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/sci/ - Science & Math


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9183592 No.9183592 [Reply] [Original]

What are you studying today, /mg/? Have you discovered something interesting in math recently?

Previous thread >>9162159

>> No.9183608
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9183608

>>9183592
struggling with some topology stuff and it was only the first day.

>> No.9183621

>>9183608
Topology is like the height of thinking in abstraction m8

But once you get through it you can read all kinds of good math.

>> No.9183628

So get this.

I finish my MS Math this December and my school has a reputable Biostats program. I've always been interested in Stats and only did Math cause we don't have Stats here.

I talked to the grad program coordinator today and he told me I couldn't take Survival Analysis in the Spring because I hadn't taken their basic intro statistics course. I looked up the class online and it's all like the basic shit you learn in any undergrad introductory statistics course.

I was like what the fuck how do you treat someone with a MS Math like that.

Looks like I'm not going to do another degree after this.

>> No.9183629

>>9183621
The best thing I like about this book we're using is how it has a lot of applied stuff. Viewing physics mechanics from a topological viewpoint is super interesting.

>> No.9183633

>>9183629
Damn...what book?

We just used Munkres at my school.

>> No.9183648
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9183648

>>9183633
Adams.
There's an entire section on robotics as well.

>> No.9183656

>>9183608
>>9183621
>If you want a good topology text go for Janich, also what about topology is making you struggle? It's at the foundation of a hell of a lot of math so it's pretty important to work through it.
>>9183628
That sounds pretty shitty, is it possible to take an online course to substitute the credit? Are there courses you could petition as a substitute (like if you demonstrate that a class you took prior covers the same or more material?). Try asking whoever is teaching the course and they may be able to help you out?
>>9183629
>Viewing physics mechanics from a topological viewpoint is super interesting.
What physics topics are you covering? I imagine it's not TQFTs, GR, or condensed matter theory so seeing topology in physics seems a bit odd if not those subjects.

>> No.9183666

>>9183628
>I finish my MS Math this December and my school has a reputable Biostats program. I've always been interested in Stats and only did Math cause we don't have Stats here.
You don't have stats there but you have biostats?

>> No.9183672

>>9183592
Just got some extra points for my introductory economy class for knowing about homogeneous functions, so that's nice.

>> No.9183696

>>9183648
Have fun bruh. I struggled with topology, too until we had to prove that RP^2 is homeomorphic to a disk glued to the boundary of a Mobius strip. Then something clicked and I started doing really well for the rest of that course.

>> No.9183713

>>9183592
https://youtu.be/8dKTYIJK5ek?t=3150

probably best to copy paste the url

>> No.9183781

Just started multivariable calculus today. Its cool

>> No.9183821

/out/ fag here. I've been messing around with the growing degree days concept in the garden. For those who don't know what I'm talking about, G.D.D. is a cumulative measure of temperature used to predict the life cycle of plants & animals.

I recently discovered that it takes around 190 G.D.D. for a sweet potato to form root nodules and to extend roots. Next, I want to track how many G.D.D. it takes for an alpine strawberry flower to become a ripe fruit. Hopefully, I can use the results to predict when I'll need to harvest without having to physically check the plants constantly.

>> No.9183830

>>9183821
Wouldn't you still need to check periodically to make sure there aren't any issues with development, insects, or soil conditions, I imagine there would be some extra unknowns that would fudge things a bit. What might be nice and fairly easy to do is make a program to keep track of local weather conditions (you could make a basic setup as well, get a some basic equipment to measure moisture, temperature, sunlight, and soil conditions in your garden, then have your computer track these variables to predict when the veggies should be ready to harvest and if there are any unforeseen issues.)

>> No.9183849

>>9183781
No it's not

>> No.9183920

recommend LA books for a retard pls

>> No.9183936

>>9183849
Go on and me about your superior intellict because youre studing graph-knot-ring-bullshit-fag theory and how much of a brainlet I am

>> No.9183947

>>9183656
Classical Mechanics is pretty much all symplectic geometry/topology at its roots.

>> No.9183995

>>9183666
Nope. Pretty retarded, right? It's because we have a prestigious med school so everything public health is top-tier and well-funded.
We're just now getting a data science program starting in Spring, and you can apparently use Biostats courses as electives for that. So that's what I'm gonna do I think.
Maybe I won't know as much about theoretical stats/probability (which is where my heart is) but I can graduate fast af and make a ton of money eh.

>> No.9184009

>>9183628
tell him you took the course and that you're better than his students

i have to take out of department classes in my grad school and you have to bully around the non math brainlets

>> No.9184015

>>9183947
>Classical Mechanics is pretty much all symplectic geometry/topology at its roots.
Mostly the symplectic geometry, the topology is the background material in which the stage is set. Topology does come up more in electrodynamics though, with the subject being likely the first intro to gauge theory/cohomology for most physics students.

>> No.9184053
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9184053

I've been studying language theory and mathematical logic.

I seriously doubt authors on language theory books know how to prove a language has certain property because none of them writes a single proof of what they affirm; for instance, one author defines a language and then states that in each word of that language the number of a's is the same as that of b's and his two lines proof basically restates what he is supposed to prove.

>Mfw I see one of those "proofs"

>> No.9184074
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9184074

Seeing this thread every week just reminds me of how much of a brainlet I am

I wish I could take maths but my passion is in mechanical engineering

>> No.9184075

>>9183781
Personally least favorite of all math courses for me and my lowest grade by far lol. Best of luck

>> No.9184115

>>9183592
https://www.youtube.com/watch?v=Ww4r9lN-2Xo

>> No.9184137

>>9183592
I'm trying learn about modal logic, multimodal logic, and epistemic logic.

It's basically about multiple agents having different knowledges about the same world.

Shit gets crazy when you get statements like A knows that B knows that A knows...

Then a temporal parameter is added where agents can exchange knowledge.

Most of the theory is built around having idealized agents that are perfectly logical and always truthful.

I'm trying to see what happens when you allow the agents to have imperfect logic and also allow them to lie because, let's be real, people can be dumb and people can lie.

>> No.9184141

>>9184115
From the ytmnd days
https://www.youtube.com/watch?v=2h6seJ3xjWA

>> No.9184267

>>9183629
How do you use it in classical mech? The only way is to use symplectic manifolds but that seems pretty stretched for an introductory class.

>> No.9184270

>>9184015
yeah but you need to define all sorts of things like tensors, differential forms, differential manifolds and so on to get to cohomology to get to the homogeneous Maxwell's equations [math]dF=0[/math].

>> No.9184279
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9184279

Need some hints on proving that:
>G is a group with |G|>1 and the only subgroups of G being {G, {e}}
>Prove that |G| is a prime and thus finite
I CANNOT use cyclic groups in the proof.

>> No.9184302

>>9183656
>I imagine it's not TQFTs, GR, or condensed matter theory so seeing topology in physics seems a bit odd if not those subjects.
Topology plays a role in string theory, but then again what doesn't?

>> No.9184309

>>9184279
Can you use Lagrange's theorem?

>> No.9184339

>>9183920
Strang

>> No.9184342 [DELETED] 

>>9184279
>>>/r/ibbit/

>> No.9184343

>>9184074
There is a lot of years of pure math you can do that are still applicable to the real world. Algebraic geometry applied to coding is one that comes to mind, or topological optimization.

>> No.9184348
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9184348

>>9184309
nope

>>9184342
t-thanks

>> No.9184357

>>9184348
WHY can you not use cyclic subgroups in the proof?

>> No.9184358

>>9184279
take any element [math]x \neq e[/math] from G
[math]\{ e, g, g^2, g^3, \dots \}[/math] is a subgroup of G

>> No.9184359

>>9184358
he said no cyclic groups

>> No.9184360

>>9184358
it's not if it is infinite
there are no inverses

>> No.9184361

>>9184279
The axiom of choice implies that |G| is actually not finite.

>> No.9184370

>>9184357
My professor specifically asked us to prove it "using everything we learnt this semester so far"
And we just started on abstract algebra

>> No.9184373
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9184373

>>9184370
which did not include cyclic groups at the date the question is posted

>> No.9184374

>>9184370
how about you start listing the topics you've learnt so far you retard

>> No.9184376

>>9184374
see >>9184373

>> No.9184377

>>9184376
cyclic groups are literally the first example of groups, are you meaning to tell me in this class you've only covered the definition of a group?

>> No.9184378

>>9184377
This question was posted 1 lecture before we started on cyclic groups

>> No.9184390

>>9184378
maybe so you could have scratched your head and derived it yourself, or so that you would do it after you did that lecture
>>9184279
well it's basically going to implicitly use cyclic groups, but:

Suppose G has more than one generator. Then each generator alone can create a different subgroup by multiplying it by itself. But the only subgroups are 0 and G, so either the generator is [math]e[/math] which by definition is not a generator, or the generator generates the whole group G. Therefore there is only one generator by contradiction.

Now if G is infinite, then if g is the generator, then g^2 is another generator. We can't have that so contradiction to G infinite.

So obviously, since g generates G, we have that |g|=|G|. Suppose |g|=pn with p prime. Then we can form a subgroup using the element g^p. But this subgroup must be either G or 0. If 0, then |G|=p, a prime. If G, then n=1 and hence |G|=p.

QED

>> No.9184392

>>9184279
Suppose first that [math]|G|= \infty[/math], and choose [math]g \neq e[/math] such that [math]g^2 \neq e[/math], and define [math]H = \{ g^{2n}\ |\ n \in \mathbb{Z} \}[/math] to get a subgroup [math]0 < H < G[/math], thus contradicting the assumption, so [math]|G| < \infty[/math]. Let then [math]p[/math] be a prime number dividing [math]|G|[/math]. We can then use Cauchy's theorem:
>Given a finite group [math]G[/math] and a prime number [math]p[/math] dividing the order of [math]G[/math], then there exists a subgroup of order [math]p[/math] in [math]G[/math]
Let [math]H[/math] be the subgroup given by Cauchy. Clearly [math]0 < H[/math], so we must, by assumption, have [math]H=G \Rightarrow |G|=|H|=p[/math].

>> No.9184398 [DELETED] 

>>9184279
>>9184348
>>9184373
>>>/r/ibbit/

>> No.9184400

>>9184370
Well, you learned about subgroups.
Show that for any g in G, Cg:={g^n ; n in Z} is a subgroup.

Now for g=/=e (you can do that since |G|>1) you obviously have that Cg is different from {e}. Therefore, from the hypothesis it must be that ****** Cg=G *****.

If G(=Cg) was infinite then you get that for k=/=0,1 Cg^k={(g^k)^n ; n in Z} is a proper and non-trivial subgroup of Cg(=G). From the hypothesis, this cannot happen. Therefore G must be finite.

Now let |G|=:n for some n>1.
Suppose n was not prime, say n=st with both s and t different from 1.
Take the element g^s.
{(g^s)^i ; i in Z} is a subgroup of Cg(=G).
Show that this subgroup is different from G and {e}.
.
.
.

>> No.9184406

Consider the set of all 6x6 complex matrices such that they have 2 eigenvalues λ and μ, where λ has algebraic multiplicity 4 and geometric multiplicity 2, and μ has algebraic multiplicity 2 and geometric multiplicity 1.

How does one clasify these matrices up to similarity?

My attempt:
We find the possible Jordan forms.
The μ part can be either {{1,1},{0,μ}} or {{μ,0},{0,μ}} and the only acceptable one is the first one because the second one gives geometric multiplicity 2.
Therefore, the number of similarity classes, are completely determined by the λ part.
Now for the λ part we have 5 possible cases:
a 4x4 block,
a 3x3 block and a 1x1 block,
a 2x2 two 1x1s, two 2x2s,
four 1x1s (diagonal).
The diagonal case gets rejected for the same reason as {{μ,0},{0,μ}}.
But, what about the other cases?

>> No.9184412

So recently one of the exces at IBM talking about watson and that in the future there wont be programmers or people who program computers, instead they will teach and train computers like students.
Anyways, I've been meaning to learn more about what they use to be able tho do things like this and wondering would I if I work through one of the popular textbooks with either "machine learning" or "data mining" in the title or is there something more to it besides that?

>> No.9184414

>>9184406
Someone answered >>9184408
thanks anyway

>> No.9184431

From the last thread:
What is the value of [math]^ii[/math]

>> No.9184454

>>9184302
Research grants.

>> No.9184468

>>9184431
What does that notation mean?

>> No.9184470

>>9184468
tetration

>> No.9184658

Hey guys, [math]\mathbb{Z} _{12}[/math] (mod 12) is NOT a manifold right? just making sure..

>> No.9184661

>>9184658
Z_{12} with what topology?

>> No.9184683

>>9184661
well idk...how would you define open sets on [math]\mathbb{Z}_{12}[/math]? Can you even do that?

>> No.9184696

>>9184683
Trivial or discrete topology for example.

>> No.9184698

>>9184658
It is a zero dimensional manifold with the obvious topology.

>> No.9184703
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9184703

>>9184683
Coinduced by the projection [math]\pi \colon \mathbb{Z} \to \mathbb{Z}_{12}[/math], [math]n \mapsto [n]_{12}[/math]. You can show this makes it a discrete space of 12 points, so it is
>Hausdorff
>second countable
>every point has a nbd homeomorphic to [math]\mathbb{R}^0[/math]
making it a 0-manifold.

>> No.9184713

>>9184703
wow...so it can even be made into a smooth manifold?

>> No.9184772

>>9184703
im asking because i wanna know if its possible for [math](\mathbb{Z}_{12},+)[/math] to be a Lie group.

>> No.9184773

>>9183592
Thermodynamics in biochem

Pretty much plug and chug

>free energy

>> No.9185057
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9185057

What do you lads think of Khan Academy? I've been using to to learn differential and integral calculus while supplementing the lack of proof and whatnot with books and I find it's going well. I'm preparing for 1st year of uni in a few weeks and I'm supposed to be familiar with diff and integral calculus as well as basic proofs. Is this the right route?

>> No.9185068

>>9185057
khan academy is great for proofless math so its certainly a good tool for learning calculus
my gut tells me that analysis is not a good entry point for learning how to prove stuff
at my uni our discrete math class served as a de facto intro to proofs, a subject much more managable at least in it's elementary phase

>> No.9185146

>>9184431
please respond

>> No.9185395

>>9184270
It's cumbersome but worth it as it gives you info on the properties of the solutions of your equations
>>9184431
>>9185146
[math]^ii=e^{-\frac{1}{2}\pi}[/math]

>> No.9185426

Trying to get some intuition for Kan extensions. They subsume pretty much all of algebraic mathematics and that makes them really fun

>> No.9185581
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9185581

How much point-set topology should I know before starting out with algebraic topology? I should be mostly fine on the categorical/algebraic part, but I haven't studied topology yet.
And what are some good introductory books on it?

>> No.9185683

>>9183849
Faggot

>> No.9185691

>>9185581
You should know about continuity, homeomorphisms, compactness, and connectedness. I would suggest reading Munkres chapters 2, 3, and 4 for this.

>> No.9185692

>>9185683
>Faggot
Why the homophobia?

>> No.9185699

>>9185691
This, although topology is like linear algebra, it never hurts to know more random topology facts.

Also don't skip the quotients section/chapter. It's not optional if you're doing algebraic topology.

>> No.9185735 [DELETED] 

>>9183608
>>9183621
>>9183656
https://twitter.com/BlockWintergold/status/909871741584809986

>> No.9185812

9185692
>>>/lgbt/
>>>/r/taiwan
>>>/r/dogs/

>> No.9185819

>>9184431
what is i^i?

