In this segment, we'll talk about quadratic spline interpolation; we will only concentrate
on the theory. If you want to see an example of quadratic spline interpolation,
just look at the link and, given in the description for the playlist.
So, before we talk about linear spline, quadratic spline interpolation, let's recall what we learned
in the linear spline interpolation. In linear spline interpolation, we took the data points
which were given consecutively from one point to another and simply drew a straight line, and one
of the things which you are automatically seeing is that the, look at this interior data point
where the slope, for example, coming this way is negative going this way is positive. And what,
and that is an artificial change in slope which is taking place. So, the, there's a discontinuity in
the slope at the interior data points. So, how do we go about avoiding that? That's something which
we have to think about. The other thing is that, if you look at this particular
straight-line approximation of the data from this point to this point, it only depends on
the values of x naught and y naught and x1 and y1. Does not depend on any sub, the other values.
So, that also is sort of a pitfall of the linear spline interpolation. So, how do we overcome that?
We can overcome that by using something called quadratic spline interpolation,
or cubic spline interpolation. In this particular
segment, we are only going to talk about quadratic spline interpolation.
So, let's look at the theory of the quadratic spline interpolation. So, you have n plus one data
points given to you, and what you want to do is you want to fit an interpolating quadratic spline
through the data. So, what that simply means is that you want to have n such splines,
n such quadratics, not splines, quadratics going through these
n plus one data points as shown here. So, like for example, if you look at the first two data points
right here, x naught comma y naught and x1 comma y1, what you have is a second-order polynomial
quadratic going through those two points. And same thing is the case for the next two, and so on and
so forth all the way up to the last one, which is x sub n minus 1 and y sub n minus 1 x sub n and y
n. What you are finding out that you have a second order polynomial, or a quadratic going through
that as well. So, what we are finding out here is that we are simply using quadratics
to approximate the curve, approximate the data from x naught to x n, and each of these
quadratics have three unknowns like you have a1 b1 c1 or a2 b2 c2. So, you'll have n of the a,
a's, a’s n of the b's and n of the c's, so that means that hey, you have three n unknowns.
So, we have three n unknowns, it makes sense that hey, somehow, we have to find three n
equations. Now, it's not easy like in linear spline interpolation, where just by looking
at the two consecutive points we're able to find the unknowns because in this case, we have only we
know that each this spline here is going through two points. That will only set up two equations,
but we have three unknowns. So, let's see how we go about trying to address the question that hey,
if we have three n unknowns how do we set up the three n equations,
so that has to be able to solve three and simultaneous linear equations?
So, the first thing which you have to recognize, is that each quadratic goes through two
consecutive data points. What does that mean? So. we'll take one example, and then we'll generalize
it. So, let's suppose we look at this first quadratic, we have a1 x squared plus b1 x plus c1,
and what you're finding out that hey, it goes through x naught, y naught and it also goes
to the point x1, y1 right here. So, if it is going through those points that means that a1
x naught squared plus b 1 x naught plus c1 should be equal to y naught, and we call it the value of
the function at x naught. Same thing that a1 x1 squared plus b1 x1 plus c1 should be equal to y1,
and that's nothing but the value of the function at x1.
So, we are setting up two equations for each of the quadratic, which is going through two
consecutive data points. So, that'll be the same case here for the second quadratic also, you will
be able to set up two equations, and so on and so forth, `till you reach the last quadratic,
where you can also set up two equations two unknowns because it goes through two points. So,
let's go and see that how those look and how many equations are we going to set up that way.
So, in the previous slide, we saw that how we derived
that the each of the splines is going through two consecutive points. So, we, right here
is the is what we had derived on the previous slide. So, this will be the case for all other
quadratics going from the second quadratic all the way to the nth quadratic. So, if you
look at the general formula, or general equations which you're going to get for the i’th quadratic,
it will be this one, and this one. And there will be n such pairs of equations available to you.
This is the one which is going through the last quadratic, it is going through the point x sub n y
sub n minus 1, and through the point x sub n y sub n. That's why we are getting an equation like this
one an equation like this one. So, if you look at what are the limits of i, you will say i goes from
1 to n because hey, this is the first quadratic and this is the nth quadratic. So, each of the
quadratics results in two equations. So, that means that you have two n equations which you will
be able to write down from the premise that each quadratic is going through two consecutive points.
Since the total number of equations which we have to develop is 3 n, so that implies that we have to
develop n more equations, and let's go and see where those n equations are going to come from.
So, we have to develop n more equations, so as to be able to set up our total of 3n equations and
recall from the linear spline interpolation that what we found out is at the interior data points,
we are not getting continuity in the slope, or continuity in the first derivative. So,
maybe by using the quadratic spline interpolation, we can ensure that happening now that the first
derivative will be continuous at the interior data points. So, let's go and take this part right
here. So, if you look at that part there what we have here is that this is the first quadratic,
and this one is the second quadratic.
