Anonymous at Mon, 23 Sep 2024 14:21:31 UTC No. 16394639

Is there an infinite number of prime numbers that have the form 10^n+1 where n is a positive integer?

🗑️ Anonymous at Mon, 23 Sep 2024 14:40:01 UTC No. 16394662

>>16394639
For [math]1+10n[/math] to be prime [math]n[/math] must be a power of a 2 since if [math]n[/math] is odd you can show the number is composite. This makes it similar to Fermat numbers, and the question of whether there are an infinite number of Fermat primes is still an open question.

Anonymous at Mon, 23 Sep 2024 14:41:12 UTC No. 16394665

>>16394639
For [math]10^n+1[/math] to be prime [math]n[/math] must be a power of a 2 since if [math]n[/math] is odd you can show the number is composite. This makes it similar to Fermat numbers, and the question of whether there are an infinite number of Fermat primes is still an open question.

Anonymous at Mon, 23 Sep 2024 15:11:19 UTC No. 16394710

Can someone of you plant nerds explain to me how the battlefields in Eastern Ukraine aren't completely overgrown fields of heath by now? Posting pic is pointless, they just look like regular agricultural fields off season.
They haven't been sown and harvested in 2.5 years.
They also aren't like mowed or anything, because they often are mined or targeted. In fact, it's a death trap to even move on them, which is why all advancing occurs along tree lines.
Yet despite no human setting foot on them since early 2022, they look freshly mowed.

I tried to ask in the /an/ plant thread, but I got ignored by the roasties who were just talking about potted plants. I guess the topic was to testosterone-ladden for them.

Anonymous at Mon, 23 Sep 2024 15:55:06 UTC No. 16394765

how to prove by definition that lim (3x+4y)=18, when x,y ->2,3?

Anonymous at Mon, 23 Sep 2024 16:10:53 UTC No. 16394789

>>16394710
post a single image of these "freshly mowed" war fields that doesn't actually depict 3 foot grass

Anonymous at Mon, 23 Sep 2024 16:54:33 UTC No. 16394867

>>16394789
I hope you don't think I meant literally freshly mown. This would be extremely autistic by you. I meant it as another way to convey "short grass".

Anonymous at Mon, 23 Sep 2024 16:58:04 UTC No. 16394873

>>16394765
Use the triangle inequality.

Anonymous at Mon, 23 Sep 2024 22:32:19 UTC No. 16395427

Why the uncertainty of 3*(10±0.5) is different than the uncertainty of (10±0.5)+(10±0.5)+(10±0.5)? I mean, I can do the math, but why is it like that? I guess it has something to do with the independence of the variables, but I still don't completely understand why.

Anonymous at Mon, 23 Sep 2024 23:20:32 UTC No. 16395481

>>16395427
same reason why rolling one die has a distribution that looks like top, but rolling three dice has a distribution that looks like bottom. the more variables you have, the more likely they are to cancel each other out and result in a more narrow distribution.

Anonymous at Tue, 24 Sep 2024 08:13:11 UTC No. 16395940

I asked my (busy) professor about an issue with my assessment extension, and he replied, "Should be ok". Should I send him a thank you email? He's the director of graduate research in addition to teaching our undergrad module, so I'm not sure whether he'd be annoyed to get a useless email on top of the load he already has to deal with. Not really good with social situations - t.autist.

Anonymous at Tue, 24 Sep 2024 12:49:24 UTC No. 16396195

>>16394665
What about the next one after 101?

Anonymous at Tue, 24 Sep 2024 13:18:52 UTC No. 16396215

>>16394665
more specifically, they belong to a classification known as generalised Fermat primes
https://oeis.org/wiki/Generalized_Fermat_numbers#Generalized_Fermat_primes_a.5E.282.5En.29_.2B_1

Anonymous at Tue, 24 Sep 2024 15:13:27 UTC No. 16396298

>>16395940
yes, nothing wrong with thank yous

Anonymous at Tue, 24 Sep 2024 15:34:20 UTC No. 16396340

[eqn]\pi \varepsilon^2 = \int_{B_\varepsilon} dS[/eqn]Is this correct - specifically the notation - if [math]B_{\varepsilon}[/math] is understood as a subset of a the two dimensional manifold [math]\mathbb{R}^2[/math] in three dimensional space?

