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Now
in the previous lecture we have introduced the concept of an interpolating polynomial,
we were deriving the Lagrange interpolating polynomial which fits a given data. Let us
just revise what we have done last time; we were trying to derive the Lagrange interpolating
polynomial.
The data that is given to us is of the form x f(x), some point x0 f at x0, x1 f at x1
so on xn f at xn, so we have got a set of n plus 1 data points, data points or data
values which we have here. We have written the Lagrange interpolating polynomial as a
linear combination of these ordinates, we shall call this xi as the abscissa of these
data, these are the abscissa and f at xi are the ordinates. Now this representation is
obvious, because we can represent x f(x) as a point in the 2 dimensional coordinate plane,
therefore we can we shall call them as abscissas and these are the ordinates. Now we write
these polynomial, polynomial of degree n at x as a linear combination of f(x0) f(x1) f(xn)
that is sum l0(x), f of x0, l1(x) f of x1 plus so on ln(x) f at xn. Then we have mentioned
that, since pn(x) is a polynomial of degree n all li(x) should also be polynomials of
degree n. Therefore these are all polynomials of degree n, pn(x) of course finally may turn
out to be a polynomial of degree less than or equal to n, polynomial of degree less than
equal to n. If the leading terms cancel here, then it is going to be a lower order polynomial.
For example you might have given a hundred data but if they are all lying on straight
line, so the hundred data will be representing a just a linear polynomial and not a polynomial
degree 99. Therefore this p n x may finally be a polynomial of degree less than equal
to n, while l0 l1 ln(x) will be polynomials of degree n only. Now then we have shown that
the, these polynomials li(x) should satisfy the condition li(xj) is equal to 0, for i
not equal to j and this should be equal to 1, for i is equal to j, under this condition
this polynomial exactly fits the, fits the given data, so in this case, in this case
pn(x) fits the data. These polynomials are called the Lagrange fundamental polynomials;
these are called the Lagrange fundamental polynomials. Now the next step is how we construct
the, the polynomials of degree n, which satisfy this particular property and the writing of
such a polynomial is really very trivial, because it should be 0 when I substitute x
is equal to xj and when I substitute i is equal to j, that is xi it should have value
1.
That means I can now define a polynomial li(x) is equal to, in the numerator I will have,
I will have a numerator and denominator (x minus x0) all the products (x minus xi-1)
(x minus xi+1) (x minus xn) and in the denominator I will have (xi minus x0) (xi minus xi-1)
(xi minus xi+1) (xi minus xn) that is i is equal to 1 2 3 so on n. Now in the numerator
we have written all the products except the product here, that is (x minus xi) and that
belongs to this suffix i. Now let us substitute x is equal to x0, it is 0 because numerator
has (x minus x0), so for all xi this is going to be 0 except when x is equal to xi, when
x is equal to xi the numerator is same as the denominator, therefore it cancels. Therefore
this satisfies the property that li(xj) is equal to 0, for i not equal to j and this
is equal 1, for i is equal to j. Therefore this is itself the Lagrange fundamental polynomial
that is required in the construction of the Lagrange interpolating polynomial.
Now we can put this in a simple notation by just using a notation, so let us set the w(x)
as the product of all the factors given in the problem, so these are the products using
all the abscissas (x minus x0) (x minus x1) so on (x minus xn). Then let us find out what
is the derivative of this with respect to x but at xi. These are n plus 1 factors. Therefore
if I differentiate there will be n plus 1 terms. There will be only 1 term which will
not contain, I will write down this particular factor here also (x minus xi), let us write
this (x minus xi). When I differentiate this, there will be only 1 term which will not contain
this, which is the derivative of this term, because these are all linear factors. When
I substitute xi all of them would vanish except 1 term which is containing this 1 and that
term will be, when I differentiate whole the thing this goes, I substitute the remaining,
so whatever you have here is (x minus x0) (xi minus x1) the previous one is (xi minus
xi-1) the next one (xi minus xi-1) and the last term (xi minus xn).
So all the terms would disappear except one term, which was the derivative using this
term particular factor that we have. We see that the denominator of li(x) is simply w
prime xi, this is same as this denominator, therefore li(x) in simple notation can be
written as w(x), I have to remove (x minus xi) so I will divide (x minus xi), therefore
w(x) upon (x minus xi) is the numerator and the denominator is w prime of xi, so I can
simply write this Lagrange fundamental polynomials in this particular notation. When once these
li(x) are determined, I will substitute in the polynomial to find Lagrange interpolating
polynomial, a summation i is equal to 0 to n li(x) f of xi, so I can substitute these
fundamental polynomials in this and simplify this to get my Lagrange interpolating polynomial.
