Question

Fake calculus tutorial thingy

Answer 1

Ok so here's our number n and it is made up of numbers multiplied together. \[\Large n =ab\] We can take the derivative and it follows the leibniz rule : \[\Large n' = a'b+ab' \] Now I'll say with respect to any one of these, a'=1, b'=1 so \[\Large n' = a+b \] But what if b was made up of other stuff? This allows us to say that if b=cd then \[\Large n=a'b+ab' = a'cd+a(cd)'= \\ \Large n' = a'cd+ac'd+acd' \\ \Large n' = n \left( \frac{1}{a}+\frac{1}{c}+\frac{1}{d} \right)\] just to give some alternate ways to play around with this. Now that's probably normal to you so far, but here's where my stuff starts to come in.

Answer 2

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Answer 3

So let's get into something interesting, I'm going to introduce double and triple derivatives that are separate from doing a single derivative. But they are linked. To show what I mean, I will use parenthesis to show single derivatives separate from double derivatives. \[\Large n'' \ne (n')' \text{ one double is not two single}\]\[\Large n''' \ne ((n')')' \text{ one triple is not three single}\]\[\Large (n'')' = (n')'' \ne ((n')')' \ne n''' \text{ order doesn't matter!}\]

So what's a double or triple derivative? Simple, instead of taking one out, it takes two or three out at a time: \[\Large n = abc \\ \Large n'' = (ab)''c+(ac)''b+(bc)''a = c+b+a\]\[\Large n''' = (abc)''' = 1\] So now let's test the rule above and show that this is a linear operator \[\Large (n'')' = (n')''\] \[\Large n' = ab+bc+ac \\ \Large (n')'' = (ab+bc+ac)'' = \\ \Large (ab)'' +(bc)''+(ac)'' = 3\] \[\Large n''= a+b+c \\ \Large (n'')' = (a+b+c)' = a'+b'+c' = 3\] Interesting to note here is \[\Large (n'')'=(n')''= 3n'''\] And this is quite interesting.

There is a nice set of relationships going on that I will reveal soon. =)

Answer 4

The first relationship that is nice is this: \[\Large n=ab \\ \Large n^{[k]}= \sum_{i=0}^k a^{[k-i]}b^{[i]}\] What does this mean? These square brackets represent the number derivative inside, so look at an example: \[\Large n'''' = a''''b+ a'''b'+a''b'' +a'b'''+ab''''\] Simpler than the binomial theorem, and really easy to remember.

The second relationship I'd like to share is this \[\Large \left(a^b \right)^{[k]}=\binom{b}{k}a^{b-k}\] This can easily be shown by the derivative rules explained earlier, but I will state it here for fun. A quick example: \[\Large (a^5)''''=a+a+a+a+a=5a=\frac{5!}{(5-4)!4!}a^{5-4}\]

The third relationship I'd like to share is quite important. It uses the multinomial theorem. I will write it without multiindicies to keep it simple, but know we could add further than i,j and k. \[\Large \binom{i+j+k}{i,j,k}n^{[i+j+k]} =((n^{[i]})^{[j]})^{[k]}\] Her is an example of using it: \[\Large \binom{1+2}{1,2}(abcd)^{[3]} = \frac{3!}{1! 2!}(a+b+c+d) \\ \Large ((abcd)'')'=(ab+ac+ad+bc+bd+cd)' = \\ \Large (ab)'+(ac)'+(ad)'+(bc)'+(bd)'+(cd)' = \\ \Large 3(a+b+c+d)\] The more important part of this formula is not so much this but that it also allows you to see that since addition is commutative, so is the order of differentiation.

The fourth relationship is probably my favorite this: \[\Large f(x)= \sum_{i=0}^k n^{[i]}x^i\]\[\Large f(-x)=\prod_{i=0}^k(r_i-x) \\ \implies f(0)=n\]\[\Large n=abc \\ \Large f(x)=n+n'x+n''x^2+n'''x^3 \\ \Large \text{ roots of f(-x) are a,b, and c.}\]

Answer 5

So yeah, that's about it. Basically these are all rules for playing around with coefficients on a polynomial in a way that feels a lot like calculus.

Answer 6

So if I know all the roots of a polynomial as p,q,r,s,t,u,v and want the coefficient on the x^4 term I could do this: \[\Large (pqrstuv)''''\] That's the answer. Well ok, so I haven't calculated it, but that's what all that stuff above is describing. =P

Answer 7

One way to calculate this using the rules is notice: \[\Large \binom{1+1+1+1}{1,1,1,1}n''''=(((n')')')'\] and work it out this way by just going through and taking out one at a time. Of course that's going to be a ridiculous method for this problem but you could do it. You should just take out all possible combination of 4 terms: \[(pqrstuv)''''=pqr+pqs+pqt+pqu+pqv+prs+prt+puv+ \cdots\] Or get a computer... lol.

