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/math/ - Mathematics

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File: question.png ( 1.95 KB , 404x273 , 1731228993157.png )

Image
  No. 1012
Reply
>>1013
Can any given probability be reduced to a simple judgment, zero or one?
This would mean that we could reduce any distribution to the binomial distribution
>>
  No. 1013
>>1012
no

File: xiom.png ( 380.54 KB , 1080x980 , 1663002740804.png )

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  No. 299
Reply
>>604 >>709 >>711 >>732 >>767 >>779 >>876 >>949
When did you realize that the reals are fake?
Mathematics can do without infinities.
46 posts and 3 image replies omitted. Click here to view.
>>
  No. 951
>>949
Without ever considering the concept of infinity it isn't really feasible. Perhaps as the least positive real e such that x^x<y^y if 1/e<x<y you might like.
>>
  No. 954
Only one thing in the universe is infinite and that's the coping and seething of finitistcels
>>
  No. 955 >>956
In order to be mathematics, Ultrafinitist and finitists need to make a logical coherent system.
They can't. It's about the physics to deceide whether the world as a whole is infinite or finite in nature.
As far as I know, our informations doesn't allow such a inference. The topic is closed at this moment.
>>
  No. 956
>>955
I'd say finitism is moreso a philosophical perspective than anything very empirical.
>>
  No. 977

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.
13 posts and 1 image reply omitted. Click here to view.
>>
  No. 605
nice
>>
  No. 626

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

Image
>>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 )

Image
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}}}
>>
  No. 1005

File: futaba.jpg ( 25.7 KB , 414x414 , 1717182282102.jpg )

Image
  No. 798
Reply
>>834 >>862
Solve these problems if you're not a retard:

Let
⟨a⟩=(a1,a2,…)\langle a \rangle = (a_1, a_2, \ldots)
denote an infinite sequence of positive integers.
→\rightarrow
Prove that there is no
⟨a⟩\langle a \rangle
such that
gcd⁡(ai+j,aj+i)=1\gcd(a_i + j, a_j + i) = 1
for all
i≠ji \neq j
.
Let
p≠2p \neq 2
be a prime.
→\rightarrow
Prove that there is an
⟨a⟩\langle a \rangle
such that
gcd⁡(ai+j,aj+i)=p\gcd(a_i + j, a_j + i) = p
for all
i≠ji \neq j
.
>>
  No. 834 >>860
>>798
cool homework bro

1. if the problem didnt have the +i +j bit then the solutions would just be permutations of the primes
a things position on a list is set, so we can just subtract its number to be re added later
therefore, because the naturals will always overtake the primes, any valid list must contain negative numbers

2
prime infinity
never let rules as intended into your heart
>>
  No. 860 >>865
>>834

what the fuck
>>
  No. 862
>>798
Okay, what now.
>>
  No. 865
>>860
i wrote that at like 7 am
it had relative coherance at the time
>>
  No. 999
>> 798
(A) The problems suggests that
p=2p = 2
is the problem. If we consider
gcd⁡(a2i+2j,a2j+2i)=1\gcd(a_{2i} + 2j, a_{2j} + 2i) = 1
, we see that at least one of
a2ia_{2i}
,
a2ja_{2j}
has to be odd — say,
a2ia_{2i}
. Then
gcd⁡(a2i+2j+1,a2j+1+2i)=1\gcd(a_{2i} + 2j + 1, a_{2j + 1} + 2i) = 1
implies that
aia_i
is odd for odd
ii
. But then
gcd⁡(a2i+1+2j+1,a2j+1+2j+1)=1\gcd(a_{2i + 1} + 2j + 1, a_{2j + 1} + 2j + 1) = 1
is a contradiction since both sides are divisible by
22
.
(B) Just let
ai=pi−i+1>0a_i = pi - i + 1 > 0
, which gives
gcd⁡(pi−i+1+j,pj−j+1+i)=gcd⁡(p(i+j)+2,pj−j+1+i)\gcd(pi - i + 1 + j, pj - j + 1 + i) = \gcd(p(i + j) + 2, pj - j + 1 + i)
and
p∤p(i+j)+2p \nmid p(i + j) + 2
.

File: books.png ( 814.06 KB , 1336x729 , 1683435673797.png )

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  No. 400
Reply
>>401 >>402 >>404 >>489
Request and deliver any books or book recommendation in this thread.

OP starts:
>Algebraic Topology, Allen Hatcher
https://pi.math.cornell.edu/~hatcher/AT/AT+.pdf
You really shouldn't go with any other book. Stick with Hatcher, even when you feel lost.
9 posts and 2 image replies omitted. Click here to view.
>>
  No. 857
>>855
Thanks.
>>
  No. 858 >>873
>>852
NTA but while I really do like his treatment of analysis (i.e. chapter 3 and onwards), I think his introduction isn't that great, specifically chapter 2 (I actually think chapter 1 is great for getting used to basic set theory and learning to handle quantifiers). Especially the problems (e.g. proving Liouville's theorem) are far too difficult compared to what you got in chapter 1. For introductory stuff, I prefer Amann Escher.
>>
  No. 873 >>874
>>858
Test
>>
  No. 874
>>873
What are you testing?
>>
  No. 996
Does anyone know of any books that explain quotient vector space and filtration of nilpotent endomorphism?

