Determaning a function from a progression?

  • Context: Undergrad 
  • Thread starter Thread starter christian0710
  • Start date Start date
  • Tags Tags
    Function
Click For Summary
SUMMARY

The discussion focuses on deriving a mathematical function from a progression defined by terms involving factorials and products. The progression is expressed as x(x-n)(x-n-1)... for increasing values of n. The user successfully identifies that the function can be represented as x multiplied by the product notation ∏k(x/k - 1), leading to the conclusion that the function can be expressed using the Gamma function: x(x-1)(x-2)...(x-n) = Gamma(x+1) / Gamma(x+1-n). This formulation effectively captures the relationship between n and the progression.

PREREQUISITES
  • Understanding of factorials and their properties
  • Familiarity with product notation (∏) in mathematics
  • Knowledge of the Gamma function and its applications
  • Basic algebraic manipulation skills
NEXT STEPS
  • Research the properties and applications of the Gamma function in advanced mathematics
  • Explore the concept of Pochhammer symbols and their significance in combinatorial mathematics
  • Learn about the derivation and applications of product notation in mathematical functions
  • Study sequences and series to understand their convergence and divergence properties
USEFUL FOR

Mathematicians, students studying advanced calculus, and anyone interested in combinatorial mathematics and factorial functions.

christian0710
Messages
407
Reaction score
8
Hi,
If i have a progression that looks like thisn | progression | Mathmatical formula
--------------------------------------
1 | x(x-1) .....| x(x-n)
2 | x(x-2)(x-1) .....| x(x-n)(x-n-1)
3 | x(x-3)(x-2)(x-1)...| x(x-n)(x-n-1)(x-n-2))

Every time n increases by one a new term (x-n) is added to the previous terms. How do I turn this into a function of n? I know that as n increases then we get (x-(n-1))*(x-(n-2)) ... (x-(n-(n-1)))) n times but how do you express this as a mathematical function? A function that adds keeps adding n terms (x-n)(x-n-1)(x-n-2)etc... all the way down to the last term (x-n-n-1)

Can anyone help me or give me a hint about how to construct a function of n, such that the above progression holds true for any n we put into the function?

I know that x! = x*x-1*x-2*--*x-x+1

but saying f(n)= (x-n)! does not make sense since we have two variales (x and n) , I'm stuck and would really appreciate a hint :)
 
Last edited:
Mathematics news on Phys.org
Wait I think i might have igured it out - or am i doing the math wrong?

I know that
∏k from 1 to n is defined as ∏k = 1*2*3*4...n

Therefore it must be true that

k(x/k - 1) going from 1 to n is defined as

∏k(x/k-1) =( (1*x)/1 -1) * ((2*x)/2-2) * ((3*x)/3-3) = (x -1)(x-2)(x-3) etc.. p to n :) Is this corret usage of the mathematical multiplication operator ∏?

So the function for this progression is
x*∏k(x/k-1)
 
See the Pochammer symbol for this function and the related closed form thanks to the Gamma function :
x(x-1)(x-2)...(x-n)=Gamma (x+1) / Gamma(x+1-n)
 

Similar threads

  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 16 ·
Replies
16
Views
2K
  • · Replies 4 ·
Replies
4
Views
3K
  • · Replies 2 ·
Replies
2
Views
1K
  • · Replies 12 ·
Replies
12
Views
2K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 28 ·
Replies
28
Views
5K
  • · Replies 1 ·
Replies
1
Views
1K
  • · Replies 6 ·
Replies
6
Views
3K