J Function Properties A9A8D0

1. **Stating the problem:** We have a function defined for any positive integer $n$ as $$J(n) := n \times \prod_{p|n} \frac{1 + p}{p}$$ where the product is over all prime divisors $p$ of $n$. We need to evaluate $J(n_i)$, $\sigma(n_i)$ (sum of divisors), and $\mu(n_i)$ (Möbius function) for three chosen numbers $n_1, n_2, n_3$ with specific properties, then explain properties of $J$ and its relation to convolution. --- 2. **Choosing numbers:** - $n_1$: prime power, e.g. $27 = 3^3$ (prime power of 3) - $n_2$: composite, squarefree, not prime power, e.g. $30 = 2 \times 3 \times 5$ - $n_3$: composite, not prime power, not squarefree, e.g. $36 = 2^2 \times 3^2$ --- 3. **Recall definitions:** - $\sigma(n)$ is sum of divisors of $n$. - $\mu(n)$ is Möbius function: - $\mu(n) = 0$ if $n$ has squared prime factor - $\mu(n) = (-1)^k$ if $n$ is product of $k$ distinct primes - $J(n) = n \times \prod_{p|n} \frac{1+p}{p}$ --- 4. **Calculate for $n_1 = 27 = 3^3$:** - Prime divisors: $p=3$ - $$J(27) = 27 \times \frac{1+3}{3} = 27 \times \frac{4}{3} = 27 \times \frac{4}{3} = \cancel{\frac{27}{3}} \times 4 = 9 \times 4 = 36$$ - Divisors of 27: $1, 3, 9, 27$ - $$\sigma(27) = 1 + 3 + 9 + 27 = 40$$ - Since $27 = 3^3$ has squared prime factor, $\mu(27) = 0$ --- 5. **Calculate for $n_2 = 30 = 2 \times 3 \times 5$ (squarefree):** - Prime divisors: $2, 3, 5$ - $$J(30) = 30 \times \frac{1+2}{2} \times \frac{1+3}{3} \times \frac{1+5}{5} = 30 \times \frac{3}{2} \times \frac{4}{3} \times \frac{6}{5}$$ - Simplify stepwise: $$30 \times \frac{3}{2} = \cancel{30} \times \frac{3}{\cancel{2}} = 15 \times 3 = 45$$ $$45 \times \frac{4}{3} = \cancel{45} \times \frac{4}{\cancel{3}} = 15 \times 4 = 60$$ $$60 \times \frac{6}{5} = \cancel{60} \times \frac{6}{\cancel{5}} = 12 \times 6 = 72$$ - Divisors of 30: $1, 2, 3, 5, 6, 10, 15, 30$ - $$\sigma(30) = 1 + 2 + 3 + 5 + 6 + 10 + 15 + 30 = 72$$ - Since 30 is product of 3 distinct primes, $\mu(30) = (-1)^3 = -1$ --- 6. **Calculate for $n_3 = 36 = 2^2 \times 3^2$ (not squarefree):** - Prime divisors: $2, 3$ - $$J(36) = 36 \times \frac{1+2}{2} \times \frac{1+3}{3} = 36 \times \frac{3}{2} \times \frac{4}{3}$$ - Simplify stepwise: $$36 \times \frac{3}{2} = \cancel{36} \times \frac{3}{\cancel{2}} = 18 \times 3 = 54$$ $$54 \times \frac{4}{3} = \cancel{54} \times \frac{4}{\cancel{3}} = 18 \times 4 = 72$$ - Divisors of 36: $1, 2, 3, 4, 6, 9, 12, 18, 36$ - $$\sigma(36) = 1 + 2 + 3 + 4 + 6 + 9 + 12 + 18 + 36 = 91$$ - Since 36 has squared prime factors, $\mu(36) = 0$ --- 7. **Summary table:** | $n$ | $J(n)$ | $\sigma(n)$ | $\mu(n)$ | |-----|--------|-------------|----------| | 27 | 36 | 40 | 0 | | 30 | 72 | 72 | -1 | | 36 | 72 | 91 | 0 | --- 8. **(b) Why $J(n)$ is always integer and multiplicative:** - Each factor $\frac{1+p}{p} = 1 + \frac{1}{p}$ is rational but when multiplied by $n$, the denominator $p$ cancels with the $p$ in $n$ because $p|n$. - Hence $J(n)$ is integer. - $J$ is multiplicative because for coprime $m,n$, prime divisors split and product factors multiply independently. - $J(n)$ and $\sigma(n)$ differ when $n$ is not squarefree or not prime power; for squarefree $n$, $J(n) = \sigma(n)$. --- 9. **(c) $J(n)$ as convolution of $\mu^2(n)$ and $i(n) := n$:** - $\mu^2(n)$ is indicator of squarefree numbers (1 if squarefree, 0 otherwise). - The Dirichlet convolution $(\mu^2 * i)(n) = \sum_{d|n} \mu^2(d) i(n/d)$ sums $i(n/d)$ over squarefree divisors $d$. - This matches the formula for $J(n)$, which weights $n$ by factors from prime divisors. - Hence, $J(n) = (\mu^2 * i)(n)$. --- **Final answers:** - $J(27) = 36$, $\sigma(27) = 40$, $\mu(27) = 0$ - $J(30) = 72$, $\sigma(30) = 72$, $\mu(30) = -1$ - $J(36) = 72$, $\sigma(36) = 91$, $\mu(36) = 0$ - $J(n)$ is integer and multiplicative because denominators cancel with factors in $n$ and prime divisors split over coprime factors. - $J(n)$ equals $\sigma(n)$ for squarefree $n$, differs otherwise. - $J(n)$ equals convolution of $\mu^2$ and $i$ by definition of squarefree indicator and sum over divisors.