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10163279 No.10163279 [Reply] [Original] [archived.moe]

Post mathematical epiphanies you've had.

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

>> No.10163298
File: 117 KB, 1200x1400, immortalitygraph.png [View same] [iqdb] [saucenao] [google] [report]

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

>> No.10163366

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)

>> No.10163377
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After calculating the facts
Adding the evidence, subtracting the lies
the truth equaled, that Adolf Hitler was right !

>> No.10163384
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>> No.10163386
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>> No.10163389

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.

>> 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.

>> No.10163402

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

>> No.10163424
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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.

>> No.10163436
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All light travels in a straight line... Sorry?

>> No.10163441

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

>> No.10163514

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


>> No.10163516

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

>> No.10163518

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

>> No.10163525
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Quality post. Pretty sure that's how Leibniz did it.

>> No.10163533

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

>> No.10163556
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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?

>> No.10163558

. 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.

>> No.10163582

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?

[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

>> No.10163583

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

>> No.10163589

Oh, that makes sense now

>> No.10163610

N = 1

>> 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.

>> No.10163654

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?

>> No.10163662

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.

>> No.10163671
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[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.


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.

>> No.10163706

>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?

>> No.10163719

Reminds me of a "Primer on Infinitesimal Calculus"

>> No.10163732

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

>> No.10163752

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

>> No.10163763

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])

>> No.10163768

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.

>> No.10163779 [DELETED] 
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>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.

>> No.10163818

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.

>> No.10163820

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

>> No.10163833

Isn't this nonstandard analysis?


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

>> No.10163849
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Real Numbers in the Neighborhood of Infinity

>> No.10163870

Basado y rojoempastillado

>> No.10163877

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

>> No.10163893

>4 ways

>> No.10164049


>> No.10164084

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.

>> No.10164086
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>> No.10164088
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>> No.10164091
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>> No.10164103

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.

>> No.10164105


>> No.10164108

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

>> No.10164110

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

>> No.10164294


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

[/math] ?

>> No.10164332

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.

>> No.10164337

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

>> No.10164342


>> No.10164362

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

>> No.10164388

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

>> No.10164440 [DELETED] 

>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

>> No.10164445 [DELETED] 

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

>> No.10164453
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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

>> No.10164807

I learned that bump functions are pretty useful.

>> No.10164835

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.

>> No.10164844

Based Jon

>> No.10164985

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.

>> No.10166232


>> No.10166259

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

>> No.10166282
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[math]Hitler\neq Wrong[/math]

>> No.10166291
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How do I learn to do mathematics as a complete beginner?

>> No.10166339

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

>> No.10166533

>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.

>> No.10166538

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

It's all numbers

>> No.10166617

they should, everything is geometry, retard.

>> No.10166639

you can't do 0+1 geometrically idiot.

>> No.10166645

d is a operator on x, not a variable

>> No.10166648

P(x)=0 does not mean its impossible

>> No.10166654

>calc 1-3
>linear alg
>differential equations
>intro to stats
>real analysis
>complex analysis
>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

>> No.10166678

>d is a operator on x

what does this mean?

>> No.10166691

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

>> No.10166699

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

separating the d from x makes it meaningless

>> No.10166702

>Can only ever be positive


>> No.10166998

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

>> No.10169077

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

>> No.10169099

You'll find that explanation in section of implicit derivation

>> No.10169105

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

>> No.10169168

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

>> No.10170332


>> 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.

>> No.10171136 [DELETED] 
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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.

>> No.10171236


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

>> No.10171366

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

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