Binomial identities,combinatorial, equivalence

In summary, the conversation discusses an expression derived in Boson calculus for the norm squared in Quantum Physics by Biedenharn and Louck. The author provides an expression for the norm squared as (k+j+1)!*(k-j)!/(2*j+1), which is equivalent to the more complicated expressions provided by the speaker. The speaker also includes two extra variables, m and mp, in their expressions, but these variables drop out in the simplified expression provided by the authors. The speaker asks how these variables drop out and what binomial/combinatorial identities are used to simplify the expression. The expressions provided by the speaker can be verified to reduce to the same value as the simplified expression for any choices of integer variables within certain constraints
  • #1
zmth
29
1
NOte this is not a homework nor related to any course nor any test problem etc. - entirely my own interest and study.
Re\: text by Biedenharn and Louck "Angular momentum in Q.Physics" . I derive an expression for the norm squared wrt a certain expression in Boson calculus. You don't really need to get into this in order to answer my question. Anyway below in the following is my two summation index form and other equivalent expressions later. Now the authors don't derive how they arrived at their expression but just give the answer as

(k+j+1)!*(k-j)!/(2*j+1)

and in fact this is equivalent(equal) to my more complicated expressions and yes my expressions also have two extra variables or call them parameters if you like which are m and mp in addtion to j and k . Now my question is why is it that the m and mp drop out of my expressions and what binomial/ combinatorial identities etc. are used to get from my multiple summation expressions with the two extra m and mp terms to his ofcourse much simpler reduced and preferred expression again being (k+j+1)!*(k-j)!/(2*j+1) with no summations and without m and mp. Now we begin with my equivalent expressions of which anyone can verify reduce exactly to the author's answer for any and all choices of integer variables under the constraints .Here for simplicity just consider k and j positive integers where k >= j ; m and mp can be positive or negative with maximum abs. value j so
-j <= m,mp <= j and in these latter parameters case you may notice like pertaining or reference to quantum angular momentum to which these pertain and actually k,j,m,mp could all be half an odd integer but to keep it simpler just assume they are all integer. Here are my expressions.

(j-m)*(j+m)*(j-mp)*(j+mp)*sum_{s=max(0,mp-m)}^{min(mp,-m)+k} *(sum_{sp=max(mp-m,0,s-k+j)}^{min(j-m,j+mp,s)}(-1)^(s-sp)* binom(k-j ,s-sp)/((j+mp-sp)!*sp!*(m-mp+sp)!*(j-m-sp)!))^2*(k-m-s)!*(k+mp-s)! *s!*(m-mp+s)!

Or equivalently using 3 summation indexes expanding the square written as:

(j+mp)!*(j-mp)!*(k-j)!^2*sum_{s=max(0,mp-m)}^{min(mp,-m)+k} *sum_{sp=max(mp-m,0,s-k+j)}^{min(j-m,j+mp,s)}(-1)^(s+sp)*binom(s,sp)*binom(k-m-s,j-m-sp)*binom(j+m,j+mp-sp) *sum_{s1=max(mp-m,0,s-k+j)}^{min(j-m,j+mp,s)}(-1)^(s+s1)*binom(s,s1)*binom(k-m-s,j-m-s1)*binom(j+m,j+mp-s1)

OR as :

(j+m)!*(j-m)!*(k-j)!^2*sum_{s=max(0,mp-m)}^{min(mp,-m)+k} *sum_{sp=max(mp-m,0,s-k+j)}^{min(j-m,j+mp,s)}(-1)^(s+sp)*binom(s,sp)*binom(k+mp-s,j+mp-sp)*binom(j-mp,j-m-sp) *sum_{s1=max(mp-m,0,s-k+j)}^{min(j-m,j+mp,s)}(-1)^(s+s1)*binom(s+m-mp ,s-s1)*binom(k-m-s,j-m-s1)*binom(j+mp,s1)

Here binom(a,b) means the usual a!/b!/(a-b)! with a slightly different interpretation when some or one of the variables are negative integers.That is all 3 of these are all equivalent as you may verify by trials esp. if you have Macsyma(unfortunately they have been out of business for quite a few years now) but I assume Maple or some other symbolic math software could also do. In fact I could go on and on writing numerous other different looking but equivalent 3 sum index expression forms.

