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File: Linear-Differential-Equation.png ( 667.17 KB , 2000x2068 , 1655761703189.png )

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  No. 207
Reply
>>626
Post diff eqs and their solutions.
12 posts and 1 image reply omitted. Click here to view.
>>
  No. 585
Here's a fun one:
∂tu+(u⋅\nablda)u−νΔu+\gradp=f\partial_t u+(u\cdot\nablda)u-\nu\Delta u+\grad p=f

div  u=0\mathbf{div}\;u=0

I'll leave it to the next person to post the solution.
>>
  No. 605
nice
>>
  No. 626

File: FTDC9FGXwAEZ211.jpg ( 297.09 KB , 2048x1616 , 1705954986668.jpg )

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>>207
how did i not see this lol i have a wide collection

Consider
y′′−2y=ln⁡(x)y''-2y=\ln(x)
.

Let's first solve for the complementary solution, which is the general solution to the complementary equation
y′′−2y=0y''-2y=0
.

The characteristic polynomial for the complementary equation is
λ2−2=0\lambda^2-2=0
, so
λ=±2\lambda=\pm \sqrt{2}
. Hence, the complementary solution is
yc(x)=Ay1(x)+By2(x)=Ae−x2+Bex2y_c(x)=Ay_1(x)+By_2(x)=Ae^{-x\sqrt{2}}+Be^{x\sqrt2}
for some constants
AA
and
BB
. Now let's find the particular solution using variation of parameters. Using the two basis solutions of the complementary solution,
y1(x)y_1(x)
and
y2(x)y_2(x)
, we construct the Wronskian and get
W(e−x2, ex2)(x)=∣e−x2ex2(e−x2)′(ex2)′∣=22W\left(e^{-x\sqrt{2}},\, e^{x\sqrt2}\right)(x) = \begin{vmatrix} e^{-x\sqrt{2}} & e^{x\sqrt{2}}\\ \left(e^{-x\sqrt{2}}\right)' & \left(e^{x\sqrt{2}}\right)' \end{vmatrix} = 2\sqrt2


Hence, for the particular solution
yp(x)=u1(x)y1(x)+u2(x)y2(x)y_p(x)=u_1(x)y_1(x)+u_2(x)y_2(x)
, we have,
u1(x)=−∫ln⁡(x)y2(x) dxW(e−x2, ex2)(x)=−∫ln⁡(x)ex2 dx22=14Ei⁡(x2)−14ex2ln⁡(x) u_1(x) = - \int\frac{\ln(x)y_2(x)\text{ d}x}{W\left(e^{-x\sqrt{2}},\, e^{x\sqrt2}\right)(x)} = -\int\frac{\ln(x)e^{x\sqrt2}\text{ d}x}{2\sqrt2} = \frac{1}{4}\operatorname{Ei}\left(x\sqrt2\right)-\frac14 e^{x\sqrt2}\ln(x)

u2(x)=∫ln⁡(x)y1(x) dxW(e−x2, ex2)(x)=∫ln⁡(x)e−x2 dx22=14Ei⁡(−x2)−14e−x2ln⁡(x) u_2(x) = \int\frac{\ln(x)y_1(x)\text{ d}x}{W\left(e^{-x\sqrt{2}},\, e^{x\sqrt2}\right)(x)} = \int\frac{\ln(x)e^{-x\sqrt2}\text{ d}x}{2\sqrt2} = \frac{1}{4}\operatorname{Ei}\left(-x\sqrt2\right)-\frac14 e^{-x\sqrt2}\ln(x)


The particular solution then is nothing but
yp(x)=u1(x)y1(x)+u2(x)y2(x)=(14Ei⁡(x2)−14ex2ln⁡(x))e−x2+(14Ei⁡(−x2)−14e−x2ln⁡(x))ex2=14e−x2(e2x2Ei⁡(−x2)+Ei⁡(x2)−2ex2ln⁡(x)) y_p(x) = u_1(x)y_1(x)+u_2(x)y_2(x) = \left(\frac{1}{4}\operatorname{Ei}\left(x\sqrt2\right)-\frac14 e^{x\sqrt2}\ln(x)\right) e^{-x\sqrt2} + \left(\frac{1}{4}\operatorname{Ei}\left(-x\sqrt2\right)-\frac14 e^{-x\sqrt2}\ln(x)\right) e^{x\sqrt2}\\ = \frac14 e^{-x\sqrt2} \left( e^{2x\sqrt2}\operatorname{Ei}\left(-x\sqrt2\right) + \operatorname{Ei}\left(x\sqrt2\right) - 2e^{x\sqrt2}\ln(x) \right)

