## Evaluating right-hand, left-hand limits...

I'm trying to get the RHL of exp1:=(2/(1+e^(-1/x)) as x->0+

and have l2:=limit(exp1,x=0,right) but that isn't giving me a value. How do I correct this?

## Double integral evaluation...

I've got the following double integral over a region A:

e^(1/x*y)/(y^2)*(x+1)^2 where A={(x,y):1/2<=x*y<=2,1<=x<=3}

to evaluate this:

I've tried :

int6:=int(int(e^(1/x*y)/((y^2)*(x+1)^2),x=1..3),y=(1/2)..(2/3));

since the largest lower bound and smallest upper bound for y based on 1/2<=xy<=2 are 1/2 and 2/3 respectively.

This statement however, only evaluates the inner integral; is my approach correct?

## How to use testeq to verify solutions of an equati...

I've got a pair of equations :

x^3-4x=y and y^3-4y=x

which I've defined as eqns:={,}

and 9 solutions as solns1:={,,..}

and being stored as s1,s2,..s9

when I run a command such as testeq(subs(s1,eqns[1])=subs(s1,eqns[2])

I get an error of passing invalid arguments into testeq. What I essentially need to show is that on substituting for x,y from each s1,..s9; both equations get the same result. What am I doing incorrectly?

I've also noticed that just subs(s1,eqns[1]) returns an equality; I don't quite understand why

## solving systems of equations...

I'm given the following two equations:

x^3-4x=y, y^3-4y=x

to solve the system, I've just used

eqns:={x^3-4x=y,y^3-4y=x};

vars:={x,y};

solns:=solve(eqns,vars);

and have obtained only four solutions when I should instead get 9. Is there a mistake in my approach?

## Review code from .mla file...

Hi all,

I google and found a program using C# connect with Maple. The Maple file is mla file - a pakage library type of Maple. I want to review data structure and all interfaces funtions to understand the way to implement this features.

Regards,

Quan Nguyen

## How to evaluate the double integral in Maple ?...

Hello everyone, I am having problem while trying to evaluate the double integral written in the file attached double_inegral.mw. Please help me out.

## LU Decomposition of n-by-n Matrix...

>restart:
>with(LinearAlgebra):
>with(Student[NumericalAnalysis]):
s := {E[1], E[2], E[3]};
v := {x[1], x[2], x[3]};
A := GenerateMatrix(s, v);
B := augment(A)
Then what i do that for any matric i can use same program.

## Email notifications gone on Christmas vacation?...

It appears that email notifications have gone on Christmas vacation.
When will they be back?

## plots[display] bounding display - can it be alterr...

I have a statement:

print(plots[display](seq(colr2[i],i=1..numplayers, xtickmarks=0,axes=NONE,view=[0..5*numplayers,0..1])):

which outputs a horizontal display of the various "colors" of the players, and it does so.

However, the height of the output is quite big and makes the output look "chunky".  Inserting the option "scaling=constrained" makes the width of the output much narrower - but there is a bounding box of the output (which is not actually displayed.)  It is displayed when the mouse cursor is clicked on the display.  I was hoping the view= [..] option would provide a solution, but this does not alter the exterior bounding rectangle.

Is there any means of modifying the dimensions of the rectangular display through Maple code?

Thanks,

David

## seq output - unexpected output for ii:=3...

Description of program: To try to find 'similar shaped' words.  eg the word 'qwirky' is similar to 'ywivhg' because q, y and g have downstrokes and h, k have an 'upstroke'.  ..at present the 'words' are not words in the dictionary sense.

