christian0710
- 407
- 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 :)
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: