/sci/ - Science & Math

[Anonymous No.6137526]

e^{i \pi} + 1 = 0

ITT Beautiful math

[Anonymous No.6137528]

>>6137526
Fuck you too LaTeX...

[Anonymous No.6137555]

0.999...=1

[_. srx. _ No.6137559]

Math is so beautiful.

[Anonymous No.6137561]

1/0 = ∞

[Anonymous No.6137577]

1/∞=0
0x∞=1
1/0=∞

n=random number
n/∞=0
0x∞=n
n/0=∞

n=1 , all number are equal

[Anonymous No.6137578]

>>6137577
division with 0 is not defined

[Anonymous No.6137579]

<span class="math"> q=\exp(\theta\mathbf{\hat{u}})[/spoiler]

[Anonymous No.6137583]

>>6137578
>n=random number
>n/∞=0
>0x∞=n
n=1=2=3
still works

[Anonymous No.6137606]

>>6137578
is right.
>>6137561
>>6137577
You can't just divide by or infinity. You can, however, divide by n and take the limit as n approaches 0 or infinity.
lim┬(n→∞)〖1/n〗=0
lim┬(n→0)〖1/n〗=∞

[Anonymous No.6137616]

2+2=4

[Anonymous No.6137632]

>>6137526

all c such that iteration z' = z*z + c remains bounded over the complex plane

[Anonymous No.6137655]

>>6137632
poor tattooer...
it must have been really boring and difficult to do this.

[Anonymous No.6137658]

>>6137606
essentially if we map numbers to sets of equivalent power dividing by zero meant equal distribution of Rset members to Members of the empty set.
>do you see the paradox here?

[Anonymous No.6137665]

Solution to ln(x)=x

[Anonymous No.6137685]

<span class="math">\int _{M} \mathrm{d} \omega = \int _{ \partial M } \omega [/spoiler]

[Anonymous No.6137691]

<div class="math">\int _{M} \mathrm{d} \omega = \int _{ \partial M } \omega </div>

[Anonymous No.6137689]

>>6137606
Thanks for the Calculus I lesson.
This thread is pretty bad.

[Anonymous No.6137696]

>>6137655
I think somebody neglected to mention the point to the artist

[Anonymous No.6137709]

>>6137691
I'd like this one better if every textbook ever didn't define manifolds at the hands of open sets, thus excluding ones with a boundary.

[Anonymous No.6137718]

>>6137709
but a manifold with a boundary is just an open manifold along with the boundary, which are both manifolds.

[Anonymous No.6137724]

2+2=4
2x2=4
fascinating when I was 12

[Anonymous No.6137727]

>>6137724
<span class="math"> 2^2 = 4 [/spoiler]
<span class="math"> 2\uparrow\uparrow 2= 4 [/spoiler]

[Anonymous No.6137740]

>>6137727
What do those lines mean?

[Anonymous No.6137744]

>>6137740
http://en.wikipedia.org/wiki/Knuth%27s_up-arrow_notation

[Anonymous No.6137769]

>>6137744
Thanks.

[Anonymous No.6137783]

Sum from n=1 to infinity of 1/n^2 = pi^2/6

completely mind blowing the first time heard about it, unintuitive answer sparked my curiosity

[Anonymous No.6137784]

e(x)=e'(x)

[Anonymous No.6137924]

>>6137783
http://www.wolframalpha.com/input/?i=sum+i%3D1+to+inf+of+1%2Fn%5E2

Nigga that shit diverges.

[Anonymous No.6137933]

>>6137924
>sum i=1 to inf of 1/n^2
>i=1
>1/n^2
>i
>n

[Anonymous No.6137936]

>>6137924
If you sum on 'i' of course it does anon

[Anonymous No.6137937]

>>6137924
Dude, you index is fucked, your series just adds a constant over and over again.

i = 1 should be n = 1

[Anonymous No.6137944]

>>6137933
>>6137936
>>6137937

Well I derped pretty hard there. My bad.

