/sci/ - Science & Math

[Anonymous No.10131635]

How can I discover multivariable calculus and linear algebra on my own? Does anyone know of a collection of motivating problems that could lead me to discover most of what is teached in these courses? What I'm thinking is more or less something like Martin Gardner's Aha! Insight book where there are lots of puzzles that solving them could indeed lead for one to reinvent the wheel and even discover the theories behind such subjects if one backlinks enough and tries to think of different ways to represent the information. I just really dislike being spoiled when it comes to mathematics, I would like to be able to make my own research and discover all of it just for fun's sake. So what do you think?

[Anonymous No.10131682]

>>10131635
So you don't want to be explicitly spoonfed, you want to be implicitly spoonfed?

[Anonymous No.10131701]

It's impossible to do it as is, if you were truly discovering the things for the first time you would look at the concepts differently and give them different names, and you'd have to make your own definitions and such. These fields get established only after some people, usually experts, take the time to make notation, definitions, wording, etc. uniform as to make it simpler and you're learning that simplified version today that doesn't result from the original papers

[Anonymous No.10132052]

>>10131701
Precisely!! I want to make my own concepts, notions, and give the my own names and definitions. And hopefully look at them differently though some previous percursory glances at Wikipedia's pages (especially some of the images) on the topics might kind of bias me. But hopefully there will be opportunities were I will give them my own names and notations. And so even if not completely discovering everything from scratch still, moderately discovering things from scratch.

[Anonymous No.10132068]

>>10131682
Lol underrated

[Anonymous No.10132078]

>>10132068
lmao
>>10131682
More or less.. but not really, it's mostly some problems just to orient myself slightly towards discovering linear algebra and multivariable calculus specifically (like I don't even know what kind of problems to think about)

[Anonymous No.10132129]

>>10132078
>linear algebra
You have a linear system and you want to solve it.
>calculus
Not happening. Fundamental Theorem is the sort of autistic revelation that happens once in a lifetime.
You might be able to come up with differentiation tho.
Imagine you have a ball moving through space, and we have its position at each instant in time. We want to know its speed at each instant.

[Anonymous No.10132137]

>>10131635
But why? Honestly.
The whole point of recording these results is so you DON'T need to fumble around rediscovering them.
Unless you're spooked about shaky foundations like logicians of the early 1900s, don't bother.
>>10132078
>like I don't even know what kind of problems to think about
You're not going to rebuild the theory from scratch if you don't even know what questions you want it answer.

[Anonymous No.10132141]

>>10132129
>Fundamental Theorem is the sort of autistic revelation that happens once in a lifetime.
You could derive it from ordinary summation.
Sum of terms gives partial sum.
Difference of two consecutive partial sums gives a term.

[Anonymous No.10132144]

>>10132129
The fundamental theorem? Hehe are you kidding me? First, that's single variable calculus and I actually inferred it on my own when I was much younger (using summation syntax though). What I'm asking for is mostly multivariable calculus of which I know very little except that it has to do with partial derivatives and that there is some inverted triangle symbol and that it's in three or more dimensions.
> You have a linear system and you want to solve it.
Well I guess that's something I could think about

[Anonymous No.10132146]

>>10132141
No, no, showing it is really easy. Noticing it is weird.

[Anonymous No.10132183]

>>10132137
It's certainly not for utility, I'm just doing it for fun! :) It makes me happy to come up with new theories.
> You're not going to rebuild the theory from scratch if you don't even know what questions you want it answer.
That's why I'm asking your help, I want to have a little bit of guidance on what to search for.

[Anonymous No.10132194]

to discover calculus, start by trying to discover the newtonian laws of physics. for performing this task, newton found sitting under apple trees helpful. perhaps you will find it helpful as well.

[Anonymous No.10132211]

>>10132194
not single variable calculus lol also it's not that difficult to find newtonian laws of physics thanks to the technology of today, one can just look up various simulations of physical phenomena like collisions and work from there to find something like the newtonian laws. (Which I actually did one day while bored though didn't reach newtonian laws but I still reached some interesting results) I'm mostly asking for multivariable calculus, perhaps it can also be related to physics.. I wonder why I didn't think of that, maybe I could look some physics simulation and derive a multivariable calculus theory from that, perhaps one of electromagnetic fields or of waves though I wonder how I'd measure things.. maybe I could use Fiji?

[Anonymous No.10132370]

>>10131635
Douche

[Anonymous No.10132418]

>>10132370
Sure maybe I, OP is a douche, but that doesn't change anything, my point is something that could be quite helpful to all anons, I'm searching for a collection of problems from which one could discover multivariable calculus and linear algebra with enough autonomy, ability to generalize and a good deal of intuition and imagination

[Anonymous No.10134043]

>>10132418
bump

[Anonymous No.10134078]

>>10131635
I don't mean this to sound shitty, but, at least as far as the very basic of multivariable calculus, all the calculus you already know sort of should be a sufficient motivation.

