11
$\begingroup$

My grandson and I were playing around with some calculus, using Mathematica to find maxima, minima, & points of inflection for various algebraic functions (e.g., $y = 2 x^3 + 3 x^2 + 5 x + 2$, etc.), We would typically solve $y' = 0$ to find the maxima & minima and then use $y'' = 0$ to see if they were maxima, minima, or POI. This all worked fine! Then, grandson suggests we try it on a trig function such as Sin[x].

Here's our results:

Solve[y'[x]==0,x] 

gave us

{{x -> ConditionalExpression[-(π/2) + 2 π C[1],  C[1] ∈ Integers]},
 {x -> ConditionalExpression[π/2 + 2 π C[1],  C[1] ∈ Integers]}}

and

Solve[y''[x]==0,x] 

gave us

{{x -> ConditionalExpression[2 π C[1], C[1] ∈ Integers]},
 {x -> ConditionalExpression[π + 2 π C[1], C[1] ∈ Integers]}}

We have no idea what this is telling us.

Can someone interpret what this means and how we would use this to evaluate maxima, minima, or POI?

Thanks!

$\endgroup$
  • 4
    $\begingroup$ +1 for going on explorations with your grandson. $\endgroup$ – Henrik Schumacher Apr 27 '18 at 18:57
  • $\begingroup$ See the documentation ConditionalExpression. The ConditionalExpression has a value only when the condition evaluates to True. If the condition evaluates to False then the expression is undefined. $\endgroup$ – Bob Hanlon Apr 27 '18 at 19:21
  • 1
    $\begingroup$ Coolest granddad ever. $\endgroup$ – ibeatty May 1 '18 at 18:54
11
$\begingroup$

This is the way Mathematica handles infinitely many solutions. C[1], C[2] etc. are generic constants that Mathematica uses to express parameterized families of solutions.

Read

x -> ConditionalExpression[-(π/2) + 2 π C[1], C[1] ∈ Integers]

as

$$x \in \left\{ - \frac{\pi}{2} + 2 \, \pi \, n \mid n \in \mathbb{Z} \right\}.$$

So $x$ can be $- \frac{\pi}{2}$ plus a multiple of $2 \pi$.

{
   {x -> ConditionalExpression[-(π/2) + 2 π C[1], C[1] ∈ Integers]}, 
   {x -> ConditionalExpression[π/2 + 2 π C[1], C[1] ∈ Integers]}
}

means that

$$x \in \left\{ - \frac{\pi}{2} + 2 \, \pi \, n \mid n \in \mathbb{Z} \right\} \cup \left\{ \frac{\pi}{2} + 2 \, \pi \, n \mid n \in \mathbb{Z} \right\}$$

That makes sense for $y = \sin$ for $y' =\cos$ and this is precisely the zero set of $\cos$.

So, this set contains all candidates for minima and maxima of Sin. In order to figure out what is what, just plug x into Sin and replace by the output of Solve[Sin'[x] == 0, x]:

Simplify[
 Sin[x] /. Solve[Sin'[x] == 0, x]
 ]
 {
   ConditionalExpression[-1, C[1] ∈ Integers], 
   ConditionalExpression[1, C[1] ∈ Integers]
 }

This tells you: The elements of the first set are minimizers, the elements of the second sets are all the maximizers.

Restricting to intervals

If you are looking for critical points within a given interval, say $[0, 2\pi]$, then you can tell Solve to look only there by using

sol = Solve[{Sin'[x] == 0, 0 <= x <= 6 π}, x]

{{x -> π/2}, {x -> (3 π)/2}, {x -> (5 π)/2}, {x -> ( 7 π)/2}, {x -> (9 π)/2}, {x -> (11 π)/2}}

Second order optimality conditions

Necessary for a local minimum is also $\sin''(x) \geq 0$. We can check that with

Sin''[x] >= 0 /. sol

{False, True, False, True, False, True}

So, only the second, fourth, sixth solution in sol can be local minimizers.

A sufficient condition for a local minimizer is $\sin'(x) = 0$ and $\sin''(x)>0$. Let's check that:

Sin''[x] > 0 /. sol

{False, True, False, True, False, True}

So our three candidates are definately local minimizers.

Including necessary conditions for boundary points

But beware: Minimizers and maximizers of a function restricted to a closed interval can also be located on the boundary of the intervals! You have to check them seperately.

For finding minimizers on a closed interval, you can also incorporate the check for the boundary points into the system that gets handed over to Solve like this

y = x \[Function] Sin[x] + x/2;
a = -1;
b = 2 Pi;
Plot[y[x], {x, a, b}]
necessaryconditions = (a <= x <= b) && ((y'[x] == 0) || (x == a && y'[x] >= 0) || (x == b && y'[x] <= 0));
sol = Solve[necessaryconditions, x]

enter image description here

{{x -> -1}, {x -> (2 π)/3}, {x -> (4 π)/3}}

In necessaryconditions, you have to read && as "and" and || as "or". So these conditions are: $x \in [a,b]$ and $x$ needs to satisfy at least one of the following conditions:

1.) $y'(x) = 0$

2.) $x$ is the left boundary point of the interval and $y$ is not descending at this point ($y'(x) \geq 0$).

3.) $x$ is the right boundary point of the interval and $y$ is not ascending at this point ($y'(x) \leq 0$).

Final remark

You may also try Reduce which might return more readable results:

Reduce[necessaryconditions, x]

x == -1 || x == (2 π)/3 || x == (4 π)/3

And once again, our example from the beginning:

Reduce[Sin'[x] == 0, x]
C[1] ∈ Integers && (x == -(π/2) + 2 π C[1] || x == π/2 + 2 π C[1])
$\endgroup$
  • $\begingroup$ So, if I understand you, -1 & +1 are the minima and maxima..., which is the correct solution. But, how do I get it to tell me the values of x for which that is true (over some interval?)? $\endgroup$ – OilerMan Apr 27 '18 at 20:13
  • 2
    $\begingroup$ @OilerMan -Specify a range of interest: sol = Solve[{y'[x] == 0, -6 Pi <= x <= 6 Pi}, x] and evaluate the function and its derivatives with the solution: {x, y[x], y'[x], y''[x]} /. sol $\endgroup$ – Bob Hanlon Apr 27 '18 at 20:53

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.