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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.

How do I construct the seuqence 1/16 , 1/32 , 1/64 , 1/128 , 1/256 in maple?

 

What's the syntax?

I looked at the examples in here:

http://www.maplesoft.com/support/help/maple/view.aspx?path=seq

 

But didn't find something similar.

 

 

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