If f(x)=0 has multiple roots,how is N-R method affected?How can N-R formula be modified to avoid this problem
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.
For simplicity sake, say that $\displaystyle f(x)$ is a differenenciable function. Select an interval $\displaystyle [a,b]$ such as, $\displaystyle f(a)f(b)<0$, i.e. they have opposite signs, and $\displaystyle f'(x)>0\mbox{ or }f'(x)<0 \mbox{ but not both}$. 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.Originally Posted by bobby77
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Here is a list of steps.
1)Find an interval $\displaystyle [a,b]$ such as, $\displaystyle f(a)f(b)<0$. (Gaurentees existence of a root).
2)Confirm that $\displaystyle f'(x)$ 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 $\displaystyle [a,b]$.
4)Algorithm will converge to that solution on that interval..