Now let us construct few simple cases, the first case we shall construct is, let us take
a linear interpolating polynomial, linear interpolating polynomial, therefore we are
considering any 2 points arbitrarily, so let us suppose we are considering the points,
x0 f0 x1 f1, so we are considering the case of only 2 points x0 f0 x1 f1. Therefore our
fundamental polynomial should be, now the factors w x will be all the factors (x minus
x0) (x minus x1) there are only two 2 abscissa therefore there will be only 2 factors. Therefore
our l0(x) must be equal to, I must skip in the numerator the factor corresponding to
this suffix that is our x0, so I must skip (x minus x0), so I will have (x minus x0)
and what is the denominator? The denominator is the numerator evaluated at xi that is the
point with the suffix, so I will have to evaluate this numerator at x0 that is (x0 minus x1).
Now let us write down l1(x), now I must skip in the numerator (x minus x1) so what is left
out is (x minus x0), evaluate it at x1, (x1 minus x0). Therefore our polynomial of degree
1, polynomial degree 1 will be l0(x) into f of x0, l1 of x f at x1 this will be the
required polynomial. Now let us take a simple example along with this, so let us take this
as example. Now I will number it so that I want to use this in the later steps, so I
will use them as number 1 and this, construct the linear polynomial which fits the data
x f x 1 2 2 5. Predict the value at 1 point 5, predict the value using this linear polynomial
what would be the value at x is equal to 1 point 5.
Now here our x0 is 1, x1 is equal to 2, f0 is 2, f1 is equal to 5. Therefore our l0(x)
is x minus, let us write down the formula again, (x minus x1) divide by (x0 minus x1),
therefore this is (x minus 2) divided by (1 minus 2), so that will be minus of (x minus
2). l1 x is (x minus x0) divide by (x1 minus x0) so that is (x minus 1) from here and (2
minus 1) that is equal to (x minus 1). Therefore our required polynomial is p1(x) is l0(x)
f0 plus l1(x) f1, let us substitute the values l0(x) is minus (x minus 2) f0 is 2, l1(x)
is (x minus 1) f1 is 5. This is 5 x minus 2 x that is 3 x, this is minus 5 plus 4 so
this is minus 1. Therefore this is a polynomial which fits exactly the given data. Now we
want to predict the value of f at 1 point 5 is what is being asked, so I would give
the estimate of this as p1 at 1 point 5 that is 3 into 1 point 5 minus 1 that is equal
to 3 point 5. So this will be the value, the predicted value of the function f at the point
1 point 5. Now we will little latter see how good this value is or what is the error in
do you use in this particular approximation.
Now let us take up what would be the quadratic interpolating polynomial, so similarly let
us derive the, sorry, the quadratic interpolating polynomial. I want to use 3 points and construct
a polynomial of a degree 2 which passes through this, that means I have got the data as x
f(x), x0 x1 x2, f0 f1 f2. Therefore I can write down the Lagrange fundamental polynomials
immediately, the numerator should miss (x minus x0), so I will have the remaining factors
(x minus x1) (x minus x2), the denominator is evaluation at x0, so this is (x0 minus
x1) (x0 minus x2) and l1(x) is, the numerator should skip now (x minus x1), so I will have
(x minus x0) (x minus x2) and I evaluate it at x1 for the denominator, so this will be
(x1 minus x0) (x1 minus x2) and the third fundamental polynomial is l2(x), I skip now
the numerator (x minus x2), so I will have (x minus x0) (x minus x1), I evaluate it at
x2, (x2 minus x0) (x2 minus x1). When once these polynomials are evaluated I immediately
write down the polynomial of degree 2, x is equal to l0(x) f0, l1(x) f1, l2(x) f2. Now
let us also illustrate an example for this so that we can discuss about the error, so
let me write down the example.
Now fit an interpolating polynomial for the given data, so let us give this data as x f(x), I will take it as 1, 2, 2, 5, 4,
17.
Then will say predict the value of f(x) at x is equal to 3. Now what is very important to note here is, these data points
are arbitrary for data points that means they are not having any step, a particular step
length or anything, you can see the distance between these two is 1, the distance between
these two is 2, so these are really a random set of points. They, you can particularize
with the case of when you have a equispace data but otherwise this is an arbitrary data.
