mmcdara

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These are questions asked by mmcdara


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))

proc (u, v) options operator, arrow; exp(-((-ln(u))^theta+(-ln(v))^theta)^(1/theta)) end proc

(1)

# How to prove this equality for any n > 0?

'F(u^(1/n), v^(1/n))^n' = 'F(u, v)'

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;

(exp(-(((-ln(u))^theta+(-ln(v))^theta)*n^(-theta))^(1/theta)))^n-exp(-((-ln(u))^theta+(-ln(v))^theta)^(1/theta))

(3)

 

 

Download Stable.mw


Thanks for your help.

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)

x = Pi

(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)

x = RootOf(f(_Z)-f(Pi))

(2)

# Finally replace f by a bijective function with no analytic expression.

s = solve(erf(x)=erf(Pi), x) assuming x::real

s = RootOf(erf(_Z)-erf(Pi))

(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?


Download solve_erf.mw

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

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 ?

Thanks in advance.

restart:

kernelopts(version)

`Maple 2015.2, APPLE UNIVERSAL OSX, Dec 20 2015, Build ID 1097895`

(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))]);

0.4185692792e2173*t^2499*exp(-5000.000000*t^(1/2))

 

 

# 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

# adaptive=true option doesn't help

plot(f, t=0.9..1.12, adaptive=true):  #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))

 

 

Download NoPlot.mw

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);

and received this error

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?

Thanks in advance

If_and_piecewise.mw

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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