• Nov 5th 2008, 10:39 PM
Naena
Hi everyone.

Ive already posted this problem in the "Calculus" section like week ago, but nobody replied and i assumed that ive chosen a wrong section.

My problem is:
Find the local minima of the polynomial f(x)=ax^3+bx^2+cx+d, which is made by cubic interpolation from these values:

Suppose we have x1, x2, x3
and f(x1), f(x2), f(x3)
and f ' (x1) -> f derived in x1,

where x1,x2,x3 is in R
and x1<x2<x3,
and f(x1)!=f(x2)!=f(x3)
and f ' (x1) < 0

I should find a proof which showes, that the desired minima=x0 is always equal to:

x0=x1 + (x3 - x1) * { [1/3(w+z) - f1'] / [w+f13-f1'] }

where f1:=f(x1)
f2:=f(x2)
f3:=f(x3)

f12:=[f(x2) - f(x1)] / [x2 - x1]
f13:=[f(x3) - f(x1)] / [x3 - x1]
f23:=[f(x3) - f(x2)] / [x3 - x2]

f1' := f ' (x1)

v:= {[x3-x1]/[x2-x1]} * {f1' +f23 - f13 - f12}

z:= v - f13 + f1'

w:= sqrt { (z^2) - (3f1' * v) } ,

where sqrt(a) means a^(1/2) of course.

Naena
• Nov 5th 2008, 11:42 PM
CaptainBlack
Quote:

Originally Posted by Naena
Hi everyone.

Ive already posted this problem in the "Calculus" section like week ago, but nobody replied and i assumed that ive chosen a wrong section.

My problem is:
Find the local minima of the polynomial f(x)=ax^3+bx^2+cx+d, which is made by cubic interpolation from these values:

Suppose we have x1, x2, x3
and f(x1), f(x2), f(x3)
and f ' (x1) -> f derived in x1,

where x1,x2,x3 is in R
and x1<x2<x3,
and f(x1)!=f(x2)!=f(x3)
and f ' (x1) < 0

I should find a proof which showes, that the desired minima=x0 is always equal to:

x0=x1 + (x3 - x1) * { [1/3(w+z) - f1'] / [w+f13-f1'] }

where f1:=f(x1)
f2:=f(x2)
f3:=f(x3)

f12:=[f(x2) - f(x1)] / [x2 - x1]
f13:=[f(x3) - f(x1)] / [x3 - x1]
f23:=[f(x3) - f(x2)] / [x3 - x2]

f1' := f ' (x1)

v:= {[x3-x1]/[x2-x1]} * {f1' +f23 - f13 - f12}

z:= v - f13 + f1'

w:= sqrt { (z^2) - (3f1' * v) } ,

where sqrt(a) means a^(1/2) of course.

Naena

I think the main reason you got no response last time was that your explanation of the problem is very difficult to follow.

CB
• Nov 6th 2008, 11:46 PM
Naena
Trying to simplify it
Ok, i will try to put it in a simpler way:

Lets say we have x1, x2, x3 from R and also their functional values f1, f2, f3 from R. That means we have three points in R^2.
For example if x1=2, x2 = 5 , x2 = 12 and f1= 15, f2 = 7, f3 = 13, then these are our three points:
(2,15) (5,7) (12,13)
Now lets say we also know the value of the f1' , which is f derived in x1.
For example, if f= -7x^3 + 2x^2 + x + 8 , then f1' = f' (x1) = -21(x1)^2 + 4(x1) + 1 = -75
we also know that this derivation f1' is lower than zero.
Now, lets say that we want to find a function f, which looks like: ax^3 + bx^2 + cx + d , where a,b,c,d are constants from R.
This function should be made from our points (x1,f1), (x2,f2), (x3,f3) and it should be the "best" function , that we can make from them.
This method of aproximation is called interpolation: Interpolation - Wikipedia, the free encyclopedia
Normally, if we had 4 points, it would be very easy to perform a standard polynomial interpolation of third degree: Polynomial interpolation - Wikipedia, the free encyclopedia
On this wiki site Polynomial interpolation - Wikipedia, the free encyclopedia
is algorithm "Constructing the interpolation polynomial". But we cant use it, because it needs n+1 points for the polynomial of n-th degree, and we have only 3 points for 3-th degree.
So we have only 3 rows in our matrix, but in my opinion, we could add the last one as the 3x1^2 + 2x1 + x1 + 0 = f1' .
So now we have a 4x4 matrix which we should be able to solve:
(x1^3 x1^2 x1 1) (a) = ( f1 )
(x2^3 x2^2 x2 1) (b) = ( f2 )
(x3^3 x3^2 x3 1) (c) = ( f3 )
(3x1^2 2x1 1 0) (d) = ( f1')
or written different:
x1^3 + x1^2 + x1 + 1 = f1
x2^3 + x2^2 + x2 + 1 = f2
x3^3 + x3^2 + x3 + 1 = f3
3x1^2 + 2x1 + 1 + 0 = f1'

so we solve it ( of which i wasnt capable ) and find a,b,c,d

now we have our desired polynomial f(x) = ax^3 + bx^2 + cx + d.
But our task is to find its minimum, so we make its first derivation, and find when its equal to zero:
3ax^2 +2bx + c = 0
which means
xa=[-2b+sqrt(4b^2-12ac)] / 6a
or
xa=[-b+sqrt(b^2-3ac)]/3a

therefore:
xb=[-b-sqrt(b^2-3ac)]/3a

but our minimum is xb, not xa, because first derivation of f in x1 -> f1' is lower than zero.
but we already know a, b, and c, so we find xb and we are done.
BUT!
:)
It should be in exactly this shape:
xb=x1 + (x3 - x1) * { [1/3(w+z) - f1'] / [w+f13-f1'] }
where:
f12:=[f(x2) - f(x1)] / [x2 - x1]
f13:=[f(x3) - f(x1)] / [x3 - x1]
f23:=[f(x3) - f(x2)] / [x3 - x2]
z:= v - f13 + f1'
w:= sqrt { (z^2) - (3f1' * v) } ,

v:= {[x3-x1]/[x2-x1]} * {f1' +f23 - f13 - f12}

Which is kind of problem for me.

If u guys can solve it and prove that it really looks like this... respect...

I tried to put it in a simpler way, hope it helped, but if you still find something confusing or unclear, please - dont hesitate and reply, everything is helpful.

Thanks,
Naena