Processing math: 100%
Ask Your Question
1

Factoring out complex exponentials

asked 8 years ago

NahsiN gravatar image

updated 8 years ago

Hi, If I have an expression as follows 38Ey2¯E+ye(iωt+3ikx)+38Ey2¯Eye(iωt+ikx)+34EyE+y¯E+ye(iωt+ikx)+34EyE+y¯Eye(iωtikx)+38E+y2¯E+ye(iωtikx)+38E+y2¯Eye(iωt3ikx)

How do I factor out a complex exponential eiωtikx from the expression above using a command?

Preview: (hide)

Comments

Maybe: fc = e^(i*omega*t - i*k*x); f = ((f/fc).expand())*fc; f

mforets gravatar imagemforets ( 8 years ago )

1 Answer

Sort by » oldest newest most voted
0

answered 8 years ago

mforets gravatar image

updated 8 years ago

An example:

Let f=ae(iωt+3ikx)+b1e(iωt+ikx)+b2e(iωt+ikx)+c1e(iωtikx)+c2e(iωtikx)+de(iωt3ikx). Then

var('omega t k x a b1 b2 c1 c2 d ')
f = a*exp(I*omega*t+3*I*k*x) + b1*exp(I*omega*t+I*k*x) + b2*exp(I*omega*t+I*k*x) + c1*exp(I*omega*t-I*k*x) + c2*exp(I*omega*t-I*k*x) + d*exp(I*omega*t-3*I*k*x) 
fc = exp(I*omega*t-I*k*x)
g = ((f/fc).expand())*fc

produces g=(ae(4ikx)+b1e(2ikx)+b2e(2ikx)+de(2ikx)+c1+c2)e(iωtikx), and we can recover the term inside the parentheses with g.operands()[0].

Preview: (hide)
link

Your Answer

Please start posting anonymously - your entry will be published after you log in or create a new account.

Add Answer

Question Tools

1 follower

Stats

Asked: 8 years ago

Seen: 484 times

Last updated: Feb 11 '17