So let us write down the fundamental polynomials, so I will have x, l0(x) is equal to, our x0
is 1, x1 is 2, x2 is 4, so I will have there is (x minus 2) into (x minus 4) divided by
(1 minus 2) (1 minus 4), x0 is 1 so I am substituting x0 is 1 (1 minus 2) (1 minus 4), so this gives
minus 3 minus 1 so this is plus 1 upon 3 (x minus 2) (x minus 4). Similarly I construct
l1(x), now we would miss x1 that is 2, factor corresponding to 2. So I will have the numerator
(x minus 1) (x minus 4) we have (2 minus 1) (2 minus 4), this is (2 minus 1) (2 minus
4), so this gives you 1, this gives you minus 2, so this gives you minus half (x minus 1)
(x minus 4) and the third polynomial is l2(x), now will have to skip my x2 that is 4 the
factor corresponding to 4, so I will have here (x minus 1) into (x minus 2) divide by
(4 minus 1) (4 minus 2) this gives us 3, this gives us 2, so I will have here 1 upon 6 (x
minus 1) into (x minus 2). Therefore the required polynomial is p2(x) that is, I am l0(x) f0
that is 1 by 3(x minus 2) (x minus 4) into 2, that is f0 is 2 minus half (x minus 1)
(x minus 4) f1 is 5, plus 1 by 6, plus 1 by 6 (x minus 1) (x minus 2) and f2 is 17.
Now I can collect the coefficients from all these, this I will leave it as a simple exercise
for you, this comes out to be x square plus 1. The coefficient of x cancels throughout
and I simply have this is the x square plus 1, so that we can predict the value of at
3 as simply equal to 10. Now this is the polynomial we can just check it again back whether we
have done it correctly or not, just substitute x is equal to 1, this is 2, when x is equal
to 2 this is 4 plus 1, 5, x is equal to 4 that is 16 plus 1, 17. So this polynomial
is fitted this data exactly. So this is a verification that we can cross check because
we are exactly fitting the given data and this is the predicted value. Now as I mentioned
earlier the polynomial that we are constructing is fitting the data exactly but when we predicted
at a particular value, not the point which is given there, we do not know how accurate
this particular value, this 10 is. Therefore we must define, what is the error in our interpolated
values? which we can also call as the truncation error.
So I would like to derive a formula for the error of interpolation, so let us call this
as error of interpolation, this is also called the truncation error, this is also called
the truncation error. As I mentioned there is no error at the nodal points because the,
which is we are fitting the data exactly, there is there is no, therefore no error at
x0 x1 x2 xn, the error arises only when we are interpolating at a particular another
point. So let us denote this f(x) minus p(x), p(x) is a polynomial of any degree 1 2 or
3 or anything, this is our approximation and let us denote this as error, error polynomial
degree n of x. So that means we are now writing in general your p(x) as pn(x) it could be
n is 1 its linear, n is 2 quadratic polynomial or otherwise its a degree n. Now remember
our data is a, data is what is this data given to us it is xi fi, this is the data given
to us, this is the data given to us. Now as I said there is no error at the nodal points
that means if I put x is equal to xi, this obviously 0, f(xi) minus p(xi) is 0 because
that is the data is exactly fitted. So error at the abscissa xi is equal to f(xi) minus
p at xi is equal to 0, because we are fitting the data exactly. Now to generalize the problem
let us denote our first point x0 as a, last point xn as b, so that we are talking of the
interval (a, b) a general interval (a, b). Now we are interpolating at an arbitrary point
in between, so let us choose x an arbitrary point contained in this interval (a, b) therefore
x is an arbitrary point and therefore x is not equal to xi obviously this is not a this point.
Now I define a new function g(t) so I will define a function g of t, t is a new variable.
I will write this as g of t is equal to f of t minus p of t, let us put it in a bracket
minus f(x), minus f(x) minus p(x) into (t minus x0) (t minus x1) (t minus xn) divided
by (x minus x0) (x minus x1) (x minus xn). Now we will understand why we are constructing
such a function because I want to get from this, what will be the expression for this
En is equal to f(x) minus p(x), this function would enable me to find such a quantity. Now
let us look at the function g(t), g(t) is containing f(t) p(t) f(x) p(x), our function
we are expecting that the, whatever the data that is given is representing continuous function.