I have posed several questions throughout the program, but will focus on the most puzzling which is in the last section of the program, reproduced below.  The variable ii is set at 1 - relating to the first letter, q, of the 'word' qwirky. It correctly outputs the list, listy[4], which contains the letter q.  Similarly if ii is set to 2 it ouputs listy[8] - w.  However when ii is set to 3, it ouputs the number 7, when listy[7] (=[i]) was expected.  Values of ii from 4 to 6 (thelength of 'qwirky') work as expected.  Ideally I would like to dispense with  the ii:=1 statement, and uncomment the for .. do loop:

> for ii from 1 to nops(letters) do
> seq(listy[listsuffix[  Ord(letters[ii] )-96] ][kk], kk=1..nops(listy[listsuffix[Ord(letters[ii] )-96] ]) );
> end do:

Another question was how to incorporate the sequence, seq() statement in a printf statement - but that's enough for now!!:-)

After reading thris through and seeing theTags, I see I have used a variable 'letters' in my program - perhaps this is causing problems?  I initially thought Maple had a bug in it - but probably a common thought by students!:-)

##End part of program

> ii:=1:   #####Try putting equal to 2, 3, 4, 5, 6
> #ii:=2 returns w - which is correct, but ii:=3 returns 7 when i is expected. Values 4 to 6 work OK
> printf("List of the listy which contains the particular letter, %d  of 'qwirky'  ie the letter %s",ii, letters[ii]);
> printlevel:=5:
> printf("Nops(word)=%d  nops(letters)=%d\n",nops(word),nops(letters));
> #for ii from 1 to nops(letters) do
> seq(listy[listsuffix[  Ord(letters[ii] )-96] ][kk], kk=1..nops(listy[listsuffix[Ord(letters[ii] )-96] ]) );
> #end do:

Thanks, David S

START of MAPLE PROGRAM here

# Similar Words - idea is to create words which look similar in shape to other words
> restart:
> with(StringTools):
> #abcdefghijklmnopqrstuvwxyz
> #a  c, e, o, s, u, x, z
> #b  d, h, k,
> #f  l, t
> #g  j, p, q, y
> #i
> #m  n
> #r  v
> #w
>
> #  Take a word eg 'qwirky' & make words which appear similar -
>
> word:="qwirky":
> numLetters:=length(word):
> printf("Length of word is %d",length(word));
> printf("Why is the length of x+2*y equal to %d - as opposed to just 5?",length(x+2*y));
> printf("Is it because x and y each take up 3 'storage locations' - and one each for +2* ?");
>
> # Help says:For other objects, the length of each operand of expr is computed recursively and added to the number of words used to represent expr. In this way, the measure of the size of expr is returned.
> #...but I'm not much wiser!:-(
> #I ended up using 'nops' instead, but not sure which is better.
> listy[1]:=[a,c,e,o,s,u,x,z]:
> #printf("Number of letters in listy[1] seq(%s, kk=1..nops(listy[1]))   is %d\n",seq(listy[1][kk], kk=1..nops(listy[1])));
> printf("The previous printf statement - commented out - does not work.  The one below does\n - but that's because I've put in eight 'per cents'\n - I'd like the program to work this out.\n");
>
> printf("Number of letters in listy[1] %s,%s,%s,%s,%s,%s,%s,%s is %d\n",seq(listy[1][kk], kk=1..nops(listy[1])  ),nops(listy[1]) );
> listy[2]:=[b, d, h, k]:
> listy[3]:=[f, l, t]:
> listy[4]:=[g, j, p, q, y]:
> listy[5]:=[m, n]:
> listy[6]:=[r, v]:
> listy[7]:=[i]:
> listy[8]:=[w]:
> #Which list number, for each letter of the 26 letters?
> #Nos below correspond to a, b,c... ..z
> listsuffix:=[1,2,1,2,1,3,4,2,7,4,2,3,5,5,1,4,4,6,1,3,1,6,8,1,4,1]:
> #printf("listsuffix[12]=%d",listsuffix[12]);
> #printf("listy[4][2]=%s",listy[4][2]);
> printf("Ascii number of Second letter in list 4 is %d",Ord(listy[4][2]));
> printf("Second letter in list 4 is %s",Char(Ord(listy[4][2])));
> #printf("listsuffix[listy[4][2]]=%d",listsuffix[listy[4][2]]);
> #Split into a list of letters
> letters:=convert(word, list):
> #Consider the letters close to the letters - go through all combinations and check to see if any match words in the dictionary.
> #printf("1");
> #printf("%d",Ord(letters[2]));
> #printf("2");
> #printf("Suffix number is %d",listsuffix[Ord(letters[2])-71]);
> #printf("3");
> for i from 1 to length(word) do
> #See which list letter belongs to
> #printf("Letter number %d, %A belongs to list %d\n",i,letters[i],listsuffix[Ord(letters[i])-96]);
> printf("Ascii number of letter number %d, %s in list %d is %d\n",i,letters[i],listsuffix[Ord(letters[i])-96],Ord(letters[i]));
>
> #printf("Ascii number of letter number %d, %s is %d in list %d which is \n",i,letters[i],Ord(letters[i]),listsuffix[Ord(letters[i])-96]);
> #printf("Below may not work");
> #printf("Ascii number of letter number %d, %s is %d in list %d which is seq(%s, kk=1..nops(listy[listsuffix[Ord(letters[ii] )-96] ]) ))\n",i,letters[i],Ord(letters[i]),listsuffix[Ord(letters[i])-96],seq(listy[listsuffix[  Ord(letters[ii] )-96] ][kk], kk=1..nops(listy[listsuffix[Ord(letters[ii] )-96] ]) ) );
>
> #Check out all letters in list, listy[listsuffix[Ord(letters[i])-96]
>
> #seq(  listy[  listsuffix[  Ord(letters[i] )-96]  ], 1..length(listy[                listsuffix[  Ord(letters[i] )-96]);
> #seq(listy[listsuffix[  Ord(letters[ii] )-96] ][kk], kk=1..nops(listy[listsuffix[Ord(letters[ii] )-96] ]) )
> end do:
> #seq(listy[2][kk], kk=1..4);
> #List of listy[2]
> printf("List of listy[2] below:");
> seq(listy[2][kk], kk=1..nops(listy[2]));
> #List of the listy which contains the particular letter of 'qwirky'
> ii:=1:   #####Try putting equal to 2, 3, 4, 5, 6
> #ii:=2 returns w - which is correct, but ii:=3 returns 7 when i is expected. Values 4 to 6 work OK
> printf("List of the listy which contains the particular letter, %d  of 'qwirky'  ie the letter %s",ii, letters[ii]);
> printlevel:=5:
> printf("Nops(word)=%d  nops(letters)=%d\n",nops(word),nops(letters));
> #for ii from 1 to nops(letters) do
> seq(listy[listsuffix[  Ord(letters[ii] )-96] ][kk], kk=1..nops(listy[listsuffix[Ord(letters[ii] )-96] ]) );
> #end do:

#####  END of program.   Below is output for Maple 7   ##########

Warning, the assigned name Group now has a global binding

Length of word is 6
Why is the length of x+2*y equal to 9 - as opposed to just 5?
Is it because x and y each take up 3 'storage locations' - and one each for +2* ?
The previous printf statement - commented out - does not work.  The one below does
- but that's because I've put in eight 'per cents'
- I'd like the program to work this out.
Number of letters in listy[1] a,c,e,o,s,u,x,z is 8
Ascii number of Second letter in list 4 is 106
Second letter in list 4 is j
Ascii number of letter number 1, q in list 4 is 113
Ascii number of letter number 2, w in list 8 is 119
Ascii number of letter number 3, i in list 7 is 105
Ascii number of letter number 4, r in list 6 is 114
Ascii number of letter number 5, k in list 2 is 107
Ascii number of letter number 6, y in list 4 is 121
List of listy[2] below:
b, d, h, k
List of the listy which contains the particular letter, 1  of 'qwirky'  ie the letter q
{--> enter printf, args = "Nops(word)=%d  nops(letters)=%d\n", 1, 6
Nops(word)=1  nops(letters)=6
30
<-- exit printf (now at top level) = }
{--> enter StringTools:-Ord, args = "q"
113
<-- exit StringTools:-Ord (now at top level) = 113}
{--> enter StringTools:-Ord, args = "q"
113
<-- exit StringTools:-Ord (now at top level) = 113}
{--> enter StringTools:-Ord, args = "q"
113
<-- exit StringTools:-Ord (now at top level) = 113}
{--> enter StringTools:-Ord, args = "q"
113
<-- exit StringTools:-Ord (now at top level) = 113}
{--> enter StringTools:-Ord, args = "q"
113
<-- exit StringTools:-Ord (now at top level) = 113}
{--> enter StringTools:-Ord, args = "q"
113
<-- exit StringTools:-Ord (now at top level) = 113}
g, j, p, q, y

