/sci/ - Science & Math

[Anonymous No.10163279]

Post mathematical epiphanies you've had.


I haven't had any so far. i learn math by brute force.

[Anonymous No.10163298]

>>10163279
If the probability of dying per unit of time λ approaches 0 as t approaches ∞, it's possible to live forever.

[Anonymous No.10163366]

>>10163279
the fact that the more money you have, the easier you have it in life. which is obvious, but it becomes even more obvious if you experience it (say, by being paid high wages)

[Anonymous No.10163377]

>>10163279
After calculating the facts
Adding the evidence, subtracting the lies
the truth equaled, that Adolf Hitler was right !

[Anonymous No.10163384]

>>10163377

[Anonymous No.10163386]

>>10163384
>>10163377

[Anonymous No.10163389]

>>10163279
Think of the mapping between real numbers [math]x \mapsto x^2 [/math]. If we slightly change the input, say, a [math]dx[/math] amount, we can say that the same happens to [math]x^2[/math]. Our objective is finding the ratio of the output change and the input change, when the input is changed in the slightest way possible.

This is, [math]x + dx \mapsto x^2 + d(x^2)[/math]

But we know that [math]x + dx \mapsto (x+dx)^2 [/math]

Therefore we have come to the equality [math](x+dx)^2 = x^2 + d(x^2)[/math]
[math]x^2 + 2xdx + (dx)^2 = x^2 + d(x^2)[/math]
[math]2xdx + (dx)^2 = d(x^2)[/math]

Clearly, we don't want our slight change to be exactly zero. Therefore we can divide by it without problem. We end up with

[math]2x + dx = \frac{d(x^2)}{dx}[/math]

Here's when the fun part comes. Here, we got the ratio between the change of the output caused by the (slight) change of the input. Let's make the change of input the slightest possible. Mathematically you say that [math]dx \rightarrow 0[/math]. Since [math]d(x^2)[/math] depends on [math]dx[/math] we're not able to say that [math]\frac{d(x^2){dx}[/math] goes to infinity as [math]dx[/math] goes to zero. But we can say that [math]2x + dx[/math] approaches to [math]2x[/math]

Therefore, [math]\frac{d(x^2)}{dx} = 2x[/math]


Obviously, this is just a rephrasing of derivatives. But at a time where everyone tried to teach me the fabulous 'geometrical interpretation' and saying stuff like "[math]dy[/math] and [math]dx[/math] are just symbols!! [math]\frac{dy}{dx}[/math] is not a fraction per se!", I came up with this idea from which I started deriving everything and still keeping the intuition fresh. Everytime I used this to explain derivatives to people they would understand immediately (with a medium grasp of what functions are obviously). Later on I got to know 'Calculus Made Easy' which explains this in the same way lol.

[Anonymous No.10163401]

I have to admit that the main problem with this method is that you end up with different results depending on the moment you let [math]dx[/math]
approach zero. You could end up getting that the derivative of sin(x) and e^x is zero. Fuck.

[Anonymous No.10163402]

>>10163279
literally everything can be described by numbers. Every time i think about it, i see that green numbers from matrix everywhere

[Anonymous No.10163424]

This one triggers sci but it's true none the less.

The speed of light in a vacuum is variable depending on the local gravity.

This is the exact opposite of mainstream science which declares that gravity bends "spacetime" which isnt a actual real thing. I say the physical path of light curves in reaction to gravity.

[Anonymous No.10163436]

>>10163424
All light travels in a straight line... Sorry?

[Anonymous No.10163441]

>>10163424
Gravity doesn't change the speed of light. It changes the path and the frequency.

[Anonymous No.10163514]

>>10163279
You can triangulate a square, but you can't square a triangle.

D:

[Anonymous No.10163516]

PI = 4 in kenimatic situations like an orbit or lap of a track.

[Anonymous No.10163518]

>>10163389
lmao this is a much better explanation than any other I've heard.

[Anonymous No.10163525]

>>10163389
Quality post. Pretty sure that's how Leibniz did it.

