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

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Discussion Overview

The discussion revolves around the simplification of the expression e^{-ikx} + e^{ikx}. Participants explore different interpretations and transformations of this expression, including its relation to trigonometric functions and potential simplifications.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants assert that e^{-ikx} + e^{ikx} simplifies to 2cos(kx), challenging the idea that it simplifies to 1 + e^{ikx}.
  • Others express confusion about exponent manipulation and suggest that changing e^{ikx} to its sine and cosine components is necessary for simplification.
  • One participant proposes an alternative form, e^{-ikx}(1 + e^{2ikx}), as a potentially simpler representation.
  • Several participants express that they arrived at the conclusion of 1 + e^{ikx} through their calculations, indicating a misunderstanding or error in their reasoning.
  • Another participant points out that the original expression was edited, which may have contributed to the confusion regarding the signs in the terms.
  • One participant confirms the correctness of the steps leading to the expression for 2cos(t) but questions how to revert it back to exponential form.

Areas of Agreement / Disagreement

Participants do not reach a consensus on the simplification of the expression, with multiple competing views on how to interpret and manipulate the terms involved.

Contextual Notes

Some participants express uncertainty regarding the manipulation of exponential forms and their relationship to trigonometric identities. There are unresolved issues related to the assumptions made during simplification and the potential for errors in reasoning.

BadAtMath6
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Can someone please show me why e[itex]^{-ikx}[/itex] + e[itex]^{ikx}[/itex] simplifies to 2e[itex]^{ikx}[/itex] instead of 1+e[itex]^{ikx}[/itex]??
 
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BadAtMath6 said:
Can someone please show me why e[itex]^{-ikx}[/itex] + e[itex]^{ikx}[/itex] simplifies to 2e[itex]^{ikx}[/itex] instead of 1+e[itex]^{ikx}[/itex]??
It doesn't at all. It is equal to 2cos(kx).
 


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


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?

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


BadAtMath6 said:
Ok, yes, that's true and I can get there if I change e[itex]^{ikx}[/itex] to cos(kx) + i*sin(kx). But I'm having exponent issues if I don't change it into sine and cosine (i.e. [itex]\frac{1}{e^{ikx}}[/itex] + [itex]\frac{e^{ikx}}{1}[/itex] ). Wouldn't that simplify to 1 + e[itex]^{ikx}[/itex]?

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


symbolipoint said:
I worked through the initial expression and also came to [itex]1+e^{ikx}[/itex]

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


BadAtMath6 said:
Can someone please show me why e[itex]^{-ikx}[/itex] + e[itex]^{ikx}[/itex] simplifies to 2e[itex]^{ikx}[/itex] instead of 1+e[itex]^{ikx}[/itex]??

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


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.
 


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

[tex]e^{- i t} + e^{i t}[/tex]

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

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