Thus the complete general solution is given by
y(x)=yc(x)+yp(x)=Ae−x2+Bex2+14e−x2(e2x2Ei⁡(−x2)+Ei⁡(x2)−2ex2ln⁡(x)) \boxed{y(x) = y_c(x) + y_p(x) = Ae^{-x\sqrt{2}}+Be^{x\sqrt2} + \frac14 e^{-x\sqrt2} \left( e^{2x\sqrt2}\operatorname{Ei}\left(-x\sqrt2\right) + \operatorname{Ei}\left(x\sqrt2\right) - 2e^{x\sqrt2}\ln(x) \right)}
>>
  No. 646
>>210
>>212
>>213

Honestly, I never understood why we defined the notion of an "elementary" function the way we did. An elementary function from what I've seen is completely restricted to polynomials/rational powers, exponentials, and logarithms (trig/hyperbolic trig and their inverses are expressible in exponentials/logs), which is extremely limited.
>>
  No. 655

File: GFv9PYsakAA1vJW.jpg ( 1.28 MB , 1400x1900 , 1708564466975.jpg )

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We have
y′=(1+x) y+xy2y'=(1+x)\,y+xy^2
.

Rewriting this gives
y′−(1+x) y=xy2y'-(1+x)\,y=xy^{2}
. This is a standard Bernoulli diffeq form
y′+a(x) y=b(x) yαy'+a(x)\,y=b(x)\,y^{\alpha}
, which is solved by performing the substitution
ϕ=y1−α\phi=y^{1-\alpha}
. Using this substitution, which for our specific case is
ϕ=y1−2=y−1\phi = y^{1-2} = y^{-1}
, we have
−y′y2+x+1y=−x⟺ϕ′+(x+1) ϕ=−x\displaystyle -\frac{y'}{y^2} +\frac{x+1}{y} = -x \Longleftrightarrow \phi' + (x+1)\,\phi = -x
.

To make this substitution more rigorous (since it is less obvious), differentials can be used. Also, to account for this division and ensure we miss no solutions, we make sure to check when the denominators are equal to 0, which is
y=0y=0
. Since plugging in
y=0y=0
satisfies the diffeq, this is a constant solution. Then, using the integrating factor
exp⁡(∫x+1 dx)=ex22+x\exp\left(\displaystyle\int x+1\text{ d}x\right) = e^{\frac{x^2}{2}+x}
, we get
ϕ′ex22+x+(ex22+x)′ϕ=−xex22+x∫(ϕ ex22+x)′dx=∫−xex22+x dxϕ ex22+x=π2eerfi⁡(x+12)−ex22+x+Cϕ=π2eerfi⁡(x+12)−ex22+x+Cex22+x \begin{align*} \phi'e^{\frac{x^2}{2}+x} + \left(e^{\frac{x^2}{2}+x}\right)'\phi &= -xe^{\frac{x^2}{2}+x}\\ \int\left(\phi \,e^{\frac{x^2}{2}+x}\right)'\text{d}x &= \int-xe^{\frac{x^2}{2}+x}\text{ d}x\\ \phi \, e^{\frac{x^2}{2}+x} &= \sqrt{\frac{\pi}{2e}}\operatorname{erfi}\left(\frac{x+1}{\sqrt{2}} \right)-e^{\frac{x^{2}}{2}+x}+C\\ \phi &= \frac{\sqrt{\frac{\pi}{2e}}\operatorname{erfi}\left(\frac{x+1}{\sqrt{2}} \right)-e^{\frac{x^{2}}{2}+x}+C}{e^{\frac{x^2}{2}+x}} \end{align*}