[Anonymous No.10163533]

>>10163298
but the probability of having not died by that point also approaches 0

[Anonymous No.10163556]

>>10163389
Pre-calc brainlet here. Wouldn't [math](x+dx)^2[/math] also square the input [math]d[/math] and change the value of the output?

[Anonymous No.10163558]

>>10163279
. The dual basis is there to help make do with the absence of orthonormal bases in plain vector spaces. In a euclidean space E, an orthonormal basis is one that corresponds to its dual basis via the identification of E with E* given by the inner product.
. Any character of a group actually factors through its abelianization, hence for a nonabelian group, one cannot hope to get all the information about it via its characters. That is why people are interested in representations of dimension > 1.

[Anonymous No.10163582]

>>10163401
I love my explanation on the derivative but I don't want it to fall apart on these cases. If any anon knows what's happening I'd appreciate having help on this. For example
[math]x \mapsto e^x[/math]
[math]x + dx \mapsto e^{x+dx} = e^x + d(e^x)[/math]
[math]e^x \cdot e^{dx} = e^x + d(e^x)[/math]
taking [math]dx \rightarrow 0[/math] we end up with
[math]e^x = e^x + d(e^x)[/math]
[math]0 = \frac{d(e^x)}{dx}[/math]
We didn't end up with any "indeterminate" forms as far as I know so the procedure should be correct
But if you were to do this instead
[math]e^x \cdot e^{dx} - e^x = d(e^x)[/math]
[math]e^x \cdot (e^{dx} - 1) = d(e^x)[/math]
[math]e^x \cdot \frac{(e^{dx} - 1)}{dx} = \frac{d(e^x)}{dx}[/math]
Taking [math]dx \rightarrow 0[/math] now yields
[math]e^x = \frac{d(e^x)}{dx}[/math]

What the fuck?

>>10163556
[math]dx \neq d \times x[/math]. It should be read as a term on its own, not as a multiplication, if that's what you mean. I'm not so sure what's your point though

[Anonymous No.10163583]

>>10163556
d is an operator that means "a very small piece of", not a variable

[Anonymous No.10163589]

>>10163582
>>10163583
Oh, that makes sense now

[Anonymous No.10163610]

P=NP
N = 1

[Anonymous No.10163633]

I was thinking about inscribed polygons, which I'd never heard of before. I realized that all rectangles were inscribed polygons, which led to me figuring out Thales's theorem.

[Anonymous No.10163654]

>>10163389
>>10163582
How is that derivative better than
[eqn]\frac{df}{dx} = \lim_{dx \to 0}\frac{f(x+dx)-f(x)}{dx}[/eqn]
It's literally the same idea of taking the limit as the change in x goes to 0 of the ratio of a change in f(x) over change in x, except it actually works. How are you gaining more intuition when it's the same idea?

[Anonymous No.10163662]

>>10163654
I didn't say it's diffferent. I just said it's a rephrasing. Most times that definition comes from a geometrical interpretation that didn't suit for me at the time.

You can define it that way and then plug in your favourite f(x) function, or you can calculate the derivative "from the beginning" of the intuition process as I pointed out.

[Anonymous No.10163671]

[math]\frac{d(\exp{(dx)}-1)}{dx}\ \ \text{as}\ \ \text{d}x\to0[/math] is undefined. You're basically just using the definition of the derivative in very notationally abusive way.
So:

[math]f(x):x\mapsto\exp(x)[/math]

by definition, [math]\text{d}f(x)=f(x+\text{d}x)-f(x)[/math] where [math]\text{d}x[/math] is very small (it will go to zero).

So, [math]\frac{\text{d}}{\text{d}x}\exp{(x)}=\frac{\exp(x+\text{d}x)-\exp(x)}{\text{d}x}=\frac{\exp(x)\exp(\text{d}x)-\exp(x)}{\text{d}x}=\exp(x)\frac{\exp(\text{d}x)-1}{\text{d}x}\ \ \text{as}\ \ \text{d}x\to0[/math]

This IS an indeterminate form.