You know how to measure instantaneous rates of change and areas under curves, aren't you naturally curious to extend the ideas of derivatives and integrals to multiple variables?

I'm sure it wouldn't be too hard to contrive a problem that basically "leads" you to these conclusions, but to get started, you should try experimenting with what you think some natural extensions of, say, the derivative to multiple-variables would be.

I think you'll surprise yourself with how close your thoughts could strike to the truth, without any motivating examples.

[Anonymous No.10134392]

>>10131635
>teached

[turd-sucker No.10134400]

Sorry OP but you are simply retarded. IT's a fact. I can prove it philosophically. It's an objective fact. It's a universal. It's one of Plato's ideas, man, your stupidity that is. Your retardation is an abstract idea, I can detect it, I am acquainted with it, you stupid fucker!

you
are
a
bee
drone

[Anonymous No.10134499]

>>10134078
OP here
Well first what would it mean for a derivative to be multi-dimensional? Say you have some balloon and you want to find the derivative at some X, or an approximation of a function f, or the rate of change of that function f in relation to whatever is X. One could perhaps think of a plane as the derivative of a function at a point in a three dimensional balloon. Now of course, it follows that in order to construct such a plane for a balloon one would have to think of how to construct a plane in first place. If a line is y = m(x-x0)+ y0, then similarly a plane should be able to tell us all the points that lie within it, through the modification of its parameters, unlike the simple line, any of its points could be modified in such and such ways so as to lie not just vertically across the spine of the plane but also horizontally across the weft (and hence warp and weft). Maybe from our initial x and y, we could come up with a z, perhaps in some z-y plane in relation to a y0 such that z = n(y-y0) + z0, now since n is always the same yet y0 changes depending on the y curve then perhaps we could rewrite this as: z = n(y-y1) + z0, and the other equation as y1 = m(x-x0) + y0. Then perhaps z = n(y-(m(x-x0)+y0)) + z0; z = n(y-y0) - nm(x-x0) + z0 that is.. z - z0 = n(y-y0) - nm(x-x0); n((y-y0) - m(x-x0)) = z - z0; n = z-z0/((y-y0) - m(x-x0)) = z-z0/(y1-y0). In essence there is a point of origin x0, y0, z0, it has two slopes: n and m. Perhaps you could speak of some function f such that it takes two arguments x and y and produces a z, and you could perhaps take a derivative at xy plane and xz plane and then use this information to get the derivative as 3-dimensional. Though is the derivative really a plane, I guess we could use these ideas to prove that.. (or disprove it). I wonder if there might be a way to simplify this equation: z = n(y-y0) - nm(x-x0) + z0 (and what is its geometric meaning) But well this would only be the beginning (to be continued)

[Anonymous No.10134501]

>>10134392
lmao not american
>>10134400
I would be glad to hear such a proof

[for the pedants out there No.10134510]

>>10134501
I mean native english speaker

[Anonymous No.10134561]

For multivar just try to extend the ideas of calculus into more than one dimension.

For LA think about numbers which exist in more than one dimension and then use the tools you develop to examine those as a method for solving linear systems

[for the pedants out there No.10134585]

>>10134499
Maybe I could do it this way:
Say you have a point at x0, y0, z0; say we're moving on that plane, then you add delta x times m to x0 to obtain a first z, then you add another delta y times n to obtain the second z and hence arrive at that point.. something like z = m(x-x0) + n(y-y0) (and unlike the other case x-x0 is not equal to y)

[for the pedants out there No.10134588]

>>10134561
I'll try!

[Anonymous No.10134591]

>>10132146
This

[Anonymous No.10134695]

Khan Academy is how I learned a lot of interesting mathematics, however, if you want to pursue specific mathematics you're just gonna have to do research.

[for the pedants out there No.10134889]

>>10134499
m and n will be mono-argument derivatives because they both occur as transformations from one change to the other, say from x to y, x to z, or y to z. that is the mono-argument derivative of f(x,y,z) with respect to x is d f(x,y,z)/dx = [f(x+h,y,z) - f(x,y,z)]/h.. And similarly for bigger dimensions. Then there is a multi-argument derivative of f(x,y,z) with respect to x and y which would be d f(x,y,z)/d(x,y) = [f(x+ah, y+bh, z) - f(x,y,z)]/(ah+bh). (it can be any number smaller than the number of arguments but bigger (or equal) to one) And finally the all-argument derivative of f(x,y,z).. d f(x,y,z) = [f(x+ah, y+bh, z+ch) - f(x,y,z)]/(ah+bh+ch). Sounds strange though to divide it by (ah+bh) and (ah+bh+ch), anyway the conjecture is that' the all-argument derivative is equal to the addition of multi-argument derivatives such that the union of the disjoint respects of the multi-argument derivatives equals to the all-argument derivative respects. For example, my idea is something like this:

d f(x,y,z)/d(x,y,z) = a*d f(x,y,z)/d(x,y) + b*d f(x,y,z)/d(z)
I will have to think about how that works... (here the union of the disjoint respect: d(x,y) U d(z), eqauls to the the respects of the all-derivative: d(x,y,z)