>> No.9185832

Book recommendations for stochastic processes? (Nothing super advanced, intro grad level)

>> No.9185910

>>9185395
that is [math]i^i[/math] if i'm not mistaken >>9185819

>> No.9185968

>>9183920
Bump

>> No.9186076

>>9184454
found the anti-intellectual conservitard

>> No.9186088

>>9184302
>>9186076
Fuck off to >>>/r/physics

>> No.9186250
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9186250

>tfw going to switch over to study stats and probability instead of pure math
Too worried about getting a job and earning money

>> No.9186274

>>9183592

Trying to wrap my head around Igusa local zeta functions. Its difficult, cause with these kinds of niche topics there's not a lot of literature. Also the lit. that exists is written for those 'in-the-know', so you have to piece things together from multiple sources.

>> No.9186311

>>9183592
This year ill finish my master and submit for publication a combinatorial style shit-paper on representation "theory" i got sick of already. What kind of math should i pick for my phd that rather deep and beautiful but have as lowest amount of calculations and hand-work as possible? Some ncatlab/homotopy stuff? Or more geometr-ish fields will do better?

>> No.9186322

>>9186311
Algebraic geometry.

>> No.9186350

>>9186311
number theory or mathematical logic, particularly model theory

>> No.9186353

>>9186350
>Number theory
>low number of calculations
?

>> No.9186428
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9186428

can [math](\mathbb{Z}_{12},+_{\mathbb{Z}})[/math] be turned into a Lie group?

>> No.9186451

>>9186428

Sub-group of the circle group consisting of the 12th roots of unity?

>> No.9186563

>>9186451
wait what?

>> No.9186695

>>9186322
Of what kind?

>> No.9186708

>>9186250
why not just applied math?

>> No.9186711

>>9186076
holy shit could you be more assblasted?

>> No.9186745

Aww yeahhh, just bought Stewart's "Calculus: early transcendentals." Very stoked. Never took calc in undergrad.

>> No.9186747

>>9186745
i'm teaching based on that text right now. gl!!

>> No.9186860

>>9186745
Isn't Steward's book too handwavy?

>> No.9187086
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9187086

>>9186708
Applied math is given at the nearby engineering university and the statistics education here is pretty great since it is based on a math undergrad. I might take some applied math courses like dynamicals systems in biology, dynamicals systems and optimal control theory, and a course in ODEs, wherever I can fit them in, I just feel spent after doing logic, complex analysis, abstract algebra and combinatorics, and feeling like I still can't do anything that would land me a job

>> No.9187392

>>9186745
have fun anon. remember to post any questions to >>>/sqt/

>> No.9187414

brainlet here. starting cc intermediate algebra in a week
any tips for studying that you smart people can add.
I always find math "boring" :(

>> No.9187451

>>9183592

This guy predicts the end of life support on Earth via CO2 poisoning give or take 2100 AD

http://advances.sciencemag.org/content/3/9/e1700906/tab-pdf

>> No.9187458

>>9184390
fucking beautiful

>> No.9187463

>>9183592
>tfw go to great math school as an undergrad but fuck around and get nothing out of it
>tfw want to go back to school for a masters or phd in math but it's been YEARS since maths and forgot most of the little bit that i managed to learn
but you guys are motivating me to pick up the books and make some progress, at least!

h e l p

>> No.9187592

>>9187086
That's to be expected. You are a brainlet after all.

>> No.9187647

best textbook to refresh my differential equations and stuff?

>> No.9187651 [DELETED] 
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9187651

>>9186076
>mfw coming back to my office finding mathematicians and string theorists trying to steal my research grant

>> No.9187820

>>9184713
Yes, a discrete space is a smooth manifold of dimension 0. This is not some pathology, it's useful to have this definition.

>>9184772
Yes, it's a closed subgroup of the circle.

>> No.9188091
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9188091

Will start college next month. As of now i cant prove that sqrt(2) is not a rational number. How fucked am i?

>> No.9188108

>>9183920
Kostrikin

>> No.9188148

>>9183920
Axler

>>9184339
Strang is just bad.

>> No.9188159

>>9187647
Start with an ODE book. Idk they're all pretty good. I found videos really helpful for diffEQs.

>> No.9188183

>>9188148
if he's a retard as he so claims, a more computational and example heavy book will make it easier to get a solid grasp, rather than taking it out of a more proof/theoretical heavy book

>> No.9188191

>>9188183
I think that he won't undestand anything from Strang's book except computing by memorizing techinques he doesn't know why they work (and he will forget them a month after). It will confuse him way more than an actual Linear Algebra textbook imo.
The guy seems like he wants to understand and learn. He won't understand or learn anything from Strang.

>> No.9188198

>>9187414
Pose questions after each section before you go on. The motivation to uncover more should arise in order to solve your current sticking point.

Math is extremely boring until you are like, 'wait a minute, i need something that would allow me to do this/that/her'. Use the current day's section to 'invent' a solution to day: now-1's question.

>> No.9188207

am physics major studying GR & some topology on the side.

Am i fucking myself over by always trying to picture a physical or geometric scenario when considering certain theorems/lemmas?

I feel like I may not be trying to tackle arbitrary concepts while respecting the arbitrary-ness of the system.

How should I be thinking about more arbitrary systems and statements in topology/diff. geometry?

>> No.9188219

>>9188207
always attempt to picture every theorem or lemma geometrically, or picture it whatever way you can. It is the most powerful way to obtain understanding.

>> No.9188236

>>9188207
Do whatever helps you further your understanding, just make sure that understanding is actually correct and you should be fine.

>> No.9188283

>>9188207
what specific theorems/lemmas do you have in mind?

>> No.9188290

>>9186428
yes, with the discrete topology

>> No.9188512 [DELETED] 

Let [math]0 \to M' \to M \to M'' \to 0\[/math] be a SES of free modules (assume finite rank if necessary).

Show there is a filtration of the the rth graded component of the symmetric algebra [math]{S^r}\left( M \right) = {F^0} \supseteq {F^1} \supseteq ... \supseteq {F^r} \supseteq {F^{r + 1}} = 0[/math] such that [math]{F^p}/{F^{p + 1}} \cong {S^p}\left( {M'} \right) \otimes {S^{r - p}}\left( {M''} \right)[/math].

>> No.9188513

>>9188512
bummer

>> No.9188516

Let [math] 0 \to M' \to M \to M'' \to 0 [/math] exact sequences of free modules over commutative ring. Assume finite rank if necessary.

Show there is a filtration [math] {S^r}\left( M \right) = {F^0} \supseteq {F^1} \supseteq ... \supseteq {F^r} \supseteq {F^{r + 1}} = 0 [/math] of the component of the symmetric algebra, such that...

[math] {F^p}/{F^{p + 1}} \cong {S^p}\left( {M'} \right) \otimes {S^{r - p}}\left( {M''} \right) [/math]

>> No.9188518

>attempted to study measure theory
WHAT THE FUCK IS THIS FUCKING AUTISM????

>> No.9188519

>>9188108
This or Artin

>> No.9188531

>>9188518
what book ?

>> No.9188544
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9188544

>>9188516

>> No.9188561

>>9188531
It's a Greek one. It's decent and well written, but holy shit it's really dry. I can't see how it could be less dry though, it seems like the subject itself is like that.

>> No.9188576

>>9188544
that is not helpful or relevant

>> No.9188648

What are two distinct square roots of i?

>> No.9188655

>>9188648

1,i,-1,-i = 4th roots of unity
Root two of those = eighth roots of unity
The answer is exp(2pi * i * (1/8)) and exp(2pi * i * (5/8)) b/c both 1 and 5 give 2 when multiplied by 2 modulo 8.

>> No.9188671

>>9188655
wow, you know the basics of modular arithmetic

>> No.9188688

>>9188671
its useful to find complex radicals

>> No.9188695

>>9188688
are sqrt(i) and -sqrt(i) two distinct roots of i? if not,w hy

>> No.9188734

>>9188219
>>9188236

Thanks. The way most are presented, either by the prof or in the book leave me to believe most people also try to picture a geometric representation of whatever the statement is saying.

