Message Boards Message Boards

Impose assumptions in binomial coefficient?

GROUPS:

Hello everyone. How can I impose constraints in the Binomial[n,m] formula in Mathematica? For example, I'd like to get Binomial[-1,m]=-1.

Thank you for your help.

POSTED BY: Gennaro Arguzzi
Answer
1 month ago

Do you only want that for positive odd integers and negative even integers? What kind of number is m? Real, complex, rational, or integer? Has m been assigned a value?

POSTED BY: Jim Baldwin
Answer
1 month ago

Hi @Jim Baldwin , m is a generic natural number.

POSTED BY: Gennaro Arguzzi
Answer
1 month ago

I suppose you know that all of the following statements are true:

Binomial[-1, 0] == 1
Binomial[-1, 1] == -1
Binomial[-1, 2] == 1
Binomial[-1, 3] == -1
Binomial[-1, 4] == 1

So while I don't know why you would want to change the usual definition, you could always use a replacement rule at the end of your calculations:

x + 2 Binomial[-1, m] /. Binomial[-1, m_] -> -1
(* -2+x *)
POSTED BY: Jim Baldwin
Answer
1 month ago

@Jim Baldwin your solution is the solution of Binomial[-1, -1], not of Binomial[-1,m] with m generic natural number.

POSTED BY: Gennaro Arguzzi
Answer
1 month ago

Would you please elaborate? m is undefined and with the above replacement rule, anytime the term Binomial[-1,h] is seen that term is replaced by -1.

If m is defined as a specific value, then that replacement rule won't ever change anything because Binomial[-1,m] gets evaluated prior to the replacement rule.

h = 4
Binomial[-1, h] /. Binomial[-1, m_] -> -1
(* 1 *)

So if the variable m is already defined, then you'll need to write your own function:

bin[n_, m_] := If[n == -1 && IntegerQ[m], -1, Binomial[n, m]]
bin[10, 3]
(* 120 *)
bin[-1, 4]
(* -1 *)
bin[-1, 5]
(* -1 *)
POSTED BY: Jim Baldwin
Answer
1 month ago

@Jim Baldwin I can obtain the same result with the simple built-in function Binomial[n,m]. Maybe is not possible to "prove" that Binomial[-1,m]=-1 for each m natural number.

In other words, I'd like to "prove" that the following two solutions are equal (see result):

http://www.wolframalpha.com/input/?i=InverseZTransform%5B1%2F(1-a*(z%5E-1))%5D+and+InverseZTransform%5Bz%2F(z-a)%5D

POSTED BY: Gennaro Arguzzi
Answer
1 month ago

I certainly didn't interpret your original question as asking for a proof. Consider the definition of a binomial coefficient when one of the values is negative:

$$\binom{-n}{m} = (-1)^m \binom{m+n-1}{m}$$

See Wiki - Binomial coefficient.

So when $n=1$ we have $\binom{-1}{m}=(-1)^m$. That should be enough to show the two equations are equal. (And there is the appearance that you really weren't looking for a proof but rather if there was an inconsistency in WolframAlpha. No problem in that. There are some inconsistencies. But it would be better to state that explicitly. You'll get more and better help that way.)

POSTED BY: Jim Baldwin
Answer
1 month ago

Group Abstract Group Abstract