Therefore this everything is continuous here, therefore g(t) is also a continuous function,
so g(t) is a continuous function, g(t) is a continuous function.
Now let us just look at the property of this g(t). Let us set t is equal to x in this,
let us put t is equal to x. Now before we look at the first one, let us first look at
this last factor, when I put t is equal to x this is same as the denominator, all the
terms are same as, therefore they cancel of and will have 1. When t is equal to x this
is [f(x) minus p(x)] minus [f(x) minus p(x)] into 1, therefore this will be simply g of
x is equal to 0. This is same as this and we have this is equal to 0. Now let us set
t is equal to xi, set t is equal to xi that is our points x0 x1 x2 xn. Then g at xi is
equal to f at xi minus p at xi that is equal to 0 by definition, that is our f(xi) minus
p(xi) is 0 but here when I put this any one of this xi's the numerator is 0 because (t
minus x0) (t minus x1) (t minus xn) when I use any value for t is xi, one of the factors
is containing (x minus xi), so this numerator is going to be 0 but this is also 0, so therefore
this is also equal to 0, these are, i is equal to 0 1 2 n.
Therefore g(t) is a function of t which vanishes at n plus 1 plus 1, n plus 2 points. Therefore
g(t) vanishes at n plus 2 points, n plus 2 points x0 x1 xn and x, and at x of course,
x is a point inside (x0, xn) that is our point (a, b), x0 point is an interior point. g(t)
is continuous on (a, b) differentiable, we assume that differentiability also required
number of times, g(t) is differentiable on the open interval (a, b). Now I have a function g(t) which is continuous on (a, b)
differentiable on (a, b) and vanishes at n plus 2 points. Now I would like to apply a
theorem which you have studied in your first semester that is your Rolle's Theorem of a
continuous function.
Now what is the Rolle's theorem, I want to use this Rolle's theorem, now these are our
abscissas, now Rolle's theorem states that if I have, we have a function, let us a remove
it some h(x), let us suppose we have a function h(x), h of a is equal to 0, h of b is equal
to 0, in fact they need not be equal to 0, it will simply h of a is equal to h of b and
h is continuous on (a, b) differentiable on (a, b), then the Rolle's theorem states that
there exists at least 1 point between (a, b) where h prime or the derivative of h will
be equal to 0. So then the Rolle's theorem states that there exists a point at which
h dash of c is equal to 0, where c is a point lying between a and b.
Now if you just remember what this was, this really states that if suppose you have a function
like this, h of a is 0, h of b is 0 then what it states is that there is at least one point,
at least this slope over the tangent will be parallel to x axis, h dash of c is equal
to 0. Now I would like to apply this Rolle's Theorem on this
So if you look at these two points x0 x1, g(t) vanishes at both this points g(x0) is
0 g(x1) is 0, g(t) is continuous g(t) is differentiable therefore there exist a point here some c1
at which your g dash of c1 is equal to 0. Now between x1 and x2, g(x1) is 0 g(x2) is
0 therefore again there exists a point c2 at which g prime at c2 is equal to 0. Now
if I go to the last point here, last interval again g at xn-1 is 0, g at xn is equal to
0, therefore there exists a point cn at which g prime at cn is equal to 0. Now let us imagine
that g prime is some new function, some new function h(x). If I take this as new function,
now I can apply the Rolle's theorem repeatedly, now I have a function h(c1) is 0, h(c2) is
equal to 0, g dash is continuous, it is differentiable therefore there exists a point between c1
and c2 such that the second derivative is equal to 0. Now if I apply repeatedly then
there are n plus 2 points, therefore repeated application Rolle's Theorem would state that
n plus 1, 1 less than that; that is n plus 1th derivative of g will be 0 at 1, at least
1 point between x0 to xn.
