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

>>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.

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.

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

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

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>>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".

>>16394765
Use the triangle inequality.

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.

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>>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.

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.

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>>16394665
What about the next one after 101?

>>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

>>16395940
yes, nothing wrong with thank yous

[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?

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>>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.

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.

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>>16394407

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

>>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.

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

>>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.

>>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?

>>16396677
yes

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

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

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]

>>16396686
look up the gamma function

>>16396682
Yes

>>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.

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

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

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

[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]?

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

>>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.

>>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.

>>16396708
Very nice

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

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?

>>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

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

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.

>>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.

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?

>>16398877
Yeah, that was a science question.

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

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

>>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.

>>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.

>>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

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

>>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.

>>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)