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In this segment, what we're going to do is we're going to
talk about differentiation of discrete data
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and we're going to look at an example by using Newton's divided difference polynomial.
So we're going to use the Newton's divided difference polynomial to be able
to calculate the derivative of a function which is given at discrete data points.|So let's
go ahead and look at what example we are looking for.|So let's suppose if you have an
aircraft landing, and you are given the times and the . . .
and the location, and what we want to be able to do is we want to be able to find out what the velocity will
be at a particular time.|So let's suppose somebody is telling you that, hey, this is time in seconds and
x in meters, and says that, hey, at different times, this is what we are finding
during emergency landing, what is happening . . . at what location the aircraft is
at different times, 2.5.
So at 0 it's at location 20,
at 0.4, it is at location 71, at 1 it's 110, then 161
at 1.75, and 178 at 2.5.|Can you go ahead and find
out what the velocity at 1.75 is?|So, you can very well see that if
the location is given to use, the velocity would be simply the derivative of the function x with
respect to time, and that's what we want to be able to calculate, we want to be able to calculate what dx by
dt is at 1.75, that's what that means.
And what we're going to do is we're going to use Newton's divided difference polynomial method to do so, and,
so if we are calculating 1.75, so let's suppose we're going to use second-order polynomial, second-order
Newton's divided difference polynomial.|So, if you're going to use second-order divided difference polynomial,
we've got to choose three points, and since this is 1.75, we'll choose 1,
1.75, and 2.5, those are the three points which we will choose for
doing the interpolation, and then differentiating it with respect to time to be able to
calculate what the velocity at 1.75 is.|So we are choosing three points, so let's suppose
we call t0 to be 1, the corresponding value of x which we are getting here
is 110, then t0 . . .
sorry, t1 is 1.75,
and the corresponding value of x is 161,
t2 is 2.5, and the value of x2
is 178.|So we've got these three data points which you're going to choose
to use a second-order polynomial, Newton's divided difference polynomial,
to be able to calculate the derivative of the second-order polynomial to calculate the velocity at
1.75.|So, as we know that the second-order divided difference
polynomial will look like this, f2 of t will turn out to be
f of the divided difference at
t0, so it will be t0 here, then f
at t1, comma t0, or I should call it
t0, comma, t1 . . . t0, comma, t1,
times . . . times t minus t0,
plus the divided difference now of t0,
t1, t2, times t minus t0, times t minus
t1.|But I'm interested in calculating the derivative of this, so f2
prime of t, this will be 0 because it's a constant, and here I'll get a 1, so it will be f
of the divided difference at t0 and t1,
t0 and t1, plus f at
t0, t1, t2, and what I'm going to
do is rather than expand this, I'm going to use the derivative of two . . . of a
product, and I'll get t minus t0 from one, and
t minus t1 from the other one, which basically tells you that you don't have to do any kind of symbolic
manipulation to be able to calculate what the derivative of the function is if you know how to expand
the derivative without having to do so.|So let's go ahead and see what we need to . . .
what we need to do in order to be able to get this.|So we know that
the divided difference t0, t1 will be nothing but the
value of the velocity at t1 minus the value of the velocity at t0, divided by t1 minus
t0, that's how that is defined.|So this one will turn out
to be equal to . . . not velocity,
but the location . . . the location.
And what is the location at t1?|It's 161.|The location at t0
is 110, divided by t1 minus t0, which is
1.75 minus 1, and this number here turns out to be 68.
So that's what we get for the divided difference of the . . .
of f with t0 and t1.|So let's go ahead and calculate what f of t0, comma,
t1, comma, t2 is, what that is.|So, f of t0,
comma, t1, comma, t2, will be nothing but the location
x2 minus t1, divided by t2 minus t1, so it's
the divided difference between t2 and t1, minus the divided difference between t1
and t0,
divided by t2 minus t1, so that's what you get there.
So if we do the substitution now, we get this as the
value of the divided difference, 178 minus 161,
because that's the value of x at those points, 2.5 minus
1.75, that's what we get there, minus 161 minus
110, divided by 1.75 minus 1,
divided by the difference between t2 and t1, which turns out to be this, and this number here
turns out to be -30.22.
So having calculated these divided differences, that means that
our velocity expression which we have will be approximately
equal to the divided difference between t0, t1, which is 68, plus this
quantity here, which is -30.22, times whatever we have
which we calculated which was t minus t0, plus t minus
t1.|So we get here 68 plus
-30.22, t is . . . t is t,
t0 is 1,
and t1 is 1.75.
So that's what we get as the expression for . . . for the velocity now, and this is
valid between the points 1 and 2.5.
Now, we are asked to calculate what the velocity is at 1.75, so all we have to do is
to substitute 1.75 in there, you get this,
1.75 minus 1, plus 1.75
minus 1.75 here, and this value here turns out to be 45.33
meters per second.|So that's how we are using
the Newton's divided difference polynomial method to develop the derivatives of
functions which are given at discrete data points, and that's an example of that.|And that's the
end of this segment.
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