# calculus- plz help..

• Aug 13th 2006, 12:06 PM
bobby77
calculus- plz help..
If f(x)=0 has multiple roots,how is N-R method affected?How can N-R formula be modified to avoid this problem
• Aug 13th 2006, 12:30 PM
dan
is N-R for Newton's method ? i.e. x_(n+1) = x_n-(f(x_n)/(f'(x_n))
dan
• Aug 13th 2006, 11:35 PM
Glaysher
If you mean Newton-Raphson then multiple roots is a common problem of any iterative method. Different starting values may give different solutions. Sketching the graph gives you a good idea of what starting values to try but the shape of the graph will determine how good N-R is and in some situations it doesn't work at all.
• Aug 14th 2006, 05:37 AM
dan
if f(x)=0 then f'(x)=0 so N-R method would give you x_n-(0/0) so i dont think you can use it
• Aug 14th 2006, 06:03 AM
Glaysher
That isn't true

eg f(x) = 2x - 2

= 0 at x = 1

f'(x) = 2 at all points not 0
• Aug 14th 2006, 06:10 AM
dan
you said f(x)= 0 i.e f(x)= x-x or say f(x)=(x/(x^2))-(1/x) in any case f'(x)=0
• Aug 14th 2006, 07:40 AM
Glaysher
He means solutions to the equation f(x) = 0 not define f(x) = 0. f(x) has not been defined. He wants a refined method for a general solution.
• Aug 14th 2006, 07:48 AM
ThePerfectHacker
Quote:

Originally Posted by bobby77
If f(x)=0 has multiple roots,how is N-R method affected?How can N-R formula be modified to avoid this problem

For simplicity sake, say that $f(x)$ is a differenenciable function. Select an interval $[a,b]$ such as, $f(a)f(b)<0$, i.e. they have opposite signs, and $f'(x)>0\mbox{ or }f'(x)<0 \mbox{ but not both}$. Then by the intermediate value theorem there is a solution on this interval. Furthermore, since the function is either strictly increasing or strictly deceasing (but not both) implies there cannot be another solution on this interval. Thus, if the function satisfies the condition for the Newton-Raphson algorithm then it will converge to that root.
----
Here is a list of steps.
1)Find an interval $[a,b]$ such as, $f(a)f(b)<0$. (Gaurentees existence of a root).
2)Confirm that $f'(x)$ has only one sign. If not try chaning step #1 slightly to adjust step #2. (Gaurentees uniquesness of a root).
3)Select any point on $[a,b]$.
4)Algorithm will converge to that solution on that interval..
• Aug 14th 2006, 07:50 AM
dan
Quote:

Originally Posted by Glaysher
He means solutions to the equation f(x) = 0 not define f(x) = 0. f(x) has not been defined. He wants a refined method for a general solution.

a ray of light... thanks i missed his point from the start ;)
dan