Can I be certain of this?

[Jack]

Let S be a set of real numbers. Can I be ceratin that the expression

sum_(k in S) k

denotes the sum of all k for which k in S, including the sum b+c, where b
and c are elements of S for which b = c?

With thanks.

[Jussi Piitulainen]

No. Different elements of a _set_ are never the same.

The usual thing is to have an index set, say K, and sum the numbers
a_k for k in K. Then you can have a_j = a_k even for j != k:

sum_(k in K) a_k

No need to mention an explicit set. The subscripts themselves will do,
with the convention that the index variable ranges over integers
unless otherwise specified:

sum_(0 <= k < n) a_k

[Jack]

"Jussi Piitulainen" <jpii...@ling.helsinki.fi> wrote in message
news:qot4nfa...@ruuvi.it.helsinki.fi...

> Jack writes:
>
>> Let S be a set of real numbers. Can I be ceratin that the expression
>>
>> sum_(k in S) k
>>
>> denotes the sum of all k for which k in S, including the sum b+c,
>> where b and c are elements of S for which b = c?
>
> No. Different elements of a _set_ are never the same.
>

I see; I didn't know that some concept of differentness was implied. I had
initially gone for a set A of ordered pairs, then

sum_((a, b) in A) a

... so, just to be clear, would I be achieving my objective, as I originally
stated it, that way?

> The usual thing is to have an index set, say K, and sum the numbers
> a_k for k in K. Then you can have a_j = a_k even for j != k:
>
> sum_(k in K) a_k
>
> No need to mention an explicit set. The subscripts themselves will do,
> with the convention that the index variable ranges over integers
> unless otherwise specified:
>
> sum_(0 <= k < n) a_k
>

So how preceisely would I do it? Something like this:
'Let K be an index set for (of?) a finite set S of real numbers. Then

sum_(k in K) a_k = < something to do with S > '

The fact that you say 'No need to mention an explicit set' makes me think
I'm not saying it very well, above.

With thanks.

[William Elliot]

On Sun, 14 Apr 2013, Jack wrote:
> "Jussi Piitulainen" <jpii...@ling.helsinki.fi> wrote in message
> >

> >> Let S be a set of real numbers. Can I be ceratin that the expression
> >>
> >> sum_(k in S) k
> >>
> >> denotes the sum of all k for which k in S, including the sum b+c,
> >> where b and c are elements of S for which b = c?
> >
> > No. Different elements of a _set_ are never the same.
>
> I see; I didn't know that some concept of differentness was implied. I
> had initially gone for a set A of ordered pairs, then
>
> sum_((a, b) in A) a
>
> ... so, just to be clear, would I be achieving my objective, as I
originally
> stated it, that way?
>
> > The usual thing is to have an index set, say K, and sum the numbers
> > a_k for k in K. Then you can have a_j = a_k even for j != k:
> >
> > sum_(k in K) a_k
> >
> > No need to mention an explicit set. The subscripts themselves will do,
> > with the convention that the index variable ranges over integers
> > unless otherwise specified:
> >
> > sum_(0 <= k < n) a_k
>
> So how preceisely would I do it? Something like this:
> 'Let K be an index set for (of?) a finite set S of real numbers. Then
>
> sum_(k in K) a_k = < something to do with S > '
>

Let K be a finite set and s a function over K into S.
What is sum{ s(j) | j in K }?

> The fact that you say 'No need to mention an explicit set' makes me think
> I'm not saying it very well, above.

Now add your extra requirements onto my statement.

[Jussi Piitulainen]

Jack writes:
> Jussi Piitulainen wrote in message

> news:qot4nfa...@ruuvi.it.helsinki.fi...
> > Jack writes:
> >
> >> Let S be a set of real numbers. Can I be ceratin that the
> >> expression
> >>
> >> sum_(k in S) k
> >>
> >> denotes the sum of all k for which k in S, including the sum b+c,
> >> where b and c are elements of S for which b = c?
> >
> > No. Different elements of a _set_ are never the same.
>
> I see; I didn't know that some concept of differentness was
> implied. I had initially gone for a set A of ordered pairs, then
>
> sum_((a, b) in A) a
>
> ... so, just to be clear, would I be achieving my objective, as I
> originally stated it, that way?

If all (a, b) are distinct, then yes, that will do it. If each b in
those pairs is distinct, this is the same thing that I'm suggesting:
each b serves as an index to an element that you want in the sum.

