Can Non-Differentiable Functions Affect Solving for Roots of a Function?

  • Thread starter Thread starter eljose
  • Start date Start date
  • Tags Tags
    Function Roots
Click For Summary
Non-differentiable functions, like the floor function or modulus, complicate root-finding methods such as Newton's or fixed-point methods due to their discontinuities. When dealing with piecewise differentiable functions, it is advisable to analyze each segment individually, although convergence issues may arise with methods like the secant or fixed-point if the root lies at a discontinuity. The discussion also explores the potential for constructing a complex-valued generalization of the floor function and its implications for complex differentiability. However, if a root exists at a point of discontinuity, little can be done to ascertain behavior at that point based on surrounding values. Overall, the challenges of non-differentiable functions in root-finding highlight significant limitations in traditional methods.
eljose
Messages
484
Reaction score
0
Let,s suppose we wish to calculate the roots of a function f(x) f(x)=0, of course you will say.."that,s very easy doc...just try Newton Method, fixed-point method or other iterative method"..the main "problem" we have is if f(x) includes non-differentiable functions such us the floor function [x] or the modulus of x |x| then how could we manage to solve it?..for example get the x values that satisfy g(x)-[g(x)]=0

the problem is that [g(x)] is not differentiable for certain values of x...how could we solve that?...:confused: :frown: :rolleyes:
 
Physics news on Phys.org
Well, if you have a piecewise differentiable function, it seems an obvious thing to do is to work with each piece individually.
 
Or, if you really don't want to use the derivative at any point, fixed point or other methods will- the secant method or midpoint method, for example, still work.
 
But still we have the same problem or if x=r is a root of [g(x)]-f(x)=0 but at the point x=r the function [g(x)] has a discontinutiy...and for the secant method..is still valid for piecewise continuous and differentiable functions?, the method of fixed point, unless we are near we may have serious convergence problems to obtain the x so f(x)=x, another question..what would happen if f(x) is nowhere differentiable?..thanks.
 
Is it possible to construct a complex-valued function g(w), w a complex variable, that generalizes the floor-function to the complex plane similar to the way that the gamma function of a complex variable z is a generalization of the factorial function to the complex plane? If so, and one can find it, one might ask "at what points in the complex plane is the generalized floor function g(w) complex differentiable in the complex plane?". It may be such that one can find an analytic continuation that extends the complex-valued function g(w) to the complex plane. Should you manage to get a closed form expression for g(w) at all points g(w) is defined, you can try to apply the argument principle to determine whether a region containing a portion of the real axis contains a zero of the function in question. One can then hone in on the zero of the function by interval halving. There are a lot of ‘ifs’ in all of this :) Is any of this plausible? If not, why not?

Inquisitively,

Edwin
 
To answer your specific question: "what if x= r satisfies f(r)= 0 but is a point of discontinuity of f?" Not much you can do! Since f can by any function at all for x not equal to r, information at any other x can't tell you anything about what happens at r.
 
There are probably loads of proofs of this online, but I do not want to cheat. Here is my attempt: Convexity says that $$f(\lambda a + (1-\lambda)b) \leq \lambda f(a) + (1-\lambda) f(b)$$ $$f(b + \lambda(a-b)) \leq f(b) + \lambda (f(a) - f(b))$$ We know from the intermediate value theorem that there exists a ##c \in (b,a)## such that $$\frac{f(a) - f(b)}{a-b} = f'(c).$$ Hence $$f(b + \lambda(a-b)) \leq f(b) + \lambda (a - b) f'(c))$$ $$\frac{f(b + \lambda(a-b)) - f(b)}{\lambda(a-b)}...

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 4 ·
Replies
4
Views
1K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 22 ·
Replies
22
Views
2K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 8 ·
Replies
8
Views
763
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 6 ·
Replies
6
Views
3K
  • · Replies 3 ·
Replies
3
Views
4K
  • · Replies 6 ·
Replies
6
Views
3K