The issue is the mathematical(quantum) physics of the problem gives me directly these two and/or 3 index summations forms and the authors give no hint as how they arrived at their reduced expression. ANd yes again there are more variables, here m and mp, in my summations versions than the reduced simple form from which somehow m and mp drop out or cancel out in the reduced form. This is my question - why and how ? I can't for the life of me show how to get from the double or triple summation forms to the simple reduced (no summations) form which again is :

(k+j+1)!*(k-j)!/(2*j+1)

and if one does it correctly substituting and evaluating my multiple sum versions one will get exactly this value for any and all choices of integer variables within the prescribed constraints. If someone thinks they may have a solution or can see thru the "forest" but does not get values that agree then i could send a .tex or .dvi or even .pdf files which may show your errors in interpreting my ascii only attempt at writing my mathematical expressions.
 
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  • #2
zmth said:
NOte this is not a homework nor related to any course nor any test problem etc. - entirely my own interest and study.
Re\: text by Biedenharn and Louck "Angular momentum in Q.Physics" . I derive an expression for the norm squared wrt a certain expression in Boson calculus. You don't really need to get into this in order to answer my question. Anyway below in the following is my two summation index form and other equivalent expressions later. Now the authors don't derive how they arrived at their expression but just give the answer as

(k+j+1)!*(k-j)!/(2*j+1)

and in fact this is equivalent(equal) to my more complicated expressions and yes my expressions also have two extra variables or call them parameters if you like which are m and mp in addtion to j and k . Now my question is why is it that the m and mp drop out of my expressions and what binomial/ combinatorial identities etc. are used to get from my multiple summation expressions with the two extra m and mp terms to his ofcourse much simpler reduced and preferred expression again being (k+j+1)!*(k-j)!/(2*j+1) with no summations and without m and mp. Now we begin with my equivalent expressions of which anyone can verify reduce exactly to the author's answer for any and all choices of integer variables under the constraints .Here for simplicity just consider k and j positive integers where k >= j ; m and mp can be positive or negative with maximum abs. value j so
-j <= m,mp <= j and in these latter parameters case you may notice like pertaining or reference to quantum angular momentum to which these pertain and actually k,j,m,mp could all be half an odd integer but to keep it simpler just assume they are all integer. Here are my expressions.

(j-m)*(j+m)*(j-mp)*(j+mp)*sum_{s=max(0,mp-m)}^{min(mp,-m)+k} *(sum_{sp=max(mp-m,0,s-k+j)}^{min(j-m,j+mp,s)}(-1)^(s-sp)* binom(k-j ,s-sp)/((j+mp-sp)!*sp!*(m-mp+sp)!*(j-m-sp)!))^2*(k-m-s)!*(k+mp-s)! *s!*(m-mp+s)!

Or equivalently using 3 summation indexes expanding the square written as:

(j+mp)!*(j-mp)!*(k-j)!^2*sum_{s=max(0,mp-m)}^{min(mp,-m)+k} *sum_{sp=max(mp-m,0,s-k+j)}^{min(j-m,j+mp,s)}(-1)^(s+sp)*binom(s,sp)*binom(k-m-s,j-m-sp)*binom(j+m,j+mp-sp) *sum_{s1=max(mp-m,0,s-k+j)}^{min(j-m,j+mp,s)}(-1)^(s+s1)*binom(s,s1)*binom(k-m-s,j-m-s1)*binom(j+m,j+mp-s1)

OR as :

(j+m)!*(j-m)!*(k-j)!^2*sum_{s=max(0,mp-m)}^{min(mp,-m)+k} *sum_{sp=max(mp-m,0,s-k+j)}^{min(j-m,j+mp,s)}(-1)^(s+sp)*binom(s,sp)*binom(k+mp-s,j+mp-sp)*binom(j-mp,j-m-sp) *sum_{s1=max(mp-m,0,s-k+j)}^{min(j-m,j+mp,s)}(-1)^(s+s1)*binom(s+m-mp ,s-s1)*binom(k-m-s,j-m-s1)*binom(j+mp,s1)

Here binom(a,b) means the usual a!/b!/(a-b)! with a slightly different interpretation when some or one of the variables are negative integers.That is all 3 of these are all equivalent as you may verify by trials esp. if you have Macsyma(unfortunately they have been out of business for quite a few years now) but I assume Maple or some other symbolic math software could also do. In fact I could go on and on writing numerous other different looking but equivalent 3 sum index expression forms.

