Items tagged with seq

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I run two small computations to test the improvement that should occur when using several cores (8 cores in my computer):

N:=100000.:

st:=time():

w:=seq(sqrt(i),i=1. ..N):

time()-st;

and:

with(Threads):

st:=time():

w:=Seq(sqrt(i),i=1. .. N):

time()-st;

And the results are: 23.6 sec with seq, and 118.4 sec with Seq!

Seriously?

How can these results be explained?

Thank you in advance

 

hi every one,i want to create some kind of numbering for index, how can i generate them using "seq" command or "for loop" ?
 

restart; with(LinearAlgebra):

 

for i to 3 do
for j to 3 do
N[i||j];print(%);od:od:

N[i1]

 

N[i2]

 

N[i3]

 

N[i1]

 

N[i2]

 

N[i3]

 

N[i1]

 

N[i2]

 

N[i3]

(1)

# i want to create the below numbers, how can i do this using for loop or seq command ?

N[11]

N[11]

(2)

N[12]

N[12]

(3)

N[13]

N[13]

(4)

N[21]

N[21]

(5)

 


 

Download problem.mw

hey is there a easy way to make maple solve an equation with similar different variables,

for eksample 

defining the different variables

x_0:=1       x_1:=4           x_2:=10

setting up and equation

solve x * 5

recieving answers for all defined x variables

= [5,20,50]

 

 

Hi,

I'm currently writing my thesis and actually haven't used Maple before.

I've got the following problem with converting the results of a sequence:

seq(Optimization:-Maximize(function with 2 variables), parameter:0..1,0.1)

The results are ok, but I can't convert the values to a spreadsheet with 4 columns (parameter value, maximum value, value of variable 1, value of variable 2).

Thank you!

Best regards

How do I define this sequence a(n) in Maple?

a(n) = 1/n for n=odd, a(n) = -1/n^2 for n = even.

Thanks!

When using seq function below in the second call, it does not generate a sequence of functions with 'a' being 1, 2, and 3, and I had expected. 

First seq function call is just to show that it works without the function "x ->" wrapping.

I could of couse use unapply as in the third call, but I had expected the second call to work.

Am I doing anything wrong, or is this a Maple bug?

 

Hello,

I have a sequence of functions :  solution[i] , i = 1..n

I have a sequence of times:          Time[i], i = 1..n+1

I need help to plot in the same graph:

 plot(  Sol[1] , t = Time[1]..Time[2] )  

 plot(  Sol[2] , t = Time[2]..Time[3] )  

 plot(  Sol[3] , t = Time[3]..Time[4] )  

etc...

plot(  Sol[n] , t = Time[n]..Time[n+1] )  

Thank you

 

 

 

Hello

Any idea about the summation of Fibonacci sequence

 

Fibonacci.mw

 

Best regards

 

Hello everybody.

I have a function:

f(x,y)=GAMMA(y, -ln(x))/GAMMA(y)

seq(sum(f(x, y), y = 0 .. 1), x = 0 .. 5)

 

and I got a error message:

Error, (in ln) numeric exception: division by zero ??
This is normal behavior in seq function or Bug?

 

but  when I'm first calculate the sum sol := sum(f(x, y), y = 0 .. 1) -> x,

and evalf([seq(sol, x = 0 .. 5)]) ->[0., 1., 2., 3., 4., 5.] works fine.

 

Seq-division_by_zero.mw

Mariusz Iwaniuk

I have the following command.

with(StringTools);
message := `Kajian ini mempunyai tiga objektif pertama seperti yang ditunjukkan dalam bahagian 1.11. Objektif tersebut harus`;

m := convert(message, bytes);

block := map(convert, m, binary);
block := map2(nprintf, "%08d", block);
block := map(proc (t) options operator, arrow; [seq(parse(convert(t, string)[i]), i = 1 .. length(convert(t, string)))] end proc, block);

block := [[0, 1, 0, 0, 1, 0, 1, 1], [0, 1, 1, 0, 0, 0, 0, 1], [0, 1, 1, 0, 1, 0, 1, 0], [0, 1, 1, 0, 1, 0, 0, 1], ........]

with(Bits);
for i to l do
for j from 3 to 7 do
block[i][j] := 1-block[i][j];  //used to flip bit in between 3rd to 7th bit in a block
end do;
c_block[i] := block[i];
end do;
c_block1 := [seq(c_block[i], i = 1 .. l)];

Error, assigning to a long list, please use Arrays

May i know how to solve this problem? I need to change some bit in a list but receive error when there is more than 100 elements in a list. Thank you.

