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Thread: [SOLVED] Squarefree and Multiplicative Functions

  1. #1
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    [SOLVED] Squarefree and Multiplicative Functions

    Problem:
    Let f be the arithmetic function given by f(n) = n if n is square-free and f(n) = 0 if n is not square-free.
    (a) Prove that f is a multiplicative function.
    (b) Is f completely multiplicative?
    (c) Let F be the summatory function of f: $\displaystyle F(n) = \sum_{d | n} f(n)$ . Find $\displaystyle F(p^{a})$ for any a. Calculate F(144).
    ================
    Attempt:
    A square-free number is a number that its unique prime factors do not repeat. 4(which is 2*2), 8(which is 2*2*2), 9(which 3*3).. are not square-free.

    So, for part (a), I would like to show for some m, n, then
    f(mn) = f(m)f(n) by the definition of multiplicative.

    $\displaystyle f(x)=\begin{cases}
    n&\text{if } n \text{is square-free}\\
    0&\text{if } n \text{is not square-free} \end{cases} $

    In order to show this, I broke it into cases.
    If m, n are not square free then f(mn) = 0 = f(m)(n)
    If either m or n are not square free then f(mn) = 0, so either f(m) = 0 or f(n) = 0.
    If both m and n are square free and $\displaystyle m \neq n $, then f(mn) = m*n=f(m)f(n)

    (b) I say this is not completely multiplicative since if m = n, then f(mn) = f(mm) = f(nn) = not a square free number; thus, f(mn) = 0.

    (c) $\displaystyle F(p^{a}) = F(p_{1}^{a1})...F(p_{t}^{at})$, but I don't know how to derive a formula for the summatory function F(n). I know that since small f is multiplicative, then so is big F.
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  2. #2
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    The first two parts look good. For the last part, I think at first, consider p as a prime.

    What is $\displaystyle \sum_{d|n} f(d)$ in this case?
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  3. #3
    Super Member PaulRS's Avatar
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    Quote Originally Posted by Paperwings View Post
    (c) $\displaystyle F(p^{a}) = F(p_{1}^{a1})...F(p_{t}^{at})$, but I don't know how to derive a formula for the summatory function F(n). I know that since small f is multiplicative, then so is big F.
    You meant to say, if $\displaystyle n=p_{1}^{a1}...p_{t}^{at}$ then $\displaystyle F(n) = F(p_{1}^{a1})...F(p_{t}^{at})$ ( where all the $\displaystyle p_i$ are different primes)

    Now $\displaystyle F(p^a)=f(p^0)+f(p^1)+...+f(p^a)$ (we sum over all the positive divisors of $\displaystyle p^a$).

    So $\displaystyle
    F\left( {p^a } \right) = \left\{ \begin{gathered}
    1 + p{\text{ if }}a \geqslant 1 \hfill \\
    1{\text{ if }}a = 0 \hfill \\
    \end{gathered} \right.
    $ (using the definition of f)
    Last edited by PaulRS; Dec 13th 2008 at 07:50 AM.
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  4. #4
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    Thank you Simon and Paul.

    However, I don't understand the first line (should it be just 1 + p if $\displaystyle a \geq 1 $):

    $\displaystyle

    F\left( {p^a } \right) = \left\{ \begin{gathered}
    1 + p + p^2 {\text{ if }}a \geqslant 2 \hfill \\
    1 + p{\text{ if }}a = 1 \hfill \\
    1{\text{ if }}a = 0 \hfill \\
    \end{gathered} \right.
    $

    since for example: $\displaystyle F(8) = F(2^3) = f(1) + f(2) + f(4) + f(8)$.
    By definition, since 4 and 8 are not square free, then f(4) = 0 and f(8) = 0.

    Then,

    $\displaystyle F(8) = f(1) + f(2) = 1 + 2 = 3 $.
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  5. #5
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    It should be, $\displaystyle 1+p$, it was a typo I think
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  6. #6
    Super Member PaulRS's Avatar
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    Yes, you are right, I put it as if It'd been cube-free instead of square-free.
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  7. #7
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    Ah, ok. Thank you for the help.
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