.
In this segment, we'll talk about
cubic interpolation.|Again, the cubic
interpolation we're going to be doing will be under the direct method.
So in this case, cubic interpolation, again you might be given several data
points, tens, or hundreds, but in cubic interpolation what you're going to do you're going to choose
four data points.|You're going to choose four data points like this,
and then draw the cubic interpolant there.|So through these
four data points you are choosing a cubic interpolant, which will be, let's suppose
f3 of x, and the form of that will be a0, plus a1 x,
plus a2 x squared, plus a3 x cubed.
So the bottom line about cubic interpolation is to find out what these four coefficients are, and those four
coefficients will be found out from the values which you will have at these four data points,
and then you can find out the value of the interpolant at any point between the lowest
value and the highest value of x which you have for those four data points.
So let's go ahead and take an example and see how this all works.
Let's suppose somebody gives me the upward velocity of a rocket
as a function of time.
So they've given me six data points, 0, 0, 10, 227.04,
15, 362.78,
20, 517.35,
22.5, 602.97,
30, 901.67.
So we're given six data points, and what we want to do is we want to be able to find
out what the value of the velocity at 16 is by using cubic
interpolation.|So if you are trying to find the value of the velocity at, you've got to first choose four data
points to be able to find the interpolant, so you want to choose the closest data points which are to
16.|So 15 is 1 away, 20 is 4 away, 22.5
is 6.5 away, and 10 is 6 away, so what you have to do is
you have to choose the four closest points to 16 on this time column,
and this is 6 away, 1 away, 4 away, and 6.5
away, the other ones are further away, so you're not going to choose those, and at the same
time, this number 16 has to be bracketed between 10, 15, 20, and 22.5,
which it is, because 16 is between 15 and 20.
So what you're going to do is you're going to take these four values here to find your cubic
interpolant.|So if you have your velocity now given as a0,
plus a1 t, plus a2 t squared, plus
a3 t cubed now, so you've got a cubic interpolant now.
So you're going to put the value of the velocity at 10,
which is a0, plus a1 times 10, plus a2 times 10
squared, plus a3 times 10 cubed.
Then you've got the value of the velocity at
15.
.
Then we've got the value of the velocity at 20, which is 517.35, which will
be a0, plus a1 times the value at 20,
plus a2 20 squared, plus a3
20 cubed, so that gives you the third equation.|Now, the fourth equation
is going to come from the value of the velocity at 22.5.
.
So all you are doing is substituting the value
of time in the cubic expression, you're substituting the value of time in
this cubic expression at these different times of 10, 15, 20, and 22.5,
and setting up the equations.|So this is going to give you four equations,
four unknowns.|And as homework, what I will ask you to do is go ahead and set it up
in the matrix form.|Set it up in the matrix form.|I'm
not going to show you how to set it up in the matrix form in this segment, but go ahead and set it up the matrix form,
and then go ahead and solve it by using Gauss elimination, or
LU decomposition method, or just simply your high school algebra method.
And what's going to happen is when you have four equations, four unknowns, you'll be able
to find out what the values of a0, a1, a2, and a3 are.|So this is what
I get after I solve these four equations, four unknowns, I get a0
equal to -4.3810,
a1 equal to 21.289,
a2 equal to 0.13065,
and a3 equal to 0.0054606.
So those are the values of the constants I get from
my third-order interpolant there, which basically means is that I can substitute
these values in the coefficients of the third-order polynomial, which is a0,
plus a1 t, plus as t squared, plus
a3 t cubed.|So that means it is
-4.3810, plus a1, which is
21.289
times t, plus 0.13065
times t squared, plus a3, which is
0.0054606 t cubed.
And keep in mind that this particular interpolant is now valid
between the values of time, t equal to 10 and 22.5,
because those are the lowest value of t and the highest value of t which you
used to . . . which you used to calculate this third-order interpolant.
interpolant.|However, we are interested in calculating the value of the velocity at 16.|When I substitute
the value of the velocity at 16 there, I'm going to get 392.06
meters per second.|All I'm doing is substituting the value of time equal to 16 here,
16 squared here, 16 cubed here, and I'll get 392.06
meters per second.|Now, how do I know how many significant I can trust
in this solution, and that should be based on what I got from the previous interpolation.
So I'm going to write that down, the value of the velocity
at 16, which I got from quadratic interpolation, was
392.19, that's what I got from quadratic interpolation,
and the value of the velocity at 16
I got to be 392.06, and this is what I got from
cubic interpolation.|So I can, in order to be able to calculate my relative approximate error,
I can choose this as my present approximation and this as my previous approximation.
So in that case, what I do is I get the relative approximate error
to be the current approximation, which is from the cubic interpolation, minus
392.16, which is from my quadratic interpolation, divided by
392.06, which is from the cubic interpolation, and I multiply it
by 100 in order to get the value in terms of percentages, and this value here turns
out to be 0.033 percent.
So what does this tell me?|It is that it is less than or equal to 5 percent,
it is less than or equal to 0.5 percent, and it's also less than or equal to 0.05
percent, so 0.05 percent.|I cannot go
beyond that, I cannot go to 0.005 percent.|So this tells me that
since this is 0.05 percent, at least
three significant
digits are correct.
So it's telling me that at least three significant digits are correct,
so in the number 392.06, I can trust 3, I can trust
9, and I can trust 2, those are the three significant digits which I can trust in that
solution.|And that's the end of this segment.
.