Not sure why I need to invoke the simplify command here, shouldn't the immediate simplification occur?

simplify(%)
                   

However, it does work algebraically

My goal is to generate incremental values of a function to plot using dataplot. Why does the first example generate numeric values in the Vector and the second example does not?

fenq := f -> 2.1^(f+1):
pf := Vector(5, fenq)
                             [  4.41   ]
                              [  9.261  ]
                         pf := [ 19.4481 ]
                              [40.84101 ]
                             [85.766121]
fseqn := (Se, f) -> Se^(f+1):
psf := Vector(5, fsenq(2.1, j))
 

I can plot the first data set using dataplot(pf). I cannot plot the second data set.

Hi, I have been looking at various questions which are similar. But, I could not exactly what I am looking for.

https://www.mapleprimes.com/questions/99679-Removing-Elements-In-A-List

In the above link it is useful to remove elemets of specific type(identical in two lists). I am looking for removing elements from list with specific Index.

L := [1,1,a,c,1];
indexList:=[3,5]

Now based on IndexList I would like to remove the elements from L. That is I want to delete L[IndexList[1]] and L[indexList[2]]. I have tried to do it using

remove(has, L, L[indexList]);  

It is working when there are no identitical elements. In my case it fails to delete exact exlemens.

subs(seq(A[indexList[i]] = NULL, i = 1 .. nops(indexList)), A); Both are working in same way. Not suitable when having identical elements in a list
How one could delete(remove) elements based on index??

Hi,

I'm very new at maple and trying to understand its engine, but it is really hard sometimes when it doesn't make sense what it is doing. I have a function that gives me

com2(e,e,f)=eh+he

and then I would like to substitute eh into <1,0> and he into <0,1>, so I write

subs({eh=<1,0>, he=<0,1>}, com2(e,e,f))

which gives me the following answer

eh+he

Here comes the confusing part. When I plug in eh+he directly into the subs thing I get

<1,1>

Why are these two outputs different from each other? To clarify, why is subs({eh=<1,0>, he=<0,1>}, com2(e,e,f))=/=subs({eh=<1,0>, he=<0,1>}, eh+he)?

Is there a way to program this:

I have to write an equation like this (x1+x2)*(x1+x3)*(x2+x3)=0. Left side is a combination, k=2 and n=3 so the number of terms is 3. For small number of terms this is not a problem to type, but if I have x1,x2,x3,x4,x5,x6, the number of combinations is 15 so it is complicated. Is there a way how to program something like this in Maple or I have to do it manually?

Thanks!

Hello,

I have two function :

And

A={g(R,L)=1}=

(I will call that L : L0=h(R))

So I want to represent f(A) in 3D

So I tried 

It does not work, and I do not see how I could do otherwise...

Thank's in advance !

hi.

According to the fhgure attaceh how i can gain the equation (2-27) . I write the equation (2-26) in maple but I couldnot to gain that result.

If possible to reach equation via maple?

Thanks

diff.mw
 

Download diff.mw

The command print placed within a while loop does not produce any printout. How do I  make MAPLE to produce  it?

I am trying to animate two variables in a 3D plot. Basically I would like to combine these two animations into one simultaneous animation:

animate(plot3d, [[y, x, (1/3)*Pi], x = 0 .. 2*Pi, y = 0 .. R, coords = spherical, scaling = constrained], R = 0 .. 5)

and

animate(plot3d, [[y, x, (1/3)*Pi], x = 0 .. p, y = 0 .. 5, coords = spherical, scaling = constrained], p = 0 .. 2*Pi)

so that both animations start and end simultaneously. Is this possible?

Thanks in advance.

Hi
I appreciate your help forfinding the best fitting line for a discrete series.

Starting with 1978 as base year and counting by two's the five year overage global temperature. where 

the temperature is given by the data : x[0],...,x[12] 

1) Find the best  fitting line : x[n+1]=a*x[n]+b
2) Assuming we can extrapolate, find the predicted value x[30]

Fitting_best_line.mw

Many thanks for your help.

