Question

Totient function

Answer 1

@dan815

$$\varphi(p^k) = p^k(1-\tfrac{1}{p}) $$

Let's just look at the simplest case where k=1 then we have:

$$\varphi(p) = p-1$$

Makes sense, all the numbers less than a prime are relatively prime to it.

Relatively prime just means that \(\gcd(a,b)=1\).

Answer 2

Basically just playing around with this if you're curious @imqwerty

Here's a dream I had last night:

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Define the backwards difference:

$$\Delta f(k) = f(k)-f(k-1)$$

Check this out:

$$\Delta \sigma(p^k)= \sigma(p^k)-\sigma(p^{k-1}) = p^k$$
$$\Delta p^k = p^k-p^{k-1} = \varphi(p^k)$$

So the second 'derivative' of the sum of divisors function is the totient function. Weird.

$$ \Delta^2 \sigma(p^k) = \varphi(p^k)$$

Answer 3

thanks :)
i've understood this concept B)

Answer 4

Anywho, I guess I'll just leave this here.

This appears to generalize if we take this to be the full multiplicative derivative:

$$\Delta = \prod_i \Delta_{p_i}$$

Where we have refined the backwards difference to be the partial derivative:

$$\Delta_{p_i} f(p_i^{k_i}) = f(p_i^{k_i}) - f(p_i^{k_i-1})$$

This allows us to say more generally:

$$\Delta \sigma(n) = n$$
$$\Delta n = \varphi(n)$$
$$\Delta^2 \sigma(n) = \varphi(n)$$

Some ideas to play around with,

$$n = \sum_{d|n} \varphi(d) = \sum_{d|n} \Delta d $$
$$\sigma(n) = \sum_{d|n} d = \sum_{d|n} \Delta \sigma(d) $$

If that doesn't interest you in trying to play with this more, then I don't know what would. lol. Looks weirdly like the fundamental theorem of calculus but with arithmetic functions lol.