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Today, first we are going to look at a number of computations needed to evaluate polynomial
pn interpolating polynomial in the Newton's form. Next we will look at cubic hermite interpolation
where not only function values are interpolated, but the derivative values are also interpolated.
So, we will look at the existence and uniqueness of a cubic hermite polynomial, the error in
it and then I will state a result about the convergence of interpolating polynomial.
After that we are going to look at piece wise polynomial. So, in this lecture we will consider
mainly piece wise linear polynomial. So, a polynomial in Newton's form interpolating
polynomial it is given by pn( x) is equal to f (x0) plus divided difference based on
x0 x1 into x minus x0 plus divided difference based on x0 x1 xn into x minus x0 x minus
x(n minus 1). So, it is a polynomial of degree less than
or equal to n which matches with f at xj j going from 01 upto n. So, it is of the form
a0 plus a1 (x minus x0 ) plus a n x minus x0 into x minus x (n minus 1).
So, in order to compute the coefficients a0 a1 an we will need to calculate the divided
differences. So, we will first look at the number of computations needed to calculate
these divided differences and then we will consider an efficient way of evaluating pn
at a point once we have computed a0 a1 up to an.
Now, the recurrence formula for the divided difference based on k plus 1 points is given
by f (x1 x2 xk) minus f (x0 x1 x (k minus 1) ) . So, here are k plus 1 points. So, here
are k points divided by xk minus x0. So, in case we have the divided differences based
on k points available then what we need is one subtraction here ,one subtraction here.
So, total two subtractions and then one division now this is for one divided difference.
So, let us look at the divided difference table the divided difference table is given
by: These are our points x0 x1 x2 xn these are
the function values then f (x1) minus f(x0) divided by x1 minus x0 will give us this divided
difference and so on ,we continue . So, the total number of divided differences which
we will need to calculate will be in first column there will be n divided differences
,in the second column there will be n minus 1 and in the last column 1.Thus, total is
n into n plus 1 by 2 divided differences.
And for each divided difference we need two subtractions and one division and thus the
total cost for the divided difference table will be, for each divided difference 2 subtraction
and 1 division and we need to calculate total n into n plus 1 by 2 divided differences.
So, it is going to be n into n plus 1 subtractions plus n into n plus 1 by 2 divisions.
So, what is important is the power is going to be of the order of n square. Now, once
we have calculated the divided difference we have got the coefficients a0 a1 an and
now, let us look at the number of computations for evaluating our interpolating polynomial.
So, let me first look at a polynomial of degree 2. So, p2 (x) is a 0 plus a1 (x minus x0)
plus a2(x minus x0 )( x minus x1) .Let us ignore these subtractions x minus x0 then
x minus x0 x minus x1.Assume that they are already done. So, here if we directly do it
there will be one multiplication here, two multiplications here so two multiplications
plus two editions. Now, in Horner's scheme what we do is we write
the same polynomial in the form a2 into x minus x1 plus a1 into x minus x0 plus a0.
So, it is the same polynomial, I am just writing differently. Now I will have 1 multiplication
here 1 addition here then I will have one multiplication here. So, instead of three
multiplications I will have two multiplications and there will be two additions as before.
So, for the quadratic polynomial instead of three multiplications we have got two multiplications,
but for a polynomial of degree n the saving is going to be much more.
So, here if you have got pn to be a polynomial in the Newton form. Here ,if I evaluate pn
directly I will have one multiplication here, the next one, next term will have two multiplications
and last term will have n multiplications. So, total n into n plus 1 by 2 multiplications
and n additions. Now, as we did before, if we consider the algorithm bn is equal to an
and then for j equal to n minus 1 up to 0 you are reducing the values of j to be bj
is equal to aj plus bj plus 1 into x minus x j then when we reach b0 that is going to
be pn(x) here you see that you are doing it for j is equal to n minus 1 up to 0.
So, there are going to be n multiplications and n additions. So, instead of n square by
two multiplications you have reduced it to n multiplications. So, that is a considerable
saving and that is known as "Hornor's scheme". Now, next what we want to do is we want to
consider cubic hermite interpolation. So, this is part of what is known as oscillatory
interpolation .What we are going to do is consider a function defined on closed interval
a ,b and let us try to find a polynomial of degree less than or equal to 3 which interpolates
the given function at 'a' and 'b' the two end points and also the polynomial it is derivative
should approximate or if p3 is the polynomial p3 dash at 'a' should be equal to f dash at
'a' where dash means derivative and p3 dash at 'b' is equal to f dash (b).
