Re: Function to solve polynomial

• To: mathgroup at smc.vnet.net
• Subject: [mg71251] Re: Function to solve polynomial
• From: "dimitris" <dimmechan at yahoo.com>
• Date: Sun, 12 Nov 2006 06:47:32 -0500 (EST)
• References: <ej42av\$r0u\$1@smc.vnet.net>

```Why you don't want to use the built in functions?
As regards you solution I think you want to use the sample points of
Plot.

Anyway here is a user defined function that use the points of the Plot
function.
But I mention that it is preferable to use Solve and NSolve here since
you deal with
a polynomial fuction!

First we will see the immediate steps of the routine and after we
collect the steps to
define the function

(*I have converted everything in InputForm)

Off[General::spell1]
(*Initialization Cell*)

f[x_] := (x - 4)*(x + 2)*(x - 1)*(x + 5)*(x - 7)                (*your
function*)

plot = Plot[f[x], {x, -10, 10}, DisplayFunction -> Identity];       (*a

points = Cases[plot, {(x_)?NumberQ, (y_)?NumberQ}, Infinity];
(*the points used by the Plot function in order to construct the plot*)

seeds=Position[Times@@@Partition[points[[All,2]],2,1],x_/;x=0]
(*find where the function changes sign*)
{{15},{45},{75},{105},{133}}

samples = Extract[Partition[points[[All,1]], 2, 1], seeds]
(*between this points in x axis there is a change in sign of f[x])
{{-5.05272770135449, -4.849437338398854}, {-2.180144395814093,
-1.9631516192595024}, {0.7731220173928826,
1.1920904288822671}, {3.7479748978290406, 4.137090696764887},
{6.820031559876605, 7.036278769183953}}

(FindRoot[f[x] == 0, {x, #1[[1]], #1[[2]]}] & ) /@ samples
(*use these points as initial guesses for the FindRoot
algorithm-secant method*)

{{x -> -5.}, {x -> -2.}, {x -> 1.}, {x -> 4.}, {x ->
6.999999999998611}}

Now the routine

Clear[f]

FindAllCrossings1D[f_,{x,a_,b_},opts___]:=Block[{
plotData},plotData=Cases[Plot[f,{x,
a,b},DisplayFunction\[Rule]Identity],{x_?NumberQ,y_?
NumberQ},8];

x/.FindRoot[f,{x,#[[1]],#[[2]]},opts]&/@(Extract[Partition[plotData[[
All,1]],2,1],Position[Times@@@Partition[plotData[[All,
2]],2,1],x_/;x=0]])]

plotFunctionRoots[f_, {x_, a_, b_}, roots_, opts___] :=
Block[{ff}, ff = Function[x, Evaluate[f]]; Plot[f, {x, a, b}, opts,
Epilog -> {Red, PointSize[0.02], (Point[{#1, ff[#1]}] & ) /@
roots}]]

f[x_] := (x - 4)*(x + 2)*(x - 1)*(x + 5)*(x - 7)

sols = FindAllCrossings1D[f[x], {x, -10, 10}]
{-5., -2., 1., 4., 6.999999999998611}

plotFunctionRoots[f[x], {x, -10, 10}, sols];

Here is another routine that makes use of the package
NumericalMath`IntervalRoots`

Clear[f]

Needs["NumericalMath`IntervalRoots`"]

FindAllCrossings3[f_, {x_, a_, b_}, {tol_, opts1___}, opts2___] :=
(FindRoot[f, {x, #1[[1]], #1[[2]]}, opts2] & ) /@
List @@ IntervalBisection[f, x, Interval[{a, b}], tol, opts1]

plotFunctionRoots[f_, {x_, a_, b_}, roots_, opts___] :=
Block[{ff}, ff = Function[x, Evaluate[f]]; Plot[f, {x, a, b}, opts,
Epilog -> {Red, PointSize[0.02], (Point[{#1, ff[#1]}] & ) /@
roots}]]

sols = x /. FindAllCrossings3[f[x], {x, -10, 10}, {0.1, MaxRecursion ->
10}]
{-5., -2., 1., 4., 6.999999999999992}

plotFunctionRoots[f[x], {x, -10, 10}, sols, ImageSize -> {400, 300},
Frame -> {True, True, False, False}, Axes -> {True, False}];

Here are some relevant links inside this forum...

There is also a solution in the following book

http://www.amazon.com/Mathematica-Action-Stan-Wagon/dp/0387986847/sr=1-2/qid=1162830266/ref=sr_1_2/103-5414680-6091022?ie=UTF8&s=books

a1b2c3d4 wrote:
> I am trying to write a function that will help me solve the polynomial f(x)=(x-4)(x+2)(x-1)(x+5)(x-7) (NSolve and Solve functions not preferred)
>
> I can only think of this function:
> solution := {Plot[{(x -4)(x + 2)(x - 1)(x + 5)(x - 7)}, {x, -10, 10}, AxesLabel -> TraditionalForm /@ {x, y}]}
>
> Does anyone have better knowledge than me in writing an automated program to solve this polynomial? Say For, While or Do Loop?
>
> Your help will be greatly appreciated.

```

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