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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 

 

Hello,

I have a little question which is in the title.

It seems to me at the view of this example :

[x[i], y[i]] $ i=1..4;

seq([x[i], y[i]],i=1..4);

May you give me your feedback to be sure ?

Thank you 

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

 

Hello,

 

I try to get the first ten fractions of Pi with the commands "seq" and "nthconver".

I tried this:

a:= cfrac(Pi);
CFRAC([3, 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, ...])
sec(nthconver(a,i),i=1..10);

 

But it doesn't work.

Who can help me?

Hello

I have such problem:

I calculated rather complicated function, and i want to obtain the array or list of data with definite step.

For this purpose I used next expression:

Out:=[seq(t,F(t),t=0..tmax,step]

I need minimum 1000 points, but calculation for the 100 points requires about 300 second, for 1000 - it will be about 3000 seconds. It's too long for me.

 

How could I accelerate this process?

 

P.S.

Plotting of my function requires about 90 seconds for the entire range. So, this is very strange, thad converting to data list requres so much time.

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