Anonymous at Tue, 24 Sep 2024 15:55:19 UTC No. 16396382

>>16394407
Is Khan Academy a good way to learn math?

I'm gonna work as a store clerc for a year before I start college and I was thinking of using Khan Academy to prepare for the math I'll encounter.

Anonymous at Tue, 24 Sep 2024 16:42:08 UTC No. 16396455

how to prove by definition that lim x,y->3,1 y/(2x-y)=1/5, I got to 2|x-3|+6|y-1| / |5(2x-y)|, I don't know how to find a lower bound for the denominator.

Anonymous at Tue, 24 Sep 2024 19:13:43 UTC No. 16396613

>>16394407

Garrote at Tue, 24 Sep 2024 19:16:14 UTC No. 16396618

I wonder if it's possible to derive all the theory (of how electricity works) just from the force law and DEs.

Anonymous at Tue, 24 Sep 2024 19:31:21 UTC No. 16396630

>>16396618
>force law and DEs.
are you referring to the Coulomb force and Maxwell's equations? if so, and by "all the theory" youre referring to the calssical theory, then you also need the Lorentz force, but then youre done, there's nothing left to derive, its a complete model.

Anonymous at Tue, 24 Sep 2024 20:05:20 UTC No. 16396654

>>16396613
I never put numbers in math mode. 7 instead of $7$. Don't think it makes a difference.

Anonymous at Tue, 24 Sep 2024 20:14:33 UTC No. 16396669

>>16394765
If handling two variables at a time is too much, try proving that [math]\lim_{(x,y) \rightarrow (2,3)} 3x = 6[/math] and [math]\lim_{(x,y) \rightarrow (2,3)} 4y = 12[/math], and then take the sum.

Anonymous at Tue, 24 Sep 2024 20:23:16 UTC No. 16396677

>>16395940
Should be ok.

>>16396455
>how to prove by definition
Wait, you mean doing it the hard way, without using even the most basic theorems?

Anonymous at Tue, 24 Sep 2024 20:24:29 UTC No. 16396679

>>16396677
yes

Anonymous at Tue, 24 Sep 2024 20:25:58 UTC No. 16396682

>>16394407
the number. 41 is a parameter, right, since it is representative of the whole? (dont worry about symbol)

Anonymous at Tue, 24 Sep 2024 20:26:43 UTC No. 16396684

>>16396679
Then, recall the proof of the limit of a quotient, and do the exact same thing but applied to that specific function.

Anonymous at Tue, 24 Sep 2024 20:27:15 UTC No. 16396686

How does a sum become an integral in Riemann zeta?
[eqn]\zeta(s) = \sum_{n=1}^{\infty}\frac{1}{n^s} = \frac{1}{\Gamma(s)}\int_0^{\infty} \frac{x^{s - 1}}{e^x - 1} \mathrm dx[/eqn]

Anonymous at Tue, 24 Sep 2024 20:33:24 UTC No. 16396691

>>16396686
look up the gamma function

Anonymous at Tue, 24 Sep 2024 20:34:22 UTC No. 16396696

>>16396682
Yes

Anonymous at Tue, 24 Sep 2024 20:43:39 UTC No. 16396708

>>16396686
Use a geometric sum
[eqn]
\frac{x^{s-1}}{e^{x}-1} = \sum_{k = 1}^{\infty} e^{-kx} x^{s-1}.
[/eqn]
Because [math]f_{k}(x) = e^{-kx} x^{s-1}[/math] are Lebesgue integrable functions on [math]\mathbb{R}_{\, \geq\, 0}[/math],
[eqn]
\sum_{k = 1}^{m} \int_{\mathbb{R}_{\, \geq\, 0}} dx\, f_{k}(x) = \Gamma(s) \sum_{k = 1}^{m} \frac{1}{k^{s}}
[/eqn]
when substituting [math]u = kx[/math] and [math]du = k \cdot dx[/math]. This is bounded for [math]m \to \infty[/math]. Using the monotone convergence theorem finally
[eqn]
\int_{0}^{\infty} dx\, \frac{x^{s-1}}{e^{x}-1} = \Gamma(s)\, \zeta(s)
[/eqn]