by: MaplePrimes

Earlier this morning, we published an update to MaplePrimes that includes two new features as well as some significant efficiency improvements. Highlights of this update include:

Of all the feature requests that we receive, this has been the most popular, and we're excited to see how it is used! Simply, this new capability allows a member to select the 'best answer' from a list of many. This best answer, selected by clicking on a new trophy icon, will always always appear at the top of the list of answers.

Tag Management

Improving how tags are managed within MaplePrimes has been another long time request from members, and we have made some significant improvements. The first thing that many will notice is that the interface for adding tags to a post or question is now easier to use and (hopefully) faster.

More importantly, however, are the new management features. For starters, only members with a reputation of at least 100 can add brand new tags. If your reputation is less than 100, you can only select from the extensive list of pre-existing keywords. In addition, members with a reputation of at least 50 can now manage tags for existing content by clicking on the new 'Manage Tags' button. It is our hope that this feature will help increase the value of tagging within MaplePrimes by optimizing the commonly-used keywords, and we encourage everyone to take advantage of this new capability!

Efficiency and Stability Improvements

You should notice significant speed increases across all pages within MaplePrimes. On average, page load times have sped up by a factor of 2x, and many times higher in some cases. In addition, interface changes and optimizations have dramatically reduced the file size of many pages, which will make them more responsive in general.

As always, thank you for making MaplePrimes the amazing community that it is! We are hopeful that you will enjoy these improvements, and we look forward to your comments and suggestions.

## mapleprimes update - links broken pics missing

MaplePrimes

Looks like mapleprimes was just updated.  I'm finding links broken and pictures missing.

## limit evaluation...

> restart:
> m:=2; k:=1.0931; a:=k-m; b:=k+m-1;
m := 2
k := 1.0931
a := -0.9069
b := 2.0931
> z:=(k*m)/10^(0.1*10);
z := 0.2186200000
> simplify(((10^(0.1*yo))^((b-a+2*p-1)/2)*z^((b-a+2*p+1)/2)*GAMMA((1-(b-a+2*p))/2))/(p!*GAMMA(p-a+1)*GAMMA(1+((1-(b-a+2*p))/2))));
1 /
---------------------- \0.3183098861 sin(3.141592654 p + 3.141592654) exp(
GAMMA(p + 1.906900000)

-3.040840432 - 1.520420216 p + 0.2302585095 yo) (exp(0.2302585095 yo)) GAMMA(
\
-1. - 1. p)/
> [seq(limit(.3183098861*sin(3.141592654*p+3.141592654)*exp(-3.040840432-1.520420216*p+.2302585095*yo)*(exp(.2302585095*yo))^p*GAMMA(-1.-1.*p)/GAMMA(p+1.906900000),p=k),k=0..10)]
Warning, inserted missing semicolon at end of statement, ...=k),k=0..10)];
[ / 1 /
[limit|---------------------- \0.3183098861 sin(3.141592654 p + 3.141592654)
[ \GAMMA(p + 1.906900000)

exp(-3.040840432 - 1.520420216 p + 0.2302585095 yo) (exp(0.2302585095 yo))

\ \ / 1 /
GAMMA(-1. - 1. p)/, p = 0|, limit|---------------------- \0.3183098861 sin(
/ \GAMMA(p + 1.906900000)