So if I apply this Rolle's theorem repeatedly, so let us say applying Rolle's theorem repeatedly,
repeatedly, we get the n plus 1th derivative at an arbitrary point zhi in the interval,
this is equal to 0, where zhi is a point between a and b. Remember this is derivative with
respective to t, we are talk of a function of t. Now let us differentiate this n plus
1 times, g(n+1) of t, so I have to differentiate this, f(n+1) of t, n plus 1 of t, p(t) is
only a polynomial of degree n but differentiating it n plus 1 times, therefore the derivative
of this will be going to be 0, minus, this is a function of x, so that stays as it is
f(x) minus p(x). Now if you look at this one, this, there are n plus 1 factors therefore
this is a polynomial f degree n plus 1 in t. Therefore if I differentiate it n plus
1 times i will get factorial n plus 1, n plus 1 into n into n minus 1 and so on, differentiate
it n plus 1 times, so I would get here factorial n plus 1 in the numerator. Differentiating
the numerator n plus 1 times, its a polynomial of degree n plus 1. The denominator stays
as it is, that is your (x minus x0) so on (x minus xn). Now this expression is 0 at
zhi therefore g(n+1) of zhi is equal to 0 and that will be equal to, let us now put
here f(n+1) of zhi minus, there is no t here, so this stays as it is, so [f(x) minus p(x)]
n plus 1 factorial. Now let us revert back this to your notation which we started with,
this was w(x) this is the product of all the factors that are there in the given problem,
that is (x minus x0) (x minus 1) (x minus xn) that is w(x). Now since this is equal
to 0, I can now take this to the left hand side and find out what is the value of f(x)
minus pn(x).
Therefore f(x) minus pn(x), so I am bringing to this side, so this goes up w(x), this comes
down n plus 1 factorial and this is f(n+1) of zhi and we started the definition as error
of interpolation is f(x) minus p(x) is equal to error at this one, therefore whatever we
have derived now is nothing but the error of interpolation and this is your error of
f(x), this is your error of interpolation. Now since this zhi is unknown to us it is
not possible for us to write this expression but we can bound it, so let us find out what
is the bound of this, let us find what is the magnitude of this is. n is a number, so
I can write it outside, this is magnitude of w(x) into magnitude of f(n+1) of zhi. Now
zhi is unknown, therefore what I will do is, I will take the maximum possible value of
f(n+1) of x, so I would therefore write this is less than or equal to 1 upon n plus 1 factorial,
maximum of w(x) of course in the interval, in the interval on, let us write down on (a,
b) maximum on (a, b), maximum f(n+1) of f(x). We can find the maximum magnitude of f(n+1)(x)
using any procedure that we know. The maximum of w(x) I can always find it and that will
give me the bound of the error, that means whatever interpolation that we are doing linear
quadratic or any approximation, we shall be able to say what is the maximum possible error
we are committing in that particular problem.
Now let us first have you look at what would be the error in the linear interpolation,
let us take the error in linear interpolation. Therefore we are having only 2 points (x0,
f0) and (x1, f1) these are the only 2 points that we have here and therefore here the case
n is equal to 0, it corresponds to the case n is equal to 0. Therefore I can write down
E1 from here, E1 from here, w(x) is the product of all the factors that is (x minus x0) (x
minus x1), now n is equal to 1, there are 2 points, n is 1, divided by 2 factorial,
1 plus 1 2 factorial and this is f double prime of zhi.
Now then let us get this bound, therefore magnitude of E1 would be less than equal to
1 by 2 maximum of (x minus x0) (x minus x1) into maximum of f double dash of x. Now I can find the maximum of this because
the quadratic polynomial, I can just set it equal to, if I put this as g(x), I can find
out what is the derivative of this, derivative of this is equal to (x minus x0) plus (x minus
x0). Just differentiate it as product (x minus x1) into 1 (x minus x0). Set this is equal
to 0 or ordinary maxima minima. It is a function of single variable, so differentiate first
derivative, set it equal 0, find out the points, set with the critical points we find it out
and at all those points, find out what is the maximum. So the maximum value of all those,
we will take as the required maximum. Therefore this gives 2 x is equal to (x0 plus x1) or
x is equal to (x0 plus x1) divided by 2. Therefore this is maximum at the middle point (x0 plus
x1) by 2 and let us find out what is the value.((x minus x0) into (x minus x1) will be (x0 plus
x1) by 2 minus x0) ((x0 plus x1) by 2 minus x1). This is (x1 minus x0), this is (minus
x0 minus x1), so I put a minus sign outside and put (x minus x0) whole square by 4, there
is a 2 here, there is a 2 here, I will have 4 here and since there are only 2 points given
to us x0 and x1 here, let us take this distance as equal to h. So that means I can just write
this as h square by 4, h is the distance x1 minus x0. No no no, I have not said its maximum,
I just evaluated it, I will have to write down magnitude. So if you want, now if you
want in to, you will have to write down the magnitude. The magnitude when once you put
this magnitude, this will be magnitude; this will become h square by 4. I thought I will
do it in next step but since you have asked, what we are talking here is, maximum of magnitude,
maximum magnitude, so this will give you the maximum magnitude is this one and magnitude
is this. Therefore the maximum magnitude is x square by 4, therefore let us put back this
value in this expression, so let us put it there.