>>9188283

Nothing too specific so far, it's more so when I'm trying to think about how certain operations may behave in higher dimensions regardless of what group i'm working with. I think I just need to work on more problems honestly, that way I can actually get a glimpse of the behavior in certain regimes that I can't picture as well.

>> No.9188737

>>9183592
Been studying finite automata and theory of computation, it's so interesting to see for instance how an accepting automaton can be generated from a regular grammar or even just a regular expression

Can't wait to be able to apply this in compilers and perhaps some natural language processing stuff

>> No.9188739

The way the protein coat of a virus is arranged is a study in topology:

https://en.wikipedia.org/wiki/Capsid

find out the 'method' of arrangement using thermodynamics and entropy and you can predict the 3d structure of the capsid, once you get that its nobel prize for sure since you can prevent ANY virus from binding to receptor by then fucking up its protein structure

>> No.9188782

>>9188695
>are sqrt(i) and -sqrt(i) two distinct roots of i?
obviously

>> No.9188793

>>9188695
http://www.wolframalpha.com/input/?i=solve+z%5E2%3Di+,+z

Also, sqr(i) is a SET, not just one number. sqrt(i) = { (1+i)/sqrt(2) , -(1+i)/sqrt(2) }

To understand how it works, all I'm gonna say that multiplication of two complex numbers is adding their angles and multiplying their lengths.

>> No.9188795

>>9188793
>sqr(i) is a SET
Wrong

>> No.9188822

>>9188795
yes it is a set
you are probably talkinf about the principal square root

>> No.9188826

>>9188822
>yes it is a set
Wrong.

>you are probably talkinf about the principal square root
sqrt is a function from C to C, it can't output sets

>> No.9188832

>>9188826
>sqrt is a function from C to C, it can't output sets
That's like saying sqrt can't output sets, therofore it can't output sets.

The definition of z^(1/n) is exp(1/n logz) and log is a multivalued function.

>> No.9188837

>>9188826
sqrt is a function (hence a mapping between sets) that maps from C to (C, C). In addition, all mappings have the form a -> (b, -b), where a and b are both complex numbers.

>> No.9188838

>>9188832
>multivalued function.
No such thing.

A function is single valued by definition, so a multivalued function is a contradiction.

>> No.9188839

>>9188837
>(C, C)
Define "(C,C)".

>> No.9188840

Is anyone familiar with sequences of rational numbers that approximate irrational numbers? I stumbled on one that where the sign of the terms alternates but the difference between the absolute value of the approximating rational number and the target is monotone decreasing.

Any ideas?

>> No.9188842

>>9188840
>Any ideas?
Ideas about what? You didn't state any problem.

>> No.9188847

>>9188838
At least you could have could have made a wikipedia search before posting this.
https://en.wikipedia.org/wiki/Multivalued_function
https://en.wikipedia.org/wiki/Complex_logarithm

>> No.9188848

>>9188838
No it's not. You can just have the second element of each ordered pair in the set representing your function be itself a set.

>>9188839
I actually meant ordered pair there, but that's wrong. A proper definition would be C -> {C, C}.

You could in fact have a function that maps a single element to infinite values if you wanted to.

>> No.9188850

>>9188848
>C -> {C, C}.
The element on the right is a multi-set, which has no place in mathematics.

Also such a function would be constant, sending every complex number to the set of all complex numbers, very boring.

>> No.9188853

>>9188847
>https://en.wikipedia.org/wiki/Multivalued_function
Irrelevant to the fact that multivalued functions do not exist.

>https://en.wikipedia.org/wiki/Complex_logarithm
Irrelevant to the fact that sqrt is a function, and hence is single-valued.

>> No.9188854

>>9188848
>A proper definition would be C -> {C, C}.
What does {C,C} even mean?
Just write C-->P(C) (the power set of C)

>> No.9188856
File: 3 KB, 698x1284, n8umjWj.png [View same] [iqdb] [saucenao] [google]
9188856

>>9188853

>> No.9188857

>>9188848
>You can just have the second element of each ordered pair in the set representing your function be itself a set.
Then you're not talking about sqrt, which is a function from C to C.

>> No.9188859

>>9188856
If Wikipedia had a page for "square with three sides" would you believe they exist?

>> No.9188875

>>9188842

Ideas regarding some "standard" approximating sequence of rational numbers that behaves like this.

>> No.9188880

>>9188859
holy shit you are SOOOO fucking retarded....
z^(1/2) is NOT A FUNCTION in the ordinary sense
log(z) IS NOT A FUNCTION in the ordinary sense
you need to make a BRANCH CUT to turn it into a function in the ordinary sense
The PRINCIPAL SQUARE ROOT is z^(1/2) when you make a BRANCH CUT at the left side of the real axis.

fuck off

>> No.9188883

>>9188880
>holy shit you are SOOOO fucking retarded....
Please refrain from ad hominems.

>z^(1/2) is NOT A FUNCTION in the ordinary sense
>you need to make a BRANCH CUT to turn it into a function in the ordinary sense
>The PRINCIPAL SQUARE ROOT is z^(1/2) when you make a BRANCH CUT at the left side of the real axis.
Do you see the contradiction here?
>the principal square root (a function according to you) is z^(1/2) (which is not a function according to you)

>> No.9188886

>>9188516
This is way easier than I thought, nevermind.

>> No.9188891

>>9188883
>Do you see the contradiction here?
no, where is it?
I told you this:
>>9188822

>the principal square root (a function according to you) is z^(1/2)
that's not according to me

>> No.9188893

>>9188891
>the principal square root (a function according to you) is z^(1/2)
>that's not according to me
But you said that right here: >>9188880
"The PRINCIPAL SQUARE ROOT is z^(1/2)"

>> No.9188904

>>9188850
See below where I go into more detail. I was just using C as a shorthand here. If you want to be super pedantic then:

Let sqrt be the function defined by the Cartesian product [math]\mathbb C \times \{a, b\}[/math] where [math]a, b \in \mathbb C[/math].

>>9188854
That's an issue since you don't define the cardinality. By {C, C} I mean a set with two elements both of which are complex numbers.

>>9188857
If you're just talking about the principle square root, then yes.

>> No.9188908

>>9188883
So, we should change the 150 years old, traditional nomenclature of complex analysis, just to sooth your autistic need for uniform categorization?

>> No.9188909

>>9188893
No, I told you
>The PRINCIPAL SQUARE ROOT is z^(1/2) when you make a BRANCH CUT at the left side of the real axis.
by which I meant that the PRINCIPAL square root of z is this:
exp(1/2 Log(z)), where Log(z) is the PRINCIPAL VALUE of log(z).

https://en.wikipedia.org/wiki/Complex_logarithm

>> No.9188910

>>9188904
>Let sqrt be the function defined by the Cartesian product C×{a,b} where a,b∈C.
In what world is this definition of sqrt useful, where every complex number has the same two square roots a and b?

>> No.9188911

>>9188893

NO, you're NOT understanding me here. FUCK SAKE. THE VERY DEFINITION OF LOG INVOLVES THE PRINCIPAL SQUARE ROOT. Do you understand what a principal branch is?

>> No.9188912

>>9188904
>By {C, C} I mean a set with two elements both of which are complex numbers.
Stop making up notation.

>> No.9188913

>>9188908
>So, we should change the 150 years old, traditional nomenclature of complex analysis, just to sooth your autistic need for uniform categorization?
You're the one confused by nomenclature here, not I.

>> No.9188915

>>9188911
>NO, you're NOT understanding me here.
It's hard to understand gibberish.

>FUCK SAKE.
Please refrain from unnecessary profanities.

>THE VERY DEFINITION OF LOG INVOLVES THE PRINCIPAL SQUARE ROOT. Do you understand what a principal branch is?
Yes, and it has nothing to do with the fact that sqrt is a function, and hence only attains a single value for each input.

>> No.9188926

>>9188853
>Irrelevant to the fact that multivalued functions do not exist.

