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Re: Re: Find many Roots

  • To: mathgroup at smc.vnet.net
  • Subject: [mg32107] Re: [mg32103] Re: Find many Roots
  • From: Andrzej Kozlowski <andrzej at tuins.ac.jp>
  • Date: Thu, 27 Dec 2001 03:34:08 -0500 (EST)
  • Sender: owner-wri-mathgroup at wolfram.com

In fact in this particular case one does not really need to look at the 
graph. It is enough to observe that function:

f[x_] := Tan[x] - 1/x

is undefined at

In[2]:=
l=Prepend[Table[(k*Pi)/2,{k,1,24/Pi,2}],0]

Out[2]=
{0, Pi/2, (3*Pi)/2, (5*Pi)/2, (7*Pi)/2}

Moreover it's derivative:

In[3]:=
D[f[x],x]

Out[3]=
1/x^2 + Sec[x]^2

is strictly positive wherever the function is defined. It follows that 
there can be just a single root in each of the open intervals that make 
up the domain. Thus we can find all the roots as follows:

In[4]:=
FindRoot[Evaluate[f[x]==0],{x,#}]&/@(Plus@@@Partition[N[l],2,1]/2)

Out[4]=
{{x -> 0.8603336747802416}, {x -> 3.4256184594916825},
   {x -> 6.437298362730657}, {x -> 9.529334419122055}}

Andrzej Kozlowski
Toyama International University
JAPAN
http://platon.c.u-tokyo.ac.jp/andrzej/


On Tuesday, December 25, 2001, at 01:44  PM, Tom Burton wrote:

> Hello,
>
> On Mon, 24 Dec 2001 08:33:36 +0000 (UTC), in comp.soft-
> sys.math.mathematica you wrote:
>
>> Tan[x]==1/x, {x,0,12}]
>
> For this particular problem, I would select:
>
> Step 1: Find the approximate roots visually:
>
> p1 = Plot[Tan[x]-1/x, {x,0,12}, PlotRange -> {-1,1}];
>
> approx = {1, 3, 6, 10};
>
> Step 2: Refine these approximations:
>
> answers = FindRoot[Tan[x] == 1/x, {x,#}] & /@ approx
>
> Step 3: Verify solutions
>
> (Tan[x] - 1/x) /. answers
>
> Step 4: Check for missing solutions
>
> Show[ p1, Graphics[{Hue[0], PointSize[0.03], Point[{x,0}] /. answers}] ]
>
> I higly recommend a visual method, but iff you need a more automated 
> method, I would suggest something like
>
> answers = FindRoot[Tan[x] == 1/x, {x,#}] & /@ Range[1,12]
>
> xx = Select[
>          Union[ x /. answers, SameTest -> (Abs[#1-#2]<10^-6 &) ],
>          0 <= # <= 12 &]
>
> Tom Burton
>
>
>



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