The Differential Operator (1 of 2: Introduction to notation)

[HannahBarta]

I love the way you explain concepts. I just got my bachelor's degree in mathematics (at a university in America), and I really appreciate going over some of the basics with you again!

[frenkyb]

I know a lot of people say this, but I will say it again: I would to have a teacher like you when I was still in school. I still have interest in math and your explanations are wonderful. Your positive energy and positive attitude makes things just wonderful to watch. Thanks a lot, Eddie.

[TomAustinIII]

Seeing math clearly and recognizing the significance of specific parts of a function have helped me to invest better

[rrrbb]

I prefer the Johnathan Bartlett approach, because it's vastly more amenable to automated computation; of the kind that shows up in AI, which does implicit diff on a large number of variables. Go even further, and define the operator d[] as an actual function for implicit differentiation. Never combine them like (d/dx). Apply implicit diff FIRST. You get a new identity that is usually overlooked. Standard second derivative Leibniz notation is literally wrong; and that's where all the confusion arises: d[C] = 0 is pronounced "C is constant". d[x] = C is pronounced "x is a line with slope C". d[d[x]] = d^2[x], because d[] is a real implicit diff operator. When d^2[x]=0 it is pronounced "x is a line". d[y]/d[x] means that you are dividing the implicit diff of y by the implicit diff of x. BEWARE OF ASSUMING THAT d^2[x]=0. Your second derivative comes out wrong when it is not. So, the true second derivative must be calculated in a nested manner. Unfortunately, it's kind of nasty. But you are back to a notation in which the algebra works correctly. You can even invert to find d[x]/d[y] with no problem when you do it this way.

d[ d[y] / d[x] ] / d[x] = d^2[y]/(d[x]^2) - (d[y]/d[x]) * ( d^2[x]/(dx)^2 )

Note the subtracted term, that is proportional to the first derivative! This means that you cannot call the following "the second derivative of y with respect to x":

d^2[y] / (d[x]^2)

Because it assumes that d^2[x]=0, which means "x is a line". It's usually the case for f[t] that t is a line. But d[] is a true operator on binary operations:

d[a + b] = d[a] + d[b]
d[a - b] = d[a + (-1 * b)]
d[a * b] = b * d[a] + a * d[b]
d[a / b] = d[a * (b^{-1})]
d[a ^ b] = b * a^{b-1} * d[a] + log_e[a] * a^b * d[b]
d[ log_a[b] ] = COMPLEXUNUSED d[a] + (1/b * 1/log_e[b]) d[b]
S[d[f]] = f - f0

That is basically ALL of calculus, because you can derive all the special cases. It works very naturally with multi-variables.

I left out the complicated log case where the base of a log is not constant; but I think I got it calculated right elsewhere, for completeness -- BECAUSE COMPUTERS WANT TO RECURSE DOWN BINARY OPERATORS. I think it's omission is a bug in the notation IMHO. Integration is just the act of making Sd cancel, kind of like inverses.

The point of this is to make this algebraic to facilitate automatic manipulation. It's a different point of view in a lot of ways:

assume:
d[t] = 1
d^2[t] = 0
d[g] = 0
d[v] / d[t] = g
d[p] / d[t] = v
p : meters
t : seconds

precompute anti-derivative:

if d[n]=0, m+1 = n:
d[x ^ n] = n * x^{n-1} d[x]
= d[x^{m+1}] = (m+1) x^m d[x]
1/(m+1) d[x^{m+1}] = x^m d[x]

then this is all blind algebra that a computer can do. Note that integration is making Sd cancel as an explicit goal.

d[v] = g * d[t]
S d[v] = S[ g * d[t]]
= g S[ t^0 * d[t]]
= v - v_0
= g S[d[t^1]]
= g t
v = v_0 + g t

d[p] = v * d[t]
= (v_0 + g t) * d[t]
S d[p] = S[ (v_0 + g t) d[t] ]
= v_0 t + g S[ t d[t]]
= v_0 t + g S[d[(1/2)t^2]]
= v_0 t + (1/2) g t^2
= p - p_0
p = p_0 + v_0 t + (1/2) g t^2

Of course the definition of the d[] operator needs a limits justification, but the notation should be stripped of ambiguity so that computer code can work with it (and also more easily handle discrete cases along-side it).

Multi-variable calculus, and differential equations should not be significantly harder than the core calculus. It's the notation that makes it hard. Computers are good at expanding and collecting terms, and the notation should facilitate working this way. If you ever start to work on a machine learning project, you will literally be doing million-variable calculus via Autodiff algorithms that tweak a measured error, to guide the AI towards the desired results.

[thewhyquestion]

Thank you sir your knowledge is awesome

[memo]

Thanks a lot for explaining that it isn't really a fraction but we're just treating it like so. My math teacher always says it's not a fraction yet when we do algebra he treats it like one and it was confusing me a lot.

[MrAnperm]

What’s the best way to go about self learning maths to be able to study engineering here in Australia? At the moment I’m relying on one text book and YouTube videos. I’ve been out of school for years and have forgotten most of the maths I learned in high school. All but arithmetic.

I feel I need some learning structure because otherwise I’m going to have plenty of holes in my knowledge before undertaking an engineering degree.

[fredrickmutisya]

This looks differential calculus not ODE.., I like the explanation and elaboration.

[TheFarmanimalfriend]

The dy/dx is an image. As such it can not be manipulated like a normal mathematical expression.

[aerrhh]

Do you teach differential and integral calculus!?

[aizhanserik]

Are the differencial operators monotone or antimonotone?