
#1
Nov312, 04:44 AM

P: 354

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? 



#2
Nov312, 05:11 AM

Sci Advisor
HW Helper
Thanks
P: 26,167




#3
Nov312, 05:44 AM

P: 354

it would be: A(coskx +isinkx)+B(coskxisinkx)
which's (A+B)coskx+i(AB)sinkx .. A'coskx+iB'sinkx where's did the "i" go? 



#4
Nov312, 06:04 AM

Sci Advisor
HW Helper
Thanks
P: 26,167

Some trigonometric, exponential thing?i told you they won't both be real! 



#5
Nov1812, 03:11 AM

P: 354

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. 



#6
Nov1812, 04:09 AM

Sci Advisor
HW Helper
Thanks
P: 26,167

Hi M. next!
You use cos and sin, or real exponentials, if you're only interested in real solutions, 



#7
Nov1812, 04:44 AM

P: 354

Thanks, am grateful



Register to reply 
Related Discussions  
Integration of exponential and trigonometric forms  Calculus & Beyond Homework  10  
Is this an ego thing, a dishonesty thing, or a smart thing to do? (concerning grades)  Academic Guidance  10  
limit of a trigonometric thing  Calculus & Beyond Homework  3  
Exponential integral with trigonometric argument  Calculus  7  
Polynomial, trigonometric, exponential and fractal curves  General Math  1 