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Hello,
it's me again, the guy with the Hermite-interpolation.
I now decided to post a concrete task so maybe it's easier to answer.
The following points and the accompanying gradients are given:
x0=0 f(x)=3 f'(x)=0
x1=3 f(x)=3 f'(x)=27
x2=-2 f(x)=-57 f'(x)=72
I now would like to find out a polynomial function with goes through all those mentioned points and the same gradient in those points.
Thanks a lot in advance.
guy.
I don't know anything about that Hermite-something. But I played with your question using what little I know about points, gradients and characteristics of some known equations/functions. And I found an equation/function that you're looking for.Quote:
Originally Posted by guuy
Graphing those ordered pairs on the same x,y plane, I saw that the graph of the unknown function is rising steeply at point (-2,-57), then it level off at (0,3), then goes down after (0,3), and then rises steeply again at (3,3).
That looks like a vertical S-curve. A cubic curve in x.
So I played with:
f(x) = Ax^3 +Bx^2 +Cx +D -----(1)
So,
f'(x) = 3Ax^2 +2Bx +C -----------(2)
At (x=0, f'(x)=0), in (2):
0 = 3A(0) +2B(0) +C
C = 0 --------------------------***
So, (2) becomes
f'(x) = 3Ax^2 +2Bx ------------(2a)
At (x=3, f'(x)=27), in (2a):
27 = 3A(3^2) +2B(3)
9 = 9A +2B --------------------(3)
At (x=-2, f'(x)=72), in (2a):
72 = 3A((-2)^2) +2B(-2)
72 = 12A -4B
18 = 3A -B
B = 3A -18
Plug that into (3),
9 = 9A +2(3A -18)
9 = 9A +6A -36
A = 45/15 = 3 ---------------------------***
So, (2a) becomes f'(x) = 9x^2 +2Bx -----------(2b)
Back to (x=3, f'(x)=27), in (2b):
27 = 9(3^2) +2B(3)
27 = 81 +6B
B = (27-81)/6 = -9 ---------------------***
So (1) becomes f(x) = 3x^3 -9x^2 +D -----(1a)
At (x=0, f(x)=3), in (1a):
3 = 3(0) -9(0) +D
D = 3 ---------------------------------***
So (1a) becomes f(x) = 3x^3 -9x^2 +3 .
And, f(x) = 3x^3 -9x^2 +3 is the function you are looking for.
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I checked that on all the given ordered pairs and it checked allright.