Here is my command

 

> teksbiasa:=`Kriptografi`;

teksbiasa:=Kriptografi

>len:=length(teksbiasa);

len:=11

>nilaiASCII:=convert(teksbiasa, bytes);

nilaiASCII:=[75,114,105,112,116,111,103,114,97,102,105]

>L:=[seq(i,i=nilaiASCII)]

L:=[75,114,105,112,116,111,103,114,97,102,105]

 

Anyone know how i need to write the command to add the lenght of the text (len) into each of the number in nilaiASCII?

What is want to get is:

[86,125,116,123,127,122,114,125,108,113,116]

Thank you~=]]

[75,114,105,112,116,111,103,114,97,102,105]

Hi,

So I'm trying to generate a .dat file that has x in the first column, then the output for y1 in the next and y2 in the following, where y1 and y2 are functions of x so the data file would look something like this from x=-5..5 in equal steps of 0.01.

-5          12           8
-4.99     7             5

etc

I'm struggling to get my head around how to do this, i understand i should use the seq function in maple and save it in array of sorts and then use writedata, but im not sure how to piece it together for having 3 columns

 

thanks in advance for any help!

I am trying to do some algebra with the derivatives of some variables within a program. As a result i need to relable them before i feed them into solve.

To relable them i create vectors and use subs. one of these vectors behaves differently within a proc to how it behaves outside it. This is weird.

TimefullBehavesFunny := proc (nPars, nVars)
local nDiffs, timefull, timeless;
nDiffs := nPars;
timefull := [seq(dx[j, i] = diff(x[i](t), `$`(t, j)), i = 1 .. nVars)];
timefull := [seq(op(timefull), j = 0 .. nDiffs), seq(x[i] = x[i](t), i = 1 .. nVars)];
timeless := `~`[`=`](`~`[rhs](timefull), `~`[lhs](timefull));
timefull, timeless, nDiffs;
end proc

When i run the above for (3,3) i get a differet result to when i run the following
nVars:=3;
nDiffs := 3;
timefull := [seq(dx[j, i] = diff(x[i](t), `$`(t, j)), i = 1 .. nVars)];
timefull := [seq(op(timefull), j = 0 .. nDiffs), seq(x[i] = x[i](t), i = 1 .. nVars)];
timeless := `~`[`=`](`~`[rhs](timefull), `~`[lhs](timefull));

and similarly for other numbers.

any ideasas to why?

 

Hi

    The following code displays values of the subscripted variables slf[], a positive integer, and a string variable filler[] which is just a set of spaces depending on how big the corresponding value of slf[] s.  This is to make the printout lined up nicely.  

printf("%s%d%s%d%s%d%s%d%s%d%s%d%s%d%s%d%s%d%s%d\n", filler[1],slf[1],filler[2],slf[2],filler[3],slf[3],filler[4],slf[4],filler[5],slf[5],filler[6],slf[6],filler[7],slf[7],filler[8],slf[8],filler[9],slf[9],filler[10],slf[10]);

   This works fine,but thought there might be a better way.  I tried:

printf("%s%d%s%d%s%d%s%d%s%d%s%d%s%d%s%d%s%d%s%d\n",seq(filler[k],slf[k],k=1..10));

but this came up with an error message.  Is there  a way of doing this more efficiently?

thanks,   David 

 

Dear all,

Thank you for helping me  to generate a table of values of f(x) starting with x=0 to 100 in steps of 1, that is for x=0,1,2,3,...,100.

 

I tried:

f:=x->2*sqrt(3)*a1*a2*(sum(pochhammer(1/3,k)*3^k*x^(3*k)/(3*k)! ,k=0..infinity)*sum(pochhammer(2/3,k)*3^k*x^(3*k+2)/(3*k+2)!  ,k=0..infinity)-sum(pochhammer(2/3,k)*3^k*x^(3*k+1)/(3*k+1)!  ,k=0..infinity)*sum(pochhammer(1/3,k)*3^k*x^(3*k+1)/(3*k+1)!  ,k=0..infinity));

tab_values:=[evalf(simplify(seq(Ni1(xx),xx=0..100)))];

But I the result is amazing.... I don't understand the problem.

Thanks

 

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