Normalize Function: F(theta)=2*e(-theta)*sin(2*theta)

[quick]

i need to normalize(F/Fmax) the function:

F(theta)=2*e(-theta)*sin(2*theta)

where theta is <= pi/2 and F(theta) is 0 otherwise.

theta can basically go to negative infinity which would make Fmax very large.

 

[HallsofIvy]

What do you mean by "normalize" here? Divide by its integral (over [itex]-\infty< \theta< \infty[/itex]) so that the integral becomes 1? Or, since you talk about "Fmax", divide by the maximum value? Do you have any reason to think that this function has either a finite integral or a finite maximum?

 

[tacman]

Normalizing a function means to scale it so that its maximum value is equal to 1. In this case, we can use the maximum value of F(theta) to normalize it. To find the maximum value of F(theta), we can take the derivative and set it equal to 0 to find the critical points.

Taking the derivative of F(theta):

F'(theta) = -2*e(-theta)*sin(2*theta) + 4*e(-theta)*cos(2*theta)

Setting it equal to 0 and solving for theta:

-2*e(-theta)*sin(2*theta) + 4*e(-theta)*cos(2*theta) = 0

2*e(-theta)*(-sin(2*theta) + 2*cos(2*theta)) = 0

sin(2*theta) = 2*cos(2*theta)

tan(2*theta) = 2

2*theta = arctan(2)

theta = arctan(2)/2

Since theta is <= pi/2, the maximum value of F(theta) occurs at theta = arctan(2)/2.

Now, we can normalize F(theta) by dividing it by its maximum value:

F(theta)/Fmax = (2*e(-theta)*sin(2*theta)) / (2*e(-arctan(2)/2)*sin(2*arctan(2)/2))

Simplifying:

F(theta)/Fmax = (e(-theta)*sin(2*theta)) / (e(-arctan(2)/2))

Therefore, the normalized function is:

F(theta)/Fmax = e(-theta)*sin(2*theta) / e(-arctan(2)/2)

F(theta)/Fmax = e(-theta+arctan(2)/2)*sin(2*theta)

This normalized function will have a maximum value of 1 and will be scaled accordingly.