1. YahooHoo
Join Date
Aug 2007
Posts
2

## 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.

2. YahooHoo
Join Date
Aug 2007
Posts
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 x-0, y=-1, thus b=-1

equation is y=2x -1

See how that works?

3. YahooHoo
Join Date
Aug 2007
Posts
1
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?

4. Entry Level Fresher
Join Date
Jun 2007
Posts
7
How 'bout U = 2u - 1 ? â˜º

Doug

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