More details in the 8th edition of Otto Forster's Analysis 3 § 5.

Anonymous at Wed, 25 Sep 2024 01:03:30 UTC No. 16396922

Does mass create a "mass field"? And if so, can a "mass flux" be measured?

Anonymous at Wed, 25 Sep 2024 01:17:55 UTC No. 16396929

>>16396922
> Does mass create a "mass field"?
Yeah, it's called gravity.

> And if so, can a "mass flux" be measured?
See above.

Anonymous at Wed, 25 Sep 2024 07:22:24 UTC No. 16397173

[math]G[/math] is a finitely generated and residually finite group (meaning the intersection of all its normal finite index subgroups is trivial). Suppose [math]N \lhd G[/math] is finitely generated and for every [math]\alpha > 0[/math] there is some [math]N \le H \le G[/math] with [math][G:H] > \alpha[/math].

Given [math]\alpha > 0[/math], can one find some [math]K \le G[/math] such that [math][G : NK][/math] and [math][N : N \cap K][/math] are both [math]> \alpha[/math]?

Anonymous at Wed, 25 Sep 2024 07:52:30 UTC No. 16397203

>>16396684
Just said I can't use any other proofs, but just by definition.

Anonymous at Wed, 25 Sep 2024 08:20:47 UTC No. 16397220

>>16396455
Just the way you do it for functions of a single variable. Here's a sketch. Write [math]\left |\frac{y}{2x-y} - \frac{1}{5} \right | = \frac{2}{5} \left |\frac{3y-x}{2x-y} \right |[/math]. Note how you can easily upper bound the numerator using the triangle inequality (I'll leave those details to you). It remains to lower bound the denominator. Intuitively, if [math](x,y)[/math] is "close" to [math](3,1)[/math] then the expression [math]|2x-y|[/math] is "close" to [math]5[/math]. Concretely, suppose [math]|x-3|^2 + |y-1|^2 < \frac{1}{4}[/math], say. Then in particular [math]\max\{|x-3|, |y-1|\} < \frac{1}{2}[/math] which implies [math]x > \frac{5}{2}[/math] and [math]\frac{1}{2} < y < \frac{3}{2}[/math]. By the so-called "reverse" triangle inequality we get [math]|2x-y| \ge |2x| - |y| = 2x - y > \frac{7}{2}[/math]. Take it from here.

Anonymous at Wed, 25 Sep 2024 21:25:09 UTC No. 16397976

>>16397203
>Just said I can't use any other proofs
You didn't but, sorry anyways, I didn't know you were this retarded. You see, said proof gives you an algorithm to prove any limit of that form; if your brain is so small you can't prove it by yourself, then you can follow the steps given by the proof to get to the desired result from the definition.

Anonymous at Thu, 26 Sep 2024 02:15:27 UTC No. 16398348

>>16396708
Very nice

Anonymous at Thu, 26 Sep 2024 14:43:46 UTC No. 16398842

Why can't I post on /sci/? Smdh.

Anonymous at Thu, 26 Sep 2024 14:46:05 UTC No. 16398846

I've derusted a load of tools in vinegar. I was going to keep the ferric acetate solution for ebonising wood, but without thinking I poured baking soda solution in.
tl;dr what is ferric acetate + baking soda, and more importantly, do I still have a reason to hoard this solution, which is now topped with a brown sludge?

