Asked by Omur
on 18 Oct 2012

Hello community, I'm trying to create a cubic polynomial function, I know these: y(0)=1 y(30)=0.5 dy(0)/dx=-1 dy(30)/dx=-1 Can I determine the values for coefficients ?

Answer by Dr. Seis
on 18 Oct 2012

Edited by Dr. Seis
on 18 Oct 2012

Accepted Answer

Yes. Just in case this is homework, I will give an example to get you started.

Let's say you have the case below (I will start with the case where we do not have derivatives at a location).

We are trying to solve for A, B, and C

y(x) = A*x^2 + B*x + C

Knowns:

x1 = 0; y1 = 1;

x2 = 3; y2 = 6;

x3 = 7; y3 = 7;

So we can construct an inverse problem of the form G * m = d, where:

G = [x1^2,x1,1; x2^2,x2,1; x3^2,x3,1];

d = [y1; y2; y3];

We determine m (equal to [A; B; C] in our example) by:

m = G\d;

And we get:

plot(-1:10,polyval(m,-1:10),'r-', [x1,x2,x3],[y1,y2,y3],'k+')

Dr. Seis
on 18 Oct 2012

Almost... since you will need to expand to a 3-order polynomial, I will give you the G for a 2-order (which would give you the plot immediately above if dydx2 = +1). In your case, your y and y' info are at the same 2 x locations... so:

| x1^2, x1, 1 | | A | | y1 |

| x2^2, x2, 1 | | B | = | y2 |

| 2*x1, 1, 0 | | C | | dydx1 |

| 2*x2, 1, 0 | | dydx2 |

You still need to have contribution from your "B", which is why the second column of the last two rows have the 1. Once you expand to a 3-order polynomial for your data, you should get something like:

Dr. Seis
on 18 Oct 2012

Omur
on 18 Oct 2012

Thank you so much !

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Answer by Suzie Oman
on 8 Jul 2017

This answer doesn't incorporate the derivative. How would this be solved taking into account the derivative?

What about in 3 dimensions?

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