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Bart
science forum beginner

Joined: 07 Jul 2005
Posts: 39

Posted: Wed Jun 21, 2006 8:36 am    Post subject: trigonometric identity

In order to understand a certain step in a proof, I would like to
proove the following identity:

exp(-Pi*I*N*theta)*sum_{k=1}^{N}exp(2*Pi*I*k*theta)
= sin(Pi*N*theta)cot(Pi*theta)

where I is the imaginary unit. However, I can't get there
Can anybody give me some hints?

Thanks,
Bart
Robert Low
science forum Guru

Joined: 01 May 2005
Posts: 1063

Posted: Wed Jun 21, 2006 8:42 am    Post subject: Re: trigonometric identity

Bart wrote:
 Quote: In order to understand a certain step in a proof, I would like to proove the following identity: exp(-Pi*I*N*theta)*sum_{k=1}^{N}exp(2*Pi*I*k*theta) = sin(Pi*N*theta)cot(Pi*theta) where I is the imaginary unit. However, I can't get there Can anybody give me some hints?

You might think of the sum as sum_{k=1}^N (exp(2pi i theta))^k
and use the formula for the sum of a (finite) geometric series.
Bart
science forum beginner

Joined: 07 Jul 2005
Posts: 39

Posted: Wed Jun 21, 2006 11:22 am    Post subject: Re: trigonometric identity

On 2006-06-21, Robert Low <mtx014@coventry.ac.uk> wrote:
 Quote: Bart wrote: In order to understand a certain step in a proof, I would like to proove the following identity: exp(-Pi*I*N*theta)*sum_{k=1}^{N}exp(2*Pi*I*k*theta) = sin(Pi*N*theta)cot(Pi*theta) where I is the imaginary unit. However, I can't get there Can anybody give me some hints? You might think of the sum as sum_{k=1}^N (exp(2pi i theta))^k and use the formula for the sum of a (finite) geometric series.

Hmm... strange... I have used the exponential sum formula (2) from
http://mathworld.wolfram.com/ExponentialSumFormulas.html (note
that my sum runs from 1 to N and the sum on this site runs from 0
to N-1, so be careful here!) and I arrive at

exp(-Pi*I*N*theta)*sum_{k=1}^{N}exp(2*Pi*I*k*theta)
= sin(Pi*N*theta)*cot(Pi*theta) + I*sin(Pi*N*theta)
^^^^^^^^^^^^^^^^^

So apparently i have an extra term there, namely I*sin(Pi*N*theta).

Did I make a mistake, or is the original equality as I posted it
not valid?

Thanks,
Bart
Robert Low
science forum Guru

Joined: 01 May 2005
Posts: 1063

Posted: Wed Jun 21, 2006 11:37 am    Post subject: Re: trigonometric identity

Bart wrote:
 Quote: On 2006-06-21, Robert Low wrote: Bart wrote: In order to understand a certain step in a proof, I would like to proove the following identity: exp(-Pi*I*N*theta)*sum_{k=1}^{N}exp(2*Pi*I*k*theta) = sin(Pi*N*theta)cot(Pi*theta) You might think of the sum as sum_{k=1}^N (exp(2pi i theta))^k and use the formula for the sum of a (finite) geometric series. ... I arrive at exp(-Pi*I*N*theta)*sum_{k=1}^{N}exp(2*Pi*I*k*theta) = sin(Pi*N*theta)*cot(Pi*theta) + I*sin(Pi*N*theta)

I get that too, so either we both made similar mistakes or
the original statement is wrong. Maybe it was just the real
part of the expression you were meant to compute?
Bart
science forum beginner

Joined: 07 Jul 2005
Posts: 39

Posted: Wed Jun 21, 2006 12:02 pm    Post subject: Re: trigonometric identity

On 2006-06-21, Robert Low <mtx014@coventry.ac.uk> wrote:
 Quote: I get that too, so either we both made similar mistakes or the original statement is wrong. Maybe it was just the real part of the expression you were meant to compute?

For as far as I understand the proof, I do not only need the real

The paper defines:

e(x) = exp(2*Pi*I*x)

A_q = (2*q+1)!/((q!)^2)

K_q(theta) = 1/N * sum_{k=1}^{N} A_q*(k/N)^q*(1-k/N)^q*e(k*theta)

And then states that

e^{-Pi*I*N*theta}K_0(theta) = 1/N*(sin(Pi*N*theta)*cot(Pi*theta))

For as far as I interpret all this, the extra term is somehow
forgotten here... right?

Best wishes,
Bart
Robert Low
science forum Guru

Joined: 01 May 2005
Posts: 1063

Posted: Wed Jun 21, 2006 1:12 pm    Post subject: Re: trigonometric identity

Bart wrote:
 Quote: On 2006-06-21, Robert Low wrote: I get that too, so either we both made similar mistakes or the original statement is wrong. Maybe it was just the real part of the expression you were meant to compute? The paper defines: e(x) = exp(2*Pi*I*x) A_q = (2*q+1)!/((q!)^2) K_q(theta) = 1/N * sum_{k=1}^{N} A_q*(k/N)^q*(1-k/N)^q*e(k*theta) And then states that e^{-Pi*I*N*theta}K_0(theta) = 1/N*(sin(Pi*N*theta)*cot(Pi*theta)) For as far as I interpret all this, the extra term is somehow forgotten here... right?

Beats me. As I said, it's always possible I messed up the algebra.
If so, I'm sure somebody will point it out fairly soon.

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