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12:22 PM
@J.M. Are you around?
 
InputString["A", "B", WindowSize -> {400, 300}]
has choice buttons
InputString["A", "B", WindowSize -> {All, 300}]
doesn't have
#dailyFun
 
12:41 PM
@Szabolcs How might I help you?
 
@J.M. Hi @J.M.! Just 1 min ...
@J.M. I was wondering if you have any ideas on how to solve this problem analytically (not numerically)
 
Fire away, but do you have a contingency if I tell you that it is analytically intractable?
 
I have a 6th order polynomial which has a single maximum. It only has even power terms plus a linear one. The coefficient of the linear term is a parameter. Example: $-(x^6 - 3/2 x^4 + x^2 + a x)$
First problem: for what values of the parameter does it have more than one maxima? In fact it will have two. I solved this.
Second problem: I have the interval for $a$ where it has two maxima. Now I want the specific value of $a$ where this function has the same value in those two maxima.
@J.M. Well, then I won't feel so bad about doing it numerically :)
@J.M. I did just ask the same in the math channel BTW
 
@Szabolcs well, you could do something like the rule of signs on the derivative so that you can count how many roots it has. Then, a check with the second derivative to ensure that the extrema you got are actually maxima.
@Szabolcs this will involve the second derivative of your polynomial, too.
 
1
Q: Automate standard comments?

LLlAMnYPFrom a comment exchange under a closed question: I'm voting to close this question as off-topic because the issue it raises is not a Mathematica issue but a mathematical one. That it is formulated in terms of Mathematica is not sufficient to make it an appropriate question for Mathematica.SE....

 
 
2 hours later…
2:33 PM
@Szabolcs Find the double tangents without a; then pick a so that the tangents are horizontal:
f0 = -(x^6 - 3/2 x^4 + x^2);
$a = Factor[((f0 /. x -> p) - (f0 /. x -> q))/(p - q)];
doubleT = {Factor[((D[f0, x] /. x -> p) - (D[f0, x] /. x -> q))/(
     p - q)] == 0,
   (D[f0, x] /. x -> p) - $a == 0,
   p < q};
dtsol = Solve[doubleT, {p, q}, Reals];
dtsol = Select[
   dtsol,
   AllTrue[D[f0, {x, 2}] /. {{x -> p}, {x -> q}} /. #, Negative] &];
asol = {a -> $a} /. dtsol;
The Select is really necessary, since the DTs occur only at maxima in this case.
The idea is this: Let p, q be the abscissas of the double tangents. Then the slopes of the tangents are the same, and the slope of the line through the points of tangency is also equal to this.
 
3:04 PM
@MichaelE2 Neat trick! I forgot you could use a divided difference for this.
 
3:32 PM
@J.M. Thanks! :)
BTW, I meant Select is not really necessary....Somehow the motor cortex in my brain feels words of three letters or less are not worth typing.
 

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