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Some trigonometric, exponential thing?

by M. next
Tags: exponential, thing, trigonometric
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M. next
#1
Nov3-12, 04:44 AM
P: 378
How can we say:

f(x)=A'sin(kx)+B'cos(kx)

or equivalently

f(x)=Ae[itex]^{ikx}[/itex]+Be[itex]^{-ikx}[/itex]??

How are these two equivalent knowing that e[itex]^{ix}[/itex]=cosx+isinx

I don't get this?
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tiny-tim
#2
Nov3-12, 05:11 AM
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Hi M. next!
Quote Quote by M. next View Post
How can we say:

f(x)=A'sin(kx)+B'cos(kx)

or equivalently

f(x)=Ae[itex]^{ikx}[/itex]+Be[itex]^{-ikx}[/itex]??

How are these two equivalent knowing that e[itex]^{ix}[/itex]=cosx+isinx

I don't get this?
They won't both be real.

Try Euler's formula
what do you get?
M. next
#3
Nov3-12, 05:44 AM
P: 378
it would be: A(coskx +isinkx)+B(coskx-isinkx)
which's (A+B)coskx+i(A-B)sinkx
.. A'coskx+iB'sinkx
where's did the "i" go?

tiny-tim
#4
Nov3-12, 06:04 AM
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Some trigonometric, exponential thing?

Quote Quote by M. next View Post
it would be: A(coskx +isinkx)+B(coskx-isinkx)
which's (A+B)coskx+i(A-B)sinkx
so B' = i(A-B)
i told you they won't both be real!
M. next
#5
Nov18-12, 03:11 AM
P: 378
Sorry, i didn't check the site from since, I had some connection difficulties.
So, my final question, can this be done? Is the exponential form an alternative for the known trigonometric one?
And why do I use it? Why not keep it in trigonometric form. I am working on potential wells, free particles and so, if this information would help you answer my question.
tiny-tim
#6
Nov18-12, 04:09 AM
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Hi M. next!
Quote Quote by M. next View Post
Is the exponential form an alternative for the known trigonometric one?
And why do I use it? Why not keep it in trigonometric form.
Yes, they're equally valid alternatives.
You use cos and sin, or real exponentials, if you're only interested in real solutions,

but you use complex exponentials if you're interested in complex solutions.
M. next
#7
Nov18-12, 04:44 AM
P: 378
Thanks, am grateful


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