For assistance computing the derivative f '(x), try the Derivative Calculator.
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Square roots of 2: f (x)=x2-2 f '(x)=2x |
Approximating Pi: f (x)=tan(x/4)-1 f '(x)=sec2(x)/4 (x0=5, n=5) |
Which root? f (x)=sin(x) f '(x)=cos(x) |
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Horizontal tangent: f (x)=x2-2 f '(x)=2x x0=0 |
Diverging: f (x)=4arctan(x) f '(x)=4/(1+x2) (x0=1.5, n=4) |
Multiple root: f (x)=x3-3x+2 f '(x)=3x2-3 (double root at x=1) |
Under certain rather technical conditions, Newton's Method can be guaranteed to converge quickly to a root xr of f (x), as long as x0 is sufficiently close to xr. However, Newton's Method encounters problems for x values near where f '(x)=0 or f ''(x)=0.
One example of a problem happens when f '(x0)=0 (see the horizontal tangent example above) -- in this case, the tangent line is horizontal, so it has no x-intercept, and there is no x1.
If x0 is not close enough to the root xr, then Newton's Method may not converge at all (see the diverging example above). Or Newton's Method may converge, but not to the expected root (see the "which root?" example above).
If the root xr satisfies f '(xr)=0, then xr is a multiple root, and even if Newton's Method converges, it will converge more slowly (as in the multiple root example above).