A derivative for whole numbers
Tonight's idea
In calculus the product rule says \((fg)' = f'g + fg'\). Differential algebra turns this around. It forgets limits and slopes and calls an operation a derivative if it obeys the right algebraic rules: the sum rule and, above all, the product rule. Tonight, keep only the product rule and try it on whole numbers. Primes are the atoms of multiplication, so let each prime play the part of \(x\), with derivative \(1\).
The arithmetic derivative: \(D(p) = 1\) for every prime \(p\), and \(D(ab) = a\,D(b) + b\,D(a)\) for all positive whole numbers \(a\) and \(b\).
Every whole number above \(1\) is a prime or a product of two smaller whole numbers, so these two lines reach every positive whole number. One worry: \(12 = 2 \times 6 = 3 \times 4\). Do both splittings give the same \(D(12)\)? They do, for every number, and the stretch shows why.
A warning: \(D\) does not obey the sum rule (exercise 1), so it is a made-up cousin of the derivative, not the real thing.
Watch one: \(D(6)\) and \(D(12)\)
Start with products of two primes. \(D(4) = D(2 \times 2) = 2 \times 1 + 2 \times 1 = 4\) and \(D(6) = D(2 \times 3) = 2 \times 1 + 3 \times 1 = 5\). Now build on those:
\[D(12) = D(2 \times 6) = 2\,D(6) + 6\,D(2) = 2 \times 5 + 6 \times 1 = 16.\]
Split \(12\) the other way and the same number comes out:
\[D(12) = D(3 \times 4) = 3\,D(4) + 4\,D(3) = 3 \times 4 + 4 \times 1 = 16.\]
Notice \(D(4) = 4\): the number \(4\) is its own derivative, as \(e^x\) is in calculus. Notice also that \(D(12) = 16\) is bigger than \(12\), while \(D(6) = 5\) is smaller than \(6\).
Your turn
- (a) Put \(a = b = 1\) in the product rule. What must \(D(1)\) be? (b) Compare \(D(2 + 3)\) with \(D(2) + D(3)\). Does \(D\) obey the sum rule?
- Find \(D(8)\), \(D(9)\) and \(D(30)\). Work out \(D(30)\) twice, from \(5 \times 6\) and from \(2 \times 15\), and check that the two agree.
- Let \(p\) be a prime. Find \(D(p^2)\) and \(D(p^3)\), guess a formula for \(D(p^k)\) with \(k \ge 1\), and prove it by going from \(p^k\) to \(p^{k+1} = p \times p^k\). Which rule from calculus does it copy?
- Use exercise 3 to find every prime power \(p^k\) (with \(k \ge 1\)) that is its own derivative, \(D(p^k) = p^k\). Write out the three smallest.
Stretch: a logarithm in disguise
In calculus \(f'/f\) is the derivative of \(\ln f\), and it turns products into sums. Let \(L(n) = \dfrac{D(n)}{n}\). (a) Divide the product rule by \(ab\) to show \(L(ab) = L(a) + L(b)\). (b) Deduce that if \(n = p_1^{k_1} p_2^{k_2} \cdots p_r^{k_r}\) with distinct primes \(p_i\), then \(L(n) = \dfrac{k_1}{p_1} + \dfrac{k_2}{p_2} + \cdots + \dfrac{k_r}{p_r}\). (c) Show that the numbers you found in exercise 4 are the only positive whole numbers with \(D(n) = n\).
Answers
- (a) \(D(1) = D(1 \times 1) = 1 \times D(1) + 1 \times D(1) = 2\,D(1)\), so \(D(1) = 0\). The rule forces it, just as a constant has derivative zero. (b) \(D(2 + 3) = D(5) = 1\) because \(5\) is prime, but \(D(2) + D(3) = 1 + 1 = 2\). So \(D\) does not obey the sum rule.
- \(D(8) = D(2 \times 4) = 2\,D(4) + 4\,D(2) = 2 \times 4 + 4 \times 1 = 12\). \(D(9) = D(3 \times 3) = 3 \times 1 + 3 \times 1 = 6\). From \(5 \times 6\): \(D(30) = 5\,D(6) + 6\,D(5) = 5 \times 5 + 6 \times 1 = 31\). From \(2 \times 15\): first \(D(15) = 3 \times 1 + 5 \times 1 = 8\), then \(D(30) = 2 \times 8 + 15 \times 1 = 31\). They agree, and \(31\) is prime, so differentiating once more gives \(1\).
- \(D(p^2) = p \times 1 + p \times 1 = 2p\) and \(D(p^3) = p\,D(p^2) + p^2 D(p) = 2p^2 + p^2 = 3p^2\). Guess: \(D(p^k) = k\,p^{k-1}\), true for \(k = 1\) since \(D(p) = 1\). If it holds for \(k\), then \(D(p^{k+1}) = p\,D(p^k) + p^k D(p) = k\,p^k + p^k = (k + 1)\,p^k\), so it holds for \(k + 1\), and so for every \(k \ge 1\). It copies the power rule \((x^k)' = k\,x^{k-1}\). Check: \(D(8) = 3 \times 2^2 = 12\) and \(D(9) = 2 \times 3 = 6\).
- \(D(p^k) = p^k\) means \(k\,p^{k-1} = p^k\). Divide by \(p^{k-1}\): \(k = p\). So the answers are exactly the numbers \(p^p\) with \(p\) prime: \(2^2 = 4\), \(3^3 = 27\), \(5^5 = 3125\), then \(7^7 = 823\,543\), and so on. Check: \(D(27) = 3 \times 3^2 = 27\).
Stretch. (a) \(\dfrac{D(ab)}{ab} = \dfrac{a\,D(b)}{ab} + \dfrac{b\,D(a)}{ab} = \dfrac{D(b)}{b} + \dfrac{D(a)}{a}\). (b) By exercise 3, \(L(p^k) = \dfrac{k\,p^{k-1}}{p^k} = \dfrac{k}{p}\). Use (a) to add up the prime powers in \(n\). This formula never mentions a splitting, and \(D(n) = n\,L(n)\) obeys both lines of the definition, which is why every splitting gives the same answer. (c) \(D(n) = n\) means \(L(n) = 1\), which rules out \(n = 1\), where \(L = 0\). For one prime, \(k/p = 1\) gives \(n = p^p\). Suppose there were two or more primes, and multiply \(\frac{k_1}{p_1} + \cdots + \frac{k_r}{p_r} = 1\) by \(P = p_1 p_2 \cdots p_r\). Then \(P\) and every term except the first, \(k_1 P / p_1\), are multiples of \(p_1\), so \(p_1\) divides \(k_1 P / p_1\). But \(P / p_1\) is a product of other primes, so \(p_1\) divides \(k_1\), and \(k_1 \ge p_1\). Then \(k_1 / p_1 \ge 1\) already, and the other terms push the sum above \(1\). So only \(p^p\) works.
One more link: for primes \(p\) and \(q\), \(D(pq) = p + q\). Goldbach's conjecture, still unproven, says every even number from \(4\) up is a sum of two primes. If that is true, every even number from \(4\) up is the derivative of something.