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Vandermde identity for reals?
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ME1130
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Joined: 05 Feb 2006
Posts: 5

PostPosted: Sat Feb 11, 2006 5:25 pm    Post subject: Vandermde identity for reals? Reply with quote

Does the Vandermde identity hold for positive reals if the combinations function is defined using the gamma function?
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Ken Pledger
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Joined: 04 May 2005
Posts: 268

PostPosted: Wed Feb 15, 2006 2:43 am    Post subject: Re: Vandermde identity for reals? Reply with quote

In article <11us7gl1f4vm8f4@corp.supernews.com>,
"ME" <abcdefg@nonodock.net> wrote:

Quote:
Does the Vandermde identity hold for positive reals if the combinations
function is defined using the gamma function?


Apparently nobody else has yet commented on this, so I'll try.

It's not easy to see exactly what you mean. Writing (n;k) for
the binomial coefficient n!/((k!)(n-k)!), I take it that you're
referring to Vandermonde's formula

(a+b; r) = (a;r) + (a; r-1)(b;1) + (a; r-2)(b;2) + .... + (b;r).

If a and b are natural numbers, there are at least two proofs of
that. You can count the selections of r things from a+b things of
which a are red and b are blue; or else you can compare the
coefficients of x^r in the expanded version of

(1 + x)^(a+b) = ((1 + x)^a)((1 + x)^b).

After that, it's easy to generalize. The Vandermonde formula is a
polynomial equation which is true for infinitely many values of its
variables a and b (namely, all the natural numbers). That's enough
to make it true for all real numbers a and b, by a standard theorem
about polynomials.

However, r is still a natural number, and the R.H.S. of the
formula is still a sum of r+1 terms. If you wanted to let r be,
say, 7/2, what exactly would the formula be? I can't see how the gamma
function would help make sense of it.

Ken Pledger.
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Ken Pledger
science forum Guru Wannabe


Joined: 04 May 2005
Posts: 268

PostPosted: Wed Feb 15, 2006 2:50 am    Post subject: Re: Vandermde identity for reals? Reply with quote

In article <ken.pledger-6F389B.15433315022006@bats.mcs.vuw.ac.nz>,
Ken Pledger <ken.pledger@mcs.vuw.ac.nz> carelessly wrote:

Quote:

.... Writing (n;k) for
the binomial coefficient n!/((k!)(n-k)!), ...

In this context I should have written

n(n - 1)( n- 2)....(n - k + 1) / (k!).

Sorry about that.

Ken Pledger.
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ME1130
science forum beginner


Joined: 05 Feb 2006
Posts: 5

PostPosted: Wed Feb 15, 2006 10:52 pm    Post subject: Re: Vandermde identity for reals? Reply with quote

Yes, this is the identity that I am referring to. I stumbled across the Chu
Vandermode identity which can be found at
http://mathworld.wolfram.com/Chu-VandermondeIdentity.html
see equation 4 . "X" and "a" can be reals but "n" must be an integer. What n
is a real and the last term in the series is multiplied by the fractional
part of "n" ? Or is there a way to integerize the "n" my increasing "x+a" by
some amount. i.e.. there exists some b such that
Comb(b*(x+a),Nprime)=Comb(x+a,n) where nprime is n rounded down to the
nearest integer.

"Ken Pledger" <ken.pledger@mcs.vuw.ac.nz> wrote in message
news:ken.pledger-6F389B.15433315022006@bats.mcs.vuw.ac.nz...
Quote:
In article <11us7gl1f4vm8f4@corp.supernews.com>,
"ME" <abcdefg@nonodock.net> wrote:

Does the Vandermde identity hold for positive reals if the combinations
function is defined using the gamma function?


Apparently nobody else has yet commented on this, so I'll try.

It's not easy to see exactly what you mean. Writing (n;k) for
the binomial coefficient n!/((k!)(n-k)!), I take it that you're
referring to Vandermonde's formula

(a+b; r) = (a;r) + (a; r-1)(b;1) + (a; r-2)(b;2) + .... + (b;r).

If a and b are natural numbers, there are at least two proofs of
that. You can count the selections of r things from a+b things of
which a are red and b are blue; or else you can compare the
coefficients of x^r in the expanded version of

(1 + x)^(a+b) = ((1 + x)^a)((1 + x)^b).

After that, it's easy to generalize. The Vandermonde formula is a
polynomial equation which is true for infinitely many values of its
variables a and b (namely, all the natural numbers). That's enough
to make it true for all real numbers a and b, by a standard theorem
about polynomials.