So, we will be interpolating function values and the derivative values at the two end points
and that is cubic hermite interpolation. We will first prove its existence and then we
will prove its uniqueness and after that we will look at the error .Now, the way we are
going to write this p3 it is if our points x0 x1 x2 x3 if these are distinct points then
we know how to write p3 (x) in the Newton form.
Now, the difference here is it will be instead of distinct points you will have x0 is equal
to x1 is equal to a. So, x0 and x1 are identical and x2 and x3 are identical, but we will try
to write a similar formula as in the case of distinct point and see whether it works.
So, assume that f to be defined on interval a,b taking real values and assume that it
is it to be differentiable ((??)CHECK) . is to find a polynomial p3 of degree less than
or equal to 3 such that p3 at 'a' is equal to f of a ,p3 at 'b' equal to f of b .The
derivative of p3 at 'a' is equal to f dash of a and derivative of p3 at 'b' is equal
to f dash of b.
Now, let us recall that if x0 x1 x2 x3 are distinct points then the interpolating polynomial
p3 (x) is given by value of f at x0 plus divided difference of f based on x0 x1 into x minus
x0 then divided difference based on x0 x1 x2 into x minus x0 x minus x1 and the last
term is divided differences based on the four interpolation points multiplications by x
minus x0 x minus x1 x minus x 2. Suppose x0 is equal to x1 is equal to 'a'
and x2 is equal to x3 is equal to 'b' then this f of x0 x1 that is going to be f of a,a
and then we have, we are defining it to be equal to f dash of a.
Here this divided differences will be f of a, a, b. So, that will be f of a ,b 'a' and
'b' are distinct minus f of a ,a which is f dash 'a divided by 'b' minus 'a'. So, f
of a ,a make sense f of a, a, b make sense and f of a, a, b, b will be defined in a similar
manner.
So, here is a divided difference table 'a' repeated twice 'b' repeated twice these are
the function values the divided difference correspond into this will be f dash of 'a'
the divided difference based on a b will be f of a ,b and divided difference based on
b, b that is f dash b. Next f of a, a, b is f of a ,b minus f dash
'a' divided by 'b' minus 'a'. Similarly f of a, b, b will be f dash 'b' minus f of a
,b divided by 'b' minus 'a' and the last term will be given by f of a ,b, b minus f of a,
a, b divided by 'b' minus 'a'. So, this is divided difference table corresponding to
'a' repeated twice 'b' repeated twice.
So, here one needs to be given function value at a function value at 'b' derivative of f
at 'a' and derivative of f at b'' and then our p3 (x) is f of a plus f of a, a into x
minus a plus f of a, a, b into x minus a2 .
So, this I write, but now I want to show that this is our desired polynomial which interpolates
the function and derivative values at two end points. So, first of all put x is equal
to a when I put x is equal to a there is x minus a term here x minus a term here x minus
a term here. So, p3 at a will be equal to f at 'a'.
Take the derivative of p3. So, this being a constant term its derivative will be 0.
So, you will have f of a, a then derivative of this will be 2 times x minus a into f of
a,a ,b plus this divided difference and then by product tool this will be 2 into x minus
a into x minus b plus x minus a2 . When I put x is equal to 'a' there is x minus
a term here x minus a term here and x minus a term here. So, p3 dash at 'a' is equal to
f dash at 'a'. So, this p3 it interpolates the function value at 'a' and its derivative
at 'a'.
Now, let us look at the case when x is equal to b. So, when I put x is equal to b ,p3 of
b will be f of a this is f dash a x is equal to b.(??CHECK) So, b minus a plus this will
be s a b minus f dash a by b minus a into b minus a square. So, one b minus a will get
cancelled then f dash a into b minus a gets cancelled f of a, b is f of b minus f of a
upon b minus a. So, this will be f of a plus f of b minus f of a. So, it is equal to f
of b.