I'll leave it to you to show that [math]\frac{\exp(z)-1}{z}=1\ \ \text{as}\
z\to0[/math], which would complete the proof that [math]\frac{\text{d}}{\text{d}x}\exp{x}=\exp(x)[/math]. Hint: use the definition of Euler's number and make a substitution.

[Anonymous No.10163706]

>>10163662
>I didn't say it's diffferent
Oh, so you thought the way you showed is equivalent? It's not, it's different and it doesn't work
>Most times that definition comes from a geometrical interpretation that didn't suit for me at the time.
No? It's the exact same intuition process as the one you said: "the ratio of the output change and the input change, when the input is changed in the slightest way possible"
>You can define it that way and then plug in your favourite f(x) function, or you can calculate the derivative "from the beginning" of the intuition process as I pointed out.
I can work out the derivative of a function "from the beginning" using only intuition and capture the correct definition: You take a small change in the output resulting from a small change in input, and divide it by that small change in input. Then you see what happens when that small change in input goes to 0. How is that not the exact same intuition process as the one you showed?

[Anonymous No.10163719]

>>10163389
Reminds me of a "Primer on Infinitesimal Calculus"

[Anonymous No.10163732]

>>10163298
Taking a non finite set defined as humans who are living as you do, there will be a time where the probability will shrink the set to 0

[Anonymous No.10163752]

>>10163582
Dude, for the second one, you get e^dx - 1 = 0 and you get the same reult as your firt try

[Anonymous No.10163763]

>>10163279
I've discovered thousands of things:
The gradient derivative
Topology (from the analysis of non-euclidean geometry)
The algebra of Regular Expressions and their correspondance to graphs
A theory to find the totality of all paths of an automata
The bayes theorem
The unit cube
Ways to color a specific graph
Ways to color the cube
The fundamental theorem of calculus
The summation of the simplest arithmetic progression
A recursive equation for the summation of the powers of the simple arithmetic progression
Interpenetration of planes and three-dimensional shapes
The laws of modular arithmetic
The partial derivative
The classification of three-dimensional symmetric groups
Certain things about cycles
Some basic stuff about statistics (like correlations)
Stuff about knot theory
A theory about probability
Some analysis about the properties of the mandelbrot set
A relationship between some probability problems and turing machines
A relationship between abstract rewriting machines and the universal property of groups
Countless of solutions to silly puzzles in video games and the like.
Though you know honestly I don't read much about math or have a math education, I just find some interesting problem and come up with several creative solutions to the problem (problems like those found in some video game (as in Zelda) or the game of life or a simulator of fractals or in a website like this one or just for the sake of making some joke [like the time I tried to formalize the real numbers because it would be funny to come up with an enormous mathematical structure just to prove that something like 1 + 1 = 2])

[Anonymous No.10163768]

>>10163582
This is actually fairly easy to remedy, just use the definition of the exponential function.
[math]e^x \cdot e^{dx}=e^x\cdot {\displaystyle \sum _{k=0}^{\infty }{\frac {(dx)^{k}}{k!}}}[/math] so [math]e^x \cdot e^{dx} - e^x= e^x(e^{dx}-1)=e^x\left({\displaystyle \sum _{k=0}^{\infty }{\frac {(dx)^{k}}{k!}}}-1\right)=e^x\left({\displaystyle \sum _{k=1}^{\infty }{\frac {(dx)^{k}}{k!}}}\right)[/math]
Dividing this all by dx we get that
[math]e^x\left({\displaystyle \sum _{k=1}^{\infty }{\frac {(dx)^{k-1}}{k!}}}\right)[/math] as we take dx to be zero all terms in the series go to zero except the leading term 1, this completes what you wanted to show.

[Anonymous No.10163779]

>>10163768
>the definition of the exponential function
but that's not the definition, anon. The Maclaurin of e^x is an identity that needs to be proven, not a definition. The definition of [math]\exp(x)[/math] is Euler's constant, that is, [math]\lim_{n\to\infty}(1+
frac{1}{n}^{n}[/math] raised to the [math]x[/math] power.