Hi everybody, i want to plot the intersection volume of three inequality, how can i do it?

 

Download plot.mw

The procedure presented here does independence tests of a contingency table by four methods:

1. Pearson's chi-squared (equivalent to Statistics:-ChiSquareIndependenceTest),
2. Yates's continuity correction to Pearson's,
3. G-chi-squared,
4. Fisher's exact.

(All of these have Wikipedia pages. Links are given in the code below.) All computations are done in exact arithmetic. The coup de grace is Fisher's. The first three tests are relatively easy computations and give approximations to the p-value (the probability that the categories are independent), but Fisher's exact test, as its name says, computes it exactly. This requires the generation of all matrices of nonnegative integers that have the same row and column sums as the input matrix, and for each of these matrices computing the product of the factorials of its entries. So, there are relatively few implementations of it, and perhaps none that do it exactly. (Could some with access check Mathematica please?)

Our own Joe Riel's amazing and fast Iterator package makes this computation considerably easier and faster than it otherwise would've been, and I also found inspiration in his example of recursively counting contingency tables found at ?Iterator,BoundedComposition. 

ContingencyTableTests:= proc(
   O::Matrix(nonnegint), #contingency table of observed counts 
   {method::identical(Pearson, Yates, G, Fisher):= 'Pearson'}
)
description 
   "Returns p-value for Pearson's (w/ or w/o Yates's continuity correction)" 
   " or G chi-squared or Fisher's exact test."
   " All computations are done in exact arithmetic."
;
option
   author= "Carl Love <carl.j.love@gmail.com>, 27-Oct-2018",
   references= (                                                           #Ref #s:
      "https://en.wikipedia.org/wiki/Pearson%27s_chi-squared_test",         #*1
      "https://en.wikipedia.org/wiki/Yates%27s_correction_for_continuity",  #*2
      "https://en.wikipedia.org/wiki/G-test",                               #*3
      "https://en.wikipedia.org/wiki/Fisher%27s_exact_test",                #*4
      "Eric W Weisstein \"Fisher's Exact Test\" _MathWorld_--A Wolfram web resource:"
      " http://mathworld.wolfram.com/FishersExactTest.html"                 #*5
   )
;
uses AT= ArrayTools, St= Statistics, It= Iterator;
local
   #column & row sums: 
   C:= AT:-AddAlongDimension(O,1), R:= AT:-AddAlongDimension(O,2),
   r:= numelems(R), c:= numelems(C), #counts of rows & columns
   n:= add(R), #number of observations
   #matrix of expected values under null hypothesis (independence):
   #(A 0 entry would mean a 0 row or column in the original, which is not allowed.)
   E:= Matrix((r,c), (i,j)-> R[i]*C[j], datatype= 'positive') / n,
   #Pearson's, Yates's, and G all use a chi-sq statistic, each computed by 
   #slightly different formulae.
   Chi2:= add@~table([
       'Pearson'= (O-> (O-E)^~2 /~ E),                     #see *1
       'Yates'= (O-> (abs~(O - E) -~ 1/2)^~2 /~ E),        #see *2
       'G'= (O-> 2*O*~map(x-> `if`(x=0, 0, ln(x)), O/~E))  #see *3
   ]), 
   row, #alternative rows generated for Fisher's
   Cutoff:= mul(O!~), #cut-off value for less likely matrices
   #Generate recursively all contingency tables whose row and column sums match O.
   #Compute their probabilities under independence. Sum probabilities of all those
   #at most as probable as O. (see *5, *4)
   #Parameters: 
   #   C = column sums remaining to be filled; 
   #   F = product of factorials of entries of contingency table being built;
   #   i = row to be chosen this iteration
   AllCTs:= (C, F, i)->
      if i = r then #Recursion ends; last row is C, the unused portion of column sums. 
         (P-> `if`(P >= Cutoff, 1/P, 0))(F*mul(C!~))
      else
         add(
            thisproc(C - row[], F*mul(row[]!~), i+1), 
            row= It:-BoundedComposition(C, R[i])
         )
      fi      
;
   userinfo(1, ContingencyTableTests, "Table of expected values:", print(E));
   if method = 'Fisher' then AllCTs(C, 1, 1)*mul(R!~)*mul(C!~)/n!
   else 1 - St:-CDF(ChiSquare((r-1)*(c-1)), Chi2[method](O)) 
   fi   
end proc:

