Question

Arithmetic Derivative

Answer 1

@Kainui

Answer 2

\[
k=ab \implies k' = a'b+ab'
\]When \(k\) is prime, \(k'=1\).

Answer 3

We established that \[
(k^n)' = k'nk^{n-1}
\]

Answer 4

For every prime \(p\) such that \(p|k\): \[
k' = k\sum_p\frac 1p
\]

Answer 5

Rules about primes shouldn't matter too much, generalizing to composite is better, in my opnion.

Answer 6

We also established a couple of fun facts: \[(p^p)' = p^p\] and \[(n!)' = n! \sum_{k=1}^n \frac{k'}{k}\]

Answer 7

What would \(
(-1)'
\) be? It would extend us to integers.

Answer 8

I think we should think more generally than (-1)' and try to look at \[\large (e^{i \theta})'\]

Answer 9

Well, \(-1\) is tough because a negative number by a positive number is negative.

Answer 10

Actually maybe we should consider looking at just simply fractions first, they seem like they might be fairly interesting. For instance, look at this pattern that begins to emerge when I write their derivatives out. \[\large 1' \ 2' \ 3' \ 4' \ 5' \ 6' \ 7' \ 8' \ 9' \ 10' \\ \large 0 \ \ 1 \ \ 4 \ \ \ 1 \ \ 5 \ \ 1 \ 12 \ \ 6 \ \ 7 \ \ 1\]
It sort of begins to count 4, 5, 6, 7 with apparently a random 12 and some 1's in there while completely avoiding 2 and 3. Maybe this isn't really a pattern? But it might give us an idea of how to interpolate?

Answer 11

Oh yeah, remember the formula: \[
\frac{a'}{a}+ \frac{b'}{b} = \frac{a'b+ab'}{ab} = \frac{c'}{c}
\]

Answer 12

\[
1 \sim c\\
p \sim x\\
c \sim f(x)
\]

Answer 13

\[
\frac{(-2)'}{-2}+\frac{(-2)'}{-2} = \frac{-2(-2)' + -2(-2)'}{4} = \frac{-4(-2)'}{4}
\]We should expect that \[
\frac{(-2)'}{-2}+\frac{(-2)'}{-2} = \frac{4'}{4}
\]This iimplies \[
-4(-2)' = 4'
\]

Answer 14

This implies that \((-2)' = -1\)

Answer 15

\[1 = (-1)(-1)\] implies \[0 = (-1)'(-1)+(-1)(-1)'\]\[0=-2(-1)'\]\[0=(-1)'\]

Answer 16

\[
(-a)' = (-1)'a +(-1)a' = -a'
\]

Answer 17

Yeah you beat me to it haha.

Answer 18

Next is:\[
\left(\frac{1}{a}\right)'
\]I know the answer, but I wonder how we can get it using our rules.

Answer 19

Try something like 4 = 12/3

Answer 20

it should be consistent.

Answer 21

\[
(a^{-1})' = -a'a^{-2}
\]

Answer 22

\[
\left(\frac 1a\right)' = -\frac{a'}{a^2}
\]

Answer 23

\[1= a *a^{-1}\]\[0 = (a)'a^{-1}+a(a^{-1})'\]

Answer 24

\[
\left(\frac ab\right)' = (ab^{-1})' = a'b^{-1} + a(b^{-1})'
\]

Answer 25

Yeah, it's all consistent with what we had before with (k^n)' like you wrote at the beginning.

Answer 26

\[
= \frac{a'}{b} + \frac{-ab'}{b^2} = \frac{a'b-ab'}{b^2}
\]

Answer 27

Next step is rational exponents.

Answer 28

We would expect:\[
a^n = a'na^{n-1}
\]to still hold true.

Answer 29

\[
(\sqrt{a})' = a'\frac12a^{-1/2} = \frac{a'}{2\sqrt{a}}
\]

Answer 30

What is weird is how similar it is to derivative, yet anti-derivative has no meaning.

Answer 31

\[
(a+b)' = a' (1+b/a) + b'(1+a/b)
\]

Answer 32

\[a=a^{k/k}=(a^k)^{1/k}\]\[\large a' = \frac{1}{k}(a^k)^{\frac{1}{k}-1}(a^k)'=\frac{1}{k}(a^k)^{\frac{1}{k}-1}*ka^{k-1}a'\] \[\large 1=(a^k)^{\frac{1}{k}-1}a^{k-1}\]\[\large 1=a^{1-k +k-1}\]

Answer 33

\[
a' + \frac{a'b}{a} + b' + \frac{ab'}{b} = \frac{a'ab + ab^2+abb'+a^2b'}{ab} = (a+b)'
\]

Answer 34

Did I mess this last one up?

Answer 35

It looks like it could be useful as hell.

Answer 36

Yeah I'm trying to figure it out, I have been wanting that for a while now haha

Answer 37

What I did was \(a+b = a(1+b/a)=b(1+a/b)\)

Answer 38

If let \(b=1\), then we get \[
(a+1)' = \frac{a'a+a}{a}
\]

Answer 39

Oh no, did I crack the code?

Answer 40

I messed up, because it would mean \[
(a+1)' = a' + 1
\]which is not true.

Answer 41

Oh, I know what I did wrong now.

Answer 42

I wasn't consistent with my factoring.

