.
In this segment, we're going to talk
about the backward divided difference,
which is also called
BDD, method, or scheme,
to find an approximate value of the
derivative of a function, so that's what we're going to do in this particular segment.
Again, what we're going to do is, we're going to look at it graphically, that if I have a
function, f of x,
as a function of x, and somebody's telling me that, hey, I want you to calculate the
value of the derivative of the function at this particular point x.
So, rather than choosing a point ahead of x, which is done in the forward divided difference
scheme, I'm going to choose a point which is behind x, or
less than x, which will be x minus delta x, which basically implies that the distance
between the two points is delta x.
I'm going to draw a line, a secant line, between
the values of the function at x and x minus delta x, so this particular
point here is the . . . the coordinate of that is x minus delta x and the value of the function
at that particular point.|The coordinate of this point here is the value of the function . . . of x,
comma, the value of the function at x.|So, we're going to do the
same thing here, and this is the run.|Now, if you look at this
distance here, which is the rise, is given by, what is the value of the function
here and what's the value of the function here, which is f of x, minus f of x minus
delta x, that's what you have there.|Now, if you
look at this run here, which is delta x, so we're going to go back to the definition
of our differential calculus class, is that f prime of x is defined
as delta x approaching 0, f of x plus delta x,
minus f of x, divided by delta x, right?
Now, in this case, delta x is to the right of x, or it is a bigger number
than x, but delta x, nobody's telling us delta x is a negative number or a positive number,
so what we can do is, we can say that f prime of x is approximately
equal to f of x, minus f of x minus delta x,
that's the run . . . rise which I am seeing here, in the figure, divided by
the run, which is delta x.|So here, we are assuming delta x to be a finite number,
and that's how we are able to get the derivative of the function, the approximate derivative of the function.
Somebody might say that, hey, this looks different from this here.|It's not any different, because if you choose delta x
to be a negative number, substitute minus delta x for delta x, you're going to get exactly
this particular number here, this particular expression here, with delta x
being a finite number.|Again, this is an approximate definition of f prime of x,
because delta x is a finite number, as opposed to approaching 0.|And this is the formula
for the backward divided difference scheme.|Very similar to the forward
divided difference scheme where you are choosing a point delta x ahead, now you are choosing a point
delta x to the left of x.|Let's
go ahead and take the same example and see that what kind of numbers we get for the approximate derivative
of the function.|So the example which we are taking is that we have f of x
equal to 2 e to the power 1.5 x,
we want to calculate the value of the derivative of the function at 3,
we're going to choose delta x equal to 0.1,
that's something which either is given to you, or it would be your choice if it is not given to you.
So let's go ahead and calculate what the derivative of the function at 3
is, so the derivative of the function at 3 is approximately equal to
the value of x, which is 3, minus . . . 3,
that's it, let me just go ahead and write down the definition
itself again, or the formula, so that we can
follow it, f prime of x was approximately found as f of x,
minus the value of the function at x minus delta x, divided by delta x.
So this turns out to be the value of the function at
3, minus the value of the function at 3 minus 0.1, divided
by 0.1, because that's the value of delta x, and that gives me the value of
the function at 3, minus the value of the function at 2.9, divided
by 0.1.|So, again, what you are seeing here, this is still rise over
run, because this number here is nothing but the run
between 2.9 and 3, so it's the rise between the value of the function at 2.9
and 3, and the run between the value of 3 and 2.9,
which turns out to be 0.1.|So, again, an approximate value of the . . .
value of the tangent . . . of the slope.|So
what is the value of the function at 3?|It's 2 e to the power 1.5 times 3,
minus 2 e to the power 1.5 times 2.9,
divided by 0.1, and this value here turns
out to be 250.77.
So that is the approximate value of the derivative of the function,
of this particular function at 3, so let me just write down
f prime of 3 here.|So f prime of 3 turns out to be 250.77
as the approximate value of the derivative of the function at 3.
Now, let's go ahead and see that, what is the exact value of the derivative of the function,
so as to be able to gauge how good or bad this number is, of 250.77.
So, if you look at f of x equal to 2
e to the power 1.5 x, that's the original function which was given to us,
we want to calculate the derivative of the function, that will be 2
times 1.5 e to the power 1.5 x, and the
formula which I have using from my differential equation . . . differential calculus knowledge
is this.
And this gives me 3 e to the power
1.5 x, so the derivative of the function at 3 is
3 e to the power 1.5 times 3, and the exact value turns out to be
270.05.|So that's what I get from my differential
calculus class, so let's go ahead and see that what I got from my approximate formula.
So let me repeat the numbers, f prime of 3 turns out to be approximately
equal to 250.77, by using delta x equal to
0.1, and by using the backward divided difference scheme.|That is the
approximate value which I obtained just now, delta x equal to 0.1 and with the
backward divided difference formula, but f prime of 3 is exactly equal to 270.05,
of course, up to five significant digits, with
delta x approaching 0, which is the definition from my differential calculus class.
So if I was going to calculate what the true error is in this case,
will be the true value minus approximate value.
The true value in this case is 270.05,
the approximate value is 250.77,
and this true error turns out to be 19.27.
And if you want to calculate
what the relative true error, absolute relative true error is in this case, turns out to be
7.138 percent,
that's what you will get as the relative true error, 7.138
percent.|So you can see that the error is large
by using delta x equal to 0.1, and you get a relative true error of
7.138 percent.
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