Anonymous at Thu, 26 Sep 2024 15:07:15 UTC No. 16398874

>>16394407
Yes, I have a stupid question.
Has anyone ever rotated mercury in a magnet in two different directions by running a circuit through it and then attaching two of those to an aluminum sheet and placing them both above and below said sheet to observe the aetherial vortex and subsequent anti gravitic effects it generates?
Why has no one tried that yet? Sorry i'm very stupid and simple. Plz help

Anonymous at Thu, 26 Sep 2024 15:10:00 UTC No. 16398877

>>16398874
>aetherial vortex and subsequent anti gravitic effects
this is /sci/, not /x/

Anonymous at Thu, 26 Sep 2024 15:49:29 UTC No. 16398913

If I have 3 coins all on heads, how do I calculate the probability that after flipping I will have 2 heads and 1 tail? I know it is 3/8, but I'm pretty sure there is some combinatorics stuff going on that I've forgotten.

Anonymous at Thu, 26 Sep 2024 15:56:43 UTC No. 16398921

>>16398913
> If I have 3 coins all on heads
This part is totally irrelevant and makes no difference to the result.

So ask yourself if you flip three coins, what are the chances you get HHT and how many permutations of that can you have.

🗑️ Anonymous at Thu, 26 Sep 2024 16:59:48 UTC No. 16398966

Start with an S-sided dice, S being some positive integer. When you roll the dice, the digit you get determines which sided dice you are rolling next. So for example, if you start with 10-sided dice and roll a seven, you roll a seven-sided dice next and repeat the same idea.

After N rolls with the S-sided dice, what is the probability of rolling 1?

Anonymous at Thu, 26 Sep 2024 17:02:25 UTC No. 16398969

Start with an S-sided dice, S being some positive integer. When you roll the dice, the digit you get determines which sided dice you are rolling next. So for example, if you start with 10-sided dice and roll a seven, you roll a seven-sided dice next and repeat the same idea.

After N rolls starting with the S-sided dice, what is the probability of rolling 1 based on the variables S and N?

Anonymous at Thu, 26 Sep 2024 17:12:20 UTC No. 16398979

>>16398877
Yeah, that was a science question.

Anonymous at Thu, 26 Sep 2024 17:30:54 UTC No. 16398992

>>16394407
That protohuman looks way smarter than 80% of human kids.

Anonymous at Thu, 26 Sep 2024 17:54:04 UTC No. 16399015

>>16398969
what do you mean rolling 1. On what turn?

Anonymous at Thu, 26 Sep 2024 18:00:37 UTC No. 16399021

>>16399015
I believe they mean N is the turn / number of rolls. So on the N'th roll, the probability would be 1 / S, where S = S(N). The question seems to be more about how S can change with N rather than the final roll.

Anonymous at Thu, 26 Sep 2024 18:25:09 UTC No. 16399046

>>16399015
It was poorly worded. I just mean if you do N rolls with that system then what's the probability you've rolled at least one 1.

Anonymous at Thu, 26 Sep 2024 18:32:42 UTC No. 16399050

>>16399046
> what's the probability you've rolled at least one 1.
that's not what you originally asked, and would be a very different answer

Garrote at Thu, 26 Sep 2024 18:56:43 UTC No. 16399073

Why is topology so kawaii and schway?
Metric spaces, I love them.

Anonymous at Thu, 26 Sep 2024 19:18:34 UTC No. 16399092

>>16399046
If this is what OP meant, you can manually do it with a sort of Pascal's triangle-esque method. Call this one, a cumulative Pascal's rectangle.

Let the S columns of the rectangle represent S, S-1, S-2, ..., 1 from left to right. The rows from top to bottom represent the rolls from n=0 up to row n=N . Let the first row begin with a single one the rest of the S-1 columns be 0. So, let's say S = 7, then row n=0 is [1 0 0 0 0 0 0]. The next row n=1 is then to be the cumulative sum of the n=0 row, which in the example S=7 is [1 1 1 1 1 1 1], and the row after that n=2 is [1 2 3 4 5 6 7]. The row after that is [1 3 6 10 15 21 28] But hold up! Turn it diagonally, this is just Pascal's triangle! You probably could've seen this coming if you know the cumulative aspect of the pascal's triangle.