[Anonymous No.6137960]

certainly not the most beautiful but there are some neat definite integrals out there
<div class="math">\int_{\infty}^\infty \frac{\cos x}{x^2+1}dx = \frac{\pi}{e}</div>

[Anonymous No.6137964]

>>6137960
and obviously the lower limit of integration should be negative infinity

[Anonymous No.6137971]

<span class="math">2*3=6[/spoiler]

[Anonymous No.6137974]

>>6137960
this one is pretty crazy
<div class="math">\int_{0}^\pi e^{\cos\theta}\cos(n\theta - \sin\theta)dx = \frac{\pi}{n!}</div>

[Anonymous No.6137988]

<span class="math">\forall p \in \wp, p \neq 2 \Rightarrow 2 | (p - 1) [/spoiler]

[Anonymous No.6138026]

<span class="math">\forall P, Q \in \mathbb{N}[X], P(\pi) = Q(\pi) \Rightarrow P=Q[/spoiler]

[Anonymous No.6138030]

How about:
<div class="math">\delta _{nm} = \frac{1}{2 \pi i} \oint _{S^1}z^{n - m - 1} dz</div>

[Anonymous No.6138035]

>>6138030
any significance to that function?

[Anonymous No.6138050]

>>6138035
That 'integral symbol' is physics not maths. So it's useless.

[Anonymous No.6138061]

>>6138035
Seriously. Never heard of Kkronecker's Delta? High school cofrimed

[Anonymous No.6138068]

>>6138050
>not knowing complex inegration

Cofirmed for retard

[Anonymous No.6138075]

>>6137974

How do you do this? contour integration maybe?

[Anonymous No.6138078]

>>6138068
Only physicians note it like that.

[Anonymous No.6138083]

>>6138078
>physicians

[Anonymous No.6138091]

>>6138061
I don't do physics.

But yeah that integral is trivial, just curious why anyone would bother.

[Anonymous No.6138092]

>>6138075
yeah IIRC you say it's the real part of some contour integral and that simplifies things. I haven't gone through the steps in a while now

[Anonymous No.6138094]

>>6138083
As soon as I saw that shitty integral symbol I close the book, or the web page. I know that's totally uninteresting. A true mathematician use only the standard integral symbol.

[Anonymous No.6138101]

>>6138030
>S^1

alright bro I know you're almost graduating high school now, but relax a bit

[Anonymous No.6138104]

777^[(pi*i)/ln(777)]=-1

[Anonymous No.6138105]

>>6138101
Are you really impressed by the standard notation of the circle?

[Anonymous No.6138111]

>>6138094
>physicians
>medical workers

also
>no true scotsman

[Anonymous No.6138119]

lol
The world of people using maths is divided into two categories, the mathematicians and the others (the physicians, you call them like you want, I don't care). And those who use that useless symbol for integral on a closed path are from the second category.
A true mathematician won't use a useless symbol.

[Anonymous No.6138150]

>>6138119
The first group it's actually divided in two other categories: walking autisms and normal people.
The former would complain about a fancy circle, the latter wouldn't care because it's just a fancy circle that doesn't change slightly the meaning of the formula.

[Anonymous No.6138152]

>>6138150
touché

[Anonymous No.6138182]

>>6138092
>>6138075
ok I did out the work so might as well show it

[Anonymous No.6138209]

>>6138119
>the mathematicians and the others (the physicians,
u French ?

I used to make the confusion too :
physician : médecin, en gros
physicist : physicien

They don't like when you mess it up ;)

[Anonymous No.6138219]

>>6138119
See
>>6138111

>the word you're looking for is "physicists"

[Anonymous No.6138388]

P = I2 * R

[Anonymous No.6138391]

0^0= 1

[Anonymous No.6138396]

I have always liked

P = I^2 * R

[Anonymous No.6138439]

>>6138119
Your are wrong and assblasted.

[Anonymous No.6138459]

can someone explain why the cardinality of the set of integers does not fall between the cardinality of the set of natural numbers and the cardinality of the set of reals?