However, r is still a natural number, and the R.H.S. of the
formula is still a sum of r+1 terms. If you wanted to let r be,
say, 7/2, what exactly would the formula be? I can't see how the gamma
function would help make sense of it.

Ken Pledger.
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Ken Pledger
science forum Guru Wannabe


Joined: 04 May 2005
Posts: 268

PostPosted: Thu Feb 16, 2006 4:10 am    Post subject: Re: Vandermde identity for reals? Reply with quote

In article <ken.pledger-6F389B.15433315022006@bats.mcs.vuw.ac.nz>,
Ken Pledger <ken.pledger@mcs.vuw.ac.nz> wrote:

Quote:

.... Writing (n;k) for the binomial coefficient ....
Vandermonde's formula

(a+b; r) = (a;r) + (a; r-1)(b;1) + (a; r-2)(b;2) + .... + (b;r).

.... That's enough to make it true for all real numbers a and b ....

However, r is still a natural number, and the R.H.S. of the
formula is still a sum of r+1 terms. If you wanted to let r be,
say, 7/2, what exactly would the formula be? I can't see how the gamma
function would help make sense of it.


In article <11v7c6f7r153le9@corp.supernews.com>,
"ME" <abcdefg@nonodock.net> wrote:

Quote:
.... I stumbled across the Chu Vandermode identity which can be found at
http://mathworld.wolfram.com/Chu-VandermondeIdentity.html
see equation 4 . "X" and "a" can be reals but "n" must be an integer.


Yes. That web page's notation x, a, n correspond to my a, b, r
respectively.


Quote:
What n
is a real and the last term in the series is multiplied by the fractional
part of "n" ? Or is there a way to integerize the "n" my increasing "x+a" by
some amount. i.e.. there exists some b such that
Comb(b*(x+a),Nprime)=Comb(x+a,n) where nprime is n rounded down to the
nearest integer.


My question above still stands. Remember that n+1 (which is my
r+1) is the number of terms in the series on the right-hand side, as
well as being a variable within those terms. What would you mean by a
series of four and a half terms? It makes no sense to me, with or
without the gamma function.

Ken Pledger.
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ME1130
science forum beginner


Joined: 05 Feb 2006
Posts: 5

PostPosted: Thu Feb 16, 2006 1:40 pm    Post subject: Re: Vandermde identity for reals? Reply with quote

Certainly (a+b;r) is defined over the domain where a,b and r are reals.
Right?
So, is there a generalization of the Chu-vandermode identity where r is a
real also? Instead of a sum maybe it is an integral?
"Ken Pledger" <ken.pledger@mcs.vuw.ac.nz> wrote in message
news:ken.pledger-2123F7.17102816022006@bats.mcs.vuw.ac.nz...
Quote:
In article <ken.pledger-6F389B.15433315022006@bats.mcs.vuw.ac.nz>,
Ken Pledger <ken.pledger@mcs.vuw.ac.nz> wrote:


.... Writing (n;k) for the binomial coefficient ....
Vandermonde's formula

(a+b; r) = (a;r) + (a; r-1)(b;1) + (a; r-2)(b;2) + .... + (b;r).

.... That's enough to make it true for all real numbers a and b ....

However, r is still a natural number, and the R.H.S. of the
formula is still a sum of r+1 terms. If you wanted to let r be,
say, 7/2, what exactly would the formula be? I can't see how the gamma
function would help make sense of it.


In article <11v7c6f7r153le9@corp.supernews.com>,
"ME" <abcdefg@nonodock.net> wrote:

.... I stumbled across the Chu Vandermode identity which can be found at
http://mathworld.wolfram.com/Chu-VandermondeIdentity.html
see equation 4 . "X" and "a" can be reals but "n" must be an integer.


Yes. That web page's notation x, a, n correspond to my a, b, r
respectively.


What n
is a real and the last term in the series is multiplied by the fractional
part of "n" ? Or is there a way to integerize the "n" my increasing "x+a"
by
some amount. i.e.. there exists some b such that
Comb(b*(x+a),Nprime)=Comb(x+a,n) where nprime is n rounded down to the
nearest integer.


My question above still stands. Remember that n+1 (which is my
r+1) is the number of terms in the series on the right-hand side, as
well as being a variable within those terms. What would you mean by a
series of four and a half terms? It makes no sense to me, with or
without the gamma function.

Ken Pledger.
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