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You will have to show your effort so far, and be more specific about what precisely you need help with before we can help you.

You will have to show your effort so far, and be more specific about what precisely you need help with before we can help you.
I absolutely can not do it

Which programming languages are you already familiar with?
What do you know about the bisection method?

Where precisely is your effort failing?

Which programming languages are you already familiar with?
What do you know about the bisection method?

Where precisely is your effort failing?
==> With c + +,
==> The method is applicable Pls We wish to solve the equation for the scalar variable x, Nowhere f is a continuous function.
The bisection method requires two initial points a and b and Standard and Poor That f (a) and f (b) have Opposite Signs. This Is Called a bracket of a root, for by the intermediate value theorem the continuous function f must have at least one root in the interval (a, b). The method now divides the interval in two by computing the midpoint c = (a + b) / 2 of the interval. Unless c is a root Itself - Which is very Unlikely, but possible - there are now two possibilities: either f (a) and f (c) have Opposite Signs and bracket a root, or f (c) and f (b ) Opposite Signs and brackets have a root. We select the subinterval That is a bracket, and Apply The Same bisection step to it. In this way the interval That Might contain a zero of f is reduced in width by 50% at Each step. We continue Until We have a bracket sufficiently small for our purposes.

I have some source code with c + + programming language ..
Can you help me?

You should be able to write a while loop that runs until you reach your target accuracy/bracket size, using if statements to check your conditions.

Write some code, test it on some function, and if it doesn't work, post it here. I, or someone else here, will look it over and offer suggestions.

You should be able to write a while loop that runs until you reach your target accuracy/bracket size, using if statements to check your conditions.

Write some code, test it on some function, and if it doesn't work, post it here. I, or someone else here, will look it over and offer suggestions.
this is my source code

# Include <stdio.h>
# Include <math.h>
float f (float x)
(
return x + cos (x);
)

main ()
(
float a, b, c, T, error, e, iteration, the root;
int i;
do
(
i = 1;
printf ("======================================== \ n");
printf ("Program Not Linear Equations Bisection \ n");
printf ("Version 1.0 =- -= \ n");
printf ("======================================== \ n");
printf ("Equation Function F (x) = x + cos (x) \ n");
printf ("Enter the initial guess of a ="); scanf ("% f", & a);
printf ("Input initial guesses b ="); scanf ("% f", & b);

printf ("f (a) =% f \ n", f (a));
printf ("f (b) =% f \ n", f (b));
if (f (a) * f (b) <0)
(
printf ("Enter the value of epsilon ="); scanf ("% f", & error);
printf ("iteration count -"); scanf ("% f", & iterations);
for (i = 1; i <= iterations; i + +)
(
e = abs, (b-a);
c = (a + b) / 2;
T = c + cos (c);
if (f (a) * T <0)
(
b = c;
)
else
(
a = c;
)
if (e <= error)
(
root = c;
)
)
printf ("The root of the equation x =% f \ n", c);
)
else
(
printf ("sorry calculation process is not subject to the terms \ n");
)
printf ("Do you want to repeat the y / t ="); scanf ("% s", & answer);
return 0;
)

I feel this is a very simple program, and all I want is a more complex program, like I could enter the function manually. My example program using the functions return x + cos (x);
My question how can I be able to enter the function as input.

It used to be, there are still many who need me ask again later.

thanks before

Making the program be able to understand an arbitrary input function is not easy, as it has to be able to parse all the different functions and notations.

If you wanted to focus on specific types of functions (polynomials, for example), then you could ask the user to input the coefficients of each term (power of x).