So, thus p3 interpolates the function value at 'b' and now what remains is whether it
interpolates the derivative value at point 'b'. So, this is our p3 dash x .We have seen
earlier put x is equal to b and then you will have f dash a, plus 2 times f of a, ,a, b
into b minus a plus this this will be nothing, but f of a ,b ,b minus f of a ,a, b and then
divided by b minus a. So, we are putting x is equal to b. So, no contribution from this
term and then you have got b minus a. So, thus p3 interpolates the function value
at 'b' and now what remains is the whether it interpolates the derivative value at point
'b'. So, this is our p3 dash x we have seen earlier put x is equal to 'b' and then you
will have f dash a, plus 2 times f of a, a, b into b minus a plus this this will be nothing
but, f of a, b, b minus f of a, a, b and then divided by b minus a. So, we are putting x
is equal to b. So, no contribution from this term and then you have got b minus a.
So, this was our p3 dash b now we simplify it this will be f dash a then here you have
2 times f of a, a, b into b minus a and here it is minus. So, that is going to be f of
a ,a, b with plus sign and then this plus f of a,b,b into b minus a this will be nothing,
but f dash b minus f of a,b divided by b minus a. So, it will get cancelled here and it is
f of a ,b minus f dash a. So, again b minus a gets cancelled and then you have f dash
b. So, thus the polynomial of degree less than
or equal to 3 which we wrote in an analogous fashion when the interpolation point where
distinct we saw that it is a desired polynomial that means it interpolates the function and
derivative values at 2 end points a and b.
Now, we want to show uniqueness. So, for the uniqueness we are going to use the result
from the fundamental theorem of algebra that if you have got a polynomial of degree less
than or equal to n and if it has got n plus 10 now these n plus 0; that means, counted
according to their multiplicities then such a polynomial has to be 0 polynomial .So, we
have let me recall that a simple 0 means f of c is equal to 0 f dash c not equal to 0
a zero of multiplicity m; that means, the function value and its derivatives up to m
minus 1 they are 0, but f m c not equal to 0 when m is equal to 2 we call it to be a
double zero.
Now, this is the fundamental theorem or rather consequence of fundamental theorem that a
polynomial of degree n has exactly n zeros counted according to their multiplicities
as a consequence of this a non-zero polynomial of degree less than or equal to n it has got
at most n distinct zeros and if a polynomial of degree less than or equal to n has more
than n zeros then it is going to be a zero polynomial so it is this result that we are
going to use to show the uniqueness of cubic hermite polynomial .So, let p3 and q3 be two
polynomials of degree less than or equal to 3 which interpolate the given function under
derivative and at points a and b.
So, if I look at p3 minus q3 that will be a polynomial of degree less than or equal
to 3 which will vanish at 'a' vanish at 'b' and also their derivatives will vanish at
'a' and at 'b'. So, if you are looking at p3 minus q3 it will be a polynomial of degree
less than or equal to 3 such that it has got double zero at 'a' and double zero at 'b',that
means, when we count according to the multiplicity p3 minus q3 has got four zeros, but it is
a polynomial of degree less than or equal to 3 so, that means, it has to be a zero polynomial
.So p3 minus q3 a polynomial of degree less than or equal to three which has got four
zeros that means a double zero at 'a' and double zero at 'b'. So, p3 minus q three x
has to be identically zero and that implies that p3( x) is equal to q3 (x).
So, now we have proved the existence and uniqueness of cubic hermite polynomial now we want to
look at the error in it when we had considered the interpolating polynomial and the interpolation
points being distinct points we have got an error formula the same formula is going to
hold even when the points are repeated when we proved the error formulae for interpolating
polynomial what we had used was Rolle's theorem. So, the proof is similar and what I am going
to do is, I am going to write directly the formula to proof is being very much similar
I am going to skip it.
So, when the interpolation points are distinct say p3 a polynomial interpolating at x0 x1
x2 x3 then f of x minus p3 of x is the error term the divided difference based on x0 x1
x 2 x3 x and multiplied by function w x where w x is x minus x0 into x minus x1 into x minus
x2 into x minus x3 we have got x0 is equal to x1 is equal to 'a' and x2 is equal to x3
is equal to 'b' so, that means, the error will be the divided difference and then x
minus a2 and x minus b2 . Now , let us calculate the infinity norm of f minus p3 that is the
maximum norm and that is modulus of f of x minus p3 x its maximum over the interval a,b.
This divided difference is going to be fourth derivative evaluated at some point c which
depends on x divided by 4 factorial and hence norm of f minus p3 infinity take the modulus
of both the sides and take maximum over interval a, b. So, this will be less than or equal
to norm f 4 infinity divided by a 4 factorial and then maximum of this function. Now, when
we want to calculate maximum of this function we want to calculate absolute maximum. So,
the absolute maximum is obtained by considering the two end points and the critical point.
