sand15

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These are questions asked by sand15

Hi,

This is a notional example

1/ I define a matrix through some procedure, for instance
p := k -> Matrix(3, 3, (i,j) -> i+j+k mod 3)

2/ Now I would like to animate the plot of matrices p(k) for successive values of k.
The red commend below doesn't work
N := 5:
plots[animate](Statistics[HeatMap], [ p(k) ], k=0..N, frames=N+1);


Could you please help me to fix this ?
TIA

Hi,

I naively thought that
f := t -> Some expression containing t
and
f := unapply(Some expression containing t, t)
where two different ways to define the same function f.

 

Recently I defined f and g this way
X := RandomVariable(Normal(m, s)):
U := Sample(X, 30000) ;  
f := t -> PDF(X, t);
g := unapply(PDF(X, t), t);

Could you explain me why f~(U) exectues in about 10 seconds and g~(U) in less than 0.1 s ?

Thanks in advance

 

 

Hi,
Can somebody explain me these strange results ?

 

restart:
with(Statistics):
X := RandomVariable(Normal(1.0, 1.0)):
S := Sample(X, 10):

# naively... but it's not what I was expecting for
#                maybe a misuse of the 'normalize=truefalse' option ????
Likelihood(Normal(1.0, 1.0), S);
       # a strange answer

# 2nd way
A := Likelihood(Normal(m, 1.0), S):
evalf(subs(m=1.0, A));
   # this is the good answer
   0.004387

# third way
Likelihood(Normal(1.0, 1.0), S, 'normalize=false');
   4.48e-7    # a wrong answer

 

Thanks for your enlightenment

Hi everyone,

I have a x -> y = f(x) function from R to R (you may suppose f is C(infinity)),  given by an explicit relation.
This function is not strictly monotonic over R.

I want to construct the global inverse of f over R by putting "side by side" local inverse functions.
Let a__0, ..., a__n values of x such that:

  1. -infinity =a__0 < a__1 < ... a__(n-1) < a__n = + infinity
  2. f is monotonic over ] a__p, a__(p+1) [   for each p=0..n-1

The idea is to define the global inverse g of f over R by
g := y ->  piecewise(y < f(a__1), g__0(y), ..., y < f(a__n), g__(n-1)(y))
where g__p(y), is the inverse function of the restriction of f to ] a__p, a__(p+1) [
 

Toy problem
f := x ->1-(1-x)^2;
x__1 := solve(diff(f(x), x);
y__1 := f(x__1);

# I thought one of these commands could work (but they don't return me a single branch as I had expected)
solve(f(x)=y, x) assuming y < y__1;
solve({f(x)=y, x < x__1}, x);

How can I obtain the inverse of a function f over an interval where f is bijective ?


TIA

 

 

 

I wrote this simple set of instructions  (Maple 2015 and Maple 2016)

restart:
p := x -> sum('a__||k' * x^k, k=0..5):
p(x);    # returns a0+ a1x +  ... + a5x5

p(1);   # returns a0+ ... + a5

p(0);   # returns 0 ...  not a0


Probably so huge a mistake that I can't see it !?!?!?

Could you please help me to fit it ?

TIA

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