Curve Fitting to Cubic poly

[markraz]

Hi, I have a yes/no question with regards to curve fitting. This example shows curve fitting to set of points

(reference https://math.stackexchange.com/questions/1882772/curve-fitting-with-derivatives ( see second post))

It shows how to fit points using 1st and 2nd derivatives. Can this methodology be extended to use 1st 2nd and 3rd derivatives??

Thanks in advance

 

[Deleted member 4993]

Hi, I have a yes/no question with regards to curve fitting. This example shows curve fitting to set of points

(reference https://math.stackexchange.com/questions/1882772/curve-fitting-with-derivatives ( see second post))

It shows how to fit points using 1st and 2nd derivatives. Can this methodology be extended to use 1st 2nd and 3rd derivatives??

Thanks in advance

View attachment 21504

Yes you can but in this case (cubic poly) it would be useless.

 

[HallsofIvy]

Yes. A linear function, y= ax+ b, has two coefficients, a and b, that need to be determined so you need to equations. One way to do that is to use two different points- a line is determined by two points- so you have $\displaystyle y_0= ax_0+ b$ and $\displaystyle y_1= ax_1+ b$ to solve for a and b. But knowing the value and slope (derivative) at a single point also gives two equations: $\displaystyle y_0= ax_0+ b$ and $\displaystyle a= a_0$ gives $\displaystyle y_0= a_0x_0+ b$ so $\displaystyle b= y_0- a_0x_0$. The equation of the linear function with value $\displaystyle y_0$ and slope $\displaystyle a_0$ at $\displaystyle x= x_0$ is $\displaystyle y= a_0x+ y_0- a_0x_0$.

Similarly, a quadratic, $\displaystyle y= ax^2+ bx+ c$ has three coefficients so we need three equations. We can get three equations using three points- a parabola is determined by three points- or using the value, first derivative, and second derivative at one point. For example, if $\displaystyle y(x_0)= y_0$, $\displaystyle y'(x_0)= m_0$, and $\displaystyle y''(x_0)= m_1$ then
$\displaystyle y(x_0)= ax_0^2+ bx_0+ c= y_0$
$\displaystyle y'(x_0)= 2ax_0+ b= m_0$ and
$\displaystyle y''(x_0)= 2a= m_1$.
Three equations we can solve for a, b, and c.

Finally, a cubic, $\displaystyle y= ax^3+ bx^2+ cx+ d$ has four coefficients so we need four equations. We can get those equations by using four points- a cubic is determined by four points- or using the value, first derivative, second derivative, and third derivative at one point. For example, if $\displaystyle y(x_0)= y_0$, $\displaystyle y'(x_0)= m_0$, $\displaystyle y''(x_0)= m_1$, and $\displaystyle y'''(x_0)= m_2$ then we have
$\displaystyle y(x_0)= ax_0^3+ bx_0^2+ cx_0+ d= y_0$
$\displaystyle y'(x_0)=3ax_0^2+ 2bx_0+ c= m_0$
$\displaystyle y''(x_0)= 6ax_0+ 2b= m_1$ and
$\displaystyle y'''(x_0)= 6a= m_2$.
Four equations we can solve for a, b, c, and d.

Of course, that can be extended to polynomials of any degree, n. Such a polynomial has n+ 1 coefficients so we need n+ 1 equations. Those can be the value of the polynomial at n+ 1 different points or the value of the polynomial and its first n derivatives at a single point.

 

[markraz]

Yes you can but in this case (cubic poly) it would be useless.

Thanks, So do you have to use a bspline or something like it to take advantage of 3rd derivative? What I'm looking to do is a G4 spline continuity

( source https://dpt3.dptcorporate.com/Content/Help/language/Mathconc/T_MC_ContinuityClassification.htm)

 

[markraz]

Yes. A linear function, y= ax+ b, has two coefficients, a and b, that need to be determined so you need to equations. One way to do that is to use two different points- a line is determined by two points- so you have $\displaystyle y_0= ax_0+ b$ and $\displaystyle y_1= ax_1+ b$ to solve for a and b. But knowing the value and slope (derivative) at a single point also gives two equations: $\displaystyle y_0= ax_0+ b$ and $\displaystyle a= a_0$ gives $\displaystyle y_0= a_0x_0+ b$ so $\displaystyle b= y_0- a_0x_0$. The equation of the linear function with value $\displaystyle y_0$ and slope $\displaystyle a_0$ at $\displaystyle x= x_0$ is $\displaystyle y= a_0x+ y_0- a_0x_0$.

Similarly, a quadratic, $\displaystyle y= ax^2+ bx+ c$ has three coefficients so we need three equations. We can get three equations using three points- a parabola is determined by three points- or using the value, first derivative, and second derivative at one point. For example, if $\displaystyle y(x_0)= y_0$, $\displaystyle y'(x_0)= m_0$, and $\displaystyle y''(x_0)= m_1$ then
$\displaystyle y(x_0)= ax_0^2+ bx_0+ c= y_0$
$\displaystyle y'(x_0)= 2ax_0+ b= m_0$ and
$\displaystyle y''(x_0)= 2a= m_1$.
Three equations we can solve for a, b, and c.

Finally, a cubic, $\displaystyle y= ax^3+ bx^2+ cx+ d$ has four coefficients so we need four equations. We can get those equations by using four points- a cubic is determined by four points- or using the value, first derivative, second derivative, and third derivative at one point. For example, if $\displaystyle y(x_0)= y_0$, $\displaystyle y'(x_0)= m_0$, $\displaystyle y''(x_0)= m_1$, and $\displaystyle y'''(x_0)= m_2$ then we have
$\displaystyle y(x_0)= ax_0^3+ bx_0^2+ cx_0+ d= y_0$
$\displaystyle y'(x_0)=3ax_0^2+ 2bx_0+ c= m_0$
$\displaystyle y''(x_0)= 6ax_0+ 2b= m_1$ and
$\displaystyle y'''(x_0)= 6a= m_2$.
Four equations we can solve for a, b, c, and d.

Of course, that can be extended to polynomials of any degree, n. Such a polynomial has n+ 1 coefficients so we need n+ 1 equations. Those can be the value of the polynomial at n+ 1 different points or the value of the polynomial and its first n derivatives at a single point.

thanks appreciate it. This ^ is great