Interpolation Functions and their derivatives

[bugatti79]

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
<br /> 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<br /> <br />

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?

 

[mathman]

It looks like f and f' are continuous at x₁.

 

[AlephZero]

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".

 

[HallsofIvy]

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).

 

[bugatti79]

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).