Just make it a function from a subset of C to a subject of the powerset of C.

>> No.9188931

>>9188926
>subject
subset*

>> No.9188932

>>9188910
That does not mean every single complex number has the same two square roots. It's common in defining functions to write it this way. It's simply stating that the function maps from the set of complex numbers to the set defined by having two elements, themselves complex numbers. That's where the C -> {C, C} comes from that I used earlier.

Is the fact that function can map from complex numbers to sets really that hard for you to grasp?

>>9188912
That's not notation I made up.

>> No.9188935

>>9188932
>That's not notation I made up.
Find one (1) example in the literature of someone using this notation.

>> No.9188942

>>9188932
>That does not mean every single complex number has the same two square roots. It's common in defining functions to write it this way. It's simply stating that the function maps from the set of complex numbers to the set defined by having two elements, themselves complex numbers. That's where the C -> {C, C} comes from that I used earlier.
The cartesian product Cx{a,b} has elements of the form (c,a), (c,b) for all complex numbers c, hence is not a function unless a=b, in which you're back to using non-mathematical multisets.

>Is the fact that function can map from complex numbers to sets really that hard for you to grasp?
There are of course functions from C to a set of sets.

sqrt is not one of them.

>> No.9188957

>>9188935
I second this. {C,C} is a total gibberish

>> No.9188959

>>9188942
>sqrt is not one of them.
[math]r{e^{i\theta }} \mapsto {\left\{ {\sqrt {\left| r \right|} {e^{i\left( {\theta + 2n\pi } \right)/2}}} \right\}_{n \in \mathbb{Z}}}[/math]

>> No.9188961

>>9188942
Are you being thick on purpose? I guess we could define its mapping as C x C^2. Happy now?

>> No.9188966

>>9188961
>I guess we could define its mapping as C x C^2. Happy now?
No, do you know what a function is? That set includes the elements (0,(0,0)) and (0,(0,1)) and so isn't a function.

>> No.9188970

>>9188966
What we're discussing here is the mapping. But fine, what, in your view, is a proper definition of a function?

>> No.9188972

>>9188970
>But fine, what, in your view, is a proper definition of a function?
The standard one you should have learned in high school, do we really need to stoop this low?

>> No.9188974

>>9188972
Yeah we do. Give me the definition.

>> No.9188975

>>9188959
>reiθ↦{|r|−−√ei(θ+2nπ)/2}n∈Z
Sqrt is not a recursive function.

>> No.9188976

>>9188974
>Yeah we do. Give me the definition.
https://en.wikipedia.org/wiki/Function_(mathematics)#Definition

>> No.9188983

>>9188976
By that definition what I just gave you was a proper function. Maybe if I give you some examples you'll catch on.

For example:
(4, {2, -2})
(9, {3, -3})
and so on.

>> No.9188986

>>9188983
>By that definition what I just gave you was a proper function.
No, because C x C^2 includes the elements (0,(0,0)) and (0,(0,1)) and so isn't a function.

>> No.9188998
File: 46 KB, 625x579, 1505345104624.jpg [View same] [iqdb] [saucenao] [google]
9188998

>>9188986
Fine, I'll make one last attempt at defining it for you, as clearly as I can, but at this point I think you're just arguing semantics.

Let the function sqrt be a subset of the cartesian product X x Y, where X = C and Y = C^2 and where every element of X is the first component of one and only one ordered pair in the subset.

>> No.9189002

>>9188983
You don't want to define square root as a function to C^2, because 0 only has one square root. It's better to define square root from C to the power set of C, so that sqrt(b) maps to the set of complex numbers c such that c^2 = b.

>> No.9189007

>>9188998
>Let the function sqrt be a subset of the cartesian product X x Y, where X = C and Y = C^2 and where every element of X is the first component of one and only one ordered pair in the subset.
This is redundant, you could have just written
>Let the function sqrt be a subset of the cartesian product X x Y, where X = C and Y = C^2
But this doesn't add anything to your argument.

>> No.9189011

Does anyone have experience with Ch 9 of Rudin's Principles of Mathematical Analysis? It's multivariable calculus theory.
In summers my uni has a course on just Ch 9 and 10 of this book, basically, and I got a B but I still don't really understand it.

>> No.9189014

>>9189002
Yeah, that's what I was doing originally, albeit not with the absolute strictest terminology that's demanded by /sci/.

>>9189007
>But this doesn't add anything to your argument.
Yes it does, since it eliminates the only counterexample you gave, i.e. that both (0,(0,0)) and (0,(0,1)) where both members of my subset. They're not with that definition.

>> No.9189028

>>9189014
>They're not with that definition.
No, they're not. But the set [math](z,(i \pi z,0)) \subseteq \mathbb{C} \times \mathbb{C}^2[/math] is also a sqrt function by your definition. along with many many others examples.

>> No.9189030

>>9189014
>Yes it does, since it eliminates the only counterexample you gave, i.e. that both (0,(0,0)) and (0,(0,1)) where both members of my subset. They're not with that definition.
But now you can't do anything with the sqrt function that actually makes it useful. You can't even write sqrt(z)^2=z, since sqrt(z) is a set.

>> No.9189031

>>9189011
if you don't feel you understand multivariable calculus, rudin is the last place where you want to go

>> No.9189033

>>9189028
I'm not discussing the actual algorithm to find the subset, just the mapping in general. I thought that was clear.

Anyhow, you're just being dumb on purpose. I'm going to reiterate this one last time, just for you.

sqrt is a function from C to {C, C}, it can't output numbers, only sets.

>> No.9189038

>>9189033
>sqrt is a function from C to {C, C}, it can't output numbers, only sets.
Since when is sqrt a constant function?

>> No.9189048
File: 12 KB, 542x416, 1486199207575.jpg [View same] [iqdb] [saucenao] [google]
9189048

>>9189038
Since never. That's not what C -> {C, C} means. Just like R -> R isn't a constant function. I'm NOT discussing the algorithm, but the domain and co-domain. I've tried to explain that idea to you in about 4-5 different ways now, and I'm tired of trying. You're obviously just being stupid on purpose, and I'm going to stop posting before I get frustrated.

>> No.9189053

>>9189048
>That's not what C -> {C, C} means.
But a function from C to {C,C} has to send every complex number to the set of all complex numbers, i.e. f(z)=C for all z in C. There's no other element in the co-domain. How is this not a constant function?

>> No.9189058

>>9189033
>sqrt is a function from C to {C, C}
Wrong. A function is between sets. The thing on the right isn't even a set.

>> No.9189059

>>9189033
in any case, the notation "{C,C}" is fucking retarded. name ONE source using this notation besides you. and the definition of sqrt as C -> C x C is fucking retarded. not saying it's not possible, but it's useless and it makes sense only for this particular example.

>> No.9189075

can we stop it with the retarded function - not function arguing

>> No.9189077

>>9189075
>can we stop it with the retarded function - not function arguing
Definitions are important in mathematics, you can't just go around calling whatever you want a function.

>> No.9189084

>>9189077
obviously one guy is right, the other wrong, or baiting. unless you're also retarded, then why would anyone keep falling for the bait, or responding to a retard?

>> No.9189088

>>9189084
>obviously one guy is right, the other wrong, or baiting. unless you're also retarded, then why would anyone keep falling for the bait, or responding to a retard?
Retards must be taught.

>> No.9189185

>>9183592
I'm studying (against my will) basic biology; read all about DNA and RNA, speculation on RNA as first living (self-replicating) organism, Miller's lab tests, phosphodiester links and it goes on and on...

>> No.9189223

>>9183592
I'm a retard having troubles with series. My professor is in her second year and does not explain anything properly and we spend half the class in groups for some god forsaken reason, so I'll basically have to teach myself. Can someone please pint me in the direction so I can actually learn this stuff.

