max(), sum(), next()

[castironpi]

On Sep 5, 9:20 pm, "Manu Hack" <manuh... at gmail.com> wrote:
> On Fri, Sep 5, 2008 at 1:04 PM, castironpi <castiro... at gmail.com> wrote:
> > On Sep 5, 3:28 am, "Manu Hack" <manuh... at gmail.com> wrote:
> >> On Thu, Sep 4, 2008 at 4:25 PM, castironpi <castiro... at gmail.com> wrote:
> >> > On Sep 4, 2:42 pm, bearophileH... at lycos.com wrote:
> >> >> David C. Ullrich:
>
> >> >> > At least in mathematics, the sum of the elements of
> >> >> > the empty set _is_ 0, while the maximum element of the
> >> >> > empty set is undefined.
>
> >> >> What do you think about my idea of adding that 'default' argument to
> >> >> the max()/min() functions?
>
> >> >> Bye,
> >> >> bearophile
>
> >> > For max and min, why can't you just add your argument to the set
> >> > itself?
>
> >> > The reason max([]) is undefined is that max( S ) is in S.
>
> >> It makes sense.
>
> >> >The reason sum([]) is 0 is that sum( [ x ] ) - x = 0.
>
> >> It doesn't make sense to me.  What do you set x to?
>
> > For all x.
>
> But then how can you conclude sum([]) = 0 from there?  It's way far
> from obvious.

You can define sum([a1,a2,...,aN]) recursively as
sum([a1,a2,...a(N-1)])+aN.  Call the sum sum([a1,a2,...,aN]) "X", then
subtract aN.

sum([a1,a2,...a(N-1)])+aN=X
sum([a1,a2,...a(N-1)])+aN-aN=X-aN

For N=2, we have:

sum([a1,a2])=X
sum([a1,a2])-a2=X-a2
sum([a1,a2])-a2-a1=X-a2-a1

Since X= a1+ a2, replace X.

sum([a1,a2])-a2-a1=(a1+a2)-a2-a1

Or,

sum([a1,a2])-a2-a1=0

Apply the recursive definition:

sum([a1])+a2-a2-a1=0

And again:

sum([])+a1+a2-a2-a1=0

And we have:

sum([])=0.