Hello im only new here. i just wanna ask if you could solve this problem in newton-rhapson's method although the prescribe one is secant method.
the problem is, f(x)= sinx + cos(1+x^2)-1
where x is in radians and make four iterations w/ initial values of
a)Xi-1 = 1 and Xi=3
and graph the results.
11 3599
Hello im only new here. i just wanna ask if you could solve this problem in newton-rhapson's method although the prescribe one is secant method.
the problem is, f(x)= sinx + cos(1+x^2)-1
where x is in radians and make four iterations w/ initial values of
a)Xi-1 = 1 and Xi=3
and graph the results.
If I recall correctly, the secant method is for root finding; the Newton/Raphson
method approximates an integral. The Newton root finding method is easy:
first find the first derivative f'(x) of f w.r.t. x and apply the following repeatedly:
x(n+1) = x(n)-delta
delta= f(x(n))/f'(x(n))
until |delta| < epsilon, where epsilon is a small posititve value and x(0) is your
first estimation.
kind regards,
Jos
If I recall correctly, the secant method is for root finding; the Newton/Raphson
method approximates an integral. The Newton root finding method is easy:
first find the first derivative f'(x) of f w.r.t. x and apply the following repeatedly:
x(n+1) = x(n)-delta
delta= f(x(n))/f'(x(n))
until |delta| < epsilon, where epsilon is a small posititve value and x(0) is your
first estimation.
kind regards,
Jos
i forgot we are ask to find the 1st positive root. what is delta? and can you pls somehow tell me how to find the derivative?
i forgot we are ask to find the 1st positive root. what is delta? and can you pls somehow tell me how to find the derivative?
To find a positive make sure that you input a positive number as an approximation to the root.
The derivative of the function is found by basic calculus, not sure if this falls under the domain of the scripts. You must know differentiation to use the Newton-Rhapson method.
delta as defined is the ratio of f(x) and its derivative, f ' (x), evaluated at the ith approximation to the root.
If you can't differentiate a function I suggest you use the secant method or the bisection method.
Code for these algorithms can be found at www.nr.com
To find a positive make sure that you input a positive number as an approximation to the root.
That is not true; example f(x) = (x+2)*(x-10); for all x(0) < 4 and > 0 the negative
root will be found by the iteration.
kind regards,
Jos
That is not true; example f(x) = (x+2)*(x-10); for all x(0) < 4 and > 0 the negative
root will be found by the iteration.
kind regards,
Jos
My apologies, you are quite right. Entering a positive approximation will not always return a positive root.
My apologies, you are quite right. Entering a positive approximation will not always return a positive root.
No need to apologize; I was just nitpicking, that's the way I am ;-)
kind regards,
Jos
To find a positive make sure that you input a positive number as an approximation to the root.
The derivative of the function is found by basic calculus, not sure if this falls under the domain of the scripts. You must know differentiation to use the Newton-Rhapson method.
delta as defined is the ratio of f(x) and its derivative, f ' (x), evaluated at the ith approximation to the root.
If you can't differentiate a function I suggest you use the secant method or the bisection method.
Code for these algorithms can be found at www.nr.com
thanks for the help guys. my prof wants us to do this using newton rhapsons method. can i just use the given Xi and disregard the Xi-1?
thanks for the help guys. my prof wants us to do this using newton rhapsons method. can i just use the given Xi and disregard the Xi-1?
I was wrong about my assumption that the Newton-Raphson method deals with
approximation of closed integrals. And yes, if Xi-1 represents the previous
estimation you can forget about it after you've calculated Xi (read my previous
reply again).
kind regards,
Jos
I was wrong about my assumption that the Newton-Raphson method deals with
approximation of closed integrals. And yes, if Xi-1 represents the previous
estimation you can forget about it after you've calculated Xi (read my previous
reply again).
kind regards,
Jos
xi-1 is meant to be for the secant method, but in newton's method only xi is needed. can i just discard it and get the same root?and btw can i graph using c++?
xi-1 is meant to be for the secant method, but in newton's method only xi is needed. can i just discard it and get the same root?and btw can i graph using c++?
Hold on for a second; first you have to implement that Newton (+ Raphson)
iteration. Perform the following steps:
1) let h be a very small positive number
2) let i = 0
2) let x(i) be an estimation of the root you want to find
3) let f'(x) be the first derivative of f(x)
4) let delta = f(x(i)/f'(x(i))
5) let x(i+1)= x(i)-delta
6) let i = i+1
7) if |delta| > h goto step 4
if you can't find f'(x) analytically you can use an approximation:
((f(x+h)-f(x-h))/2h for it.
Graphing something is an entirely different subject; first you have to understand
how that NR method works.
kind regards,
Jos
xi-1 is meant to be for the secant method, but in newton's method only xi is needed. can i just discard it and get the same root?and btw can i graph using c++?
Graphics in C++ is an entirely different matter.
If you want to graph your function to determine an initial approximation to the root then use a graphing package such as Excel, Mathematica, Maple, MatLab, Gnuplot. If you cannot graph the function then you should start your iterations at something like x = 0.
If you want to use Newton-Rhapson it's best if you know the derivative of the function before hand. If you try numerically differentiating your function you are just introducing errors into the calculation. There are plenty of resources on the web that will teach you how to differentiate, if you don't already know. Either google differentiation or go to mathworld.wolfram.com
What code have you done so far?
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