That is, why does the continuum hypothesis exist? intuitively the integers are a superset of the naturals, and should have a larger cardinality

[Anonymous No.6138464]

>>6138459
sets have equal cardinality not based on being subsets/supersets, but based on whether they can be put in one-one correspondence with each other

[Anonymous No.6138468]

>>6138464
i just realized my prof explained the continuum hypothesis incorrectly

carry on

[Anonymous No.6138472]

>>6138091
Really nigger? You really believe that is the only instace kronecker's delta emerges? Are you truly that ignorant of the full usefulness of that symbol? You went full high schooler.

[Anonymous No.6138474]

>>6138150
This nigga

[Anonymous No.6138491]

>>6137988
I'm not shiggy with the notation, but does this say that a prime number minus 1 is even? lol

[Anonymous No.6138497]

>>6138491
LOL SO BEAUTIFUL MATH

[Anonymous No.6138498]

>>6138491
prime other than 2, yes

another remarkable result is that most primes are not congruent to zero, modulo 3

[Anonymous No.6138506]

>>6138498
kek

[Anonymous No.6138564]

>>6137691
how the fuck did you write that

[Anonymous No.6138570]

>>6138564
double click to see source

[Anonymous No.6138576]

>>6138570
thanks
and thanks for not demeaning me

[Anonymous No.6138813]

e^i*pi=0.99999999......

[Anonymous No.6138822]

>>6137526
Math is boring.

Fractal algorithms though... certain fractals describe biological and natural phenomenon in a way pure mathematics simply wouldn't be able to.

[Anonymous No.6138824]

>>6138822
>math is boring
>describes math
Oh boy

[Anonymous No.6138830]

>>6138824
But /sci/ hates algorithms?

[Anonymous No.6138841]

>>6138830
I love algorithms.
Every time I see a system of equations I just want to start running LU decomposition on it.

[Anonymous No.6138873]

>>6138822
you're an idiot

[Anonymous No.6139285]

<div class="math">\int_{0}^{\pi/2} \frac{dx}{1+\tan^{\sqrt{2}}(x)} = \frac{\pi}{4}</div>

[Anonymous No.6139384]

(1 / (1 + e ^ -z)) ' = (1 / (1 + e ^ -z)) * (1 - (1 / (1 + e ^ -z)))

[Anonymous No.6140464]

>>6137579
Well, why not lel
>>6137691
>nice
>>6137988
Wow, that's pretty profound
>>6138030
>>6138094
What
Stahp
Are you aware that half of mathematicians are applied anyway?
>>6138841
Oh I know, so terribly sexual
Also
<span class="math"> \forall x \exists c: 2^{x} \setminus { \varnothing } \to x: \forall y \in x: c(y) \in y
[/spoiler]

[Anonymous No.6140468]

>>6140464
Fuck
<span class="math"> \forall x \exists c: 2^{x} \setminus { \emptyset } \to x: \forall y \in x: c(y) \in y
[/spoiler]

[Anonymous No.6140471]

>>6140464
Bollocks

<span class="math"> \forall x \exists c: 2^{x} \setminus { \emptyset } \to x: \forall y \in 2^{x} \setminus { \emptyset }: c(y) \in y
[/spoiler]

[Anonymous No.6140474]

>>6140471
Aw shoot, there's no brackets round the <span class="math"> \emptyset [\math] as I can't in2 LaTeX very well but I'm sure y'all can survive[/spoiler]

[Anonymous No.6140475]

>>6137606
You fucking cocksucker. The limit of 1/x as x approaches 0 does not exist.

Go fucking kill yourself pleb.

[Anonymous No.6140512]

>>6137526
1+2+3+4+...= -1/12
Hehhahehhhehhah

[Anonymous No.6140514]

>>6140512
explain?

[Anonymous No.6140528]

>>6140514
http://en.wikipedia.org/wiki/1_%2B_2_%2B_3_%2B_4_%2B_%E2%8B%AF

This sum diverges when the real part of s is less than or equal to 1, but when s = −1 then the analytic continuation of ζ(s) gives ζ(−1) as −1/12.

