Interpolation Functions and their derivatives

  • Thread starter bugatti79
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  • #1
bugatti79
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Folks,

How do determine whether the derivative of a quadratic interpolation function ##ax^2+bx+c## is continous/discontinous in the context of the following

We have a a true solution approximated by 2 quadratic interpolation functions ie,

The approximation function
[itex]
f_1(x)=ax^2+bx+c, g \le x \le x_1\\ f_1(x)=dx^2+ex+f, x_1 \le x \le h

[/itex]

See attached my sketch.

Would'nt ##f_1(x)=f_2(x)## and ##f'_1(x)=f'_2(x)## at ##x_1## for the approximation function to be continous?
 

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Answers and Replies

  • #2
mathman
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It looks like f and f' are continuous at x1.
 
  • #3
AlephZero
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Whether or not you want or need the first deriviative to be continuous depends what the interpolation is used for.

A common notation is "C0-continuous" if the function is continuous but the first derivative is not (except by accident in a special case), and "C1-continuous" if the function and its first derivative are both continuous.

The word "continuous" on its own means "C0-continuous".
 
  • #4
HallsofIvy
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Whether or not you want or need the first deriviative to be continuous depends what the interpolation is used for.
True but there would be little point in using a quadratic to interpolate if we don't want the first derivative to be continuous. The point is that for f(x)= ax^2+ bx+ c, f'(x)= 2ax+b, f''(x)= 2a, a constant. If we only want to match values and don't need "smoothness", we would use piecewise linear functions. If we want to match up second derivatives, we should use piecewise cubics (cubic splines).
 
  • #5
bugatti79
792
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Whether or not you want or need the first deriviative to be continuous depends what the interpolation is used for.

A common notation is "C0-continuous" if the function is continuous but the first derivative is not (except by accident in a special case), and "C1-continuous" if the function and its first derivative are both continuous.

The word "continuous" on its own means "C0-continuous".

Well I am referring back to the finite element theory. My query is based on the authors comment as attached.
Why wouldn't the derivative of a second order lagrange interpolation function be continuous as I have shown in my original sketch.

In other words, would'nt ##f'_1(x_1)=f'_2(x_1)## hold and thus the derivatuve is continuous...?

True but there would be little point in using a quadratic to interpolate if we don't want the first derivative to be continuous. The point is that for f(x)= ax^2+ bx+ c, f'(x)= 2ax+b, f''(x)= 2a, a constant. If we only want to match values and don't need "smoothness", we would use piecewise linear functions. If we want to match up second derivatives, we should use piecewise cubics (cubic splines).
 

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