> > The usual thing is to have an index set, say K, and sum the
> > numbers a_k for k in K. Then you can have a_j = a_k even for j !=
> > k:
> >
> > sum_(k in K) a_k
> >
> > No need to mention an explicit set. The subscripts themselves will
> > do, with the convention that the index variable ranges over
> > integers unless otherwise specified:
> >
> > sum_(0 <= k < n) a_k
>
> So how preceisely would I do it? Something like this: 'Let K be an
> index set for (of?) a finite set S of real numbers. Then
>
> sum_(k in K) a_k = < something to do with S > '
>
> The fact that you say 'No need to mention an explicit set' makes me
> think I'm not saying it very well, above.

It's probably best if you take the index set as a range of integers
{1,...,n} and think of your collection of numbers as a _sequence_
(a_1, ..., a_n). Then your sum is simply

sum a_k for 1 <= k <= n

Say, if a = (1,4,1,5), then a_1 = 1, a_2 = 4, a_3 = 1, a_4 = 5, and
(sum a_k for 1 <= k <= 4) = 1 + 4 + 1 + 5 = 11. There is one term for
each k and it is not a problem that a_1 = a_3 because 1 != 3.

Any set of n elements will do, but the one above is common. You know
this, you just need to become aware of it: it's implicit in any sum
written as

n
sigma a_k
k=1

[Jack]

not saying it very well, above.
>
> It's probably best if you take the index set as a range of integers
> {1,...,n} and think of your collection of numbers as a _sequence_
> (a_1, ..., a_n).

I was told, at some point, that all sequences were infinite. Am I wrong
here?

Then your sum is simply
>
> sum a_k for 1 <= k <= n
>

What about this specific example. Let p(n) be the n-th prime. Then for all
n,

sum_(2 <= k <= n) (p(k)-p(k-1)) + 2 = p(n).

Is that OK?

With thanks.

[Jussi Piitulainen]

Jack writes:

> not saying it very well, above.
>
> > It's probably best if you take the index set as a range of
> > integers {1,...,n} and think of your collection of numbers as a
> > _sequence_ (a_1, ..., a_n).
>
> I was told, at some point, that all sequences were infinite. Am I
> wrong here?

Don't worry about the elements that are not included in the sum.

Note that your sequences of primes and their differences below _are_
naturally infinite. Do _not_ worry about p(k) - p(k - 1) when k = 1,
because k = 1 is not included in the sum. Same for all k > n.

> Then your sum is simply
> >
> > sum a_k for 1 <= k <= n
>
> What about this specific example. Let p(n) be the n-th prime. Then
> for all n,
>
> sum_(2 <= k <= n) (p(k)-p(k-1)) + 2 = p(n).
>
> Is that OK?

Yes, that's good :)

You could be explicit that n = 1, 2, 3, ..., and p(1) = 2, but it's
clear that that is what you intend.

[Jack]

Thanks. So how does this look:

For x < y, let [x, y] denote the interval of all integers, i, such that x <=
i <= y.

For any m > 1, let A be a finite set {a_1, a_2, ..., a_m} of real numbers.
For some subset K of [1, m], let R = {a_i : i in K}, with (sum_i in K)
(a_i)/|K| < (sum_(1 <= h <= m) a_h)/m. Then

(sum_(1 <= k <= m) a_k) /m > (sum_(j in [1, m] \ K) a_j) / (m - |K|).

Is all that OK? Does it need further proof/refinement (or is it even,
perhaps, false in general?)?

Cheers.

[Jack]

Actually I can see that I didn't need to introduce R, only the subset K of
[1, m].

"Jack" <nomai...@hotmail.com> wrote in message
news:95Bat.21620$mX1....@fx23.fr7...

[Frederick Williams]

Is "with (sum_i in K) (a_i)/|K| < (sum_(1 <= h <= m) a_h)/m" part of the
definition of the set R? If so then it needs to be 'twixt the { and }
like this:

let R = {a_i : i in K, with (sum_i in K) (a_i)/|K| < (sum_(1 <= h <= m)
a_h)/m}.

Thought I note your next post in which you say you don't need R anyway.

