Tue 2017-12-26 08:04:59 UTC, BDesco:
> Sorry for my late reply. Thanks for your help.
> unhold works fine but I could not find any documentation on it.
You can get the documentation of .unhold by using ".unhold?":
sage: k, n = SR.var('k n')
sage: s = sum(n^5, n, 0, k, hold=True)
sa
Tue 2017-12-26 08:03:44 UTC, BDesco:
>
> for n=3. I get binomial(-1,0) = 1
> I don't know why.
You can look at the documentation of binomial:
sage: binomial?
You will see it indicates that the definition is extended:
We extend this definition to include cases when [...]
and as a resu
Sorry for my late reply. Thanks for your help.
> unhold works fine but I could not find any documentation on it.
>
Bart
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>
> for n=3. I get binomial(-1,0) = 1
>
I don't know why.
Secondly the sum stays 20. But when I calculate the terms I get 5+5+5+7.
At least both results should be the same. Both 20 or both 22.
Thanks.
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Also I think that results other than 20 are wrong for both formulae because
the binomial part always amounts to binomial(2-n,3-n), i.e. zero, and
you're just adding 5 four times. But I'm not 100 per cent on that.
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BTW instead of unhold() you can also use n() to get the value of an
unevaluated definite sum.
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This looks like a bug, you get the sum unevaluated. You can as a workaround
evaluate it manually. I get:
print(h3(3,h3n=5).unhold())
20
Thanks for the report.
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