## 9 Badges

11 years, 86 days

## @Preben Alsholm also the "sort" command ...

@Preben Alsholm also the "sort" command will work! :)

f := b*a;
f := a b
sort(f, a);
b a

## Thanks. :)...

@nm It seems that the suggestion by Alejandro Jakubi is the best way to do this. :)

## Thank you so much! You saved me! :D...

@Alejandro Jakubi This was exactly what I was looking for! many many thanks! :)

## Not working as expected! :D...

@Alex Smith Can you give it a try for the following sample. It is not working on my system. Of course it worked for the example you mentioned above! :)

Sample.mw

## @Alex Smith So I should convert the equa...

@Alex Smith So I should convert the equation to a latex format and then import to mathtype!
Lets see how it will work!

Have you checked maple 18 for the polishment!? :D

## Thank You! :)...

@Alejandro Jakubi yes, writing a rule is option but I do prefer to this with maple bult-in functions! :)
Or some generalized procedure as you told!

so the problem is to tell maple to factor out a linear combination, like "(a*x+b*y)" out of a expression "f"! :)
Do you have any idea to this with maple's functions? or to write a simple procedure using maples functions to generalize the collect command?

## @nm You are totally right but I think th...

@nm You are totally right but I think that I have stated my problem a little bad! :)

Consider that a=(m+n) then f becomes:

f=(m+n)*b*x+(m+n)*b*y

Now, can you just factor out the (m+n)?

collect command doesn't work in this case!

## Likely it is! :D...

@Axel Vogt Of course I thought of doing it in this way! :)
But it is not clean! :D
Using this method instead of "collect(f,a)" will do the job even if "a" is also a linear combination ( for example a=m+n )!
But the collect command won't work in that case!

## @Preben Alsholm Nice trick! :)Thanks...

@Preben Alsholm Nice trick! :)
Thanks

## @nm You are totally right! :)My mind was...

@nm You are totally right! :)
My mind was just stuck in "op" command! :D

## Thank You! :)...

@Alejandro Jakubi
It seem working but you should write a proc for each kind of factorization you want everytime! :)

Don't you have any Idea about the first question above?

## Two Other Questions!...

1-Suppose that after applying the command "factor(f)" the "f "takes the form:

f=(BesselJ(0,r))*(A very lengthy term)

Is there a way to force maple to show f like below:

f=(A very lengthy term)*(BesselJ(0,r))

2-Suppose f is written as follows:

f=a*b*x+a*b*y

Can you suggest a way (without using "op" command) to write f as:

f=a*(b*x+b*y)

## Can't Vote Up! :/...

@Carl Love It is not working yet! :/
It says:

You need at least 10 reputation to vote items up

## Many Many Thanks,...

@Carl Love The select() function that you suggested, solved lots of my difficulties! :)
I wanted to vote you up but I do not have enough reputation! :D

## @Carl Love This one will also do the job...

@Carl Love This one will also do the job:

select(hastype,op([1,1],BC4),sinh(algebraic));

The reason is that always op([1,1],BC4) will give the operand of summation! Actually the operand numbers of a sum don't change!

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