/sci/ - Science & Math

>>10634550
Why do some 17 year olds look like this.

And others look like pic related?

Complete formalizations are usually not particularly readable.
There are many libraries, the biggest being Mizar (a more or less dead project now)
http://en.wikipedia.org/wiki/Mizar_system
But to interpret the syntax you'll heave to study some first-order logic.
You shouldn't read Mizar or any other formal library, but learning logic is great, of course.
Here is a fairly comprehensive literature overview to introductory logic:
http://www.logicmatters.net/resources/pdfs/TeachYourselfLogic2015.pdf

>well accepted elementary
Given you understand the language, some direct axiomatizations, e.g. of the theory of groups or rings, is extremely well understood and has numbers as a 'model'. Same with natural numbers and arithmetic (google Peano arithmetic). But then again, foundational math, trying to set up an encompassing framework, will redefine (or model) the idea of groups, ring or numbers.
Just do some logic and that will make sense.

That being said, the philosophy behind the foundations of mathematics, and the many approaches to it, are very very very far from being agreed upon.
Working mathematicans are mostly agnostic and accept the set theory written down in first order logic foundational framework, which established itself around 100 years ago.

>decimals
Decimals come pretty late, from the above perspective. Computational matters, or questions of representation, are generally often treated as an orthogonal subject. Extremely related, but somewhat of a subject of its own.

No, I don't.

Most people are plebs, and there being more clever people isn't negative - the docrine for status and success is a matra of our society producing productive and efficient drones, the clever people are generally not at all the happy ones. Enjoy your subject.

There is the "Fermat-observation" that you can algebraically differentiate polynomials in the sense that if p is the function, and you define

p(x;h) := (p(x+h)-p(x))/h

then

p'(x) = p(x;0)


e.g. if
p(x) := 2 x^2 - x + 3
then
p(x+h)-p(x) = h ((4x-1)+2h)
so that
p(x;h) = (4x-1)+2h

We don't need to take the limit, we can just set h=0.
Now it also works on
f(x) = 1 / (a+bx)
but I couldn't find much more examles.

In particular higher powers alla 1/x^n spoil it and functions like sin or exp are not approachable, as I don't have Taylor expansion available.

Can anyone characterize the expression for which this worls?

This relates to Hadamard's lemma.