## 385 Reputation

11 years, 84 days

## @acer Thanks. Why can I then use multipl...

@acer Thanks. Why can I then use multiple arguments in ?collect() without getting an error?

Try:

restart;

f := simplify(GAMMA(2*q-2)*GAMMA(-2*q+L)*hypergeom([-L+1, 2*q-2], [-L+2], 1)/((2*(L-1))*(-1+2*q)*(4*q-3)*GAMMA(L+2*q-3))+GAMMA(-2*q+L)*GAMMA(2*q-2+L)*hypergeom([1, 1, 2*q-2+L], [2, L+1], 1)/((2*(-1+2*q))*(4*q-3)*GAMMA(L+2*q-3)*GAMMA(L+1)));

collect(f, hypergeom, simplify);

collect(f, hypergeom, simplify, factor)

## @Kitonum If I have a list how do I use f...

@Kitonum If I have a list with function expressions how do I use frontend ?

o:=[f1,f2,f3]

frontend~(expand,o) or frontend~(expand,[o])

does not work.

## @acer  I can write collect(f,[k,G...

I can write

collect(f,[k,GAMMA],simplify,combine)

and each operand is simplified and combined separately. I always assumed that this is done in the order specified so in this case first the simplification and then the combination. In essence this is not precisely equivalent to what you are showing me:

collect(numer(f),[k,GAMMA],u->u/denom(f),simplify)

I expected to first collect the numerator, then do u/denom(f) for each operand and for each operand the result is simplified. In fact doing this above the denominator is completely lacking.

## @acer What goes wrong with this map u-&g...

@acer What goes wrong with this map u->u/denom(f) when I add additional arguments such as ?simplify() ?

I would have expected that each term is simplified seprately as is usually the case.

## @Mariusz Iwaniuk Ah yeah right. Could ha...

@Mariusz Iwaniuk Ah yeah right. Could have thought about it. I think this is probably even intended, while still confusing. I think it should be either 0 or 4 defaultwise with the option for both.

## I still have a similar problem with this...

I still have a similar problem with this where the "convert" trick doesn't work

restart;

((1-t)^2)^q;

assuming([combine(simplify(%))], [1-t > 0, q > 0, q < 1])

## @Carl Love Since q € (0,1) why is i...

@Carl Love Since q € (0,1) why is it only true for t € (0,Pi) ? The domain (-Pi,0) is then also mapped to (-Pi*q,0) which is not in the upper half of the domain and thus distinct.

## @acer What do you mean by "fortuito...

@acer What do you mean by "fortuitous Change and Split might allow resolution of the abs" ?

Are you talking about abs(cos(theta1)+cos(theta2)) in the denominator of the arctan of your reintegrated Q ?

Do you think this integral can be actually done analytically?

@vv Yes the precise value is expected to be 1/8 (see putnams 1992 A6)

It's a straight-forward but not very elegant way of solving it...

## @acer Then I'm wondering about the w...

@acer Then I'm wondering about the weird errors beyond the epsilon specification typical for the _MonteCarlo methods.

Besides what is the precise issue Maple has before Changing phi2 -> phi2*s.

I would assume it is the exact same, but apparently not for maple.

Why is for numerical purposes a fixed integration interval easier to handle than an interval depending on another integration variable?

## @acer Hey thanks, didn't think of re...

@acer Hey thanks, didn't think of reintegrating it symbolically ;)

## @vv Any idea why only the MonteCarlo met...

@vv Any idea why only the MonteCarlo methods seem to work?

## @Mariusz Iwaniuk Probably you are right,...

@Mariusz Iwaniuk Probably you are right, because of the phi2 integration which is antisymmetric with respect to 0..Pi and Pi..2Pi.

Replace the ingerand by its absolute value.

PS: Actually I just replaced 0..2Pi by 0..Pi and now I get a value, but only using _MonteCarlo and still not with the NAG methods.

## Good to know, but yes it was only for il...

Good to know, but yes it was only for illustration, so thanks all.

## @ecterrab Thanks for that! By the time I...

@ecterrab Thanks for that! By the time I was writing this I was using Maple2015. Now with 2018 I get the same output as you, but with 2015 the time was pretty surely on the 4th position and the expansion was not d_(x1), but some \partial(x1), at least in the 2d-visualization.

Also the metric was not collected when using the line_element command.

Overall "signature" has different meanings depending on where and how it is used? Obviously it is the signature I change using the Redefine command. But the signature in Setup has no real information, except telling me the sign of the minkowski metric in the local tetrad frame?

## @ecterrab Mind me if I reopen this threa...

@ecterrab Mind me if I reopen this thread, but I think it best fits here.

Above you explained how to change the signature; in particular changing time from 4th to 0th. However it does not seem to work for a general metric. For instance

restart;

with(Physics);

Setup(mathematicalnotation = true, signature = +---, metric = [15, 23, 2]);

Coordinates(X);

g_[]

does nothing to the signature in changing it from 4th to 0th.

Another question regarding the line-element:

It is expanded/displayed by symbols \partial(x1), \partial(x2) and so on...Is it possible to collect() for \partial(x1),\partial(x2) ?

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