[Anonymous No.6140547]

>>6140514
It is proved by using the analytic continuation of the Riemann Zeta-Function, which satisfies
<span class="math">\pi^{-frac\{1}{2}s}\Gamma(frac\{1}{2}s)\zeta(s)=\pi^{-frac\{1}{2}(1-s)\Gamma(frac{1-s}{2})\zeta(1-s)
Plug in s=-1 and with \Gamma(1)=1 and \zeta(2)=frac{\pi^2}{6} and \Gamma(-frac{1}{2})=-2\sqrt{\pi} you get the result[/spoiler]

[Anonymous No.6140554]

The classification of semisimple Lie algebras reduces to the classification of Dynkin diagrams.

[Anonymous No.6140565]

>>6140512

http://en.wikipedia.org/wiki/1_%2B_1_%2B_1_%2B_1_%2B_%C2%B7_%C2%B7_%C2%B7


huehuehue

[Anonymous No.6140583]

>>6137526
Sigh
Closes door

[Anonymous No.6140604]

http://en.wikipedia.org/wiki/Yoneda_lemma

[Anonymous No.6140619]

>>6137606
Division by infinity = infinity, no?

[Anonymous No.6140630]

https://www.google.com/search?q=golden+ratio%2Btan(e%5Ee)

[Anonymous No.6140682]

>>6140630
phi - tan(e^e) is almost sqrt(5)
wat?

[Anonymous No.6140711]

Sorry for the long post. I think I must have been beaten on the spot, but nevermind, my post is probably still interesting. So I post it. It comes in four parts.
1/4
>>6138026
Pretty nice. Works with rationals also.

>>6138468
Since he didn't answer I'll do it for him: with finite sets you compare cardinals, that is to say the number of elements.

With infinite sets, however, it can't work. Thus you have to generalize your method of comparing sets in a way that is cogent with the case of finite sets.

Here we need some vocabulary. Let A and B be two sets. If, for each element a of A, you choose one and only one element b of B and associate it to a, you define what is called an application (basically a function) from A to B. If your function is such that no element b of B is associated to two different elements of A (i.e if you know b you know the a it's associated with), it is called an injection. If your function is such taht any element b of B has at least one a associated to it (your function reaches every element of B), we call it a surjection. A function that is called an injection and a surjection is a bijection.

What is counting, then ? From each element a of A, you choose one and only one natural number n, starting from 1, in a way that a number never stands for two differents elements of A (you want to count all the elements of A). When all elements of A have been counted, you stop, and the last natural number you reached is the number of elements in, i.e the cardinal n of, A. Thus you have defined a bijection from A to the set of natural number from 1 to n (we'll note it (1,..,n)).

You can acutally do the same with infinite set. To every single element of your first set A, you associate one and only one element of your second set B. If you make sure that no element of B is associated with more than one element.

[Anonymous No.6140723]

>>6140711
2/4
Thus: if there is a bijection between A and B, they have same cardinality (even if they are infinite, one of the point of bijection is precisely to define cardinality for infinite set). If there is an injection from A to B, B has greater or equal cardinality as A, and if there is a surjection from A to B, A has greater of equal cardinality as B. As a result, all infinite sets have greater or equal cardinality as N and all finite set has strictly smaller cardinality than N.

Now be cautious: there are many way of building a correspondance between A and B. Not all of them will be bijective. If a given correspondace isn't bijective, that doesn't mean the sets aren't of the same cardinality. You'll just have to search for a better correspondance. If you can show, however, that there is no bijection between A and B, and if on top of that there is a bijection between A and a part of B, then you can say that B has a cardinal striclty greater than that of A. For instance R (set of real numbers) is strictly greater than N (natural numbers) in that respect.

[Anonymous No.6140729]

>>6140723
3/4

Now, one of the surprising results of the theory of cardinality, is that a set can have the same cardinality as one of its strict parts. For instance, N is stricly included in Z, but N and Z have the same cardinality. Having the same cardinality as one of your strict part is actually another way to describe an infinite set.

