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Hi

I need your help .. I solve a system and I get the error Warning, solutions may have been lost.

The maple code is attached.

testerror.mw

Thank you for your help.

 

Cause CodeTools:-Test will not eval the paramater:

```
f := x -> x+1;

y := 2;

CodeTools:-Test( f(1), 2); # this will work as normal

CodeTools:-Test( f(1), y); # this will not work because `y` will not eval as 2

```

So current my solution is

```
test := subs(y=2, () -> CodeTools:-Test( f(1), y));

test();

```

 

I don't know is there any more proper solution for this.

More general, is there a way to force evaluate an `uneval` parameter?

 

Hi,

I want to conduct the t- test with a

- null hypothesis of : M1-M2 >= 0

- alternative hypothesis of M1-M2 < 0

I found that I can choose the upper or lower tail option for the alternative hypothesis but I am not sure if the same as I want 

any idea ?


firstly apologies in advance for stuff in this question such as "triangle symbol",  my computer is pretty old. 


ok so i was confused a bit here, what i'm trying to do is write a maple procedure that computes Af for a given f contained in V . except we only need to correct the bug in the script below. This script demonstrates such a procedure in the case that omega is a square. The domain is given here as the negative set of a function F contained in V .  I have left in notes where/what i think we need to do but i dunno how to...

N:=10 ; # Global Var
F:=(x,y)->sgn(abs(x-N/2)+abs(y-N/2)-N/4);
Average := proc(F, f0) local f, i, j;
f := f0; # !!!!!!!!!!!!!! something is bad here...
for i to N do for j to N do
if F(i, j) < 0 then
f[i, j] := (f0[i - 1, j] + f0[i + 1, j] + f0[i, j + 1] + f0[i, j - 1])/4 ;
end if;
end do;end do;
return f;
end proc;
f0:=Matrix(N,F); # just to have something to test the procedure
Average(F,f0); # does not return the expected average, modifies f0

 

the necessary information we were given to produce this so far was..

Let N be a positive integer and [N] = {i contained in N | 1<= i <=N }  Let "Omega" C {(i,j) contained in [N] x [N] | 2<=i,j<=N-1} be a subset. Let V = R^([N]x[N]) be the vector space of real valued functions [N]x[N] -> R
and A, "triangle symbol":V->V (average) and "triangle symbole" (Laplacian) be the linear maps such that
[Af](i; j) = f(i; j)      if (i; j) not contained in "Omega"   OR

                             [f(i, j + 1) + f(i, j - 1) + f(i + 1, j) + f(i - 1, j)]/4 if (i,j) is contained in "Omega"

["traingle symbol"f](i,j) =  0 if (i,j) isnt contained in "Omega"   OR

                            ( f(i,j) - [f(i, j + 1) + f(i, j - 1) + f(i + 1, j) + f(i - 1, j)]/4 )    if (i,j) is contained in "Omega"

 Please and thank you for any help in advance <3

                           

Maple IDE team is launching a Tester Program that will allow us to incorporate more input in the product design and development process.

We are looking for regular testers for Maple IDE. Your testing complements our in-team testing of new software versions. By enlisting a diverse group of beta testers, we can see how the software performs when it's being used for normal, ordinary tasks. This lets us provide a more стабле Maple IDE with better user experience. As a member...

has testeq a bug?...

October 22 2012 awehring 55

testeq() gives inconsistent results. See this example:

Equation (1):

Y*(1-h*p^Phi)*(1-t)+sum(Y*(1-t)/(1+r)^s, s = 1 .. infinity)

computes to

Y*(1-h*p1^Phi)*(1-t)+sum(Y*(1-t)/(1+r)^s, s = 1 .. infinity)

Equation

In it's recent edition of Mathematics Today (in print and online), the UK-based Institute of Mathematics and it's Applications, compared 4 symbolic solvers: Maple 15, Mathcad 15, the student edition of Matlab v5 and the Casio CFX-9970G calculator, concluding that "Maple would be the natural choice for research mathematicians, theortetical physicists, those working in any area where mathematics is demanding or for mathematics undergraduates for whom costs are lower"

I need to test a list of numbers to find if any of them are complex. if a complex is found then remove that complete row from the list. the potential complex numbers in my list are a, b and c. b is complex i.e has a non zero imaginery part in row 1 so only row 1 should get deleted. the a b and c were originally generated from the arcsin function. Would appreciate help on this. I know it should be simple enough to do. Thanks in advance.

Can't find what the matter in recursive occurance of this problem. Anyone knows how to sort it out, thanks

k:=20:
m:=4:
p[0] := 1:
tr := 4:

omega[n]:=sqrt(k/m):

T[n]:=(2 pi/omega[n]):

> maxpoint := proc (x)
local tr, p1;

tr := evalf(x*T[n]):
p1 := piecewise(t1 <= tr, p[0]*t1/tr, p[0]):
maximize(int(p1*sin(omega[n]*(t-t1))/(m*omega[n]), t1 = 0 .. t), t = 0 .. dur) :

This is one of rank tests.
Non-parametric methods are widely used for studying populations that take on a ranked order (such as movie reviews receiving one to four stars).
The use of non-parametric methods may be necessary when data have a ranking but no clear numerical interpretation, such as when assessing preferences.
In terms of levels of measurement, non-parametric methods result in "ordinal" data.
After the introduction to the topic let's turn to an example.

How do I test a variable against itself to see if it has a value or not?

As an example

p1:=sin(x)
p2:=46

[seq(`if`(p||i=p||i,p||i,NULL),i=1..5)]  # when p||i actually equals p||i and not some value ie sequence should not include p3,p4 and p5

 

   The Kolmogorov-Smirnov test is a widespread, simple, and effective test to check the hypotheses of the form H[0]:=F[ksi](x)=F(x), where a function F[ksi](x) is the CDF of a population distribution, a function F(x) is a given continuous function (the Kolmogorov  test), and the hypotheses of the form  H[0]:=F[1](x)=F[2](x), where F[j](x), j=1,2, are the CDF of two population distributions, both are assumed to be continuous (the Smirnov test).  See the ...

Hi there,

I'm doing a lot of computations with Maple right now (currently the ancient 10 and 12 versions but maybe soon 14).

I would like to do Unit-Tests of my Software as known for instance in JUnit with Java. Is it possible or do I have to use the ASSERT(...) - statements?

Thanks in advance.

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