How to find $\operatorname{Re}\left(1+e^{i\triangle\varphi}+e^{i2\triangle\varphi}+...+e^{i\triangle\left(N-1\right)\varphi}\right) $

Question

So, actually this is a calculation I'm struggling to make in physics exercise.

I have to find $\operatorname{Re}\left(1+e^{i\triangle\varphi}+e^{i2\triangle\varphi}+...+e^{i\triangle\left(N-1\right)\varphi}\right) $ and it's supposed to be something that looks like this:

$ \frac{\sin\left(\frac{N\cdot\triangle\varphi}{2}\right)}{\frac{N\cdot\triangle\varphi}{2}} $

But I can't see how.

Here's my attempt:

Notice that we can write the sum as :

$ \operatorname{Re}\left(\sum_{j=0}^{N-1}\left(e^{i\triangle\varphi}\right)^{j}\right) $

And that's just a geometric series sum, so : $ \operatorname{Re}\left(\frac{1-e^{iN\triangle\varphi}}{1-e^{i\triangle\varphi}}\right) $

Now I'm not sure if there's easier way to find the real part, so I just wrote the complex number and tried to use algebraic manipulation:

$ \frac{1-e^{iN\triangle\varphi}}{1-e^{i\triangle\varphi}}=\frac{1-\cos\left(N\triangle\varphi\right)+i\sin\left(N\triangle\varphi\right)}{1-\cos\left(\triangle\varphi\right)+i\sin\left(\triangle\varphi\right)}=\frac{\left(1-\cos\left(N\triangle\varphi\right)+i\sin\left(N\triangle\varphi\right)\right)1-\cos\left(\triangle\varphi\right)-i\sin\left(\triangle\varphi\right)}{\left(1-\cos\left(\triangle\varphi\right)+i\sin\left(\triangle\varphi\right)\right)\left(1-\cos\left(\triangle\varphi\right)-i\sin\left(\triangle\varphi\right)\right)} $

After taking just the real terms from the product, I reached this:

$ =\frac{1-\cos\left(\triangle\varphi\right)-\cos\left(N\triangle\varphi\right)+\cos\left(N\triangle\varphi\right)\cos\left(\triangle\varphi\right)+\sin\left(N\triangle\varphi\right)\sin\left(\triangle\varphi\right)}{\left(1-\cos\left(\triangle\varphi\right)\right)^{2}+\sin^{2}\left(\triangle\varphi\right)} $

Which seems really different from the form I'm supposed to get to.

I'll write a reference to the physics calculation I'm trying to do: It's about a variation of the Double slit experiment . This variation is about wide slit experiment, say $ a $ is the width of the slit, and there are $ N $ light sources, the distance between to light waves given by $ \frac{a}{N} $ and each wave described by $ y_{j}\left(x,t\right)=A\cos\left(kx-\omega t+j\triangle\varphi\right) $. So I want to sum all of the waves and find the amplitude, thus:

$ \overline{y}\left(x,t\right)=\sum_{j=1}^{N-1}y_{j}\left(x,t\right)=A\cos\left(kx-\omega t\right)+A\cos\left(kx-\omega t+\triangle\varphi\right)+A\cos\left(kx-\omega t+2\triangle\varphi\right)+...+A\cos\left(kx-\omega t+\left(N-1\right)\triangle\varphi\right) $

So :$ A\cos\left(kx-\omega t\right)+A\cos\left(kx-\omega t+\triangle\varphi\right)+A\cos\left(kx-\omega t+2\triangle\varphi\right)+...+A\cos\left(kx-\omega t+\left(N-1\right)\triangle\varphi\right)=\operatorname{Re}\left(e^{i\left(kx-\omega t\right)}\left(1+e^{i\triangle\varphi}+e^{i2\triangle\varphi}+..e^{i\left(N-1\right)\triangle\varphi}\right)\right) $

And since I'm only interested in the amplitude, I want to calculate the sum I presented here in the first place:

$ \operatorname{Re}\left(1+e^{i\triangle\varphi}+e^{i2\triangle\varphi}+..e^{i\left(N-1\right)\triangle\varphi}\right) $

Thanks in advance.

Answer 1

It's not exactly as you think. You need a formula for the sum $1+\mathrm e^{i\theta}+\mathrm e^{2i\theta} +\dots+\mathrm e^{(n-1)i\theta}$.

Just apply first the formula for the sum of terms of a geometric progression: $$\sum_{k=0}^{n-1}\mathrm e^{ki\theta}=\frac{\mathrm e^{ni\theta}-1}{\mathrm e^{i\theta}-1}=\frac{\mathrm e^{\tfrac{ni\theta}2}}{\mathrm e^{\frac{i\theta}2}}\frac{\mathrm e^{\tfrac{ni\theta}2}-\mathrm e^{-\tfrac{ni\theta}2}}{\mathrm e^{\frac{i\theta}2}-\mathrm e^{-\frac{i\theta}2}}=\frac{\mathrm e^{\tfrac{ni\theta}2}}{\mathrm e^{\frac{i\theta}2}}\frac{2i\sin\tfrac{n\theta}2}{2i\sin\frac{\theta}2}=\mathrm e^{\tfrac{(n-1)i\theta}2}\frac{\sin\tfrac{n\theta}2}{\sin\frac{\theta}2},$$ from which you can easily deduce the real and the imaginary parts.

Answer 2

You have to add together all the real parts (I will use $t=\Delta$) $$ R=1+\cos t+\cos 2t+\cos 3t+\dots +\cos (N-1)t. $$ Now, you multiply by $\sin(t/2)$ and use the formulas to transform products into sums $$ \sin A\cos B=\frac 1{2}(\sin(A+B)+\sin(A-B)) $$ to get $$ \sin (t/2)R=\sin t/2+\frac 1{2}(\sin 3t/2-\sin t/2)+\frac 1{2}(\sin 5t/2-\sin 3t/2)+\dots +\frac 1{2}(\sin (N-1/2)t-\sin (N-3/2)t). $$ This is a telescopic sum where things cancel pairwise. You are left with $$ \sin (t/2)R=\frac{\sin t/2+\sin(N-1/2)t}{2} $$ which, after converting the sum into a product and solving, gives $$ R=\frac{\sin Nt/2\cos(N-1)t/2}{\sin t/2} $$ which differs from your answer by a factor.