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pointwise multiplication of a completely multiplicative function distibutes over convolution (Theorem)
Theorem   Let $ f$ be a completely multiplicative function and $ g$ and $ h$ be arithmetic functions. Then $ f(g*h)=(fg)*(fh)$.
Proof. Let $ n$ be a positive integer. Then
$ \displaystyle (f(g*h))(n)$ $ \displaystyle = f(n)(g*h)(n)$
  $ \displaystyle = f(n) \sum_{d\vert n} g(d)h\left( \frac{n}{d} \right)$
  $ \displaystyle = \sum_{d\vert n} f(n)g(d)h\left( \frac{n}{d} \right)$
  $ \displaystyle = \sum_{d\vert n} f\left( d \cdot \frac{n}{d} \right) g(d)h\left( \frac{n}{d} \right)$
  $ \displaystyle = \sum_{d\vert n} f(d)f\left( \frac{n}{d} \right) g(d)h\left( \frac{n}{d} \right)$
  $ \displaystyle = \sum_{d\vert n} (fg)(d)(fh)\left( \frac{n}{d} \right)$
  $ \displaystyle = ((fg)*(fh))(n)$.
$ \qedsymbol$



"pointwise multiplication of a completely multiplicative function distibutes over convolution" is owned by Wkbj79.
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See Also: arithmetic function, completely multiplicative, multiplicative function


Attachments:
formula for the convolution inverse of a completely multiplicative function (Corollary) by Wkbj79
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Cross-references: integer, positive, arithmetic functions, completely multiplicative function
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This is version 5 of pointwise multiplication of a completely multiplicative function distibutes over convolution, born on 2006-06-13, modified 2006-10-09.
Object id is 8022, canonical name is PropertyOfCompletelyMultiplicativeFunctions.
Accessed 871 times total.

Classification:
AMS MSC: 11A25 (Number theory :: Elementary number theory :: Arithmetic functions; related numbers; inversion formulas)
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name by Wkbj79 on 2006-06-13 12:41:19
I feel that the name of this entry is too vague. Does anyone have suggestions for improvement?
[ reply | up ]
  • Re: name by silverfish on 2006-06-13 13:08:58
    • Re: name by silverfish on 2006-06-13 13:11:47
      • Re: name by Wkbj79 on 2006-06-13 15:16:01
  • Re: name by silverfish on 2006-06-13 13:10:26

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