This rectangle though gives a clear answer though. The number of ways to roll a 1 at whatever n is the final column, which was 0, 1, 7, 28, etc. The total ways to roll is the sum of each row.

There's almost def a closed form for this stuff using binomials and wtv, but I already did most of the work, so finish the rest. The point is if you think of it like a multiverse thing, each action spawns multiple possible futures, whose actions spawn other futures for each successive row. But it's simple enough to just be pascal's triangle.

Anonymous at Thu, 26 Sep 2024 20:07:30 UTC No. 16399130

>>16399092
the sum of the coefficients are also a binomial on Pascal's Triangle, so assuming this is right the answer is one binomial divided by another, or (a Choose b) / (c Choose d)

Anonymous at Fri, 27 Sep 2024 11:30:08 UTC No. 16399832

Do turbines work by moving magnets to induce currents?

Anonymous at Fri, 27 Sep 2024 13:41:02 UTC No. 16399931

>>16399832
yes

Anonymous at Fri, 27 Sep 2024 13:50:46 UTC No. 16399941

>>16398846
Iron acetate + baking soda should produce iron + sodium acetate. If it's brown the iron might have rusted again, which you could reverse with more vinegar. No idea what effect sodium acetate has on wood though.

Anonymous at Fri, 27 Sep 2024 17:45:03 UTC No. 16400182

The average of e and pi. Is there anything special about this number?

Anonymous at Fri, 27 Sep 2024 17:51:04 UTC No. 16400186

L1 and L2 are the tangent lines to C1 and C2 at z respectively. I want to show that the circles intersect at the same angle at z*. How do I do that? I had a feeling I would have to draw triangles using the radii from the centers to z and z*, but I don't really know where to go from there. Would appreciate any ideas.

Anonymous at Fri, 27 Sep 2024 18:12:50 UTC No. 16400214

>>16400186
Letting c1 denote the centre of C1, we know that all points on C1 are the same distance from c1 (by the definition of a circle) - including the lengths from c1 to z and the length from c1 to z*.
The same applies to C2, denoting its centre as c2 similarly.
Furthermore, the distance from c1 to c2 is obviously constant.

So the length from c1 to z is the same as the length from c1 to z*. The length from z to c2 is the same as the length from z* to c2. And the length from c2 to c1 is obviously the same as itself.
The two triangles, having identical side lengths to each other, must be congruent, with congruent angles to match

Anonymous at Fri, 27 Sep 2024 18:15:00 UTC No. 16400215

Are africans closer too chimpanzees than others? If yes how close in comparison?

Anonymous at Fri, 27 Sep 2024 18:19:21 UTC No. 16400218

>>16400182
No, it has no interesting properties. However we still do not know if the sum of [math]e + \pi[/math] is irrational or transcendental, no proof of either exists.

Anonymous at Fri, 27 Sep 2024 18:40:36 UTC No. 16400228

>>16400218
Wasn't there some weird thing that one of e*pi or e+pi must be irrational even though neither has been proven to be irrational in particular?

Anonymous at Fri, 27 Sep 2024 18:44:44 UTC No. 16400231

>>16400228
Yes, I believe so. One of them could be rational and if that was the case the other has to be irrational. Of course the most likely scenario is both are irrational.

Anonymous at Fri, 27 Sep 2024 18:49:26 UTC No. 16400235

>>16400231
I wonder how that was proven it's kind of interesting

Anonymous at Fri, 27 Sep 2024 18:50:29 UTC No. 16400236

>>16400235
>>16400228
>>16400231
The specific reasoning is that since both π and e are known to be transcendental, [math](x-\pi)(x-e)=x^2-(\pi+e)x+e\pi[/math] cannot be algebraic, which it would be if both [math]e\pi[/math] and [math]\pi+e[/math] were rational

Anonymous at Fri, 27 Sep 2024 19:44:34 UTC No. 16400269

>>16400214
Thanks anon, but unless I'm missing something here, I don't think that's the angle I need. The angle I'm looking at here is the one made at the intersection of the lines tangent to C2 and C1 at z*. I need to show that they're the same as the angle between L1 and L2.

