
Numerical Analysis Question?
Find a linear tranformation of x=x(u) that will take the interval 0<=u<=1 onto the interval 1<=u<=1.
Show how this is done.
If this helps: Original equation is e^x using the Maclaurin expansion:
1+x+(x^2/2)+(x^3/6)+(x^4/24)+(x^5/120)+(x^6/720)
And how do you find the inverse tranformation u=u(x) that takes the interval 1<=x<=1 onto the interval 0<=u<=1
and the original should interval should read
0<=u<=1 onto the interval 1<=x<=1.

The way I do it:
Use a sketch. Locate the original interval on x and the desired interval on y using the endpoints. Now you have 2 points that describe a line. Just find the equation of the line.
Or, if you're good,
p1: (0,1)
p2: (1,1)
slope is 2/1
at x0, y=1, thus b=1
equation is y=2x 1
See how that works?

Are you sure such a transformation exists?
A linear transformation is a function L such that L(ax + by) = aL(x) + bL(y) for all x, y in your set of "vectors" and all a, b in your set of "scalars."
If your field of scalars includes the reals (which is typically does), then if we had a linear transformation L, we would have L(k) = kL(1) for all 0 <= k <= 1. In particular, L is always either nonpositive or nonnegative for all 0 <= k <= 1, so it is certainly not onto 1 <= u <= 1.
Am I missing something here? Do you mean a Moebius transformation or a conformal mapping or some other kind of function?

How 'bout U = 2u  1 ? â˜º
Doug
Tags for this Thread
Posting Permissions
 You may not post new threads
 You may not post replies
 You may not post attachments
 You may not edit your posts

Forum Rules