Now one can generalize the argument to say that a multi-derivative with respect to R (where R is a set of respects) equals to the addition of multi-derivatives with disjoint respects S, T, ..., where the union is R

[Anonymous No.10134937]

>>10131635
You can't. Not in your lifetime anyway. It took some of the greatest minds our lineage has to offer thousands of years to come up with multi variable calculus for example. If you tried to discover all this shit on your own, you would spend your whole life coming up with proofs for basic concepts.

If you decide to learn the material through textbooks and lectures, you may find that lots of material is simply too difficult.

[Anonymous No.10134956]

>>10132144
If you can infer the fundamental theorem all on your own as a kid, how the fuck can’t you understand how to generalize single variable to multivariable? Much less what a gradient is?

Troll.

[for the pedants out there No.10134968]

>>10134937
I'll prove you wrong! I'll prove you I can find the answers! :}
>>10134956
well I could.. that doesn't mean it took me zero amount of time

[Anonymous No.10134981]

>>10131635
MIT open courseware, 3blue1brown, a couple books on each subject, schaums outlines, tons of pen and paper and hours upon hours of problem solving

[Anonymous No.10135353]

>>10131635
You can do that once you get to the proof based mathematics.

[Anonymous No.10135524]

Try this: imagine you have a ball and a slope from point A to B, what is the shape of the slope that will give the shortest time taken for the ball to go from point A to B?
You might discover the calculus of variations with this question.

[for the pedants out there No.10135893]

>>10134968
>>10134889
Okay, I thikn I might have proven what I stated at the beginning...

(f(x + ah, y) - f(x,y))/h + f(x,y+bh) - f(x,y) = (f(x+ah,y+bh) - f(x,y))/h (what one has to prove) (here sqrt(a^2 + b^2) = 1)..
f(x + ah )- f(x,y) + f(x,y+bh) = f(x+ah, y+bh)
f(x+ah,y) - f(x,y) = f(x+ah, y+bh) - f(x,y+bh)
or the derivative of f(x,y) with respect to x (for ah increment) equals the derivative of f(x,y) with respect to x (for ah increment). Now I wonder how I'll take out the ah and bh from the equation, but see it works with f(x,y) = x*y which derivative is x*y' + x'*y (just like the addition rule, perhaps replacing by f(x) and g(x) instead of x and y, and perhaps since it's ah and bh (instead of h) the chain rule could help, who knows...

[for the pedants out there No.10136125]

>>10135524
I'll try that, though I wonder what might be the time a ball takes for any shape, like say for an inclined plane

>>10134981
I'll look into it

And also what do you think of my "proof"?

[Anonymous No.10136272]

>>10135893
the multiplciation rule* I mean

[Anonymous No.10137264]

>>10135893
I just solved it, as to how to make the x+ah become just x+h...
define a function g(z) = a*z where z = x/a
define a function h(z) = f(g(z), y)
that is h'(z) = (h(z+€) - h(z)) / €
that is h'(z) = (f(g(z+€), y) - f(g(z),y))/€
Or otherwisely using the chain rule:
that is h'(z) = f'(g(z),y) * f(g'(z), y)
now g'(z) = a
that is h'(z) = f'(g(z),y) * a
that is h'(z) = (f'(g(x/a)+€,y) - f'(g(x/a), y))/€*a
or in other words:
(f'(g(x/a)+h, y) - f'(g(x/a), y))/h * a = (f'(x+ah, y) - f'(x, y))/h
or
(f'(x+h, y) - f'(x,y))/h * a = (f'(x+ah, y) - f'(x,y))/h

therefore
(f(x+ah,y+bh,...) - f(x,y,...))/h = (f(x+h,y,...) - f(x,y, ...))/h * a + (f(x,y+h, ....) - f(x,y, ...))/h * b + ...
that is
(f(x+ah,y+bh,...) - f(x,y,...))/h = a*df/dx + b*df/dy + ...
or
df/d(x,y,...) = a*df/dx + b*df/dy + ...

Am I doing it right?

[Anonymous No.10137270]

>>10137264
>that is h'(z) = (f(g(z+€), y) - f(g(z),y))/€
that should be equivalent to:
h'(z) = (f(x+ah, y) - f(g(z), y))/€

>(f'(x+h, y) - f'(x,y))/h * a = (f'(x+ah, y) - f'(x,y))/h

and therefore this is true:
(f'(g(x/a)+h, y) - f'(g(x/a), y))/h * a = (f(x+ah, y) - f(x, y))/h
(without the single quotes in the previous thread)