>> No.9189229
File: 340 KB, 398x582, 1456449657749.png [View same] [iqdb] [saucenao] [google]
9189229

>>9189223
You need a french dictionary and Éléments de mathématique

>> No.9189248
File: 33 KB, 1033x719, beautiful caligraphy.png [View same] [iqdb] [saucenao] [google]
9189248

>> No.9189278

>>9189223
Just grab an analysis book and read its section on sequences and series. Rudin works fine. I'm sure whatever textbook your class uses would suffice too. Of course the best way to deal with series is to learn measure theory and integrate against the counting measure, but most books are too babby for that.
>>9189248
Why has there been so much autism about [math]\sqrt{z}[/math]? No one defines it as function over all of [math]\mathbb{C}[/math] anyway. as it would no longer be holomorphic or even continuous. Your definition still leads to it being multivalued as you haven't restricted theta to any interval (consider [math]\theta=\pm\pi[/math].) Your handwriting is also terrible and you should just learn to use tex anon.

>> No.9189304

>>9188975
The sqrt on the RHS is the usual function on the real numbers.

>> No.9189340

>>9189278
>θ=-π
disgusting

>> No.9189346

>>9189248
>tfw you define sqrt(z) but can't do any algebra with it because it outputs sets
>can't compute sqrt(z)^2
>can't compute 2*sqrt(z)
>etc.

>> No.9189350

>>9189340
It may be disgusting, but you still need a function to be well-defined.

>> No.9189353

>>9189248
Do you define your real square root like this too? Disgusting.

>> No.9189372

>>9189346
>cant do algebra with sets
hmm
>>9189353
reals are a subset of the complex numbers :^)

>> No.9189412

>>9189372
>>cant do algebra with sets
How would you?

>> No.9189424

>>9189412
Well every function can be represented as a set of ordered pairs, and we can define groups of functions where the binary operation is composition. [math]\{\{(x,x):x\in\mathbb{R}\},\{(x,-x):x\in\mathbb{R}\}\}[/math] would be isomorphic to [math]\mathbb{Z}_2[/math], for example. And yes, I know that isn't what you meant.

>> No.9189467

Can anyone recommend me some light math reading? I'm looking for anything thats not a textbook.

>> No.9189524

is there a name for a differential equation like [math] y' = f(y(x),y(c_1 x),y(c_2 x), ...,y(c_n x)) [/math]

>> No.9189535
File: 89 KB, 659x506, dasitmane.jpg [View same] [iqdb] [saucenao] [google]
9189535

>mfw /k/ommando
>mfw browsing boards high as a kite while fondling rifle
>mfw see an entire community based on math

https://www.youtube.com/watch?v=WJtjeg7gjPg

>> No.9189554

>>9183592
dudes is something like d(x,y) < ε/(1+ε) legal in an analysis proof

>> No.9189560

>>9189535
Not just here, also on various websites like math stacks/overflow and the like. There's actually a surprisingly large community for people who like the subject.

>> No.9189563

>>9189554
Assuming you just want a [math]\varepsilon[/math],
[math]\frac{\varepsilon}{1+\varepsilon}<\varepsilon[/math] for all [math]\varepsilon>0[/math].

>> No.9189569

>>9189563
i have a metric d'(x,y) defined in terms of another metric d'(x,y) = d(x,y)/(1+d(x,y) and i'm showing {xn} converges in (X,d) iff it converges in (X,d').

the proof was almost too easy and now im freaking out about that ε/(1+ε) term

>> No.9189585

>>9189569
The proof should be pretty easy. If d(x,y)<[math]\varepsilon[/math] then d'(x,y)= [math]\frac{\varepsilon}{1+\varepsilon}<\varepsilon[/math]. If d'(x,y)<[math]\frac{\varepsilon}{1+\varepsilon}[/math] then d(x,y)<[math]\varepsilon[/math] (though you should use some algebra to show this one).

>> No.9189598

>>9189585
that's exactly what i did

>> No.9189601

>>9189598
Then all you have to do is bring in the definition of convergence and you should be fine.

>> No.9189614

creamed my pants in the lecture theatre when I learned about Gauss quadrature desu

>> No.9189654
File: 87 KB, 397x231, 1505789837832.png [View same] [iqdb] [saucenao] [google]
9189654

>just started undergrad
>physics major
>calculus i
>of course i got the professor visiting from fucking romania for ONLY this fall semester
>he barely even speaks english
>doesn't even teach either, just throws shit up on the board and rambles in not-english with zero context
fucc

>> No.9189691

>>9189654
Keep calm and study Stewart's essential calculus

Also kahn academy

>> No.9189773

How can I show that:
[eqn]\frac{1}{2}arccos \left(\frac{2x-a}{a} \right) = arccos \left( \sqrt{\frac{x}{a}} \right)[/eqn]

>> No.9189789

>>9189569
it's sufficient to have [math]d(x,y) < g(\epsilon)[/math], where [math]g(x) \to 0 \text{ for } x \to 0[/math]. no need to monkey around with eps/3 in proofs etc. (it's just the squeeze theorem)

>> No.9189814

>>9189773
take cos of both sides, use double angle formula to deal with the 1/2, simpilfy, and go see wsr

>> No.9189822

>>9189467
John Stillwell's Elements of Mathematics

>> No.9189864

what is the value of [math]^ii[/math] (i tetrated by i) please respond

>> No.9189897

>>9189535
wow a science and math board talks about math what a surrrrrrrrprise
dumb /k/uck

>> No.9189930
File: 8 KB, 1617x802, the truth.png [View same] [iqdb] [saucenao] [google]
9189930

>>9183592

>> No.9189937

>>9189864
Define tetration by a complex number first and then we talk.

>> No.9189940

>>9189864
Define a complex number first and then we talk.

>> No.9189943

>>9189940
>Define a complex number first and then we talk.
https://en.wikipedia.org/wiki/Complex_number

>> No.9189969

>>9189943
Can't access that website for some weird reason.
Could you perhaps define it right here?

>> No.9189973

>>9189937
I think that's what they're asking...

>> No.9189979
File: 1.45 MB, 1348x9716, Screen Shot 2017-09-25 at 07.12.51-fullpage.png [View same] [iqdb] [saucenao] [google]
9189979

>>9189969
>Can't access that website for some weird reason.

>> No.9189981

>>9189979
How did you take a long screenshot

>> No.9189988

>>9189981
In Firefox, shift+F2 then type screenshot --fullpage

>> No.9190103

>>9189223
baby rudin

>>9189278
>Of course the best way to deal with series is to learn measure theory and integrate against the counting measure
yeah, I'm sure that's gonna make a lot of sense to a freshman

>> No.9190110

>>9183592

>> No.9190113
File: 944 KB, 4600x2400, UniversalReferenceSheet.png [View same] [iqdb] [saucenao] [google]
9190113

>>9183592
Where is the graduate edition?

>> No.9190136

>>9189937
Why? You don't have to intuitively know what raising a number to a complex power "means" to be able to do calculations with it, similarly theres no reason to need to know what tetration by a complex number "means".

>> No.9190147

>>9190136
e^ix = cos x + i sin x is a definition, not a theorem

>> No.9190156

>>9190147
Doesn't refute anything I said, thanks for posting.

>> No.9190199

>>9189614
i creamed my pants when i derived it

brainlet

>> No.9190201

>>9189773
cancel by arccos on each side

>> No.9190218

>>9190147
who told you that ?

>> No.9190242

>>9190147
It's a theorem. There is no obvious connection between a+bi and e, so you need to prove they are connected.

>> No.9190347

>>9190136
Yes I don't have to know what it "means", but I have to know what it is supposed to be, and google says jackshit about it.

>> No.9190365

>>9190147
That's not really a good definition.
It's the extension of the real exponential on C.
Like, you take the power series of exp(x), then show that it converges for any complex number, and since two holomorphic functions agreeing on at dense set are equal, that (holomorphic) extension must be unique.

