If there are (n+1) values of y corresponding to (n+1) values of x, then we can represent the function f(x) by a polynomial of degree

zaasmi

Asia Noor

@zaasmi said in If there are (n+1) values of y corresponding to (n+1) values of x, then we can represent the function f(x) by a polynomial of degree:

If there are (n+1) values of y corresponding to (n+1) values of x, then we can represent the function f(x) by a polynomial of degree

If there are values of corresponding to values of , the function can be represented by a polynomial of degree .

This is based on the concept of polynomial interpolation, specifically the Lagrange interpolation formula, where given distinct points, a unique polynomial of degree will pass through all those points.

zaasmi

@zaasmi said in If there are (n+1) values of y corresponding to (n+1) values of x, then we can represent the function f(x) by a polynomial of degree:

If there are (n+1) values of y corresponding to (n+1) values of x, then we can represent the function f(x) by a polynomial of degree

n+1
n+2
n
n-1

zaasmi

@zaasmi said in If there are (n+1) values of y corresponding to (n+1) values of x, then we can represent the function f(x) by a polynomial of degree:

@zaasmi said in If there are (n+1) values of y corresponding to (n+1) values of x, then we can represent the function f(x) by a polynomial of degree:

If there are (n+1) values of y corresponding to (n+1) values of x, then we can represent the function f(x) by a polynomial of degree

n+1
n+2
n
n-1

n

Asia Noor

@zaasmi said in If there are (n+1) values of y corresponding to (n+1) values of x, then we can represent the function f(x) by a polynomial of degree:

If there are (n+1) values of y corresponding to (n+1) values of x, then we can represent the function f(x) by a polynomial of degree

If there are values of corresponding to values of , the function can be represented by a polynomial of degree .

This is based on the concept of polynomial interpolation, specifically the Lagrange interpolation formula, where given distinct points, a unique polynomial of degree will pass through all those points.

dy/dx - = 1 - y,y(0) = 0 is an example of

In Double integration, the interval [a, b] should be divided into [c, d) should be divided into --sub intervals of size k. --subintervals of size h and the interval

The (n + 1) th difference of a polynomial of degree n is...

Let P be any real number and h be the step size of any interval. Then the relation between h and P for the backward difference is given by

In integrating $\int_{0}^{\frac{2}{2}} \cos x d x$ by dividing the interval into four equal parts, width of the interval should be

In fourth order Runge-Kutta method K 4

In fourth order Runge-Kutta method k2

What is the Process of finding the values outside the interval (Xo,x,) called?

When we apply Simpson's 3/8 rule, the number of intervals n must be

Milne's P-C method is a multi step method where we assume that the solution to the given initial value problem is known at past --equally spaced points.

The truncation error in Adam's predictor formula is ....-times more than that in corrector formula

To apply Simpson's 3/8 rule, the number of intervals be

Which formula is useful in finding the interpolating polynomial?

Rate of change of any quantity with respect to another can be modeled by

Romberg's integration method is ------ than Trapezoidal and Simpson's rule.

In integrating f, e2* dx by dividing into eight equal parts, width of the interval should be......

To apply Simpson's 1/3 rule, valid number of intervals are?

Newton's divided difference interpolation formula is used when the values of the independent variable are

If there are (n+1) values of y corresponding to (n+1) values of x, then we can represent the function f(x) by a polynomial of degree

MTH603 Assignment 1 Solution and Discussion

If there are (n+1) values of y corresponding to (n+1) values of x, then we can represent the function f(x) by a polynomial of degree

File Sharing

3

3.0k

2.8k

8.2k