# Math Help - Analysis

1. ## Analysis

Def: A polynomial f(x) with coefficients in Q (the rationals) is called a "numerical polynomial" if for all integers n, f(n) is an integer also.

I have to use induction to prove that for k > 0

that the function f(x) := (1/k!)*x*(x-1)...(x-k+1) is a numerical polynomial

I checked that this is true for k=1, but to be honest I'm not even sure what the dot dot dot means. If k=5 for say, I interpreted the dot dot dot as (1/5!)*x*(x-1)*(x-2)*(x-3)*(x-4). Is this the correct interpretation? If so it is indeed true for k=1, but nonetheless I don't know how to show that it is true for k+1.

Thanks so much.

P.S My professor sucks and it is really discouraging as a recently declared math major. So I will be on here often! Loves.

2. Originally Posted by Zero266
Def: A polynomial f(x) with coefficients in Q (the rationals) is called a "numerical polynomial" if for all integers n, f(n) is an integer also.

I have to use induction to prove that for k > 0

that the function f(x) := (1/k!)*x*(x-1)...(x-k+1) is a numerical polynomial

I checked that this is true for k=1, but to be honest I'm not even sure what the dot dot dot means. If k=5 for say, I interpreted the dot dot dot as (1/5!)*x*(x-1)*(x-2)*(x-3)*(x-4). Is this the correct interpretation? If so it is indeed true for k=1, but nonetheless I don't know how to show that it is true for k+1.

Thanks so much.

P.S My professor sucks and it is really discouraging as a recently declared math major. So I will be on here often! Loves.

Hi Zero266,

This is what I have so far, if this is not enough to help, you may need to wait for someone more knowledgeable.

If you checked the n=1 case, then you will need to assume that the n=k case is true. From there you should be able to prove that the n=k+1 case is true. (This is normally how the induction method works)

Assume $f_k(x) = \frac{1}{k!}*x*(x-1)...(x-k+1)$ is a numerical polynomial. Then
$f_{k+1}(x)=\frac{1}{k+1!}*x*(x-1)...(x-k+1)*(x-(k+1)+1)$
$=\left[\frac{1}{k!}*x*(x-1)...(x-k+1)\right]*\left[\frac{1}{k+1}(x-(k+1)+1)\right]$
=...

Now use the fact that $f_k(x)$ is a numerical polynomial. You should be able to concentrate on the factor $\left[ \frac{1}{k+1}(x-(k+1)+1)\right]$.

3. Thanks a lot. But was my interpretation of the ... correct? I'm really bad at reading math. =/ But thanks so much, I do see the light for this one.

4. Originally Posted by Zero266
Thanks a lot. But was my interpretation of the ... correct? I'm really bad at reading math. =/ But thanks so much, I do see the light for this one.
I agree with you on that.

5. Okay I'm having A LOT of difficulty with this problem. The part you suggested I focus on turned out that it is not numerical for all k which is clear if you plug some values for the free variable , x. SO i tried to take the proof in a different direction, and I discovered that that function, f_k (x) turns out to be the binomial coefficients,which was really surprising. But I have no clue what to do. A full proof would really be appreciated since i sweated this one long enough. Thanks!