>> No.9190471

Can anyone offer any advice on induction? This shit has never made me feel more stupid. I think I've got the basic idea of it down.
>Claim is your general form
>Proof is constructed in 3 parts
>Prove the base case (n=1)
>Assume the inductive step (n=k)
>Prove the hypothesis (n=k+1)
I have a lot of trouble manipulating the Algebra. I can prove it by manipulating both sides but I'm pretty sure that doesn't prove the hypothesis since it changes it.

>> No.9190510
File: 428 KB, 800x600, Dominoeffect.png [View same] [iqdb] [saucenao] [google]
9190510

>>9190471
>I have a lot of trouble manipulating the Algebra. I can prove it by manipulating both sides but I'm pretty sure that doesn't prove the hypothesis since it changes it.
What do you mean?

>> No.9190590

>>9190510
So if I'm correct, you cannot manipulate the left hand side of the equation and the right hand side of the equation. You must manipulate that right hand side to look like the left hand side or vice versa. That's what I'm having trouble with.

>> No.9190593
File: 46 KB, 288x358, 1488066200298.jpg [View same] [iqdb] [saucenao] [google]
9190593

>>9190590
Give us an example of a proof you have trouble with. I'll try to explain it to you.

>> No.9190594

>>9190590
the sides of what equation?

>> No.9190719

>>9190593
Claim: [math]a_n=2^n[/math]
Proof. Let [math]1+2+3+\cdots + 2^n = 2^{n+1}-2[/math]. Proving the base case is [math]2^1=2^{1+1}-2=2[/math] which is the first number in the sum. Now, abstracting n, we get [math]1+2+3+\cdots + 2^k = 2^{k+1}-2[/math] for any natural number [math]k[/math]. We must prove [math]1+2+3+\cdots + 2^k+2^{k+1} = 2^{k+1}-2+2^{k+1}[/math]. (I'm not actually sure I did this part correctly. My thought is, we have the left hand side of the equation which gives us a natural number. So we must add a successor to the sum on the left hand side. Because we added a successor, we must also add a successor to the right hand side in the same form.) And this is where I get lost. Because we have two forms. My thought is, we manipulate the right hand side to look like the left hand side without touching the left hand side or we manipulate the left hand side to look like the right hand side without touching the right hand side.

>> No.9190722
File: 38 KB, 549x560, 0fc13413.jpg [View same] [iqdb] [saucenao] [google]
9190722

>>9190719
What is the claim exactly? I'm sorry but I don't understand what you are trying to prove.

>> No.9190727

We tried to make a cat with Fourier...

>> No.9190736

>>9190722
I'll post the problem in its entirety.
Prove the following statement using the Principle of Mathematical Induction.
[math]\forall n\in\mathbb{N}\mbox{ }, 2+2^2+\cdots+2^n=2^{n+1}-2[/math]

How can I revise my claim then? I'm trying to say that [math]a_n=2^n,\mbox{ }\forall n\in\mathbb{N}[/math]. Is that even an appropriately formed claim?

>> No.9190738

>>9189930
Holistic math goes into this thread: >>9190624

>> No.9190762

>>9190736
where did you get 1+2+3+... from?
the 3 in particular

>> No.9190772

>>9190736
For the base we show that the formula holds for [math]n=1[/math]. We should have [math]2^1 = 2^{1+1}-2[/math] which is clearly true.

Next assume that the formula holds for some [math]k \in \mathbb{N}[/math]. This means that we take the following equality for granted:[eqn]2 + 2^2 + \dots + 2^k = 2^{k+1} - 2[/eqn]Because the above equality is true (that's our assumption), it surely remains true if we add [math]2^{k+1}[/math] to both sides:
[eqn]2 + 2^2 + \dots + 2^{k} + 2^{k+1} = 2^{k+1} - 2 + 2^{k+1}[/eqn]You can check for yourself that the right hand side does equal [math]2^{k+2}-2[/math] and that's the induction step.

>> No.9190775
File: 227 KB, 391x400, swag.gif [View same] [iqdb] [saucenao] [google]
9190775

>>9190736
Okay, let's construct your proof.
>base case
Let [math]n=1[/math]. Then [math]2^n = 2^1 = 2^2-2 = 2^{n+1}-2[/math], just like you said.
>induction
Now suppose the claim holds for some [math]k \ge 1[/math]. We then want to show it holds for [math]k+1[/math]. Now, look at these steps: [math]2 + 2^2 + \cdots + 2^k + 2^{k+1} = (2 + 2^2 + \cdots + 2^k) + 2^{k+1}[/math], and our assumption now gives [math]2 + 2^2 + \cdots + 2^k = 2^{k+1}-2[/math]. Then we plug this in to get [math](2 + 2^2 + \cdots + 2^k)+2^{k+1} = 2^{k+1} - 2 + 2^{k+1} = 2(2^{k+1})-2 = 2^{k+1+1}-2 = 2^{k+2} -2[/math], and the claim holds for [math]n=k+1[/math] as well. Does this help?

>> No.9190777

>>9190736
>I'm trying to say that an=2n,∀n∈Nan=2n,∀n∈N. Is that even an appropriately formed claim?
Saying an=2^n isn't something to be proven. you simply defined/named something.
To prove the statement using mathematical induction you do this:

For n=1:
It is obviously true that 2=2^2-2 therefore the statement is true when n=1.

Let the statement be true for a random natural number n. We want to show that it is true for n+1 as well:
We have:
2^(n+2)-2 = 2*2^(n+1) - 2
and by the inductive hypothesis this is equal to
2 * (2+2^2+...+2^n + 2) - 2 = 2^2 + 2^3 + ... + 2^(n+1) + 2^2 - 2 =
= 2 + 2^2 + 2^3 + ... +2^(n+1)
This shows that the statement is true for n+1

>> No.9190842

>>9190772
>add 2^k+1
another method would be to multiply both sides by 2, then add 2 to both sides

>> No.9190862

>>9190762
If you're referring to >>9190719 then I completely miswrote that. That's my mistake sorry.
>>9190772
Yes, I completely get the base case. My issue comes when adding the parts in other induction proofs that are the [math]2^{k+1}[/math]. I think >>9190775 showed it perfectly though.

>>9190775
I'm going to reiterate just to make sure I get this down correctly.
So when it comes to [math]2+2^2+\cdots+2^{k}[/math] we just add a [math]2^{k+1}[/math] at the end of it. Which would be:

[math]2+2^2+\cdots+2^{k}+2^{k+1}[/math]
Then we group [math]2+2^2+\cdots+2^{k}[/math]. Once that's done we can substitute it for [math]2^{k+1}-2[/math]. And now we have two forms
[math](2+2^2+\cdots+2^{k})+2^{k+1}[/math]
[math]2^{k+1}-2+2^{k+1}[/math].
We make these equal.
[math](2+2^2+\cdots+2^{k})+2^{k+1}=2^{k+1}-2+2^{k+1}[/math].

From here on out I get sort of confused. Can you only touch the right hand side of the equation and make it fall into the form of the left hand side without touching the left hand side or are you allowed to touch both? This is what I did originally. Please keep in mind I used a summation from 1 to k of [math]2^n[/math]
[math]2^k+2^{k+1}=2^{k+1}-2+2^{k+1}[/math]
[math]2^k+2^{k+1}=2^{k+1}+2^{k+1}-2[/math]
[math]2^k+2^k+2=2^k+2^k+2^2-2[/math]
[math]2^{2k}=2^{2k}[/math]

>> No.9190872

>>9190862
godfuckingdamnit. I hope you guys can read that. Sorry about the typos.