So, the critical points are the points where the derivative vanishes or it does not exist.
So, let me call this x minus a square x minus b square to be function g of x it is derivative
is given by using product rule it is 2 into x minus a x minus b square plus 2 into x minus
a square into x minus b. So, take common x minus a into x minus b and simplify. So, you
will get 4 x minus a x minus b x minus a plus b by 2.
So, we have got g of x is equal to x minus a2 x minus b2 we are trying to calculate maximum
of modulus of g x ,x belong into interval a ,b g dash at 'a' is equal to 0 g dash at
'b' is equal to 0 and g dash of a plus b by 2 is 0. So, we look at the value of g at these
three points a ,b and a plus b by 2 and whichever is the maximum that is the absolute maximum.
So, when I look at g x g at 'a' will be 0 g at 'b' will be 0and g at 'a' plus b by 2
will be b minus a by2 square and hence we will have error in the cubic hermite polynomial
to be less than or equal to norm f 4 infinity divided by4 factorial into b minus a by 2
raise to 4. So, what I am going to do is I am going to
look at a example. So, we will consider a specific example calculate it is cubic hermite
polynomial and then afterwards we will consider the convergence of interpolating polynomial.
So, let me just recall that if I want to calculate the cubic hermite interpolating polynomial
I need to prepare such a divided difference table here a is repeated. So, this entry is
f dash a b is repeated. So, this entry is f dash b and f of a b is the usual recurrence
formula once we prepare such a divided differences table then p 3 x is f a plus f dash a into
x minus a plus f of a ,a, b x minus a square and f of a, a, b, b x minus a square x minus
b this term corresponds to x minus x0 x minus x1 our x0 is equal to x1 is equal to 'a'.
So, look at function f x to be x raise four plus x cube plus x square plus x plus 1 its
derivative will be given by 4 x cube plus 3 x square plus 2 x plus 1 the value at 0
it is going to be f of 0 will be 1 f of 1 is going to be 5 this entry here this is going
to be f dash at 0 from this formulae it is 1 then this will be f of 11. So, it is five
minus one divided by 1 minus 0. So, that is 4.
This entry is f dash of 1. So, f dash of 1 is going to be equal to 10 put x equal to
1in this formula now this will be 4 minus 1 divided by 1 minus 0. So, that is 3 then
10 minus 4 divided by1 minus 0. So, that is 6 and finally, this entry is 6 minus 3 divided
by 1 minus 0. So, that is 3 now our polynomial p3 x will be 1 plus 1 into x because that
is x minus 0 plus 3 into x minus 0 square. So, that is x square plus 3 into x minus 0
square and then x minus one.
So, this is going to be cubic hermite polynomial for function f of x now let me add one more
interpolating point. Suppose, I want to find a polynomial which interpolates the given
function at 0 repeated 1 repeated and 2. So, then this divided difference table we had
already calculated. So, we just need to add these entries. So, here is 2 f of 2 is 31
this entry will be 31 minus 5 divided by 2 minus 1 this entry is 16 minus 20 divided
2 minus 1. So, it is 16 then you have 16 minus 6 divided by 2 minus 0. So, that is why 5
and 5 minus 3 divided by 2 minus 0. So, that is 1 and now p4 x will be this p3 x and then
you add one more term the coefficient of that term is 1 and then your interpolation points
are 0011. So, that is why x square into x minus 1 square
when you simplify. So, this was our p3 x 1 plus x plus 3 x cube x square into x minus
1 square will be x raise to 4 minus 2 x cube plus x square. So, you get back f of x which
is what it should be you have got a function f to be a polynomial of degree 4 and you are
considering the interpolation from a space of polynomial of degree less than or equal
to 4.
So, we have seen that the interpolation it reproduces polynomials. So, even when some
of the interpolation points are repeated the property will hold and you will get the interpolating
polynomial to be the function itself because we had a polynomial of degree 4 and we had
5 interpolation points 0 repeated twice 1 repeated twice and 2. Now, that brings me
to convergence of interpolating polynomial. So, we have f x minus pn of x to be the divided
difference multiplied by w of x where w of x is product of x minus x of j, j going from
01 up to n .pn is the polynomial which interpolates the function at n plus 1 points and which
has got degree to be less than or equal to n if your function f is sufficiently differentiable
in this case. Suppose it is n plus 1 times differentiable then the divided difference
will be f n plus 1 evaluated at some point c divided by n factorial.
