# Interpolation functions

• Dec 22nd 2012, 08:41 AM
bugatti79
Interpolation functions
I am trying to establish what interpolation functions he used to obtain the 8 approximation functions (eqn 5.3.25a) shown in the attached excerpt from the book.

He states that a cubic approximation can be used for $w(x)$ and an interpendent quadratic approximation for $\Psi(x)$ thus I write

$\Delta^e_1=w^e_1=c_1^e+c_2^ex_e+c_3^ex_e^2+c_4^ex_ e^3$
$\Delta^e_2=S^e_1=c_1^e+c_2^ex_e+c_3^ex_e^2+0$

$\displaystyle\Delta^e_3=w^e_2=c_1^e+c_2^ex_{e+1}+c _3^ex_{e+1}^2+c_4^ex_{e+1}^3$
$\displaystyle\Delta^e_4=S^e_2=c_1^e+c_2^ex_{e+1}+c _3^ex_{e+1}^2+0$

Putting into matrix form and inverting to find the c values wolfram comes up with the matrix being singular (note I am using dummy variables).

inverse[{{1,x,x^2,x^3},{1,x,x^2,0},{1,y,y^2,y^3},{1,y,y^2, 0}}] - Wolfram|Alpha

Note: This has been submitted to PF 3 weeks ago with no response
Interdependent Interpolation Element- Timoshenko Beam

Regards
bugatti
• Dec 22nd 2012, 11:15 PM
chiro
Re: Interpolation functions
Hey bugatti79.

Can you reduce it to reduced row-echelon form to show what is linearly dependent?
• Dec 23rd 2012, 09:56 AM
bugatti79
Re: Interpolation functions
Hi Chiro,

See link from Wolfram [{{1,x,x^2,x^3},{1,x,x^2,0},{1,y,y^2,y^3},{1,y,y^2, 0}}] row echelon form - Wolfram|Alpha

This indicates the functions are linearly dependent, right?
• Dec 23rd 2012, 12:15 PM
chiro
Re: Interpolation functions
Yeah it's definitely linear dependent because the last row in the last column has a 0.

The next step is to figure out exactly what other vectors the last is linearly dependent on: in other words write {1,y,y^2,0} in terms of a linear combination of the other basis vectors and this will highlight what other data it depends on.
• Dec 23rd 2012, 12:39 PM
bugatti79
Re: Interpolation functions
Quote:

Originally Posted by chiro
Yeah it's definitely linear dependent because the last row in the last column has a 0.

What is the significance of $a_{44}=0$ for linear dependence since we also have $a_{24}=0$ in the original matrix..?

Quote:

Originally Posted by chiro
The next step is to figure out exactly what other vectors the last is linearly dependent on: in other words write {1,y,y^2,0} in terms of a linear combination of the other basis vectors and this will highlight what other data it depends on.

Not sure what we are trying to establish here...? Is my interpretation of the matrix correct according to the book?

thanks
• Dec 23rd 2012, 12:46 PM
chiro
Re: Interpolation functions
For the second question you need to consider how to write {1,y,y^2,0} = a*{1,x,x^2,x^3} + b*{1,x,x^2,0} + c*{1,y,y^2,y^3} for constants a, b, and c.

For the first question you might want to calculate the symbolic determinant such that it is unequal to zero for a given value of a_24 and a_44 and see what values they have to be to get a proper inverse.
• Dec 23rd 2012, 01:02 PM
bugatti79
Re: Interpolation functions
Quote:

Originally Posted by chiro
For the second question you need to consider how to write {1,y,y^2,0} = a*{1,x,x^2,x^3} + b*{1,x,x^2,0} + c*{1,y,y^2,y^3} for constants a, b, and c.

For the first question you might want to calculate the symbolic determinant such that it is unequal to zero for a given value of a_24 and a_44 and see what values they have to be to get a proper inverse.

Hmm, ok. Why do the vectors have to be linearly independent? Ie, does a matrix of linearly dependent vectors become singular upon inverse?
• Dec 23rd 2012, 01:10 PM
chiro
Re: Interpolation functions
You can't invert a matrix if the rows or columns are collinear: you get a singular matrix if this happens.

To answer your question, yes: a matrix becomes singular when its rows or columns are linearly dependent in some way.