casperyc

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14 years, 312 days
University of Kent
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These are questions asked by casperyc

v1:=[seq(tau[lambda,i]^2,i=1..5)];

latex(v1);

 

gives:

[{\tau_{{\lambda,1}}}^{2},{\tau_{{\lambda,2}}}^{2},{\tau_{{\lambda,3}}
}^{2},{\tau_{{\lambda,4}}}^{2},{\tau_{{\lambda,5}}}^{2}]

which produces a non natural output in latex, see below.

I wonder it could be improved in the future release?

'latex' command is really useful when i have large algebric output. But having to change this tiny thing at many places is just very time consuming. I'd really appreciate if this can be fixed in updates or future versions of Maple.

 

Casper

 

Hi all,

I will use the following dummy example.

with function,f

f:=(xid,yid)->sum(x[i],i=1..xid)*sum(y[i],i=1..yid);

 

and a complicated term, myterm


myterm:=(f(3,4)+f(2,2))*f(1,1):
myterm:=expand(myterm);

 

'if' i have some previous knowledge, or know a bit of the term, i can find the structure by doing this


repar:=[f(1,1),f(2,2),f(3,4)];  # Or with more f(xid,yid) terms

tmp:=seq(repar[i]=ff[i],i=1..3);
simplify(myterm,{tmp});          # This is fine, gives me what i want

 

But, can we go further, and more 'obvious'

 

Given the fucntion f, same as before, and the same 'myterm'

can I have this
restart:
iwant:=(f(3,4)+f(2,2))*f(1,1);  # as a result, straightforward

so I dont have to go back to 'repar' and find that the terms exactly are.

 

Thanks,

 

Hi, with a list

l:=[1,1,1,2,3,3,4];

What's the best way to get the index(s) for the values equal to '1'?

Say for x=1, we want

[1,2,3]

for x=2, we want

[4]

ect.

 

Hi say I have the vector V1.

V1:=Vector([a,b,c,d,e,f,g]):

and function myfun.

 

how do i use it as the input to the function my fun, by taking away each element in turn?
myfun(V1[2..]);              # 1st element removed
myfun(V1[[1,3..]]);         # 2nd element removed
myfun(V1[[1,2,4..]]);      # 3rd element removed

and so go

is there a more efficient way?

 

Many thanks,

Say I have this list,

tmp:=[[0, 1, 2], [1, 0, 2], [1, 1, 2], [1, 2, 0]];

and the sums of each element (list),

map(x->convert(x,`+`),tmp);

 

How do I quickly pickup the elements, where they sum to 3? Like this:

wanttohave:=[[0, 1, 2], [1, 0, 2],  [1, 2, 0]];

Thanks,

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