[Anonymous No.10163818]

>>10163389
>>10163768
You know what, there is something I quite like about this way of doing things, it motivates the definition of continuity. Let me explain. In order for [math]\frac{d(x^2)}{dx}[/math] to converge to something reasonable as [math]{dx}[/math] goes to zero, we need to put some conditions on [math]d(x^2)[/math] dependent on [math]{dx}[/math]. This is necessary since we want to extract some finite number from [math]\frac{d(x^2)}{dx}[/math] so we need to understand what [math]d(x^2)[/math] does as [math]{dx}[/math] goes to zero. Well, obviously we want [math]d(x^2)[/math] to go to zero, so using our definition of limits that means that for every [math]0 < \epsilon[/math] we have that [math]|d(x^2)|< \epsilon[/math] in the limit as [math]dx[/math] goes to zero. But hang on, we've just used another limit, so if we use the definition of what a limit means for [math]dx[/math] to go to zero it means that [math]|dx|< \delta[/math] where [math]\delta[/math] is some positive real. So tying this all together, in order for [math]\frac{d(x^2)}{dx}[/math] to make sense we need [math]d(x^2)[/math] to go to zero as [math]dx[/math] goes to zero. This means that [math]|d(x^2)|< \epsilon[/math] subject to the constraint that [math]|dx|< \delta[/math] for some [math]\delta[/math]. We may call this necessary condition continuity. But it is not sufficient, take for example the function [math]|x|[/math] at zero, this is not differentiable. I think this perspective might be useful. People are familiar with polynomials, they're easy, but a generic continuous function may be pathological as fuck. I think a neat way of motivating continuity may be to start with >>10163389 and then show that in order for this process to work appropriately one must introduce a certain condition that is equivalent to continuity, thus motivating the definition like I've hopefully done above, and then showing it is necessary but not sufficient for a function to be differentiable at a point.

[Anonymous No.10163820]

>>10163279
>Post mathematical epiphanies you've had.
Basically figuring out the geometric interpretation of the determinant. Made linear algebra even more trivial and intuitive.

[Anonymous No.10163833]

>>10163389
>>10163582
Isn't this nonstandard analysis?

[math]\operatorname{st}\frac{(x+\epsilon)^2-x^2}{\epsilon}=\operatorname{st}\frac{2x\epsilon+\epsilon^2}{\epsilon}=\operatorname{st}\frac{2x\epsilon+\epsilon^2}{\epsilon}=\operatorname{st}(2x+\epsilon)=2x[/math]

[math]\operatorname{st}\frac{e^{x+\epsilon}-e^x}{\epsilon}=e^x\operatorname{st}\left(\frac{e^{\epsilon}-1}{\epsilon}\right)=e^x[/math]
since [math]\operatorname{st}\left(\frac{e^{\epsilon}-1}{\epsilon}\right)=1[/math]

[El Arcón No.10163849]

Real Numbers in the Neighborhood of Infinity
http://www.vixra.org/abs/1811.0222

[Anonymous No.10163870]

>>10163377
Basado y rojoempastillado

[Anonymous No.10163877]

>>10163279
I've synthesized sound waves using Pythagoras theorem. They looked like parts of circle on oscilloscope.

[Anonymous No.10163893]

>>10163849
>Trinity
>4 ways

[Anonymous No.10164049]

>>10163279
bump

[Anonymous No.10164084]

>>10163279
Here's one I discovered recently while brushing up on differential equations and calc II material. In Calc II and Diff-eq (which is really just an extension of calc ii anyway), we're taught to mindlessly add the "plus C" and teachers almost universally will tell you that it's just an undecided parameter. WRONG!