Working off that what you said though, I think I could draw this green triangle and use similar arguments to draw that purple triangle (which would be similar), but I don't know how to show that the two purple lines in that triangle are actually the tangent lines to C1 and C2 at z*.

Anonymous at Sat, 28 Sep 2024 10:16:10 UTC No. 16400812

does the euclidian norm preserve convexity of convex sets?

Anonymous at Sat, 28 Sep 2024 11:54:15 UTC No. 16400843

>>16400812
Yes. Every convex subset of R^n is pathconnected, and the norm function is continuous, so the image must also be pathconnected. But a pathconnected subset of R is just an interval, and thus convex.

Anonymous at Sat, 28 Sep 2024 13:42:13 UTC No. 16400932

>>16399941
Thanks lad. Feel a bit dim forgetting my hs chemistry this much. Cool.

Anonymous at Sat, 28 Sep 2024 13:43:51 UTC No. 16400937

>>16400843
thanks

🗑️ Anonymous at Sat, 28 Sep 2024 20:57:26 UTC No. 16401481

I typed a completely random six-digit number and tested if it's a prime: 618547 and it was. What is the probability that a random six digit number (or N-digit number) is a prime?

Anonymous at Sat, 28 Sep 2024 23:20:46 UTC No. 16401661

>>16401481
There's 68,906 six digit prime numbers out of
900,000 six digit numbers (no leading zeros).
That'll be 7.656%

As the primes increase, they become few and
far between as the number of digits increase,
and so the probability gets lowered.

Anonymous at Sun, 29 Sep 2024 12:14:22 UTC No. 16402118

If I wanted to make my own version of MagicMind (expensive nootropic drink) could I follow this ingredient list and mix them together to make my own /diy/ version?

Anonymous at Sun, 29 Sep 2024 16:25:42 UTC No. 16402441

I'm halfway through my linear algebra course and am currently benchmarking my knowledge, so far, polynomials destroy me, what resources will let me fine adjust my knowledge? I've got heaps of books, Strang, Axler, etc. But they don't really delve deeper into polynomials. Thanks

Anonymous at Sun, 29 Sep 2024 16:28:04 UTC No. 16402443

>>16402441
I want to specify, I want some sort of workbook that I can throw my head at ad nauseum until it sticks

Anonymous at Sun, 29 Sep 2024 20:44:05 UTC No. 16402824

The vector space of all infinite sequences is equivalent to the vector space of all functions [math]\mathcal{F}_{\mathbb{R}}(\{1, 2, 3, \dots\})[/math],
so the vector space of all infinite sequences such that [math]x_n = x_{n+2}[/math] is equivalent to the subspace
[math]\{f \in \mathcal{F}_{\mathbb{R}}(\{1, 2, 3, \dots\}) : f(n) = f(n + 2)\}[/math].
So, is the basis of this subspace just Span{[math]f(n)[/math]} or am I missing something here?

Anonymous at Sun, 29 Sep 2024 20:48:25 UTC No. 16402830

Is the number e divided by pi irrational?

🗑️ Anonymous at Sun, 29 Sep 2024 21:03:36 UTC No. 16402855

would really appreciate it if someone could walk me through part c, or at least help me get started so i can figure out the rest.

thanks

Anonymous at Sun, 29 Sep 2024 21:04:35 UTC No. 16402857

>>16402830
It's still an open question. That's the case with most combinations of e and pi you can come up with. The methods used to prove the irrationality of e and pi by themselves don't work when they are combined.