Answer 43

\[
(a+b)' = a'(1+b/a) + a(1+b/a)'
\]

Answer 44

I am playing around with the integral and it's sort of bizarre how it doesn't work quite right. \[6'=5=(\frac{1}{2}5^2)'\]

Answer 45

\[
(a+1)' = a'(1+1/a) + a(1+1/a)'
\]

Answer 46

Well I guess it does work, I was thinking as though the 1/2 is a constant... But it's not even a thing here @_@

Answer 47

\[
a'\left(\frac{1+a}{a}\right) + a\left(\frac{1+a}{a}\right) '
\]

Answer 48

I think if we can do \((a+1)'\) we win.

Answer 49

but the \(1+a\) inside the derivative means we'll probably end up at a circular equality.

Answer 50

\[(\frac{1}{2}5^2)' = -(\frac{5}{2})^2+5=-\frac{5}{4} \ne 5\] for the record.

Hmm I think I might have a way to make the derivative work. It's just so odd that it's not a linear operator but so many other properties seem to hold pretty well.

Answer 51

Where did you get \(\frac 12 5^2\)?

Answer 52

In my misguided attempt to integrate 5. I knew 6'=5 and thought I could just increase the exponent by 1 and divide by it.

Answer 53

Another interesting thing, 21'=25'=10

Answer 54

\[
(a^n)' = a'na^{n-1}
\]

Answer 55

\[
(p^n)' = np^{n-1}
\]So you can do the anti derivative if you think it comes from a prime.

Answer 56

But only \(1\) can come from a prime.

Answer 57

Are there any cases where a'=1 when a is not prime?

Answer 58

Well now that we have negative numbers floating around, maybe it's possible?

Answer 59

Well, suppose there was a composite \(a\) such that \(a'=1\).

Answer 60

It would mean that \((a^a)' = a^a\)

Answer 61

Well I'm not just saying composite, but any rational or irrational number we have access to now.

Answer 62

Oh

Answer 63

We would need to search for:\[
1 = \frac{(a^{n})'}{na^{n-1}}
\]

Answer 64

\[
1=\frac{a'a^{n-1}+a(a^{n-1})'}{na^{n-1}}
\]

Answer 65

\[(-1)' = (e^{i \pi})'\]\[0=e' i \pi e^{i \pi -1}\]\[0=e'\]

Answer 66

If only we figured out: \[
(f(n))'
\]

Answer 67

\[i' = (e^{i \pi /2})'\]\[i' = e' i \frac{\pi}{2}e^{i \pi/2 -1}\]\[i'=0\]

Answer 68

\[
i = (-1)^{1/2} = (-1)'(\ldots) = 0
\]

Answer 69

\[e' = [\sum_{n=0}^\infty \frac{1}{n!} ]'=0\]\[e' = [\lim_{n \rightarrow \infty} (1+\frac{1}{n})^n]'=0\]

Answer 70

What about \(\pi\) as a series?

Answer 71

I don't know, I think it has several kinds of infinite representations. But I'm more interested in working out some kind of sum rule. I just sorta wanted to throw those out there since they sorta came to mind.

Answer 72

There is also vectors.

Answer 73

Ohhhh speaking of which: \[\LARGE a+bi = re^{i \theta}\]

Answer 74

Perhaps gcd thing somehow involves \(aa'\).

Answer 75

For primes \(aa' = a\).

Answer 76

Interesting thought.

Answer 77

\[
\langle a',b' \rangle \cdot \langle b,a \rangle = a'b+ab'
\]

Answer 78

We could say \(\mathbf v' \) is just \(\langle v_1',v_2',\ldots\rangle\).

Answer 79

Unless you think there is a better way of defining it.

Answer 80

\[
\mathbf v\cdot \mathbf v' = v_1v_1'+v_2v_2'+\ldots
\]

Answer 81

For a vector of primes, it would just be the sum of the elements.

Answer 82

Hmmm well if it is consistent with something that seems natural. I don't know if I'm willing to say anything yet.

I'm currently looking at; \[(a+bi)' = (re^{i \theta})'=r'e^{i \theta}+r (e^{i \theta})'=r'e^{i \theta}\] So the derivative seems for complex numbers to be only dependent on the length not the direction.

Answer 83

What I'm saying is the derivative should be the derivative of the length of the vector in the same direction as the old vector.

Answer 84

And \[
r = \sqrt{a^2+b^2}
\]?

Answer 85

Yes, exactly. Or also written as \[r= [(a+bi)(a-bi)]^{1/2}\]

Answer 86

But it follows from this fairly well: \[(re^{i \theta})' = r'e^{i \theta}\] since e'=0

Answer 87

So you are saying: \[
\mathbf v' =|\mathbf v|' \frac{\mathbf v}{|\mathbf v|}
\]

Answer 88

Yeah perfect.

Answer 89

So when \(|v|\) is prime, you end up with a unit vector.

Answer 90

Yeah exactly so v=<3, 4> has a derivative v' as a unit vector in the same direction

Answer 91

And the anti derivative of any unit vector is simply to multiply it by any prime number.

Answer 92

And the derivative of any unit vector is 0 vector.

Answer 93

Oddly enough. So instead of adding +C it's like multiplying by p^1

Answer 94

I have to wonder now...

Answer 95

For all \(a\), does there exist an \(n\), such that:\[
a^{(n)} = 1
\]Or do loops exist?

Answer 96

Well, loops exist for \(p^p\)

Answer 97

Yeah, and some actually increase for instance 12'=16, 16'=32, etc?

Answer 98

\[
12=2^23 \\
16 = 2^4\\
32 = 2^5
\]