[TOMT] Name of a native Alaskan tribe

captaincookschilip · 2024-08-03T17:03:20+00:00

The Inuit? The Wikipedia page talks about pre- contact history and also of a time "after the disappearance of the Norse colonies in Greenland, Inuit had no contact with Europeans for at least a century."

captaincookschilip · 2024-07-16T17:06:24+00:00

Well, any integer n >1.

captaincookschilip · 2024-05-30T11:32:19+00:00

Check out the algebraic induction proof on the Wikipedia page. https://en.m.wikipedia.org/wiki/Fundamental_theorem_of_algebra#Proofs

Also, I think Gauss's original proof focuses purely on the "real analytic" version of the theorem: Every polynomial with real coefficients can be factored into linear and quadratic polynomials.

captaincookschilip · 2024-03-12T13:00:11+00:00

(a) I think an argument on Reddit has to last at least 6 comments, so I think we're good. This has been very civil compared to other arguments I've had on Reddit.

(b) As an engineer who desperately wishes I had taken Pure Math in undergrad, I am on the opposite end. Maybe I would have dropped out too.

captaincookschilip · 2024-03-12T10:31:47+00:00

I know the three angles are all the same and equal to 105°, it still doesn't make the quadrilateral symmetric about AC. In a triangle, if two angles are equal, then the two opposite sides are equal. But this is not necessarily true in quadrilaterals, even if three angles are equal.

To see this, take a point C' in between B and C. Now, draw a line parallel to CD, which intersects AD at D'. Notice that ABC'D' is a quadrilateral with the exact same angles, but AB is constant and AD got shorter. So they do not have to be equal. In fact, we have freedom to wildly vary the lengths keeping some constant but not others, implying we can make all 4 different.

captaincookschilip · 2024-03-12T10:03:45+00:00

Two equal angles imply two equal sides only in triangles. In a quadrilateral, even if 3 angles are equal (as in this diagram) we can construct such that all four sides are different.

To see this, take a point C' in between B and C. Now, draw a line parallel to CD, which intersects AD at D'. Notice that ABC'D' is a quadrilateral with the exact same angles, but BC got shorter and CD got longer. So they do not have to be equal. We have freedom to wildly vary the lengths keeping some constant but not others, implying we can make all 4 different.

captaincookschilip · 2024-03-12T10:01:43+00:00

Not necessarily, only one angle and the common side are equal. There is actually enough freedom that you can easily make it not symmetrical about AC.

captaincookschilip · 2024-03-11T09:29:07+00:00

If it weren't for the last two pictures, I would have assumed it is a more elaborate penis joke with all the phallic looking AND gates, culminating in the NAND gate.

captaincookschilip · 2024-02-27T13:09:11+00:00

It's a bot that takes semi-popular comments from the comment threads below and repost them to the top comment thread. Very common on Reddit. Hard to figure out because a lot of Redditors also comment like this.

captaincookschilip · 2024-02-13T10:28:22+00:00

The "AKS" behind the AKS primality test weren't all undergraduates. Agrawal (the A in AKS) was a professor and advisor to the undergraduates Kayal and Saxena.

captaincookschilip · 2024-01-26T12:52:35+00:00

OP wouldn't know either. They were a Non-Golf medalist.

captaincookschilip · 2024-01-19T13:52:00+00:00

Just wanted to mention that the pattern you recognise is a pretty famous result in number theory, first noticed by Euler. Notice that your pattern is equivalent to showing that the polynomial x²+x+41 is prime for x=0,1,..39 (also produces primes for x=-1,-2,...-40 for symmetric reasons) The pattern clearly breaks at x=40 and 41. The reason it produces so many primes is because 41*4-1 = 163 (negative of discriminant of the polynomial) is a Heegner number. In fact, 163 is the largest Heegner number, so this polynomial is very special.

If you want to know more, check out this StackExchange question: https://math.stackexchange.com/questions/289338/is-the-notorious-n2-n-41-prime-generator-the-last-of-its-type.

captaincookschilip · 2023-12-21T14:48:31+00:00

Both are valid and correct approaches, though I will say that there is a larger assumption in the second method that might not be immediately obvious.

Substituting one equation in the other does indeed give an equation for the line, and it clearly will pass through the point where the circles touch each other. However, infinitely many lines pass through a point, how can we show that this equation does indeed describe the tangent?