The worksheet below contains the code above and one problem solved by the 4 methods

Download FishersExact.mw

Hi,

I am trying to solve the equation below. First method was subbing in all the parameters into the equation and the second method is plugging all the parameters straight into the equation. As you can see below, they give different results. Can someone please help me? I've been scratching my head for the past hr trying to figure out what am I doing wrong.

Method 1:

Y1 := P*b*x*(-b^2+l^2-x^2)/(6*l*E*I)     

maxdeflection := subs(P = 60000, b = 2050, E = 16000, I = 3062827708, l = 6600, x = (50/3)*sqrt(47229), Y1)

=-(77224295347218906250/297)*sqrt(47229)

Method 2:

Y2 := (50/3)*(60000*2050)*sqrt(47229)*(6600^2-((50/3)*sqrt(47229))^2-2050^2)/(16000*(6*6600)*3062827708)

=(50426796875/1819319658552)*sqrt(47229)

Thank you!

Here's a slightly reduced form of a little module from some code that I posted recently:

KandR:= module()
local
   a, b, c, e, #parameters

   #procedure that lets user set parameter values:
   ModuleApply:= proc({
       a::algebraic:= KandR:-a, b::algebraic:= KandR:-b, 
       c::algebraic:= KandR:-c, e::algebraic:= KandR:-e
   })
   local k;
      for k to _noptions do thismodule[lhs(_options[k])]:= rhs(_options[k]) od;
      return
   end proc
;
end module:

The purpose of the module is simply to be a container for the four parameters and to provide a simple ModuleApply interface by which they can be set, reset, and/or unset. 

I very often use a procedure parameter of a ModuleApply to set a local variable of same name in the module. Because of the name conflict, thismodule needs to be used in these situations. I see this as the primary use of thismodule. In the module above, the purpose of the line 

for k to _noptions do thismodule[lhs(_options[k])]:= rhs(_options[k]) od;

is to avoid the need to explictly use the parameters yet a third time. First off, I am amazed that this works! I've had many disappointments with thismodule (which is essentially undocumented---its miniscule help page is nearly worthless). I am using Maple 2018, release 1. Another Maple 2018 user (not sure which release) reports that the above line gives an error (when executed) that thismodule's index must be a name.

Question 1: What's up with that?

[Edit: It's been determined that the problem was due to an unfortunate global assignment in that user's initialization file rather than different behavior of thismodule. So, I consider Question 1 to be completely answered, and it should be ignored.]

Question 2: The for loop is not entirely satisfying to me. Is there a better way?

[Status: Answered, see below.]

Question 3: Ideally, I'd like to explicity use the four parameters once, not two or three times. Is there a way? If I need to use a container for the parameters (such as a Record), to achieve that, I'd be happy to do that, and I wouldn't mind needing to invoke that container's name any number of times.

[Status: Answered, see below.]

Note that op and exports can be applied to thismodule to extract the module's operands. I have found this occasionally useful.

Question 4: What are some other good uses for thismodule? The one and only example given on its help page seems ridiculous to me.

The docs say that you can assign initial values to a record as shown in this screenshot:

I would expect the last two lines of output to be 1, 2. The slighly more complicated example in the docs does not work as expected either. This is the worksheet: queery.mw

I can assign to a record subsequently, but that makes for very prolix code,

Thanks for any help.