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

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:

Best Answer

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.

Having your answer selected as 'best' will reward you with 7 reputation points, as well as make you eligible to receive one of two new badges to show off your question-answering prowess!

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.

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

> 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) 
p 
-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)

p 
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)

p 
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)

p 
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)

p 
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)

p 
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)

p 
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..????

Download mmgf_part1.mw

Any help???

what are the examples of calculating devices and packages?

Just a begineer in maple and dont know how to rectify this error.

any help..???

Download c_ora.mw

Need to evaluate above summation in maple. For simplicity infinity can be replaced by some constant say 1000.

The value of parameters used are:

alpha=73.11

beta=76.11

E=4.75

help....

Hello everyone.

need to know what is the difference between use of GAMMA and gamma in maple.

thanks.

I'm used to Mathcad, and I am very new to Maple. Something I cannot figure out right now is how to define multiple elments of a matrix using a function.

Input Data

Define system dimensions as n:=2;

i:=1..n;

j:=1..n;

lambda:=Matrix(n);

Lambda:=Matrix(n);

upsilon:=vector(n); 

Minor side note: I originaly had upsilon defined using the syntax "Vector[row](ncomp)", but this was giving me an 'exponentiation' operation error, so I changed it to what it is now (basically a list/array, which I guess has different type definitions that no longer cause the error?).

lambda[1,2]:=471.0433;

lambda[2,1]:=883.7530;

upsilon[1]:=58.69;

upsilon[2]:=18.07;

The Problem I'm Having

Now I just want to define Lambda using a function to define all elements (like I would in Mathcad).

Lambda[i,j]:= (upsilon[j]/upsilon[i]) * exp (-lambda[i,j]/2853);

which gives me an extremely long error message:

Error, invalid input: exp expects its 1st argument, x, to be of type algebraic, but received Matrix(2, 2, {(1, 1) = -0.3504976272e-3, (1, 2) = -.16509955895845776, (2, 1) = -.30975332953088164, (2, 2) = -0.3504976272e-3}, datatype = float[8]

As far as I can tell (keep in mind that this is my very first Maple project) that it doesn't like lambda as a matrix? But shouldn't it just evaluate to the element? and why is (2,2) and (1,1) giving values, because it should intialize to zero, so exp(0) = 1 in these cases??

I tried to simplify further by just trying

Lambda[i,j]:=(upsilon[j]/upsilon[i]);

but then it just gives me 1, and when I look into Lambda all the elements are now 1 when only the diagnol elements should be 1 (the rest some fractional amount)??

I am at a complete loss. I thought about doing something like For i = 1 to n etc. but then it just looks like coding, which defeats the purpose of trying to make a calculation sheet documenting the procedure...

Any help and/or insights into what I am doing wrong here would be most welcomed :)

P.S. I've just noticed that there is an upload option :/ Wilson_Equation.mw.

I want to get numerical solution of the Eqs.ode(see the folowlling ode and ibc）in Maple.However,when i run the following procedure,it prompts an error "Error, (in dsolve/numeric/bvp) cannot determine a suitable initial profile, please specify an approximate initial solution". How to solve the issue? Please help me.

restart:
n := 1.4; phi := 1; beta := .6931; psi := 1

> restart;
> n := 1.4; phi := 1; beta := .6931; psi := 1;

> s := proc (x) options operator, arrow; evalf(1+(phi*exp(beta*psi)*h(x))^n) end proc;

> Y := proc (x) options operator, arrow; evalf(f-(1/2-(1/2)/n)*ln(s(x))+2*ln(1-(1-s(x))^(-1+1/n))) end proc;

> ode := diff(h(x), `$`(x, 2))+(diff(Y(x), x))*(diff(h(x), x)+1) = 0;


> ibc := h(0) = 0, ((D(h))(10)+1)*s(10)^(-(1-1/n)*(1/2))*(1-(1-1/s(10))^(1-1/n))^2 = 0;

> p := dsolve({ibc, ode}, numeric);
Error, (in dsolve/numeric/bvp) cannot determine a suitable initial profile, please specify an approximate initial solution
>

I'm not being notified anymore about responses to answers or comments to MaplePrimes. I'm assuming that I'm not the only one (?). This has happened before.

Hy all.

I want to solve this equation, with„dd” as numerical result. What do I do wrong? Thanks. Nico

restart;
TTot := 70;
TC := 17;
GM := .26;
QMax := 870;
V := 3600*GM*QMax*TTot;
eq := V = int(QMax*exp((-t+TC)/dd)*(1+(t-TC)/TC)^(TC/dd), t = 0 .. TTot);
fsolve(eq, dd);

sin(xy) = x + y

subs( y(x)=y, solve( diff(subs( y=y(x), (1) ),x), diff(y(x),x) ) );

-1

subs( x(xy)=x, solve( diff(subs( x=x(xy), (1) ),xy), diff(x(xy),xy) ) );

cos(xy)

I can not get this answer to come out correctly when using this software please help. the correct answer in the back of the book is 

1- y cos(xy)

--------------

x cos(xy) - 1

And is there a way that this program can tutor me on how to get this answer intead of spitting out the answer 

the diff tutor only allows for a one sided equation to be entered.

Hi All. Hope all is well.

Assume that we have partitioned [0,a], into N equidistant subintervals and in each subinterval we have M sets of polynomials of arbitrary form[say b_ij(t)](a.e Taylor series, or Bernstein series,…)

for Example with N=4, M=3 and by Taylor series we have:

now we want to approximate a function, asy f(t), in this interval with following form:

If we have:

(Tau is a constant number)
then: How can  we find L and Z matrices using maple? Is it any way? (or other softwares?)

Regards

Mahmood   Dadkhah

Ph.D Candidate

Applied Mathematics Department