So, norm of f minus pn will be less than or equal to norm of f n plus 1 divided by n plus
1 factorial into norm of w w is given by x minus x0 into x minus xn x varies over the
interval a ,b x0 x1 xn they are also in the interval a, b . So, x minus x j will be at
the most b minus a and hence modulus of w of x will be less than or equal to b minus
a raised to n plus 1. So, look at now the error suppose that this
n plus first derivative is less than or equal to n for all n this is the same bound for
all the derivatives then we will have m into b minus a raise to n plus 1 divided by n plus
1 factorial. Now, no matter how big b minus a is b minus a raise to n plus 1 divided by
n plus one factorial that will tend to 0 and hence, if you have got this condition that
your function f is infinitely many times differentiable and the derivatives they are bound by a constant
in that case your interpolating polynomials pn they will converge to f in the uniform
norm as n tends to infinity for example, the function f of x is equal to e raise to x sin
x the entire functions which are defined on interval a, b.
Now, this is a very strong condition. What a Weierstrass theorem tells that if you have
got f to be a continuous function then it is a uniformly approximated by polynomials.
Now, the Weierstrass theorems says that it is uniformly approximated by polynomials it
does not say by interpolating polynomial and in fact , the result is not true.
So, here you have got well known example which is known as Runge' example that f of x is
equal to 1 upon to 1 plus 25 x square x belonging to minus 1 to 1 and suppose you choose interpolation
points to be equidistant point, if you want to consider p1 you will need two points. So,
take them to be minus 11 for p2 you will need three points. So, take them to be minus 1
zero 1 for pn you will need n plus 1 points so divided minus 1 to 1 into n equal parts
and look at the partition points. Then norm of f minus p n infinity norm tends
to infinity as n tends to infinity. So, not only it does not tend to 0, but it tends to
infinity. So, then one feels that may be something is wrong with the equidistant points that
I chose interpolating points to be equidistant points and then my interpolation is interpolating
polynomials they do not converge in the norm. So, why not take the Chebyshev points those
were in some way optimal and it minimized the norm of w, but then there is a result
that no matter how you choose your interpolation points there will always exist a continuous
function for which you will not have convergence. So, here is a negative result that if I want
to approximate f by interpolation points and I am ready to increase the degree, but then
no matter how I will choose my interpolation points there will always exist a continuous
function for which I will not have convergence. So, this is the drawback of interpolating
polynomials, but we had seen that as such one should not go for higher degree polynomials
because they have got stability problem. So, then what one does is instead of increasing
the degree of the polynomial subdivide your interval a, b into n parts and on each interval
you approximate or you consider a interpolating polynomial of fix degree for example, polynomial
of degree less than or equal to 1. So, what you are doing is you are fixing the degree
of the polynomial on each subinterval and you are ready to increase the number of intervals.
Now, if you do such a thing then under relatively modest conditions you are going to have convergence.
So, from polynomials you are going to piece wise polynomials now there is always a trade
off in this case we gain in convergence, but there is a loss in smoothness of polynomials
and they are infinitely many times differentiable whereas, piece wise polynomials they will
their differentiability properties they will get reduced for example, if you want piece
wise linear polynomials then at the most you can have linear overall continuity.
If you are considering piece wise quadratic then you can have c1 continuity. So, what
I am saying is on each interval it is a quadratic polynomial or polynomial of degree less than
or equal to 2 and then overall it will be differentiable.
When you consider in the interior of sub interval there will be no problem it will be infinitely
many times differentiable, but you are going to join two quadratic polynomials when you
look at the adjoining intervals you are going have different quadratic polynomials. So,
at the joining point there is going to be you can say only or you can demand only the
c1 continuity if you say it should be a c2 continuity it will be a single polynomial
on the two intervals.
So, here is the feber's result that you choose interpolation points in any manner you want
whatever rule you do there exist a continuous function f for which the interpolation point
interpolating polynomials will not converge to f in the maximum norm now this is a negative
result there is also positive result and that result is you fix a continuous function then
there will always exist a set of points if you choose those set of points and consider
interpolating polynomials then norm of f minus pn infinity norm will tend to 0 as n tends
to infinity now there is a catch and a catch is their exist; that means, it difficult to
know like given a continuous function I do not know what there exist, but I do not have
a recipe for such interpolation point, but anyway it is a positive result that there
exist a set points for which a interpolation points they are going to converge to f in
the norm.