It's normally /used/ as an initial condition but that's glossing over what's really going on. It is shown by the fundamental theorem of calculus that the solution to any differential equation satisfies this:

[math]F(b)-F(a) = \int_{a}^{b}f(x,y(x))dx[/math]

And even worse, when using indefinite integrals, notation that's describing what is really happening gets thrown at the window. Say you're integrating some differential equation [math]\frac{dy}{dx}=f(x,y)[/math], you're taught to do this:

[math]y = \int f(x,y(x))dx + C[/math]

But what it really looks like is this:

[math]F(x) = \int_{x_0}^{x} f(x,y(x))dx + F(x_0)[/math] Notation is fucking everything! Not teaching this form and throughly explaining what it's good for just makes more rigorous study down the road that much fucking harder. The state of math education at US colleges, everybody.

[El Arcón No.10164086]

[El Arcón No.10164088]

[El Arcón No.10164091]

[Anonymous No.10164103]

>>10163533
>>10163732
Imaging if there is someone who has a 10% chance of dying in time period 1, a 5% chance of dying in time period 2, a 2.5% chance of dying in T3, a 1.25% chance of dying in T4, etc. They would only have a 20% probability of dying ever.

[Anonymous No.10164105]

>>10164103
*imagine

[El Arcón No.10164108]

>>10164105
*imagine if you put squinty skull head in front of me and I killed him

[No Context Slam Sam *(I slam u(...) No.10164110]

>>10163893
Yes that is how trinities work
>3 things each with something in common with two of the others but not the third
>as a whole, a 4th element results from the combination of these relationships

[Anonymous No.10164294]

>>10163389

why isn't it
[math]
x+dx /mapsto (d+1)^2(x)^2

[/math] ?

[Anonymous No.10164332]

>>10163279
I thought I disproved infinity (or randomness), by proposing that given a random event with infinite outcomes, such as picking a random number, would be impossible because the likely hood of that event happening would effectively be zero, because you have a finite number divided by infinity.

Then I found this shit: https://en.wikipedia.org/wiki/Almost_surely

Fucking stupid mathematicians thought of everything. In my high school brain I thought I was a revolutionary genius.

[Anonymous No.10164337]

>>10164332
This is stats 101. Probability 0 doesn't mean impossible

[Anonymous No.10164342]

>>10163610
underrated

[Anonymous No.10164362]

>>10164337
>Probability 0 doesn't mean impossible
Not true. This is entirely dependent on the cardinality of your sample space.

[Anonymous No.10164388]

>>10164362
I didn't specify the cardinality of the sample space, retard. So given only that the probability of a event is 0, one cannot deduce that said event is impossible

[Anonymous No.10164440]

>>10164337
>>10164362
>anon A asserts that P does not imply Q
>anon B criticizes anon A for asserting that P implies not Q

/sci/ is the best place for math discussion

[Anonymous No.10164445]

>>10164362
tfw [math]¬(P\implies Q) \implies (P \implies ¬Q)[/math]

[Anonymous No.10164453]

>>10163279
Permutations get real fucking big real quick, and the growth in complexity of formulating theories and proofs are astronomically bigger than the human lifespans' rate of growth and the computation power of computers, and we're hitting the wall too hard

[Anonymous No.10164807]

I learned that bump functions are pretty useful.

[Anonymous No.10164835]

>>10163279
Some stuff in basic category theory like Yoneda and friends are a fancy way of saying something is as free as possible subject to some constraint. It makes a bunch of constructrions trivial, but is kind of messy to formalize.

[Anonymous No.10164844]

>>10163849
Based Jon

[Anonymous No.10164985]

>>10163389
Congratulations, you discovered that, when a function maps from R^1 to R^1, it’s usually pointless to graph it on R^2, and that the introduction of geometry serves only to confuse people.

[Anonymous No.10166232]

bump

[Anonymous No.10166259]

>>10163279
You can multiply any two large numbers mentally by halving one and doubling the other until you get a number multiplied by two. Then add. The whole thing is addition
I know, very profound

[Anonymous No.10166282]

>>10163377
[math]Hitler\neq Wrong[/math]

[Anonymous No.10166291]

How do I learn to do mathematics as a complete beginner?

[Anonymous No.10166339]

>>10164985
>implying algebra is more fundamental than geometry
t. mindless brainlet symbol pusher

[Anonymous No.10166533]

>>10166291
>How do I learn to do mathematics as a complete beginner?