Anonymous at Mon, 30 Sep 2024 13:09:41 UTC No. 16403680

Why does the right-hand-rule exist?
Why is it that out of the combination of the x and y vector the z vector goes in one direction and not the other?
Why is it that if you go forward and right you also go up instead of down?
It just doesn't make any sense to me.

Anonymous at Mon, 30 Sep 2024 13:41:10 UTC No. 16403707

>>16403680
It's just a visual representation of the action of taking a cross product of two vectors. As for why right-handed, that's just convention. We almost always use a right-handed coordinate system (x is the horizontal axis, y is the vertical axis, z is the 'vertical' axis) so a cross product of two vectors using such a coordinate system with follow the same axes.

🗑️ Anonymous at Mon, 30 Sep 2024 14:02:32 UTC No. 16403726

>>16403707
According to pic related magnetic force goes up when the moving charge goes left.
If you want magnetic force to go down you have to either move the charge to the right or the magnetic field line away from you.
It is clearly not symmetrical, what causes this asymmetry?
The direction seems arbitrary.

Anonymous at Mon, 30 Sep 2024 14:03:58 UTC No. 16403728

>>16403707
According to pic related magnetic force goes up when the moving charge goes left.
If you want magnetic force to go down you have to either move the charge to the right or the magnetic field line away from you.
It is clearly not symmetrical, what causes this asymmetry?
The direction seems arbitrary but there must be a reason for it.

Anonymous at Mon, 30 Sep 2024 15:05:29 UTC No. 16403792

>>16403728
Again, it's just an arbitrary convention. Remember that conventional current isn't actually defined as the direction electrons flow, but of 'positive' charge. They got the sign wrong because they didn't completely understand electricity in the beginning. You could just have easily come to learn the left-hand-rule instead. tl;dr it's used because it works.

Anonymous at Mon, 30 Sep 2024 16:25:35 UTC No. 16403878

what's the difference between SU(2) and SL(2,C)?

I'm beginning QFT and I don't understand this. SU(2) is the double cover of SO(3) and SL(2,C) is the double cover of SO(1,3)

Anonymous at Mon, 30 Sep 2024 16:59:19 UTC No. 16403915

>>16403728
>It is clearly not symmetrical
thats called parity violation, and sometimes that happens in physics. we resolve it by bundling spacial dimensions with charge and/or time, which creates a stronger symmetry that hasnt been violated (yet). see: https://en.wikipedia.org/wiki/CPT_symmetry
that being said, the Lorentz force does not violate party (nor does any of classical EM); your pic actually is symmetrical, confusingly enough. you use the right-hand rule to find the magnetic (Lorentz) force, but you also use the right-hand rule to determine the direction of the magnetic field in the first place; in the integral form of Ampere's law, the direction of the closed-looped integral on the left side of the equation is (arbitrarily) determined using the RHR. you could use the LHR if you wanted to, which would cause the sign of the B field to flip, which would then cancel out the fact that you *also* must now use the LHR for the Lorentz force, and your thumb would still point in the same direction even though you used LHR instead of RHR. all the math works out the same. it doesnt look symmetric, but it is.

Barkan at Mon, 30 Sep 2024 17:08:02 UTC No. 16403928

>>16394407
This pic

Read it wisely of what I said

Anonymous at Mon, 30 Sep 2024 17:27:10 UTC No. 16403948

>>16402824
(1010101010101...)
(0101010101010...)

Anonymous at Mon, 30 Sep 2024 17:33:18 UTC No. 16403960

>>16403878
SU(2) is all 2x2 unitary matrices. SL(2,C) is the same but also includes 2x2 hermitian matrices. Those unitary matrices are for rotations, and the hermitian matrices are for boosts.

Anonymous at Mon, 30 Sep 2024 17:37:48 UTC No. 16403964

>>16403960
thanks

Anonymous at Mon, 30 Sep 2024 18:46:01 UTC No. 16404050

>>16394407
>Horizontal line multiplied vertical line equals square mile
Am I correct or being stupid?