--
When a true genius appears in the world, you may know him by
this sign, that the dunces are all in confederacy against him.
Jonathan Swift: Thoughts on Various Subjects, Moral and Diverting

[Jussi Piitulainen]

Jack writes:

> Thanks. So how does this look:
>
> For x < y, let [x, y] denote the interval of all integers, i, such
> that x <= i <= y.
>
> For any m > 1, let A be a finite set {a_1, a_2, ..., a_m} of real
> numbers. For some subset K of [1, m], let R = {a_i : i in K}, with
> (sum_i in K) (a_i)/|K| < (sum_(1 <= h <= m) a_h)/m. Then

The set A is a distraction, similar to R that you noted yourself. Just
say, for m > 1, "let a_1, a_2, ..., a_m be real numbers".

> (sum_(1 <= k <= m) a_k) /m > (sum_(j in [1, m] \ K) a_j) / (m - |K|).
>
> Is all that OK? Does it need further proof/refinement (or is it
> even, perhaps, false in general?)?

I don't know if it's true or obvious, but I think it makes sense.

Hm. It's a statement about the arithmetic means of the numbers in K
and those not in K, right? If one is greater than the arithmetic mean
of all m numbers, the other must be smaller. Is one of your relations
the wrong way around, though?

You should state that both K and [1,m]\K are non-empty. They form a
_partition_ of [1,m].

[Jack]

>> (sum_(1 <= k <= m) a_k) /m > (sum_(j in [1, m] \ K) a_j) / (m - |K|).
>>
>> Is all that OK? Does it need further proof/refinement (or is it
>> even, perhaps, false in general?)?
>
> I don't know if it's true or obvious, but I think it makes sense.
>
> Hm. It's a statement about the arithmetic means of the numbers in K
> and those not in K, right?

Yes*. In fact that is the way I originally phrased it, and wonder whether I
ought to revert to that.

* Except: remember that K is just an index set to the set whose mean is
being evaluated.

If one is greater than the arithmetic mean
> of all m numbers, the other must be smaller. Is one of your relations
> the wrong way around, though?

Yes, the second one should be a '<'. I don't know what I would need to say
by way of proof.

>
> You should state that both K and [1,m]\K are non-empty. They form a
> _partition_ of [1,m].

Ah -- useful to know. I hope it's clear that K is not necessarily a
subinterval of [1, m].
With thanks.

[Jack]

>
> Is "with (sum_i in K) (a_i)/|K| < (sum_(1 <= h <= m) a_h)/m" part of the
> definition of the set R?

Yes.

[Jack]

> You should state that both K and [1,m]\K are non-empty. They form a
> _partition_ of [1,m].

Doesn't the condition

(sum_i in K) (a_i)/|K| < (sum_(1 <= h <= m) a_h)/m

necessitate that K is nonempty? If so, surely I don't need to state it.

[Jussi Piitulainen]

That's why I said it has to be non-empty, but now I see another
possibility: since the sum is empty when K is empty, the apparent
division by zero doesn't happen and there is no problem, except maybe
the problem of the truth of your statement in that case. How do you
like the idea of your readers having thoughts like this?

And what about K = [1,m]?

I think I've seen the concept of a _proper subset_ used in such a way
that it excludes _both_ extremes. You could try that. I don't know how
well it is accepted. I _think_ it's more usual that it only excludes
the equality of the sets, but I might even be wrong here.

[Jack]

"Jussi Piitulainen" <jpii...@ling.helsinki.fi> wrote in message

news:qota9oz...@ruuvi.it.helsinki.fi...

> Jack writes:
>
>> > You should state that both K and [1,m]\K are non-empty. They form
>> > a _partition_ of [1,m].
>>
>> Doesn't the condition
>>
>> (sum_i in K) (a_i)/|K| < (sum_(1 <= h <= m) a_h)/m
>>
>> necessitate that K is nonempty? If so, surely I don't need to state
>> it.
>
> That's why I said it has to be non-empty, but now I see another
> possibility: since the sum is empty when K is empty, the apparent
> division by zero doesn't happen

I thought it would undefined, even if it did.

and there is no problem, except maybe
> the problem of the truth of your statement in that case.

Are you sure it would be within the bounds of acceptibility to have

0/0 < (sum_(1 <= h <= m) a_h)/m

?

How do you
> like the idea of your readers having thoughts like this?
>
> And what about K = [1,m]?
>
> I think I've seen the concept of a _proper subset_ used in such a way
> that it excludes _both_ extremes. You could try that. I don't know how
> well it is accepted. I _think_ it's more usual that it only excludes
> the equality of the sets, but I might even be wrong here.
>

It has been used on this NG a few times to mean the second of these.

With thanks.

