#include #include int main() { int n,i; float a[30],x,sum; printf("Enter the val ue of \'n\'\n"); scanf("%d",&n); printf("Enter \'n+1\' values\n"); for(i=0;i0;i--) sum=(sum+a[i])*x; sum=sum+a[0]; printf("Evaluated result = %f \n",sum); getche(); }
#include#includestruct polynode{float coeff;int exp;struct polynode *link;};void poly_append(struct polynode **,float,int);void display_poly(struct polynode *);void poly_multiply(struct polynode *, struct polynode *, struct polynode **);void padd(float, int, struct polynode **);main(){struct polynode *first, *second, *mult;int i,coeff,exp,high;first = second = mult = NULL;printf("Enter the highest power of polynomial 1: \n");scanf("%d",&high);for(i=high;i>0;i--){printf("Enter value for coeff for X^%d : ",i);scanf("%d",&coeff);poly_append(&first, coeff,i);}printf("\nEnter the highest power of polynomial 2: \n");scanf("%d",&high);for(i=high;i>0;i--){printf("Enter value for coeff for X^%d : ",i);scanf("%d",&coeff);poly_append(&second, coeff,i);}printf("\n\n");display_poly(&first);printf("\n");display_poly(second);printf("\n");for(i=1;i coeff = c;r -> exp = e;r -> link = temp -> link;temp -> link = r;return;}temp = temp -> link; /* go to next node */}r -> link = NULL;temp -> link = r;}}
In Java, or C, the expression is simply:i == jIf the two are equal, this expression will evaluate to true; if not, it will evaluate to false.In Java, or C, the expression is simply:i == jIf the two are equal, this expression will evaluate to true; if not, it will evaluate to false.In Java, or C, the expression is simply:i == jIf the two are equal, this expression will evaluate to true; if not, it will evaluate to false.In Java, or C, the expression is simply:i == jIf the two are equal, this expression will evaluate to true; if not, it will evaluate to false.
The word evaluate in the programming language COBOL can be used to replace a nested if statement. Instead of long statement evaluate allows one to shorten the coding required and write cleaner code.
A do-while loop guarantees the body of the loop will execute at least once. A while loop might not execute at all. // this code will execute, even though the condition test will always evaluate to false do { // stuff }while(false); // this code will never execute because the condition test will always evaluate to false while(false) { // stuff }
Constant values are expressions as well, still I don't think they are so hard to evaluate... well, in this case it is twelve
You can evaluate a polynomial, you can factorise a polynomial, you can solve a polynomial equation. But a polynomial is not a specific question so it cannot be answered.
Substitute that value of the variable and evaluate the polynomial.
You set x = 0 and evaluate the polynomial. Note that this should be "y-intercept" in the singular, not in the plural.
Evaluating a Polynomial expression using a singly linked list visit : http://myfundatimemachine.blogspot.in/2012/06/polynomial-evaluation-in-c.html
A disinfectant with a phenol coefficient of 40 indicates that the chemical agent under study is 40 times as effective as phenol.
The correlation coefficient, plus graphical methods to verify the validity of a linear relationship (which is what the correlation coefficient measures), and the appropriate tests of the statisitical significance of the correlation coefficient.
A zero of a polynomial function - or of any function, for that matter - is a value of the independent variable (often called "x") for which the function evaluates to zero. In other words, a solution to the equation P(x) = 0. For example, if your polynomial is x2 - x, the corresponding equation is x2 - x = 0. Solutions to this equation - and thus, zeros to the polynomial - are x = 0, and x = 1.
That all depends on the meaning of the context. If you want to determine the values of the polynomial function, then you need to substitute the value for the input variable of the function. Finally, evaluate it. For instance: f(x) = x + 2 If x = 2, then f(2) = 2 + 2 = 4.
Assuming the polynomial is written in terms of "x": It means, what value must "x" have, for the polynomial to evaluate to zero? For example: f(x) = x2 - 5x + 6 has zeros for x = 2, and x = 3. That means that if you replace each "x" in the polynomial with 2, for example, the polynomial evaluates to zero.
#include#includestruct polynode{float coeff;int exp;struct polynode *link;};void poly_append(struct polynode **,float,int);void display_poly(struct polynode *);void poly_multiply(struct polynode *, struct polynode *, struct polynode **);void padd(float, int, struct polynode **);main(){struct polynode *first, *second, *mult;int i,coeff,exp,high;first = second = mult = NULL;printf("Enter the highest power of polynomial 1: \n");scanf("%d",&high);for(i=high;i>0;i--){printf("Enter value for coeff for X^%d : ",i);scanf("%d",&coeff);poly_append(&first, coeff,i);}printf("\nEnter the highest power of polynomial 2: \n");scanf("%d",&high);for(i=high;i>0;i--){printf("Enter value for coeff for X^%d : ",i);scanf("%d",&coeff);poly_append(&second, coeff,i);}printf("\n\n");display_poly(&first);printf("\n");display_poly(second);printf("\n");for(i=1;i coeff = c;r -> exp = e;r -> link = temp -> link;temp -> link = r;return;}temp = temp -> link; /* go to next node */}r -> link = NULL;temp -> link = r;}}
Let's start with a first degree polynomial equation:This is a line with slope a. We know that a line will connect any two points. So, a first degree polynomial equation is an exact fit through any two points with distinct x coordinates.If we increase the order of the equation to a second degree polynomial, we get:This will exactly fit a simple curve to three points.If we increase the order of the equation to a third degree polynomial, we get:This will exactly fit four points.if we have more than n + 1 constraints (n being the degree of the polynomial), we can still run the polynomial curve through those constraints. An exact fit to all constraints is not certain (but might happen, for example, in the case of a first degree polynomial exactly fitting three collinear points). In general, however, some method is then needed to evaluate each approximation. The least squares method is one way to compare the deviations.
I presume by hexa you mean hexadecimal, which is base 16. A hexadecimal number can be considered a polynomial in a variable whose value is 16, the coefficients being whole numbers ranging from 0 to 15. The hexadecimal digits after 9 are usually written as the letters A through F (sometimes in lower case); A means 10, B means 11, and so on. Just plug in 16 for the variable and evaluate the polynomial. For example, consider the hexadecimal number 3E7. The 7 is in the units place, the E, meaning 14, is in the 16's place, and the 3 is in the 162 (256) place. So the value of this number, in decimal, is 3 x 256 + 14 x 16 + 7, which works out to be 999 (so help me, I picked those digits at random!).