File: photo_2022-03-18_17-07-07.jpg ( 48.05 KB , 1280x800 , 1662141820030.jpg )

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  No. 282
Reply
>>438
Hey /math/

I've been trying to understand spherical harmonics to grok an ML paper that represented points in euclidean space in a rotation-and-translation-invariant way (https://arxiv.org/pdf/1802.08219.pdf). I found a great textbook on SO(3) (https://www.diva-portal.org/smash/get/diva2:1334832/FULLTEXT01.pdf), but while I can kind of maybe sort of follow what's being done with them in this particular paper I fail to really get an intuition of what spherical harmonics are, and it feels like there's some pretty beautiful insight in there.

Do you have any advice or perspectives on how to intuitively grasp what these harmonics are and mean, beyond just group theory definitions?
>>
  No. 285
Harmonics are solutions of Laplace's equation in a given coordinate basis. Spherical harmonics are the solutions of Laplace's equation in spherical coordinates.
>>
  No. 438
>>282
https://youtu.be/Ziz7t1HHwBw
>>
  No. 893 >>983
fdsa
>>
  No. 982
fdsafas
>>
  No. 983
>>893
fdsafdsafdsa

File: 3t0kod.jpg ( 76.97 KB , 800x450 , 1675025207218.jpg )

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  No. 346
Reply
>>347 >>385 >>380 >>429 >>507 >>525 >>530 >>544 >>659 >>894 >>981
So, I have a cousin who wants to win medals in math competitions, especially IMO. Post resources, guides and tips for olympiads.
his prep level: he's 12yo(7th standard), has completed mathematics books upto the 10th standard level. What should be his target next? And how do I help him clear doubts? We don't have decent teachers where we live, and the internet doesn't help much
54 posts and 7 image replies omitted. Click here to view.
>>
  No. 897 >>898
>>896
in my opinion p6 seems doable
>>
  No. 898
>>897
it was one of the hardest p6s of all time
it's not "doable", as very few contestants got it right in the time limit
>>
  No. 899
>>411
Uu
>>
  No. 941
dead
>>
  No. 981
>>346
holy what

File: vector-division.png ( 12.35 KB , 1478x540 , 1726609814969.png )

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  No. 964
Reply
>>969
Any reason not to teach the algebra of Euclidean vectors like this? This would come after multiplication of vectors by scalars but before the scalar product of two vectors, and the target audience is students at the level of typical high school juniors or seniors.
1 post omitted. Click here to view.
>>
  No. 970
>>969
Finally somebody gets it
>>
  No. 971 >>972
>Introduce the structures mathematicians actually care about.
Which would include the structure in OP.
>>
  No. 972 >>973
>>971
Ah yes, my favorite structure of the map taking two vectors from a normed vector space, the second of which is nonzero, and outputting the unique scalar by which one is to scale the second such that it is closest to the first one. Certainly we should teach examples of this before the notion of a vector space is introduced, giving it a designated name.
>>
  No. 973 >>974
>>972
rather important in inner product spaces bro
albeit the scaled vector moreso than the scalar itself
the remainder is important too
>>
  No. 974
>>973
Yeah, it's an important example of orthogonal projections

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  No. 685
Reply
>>963
I know absolutely nothing about math, my knowledge is basically stuck to 7th grade stuff at best.
That said, even I can tell those wrong captcha answers are nonsensical, to the point I can easily get around verification by simply guessing the right answer.

Just a heads-up
4 posts omitted. Click here to view.
>>
  No. 726 >>733
>>725
Why?
>>
  No. 733
>>726
This board is being flooded with useless discussion
>>
  No. 735
>>725
it's okay, all these posts will be drafted to /ret/ - retards soon
>>
  No. 963
>>685
the captcha filters me :(
>>
  No. 966
I do not know whether you see me as a part of the problem.
I just comes here to discuss something about mathematical logic. Even if you hold the opinion that logic belongs merely in the realm of philosophy or whatever, you have to admit that symbolical logic like propositional logic are teached and investigated by serious mathematicans like Gödel, Hilbert etc. Hasn't even Erdos proofed something in this field?

I know, am far away from mastery but compared to the endless discussion about 3/3 = 0,999... = 1 or something, I regard myself as a good contributer.

  No. 136
Reply
>>137 >>173 >>667 >>371 >>372 >>403 >>414 >>448 >>474 >>550
Is thrembo real? It's an integer between 6 and 7.
45 posts and 6 image replies omitted. Click here to view.
>>
  No. 832 >>833
>>831
Yes but the final answer was right is all what matters for this captcha
>>
  No. 833 >>838
>>832
The final answer is 3 and in >>830 it is 2.
>>
  No. 838
>>833
no the answer is clearly 2
two integers cant represent a complex number
>>
  No. 839
sneed
>>
  No. 962
logsday is the thremboth day of the week

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