Author Topic: A proof (Read 1632 times) Tweet Share

Mao · « **on:** February 03, 2010, 01:17:53 am »

If anyone can be bothered, prove (or prove otherwise) that for any given two points $(x_1,y_1)$ and $(x_2,y_2)$ and the gradient at the first point $(x_1,y_1')$ , a cubic function of the form $f(x)=a (x-b)^3 + d$ can always be fitted. i.e. There exists at least one set of real solutions {a,b,d} such that

$\begin{align*} a (x_1 - b)^3 + d & = y_1\\ a (x_1 - b)^2 & = \frac{y_1'}{3} \\ a (x_2 - b)^3 + d & = y_2 \\ \end{align*} $

Cheers,

enwiabe · « **Reply #1 on:** February 03, 2010, 01:56:09 am »

Mao, just form the 3 equations, chuck them into a matrix and see if the determinant is non-zero

Mao · « **Reply #2 on:** February 03, 2010, 09:19:41 am »

Quote from: enwiabe on February 03, 2010, 01:56:09 am

Mao, just form the 3 equations, chuck them into a matrix and see if the determinant is non-zero

Sure, but they aren't linear

Mao · « **Reply #3 on:** February 04, 2010, 11:59:30 am »

Nevermind, I can't count. This proof was required as part of a bigger problem, but if I could count correctly, the bigger problem can be reduced to a system of linear equations, which makes this proof redundant.

Lesson learnt: LEARN TO COUNT.

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Author Topic: A proof (Read 1632 times) Tweet Share

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A proof

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