Anonymous at Mon, 30 Sep 2024 18:56:17 UTC No. 16404056

>>16404050
Ah, sorry, should have clarified, I have a worldographer hexagon map that has a perimeter of 1200/600 horinzontal/vertical and each hexagon has 6 square miles. I want the total amount of sqaure miles, but so far I´ve gotten 3 different results with my math and I got no clue which is the correct way. It´s been a decade since I´ve done any sort of serious math

Anonymous at Mon, 30 Sep 2024 19:08:56 UTC No. 16404068

>>16404056
Any area represented by square miles is telling you how many squares with side length of one mile fits inside that area. So you simply count how many squares fit.

Anonymous at Mon, 30 Sep 2024 19:23:08 UTC No. 16404092

>>16404068
So I go with 600 multiplied by 1200 and multiply it again by 6? That would be then 360000 square miles if I´m correct?

🗑️ Anonymous at Mon, 30 Sep 2024 19:28:34 UTC No. 16404103

is P = NP or nah?

Anonymous at Mon, 30 Sep 2024 19:30:44 UTC No. 16404106

>>16404103
Nah, but since we can't prove it there will always remain a small, niggling doubt. A what if ...

Anonymous at Mon, 30 Sep 2024 19:50:40 UTC No. 16404151

>>16404056
>1200/600 horinzontal/vertical
what units are those in? can you provide a picture/sketch?

Anonymous at Mon, 30 Sep 2024 21:31:31 UTC No. 16404292

>>16394407
I know perpetual motion is impossible but I can't simply say where the error is. Plus this case has the supposition the compound is made out of things in the ocean like oxygen or sodium. Does reactions are harder under more pressure?

Anonymous at Tue, 1 Oct 2024 01:15:45 UTC No. 16404530

>>16404292
> "The energy wasted to create the compound is constant"
That's your answer. Ask yourself what is replacing that energy over each cycle.

Anonymous at Tue, 1 Oct 2024 02:57:13 UTC No. 16404650

https://www.space.com/earth-mini-moon-asteroid-2024-pt5

What do they mean by "geocentric energy"? How is that defined?

Anonymous at Tue, 1 Oct 2024 03:12:18 UTC No. 16404666

>>16404650
I can't find a good link, only this page from a textbook. Basically it's the energy of the system if you shift the frame of reference to be the center of mass of the two bodies. If E < 0 then it's a bound state (orbit is achieved), if E > 0 then they fly apart.

Anonymous at Tue, 1 Oct 2024 03:40:27 UTC No. 16404692

>>16404666
Thanks Satan! But how does the mini-moon achieve negative geocentric energy? Is this energy temporarily borrowed by Earth in the form of kinetic energy and then kicked back later?

Anonymous at Tue, 1 Oct 2024 03:44:24 UTC No. 16404693

>>16404692
Yes. It's a combination of kinetic and (gravitational) potential energy. So if the object gains some energy the Earth has to lose some.

Anonymous at Tue, 1 Oct 2024 03:45:48 UTC No. 16404694

>>16404292
How is it even perpetual? The compound is being yeeted

Anonymous at Tue, 1 Oct 2024 04:26:37 UTC No. 16404726

>>16404666
Wait, is this the same thing?

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

Anonymous at Tue, 1 Oct 2024 15:52:21 UTC No. 16405225

>>16404530
Yeah the energy could be outsourced from any other place,however the depth and therefore the distance and energy can be increased ,for example imagine that to create 1kg you waste 100J but I have the machine at 1000m of depth you could generate 9700J of energy.
>>16404694
It's perpetual because it produces more energy than the system is introduced,energy always is always lesser than the original input when harvesting it.

>>16404694

Anonymous at Tue, 1 Oct 2024 16:56:33 UTC No. 16405308

>>16405225
That's not perpetual. I burn wood to power a turbine. When the wood's gone the turbine stops. That's not even remotely perpetual

Anonymous at Tue, 1 Oct 2024 17:47:33 UTC No. 16405379

>>16403964
oh, also, the S stands for special in that the determinant is equal to 1, my b.