}}$

$\displaystyle = \frac{\sin{\left[\frac{(2n+1)u}{2} + \frac{2u}{2}\right]}}{\sin{\left(\frac{u}{2}\right)}}$

$\displaystyle = \frac{\sin{\left[\frac{(2n+1)u}{2} + u\right]}}{\sin{\left(\frac{u}{2}\right)}}$

$\displaystyle = \frac{\sin{\left[\frac{(2n+1)u}{2}\right]}\cos{(u)}+\cos{\left[\frac{(2n+1)u}{2}\right]}\sin{(u)}}{\sin{\left(\frac{u}{2}\right)}}$

Now I'm a little stuck too, will keep thinking about it...
• Aug 8th 2010, 02:32 AM
Vlasev
Also sprach, you are looking for $\displaystyle 1+2T$, not just $\displaystyle T$.
• Aug 8th 2010, 02:43 AM
Vlasev
BTW, this sum is known as the Dirichlet Kernel Clicky!

EDIT: Ahh, double post lol. I should have just edited!
• Aug 8th 2010, 06:06 AM
Quote:

Originally Posted by stalker123
first up, sorry if this is the wrong forum to post but i was unsure where to post it under.

i am trying to prove that:

$\displaystyle 1 + 2* sum( cos(ku), k=1 to n) = sin( (2n+1)u/2) / sin(u/2):$
(ment to read sum( cos(ku), k=1 to n))

my induction assumption is that:
$\displaystyle 1 + 2( cos(u) +.....+ cos(ku) ) = sin( ku + u/2)/sin(u/2)$

and so i am trying to reduce:
$\displaystyle sin(ku + u/2)/sin(u/2) + 2cos( ku + u)$
to become
$\displaystyle sin( ku + 3u/2) / sin(u/2)$

this is where i am stuck. any help would be appreciated greatly

P(n)

$\displaystyle 1+2\displaystyle\huge\sum_{k=1}^ncos(ku)=\frac{sin \left[(2n+1)\frac{u}{2}\right]}{sin\left(\frac{u}{2}\right)}$

P(n+1)

$\displaystyle 1+2\displaystyle\huge\sum_{k=1}^{n+1}cos(ku)=\frac {sin\left[(2[n+1]+1)\frac{u}{2}\right]}{sin\left(\frac{u}{2}\right)}\ ?$

Proof

$\displaystyle 1+2\displaystyle\huge\sum_{k=1}^{n+1}cos(ku)=1+2\s um_{k=1}^ncosku+2cos[(n+1)u]$

If P(n) is valid, this will equal

$\displaystyle \displaystyle\huge\frac{sin\left[(2n+1)\frac{u}{2}\right]}{sin\frac{u}{2}}+2cos[(n+1)u]$

Hence, the question is

$\displaystyle \displaystyle\huge\frac{sin\left([2(n+1)+1]\frac{u}{2}\right)}{sin\frac{u}{2}}=\frac{sin\left ([2n+1]\frac{u}{2}\right)+\left(sin\frac{u}{2}\right)2cos[(n+1)u]}{sin\frac{u}{2}}\ ?$

$\displaystyle sin\left([2n+3]\frac{u}{2}\right)=sin\left([2n+1]\frac{u}{2}\right)+2cos\left([n+1]u\right)sin\frac{u}{2}\ ?$

Expanding the rightmost term

$\displaystyle 2cos\left([n+1]u\right)sin\frac{u}{2}=sin\left([n+1]u+\frac{u}{2}\right)-sin\left([n+1]u-\frac{u}{2}\right)$

$\displaystyle =sin\left([2n+3]\frac{u}{2}\right)-sin\left([2n+1]\frac{u}{2}\right)$

The inter-term relationship is proven,
hence the trail of gunpowder has been placed all the way to that bank.
"What bank?"
That bank at infinity!

All that's required is to prove the equality is true for n=1..

$\displaystyle 1+2cosu=\displaystyle\huge\frac{sin\frac{3u}{2}}{s in\frac{u}{2}}=\frac{sin\left(u+\frac{u}{2}\right) }{sin\frac{u}{2}}$

and since

$\displaystyle sinu=2sin\frac{u}{2}cos\frac{u}{2}$

and using

$\displaystyle sin(A+B)=sinAcosB+cosAsinB$

the result follows