Hi,
Finally made it back from the storm
Q1:
Dont fret over this too much. When they talk like that it's because in the more
general case we have a quadratic and not a straight line, because in ideas like
this one it is as if there is no such thing as a straight line. If it was a
straight line you wouldnt need this method
Also note that a straight line could be considered just a degenerate case of the
quadratic:
a*x^2+b*x+c=d
and with a very small 'a' we have:
bx+c=d
but to make this a quadratic we only need include a very very small 'a':
1e-12*x^2+b*x+c=d
and then instead of a straight line we have an approximation to a straight line
using a quadratic. So we can view every line as being a quadratic especially
since it will be very rare to find an actual straight line in this context.
[Strictly speaking 'a' would be very small relative to the other constants and 1e-12
is just a simple example].
Q2:
Equation 2.19 is better known as Lagrange's Interpolation Formula (for just
three points) and can be obtained in a much simpler manner.
They are just using simultaneous equations to find the unknown variables. That is
all that it is and nothing more. It doesnt matter that it is a quadratic. The
only thing that matters is that we find a, b, and c.
Using the equation:
a*x^2+b*x+c=f
we just repeat that as many times as needed. We need to find three variables so
we need to repeat this three times. For each equation we use one point (x,f) which
we usually write (x,y) but here they are using 'f' as the function value at x.
So we pick three points:
(x1,f1), (x2,f2), (x3,f3)
and we use them in the equation above forming three equations:
a*x1^2+b*x1+c=f1
a*x2^2+b*x2+c=f2
a*x3^2+b*x3+c=f3
and this is now just a set of linear equations because x1, x2, x3, and f1, f2, and f3
are all constants and we only need to find the variables a, b, and c.
So there are many methods you can use there to find a,b,c and you should get the same
results if you just solve for a,b,c and then plugging the solutions back into the
equation and then factoring, simplifying, etc. It may not be easy though.
You could also calculate the determinate they show and then set it to zero and then
solve for 'f' and then simplify, but again it may not be easy because it's a 4x4.
When they say 'conveniently' they are not suggesting a simple solution to that determinant
i think what they really just mean is that it can be written that way which is much
simpler to write than other methods might be.
In many cases however we would leave the problem in matrix form and just use a matrix
solver to do the math when needed (when we dont need an analytical result).
As a side note, Lagrange's Interpolation Formula can be obtained by noting the pattern in that three point formula. In the numerators, there are terms (x-m) where m is one of the xi's, but with one of them missing (instead of all three xi's there are only two). In the other numerators, leave out a different xi and that forms the numerators. The denominators follow a slightly different but similar pattern.
So in other words the general term looks like:
Nn*fn/Dn
where Nn is the product of differences of x minus all the other xi's not including xn, and
where Dn is the product of differences xn minus all the other xi's not including xn, so
for a four point formula we would get:
(x-x2)(x-x3)(x-x4)*f1/((x1-x2)(x1-x3)(x1-x4))
for the first term, and
(x-x1)(x-x3)(x-x4)*f2/((x2-x1)(x2-x3)(x2-x4))
for the second term and the other terms in a similar manner.
This gets you right to the formula without all the linear algebra, and also you can use it for a larger number of points for interpolating other functions as well. There's a special version called the "Five Point Formula" that uses (yes) five points (instead of the three shown in this paper), but it is special because it is especially simple and includes 5 points which is pretty good for an interpolation formula.