matlab,numerical-methods,newtons-method

According to https://de.mathworks.com/matlabcentral/newsreader/view_thread/23776, numjac is exclusively intended to be used for ODE functions of the format doty = f(t,y). And n_df should be the derivative of n_f, thus use the derivative of f in the iteration point x. Thus n_f = @(x) x - Y(i) - h*f(T(i+1),x); n_df = @(x)...

It is the inlines that causes trouble. If you are going to use something similar you should use anonymous functions, since inline functions are deprecated and about to be removed. However, when going symbolic this is completely redundant. It will only cause more trouble for you, so I strongly suggest...

c#,math,binary-search,biginteger,newtons-method

Newton's method is N(x) = x - (x^n-a)/(n*x^(n-1))=( (n-1)*x + a/(x^(n-1)))/n This should work well also with integer operations. You may have to correct the result by plusminus 1. It is important to use a good initial value since far away from the root the convergence of Newtons method for...

If I understand right, you are using Newton's method whose formula is: As far as I remember, for convergence of Newton's method (besides the requirements of continuity) two conditions should be true: 1) Derivative of function should not be zero at any of the stages. So you can insert in...

c++,c,opencl,fractals,newtons-method

Well somewhat it really helped to run the code as normal C code.. as this makes Debugging easier: so the issue were some coding issues which I have been able to solve now.. for example my pow function was corrupted and when I added or subtracted I forgot to set...

python,function,typeerror,newtons-method

I think the problem is that, per the documentation args: tuple, optional Extra arguments to be used in the function call. the args argument should be a tuple. Just putting parentheses won't do it; the syntax for tuples is the comma. For example: >>> T = 0 >>> type((T)) <type...

matlab,numerical-methods,newtons-method

Try replacing the following line u1 = u' + inv(DF)*F'; with u1 = u' - inv(DF)*F'; When I make this change, the solution I get is [60,0,-60] which does correctly solve the three equations....

When a equals 0 it causes f'(x) = 2a to go to 0 in which case you get division by 0 in this step: a = (a * a + x) / (2 * a); When f'(x) goes to 0, it indicates you are at a minimum or maximum: http://en.wikipedia.org/wiki/Newton%27s_method...

Newton's method is so x1 = f(x) - (f(x)/df(x)) should be x1 = x - (f(x)/df(x))...

numerical-methods,ode,newtons-method,numerical-stability

Usually you chose the solution that is closest to y(i). The correct solution should satisfy y(i+1)=y(i)+h*f(x(i),y(i)) + O(h²) However, for stiff problems the constant in O(h²) can be very large so that this relation is not as helpful as it seems. If L is the Lipschitz constant of f and...

python,python-3.x,newtons-method

The Raphson(5) call returns a large negative number: >>> Raphson(5) -5501.616437351696 This is passed on to the recursive call: return Q1_6_Raphson(Raphson(rx0),Iter=Iter+1) so Q1_6_Raphson() is called with -5501.616437351696 as the rx0 argument. This is then passed on to fdiv(): elif fdiv(rx0)==0: which throws the exception because that number is larger than...

You can use uniroot(...) for this. [Note: If the point of this exercise is to implement your own version of a Newton Raphson technique, let me know and I'll delete the answer.] If I'm understanding this correctly, you want to generate random samples from a distribution with probability density function...

You need to find the zeros of g(x) = f(x)-y : def source(f,y): def g(x): return f(x)-y x = NR(g, diff_param(g)) return x This returns a single x, but there may be others. To find them you need to try other initial values x0....

matlab,math,numerical-methods,newtons-method

I didn't go through all the code, but at least the first coefficient of the Jacobian should be 0.32 instead of 0.8 to match the function defined one line above. Having the wrong Jacobian can lead to sub-quadratic convergence, although still achieving convergence in some cases.

c#,algorithm,time-complexity,square-root,newtons-method

What is a good rule of thumb for the number of iterations on a number, if one exists? Newton's method has quadratic convergence, ie. at every step of the algorithm, the number of significant digits in the answer doubles. Thus the algorithm computes square roots upto D digits of...