Answer 8

So yeah, I'm pretty much done here, I am hoping someone asks a question or something about this nonsense, like how to prove these or how it works, or maybe help me come up with some fun problems to solve with this method. There's a lot of interesting tools here to play with, I don't know how useful any of it is, but I thought it was fun coming up with this.

Answer 9

I realize I might have left out some important details in my explanation since there's so much going on, so if you're stumped just say so.

Answer 10

just trying to understand the difference between "double derivative" and taking derivative two times : \[\Large n=ab \\ \Large n^{[k]}= \sum_{i=0}^k a^{[k-i]}b^{[i]}\]

for \(n = 5*7\), do we have \(\large n^{''} = 0\) and \(\large (n')' = 16\) ?

Answer 11

Ah well there really aren't any "numbers" plugged in until the end, but supposing we do plug them in: \[\Large n=5*7 \\ \Large n' = (5*7)'=5'*7+5*7' = 7+5=12 \\ \Large n'' = 1 \\ \Large (n')' = (7+5)'=7'+5'=2\]

So when I calculate the double derivative you take two things and make them 1, so that's why I was able to instantly turn it into 1. However suppose we didn't know 5 or 7 were the final things and perhaps they were composed of smaller factors that were differentiable, we can use that formula to write: \[\Large (5*7)'' = 5''*7+5'*7'+5*7''\] There's no harm in this since we see 7''=0, 5''=0 and 5'=7'=1

Answer 12

Also, we don't have to restrict ourselves to primes. The numbers can be anything.

Also, the polynomial that goes with this problem you've just come up with is: \[\Large (x+5)(x+7)=(5*7)+(5+7)x+(1)x^2 = \\ \Large n+n'x+n''x^2\]

Answer 13

one sec, basic doubt before moving to advanced stuff
\((12)' = (2^2*3)' = 2^2*3 + 2^2*1 = 16\) right ?
how did you get \(((5*7)')' = (5+7)' = 2\) ?

Answer 14

Ahh, cause it's a linear operator. I prefer it if we didn't use numbers since it confuses things and might lure you into adding them together at some point. \[\Large ((ab)')'=(a+b)' =(a)'+(b)'=1+1=2\]

Answer 15

For example, we can confuse an coefficient with the actual numbers themselves so for 12, we would write: \[\Large a=2, b=3\] \[\Large (a*a*b)' = a'ab+aa'b+aab' = 2aa'b+a^2b' \\ \Large 2ab+a^2\] Or alternatively in a single step using the power rule \[\Large (a^2b)'=2ab+a^2\] Plugging in, you are correct, but notice the distinctions between the 2 as a coefficient and 2 as our number of differentiation. \[\Large 2*2*3+2*2=16\]

Answer 16

\((a+b)' \ne a' + b'\) right ?

Answer 17

The thing is, this simplifies down to the arithmetic derivative for when we are using primes since we can always factor a number down unambiguously, but we don't need to restrict ourselves to "differentation with respect to single primes" which is what this is expanding on. The problem with the arithmetic derivative is it is unpredictable for its second derivatives and higher except in special circumstances, and it lacks any real utility. So in the "traditional" sense of doing: \[\Large 12''=16'=32\] This does not exist in this version because this requires "plugging in". My version here by contrast will always eventually terminate and give a derivative of 0 if you go high enough. \[\Large (2*2*3)'= 2*2+2*3+2*3=16 \\ \Large (2*2*3)'' = 2+2+3=7 \\ \Large (2*2*3)'''=1 \\ \Large (2*2*3)''''=0\]

Answer 18

Nope, (a+b)'=a'+b'

Answer 19

I don't see how thats possible, are we still playing by the rules of arithmetic derivatives ?

Answer 20

(5+7)' = 5'+7' = 1 + 1 = 2

(12)' = (2^2*3)' = 2^2*3 + 2^2*1 = 16

2 \(\ne\) 16

Answer 21

Not exactly, this isn't really arithmetic derivatives anymore.

Answer 22

Ohk.. let me scroll up read again lol i see why im getting confused

Answer 23

I understand your confusion. Take this to be the definitions of the derivatives: \[\Large (a+x)(b+x)=n+n'x+n''x^2 \\ \Large ab=n \\ \Large a+b=n' \\ \Large 1=n''\]

Answer 24

Further, check to see that this aligns as it should, this will make understanding the similarities easier I think. \[\Large (a+x)(b+x)(c+x)=n+n'x+n''x^2+n'''x^3 \\ \Large abc=n \\ \Large ab+ac+bc=n' \\ \Large a+b+c=n'' \\ \Large 1=n'''\]

Answer 25

interesting, you're connecting vieta's formulas with "arithmetic derivatives" xD

Answer 26

what's vieta's formulas?

Answer 27

http://mathworld.wolfram.com/VietasFormulas.html

Answer 28

i haven't completed reading above..

Answer 29

Ok I'll check it out while I read into this, might make some of this stuff useful or simpler haha