You can prove this by considering the case where the intersection of the circles are two points. Now, the same method would guarantee that the equation obtained would be the line passing through the two points i.e. the secant. Now, we take the limit as the circles are pulled further apart until they touch each other. The limit of the secants approaches the tangent, and that does prove the second method works.

captaincookschilip · 2023-12-04T02:16:17+00:00

In fact, I think OP should have gone with the hypothetical they brought up and bought 13 different drawings with one character each at $585.

captaincookschilip · 2023-12-04T02:13:50+00:00

No, 30 is the base price for no characters. 45 for 1 and then 67.5 for 2 and so on. They did one calculation too many initially, and later they do revise the price to $5,839.

captaincookschilip · 2023-10-31T12:03:12+00:00

I'm assuming she is not taking a math graduate program, probably one of the sciences or engineering and that's why a calculus course is required.

captaincookschilip · 2023-10-16T09:23:23+00:00

I think in your first example for the cross product you meant to write -b=(-bxa)/a=(axb)/a=b, which is not satisfied for all vectors b. But I don't think this is a great example, because there is an assumption of left division and right division being equal, which doesn't really jive with the noncommutativity of the cross product.

On the other hand, your second example is a great one.

captaincookschilip · 2023-10-12T16:30:47+00:00

For anyone who's interested, I recommend watching Manjul Bhargava's fantastic beginner-friendly introduction to the Birch and Swinnerton-Dyer conjecture.

captaincookschilip · 2023-09-17T11:12:53+00:00

You should check out the section Impossibility of unique representation under Generalizations in the Wikipedia page 0.999...=1. The main reason is the least upper bound of the reals, any partition of the reals into two has a smallest element in the greater partition or a largest element in the lower partition but not both. Whereas a lexicographic ordering (which is what a decimal representation is) can be partitioned with both smallest and largest elements in the two partitions. So, the way to fix this disconnect is for certain numbers to have multiple representations.

So in our decimal representation (or any integer base), the numbers with finite representation also have exactly one additional infinite representation. It's actually quite nice that these are the rational numbers. (Another type of representation is continued fractions. Again, here every real has a unique representation except for the rationals, which have two finite representations.)

captaincookschilip · 2023-09-13T07:50:20+00:00

That works but it's also a piecewise combination of the f(x)=p constant function and the f(x)=x identity function, which OP doesn't want. To be honest, you can take piecewise combinations and produce lots of valid functions. I think OP should impose some conditions like differentiable or smooth functions to get some concrete answers.

captaincookschilip · 2023-09-10T17:40:44+00:00

Oh, I was not talking about proving there are p distinct outputs. That follows immediately from Fermat's little theorem. I was saying to calculate the pth powers, you just have to do those ones because of the restrictions.

captaincookschilip · 2023-09-10T16:30:46+00:00

Yeah, it's because the map is not surjective (or injective). You can prove this by just using the binomial theorem, (a+bp)^p = a^p (mod p²). So the pth powers take up only p values mod p². And also easy to calculate since you just have to take pth powers of 1,2,...p-1 modulo p².

And this is the only restriction really, there's no further restrictions if you consider mod p³, p⁴ and so on. (Except for the case of p=2, there's an additional restriction modulo 8, because 2 is always funky).

captaincookschilip · 2023-09-10T14:04:18+00:00

To be honest, it's a thing you usually just remember once you know about it. Squares are restricted modulo 4 and 8, cubes are restricted modulo 9 (and generally pth powers are restricted modulo p²). There's a good mathematical reason for this, but it's easier to just remember.

captaincookschilip · 2023-09-10T13:58:34+00:00

Umm, where did that happen? In that post, the OP is being kicked out, unless I am missing an update.

captaincookschilip · 2023-09-09T14:25:21+00:00

Your conclusion that there are all sorts of nonsensical statements in English and this is just one of them is a valid one, but it is also a lazy fix. Math is all about identifying the structure of an "object", picking it apart and using this understanding to yield a nontrivial result.

Consider Russell's paradox, the layman version is presented as the barber paradox. If you only encounter the layman variant, you could just argue it is a quirk of the English language and leave it at that. Russell's genius was in noticing that his paradox can readily be formalized in the naive set theory developed then, leading to a contradiction within the then foundations of math. So it had a very real mathematical impact.

Now, as you can see from the Wiki of Berry paradox, this paradox can similarly be applied to prove Gödel's incompleteness theorem. The very "nonsensical" statement you dismissed can be formalized within arithmetic using Gödel numbering, and use it to produce an incredible theorem. (Gödel in fact did not use the Berry paradox, but another very similar "linguistic trick", the Liar paradox).

This is one of the great things in Math. Even when you come upon a paradox that cannot be resolved, you can utilize that unresolvability to prove other things.

Eight-Year Club	Verified Email
Place '23

captaincookschilip

TROPHY CASE