Here is an example of bijection between N and Z: you go from N to Z. To 0 you associate 0. To any even number 2n that isn't 0 you associate n. To any uneven number 2n+1 you associate -n. Every number in N has one and only one image. Any number in Z has one and only one pre-image (or antecedent) in N. N and Z are in bijection, they have same cardinal. You can do the same between N and Q (to any given x=p/q in Q (where p and q are coprime), you associate 2^p*3^q, and the unicity of decomposition in product of prime number makes sure that this is a bijective correspondance between Q and a part of N. I personally find that one really pretty).

Now we get (at last) to the continuum hypothesis. A fundamental preliminary result is that there is never a bijection between a set A and the set of its parts P(A) (the set whose elements are the parts of A). I f you admit this result, you have one important consequence: there is no bijection between N and R.

[Anonymous No.6140734]

>>6140729
4/4
How so ? You can write R as the set of numbers who have a binary development of 0s and 1s. Thus there is a bijection between R and the set of sequels of who take only 0 and/or 1 as value. But you can also build any part of N this way: for each part A of N, for each integer n in N, you decide if n is in A (in this case you associate it to 1) or not (you then associate it to 0). I other words, you have a bijection between P(N) and the sequels of 0s and 1s. Which mean you also have a bijection between P(N) and R.
Since N and R are both infinite, you can say that R is in a way "infinitely gretaer" than N. Any set that as same cardinality as N is said to have "power of the numerable" while any set that as same cardinaliy as R is said to have "power of the continuum".

Actually you can go a step further. Consider P(R) the sets of the parts of R. There is no bijection between R and P(R). Thus P(R) is infinitely greater than R. But P(P(R)) is infinitely greater than P(R). And so on and so on. With this you have an infinite sequel of sets that are each infinitely greater than the precedent.

There's a question left, however. If R (roughly equivalent to P(N)) is infitinely gretaer than N, are all sets that are infinitely gretaer than N as great as R, or can they be smaller ? Namely, is there a set A such as there is an injection from N to A and a injection from A to R but no bijection between either N and A and A and R ? This is the question of the continuum hypothesis.

The famous logician Gödel work a lot on this, but the definite proof was later given by Cohen, who got the Fields Medal for it. Cohen's result is that the continuum hypothesis can be either true or false, more precisely, the common axioms of mathematics (called Zermelo-Frankell axiomatic) allow for no definite answer. You'll have to pick more specific axioms if you want it to be true (or false). Cohen obtained his result with a groundbreaking method of reasoning named "forcing".

[Anonymous No.6140751]

>>6140734
Let's add a little bonus:

I recently heard that logician have made a leap in defining infinite cardinal by inventing the notion of unreachable cardinals (don't know if that's the English word, though). Here's the idea: I've said before that the first infinite cardinals were defined by considering N, then P(N) (which infinitely greater), then P(P(N)), and so on. For instance R has the same cardinal as P(N), to some extent you can say that R is "reachable" from N through the "parts" operator. But how about cardinals tht can't be reached that way ? We say that a set A has unreachable cardinal if, for every B that has strictly smaller cardinality than A, P(B) also has stricly smaller cardinal than A. Such a set can't be "reached" by smaller set through the "P" operator. The first example of unreachable cardinal is N. We've said that if A is infinite, than A has at least the cardinal of N. And N itself is infinite. Which mean that if A has strictly smaller cardinal than N, it is finite. And if A is finite, P(A) is finite, and so is P(P(A)) and so on. Thus N can't be reached by smaller cardinals, which are the finite cardinals. However, R is reachable, as it is equal (in cardinality) to P(N). An open question is wether there are unreachable cardinals striclty gretaer than N. From what I heard, you need an axiom for that (it can't be demonstrated but has to be postulated). I even heard (though treat that statement with due cautiousness) that admitting there are unreachable cardinals greater than N allows you to prove Fermat-Wiles theorem very quickly.

