Re: Looking for a simple function with known values


William Elliot wrote:
On Mon, 20 Feb 2006, Anderson wrote:

do you see a simple increasing function with the following constraints:
f(0) = 0
f(1) = 1/2

f'(0) = sqrt(2)/2
f'(1) = log(2)/2

No. As log 2 and ln 2 are negative, f cannot be increasing at 1.

Um, ln(2) = 0.6931

In answer to the original question:

Consider a cubic f(x) = a3 x^3 + a2 x^2 + a1 x + a0.

f(0) = a0.
f(1) = a3 + a2 + a1 + a0
f'(0) = a1
f'(1) = 3 a3 + 2 a2 + a1

a0 and a1 are given immediately by your conditions.
You can easily solve for a2 and a3.

There is nothing to guarantee that such an interpolating cubic has
to be an increasing function, but in this case it is.

- Randy

.

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