3.141592654 p + 3.141592654) exp(-3.040840432 - 1.520420216 p

p \ \
+ 0.2302585095 yo) (exp(0.2302585095 yo)) GAMMA(-1. - 1. p)/, p = 1|,
/

/ 1 /
limit|---------------------- \0.3183098861 sin(3.141592654 p + 3.141592654)
\GAMMA(p + 1.906900000)

exp(-3.040840432 - 1.520420216 p + 0.2302585095 yo) (exp(0.2302585095 yo))

\ \ / 1 /
GAMMA(-1. - 1. p)/, p = 2|, limit|---------------------- \0.3183098861 sin(
/ \GAMMA(p + 1.906900000)

3.141592654 p + 3.141592654) exp(-3.040840432 - 1.520420216 p

p \ \
+ 0.2302585095 yo) (exp(0.2302585095 yo)) GAMMA(-1. - 1. p)/, p = 3|,
/

/ 1 /
limit|---------------------- \0.3183098861 sin(3.141592654 p + 3.141592654)
\GAMMA(p + 1.906900000)

exp(-3.040840432 - 1.520420216 p + 0.2302585095 yo) (exp(0.2302585095 yo))

\ \ / 1 /
GAMMA(-1. - 1. p)/, p = 4|, limit|---------------------- \0.3183098861 sin(
/ \GAMMA(p + 1.906900000)

3.141592654 p + 3.141592654) exp(-3.040840432 - 1.520420216 p

p \ \
+ 0.2302585095 yo) (exp(0.2302585095 yo)) GAMMA(-1. - 1. p)/, p = 5|,
/

/ 1 /
limit|---------------------- \0.3183098861 sin(3.141592654 p + 3.141592654)
\GAMMA(p + 1.906900000)

exp(-3.040840432 - 1.520420216 p + 0.2302585095 yo) (exp(0.2302585095 yo))

\ \ / 1 /
GAMMA(-1. - 1. p)/, p = 6|, limit|---------------------- \0.3183098861 sin(
/ \GAMMA(p + 1.906900000)

3.141592654 p + 3.141592654) exp(-3.040840432 - 1.520420216 p

p \ \
+ 0.2302585095 yo) (exp(0.2302585095 yo)) GAMMA(-1. - 1. p)/, p = 7|,
/

/ 1 /
limit|---------------------- \0.3183098861 sin(3.141592654 p + 3.141592654)
\GAMMA(p + 1.906900000)

exp(-3.040840432 - 1.520420216 p + 0.2302585095 yo) (exp(0.2302585095 yo))

\ \ / 1 /
GAMMA(-1. - 1. p)/, p = 8|, limit|---------------------- \0.3183098861 sin(
/ \GAMMA(p + 1.906900000)

3.141592654 p + 3.141592654) exp(-3.040840432 - 1.520420216 p

p \ \
+ 0.2302585095 yo) (exp(0.2302585095 yo)) GAMMA(-1. - 1. p)/, p = 9|,
/

/ 1 /
limit|---------------------- \0.3183098861 sin(3.141592654 p + 3.141592654)
\GAMMA(p + 1.906900000)

exp(-3.040840432 - 1.520420216 p + 0.2302585095 yo) (exp(0.2302585095 yo))

\ \]
GAMMA(-1. - 1. p)/, p = 10|]
/]

why the solution is in limit approaches to form??? need to have a closed form expression. any help..????

## numeric range error...

 > restart:
 > m:=2; k:=10.1; a:=k-m; b:=k+m-1;

 > z:=(k*m)/10^(0.1*15);

 > yo:=10^(0.1*10);

 > N1:=evalf(sum(((((yo/z)^((b-a+2*p-1)/2))*GAMMA((1-(b-a+2*p))/2))/(p!*GAMMA(p-a+1)*GAMMA(1+((1-(b-a+2*p))/2)))),p=0..10));

Error, (in NumericRange) summand is singular in the interval of summation

 > N2:=evalf(sum(((((yo/z)^((b+a+2*p-1)/2))*GAMMA((1-(b+a+2*p))/2))/(p!*GAMMA(p+a+1)*GAMMA(1+((1-(b+a+2*p))/2)))),p=0..10));

 >
 >
 >