[Jussi Piitulainen]

Jack writes:

> Jussi Piitulainen wrote:
> > Jack writes:
> >
> >> > You should state that both K and [1,m]\K are non-empty. They
> >> > form a _partition_ of [1,m].
> >>
> >> Doesn't the condition
> >>
> >> (sum_i in K) (a_i)/|K| < (sum_(1 <= h <= m) a_h)/m
> >>
> >> necessitate that K is nonempty? If so, surely I don't need to
> >> state it.
> >
> > That's why I said it has to be non-empty, but now I see another
> > possibility: since the sum is empty when K is empty, the apparent
> > division by zero doesn't happen
>
> I thought it would undefined, even if it did.
>
> > and there is no problem, except maybe the problem of the truth of
> > your statement in that case.
>
> Are you sure it would be within the bounds of acceptibility to have
>
> 0/0 < (sum_(1 <= h <= m) a_h)/m
>
> ?

No, but "sum a_i/|K|" is "sum (a_i/|K|)", not "(sum a_i)/|K|". The
former is simply 0 when there are no such i; the latter is 0/0.

> > How do you like the idea of your readers having thoughts like
> > this?
> >
> > And what about K = [1,m]?
> >
> > I think I've seen the concept of a _proper subset_ used in such a
> > way that it excludes _both_ extremes. You could try that. I don't
> > know how well it is accepted. I _think_ it's more usual that it
> > only excludes the equality of the sets, but I might even be wrong
> > here.
>
> It has been used on this NG a few times to mean the second of these.

Yes, I think you can use it that way if you like. Define K as a proper
subset of [1,K].

[Jack]

"Jussi Piitulainen" <jpii...@ling.helsinki.fi> wrote in message

news:qotli8i...@ruuvi.it.helsinki.fi...

> Jack writes:
>> Jussi Piitulainen wrote:
>> > Jack writes:
>> >
>> >> > You should state that both K and [1,m]\K are non-empty. They
>> >> > form a _partition_ of [1,m].
>> >>
>> >> Doesn't the condition
>> >>
>> >> (sum_i in K) (a_i)/|K| < (sum_(1 <= h <= m) a_h)/m
>> >>
>> >> necessitate that K is nonempty? If so, surely I don't need to
>> >> state it.
>> >
>> > That's why I said it has to be non-empty, but now I see another
>> > possibility: since the sum is empty when K is empty, the apparent
>> > division by zero doesn't happen
>>
>> I thought it would undefined, even if it did.
>>
>> > and there is no problem, except maybe the problem of the truth of
>> > your statement in that case.
>>
>> Are you sure it would be within the bounds of acceptibility to have
>>
>> 0/0 < (sum_(1 <= h <= m) a_h)/m
>>
>> ?
>
> No, but "sum a_i/|K|" is "sum (a_i/|K|)", not "(sum a_i)/|K|". The
> former is simply 0 when there are no such i; the latter is 0/0.

Eh? I was careful to use the latter.

[Frederick Williams]

Then you must ensure that |K| =/= 0.

[Jussi Piitulainen]

You wrote "(sum_i in K) (a_i)/|K|".

It's not "(sum_(i in K) a_i)/|K|" nor is it careful.

This is an irrelevant tangent.

[Jack]

"Frederick Williams" <freddyw...@btinternet.com> wrote in message
news:516D99BC...@btinternet.com...

> Jack wrote:
>>
>> "Jussi Piitulainen" <jpii...@ling.helsinki.fi> wrote in message
>> news:qotli8i...@ruuvi.it.helsinki.fi...
>> > Jack writes:
>
>> >>
>> >> Are you sure it would be within the bounds of acceptibility to have
>> >>
>> >> 0/0 < (sum_(1 <= h <= m) a_h)/m
>> >>
>> >> ?
>> >
>> > No, but "sum a_i/|K|" is "sum (a_i/|K|)", not "(sum a_i)/|K|". The
>> > former is simply 0 when there are no such i; the latter is 0/0.
>>
>> Eh? I was careful to use the latter.
>
> Then you must ensure that |K| =/= 0.

I take Jussi's point that I was *not*, in fact, careful the first time I
wrote it. But as a matter of comprehending mathematical style (or perhaps,
even, correctness of expression), I would be interested to know why the
possibility of invoking the mathematically unacceptable 0/0 should not
imply, in full accordance with convention, that only possibilities that
exclude 0/0 are to be considered.

With thanks.