So magnitude of E1 is less than or equal to half, this is h square by 4 and let us denote
this by M2, I will denote M2 as maximum of f double dash x. We do not know what is f
double dash x, therefore we need to have estimate through some other source. This is equal to
h square by 8 M2. This result can be used in many problems where we need to find the
step length h to construct a table, when we take up the problem it is illustrated. Now
another important one is error in quadratic interpolation. Let us do quadratic interpolation
also.
Now the error in quadratic interpolation, let us write straight away from this last
expression, that is your error of E2 (f, x) would be less than or equal to 1 upon, n is
equal to 2 now that is n is equal to 2, 1 upon factorial 3, 1 upon factorial 3 maximum
of (x minus x0) (x minus x1) (x minus x2) into M3. I will write M3, were M3 is the maximum
of f triple dash x in the interval (x0, x2), in the interval we are talking of only (x0,
x2), maximum of (x0, x2), (x0, x2). If there is a point, critical point falls outside the
range, we are not considering that particular point, we are considering only the points
which will fall when we are finding the maxima of this, the critical point should lie between
x0 and x2, only at that point we can find the maximum. This is a cubic, again we can
differentiate it, find the value of the root, there will 2 values, 2 critical points, we
can find those critical points and then find out what is the value at this one.
Now before we take up the example 1 example 2 illustrate it, if the data given is equispaced
not randomly given, then this also simplifies into a very simple form like this. Let us
take the particular case, let us take the case of equispaced, equispaced data. Now if
it is a equispaced data, I have x0 here, I have x1 here, I have x2 here, now each is
separated by, let us say h because it is equispaced data, x1 minus x0 is equal to h, which is
same as x2 minus x1, that is the equispaced data. We shall use a simple trick to find
this maximum, let us take this middle point as my origin, some value t, I will call this
as t. Then this point behind is t minus h, this point will be t plus h, because they
are all equispaced therefore the point behind is h behind, point ahead is h ahead. Therefore
if I take the central point as my origin then this previous point is t minus h and the next
point will be t plus h, then this expression becomes very simple.
So, now what I want in that case will be the maximum of, I want then the maximum of (t
minus h) t (t plus h). (x minus x0) (x minus x1) (x minus x2) so I wanted the maximum of
this and this you can see that is a very simple expression, this is t square minus h square
into t, so this will be t cube minus t h square. Now let us call this as g of t, so that I
can find out the derivative g dash of t that is 3 t square minus h square. I set this as
0, therefore t is equal to plus minus h upon root 3. Now I want the maximum of this, therefore
I want the maximum of, now you can see because of this symmetry plus minus h here, whether
I take plus or minus h both of them are going to give the same value because one it will
this case positive negative or positive negative. So it is going to give me t square minus h
square into t that is what I want here, that is maximum of, that will be equal to magnitude
of t square is h square by 3 minus h square into t that is h upon root 3, h upon root
3. That gives me, this is minus 2 by 3 h cubed by root 3 that is 2 by 3 root 3 h cubed. Now
when once I get this maximum, I can substitute it this and then get what is the error here,
therefore error now would be less than or equal to 1 upon factorial 3 that is 1 upon
6, 2 upon root 3 h cubed M3 that is equal to h cubed by 9 root 3 M3.
Therefore this is the error of interpolation in quadratic interpolation, if it is equispaced,
whether it is equispaced or not, the linear interpolation this will be the error of approximation
but if I want the data which is not equispaced then I shall use this particular expression
to find out the maximum of the cubic. I can differentiate it and then get the values,
2 values of for x from here and find out the maximum out of those 2 values, substitute
here and find the maximum and that will give me the maximum value of the error in this
one. This, that will be the bound for the error of interpolation, if you remember the
example 1 and 2, we are finally said, predict the value at a particular point, we have predicted
the value. We shall now be able to say what would be the possible error of interpolation
when we predict the value at any one of this interprodian points either by linear polynomial
quadratic or in general by nth degree polynomial. Now we shall take the examples on this next
time.