0
$\begingroup$

After plotting the graph of $$f(x)=\frac{π}{3\sqrt{2 π}} (\frac{e^{\frac{-x^2}{18}}}{3} + \frac{e^{\frac{-(x - 5)^2}{2}}}{1} +\frac{e^{\frac{-(x - 15)^2}{8}}}{2} ) (1 + x^2)$$

I would like to verify with mathematica that $f(x)<20$, for $\vert{x}\vert>20$ . So I write in the notebook:

Solve[π/(3 Sqrt[2 π]) (E^(-x^2/18)/3 + E^(-(x - 5)^2/2) + 
        E^(-(x - 15)^2/8)/2) (1 + x^2) < 10 && Abs[x] > 20, Reals]]]

and then I get the message that

Solve::fulldim: The solution set contains a full-dimensional component; use Reduce for complete solution information.

On the other hand WolphramAlfa gives me the answer.

Can you please tell me what I did wrong here? Is there another way to check this inequality? Thanks.

$\endgroup$
  • 5
    $\begingroup$ Reduce[[Pi]/(3 Sqrt[2 [Pi]]) (E^(-x^2/18)/3 + E^(-(x - 5)^2/2) + E^(-(x - 15)^2/8)/2) (1 + x^2) < 10 && Abs[x] > 20, Reals] $\endgroup$ – Moo Oct 29 '19 at 19:19
  • 1
    $\begingroup$ @Moo, Thanks it works now...To be honest I wrote also Reduce[ ] ... but outside of Solve[ ] . $\endgroup$ – dmtri Oct 29 '19 at 19:27
  • 3
    $\begingroup$ Maximize[{f[x], Abs[x] > 20}, x, Reals] // Simplify // Quiet indicates that for Abs[x] > 20 the function is below about 3.6803 $\endgroup$ – Bob Hanlon Oct 29 '19 at 20:56
3
$\begingroup$

Updated since the question indicates one thing and the code indicates another.

Reduce can get pretty exact. Your question asks for f[x]<20, but you code tries to solve f[x]<10.

f[x_] = π/(3 Sqrt[2 π]) (E^(-x^2/18)/3 + E^(-(x - 5)^2/2) + 
    E^(-(x - 15)^2/8)/2) (1 + x^2)

If you mean f[x]<20

Reduce[f[x] < 20] // N
(*x < 12.8888 || x > 18.1476*)

which proves your premise and then some.

If you mean f[x]<10

Reduce[f[x] < 10] // N
(*x < 4.67464 || 6.01107 < x < 12.0177 || x > 19.0257*)

which also supports your premise.

$\endgroup$
  • 1
    $\begingroup$ I believe that you meant Reduce[f[x] < 10] // N; in which case, there are three intervals. The outer intervals being the ones of interest. $\endgroup$ – Bob Hanlon Oct 30 '19 at 1:17
  • 1
    $\begingroup$ Actually I keyed off your question. f[x]<20. 10 is not mentioned in your question. Is that a typo? $\endgroup$ – Bill Watts Oct 30 '19 at 3:38
  • 1
    $\begingroup$ the 20 in my comment was a constraint on Abs[x] not on f[x]. The OP's Solve constrained f[x] < 10. Since the maximum for f[x] with Abs[x] > 20 is about 3.6803 this shows that the constraint of f[x] < 10 is met. $\endgroup$ – Bob Hanlon Oct 30 '19 at 3:51
  • 1
    $\begingroup$ I looked at the statement right under the OP s top equation. I was not making any comment on your statements. His question contains f[x]<20. $\endgroup$ – Bill Watts Oct 30 '19 at 3:58
  • 1
    $\begingroup$ Sorry, I just went with his code. $\endgroup$ – Bob Hanlon Oct 30 '19 at 4:01
1
$\begingroup$

Here is "visual" verification with confirmation that your interval $|x|>20$ is contained in solution interval for which $f(x)<20$:

f[x_] := \[Pi]/(3 Sqrt[2 \[Pi]]) (E^(-x^2/18)/3 + E^(-(x - 5)^2/2) + 
    E^(-(x - 15)^2/8)/2) (1 + x^2)
np = NumberLinePlot[{RealAbs[x] > 20, 
    r = Reduce[f[x] < 20, x]}, {x, -20, 20}];
p = Plot[f[x], {x, -30, 30}, 
   GridLines -> {{r[[1, 2]], r[[2, 2]]}, {20}}, PlotRange -> Full];
Show[p, np]
IntervalMemberQ[
 Interval[{-Infinity, r[[1, 2]]}, {r[[2, 2]], Infinity}], 
 Interval[{-Infinity, -20}, {20, Infinity}]]

enter image description here

$\endgroup$
  • $\begingroup$ Nice graphics! Actually, how do you copy from notebook of mathematica and paste it to stack exchange? $\endgroup$ – dmtri Oct 30 '19 at 9:25
  • $\begingroup$ @dmtri I just use a screen capture software Jing. There are many others and mine is an old version. Just suffices for me. $\endgroup$ – ubpdqn Oct 30 '19 at 9:30

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.