>> No.9190943

>>9190471
I'll give a useful example that involves inequalities:

Show that for [math]k\geq 4[/math]: [math]2^k\geq k^2[/math]

Base case ([math]k=4[/math]): [math]2^4=16=4^2[/math]

Induction step: assume true for some [math]n\in\mathbb N[/math] : [math]2^n\geq n^2[/math]

Prove hypothesis: [eqn]2^{n+1}=2\cdot 2^n \stackrel{ID}> 2n^2=n^2+n^2\stackrel{k\geq4}> n^2+2n+1=(n+1)^2[/eqn]

QED

You have to use a justification for every equals sign or inequality (unless its obvious, obviously)

>> No.9190946

>>9190943
I meant IH not ID, and also it should be a [math]\geq[/math] not a [math]>[/math] under it

>> No.9191153

>>9188183
>>9188191
pls help my retarded self

should I go axler? rn I'm trying out Valenza's text on the subject after ditching shilov

>> No.9191187

>>9190943
>Prove hypothesis: [eqn]2^{n+1}=2\cdot 2^n \stackrel{ID}> 2n^2=n^2+n^2\stackrel{k\geq4}> n^2+2n+1=(n+1)^2[/eqn]
I don't get the justification of k≥4

>> No.9191261

>>9191187
You could use an induction argument to show that [math]n^2>2n+1[/math] when [math]n \geq 4[/math], but it is easy to see that a quadratic function will beat a linear one at some point, and that point is exactly 3, but since we're looking at numbers over 3, its irrelevant

>> No.9191442

brainlet here, how do I get interested in math like you guys, its like you're all talking another language

>> No.9191448

>>9183592
Diff Geometry.

>> No.9191463

>>9183821
garden is for turning the brain onto different context m8

>> No.9191503

>>9188091
Not at all, college doesn't assume anything past high school level stuff.

>> No.9191517

I want to do some pen n paper lambda calculus exercises, problems, blah blah. Like I'm back in undergrad calc1 grinding problems.

How do I go about this?

>> No.9191520
File: 76 KB, 850x400, iverson_quote.jpg [View same] [iqdb] [saucenao] [google]
9191520

>>9191442
>you're all talking another language
language of patterns bruh

You gotta just start learning it, and then you find interesting things people are saying in it. No one's going to hold your hand. People 500 years ago were ahead of where most people are at today, so what's to be done? You leave behind the people who don't care.

So just start my friend. Welcome, my friend.

>> No.9191564

>>9191442
It all really depends on what you want to do in life, Linear Algebra, for instance can be applied in many fields, Artificial intelligence, Quantum mechanics, mechanical engineering, are just a few of those fields that come to my mind.

Studying math however is not for everyone, you need to be able to see the beauty of it, an equation that's just memorized because a class required you to, will not trigger the awe or allow you to see it's beauty.

That said despite math being used in many fields and being the mother of all other sciences, there are some who enjoy it just for the sake of the mental stimuli, that sensation when you finally understand a subject, or when you finally solve a problem is pretty satisfying and it rewards your brain, this reward creates some sort of addiction in some.

>> No.9191568
File: 48 KB, 920x515, 920x920.jpg [View same] [iqdb] [saucenao] [google]
9191568

Took an abstract algebra exam today, forgot how to prove a function was an injection in the heat of the moment, the fuck is wrong with me?

>> No.9191570

Does anyone else hate numerical methods? They have their use and everything, but they're inefficient algorithms at best.

>> No.9191571

>>9191568
Answer it correctly for us now to make amends

>> No.9191572

>>9191571
I realized how to solve the problem while taking a leak right after the exam, fucking sucks bruh. This type of shit usually happens to me, I will forget some trivial method and everything is fucked.

>> No.9191579

>>9191572
If you can't place it right now than you have failed twice.

Memorizing for a fucking test is meaningless.
Place your answer now, and if you can't, then you need to address that you are inadequate.

>> No.9191582

>>9191579
Place what? how to show a function is injective? assume f(x) = f(x'), then show x = x'.

>> No.9191618
File: 375 KB, 900x900, 1479693352543.jpg [View same] [iqdb] [saucenao] [google]
9191618

Just started with calculus(it's the 2nd month already). For some retarded reason, I started using a different book for homeworks, and it confused the hell out of me because lecture, and homework had 0 relation.
In the end, for our 1st test, I had to redo all the homeworks, and didn't even pass the test when it was about limits. We are doing derivatives now, and I've been understanding pretty well so far.

>> No.9191625

Fellas, does [eqn] \prod\limits_{n=1}^{x} n^n =O\left(e^{e^x}\right) [/eqn]? I haven't into number theory yet but I'm curious.

>> No.9191666

>>9191625

[math]n^n = e^{n \ln n}[/math]

Sum of n ln n is less than or equal to [math]x^2 ln x[/math].

So it is [math]O(e^{x^2 \ln x}[/math] which is clearly [math]O(e^{e^x})[/math].

Obviously not a tight bound so it is not true if you meant big theta instead of big O

>> No.9191704

>>9191618
hmmm which uni?

>> No.9192001

>>9191153
Just pirate both of them man and see yourself.

>> No.9192060

Does any one know the difference between strang's, Introduction to linear algebra, and strang's linear algebra and it's applications? They seem the same table of content-wise.

>> No.9192129

>>9191582
You mentioned abstract algebra, it can sometimes be a lot easier to just show that the kernel of your homomorphism is trivial.

>> No.9192183

>>9191568
>abstract algebra
probably had to show kernel is trivial, easiest way usually

>> No.9192203

>>9188290
how exactly?

>> No.9192206
File: 214 KB, 704x480, Asuka_Charming.png [View same] [iqdb] [saucenao] [google]
9192206

How can you make [math](\mathbb{Z}_{12},+)[/math] into a Lie group using the discrete topology?

>> No.9192211

>>9192206
in the obvious way

>> No.9192249

>>9191572
it's because you're retarded.

>> No.9192400

>>9183592

Anyone know any theorems on figuring out whether a polynomial is irreducible, similar to Eisenstein's criterion and Cohn's criteron? Specifically for a polynomial in which the coefficients are divisible by increasingly higher powers of primes (e.g. n or n! etc).

>> No.9192411

>>9190147
Where the fuck did you learn that?

>> No.9192416

>>9183592
Vector analysis in Cal III
Haven't had first exam and I already wanna an hero

>> No.9192434
File: 57 KB, 635x610, Captura.png [View same] [iqdb] [saucenao] [google]
9192434

just started Discrete mathematics for my CS degree, anyone know a source I can use to learn about pic releated?

>> No.9192505

>>9192400
You can take the polynomial and take the modulo of it to some integer m which doesn't divide the leading coefficient. If the transformed polynomial is irreducible in [math] \mathbb{Z}_m[x] [/math] then the original polynomial is irreducible in [math] \mathbb{Q}[x] [/math]

>> No.9192519

>>9192434
You can use the axiomatic schemata plus modus ponens as well with some metatheorems.
You can also use the completeness theorem and prove them by truth tables.

>> No.9192534

Diagrammatic algebra right now.
Here's a link to a pdf (the basics)

>> No.9192537

>>9192534
http://www-bcf.usc.edu/~lauda/talks/diagram.pdf
I'm an idiot and forgot the link heh. Here ya go.

>> No.9192563

>>9192211
well what the fuck is the obvious way smartass?

>> No.9192574

>>9188840
Alternating around the limit always smells like continued fraction

>> No.9192580
File: 212 KB, 768x1015, themangaguidetolinearalgebra-150118123345-conversion-gate02-thumbnail-4.jpg [View same] [iqdb] [saucenao] [google]
9192580

>>9189467

>> No.9192583

>>9192580
Why didn't I have this when I learnt linear :c

>> No.9192592

>>9192563
>well what the fuck is the obvious way smartass?
look at the definition...

>> No.9192598

>>9189822
Thanks, looks interesting.

>>9192580
kek I actually own this already

>> No.9192739

>>9192505

Thanks, I assume you mean the finite field for a prime m? Or does this work with composite numbers/prime powers?

>> No.9192761
File: 165 KB, 1209x902, Fraleigh_Section 23_Excercise 37_Solution.jpg [View same] [iqdb] [saucenao] [google]
9192761

>>9192739
>Or does this work with composite numbers/prime powers?
It does

>> No.9192816

>>9192761

Thank you!

>> No.9192927

>>9192592
oh gee didnt look at that. THANKS..

>> No.9192998

>>9192927
>oh gee didnt look at that. THANKS..
No problem.