By asking people how to learn mathematics on 4chan instead of going on YouTube and starting somewhere.

[Anonymous No.10166538]

>>10166339
You're full of shit. No one does math geometrically.

It's all numbers

[Anonymous No.10166617]

>>10166538
they should, everything is geometry, retard.

[Anonymous No.10166639]

>>10166617
you can't do 0+1 geometrically idiot.

[Anonymous No.10166645]

>>10164294
d is a operator on x, not a variable

[Anonymous No.10166648]

>>10164332
P(x)=0 does not mean its impossible

[Anonymous No.10166654]

>>10166291
>calc 1-3
>linear alg
>differential equations
>intro to stats
>real analysis
>complex analysis
>topology
>probability theory
>abstract algebra
>statistical inference

are the basics in order of what to cover

then you decide which area you liked the most and focus on it

[Anonymous No.10166678]

>>10166645
>d is a operator on x

what does this mean?

[Anonymous No.10166691]

>>10163279
I realized that
[math]
\sqrt{ x }
[/math]
Can only ever be positive. Blew my fucking mind. So glad my employer is paying for this class.

[Anonymous No.10166699]

>>10166678
d is an operator that means a very very small part of x

separating the d from x makes it meaningless

[Anonymous No.10166702]

>>10166691
>sqrt(x)
>Can only ever be positive

??????

[Anonymous No.10166998]

>>10166691
I think that every root of positive numbers have two solutions of exact same value but on different side from 0, prove me wrong.

[Anonymous No.10169077]

>>10166639
I mean, vector spaces encode geometric information, and the real line is a vector space, so technically.

[Anonymous No.10169099]

>>10163389
You'll find that explanation in section of implicit derivation

[Anonymous No.10169105]

>>10163556
"dx" is a quantity as he said it's not a dot product

[Anonymous No.10169168]

>>10163366
workaholics can have lots of money, but they still put long hours in the office.

[Anonymous No.10170332]

>>10163633
Nice.

[Anonymous No.10170336]

The reason why the normal distribution has a well defined mean and the Cauchy distribution doesn't is the following: if you take the average of a sample from a normal distribution, that value will again be a normal distribution with the same mean, but rescaled standard deviation. The stddev goes down with the sqrt of the sample size, until you reach an arbitrarily narrow distribution.

If you try to do the same with the Cauchy distribution, the average of the sample is distributed as the exact same Cauchy distribution. The width stays exactly the same. Thus taking a single sample or averaging a billion samples gives you an exactly equally bad estimate of the center of the distribution.

[Anonymous No.10171136]

Not that much of an epiphany but even so: when I was taught group actions on sets and in particular (the so called) Burnside's lemma I felt I did not undestand it. It took me a little while to figure out something like the action groupoid, and drawing few sets of points with loops around them made me visualize and realize that the actions of groups are very strongly structured. Next thing I knew I was in love with the concept, and I tried (without success) to develop baby group theory (for example, first iso theorem) in that language. I hoped to get isomorphisms of groups instead of bijections but I failed, and even trying with actions of groups on groups was unsuccessful.
Another one was me trying to generalize the lifting of curves (and boxes) along covering maps, I was not so successful (I was getting to CW-complexes without knowing what they were) as I learned after a short while the general criterion. The existence of the universal cover was something strange until I decided to consider S^1, then it became clear.
Most of attempts like those revealed to be dead ends, but it was fun and I learned something. Sometimes instead it was successful, for example giving an analytic proof (of a lemma) instead of a geometric argument, the latter being not so rigorous or correct. I feel good when I finally gather the courage to attempt another proof, or follow a proof on easy examples. But this is something everybody does (right?), not so worth of a clap.

[Anonymous No.10171236]

>>10166691

are you retarded? sqrt(9) = +- 3

[Anonymous No.10171366]

>>10163763
Apparently only me and El Arcon have had non-basic discoveries (that is, not just calculus-based)