- #1

christian0710

- 409

- 9

Hi,

If i have a progression that looks like this

--------------------------------------

1 | x(x-1) .....| x(x-n)

2 | x(x-2)(x-1) .....| x(x-n)(x-n

3 | x(x-3)(x-2)(x-1)...| x(x-n)(x-n

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

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

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 this

**n | 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 :)

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