With how the initial matrix is derived, are we guaranteed that an inverse always exists (at least, in the context of this course)?
SuryaTalla22
This is a great question: if we have a determinant of zero, that means that we have either no solution or infinitely many solutions. In every other case, an inverse is guaranteed to exist.
GarciaEricS
One thing I was wondering about this slide, is why do we have this inverse here? I don't mean why does the inverse solve it, but why do we use the inverse when the system can be greatly simplified. Well one reason is that we don't actually need to do the inverse everytime, we can simply solve explicitly the first time then never invert again. This actually leaves things a bit simpler, because we don't need to do substitution before solving the rest of the system, like we would do if we first simplified. Instead, we can just plug in all the values directly. So the inverse complicates things initially but actually makes things simpler (and faster) in the end.
spegeerino
@brandonlouie The matrix is the same for any cubic Hermite interpolation; it doesn't depend on the values that you want for the polynomial, just where in particular we are evaluating the polynomials. Intuitively, any cubic polynomial can be defined by 4 "independent pieces of information," so if we always assign 4 pieces of information (like the value and derivative of the polynomial at two distinct points x, y), we're always going to be able to solve this system of equations for the coefficients of the polynomial we want, and so the matrix inverse will have to exist. The only way the inverse wouldn't exist is if we gave pieces of information that either contradicted each other (like P(0) = 1 and P(0) = 2) or two identical pieces of information, which would yield a matrix equation with no solutions or infinitely many solutions respectively.
llejj
@spergeerino thanks for the clarification, it's kinda amazing that we can interpolate with such a concise method
Alescontrela
I wonder if cubic hermite polynomials can display weird boundary effects (i.e. in the case where P' is infinity). Are there better polynomial representations we can use?
With how the initial matrix is derived, are we guaranteed that an inverse always exists (at least, in the context of this course)?
This is a great question: if we have a determinant of zero, that means that we have either no solution or infinitely many solutions. In every other case, an inverse is guaranteed to exist.
One thing I was wondering about this slide, is why do we have this inverse here? I don't mean why does the inverse solve it, but why do we use the inverse when the system can be greatly simplified. Well one reason is that we don't actually need to do the inverse everytime, we can simply solve explicitly the first time then never invert again. This actually leaves things a bit simpler, because we don't need to do substitution before solving the rest of the system, like we would do if we first simplified. Instead, we can just plug in all the values directly. So the inverse complicates things initially but actually makes things simpler (and faster) in the end.
@brandonlouie The matrix is the same for any cubic Hermite interpolation; it doesn't depend on the values that you want for the polynomial, just where in particular we are evaluating the polynomials. Intuitively, any cubic polynomial can be defined by 4 "independent pieces of information," so if we always assign 4 pieces of information (like the value and derivative of the polynomial at two distinct points x, y), we're always going to be able to solve this system of equations for the coefficients of the polynomial we want, and so the matrix inverse will have to exist. The only way the inverse wouldn't exist is if we gave pieces of information that either contradicted each other (like P(0) = 1 and P(0) = 2) or two identical pieces of information, which would yield a matrix equation with no solutions or infinitely many solutions respectively.
@spergeerino thanks for the clarification, it's kinda amazing that we can interpolate with such a concise method
I wonder if cubic hermite polynomials can display weird boundary effects (i.e. in the case where P' is infinity). Are there better polynomial representations we can use?