Also, since I'm a righteous faggot:
>>6140619
If you divide a finite number by a quantity that has infinity has a limit, the quotient will become infinitely small and have zero as a limit. Except in rare cases, infinity can only be properly understood as a limit.

Again, sorry for the long post, and >>6138468, I hope I answered your question.

[Anonymous No.6140772]

>>6138813
* <span class="math">e^i \pi =-0.99999999......[/spoiler]

[Anonymous No.6140778]

>>6140772
SHIT
[/math] e^{i \pi} =-0.\overline{9} [/math]

[Anonymous No.6140781]

>>6140778
Just forget it.

[Anonymous No.6140799]

>>6140781
It's okay, you failed at being retarded.

[Anonymous No.6140852]

Formula for the nth Fibonacci number:
<span class="math">F_n = \frac{1}{\sqrt{5}}\left ( \left ( \frac{1 + \sqrt{5}}{2} \right )^n + \left ( \frac{1 - \sqrt{5}}{2} \right )^n \right )[/spoiler]

I think it's pretty for involving both the golden ratio and it's inverse, and also that all those square roots will disappear and the result is always an integer.

[Anonymous No.6140855]

>>6140852
Formula for the nth Fibonacci number:

<span class="math">F_n = \frac{1}{\sqrt{5}}\left ( \left ( \frac{1 + \sqrt{5}}{2} \right )^n - \left ( \frac{1 - \sqrt{5}}{2} \right )^n \right )[/spoiler]

I think it's pretty neat for it involves both the golden ratio and its inverse, and also that all those square roots will disappear and the result is always an integer.

[Anonymous No.6140873]

>>6137727
<span class="math">^22[/spoiler]
Standard notation master race

[Anonymous No.6140882]

>>6140855

What is this witchcraft!?

[Anonymous No.6140887]

>>6137727
I never thought this shit could be a thing. When I was fucking around with proofs. What is it used for?

[Anonymous No.6140892]

>>6140873
I guess I meant
<span class="math"> 2\uparrow^{2}2=4[/spoiler]

[Anonymous No.6140900]

>>6140892
or even more generally
<span class="math"> 2\uparrow^{n}2 [/spoiler]

[Anonymous No.6140909]

The rigurous proof that any number multiplied by 0
is 0. I lost my shit when learned it.

[Anonymous No.6141550]

>>6137727
Isn't that called "tetration"?

[Anonymous No.6141570]

>>6140475
lrn2epsilon delta

[Anonymous No.6141687]

>>6140882
linear algebra

[Anonymous No.6141690]

>>6141550
if you call <span class="math">2\uparrow 2 [/spoiler] exponentiation

also I meant to fix this last time
>>6140900
*<span class="math"> 2\uparrow^{n} 2=4[/spoiler]

[Anonymous No.6141693]

>>6141687
because that has to do with vector spaces, right.

[Anonymous No.6141701]

>>6141693
you're embarrassing yourself

[Anonymous No.6141733]

1 + 2 + 4 + 8 + .... = -1

[Anonymous No.6141741]

>>6141733

[Anonymous No.6141744]

s = 1 + 2 + 4 + 8 + ...
s = 1 + 2(1 + 2 + 4 + 8 + ...)
s = 1 + 2s

[Anonymous No.6141755]

1 / 0 = infinity

[Anonymous No.6141757]

>>6141744
was aimed at
>>6141741

[Anonymous No.6141770]

>>6141744
Shit like this makes me feel so redundant

[Anonymous No.6141775]

>>6141770
Too bad it's wrong and you can't just dick around with infinite sums that way.

[Anonymous No.6141777]

>>6141775
It's still beautiful despite being wrong then?

[Anonymous No.6141781]

>>6141770
Fractals are beautiful man

[Anonymous No.6141787]

>>6141781
My nigga. I want to just become one and not give a shit about anything.

[Anonymous No.6141802]

>>6140909
Link?

