casperyc

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12 years, 164 days

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These are replies submitted by casperyc

@Markiyan Hirnyk I am not sure what you meant by the "first term". But it still does not work for me.

Note that my problem is not with latex, but the latex code produced from Maple. Thanks,

@Markiyan Hirnyk I dont think it's a 'correct' Maple syntax to input the variables I want.

This is completely wrong and does not make any sense.

@Carl Love I think i a sequence is 'faster'? so i will definitely use sequence for it.

@Joe Riel Yes, I want them as ascii text, with no 2D math.

val := [0,0,1]:

printf("%d%d%d\n", op(val));
001

This is fine. I wonder though, could I use a simpler way. Instead of %d%d%d, is it possible to use a single %d to represent this? as my 'val' is not fixed length. So if I have

val := [0,0,1,0]:

I would have to use

printf("%d%d%d%d\n", op(val));

Every time, I will have to adjust the number of '%d's.

 

Thanks,

 

Ghostyc

@Carl Love Yes, of course they are effectively the same. I guess at this point, Maple just can't run it with a single (using some recursive) line.

 

I might run into some more "complex" substitutions where I might just have to do the subs (or simplify) in groups then.

 

 

@Carl Love Yes, of course they are effectively the same. I guess at this point, Maple just can't run it with a single (using some recursive) line.

 

I might run into some more "complex" substitutions where I might just have to do the subs (or simplify) in groups then.

 

 

@Carl Love 

Yes, goal is to use the substitutions sstest3(4,2).

sskappa( sstest(4,2) , kappa );

in ssTwoStage the procedure basically converts s2,s8,s9,s10,s11 to the s8,s9,s10 in sstest3(4,2).

The s1 to s10 from sstest3(4,2) was constructed based on sstest3(4,2).

If I remember correctly, it does

{s8=s2*s8+s9, s9=-s2^2*s8+s10,s10=s2^3*s8+s11}

 

Simply put, I can't subs sstest3(4,2) in a single step, so I did it in two steps.

 

@Carl Love 

Yes, goal is to use the substitutions sstest3(4,2).

sskappa( sstest(4,2) , kappa );

in ssTwoStage the procedure basically converts s2,s8,s9,s10,s11 to the s8,s9,s10 in sstest3(4,2).

The s1 to s10 from sstest3(4,2) was constructed based on sstest3(4,2).

If I remember correctly, it does

{s8=s2*s8+s9, s9=-s2^2*s8+s10,s10=s2^3*s8+s11}

 

Simply put, I can't subs sstest3(4,2) in a single step, so I did it in two steps.

 

Hi Carl,

As you mentioned it, what exactly does "complexity" mean in this case?

From what I understand, the last 3 terms are ordered in order of "power". I would think "powers" are more "complex" than "plus" or "minus"?

 

Also, my current method

sskappa( sstest(4,2) , kappa );

ssTwoStage(%,4);

take less than 20 sec.

 

If you have a look at

sstest3(K,2)

for increasing K.

Are they not in the right order by "complexity" then?

 

Casper

 

Hi Carl,

As you mentioned it, what exactly does "complexity" mean in this case?

From what I understand, the last 3 terms are ordered in order of "power". I would think "powers" are more "complex" than "plus" or "minus"?

 

Also, my current method

sskappa( sstest(4,2) , kappa );

ssTwoStage(%,4);

take less than 20 sec.

 

If you have a look at

sstest3(K,2)

for increasing K.

Are they not in the right order by "complexity" then?

 

Casper

 

@Carl Love Hi Carl,

I just updated the top (original post) with attachment.

I was not sure this modfied version is "smaller" as it is the "smallest" example I am working at the moment.

But I did try to use less words to describe it. Hopefully, I had given a better explanation of what I want.

 

sstest3(4,2) is the final subs I hope to use to simplify kappa.

For now, I do it twice using

sskappa( sstest(4,2) , kappa );

ssTwoStage(%,4);

 

If my coding is correct, equation (7) is the simplified expression of kappa using sstest3(4,2).

 

Casper

Just curious, why would you need to do that? Can you not just "check" it in the normal way?

@Carl Love I will tidy up my work sheet a bit and ask a seperate question late. Meanwhile, I will work on it a bit more.

 

Many thanks.

 

Casper

@Carl Love I will tidy up my work sheet a bit and ask a seperate question late. Meanwhile, I will work on it a bit more.

 

Many thanks.

 

Casper

Hi Carl,

I am now working on something further and now the previous trick seems to have failed.

In the worksheet, I have two reparameterisation

sstest(K,C) and sstest2(K,C)

and I want to subs it into

kappa:=mykappa(4,2,modname="both",tvarp=true);

 

sskappa is a procedure that does the substitution using your earlier "reverse" trick.

kappanew:=sskappa(sstest2(4,2),kappa);
works fine. My knowledge from

Estpars(DD1,pars,10);

tells me that

sstest(4,2);
should do the reparameterisation just fine, as good as sstest2(4,2);

 

But sskappa(sstest(4,2),kappa) had failed.

I have kept sstest(4,2) in the usual order, from "simpliest" to "most complex" terms.

I wonder whether its possible to improve more?

 

Thanks!

Casper

 

work.zip

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