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From: icystorm on 14 Feb 2010 11:18
Hi,

Would someone examine my sigma notation to make sure it is technically
correct, according to numbers theory?

http://www.box.net/shared/yl9pesf7dl

If so, are you able to determine the summation of y using the
information provided?

The correct answer is 3.

If you see a better way to present Sigma in my problem, please tell
me.

Thank you!

Cheers,
J
From: hagman on 14 Feb 2010 15:34
On 14 Feb., 17:18, icystorm <icyst...(a)hotmail.com> wrote:
> Hi,
>
> Would someone examine my sigma notation to make sure it is technically
> correct, according to numbers theory?
>
> http://www.box.net/shared/yl9pesf7dl
>
> If so, are you able to determine the summation of y using the
> information provided?
>
> The correct answer is 3.
>
> If you see a better way to present Sigma in my problem, please tell
> me.
>
> Thank you!
>
> Cheers,
> J

I don't see how y_{i,j,k,m} is defined.

It might be less confusing to write something like
sum_{ (i,j,k,m) in { (i,j,k,m) | x_i <= R34e, x_j <= ... } y_{i,j,k,m}
From: icystorm on 14 Feb 2010 18:21
On Feb 14, 2:34 pm, hagman <goo...(a)von-eitzen.de> wrote:
> On 14 Feb., 17:18, icystorm <icyst...(a)hotmail.com> wrote:
>
>
>
>
>
> > Hi,
>
> > Would someone examine my sigma notation to make sure it is technically
> > correct, according to numbers theory?
>
> >http://www.box.net/shared/yl9pesf7dl
>
> > If so, are you able to determine the summation of y using the
> > information provided?
>
> > The correct answer is 3.
>
> > If you see a better way to present Sigma in my problem, please tell
> > me.
>
> > Thank you!
>
> > Cheers,
> > J
>
> I don't see how y_{i,j,k,m} is defined.
>
> It might be less confusing to write something like
> sum_{ (i,j,k,m) in { (i,j,k,m) | x_i <= R34e, x_j <= ... } y_{i,j,k,m}- Hide quoted text -
>
> - Show quoted text -

Hagman, thanks. Can you post a picture of what that looks like when
arranged with Sigma? I'm not sure I completely followed your syntax
above. I am rusty at textbook numbers theory. If I can see it
arranged, I will understand.

Thanks!

J

From: Ray Koopman on 14 Feb 2010 19:39
On Feb 14, 8:18 am, icystorm <icyst...(a)hotmail.com> wrote:
> Hi,
>
> Would someone examine my sigma notation to make sure it is technically
> correct, according to numbers theory?
>
> http://www.box.net/shared/yl9pesf7dl
>
> If so, are you able to determine the summation of y using the
> information provided?
>
> The correct answer is 3.
>
> If you see a better way to present Sigma in my problem, please tell
> me.
>
> Thank you!
>
> Cheers,
> J

The only way I can make of this is to infer additional indices
that are treated as part of the names of the variables:

i x1_{i} y1_{i} j x2_{j} y2_{j} etc.
0 0.0 0 0 0.0 0
1 1.0 1 1 1.0 1
2 99.0 1 2 49.7 1
3 157.4 1 3 78.2 1
4 106.7 1

The summand is the product y1_{i} y2_{j} y3_{k} y4_{m},

and the conditions under the summation are

i: x1_{i} <= R34e,
j: x2_{j} <= R50e,
k: x3_{k} <= R64e,
m: x4_{m} <= R87e.

The given values of R require i in {0,1,2}, j in {0,1}, k = m = 0,
and the sum is 0 because every product includes y3_{0} y4_{0},
both of which are 0. How do your get 3?
From: A N Niel on 14 Feb 2010 20:52
In article
<4883bb0e-00db-4831-a0c5-a69056f3354f(a)a16g2000pre.googlegroups.com>,
Ray Koopman <koopman(a)sfu.ca> wrote:

> On Feb 14, 8:18 am, icystorm <icyst...(a)hotmail.com> wrote:
> > Hi,
> >
> > Would someone examine my sigma notation to make sure it is technically
> > correct, according to numbers theory?
> >
> > http://www.box.net/shared/yl9pesf7dl
> >
> > If so, are you able to determine the summation of y using the
> > information provided?
> >
> > The correct answer is 3.
> >
> > If you see a better way to present Sigma in my problem, please tell
> > me.
> >
> > Thank you!
> >
> > Cheers,
> > J
>
> The only way I can make of this is to infer additional indices
> that are treated as part of the names of the variables:
>
> i x1_{i} y1_{i} j x2_{j} y2_{j} etc.
> 0 0.0 0 0 0.0 0
> 1 1.0 1 1 1.0 1
> 2 99.0 1 2 49.7 1
> 3 157.4 1 3 78.2 1
> 4 106.7 1
>
> The summand is the product y1_{i} y2_{j} y3_{k} y4_{m},
>
> and the conditions under the summation are
>
> i: x1_{i} <= R34e,
> j: x2_{j} <= R50e,
> k: x3_{k} <= R64e,
> m: x4_{m} <= R87e.
>
> The given values of R require i in {0,1,2}, j in {0,1}, k = m = 0,
> and the sum is 0 because every product includes y3_{0} y4_{0},
> both of which are 0. How do your get 3?

I interpreted it ... add the first 3 y's in column 1 (0+1+1 = 2) the
first two y's in column 2 (0+1 = 1 more) and the 0's in the last two
columns, getting 3 in all.
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