Some trigonometric, exponential thing?

M. next

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?

tiny-tim

Hi M. next!
They won't both be real.

Try Euler's formula …

what do you get?

M. next

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

so B' = i(A-B) …

i told you they won't both be real!

M. next

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

Hi M. next!
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

Thanks, am grateful