[Anonymous No.6141825]

The fact that rearranging a conditional convergent series can produce any answer you want it to. Our professor showed us with an infinite series of the series n=1 to infinity of [(-1)^(n+1)]/n you get the answer

1 - 1/2 + 1/3 - 1/4 + 1/5 -....... = A

can be rearranged by saying x = 2x-x so you get
(2-1) - 1/2 + (2/3 - 1/3) - 1/4 + (2/5 - 1/5) -..... = A

then rearrange the terms to get
(this is a completely math legal law of addition/subtraction without using infinite sets)

2 - 1 + 2/3 - 1/2 + 2/5 - 1/3..... = A

then factor out a 2 to get

2(1 - 1/2 + 1/3 - 1/4 + 1/5 -.......) = A

2*A = A

when working with these types of series, 2*1 can equal 1

mind blown, gave me the chills when I was first showed it

[Anonymous No.6141831]

>>6141733
0/10 troll

series diverges by geometric series test, cmon /sci

[Anonymous No.6141859]

>>6141825
>The fact that rearranging a conditional convergent series can produce any answer you want it to.
Too bad it's incorrect after you start fucking around with it like that.
1=1+0+0+...
=1+(-1+1)+(-1+1)+...
=(1-1)+(1-1)+(1-1)+...
=0+0+0+...
=0
Hurrrr, 1=0.

[Anonymous No.6141872]

>>6141859
I'm sorry, what series produces the result1 = 1+0 +0 + 0?

By the looks of it you don't understand what a conditional convergent series is, would you be able to explain where the numbers you chose came from? And yes, what you've described is why the rule of addition doesn't work with infinite series. So what you've said is true, so long as your numbers have a series that they're attached to and that series has an absolute value that doesn't converge. I'm confused by the "Hur" if this is absolutely correctly when these conditions are satisfied. Could you further elaborate please to help me understand please?

[Anonymous No.6141890]

>>6141872
Obviously if you have the sequence {1,0,0,...} you have the series 1+0+0+...
You almost said it yourself with
>then rearrange the terms to get
>(this is a completely math legal law of addition/subtraction without using infinite sets)
after which you said
>when working with these types of series, 2*1 can equal 1
which isn't at all surprising given that you're not using correct mathematics anymore.

[Anonymous No.6141907]

>>6141890
Sorry if I've sounded confusing, but I was asking for the series in "sigma form" if that makes sense. From n = 1 to infinity of (insert equation for series here). And yes, what you've concluded is true. I've just found it amazing that the most basic forms of learned maths taught all the way to us back in elementary school break down when we're introduced to ideas concerning infinity. This was by far my favorite moment in Calc 2, nothing gave me chills like this did.

[Anonymous No.6141911]

P = NP

[Anonymous No.6141949]

>>6141907
>the most basic forms of learned maths taught all the way to us back in elementary school break down when we're introduced to ideas concerning infinity
It's a shame Cantor received as much criticism as he did when he published his work on infinite sets, but it just goes to show that even the top mathematicians of the time had trouble dealing with infinity. It would be interesting to see what would happen if one tried to teach children these concepts from a young age, I wonder if they would build a stronger understanding about these abstract concepts or the physical world would give them an intuitive feel of mathematics that would cause these things to be surprising.
As for the series in sigma form, you have the sequence A={1,0,0,...}, so the series would be
sum from n=1 to infinity of a_n=
=(sum from n=1 to 1 of a_n) + (sum from n=2 to infinity of a_n)
=1 + (sum from n=2 to infinity of a_n),
which simplifies to
=1 + (0),
but if you try to say that the a_n is (-1+1), which equals zero, and apply various identities incorrectly you end up with incorrect results.

[Anonymous No.6141972]

>>6141802
here ya go

[Anonymous No.6142013]

>>6141949
Unfortunately for us out in the US this is a lost cause. I would love to see children taught these ideas in elementary school, but our schooling system is such a mess that all they're being forced to learn is "plug and chug" instead of abstract thinking exercises. Folks out here in the USA are against maths because of how much the ugly parts of it are forced into our curriculum/testing standards. It's disappointing to say the least