Undoing our substitution, we have, for the positive Dawson Integral
D+(x)=π2e−x2erfi⁡(x)D_+(x) = \frac{\sqrt{\pi}}2e^{-x^2}\operatorname{erfi}(x)
that
ϕ=y−1=π2eerfi⁡(x+12)−ex22+x+Cex22+x=2 D+(x+12)−1+Ce−x22−x\displaystyle \phi = y^{-1} = \frac{\sqrt{\frac{\pi}{2e}}\operatorname{erfi}\left(\frac{x+1}{\sqrt{2}} \right)-e^{\frac{x^{2}}{2}+x}+C}{e^{\frac{x^2}{2}+x}} = \sqrt{2}\,D_+\left(\frac{x+1}{\sqrt2}\right)-1+Ce^{-\frac{x^2}{2}-x}

  ⟹  y(x)=0,12 D+(x+12)−1+Ce−x22−x\displaystyle\implies \boxed{y(x)=0, \qquad \frac{1}{\sqrt{2}\,D_+\left(\frac{x+1}{\sqrt2}\right)-1+Ce^{-\frac{x^2}{2}-x}}}

File: download.png ( 2.77 KB , 220x220 , 1643169739562.png )

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  No. 118
Reply
>>124 >>160
What the hell is a quotient group and why should i care? it's just a bunch of cosets
5 posts omitted. Click here to view.
>>
  No. 160
>>118
You'll need it to understand algebraic topology
>>
  No. 643

File: 824ae33f726460269660f8be88357bcc.jpg ( 222.61 KB , 873x1225 , 1706912919666.jpg )

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>>124
>Quotients are essential tools for investigating algebraic structures. If you want to understand algebra, master this. Once you do you get the First Isomorphism Theorem which is used in countless proofs
Yeah, you use quotient groups and the FIT to construct algebraic structures out of existing ones. For example, the quotient ring R[x] / (x^2 + 1) is the complex numbers.
>>
  No. 647
More generally speaking, the idea of quotients in mathematics reaches far beyond its specific application in group theory. A quotient is a new structure built out of an existing structure by declaring certain elements of the old structure equivalent to each other.

The first time in mathematics you probably saw a quotient was when you learned about equivalent fractions. We can define addition, subtraction, multiplication, and division on numerator-denominator pairs, but that structure isn't very interesting or useful in itself, nor do the arithmetic operations have the nice properties we expect of them. Only after taking the quotient by declaring certain numerator-denominator pairs equivalent do we get the nice and useful field of fractions.

In order to make a quotient, you need:
- some type of mathematical object
- an equivalence relation on those objects (reflexive, symmetric, and transitive)
- strictly speaking optionally, but this is what makes the quotient useful: some operations on the type of object that respect the equivalence relation
(For example, if a/b and c/d are equivalent fractions, then a/b - e/f is equivalent to c/d - e/f, and similarly e/f - a/b is equivalent to e/f - c/d; thus subtraction respects the equivalence relation.)

There are several ways to construct a quotient:
- Elements of the quotient can be taken to be the equivalence classes of elements of the original structure. This is the most common method. In the example of quotient groups, cosets are the equivalence classes.
- You can choose a representative element from each equivalence class (for example, reduced fractions).
- You can assume that quotient types exist and assume rules about how quotient types work, as is done in the Lean proof assistant.

Quotients are ubiquitous in mathematics, even in constructing basic stuff like naturals -> integers -> rational numbers -> complex numbers. We just mentioned rational numbers, which are presented as a quotient even to schoolchildren. Someone already mentioned how we can construct the complex numbers from real polynomials in a variable called i by setting i^2 = -1. Integers may be constructed from pairs of natural numbers in a manner very similar to the construction of the rationals. The common Cauchy-sequence construction of the reals starts with a subset of sequences of rational numbers, then takes sequences to be equivalent if their difference approaches zero.Post too long. Click here to view the full text.
>>
  No. 652 >>653
Exercise for the reader: Show that an equivalence relation ~ on the elements of a group is respected by the group operation (meaning a~b and c~d implies ac~bd) if and only if its equivalence classes are the cosets of a normal subgroup.
>>
  No. 653
>>652
This shows that the variety (in the universal algebraic sense) of groups is what's called ideal determined.

