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Math Help - Analysis

  1. #1
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    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.
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  2. #2
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    Quote Originally Posted by Zero266 View Post
    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].
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  3. #3
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    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.
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  4. #4
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    Quote Originally Posted by Zero266 View Post
    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.
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  5. #5
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    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!
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