Can someone please show me why e[itex]^{-ikx}[/itex] +

[BadAtMath6]

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

 

[mathman]

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

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

 

[BadAtMath6]

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}?

 

[Char. Limit]

Well, neither of your answers are really correct. Your answer is already in about as simple a form as it can get... maybe this could be simpler?

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

 

[symbolipoint]

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}?

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

 

[Char. Limit]

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

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

 

[symbolipoint]

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

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}

 

[symbolipoint]

As the addition instead of the subtraction, this process:
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.

 

[Char. Limit]

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.