x*x1/2= x3/2 Derivative = 3/2 * x1/2
#include#include #include#define ESP 0.001#define F(x) (x)*(x)*(x) + (x)*(x) + (x) + 7void main(){int i=1;float x0,x1,x2;double f1,f2,f0,t;clrscr( );printf( "\nEnter the value of x0: ");scanf( "%f",&x0);printf( "\nEnter the value of x1: ");scanf( "%f",&x1);printf( "\n____________________________________________\n");printf( "\niteration\tx0\tx1\tx2\tf0\tf1\tf2");printf( "\n_____________________________________________\n");do{x2= (x0+x1)/2;f0= F(x0);f1= F(x1);f2= F(x2);printf( "\n%d %f %f %f %lf %lf %lf",i,x0,x1,x2,f0,f1,f2);if (f0*f2ESP);printf( "\n________________________________________\n");printf( "\n\nApp.root = %f",x2);getch( );}
Interpolation is usually found when studying two variables such that there is some mathematical relationship between them. The relationship need not be causal. Interpolation entails finding the value of one of the variables which corresponds to a given value of the other variable when that given value lies between two known values. Thus, if Y is y1 when X is x1 and Y is y2 when X is x2, interpolation is required to find the value of Y when X is between x1 and x2 or to find the value of X when Y is between y1 and y2.
It is x1 which is x.
Here are some examples. x1/2 = square root of x; x1/3 = cubic root of x; in general, x1/n = nth root of x. Also, x2/3 = the square of the cubic root of x, or equivalently, the cubic root of the square of x.
sqr.rtx/x= sqrt.x*sqr.rtx/sqr.rtx=x/x*sqrt.x=1/sqrt.x. x1/2 = x1/2 * x1/2 = x = 1 (x1/2) /x= 1/x1/2
no, its subtraction. if x2 is over x to the first power, the exponent at the higher value subtracts the x1. whether the x2 is above or below x1, x1 is always being subtrated from x2. X2 X(1) x __ OR __ = X or _ X(1) X2 1
Suppose you have a differentiable function of x, f(x) and you are seeking the root of f(x): that is, a solution to f(x) = 0.Suppose x1 is the first approximation to the root, and suppose the exact root is at x = x1+h : that is f(x1+h) = 0.Let f'(x) be the derivative of f(x) at x, then, by definition,f'(x1) = limit, as h tends to 0, of {f(x1+h) - f(x1)}/hthen, since f(x1+h) = 0, f'(x1) = -f(x1)/h [approx] or h = -f'(x1)/f(x1) [approx]and so a better estimate of the root is x2 = x1 + h = x1 - f'(x1)/f(x1).
The change in the y-value over the x-value, the slope, m, (y1-y2)/(x1-x2).
this is the increasing function theorem, hope it helps "If F'(x) >= 0 , and all x's are and element of [a,b], Then F is increasing on [a,b]" use Mean Value Theorem (M.V.T) Let F'(x)>=0 on some interval Let x1< x2 (points from that interval) by M.V.T there is a point C which is an element of [x1,x2] such that F(x2)-F(x1) / X2- X1 = F'(C) this implies: F(x2)-F(x1) = F'(C) X [x2-x1] F'(C)>=0 [x2-x1]>0 therefore: F(x2)>=F(x1) Therefore: F is increasing on that interval.
it equals x1 it equals x1
x1 = x
x*x1/2= x3/2 Derivative = 3/2 * x1/2
#include#include #include#define ESP 0.001#define F(x) (x)*(x)*(x) + (x)*(x) + (x) + 7void main(){int i=1;float x0,x1,x2;double f1,f2,f0,t;clrscr( );printf( "\nEnter the value of x0: ");scanf( "%f",&x0);printf( "\nEnter the value of x1: ");scanf( "%f",&x1);printf( "\n____________________________________________\n");printf( "\niteration\tx0\tx1\tx2\tf0\tf1\tf2");printf( "\n_____________________________________________\n");do{x2= (x0+x1)/2;f0= F(x0);f1= F(x1);f2= F(x2);printf( "\n%d %f %f %f %lf %lf %lf",i,x0,x1,x2,f0,f1,f2);if (f0*f2ESP);printf( "\n________________________________________\n");printf( "\n\nApp.root = %f",x2);getch( );}
(y-y1)=m(x-x1) OR we can write it y=m(x-x1)+y1
pick two points on the line with good distances from each other, get the x1,y1 value for it... do the same for x2y2 value.. x will be the horizontal axis of the graph, and y will be the vertical. plug the numbers into m=y1-y2/x1-x2
y - y1 = m(x - x1), where m is the slope of the line, and (x1, y1) is a point on the line.