3
$\begingroup$

Suppose I have the following rational function: 

pol=(354.071 x + 1136.03 x^2 + 2301.83 x^3 + 3771.25 x^4 + 5371.14 x^5 + 
   6303.09 x^6 + 6706.8 x^7 + 6518.86 x^8 + 5439.16 x^9 + 
   4614.45 x^10 + 3459.79 x^11 + 2655.16 x^12 + 2101.17 x^13 + 
   1522.11 x^14 + 1076.82 x^15 + 756.594 x^16 + 480.121 x^17 + 
   262.326 x^18 + 204.555 x^19 + 111.024 x^20 + 56.7865 x^21 + 
   27.375 x^22 + 28.2448 x^23 + 19.7196 x^24 + 5.61942 x^25 + 
   3.73647 x^26 + 0.931673 x^27 + 1.85847 x^28 + 2.7804 x^29 + 
   0.924375 x^30 + 1.84391 x^31)/(354 x + 1143 x^2 + 2320 x^3 + 
   3811 x^4 + 5441 x^5 + 6403 x^6 + 6829 x^7 + 6658 x^8 + 5571 x^9 + 
   4737 x^10 + 3560 x^11 + 2741 x^12 + 2174 x^13 + 1579 x^14 + 
   1120 x^15 + 789 x^16 + 502 x^17 + 275 x^18 + 215 x^19 + 117 x^20 + 
   59 x^21 + 30 x^22 + 30 x^23 + 21 x^24 + 6 x^25 + 4 x^26 + x^27 + 
   2 x^28 + 3 x^29 + x^30 + 2 x^31)

I am interested in this function between 0 and 1, so I plot it as: 

Plot[pol, {x, 0, 1}, PlotRange -> All]

which gives:

enter image description here

Now I want to find the maximum value of this plot, it can be seen roughly that the point exists roughly above 1. To find it I do: 

FindMaxValue[pol, {x, 0, 1}]

which thrown error as valuation from exact 0 gives 1/0 in the fraction, so I reduce the precision from 

FindMaxValue[pol, {x, 10^-3, 1}]

which gives: 

67857.7

which is plain wrong. I wonder how one solves this issue? and how one can evaluate the function between 0 and 1 rather that 10^-3.

|improve this question
$\endgroup$
  • 1
    $\begingroup$ tried FindMaxValue[{pol, 0 <= x <= 1}, {x, .5}]? $\endgroup$ – kglr Jan 21 at 12:31
  • $\begingroup$ The issue is that pol[0] is undefined as it evaluates to 0/0. The classical method to find the maximal value is to determine the first and eventually the second derivative. However, both of these do not exist at x=0. Thus most likely the algorithm fails. $\endgroup$ – gothicVI Jan 21 at 12:31
  • $\begingroup$ @kglr thanks now it works. $\endgroup$ – William Jan 21 at 12:35
  • $\begingroup$ Try pol=Simplify[pol] before using it. $\endgroup$ – Somos Jan 21 at 20:43
4
$\begingroup$ 
FindMaxValue[{pol, 0 <= x <= 1}, {x, .5}]

1.0002

FindMaxValue >> Details and Options:

  • FindMaxValue[$f$,{$x$,$x_0$,$x_1$}] searches for a local maximum in $f$ using $x_0$ and $x_1$ as the first two values of $x$, avoiding the use of derivatives.

That is, {x, 0, 1} in in FindMaxValue[pol, {x, 0, 1}] does not represent a constraint; you need to specify the constraints in the first argument.

|improve this answer
$\endgroup$
2
$\begingroup$

This might be a bit more stable in some cases:

f = Log[Numerator[pol]] - Log[Denominator[pol]];
Exp@FindMaxValue[{f, 0 <= x <= 1}, {x, .5}]

1.0002

|improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.