## 6351 Reputation

7 years, 357 days

## How to prove this equality under given c...

Maple 2015

How to force Maple to prove equality (2) under conditions cond.

 > restart:
 > # Given #     0 < u < 1 #     0 < v < 1 #     theta > 1 # # let F the function defined by: F := (u, v) -> exp(-((-ln(u))^theta+(-ln(v))^theta)^(1/theta))
 (1)
 > # How to prove this equality for any n > 0? 'F(u^(1/n), v^(1/n))^n' = 'F(u, v)'
 (2)
 > cond := u > 0, u < 1, v > 0, v < 1, theta > 1, n > 1: simplify(F(u^(1/n), v^(1/n))^n - F(u, v)) assuming cond;
 (3)
 >

## How to "simplify" a RootOf result?...

Maple

I rephrased my previous question in a more synthetic form
(there was probably a lot in it that I thought was important for understanding the problem, but I realized afterwards that it only added confusion).

The true question is yellow-highlighted in the code below

 > restart
 > # The result below seems natural: we were taught in school that exp  # being a bijective function we can get rid of it in the equality to # solve and write simply x=Pi. x = solve(exp(x)=exp(Pi), x)
 (1)
 > # But the solution method solve uses is not that natural (and I # don't really understand it). # infolevel[solve] := 10: # x = solve(exp(x)=exp(Pi), x);
 > # Replacing now exp by some undefined function f produces a # kind of "no-solution" answer: this seems quite normal because # not knowing the properties of f one cannot simply get rid of it. infolevel[solve] := 0: x = solve(f(x)=f(Pi), x)
 (2)
 > # Finally replace f by a bijective function with no analytic expression. s = solve(erf(x)=erf(Pi), x) assuming x::real
 (3)
 > # It would have seem reasonable for Maple to answer x=Pi, or # at least it is what I would have done given the properties # of the erf function. # # How can I "force" Maple to "simplify" it's RootOf result to get # x=Pi?

For those interested in the motivations of this quastion, see here Where_does_the_question_come_from.mw

The original question is here Original_question.pdf

## Why doesn't "plot" display the graph?...

Maple 2015

Do you have any idea why the graph of function f (see the attached file) is not displayed?
How can I plot it without using the

`plot([seq([t, f], t in [seq](0.9..1.12, 0.002))]);`

command ?

 > restart:
 > kernelopts(version)
 (1)
 > f := 4.185692792*10^2172*t^2499*exp(-5000.000000*sqrt(t)); # Here is a plot of f plot([seq([t, f], t in [seq](0.9..1.12, 0.002))]);
 > # How can I plot f using simply: plot(f, t=0.9..1.12);  #no graph
 > # As numelems([seq](0.9..1.12, 0.002)) = 111, I assume # that forcing numpoints to a number that at least equal # to this one could give a non null display? plot(f, t=0.9..1.12, numpoints=1000):  #no graph
 > # Last attempt by forcing a list of points where f has to be evaluated. plot(f, t=0.9..1.12, sample=[seq](0.9..1.12, 0.002))
 >

## Why doesn't `if` recognize three argumen...

Maple 2015

Here is a chunk of a more complex code

```# syntax 1

decisions := "accept", "reject":

T := 2:
# The true test is `if`(t > T, ...) where t comes from some computation.
# In order to focus on the issue I assumed t was equal to 1.
`if`(1 > T, decisions[1], decisions[2]);
"reject"
```

To get a more concise writing, I did the following

```# syntax 2

decisions := "accept", "reject":
T := 2:
`if`(1 > T, decisions);
```

`Error, invalid input: `if` expects 3 arguments, but received 2`

Why doesn't `if` recognizes that decisions is a two parameters sequence?
Is there a way to force `if` to understand syntax 2 ?

I tried replacing `if` by piecewise: while getting no error I can't understand why I got si strange results:

```piecewise(1 > T, decisions);
piecewise(3 > T, decisions);
0
"accept", "reject"
```

What mechanism does piecewise use to return these values?

If_and_piecewise.mw

## background option not correctly understo...

Maple

I often use DocumentTools:-Tabulate or  DocumentTools:-Layout do display a vector or matrix of plots instead of plots:-display. because I find the latter less practical.
But it seems that the 'background' option is not correctly managed with DocumentTools:-Tabulate or  DocumentTools:-Layout.
The attached file shows that:

• the 'background' option is correctly managed if each "view" contains a single plot,
• but not correctly as soon as at least one "view" contains morethan one plot.

DocumentTools_and_Background.mw

How can I get with DocumentTools:-Tabulate  /  DocumentTools:-Layout the same rendering I get with plots:-display?

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