This point is the first data point, let's suppose, and this one is the second data point,
and this one is the third data point which we have here. And what we are saying is that at
this common data point which we have between this spline and this spline, that the derivative is
going to be the same. So, what that means, that we have to equate the derivative of the first spline
and make it equal to the derivative of the second spline, but you've got to keep in mind
that it is not same everywhere. It's only same at the common point, which is at x equal to x1.
So, the derivative of this will be 2 a1 x plus b1, and we want to calculate it at x1,
and the derivative of the right-hand side will be 2a 2 x plus b2,
and we want to calculate the same point x1, which will ensure that the first derivative
is continuous at that interior data point. So, we get 2 a1 x1 plus b1 is equal to 2 a2 x plus b2,
and since a2 and b2 are the unknowns, we'll have to take them to the left-hand
side. We'll get 2 a2 x1 plus b1 minus 2 a2 x1 minus b2 equal to 0.
So, that is the equation which we are getting by saying that the first spline and the second spline
have a continuous slope at the common point x equal to x1. And since
we have so many interior data points, let's go and see how many such equations can we develop.
So, going back, this is the equation which we developed for the
continuity of the slope at the interior data point x1 y1, and so for the i'th interior data point,
our continuity of slope will be given by that particular equation right there,
and this one is for the last interior data point right here which is this one right here, is then
we get equation from there and this one we got from the first one. So, what you got to, what you
got to think about now is that hey, how many such equations do I have? I have n plus 1 data points,
and I have n minus 1 interior points,
because there are two exterior points. One is this one, and one is this one. Those two
are exterior points at which we cannot apply the condition that the slopes are continuous, because
there's no quadratic before this one and there's no quadratic after this one. So, we'll get only
n minus one equation so we're going to get n minus equations from equating the slopes at the
n minus one interior data point. So, so far, so far, we have two n equations coming from
that each quadratic goes through two consecutive data points, and then we have n minus equations
coming from the slopes are continuous at the interior data points from the common,
from the splines, which are connected to that point. And we get 3n minus 1 equations. So,
this leaves us a little bit of query that hey, where am I going to get that last equation from,
so that the total number of equations turns out to be 3n? So, since we have 3 equations,
3n minus 1 equations, we need 3n equations. We got to find that what that one last equation would be.
So, what we do is we choose either a1 equal to 0 or a n equal to 0. So, what do we mean by that,
is that if you look at the first spline, that is this one, and if we choose a1 equal to 0 as our
last equation, that means that we are basically assuming that the first plot, first quadratic
is a linear one actually. And if we look at the last, last quadratic, which is this one,
by assuming an equal to 0 we're basically saying the last quadratic is also a straight is,
is a straight line. So, the question arises that since we only need one more equation,
do we choose a1 equal to zero or a n equal to zero? A simple thing can be simply to check hey,
is x1 minus x naught, which is the distance between the two point first two points,
and we look at the distance between the last two points on the x-axis right here.
If this turns out to be less than, so if you find out that hey, x1 minus x-naught here is less than
or equal to what we have as the distance between the last two points on the x-axis,
then we will choose a1 equal to zero because the smaller distance. So, assuming it will be linear
in the smaller distance part might be a better assumption to make. Else, we choose a n equal
to zero. So, it's a nice if loop which is taking place based on what criteria you're going to use,
based on the criteria you're going to use. Whether a1, a1 would be 0 or a n equal to 0. So, we have 2
n equations coming from that each quadratic spline goes through two consecutive data points, we have
n minus 1 equations coming from that hey, the slopes are continuous at the interior data points,
and then we have one which is coming from this one right here, and that gives us 3n equations.
So, now we have 3n equations 3n unknowns and all we have to do is to solve these
3n simultaneous linear equations by any method
which is plausible and be able to find out what the coefficients of these quadratics are.
So, let's go and summarize again, and set up these 3n equations
here. What we said that each quadratic goes through two consecutive data points,
so these are the two equations which you're going to get the i'th, for the i'th quadratic,
and since we have n quadratics, we're going to get 2n equations from here. Then the first derivative
is same of the quadratics at the common point, so like these common points. Whatever spline
is going through that common point, you'll find out that the first derivative will be the same.
So, we'll get n minus 1 such equations because we have n minus 1 interior points
because two of them are exterior. Total number of points is n plus one, and then we said hey,
we're going to get one equation from here because either we'll assume that the first spline is,
the first quadratic is linear or we're going to assume the last quadratic to be linear. So,
we get 2n equations, plus n minus one plus one. We are reviewing it this again, we get 3n equations
we have 3n unknowns. So, we have 3n unknowns, and those 3n unknowns which we have are the a sub i's,
are the a sub i's which is going from one to n, the b sub i is going from 1 to n,
and c sub i is going from i is equal to 1 to n. And you can also look at it that hey,
since each quadratic has 3 unknowns, and since we have total of n quadratics, we'll get 3n
unknowns. So, we have 3n equations three unknowns, and we are able to find out what these
unknowns are a sub i's, b sub i's and c sub i's, and that's the theory behind
the quadratic spline interpolation, and that is the end of this segment.