# How to make this zero

## Main Question or Discussion Point

Hello,

I have the following summation:

$$\sum_{l=0}^Lb_l\exp\left(-j\frac{2\pi}{N}kl\right)$$

If

$$b_1=\cdots b_L=b$$

then by making L=N-1 the above summation will be zero. Now, if the coefficients are not equal, can we make the summation zero, and how?

Thanks

I don't think you can unless the coefficients are allowed to be negative, since the exponential function can never be negative, thus it can never add up to zero.

I don't think you can unless the coefficients are allowed to be negative, since the exponential function can never be negative, thus it can never add up to zero.
The coefficients are strictly greater than zero.

HallsofIvy
Homework Helper
I don't think you can unless the coefficients are allowed to be negative, since the exponential function can never be negative, thus it can never add up to zero.
The "j" here is the imaginary unit (called "i" by us ordinary "non-engineering folk) so your statement that "the exponential function can never be negative" is not true. For example, $e^{j\pi}= e^{i\pi}= -1$.

The "j" here is the imaginary unit (called "i" by us ordinary "non-engineering folk) so your statement that "the exponential function can never be negative" is not true. For example, $e^{j\pi}= e^{i\pi}= -1$.
That is right, j is the imaginary unit.

oh i didn't notice the imaginary unit, i tried some stuff with euler's identity, but nothing worked..

haruspex
Homework Helper
Gold Member
Hello,

I have the following summation:

$$\sum_{l=0}^Lb_l\exp\left(-j\frac{2\pi}{N}kl\right)$$

If

$$b_1=\cdots b_L=b$$

then by making L=N-1 the above summation will be zero.
Is that always true? What if k divides N? Won't all the terms reduce to bl?
Now, if the coefficients are not equal, can we make the summation zero, and how?
Thanks
If k is prime to N then the $$\exp\left(-j\frac{2\pi}{N}kl\right)$$ are just a reordering of $$\exp\left(-j\frac{2\pi}{N}l\right)$$. The sum can be rewritten as $$\sum_{l=0}^Lb_lω^l$$ where ω is a primitive Nth root of unity ($$ω=\exp\left(-j\frac{2\pi}{N}a\right)$$, where a is prime to N). Choose any m, n s.t. |m-n| is not a multiple of N/2 (easy if N odd). Each ωl can be written as a linear combination of ωm and ωn with real coefficients. So except for the special case where |m-n| = N/2, you can specify all but those two of the bl and solve to make the sum zero.
I haven't investigated the case where k is neither a multiple of N nor coprime to it, but a similar approach should work.

chiro
Hello,

I have the following summation:

$$\sum_{l=0}^Lb_l\exp\left(-j\frac{2\pi}{N}kl\right)$$

If

$$b_1=\cdots b_L=b$$

then by making L=N-1 the above summation will be zero. Now, if the coefficients are not equal, can we make the summation zero, and how?

Thanks
Do you need any other constraints other than the coeffecients being different?

As long as you don't get all the terms being zero, positive, or negative, then you are going to have a lot of different solutions.

If you had to get some kind of analytic expression, then one way to do this would be to put it into cos(x) + i sin(x) form and then use trig identities to get things in terms of exact square roots. This would be done for specific values of your variables that are supplied.

From this you do the above until you get one-degree of freedom left and use the information to set the coeffecient so that the whole expression equals zero.

An example of the square root thing is say finding cos(5pi/6) = cos(pi/3+pi/2) = cos(pi/3)cos(pi/2) - sin(pi/3)sin(pi/2) = -SQRT(3)/2.

You could generate a decomposition routine to find an analytic expression given everything inside the exponential and thus generate an exact surd form for the exponential expressions which will give you an exact relationship between all the b's. This is if you need exact solutions. If you don't, then you can use numeric analysis to get solutions that are 'good enough'.

The key thing that you will have to do is generate the results for prime denominators (i.e. cos(pi/2), cos(pi/3), cos(pi/5), cos(pi,7) and so on. Once you have these, you can then generate the exact form of every cos(pi*Q) where Q is any rational number.