File: function.png ( 258.48 KB , 512x512 , 1701966361868.png )

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  No. 549
Reply
Why are so few people interested in looking for nice function plots? It's like magic, a small formula of the x and y coordinate can make such a complex and beautiful picture. Can we start looking for such nice functions here? Use whatever tool you have (if you want I made a dirty JS tool at http://www.tastyfish.cz/functionplot.html).
>>
  No. 551
the js thing you made is genuinly pretty cool. nice job! :D

that being said, here's an interesting function plot :)

>https://mathworld.wolfram.com/TuppersSelf-ReferentialFormula.html
>>
  No. 645

File: eo.png ( 169.69 KB , 1738x1844 , 1707001997952.png )

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Here's a totally underrated graph: Try
gcd⁡(x,y)=1\gcd(x, y) = 1
on Desmos. It seems to be following the pattern of Euclid's orchard.

File: ConfusedAnimeGirl.gif ( 40.83 KB , 461x416 , 1703636547922.gif )

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  No. 569
Reply
What's the deal with local and global fields?
9 posts and 5 image replies omitted. Click here to view.
>>
  No. 597 >>598 >>620

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>>
  No. 598 >>602 >>620
>>597
Neither will a calculator solve most math problems, nor will a tank solve the rest, Alberto.
>>
  No. 602 >>620
>>598
You miss the... ehm, deeper point.
>>
  No. 620 >>621

File: たそ～💊타소.png ( 486.14 KB , 1220x1037 , 1705112459065.png )

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>>597
>>598
>>602
What if your tank has a CAS with an integrator module?
>>
  No. 621
>>620
>CAS
>close air support

>integrator module
>UAV

Lol

File: AG.PNG ( 108.23 KB , 740x599 , 1650887048646.png )

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  No. 167
Reply
>>168 >>171
Talk schemes
>>
  No. 168
>>167
>weird pentagram thing
Why is /math/ so schizo?
>>
  No. 171

File: Daniell.png ( 411.77 KB , 1675x739 , 1651214487808.png )

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>>167
I honestly have no clue what a scheme is, err the algebraic geometry thing-wise. I do know about this scheme though

File: Bijection.png ( 32.95 KB , 1200x1200 , 1664017449826.png )

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  No. 310
Reply
>>600 >>406 >>409 >>410
>Bijection
>Isomorphism
>Equivalence Relation
It's all just the same, innit?
4 posts omitted. Click here to view.
>>
  No. 409
>>310
>>406
Yup. Isomorphism is a bijection that preserves any structure we care about. For groups, it's the group structure, homeomorphisms preserve the continuity both ways, diffeomorphisms preserve both the continuity and the differentiability. Thus we would say "the notion of isomorphism for a structure" is diffeo/homeo/etc/morphism.
>>
  No. 410
>>
  No. 435
>>364
An isomorphism induces an equivalence between one structure with another. You can even give it a equality meaning using univalence axiom.
>>
  No. 553
equivalence relation is different
>>
  No. 600
>>310
functions are set homomorphisms
bijections are set isomorphisms

File: JohnvonNeumann.gif ( 187.66 KB , 490x639 , 1703776296430.gif )

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  No. 590
Reply
Happy Birthday von Neumann
>>
  No. 591
Happy Birthday to him
>>
  No. 592

File: 229f48a3e2e51980f404997bd1688f0a.jpg ( 1.79 MB , 3025x3912 , 1703801252663.jpg )

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Von Neumann helped fix the contradictions in Set Theory by rephrasing sets as classes which are members of classes.

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  No. 351
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>>376 >>490 >>492
(please move to /adv/ if this belongs there)

I started with category theory a week ago with Lane's book. It's a bit hard and there are some examples (like lie groups) that I don't quite get. Related to this I've read Fraleigh's book on abstract algebra. Is this enough background or should I wait a bit before getting into category theory?

Thanks
7 posts and 2 image replies omitted. Click here to view.
>>
  No. 568 >>570 >>573

File: CategoryCatGirl.jpg ( 281.73 KB , 1669x2160 , 1703630357817.jpg )

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>>567
Not >565 but category theory comes in handy to formulate many concepts, it's a useful language to have.

For example often one will have already established what a "morphism" is and would like to also have the notion of "isomorphism". Category theory tells one that the latter notion follows from the former while having all the desirable properties. Similiar applies to say the notion of product or coproduct.

