anonymous
  • anonymous
For any prime p, establish each of the following assertions below: \[\sigma (p! )=(p+1)\sigma((p-1)!)\] \[\tau (p! )=2\tau((p-1)!)\]
Mathematics
  • Stacey Warren - Expert brainly.com
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schrodinger
  • schrodinger
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anonymous
  • anonymous
\(\sigma(x)\) is the sum of positive divisors of \(x\), right?
anonymous
  • anonymous
Yes it is.
zzr0ck3r
  • zzr0ck3r
pfft too long since number theory. I really should read a book...

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anonymous
  • anonymous
.
anonymous
  • anonymous
well, it's trivial to show that the divisors of \(n=p_1^{n_1} p_2^{n_2} p_3^{n_3}\cdots \ p_k^{n_k}\) must be of the form \(p_1^{j_1} p_2^{j_2} p_3^{j_3} \cdots \ p_k^{j_k}\) where \(0\le j_i\le n_i\); in other words, there are \(n_i+1\) choices for the exponent of \(p_i\) and so a total \((n_1+1)(n_2+1)(n_3+1)\cdots(n_k+1)\) such combinations; in other words, \(\sigma(p_1^{n_1} p_2^{n_2} p_3^{n_3} \cdots p_k^{n_k})=(n_1+1)(n_2+1)(n_3+1)\cdots(n_k+1)\)
anonymous
  • anonymous
now, if \(p\) is prime then \(p\not|\ (p-1)!\); in other words, \((p-1)!\) as a prime factorization can be written \((p-1)!=p_1^{n_1} p_2^{n_2} p_3^{n_3}\cdots\ p_k^{n_k}\) where \(p_i\ne p\), so then \(p!=p(p-1)!=p_1^{n_1} p_2^{n_2} p_3^{n_3}\cdots\ p_k^{n_k} p^1\) so it follows \(\sigma(p!)=(n_1+1)(n_2+1)(n_3+1)\cdots(n_k+1)\cdot 2 =2\sigma((p-1)!)\)
anonymous
  • anonymous
i guess this is \(\tau\) in your lexicon? and \(\sigma\) is meant to be sum of divisors instead of just counting the # of divisors?
anonymous
  • anonymous
τ(n) is the number of positive divisors of n. σ(n) is the sum of the positive divisors of n.
anonymous
  • anonymous
okay, well the proof there is for the \(\tau\) identity
anonymous
  • anonymous
for the sum of the positive divisors of \(p_1^{n_1} p_2^{n_2} p_3^{n_3}\cdots\ p_k^{n_k}\) it's clearly $$\sum_{j_1=0}^{n_1}\sum_{j_2=0}^{n_2}\sum_{j_3=0}^{n_3}\cdots\sum_{j_k=0}^{n_k} p_1^{j_1}p_2^{j_2}p_3^{j_3}\cdots p_k^{n_k}=\prod_{i=0}^k\left(\sum_{j=0}^{n_i} p_i^j\right)=\prod_{i=0}^k\frac{p_i^{n_i+1}-1}{p_i-1}$$ using the fact \((1+a)(1+b)=1+a+b+ab\)
anonymous
  • anonymous
then the argument is pretty much identical form: $$(p-1)!=p_1^{n_1}\cdots p_k^{n_k}\\p!=p(p-1)!=p_1^{n_1}\cdots p_k^{n_k} p^1$$so$$\sigma(p!)=\prod_{i=0}^k\frac{p_i^{n_i+1}-1}{p_i-1}\cdot\frac{p^{1+1}-1}{p-1}=(p+1)\prod_{i=0}^k\frac{p_i^{n_i+1}-1}{p_i-1}=(p+1)\sigma((p-1)!)$$
anonymous
  • anonymous
so all these problems really require is knowledge of prime factorization and some basic algebra :p
anonymous
  • anonymous
oops, that earlier sum should be of \(p_1^{j_1}\cdots\ p_k^{j_k}\) not \(p_k^{n_k}\)
anonymous
  • anonymous
note that this identity gives us \(\sigma(p!)=(p+1)\sigma((p-1)!)=\dots=(p+1)!\cdot \sigma(1)=(p+1)!\)

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