       Re: Help with findroot

• To: mathgroup at smc.vnet.net
• Subject: [mg9182] Re: [mg9169] Help with findroot
• From: Steven Wilkinson <wilkinson at NKU.EDU>
• Date: Tue, 21 Oct 1997 02:02:55 -0400
• Sender: owner-wri-mathgroup at wolfram.com

```Karl,

Concerning the strange behavior of FindRoot[Sqrt[x/(1.2 10^
-4)]==-0.1*Tan[Sqrt[x/(1.2*10^
>-4)]],{x,0.1,0.1,2}] , the bad news is that it is difficult to choose
>first guesses for solutions to your equation that Newton's method
>(FindRoot[ ]) can handle. To see this let me simplify your equation to

u == -0.1Tan[u]

Tan[u] is discontinuous for values of u that are odd multiples of Pi/2.

Graph u + 0.1Tan[u] over the range {u, Pi/2, 3Pi/2}.

The solution to your equation on this range is very close to the
discontinuity at Pi/2. Newton's method works by drawing the tangent
line at your initial guess and sees where it crosses the u-axis. In
this case, unless your initial guess is very close to the actual
solution, the tangent line will not even cross the u-axis in the range
{u, Pi/2, 3Pi/2}.

There is hope, however. Note that if your initial guess is between Pi/2
and the part of the curve that has maximal bend, then you are in good
shape. Curvature is a quantity you can compute and even maximize. I
calculated that the maxima of the curvature for this function are found
at u =  + or - 0.0000312478 around each discontinuity.

Steve Wilkinson

>I'm having a problem using findroot to solve an equation.  Perhaps
>someone
> could shed some light on what's wrong.
>
>FindRoot[Sqrt[x/(1.2 10^ -4)]==-0.1*Tan[Sqrt[x/(1.2*10^
>-4)]],{x,0.1,0.1,2}]
>
>Mathematica 3. returns a value of -0.07 for x which is not anywhere
>close to correct.
> Further, I've tried several different starting values and min/max
>limits, but
> a negative answer is always returned.  Eventually I'de like to compile
>a list
> of all the roots of the equation up to, say, x=1000, but I can't even
>get one
> right now.
>
>Thanks,
>
>Karl Kevala

```

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