## 6234 Reputation

7 years, 318 days

## How to get the result -y(1) + y(2)  ?...

Maple 2015

(I would prefer a solution for Maple 2015, but answers relative to newer versions are welcome)

Is there a simple way to force the result -y(1) + y(2) without using one of these two tricks?

```# how can I get the expression of
int(diff(y(x), x), x=1..2);
/ d                  \
int|--- y(x), x = 1 .. 2|
\ dx                 /

# Trick 1
int(diff(y(x), x), x);
eval(%, x=2)-eval(%, x=1)
y(x)
-y(1) + y(2)

# Trick 2
J := Int(diff(y(x), x), x = 1..2):
value(IntegrationTools:-Parts(J, 1));
-y(1) + y(2)
```

TIA

## How to get continuous level curves?...

Maple 2015

I have a function Gpdf from IR2 to IR+ of class C1 (this comes from the way this function is built).
Although its level curves are continuous, their display show discontinuities for some level values.

The reason is that  Gpdf contains a term whose denominator vanishes and so, even if the left and right limits of Gpdf are the same at the vanishing point, the resulting plot is dicontinuous.

More details are given in the attached file Discontinuous_contours.mw.

I have tried to adjust the plotting grid, or even to superimpose contours drawn in domains containing no singularities, but I wasn't capable to get continuous drawings (see the attached file).

Do you have any idea to achieve this?

TIA

## ​​​​​​​In search for references...

Maple

Could someone provide me with the references used when implementing the Statistics:-PredictiveLeastSquares function?

TIA

## 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

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