This also allows one to develop machinery and tools in the abstarct context of category theory, e.g. homological algebra in Abelian categories, and then apply those tools in many different contexts, e.g. homological algebra for R-modules, or sheaves of R-modules, etc..
>>
  No. 570 >>571

File: 2ce513be6e4d8ace1937f2c8ad6706c0.png ( 7.71 MB , 1669x2471 , 1703638382677.png )

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>>568
Would you say that if set theory is the skeleton, then category theory is the connecting tissue?
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  No. 571 >>573

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>>570
Due to the way things end up being I'd say it's not an unfair analogy to draw. However primarily I'd say they're a priori just two different languages for describing maths.
>>
  No. 573 >>575
>>568
>Not >565 but category theory comes in handy to formulate many concepts, it's a useful language to have.

And why is it better than set theory, HOL or mereology?
(Or even some kind of formal ontology in the informatics)

As far as I understand, you could easly define a predicate "x is isomorph to y" := Iso(y,x) and work with this.
Okay, I see the advantage of a graphical picture.

What I have see in categories looks suspicies like a usual powerset.

Or I'm just to simple-minded to get it at all?

>>571
Okay.

As far as I see, the central relation in set theory is the "being part of", maybe in the HOL is more "impled".
What is the categories about?
>>
  No. 575
>>573
First of all I wouldn't say it's "better" than set theory or higher order logic.
Second of all you can't just easily define a predicate "is isomorphic to", well you can, but it's useless and you're missing the point. You want to define it in a way for it to have the desirable properties and if you just define a predicate, then you have to figure out how to piece it into the rest of the theory.
Third of all higher order logic is fundamentally different from set theory or category theory, as it's part of the deductive system of your theory, not part of the actual things you talk about. So indeed I'd say yes you're being too simple minded/delusional to get it right now.

As to what category theory is "about" is I'd say morphisms, the same way set theory is abour membership, since morphisms are what ought to be defined to speak of a category and draw the beloved arrows.

File: logica.jpg ( 94.1 KB , 566x929 , 1703673978707.jpg )

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  No. 574
Reply
hat is your opinion about the Logica Hamburgensis, anon?

What do you say about?
https://www.digitale-sammlungen.de/de/view/bsb11273280?page=7

File: question18.png ( 11.57 KB , 693x61 , 1690377094530.png )

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  No. 456
Reply
>>459 >>457 >>482
Prove you are not a Midwit.
7 posts and 2 image replies omitted. Click here to view.
>>
  No. 514 >>533
>>460
Can be done in a couple of ways; summing
⌊akm⌋⌊bkm⌋\lfloor \frac{ak}{m} \rfloor \lfloor \frac{bk}{m} \rfloor
or computing
∫01{ax}{bx}\int_0^1 \{ax\}\{bx\}
works.

For(1),We will show that
λ=∫01{ax}{bx}dx\lambda = \int_{0}^{1}\left \{ ax \right \} \left \{ bx \right \}dx
satisfied. In fact,we would show that
[ab(k−1)m,abkm]∩Z=∅\left [ \frac{ab\left ( k-1 \right )}{m} ,\frac{abk}{m}\right ]\cap\mathbb{Z}= \emptyset

then we are done. This
⟺{akm}{bkm}−m∫k−1mkm{ax}{bx}dx=O(1m)\Longleftrightarrow\left \{ \frac{ak}{m} \right \} \left \{ \frac{bk}{m} \right \} -m\int_{\frac{k-1}{m} }^{\frac{k}{m} }\left \{ ax \right \}\left \{ bx \right \} dx=\mathrm {O}\left ( \frac{1}{m} \right )
and
⟺sup⁡x∈[k−1m,km]{ax}{bx}−inf⁡x∈[k−1m,km]{ax}{bx}=O(1m)\Longleftrightarrow \sup_{x\in \left [ \frac{k-1}{m} ,\frac{k}{m} \right ]}\left \{ ax \right \} \left \{ bx \right \} -\inf_{x\in \left [ \frac{k-1}{m} ,\frac{k}{m} \right ]}\left \{ ax \right \} \left \{ bx \right \} =\mathrm {O} \left ( \frac{1}{m} \right )
It's obviously right.

■\blacksquare
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  No. 533 >>534 >>536
>>458
>>514
>mathchan
>can't even render weblatex
this is sad
>>
  No. 534 >>547
>>533
You have to use 
\[ ... \]
instead of dollars.
>>
  No. 536
>>533
Exactly just what I was thinking lmao
>>
  No. 547
>>534
>he did not read the sticky

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