# Can someone please show me why e$^{-ikx}$ +

1. Sep 25, 2011

Can someone please show me why e$^{-ikx}$ + e$^{ikx}$ simplifies to 2e$^{ikx}$ instead of 1+e$^{ikx}$??

Last edited: Sep 25, 2011
2. Sep 25, 2011

### mathman

Re: Simplifying

It doesn't at all. It is equal to 2cos(kx).

3. Sep 25, 2011

Re: Simplifying

Ok, yes, that's true and I can get there if I change e$^{ikx}$ to cos(kx) + i*sin(kx). But I'm having exponent issues if I don't change it into sine and cosine (i.e. $\frac{1}{e^{ikx}}$ + $\frac{e^{ikx}}{1}$ ). Wouldn't that simplify to 1 + e$^{ikx}$?

4. Sep 25, 2011

### Char. Limit

Re: Simplifying

$$e^{-i k x} \left( 1 + e^{2 i k x} \right)$$

5. Sep 25, 2011

### symbolipoint

Re: Simplifying

I worked through the initial expression and also came to $1+e^{ikx}$

6. Sep 25, 2011

### Char. Limit

Re: Simplifying

Show us your work. You're making a mistake here somewhere.

7. Sep 25, 2011

### symbolipoint

Re: Simplifying

I just noticed: You edited the original post for a sign change. I worked with what you wrote originally, e^(-ikx) - e^(ikx), $e^{-ikx}-e^{ikx}$

8. Sep 25, 2011

### symbolipoint

Re: Simplifying

Using t=kx,

cos(-t)+i*sin(-t) + cos(t)+ i*sin(t)
= cos(t) - i*sin(t) + cos(t) +i*sin(t)
= cos(t) + cos(t) + i*sin(t) - i*sin(t)
= 2cos(t)

Please check my understanding of these steps, since I cannot think of a way to transform that back to exponential form.

9. Sep 25, 2011

### Char. Limit

Re: Simplifying

Yup, you got that part right. The thing is, you started with the exponential form for 2cos(t). You started with

$$e^{- i t} + e^{i t}$$

and ended with 2 cos(t). What you started with IS the exponential form.