Re: Summation over non-integers
- From: "Jon Slaughter" <Jon_Slaughter@xxxxxxxxxxx>
- Date: Mon, 18 Apr 2005 01:17:06 -0500
"Virgil" <ITSnetNOTcom#virgil@xxxxxxxxxxx> wrote in message
news:ITSnetNOTcom%23virgil-4CBC2E.23321117042005@xxxxxxxxxxxxxxxxxxxxxxxxxxx
> In article <1166ckn47c5r6af@xxxxxxxxxxxxxxxxxx>,
> "Jon Slaughter" <Jon_Slaughter@xxxxxxxxxxx> wrote:
>
>> I was thinking about how to do sums that do not start and stop on
>> integral
>> values... something like
>>
>> sum(f(x),x=a..b) where a and b are not necessarily integers.
>>
>> Is anything like this done before?
>
> In computer languages they often have "FOR" loops with non-integral
> values, but then they have to include a step size, e.g.:
>
> SUM := 0;
> FOR x := a to b step c do SUM := SUM + f(x);
>
> where the siign of c must be the same as the sign of b-a, and not zero.
Yes, but I don't think that the results are the same... basicaly one gets an
integral if they use a step size(an approximation to the integral atleast,
but if one is going to use a step size then it should be independent of the
step size and hence why not take the limit?).
For example
take the limits to be a to b
sum(f(x),x=a..b)
by using a "step system", we have
sum(f(a + k*dx),k=0..n)
where dx = (b-a)/n
now let f(x) = x, a = 0
dx*sum(k,k=0..n) = dx*n*(n+1)/2 = b*(n+1)/2
but this depends on the step size, so surely we must divide by n(an avg)
atleast... (else it diverges as we increase the step size which gives us a
nonsensical result)
so we then get something like
b*(n+1)/2n... taking n->oo we get b/2
but remember, we want sum(k,k=0..b) = b*(b+1)/2 if b is an integer
this is not close.. if we multiply by dx as if it were an integral, we would
have b/2*b = b^2/2 ~ b*(b+1)/2, so it doesn't even agree if b is an integer
;/
My point is, that if we do the step size method, we are basicaly computing
an integral(since it should be independent of step size and it must make
sense)... we know that the integral and sum are related and the integral can
take non-integer limits but in general they differ even at integral
limits... so we shouldn't expect them to be equivilent at non-integral
limits.
That is:
sum(f(x),x=a..b) != Int(f(x),x=a..b) since it is not even true that it holds
for integral limits... now, maybe there is a different way to approach the
step size that it works in general, but if that is the case then one has
then come up with a way to compute sums in terms of integrals over the same
argument.
It would also mean that one can write the binomial formula
(x+y)^z = int(B(x,y,z,t),t=0..z) or similar...
While it is impossible to treat the sum with < 1 step size as an integral,
it might be possible to reformulate the method slightly so that it works(but
I would not know how to do this and not know if it is impossible or not).
I think the biggest problem is defining what one means by sum(f(x),x=a..b).
Is is sum(f(x),x=ceil(a)..floor(b) + sum(f(x),x=a..ceil(a) +
sum(f(x),x=floor(b)..b)? where the first some ofcourse must agree with the
integer version(since the limits are integral) and the others are the
"excess" that are computed in some way...
I think its natural to expect that something like sum(f(k),k=0..4.2) to be
something between sum(f(k),k=0..4) and sum(f(k),k=0..5 and closer to 4 than
5... one might also have something like sum(f(k-t),k=a+t..b+t) =
sum(f(k),k=a..b) but I have not looked into it much...
Anyways, I will try to think about it some more ;)
Jon
.
- References:
- Summation over non-integers
- From: Jon Slaughter
- Re: Summation over non-integers
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- Summation over non-integers
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