The issue is the mathematical(quantum) physics of the problem gives me directly these two and/or 3 index summations forms and the authors give no hint as how they arrived at their reduced expression. ANd yes again there are more variables, here m and mp, in my summations versions than the reduced simple form from which somehow m and mp drop out or cancel out in the reduced form. This is my question - why and how ? I can't for the life of me show how to get from the double or triple summation forms to the simple reduced (no summations) form which again is :

(k+j+1)!*(k-j)!/(2*j+1)

and if one does it correctly substituting and evaluating my multiple sum versions one will get exactly this value for any and all choices of integer variables within the prescribed constraints. If someone thinks they may have a solution or can see thru the "forest" but does not get values that agree then i could send a .tex or .dvi or even .pdf files which may show your errors in interpreting my ascii only attempt at writing my mathematical expressions.


Note(I am the poster zmth) there are errors here and I will repost. In particular left off the factorial sign in 4 places in the first expression:

(j-m)*(j+m)*(j-mp)*(j+mp)*sum_{s=max(0,mp-m)}^{min(mp,-m)+k} *(sum_{sp=max(mp-m,0,s-k+j)}^{min(j-m,j+mp,s)}(-1)^(s-sp)* binom(k-j ,s-sp)/((j+mp-sp)!*sp!*(m-mp+sp)!*(j-m-sp)!))^2*(k-m-s)!*(k+mp-s)! *s!*(m-mp+s)!

It should read.

(j-m)!*(j+m)!*(j-mp)!*(j+mp)!*sum_{s=max(0,mp-m)}^{min(mp,-m)+k} *(sum_{sp=max(mp-m,0,s-k+j)}^{min(j-m,j+mp,s)}(-1)^(s-sp)* binom(k-j ,s-sp)
/((j+mp-sp)!*sp!*(m-mp+sp)!*(j-m-sp)!))^2*(k-m-s)!*(k+mp-s)! *s!*(m-mp+s)!


Also seems there is some error in the expression:

(j+mp)!*(j-mp)!*(k-j)!^2*sum_{s=max(0,mp-m)}^{min(mp,-m)+k} *sum_{sp=max(mp-m,0,s-k+j)}^{min(j-m,j+mp,s)}(-1)^(s+sp)*binom(s,sp)*binom(k-m-s,j-m-sp)*binom(j+m,j+mp-sp) *sum_{s1=max(mp-m,0,s-k+j)}^{min(j-m,j+mp,s)}(-1)^(s+s1)*binom(s,s1)*binom(k-m-s,j-m-s1)*binom(j+m,j+mp-s1)

so just leave that out. NOw the expression after that which reads:

OR as :

(j+m)!*(j-m)!*(k-j)!^2*sum_{s=max(0,mp-m)}^{min(mp,-m)+k} *sum_{sp=max(mp-m,0,s-k+j)}^{min(j-m,j+mp,s)}(-1)^(s+sp)*binom(s,sp)*binom(k+mp-s,j+mp-sp)*binom(j-mp,j-m-sp) *sum_{s1=max(mp-m,0,s-k+j)}^{min(j-m,j+mp,s)}(-1)^(s+s1)*binom(s+m-mp ,s-s1)*binom(k-m-s,j-m-s1)*binom(j+mp,s1)


IS correct as stands.
 

Related to Binomial identities,combinatorial, equivalence

1. What are binomial identities?

Binomial identities are mathematical expressions that involve binomial coefficients, which are combinations of numbers used to represent the number of ways to choose a certain number of objects from a larger set. These identities are used to simplify and solve complex equations, and they are derived from the binomial theorem.

2. How are binomial identities related to combinatorial problems?

Binomial identities are used to solve combinatorial problems, which involve counting the number of ways to select or arrange objects from a given set. The binomial coefficients in these identities represent the number of combinations or permutations of objects, making them useful in solving combinatorial problems.

3. What is the significance of equivalence in binomial identities?

Equivalence in binomial identities refers to the fact that two expressions are equal in value, even if they may appear different. This is due to the properties of binomial coefficients, which allow for the simplification of expressions and the creation of equivalent forms.

4. How are binomial identities used in real-world applications?

Binomial identities have many applications in fields such as statistics, physics, and computer science. They are used to solve problems involving probability, combinations and permutations, and the manipulation of mathematical expressions. These identities also have practical applications in data analysis, cryptography, and algorithms.

5. Can you give an example of a binomial identity?

One example of a binomial identity is the Pascal's Triangle, which is a triangular array of numbers that follows a specific pattern. Each number in the triangle represents a binomial coefficient, and the identities derived from this triangle are used to solve a variety of mathematical problems.

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