Now, we are going to consider the piece wise polynomials. So, the first thing as I said
we are going look at a piece wise linear polynomial. So, let me first recall the single linear
polynomial if you have got f to be defined on interval a, b taking real values x0 and
x1 these are points in interval a ,b then p1 of x is equal to f of x0 plus f of x0 x1
x minus x0 this is polynomial of degree less than or equal to 1. So, p¬n polynomial of
degree less than or equal to 1 and p1 x0 is equal to f of x0 ,p1 of x1 is equal to f of
x1 the error is given by f of x minus p1 of x is equal to f of x0, x1, x into x minus
x0 into x minus x1
So, this is for one polynomial now what I am going to do is I am going to subdivide
my interval into n equal parts. So, I have interval a, b I will subdivide into smaller
interval now in each interval I will choose interpolation points. So, suppose I am taking
two interpolation points here and so on. Now, suppose these are the function values then
this is going to be linear polynomial interpolating the given function at these two points if
I look at the linear polynomial interpolating the given function at these two points it
can be something like this it depends on the function value.
So, I may not even have continuity. So, if I want continuity then let me do like this
way that this is my interval a, b. So, when I consider this point this interval take the
interpolation points to be the end point. So, I will have some function now in this
interval take these two as the interpolating points. So, this interpolation point is common.
So, you are going to have a polynomial here then a polynomial here and so on. So, this
will be piece wise linear.
Now, these are the corner points wherever function will not be differentiable if I consider
a point here then no problem. So, overall we are going to have only continuity. So,
this will be piece wise linear. So, we have say f is from a ,b to R then consider n to
be a natural number h to be b minus a divided by n and ti to be equal to a plus IhI going
from 01 up to n. So, t0 is going to be equal to a and tn is going to be equal to b and
we try to look at a function gn which is defined on interval a ,b taking real values such that
gn restricted to ti to ti plus 1 is a polynomial of degree less than or equal to 1. So, on
each interval it should be a polynomial degree less than or equal to 1 gn at ti should be
equal to f at ti i going from 0 1 up to n.
Now, the choice of these interpolation points they will guarantee that gn will be continuous
we have our function defined on interval a,b we are looking at the n plus 1 points in the
interval a ,b. So, you are going to have some points n plus 1 points if there are n plus
1 points I can fit a polynomial of degree less than or equal to n instead of that we
go to piece wise linear and then we join by straight line. So, that is the piece wise
linear interpolation.
Now, we want to look at the error in the interpolating polynomial. So, we have in case of linear
polynomial interpolation the error is given by norm f double dash infinity divided by
2 into b minus a by2 square the way we had seen that the maximum of modulus of x minus
a square x minus b square it was attended at the midpoint at a plus b by 2 . Similarly
the modulus of this function it is maximum will be again at the midpoint and then you
have got norm of f minus p 1 infinity to be less than or equal to this.
Now, when I consider the piece wise linear interpolation then you have gn on the interval
ti to ti plus 1 it is polynomial of degree less than or equal to 1. So, using this bound
you will get norm f double dash infinity by 2 b minus a is length of the interval now
the length of the interval is h. So, it is h by 2 square now what I am doing is here
I could have dominated by maximum of the second derivative on ti to ti plus 1, but I dominated
by maximum over the whole interval a, b. So, that this bound becomes independent of the
interval and we will get norm f minus g n infinity norm which is maximum over interval
a b that will be less than or equal to norm f double dash infinity by 8 h square and h
being b minus a divided by n that will tend to 0 as n tends to infinity that is about
the piece wise linear interpolation. In the next lecture we will consider other
piece wise polynomials such that piece wise quadratic interpolation then piece wise cubic
hermite interpolation and lastly, we will consider cubic spline interpolation which
consists of piece wise polynomials of degree less than or equal to 3. So, on each interval
it will be a polynomial of degree less than or equal to 3 and overall you will have c2
continuity; that means, our piece wise cubic polynomial will be 2 times continuously differentiable.
So, that we are going to do in the next lecture. So, thank you!