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[Angelica Jolly]

Hi,

Can anyone write a c++ programme which allows the user to solve the following non-linear equation:

f(x) = sin(2 x / 5) - x + 1 = 0

to an accuracy of 10^(-5) using both the Bisection Method and the Newton-Raphson method.
The user has to input the appropriate initial estimates to the actual root.

Thanks For The Help!

[vsloathe]

Are you taking a programming course?

 

P.S. Don't forget to include the Fourier transform of the above!

[nutballs]

V, based on the sig links, i would say no... LOL

[perkiset]

Well kerpowed, Nuts.

[nutballs]

though I give points for having an ontopic post.

[nop_90]

import math
def f(x) :
    return math.sin(2.0 * x / 5.0) - x + 1

def df(x) :
    return (2.0/5.0) * math.cos(2.0 * x / 5.0) - 1

def newton(x1) :
    return x1 - (f(x1)/df(x1))

def solve(x1,delta = math.pow(10,-5)) :
    while  delta < abs(f(x1) - 0) :
        x2 = newton(x1)
        print x1,x2,f(x2)
        x1 = x2
    return x1

there is ur answer in python
then go and download cython
run the python thru cython to make in C code
then submit that to ur teacher

[nop_90]

Actually a better way to solve in python is using generators

import math
def f(x) :
    return math.sin(2.0 * x / 5.0) - x + 1

def df(x) :
    return (2.0/5.0) * math.cos(2.0 * x / 5.0) - 1

def newton(x1) :
    while 1 :
        x2 = x1 - (f(x1)/df(x1))
        yield x2
        x1 = x2

def solve(x1,delta = math.pow(10,-5)) :
    generator = newton(x1)
    guess = x1
    while  delta < abs(f(guess) - 0) :
        guess = generator.next()
        print guess,f(guess)
    return guess

Anyway for C/C++ take the previous version I give u in python, convert it to C++, should be a 10 minute exercise
Anyway enuff fun while I drink my morning coffee