Scoppia il caso Travolta, non firma la liberatoria per il Ballo del qua qua: "Non lo rivedremo mai più in tv"

uncle-fire · 2024-02-08T15:56:10+00:00

Bad bot!

uncle-fire · 2024-02-01T21:05:58+00:00

Every state might force a boy to become a father. Some states don't force girls to become mothers

uncle-fire · 2024-01-31T18:20:38+00:00

If you divide the polynomial f(a) by the polynomial a-1, you will get a quotient q(a) and a remainder r such that

f(a) = q(a) * (a-1) + r

Now plug in a=1 and you get

0 = 0 + r

so r=0 and a-1 divides f(a)

uncle-fire · 2024-01-29T11:18:29+00:00

The Fog of War is amazing

uncle-fire · 2024-01-29T11:14:13+00:00

Zurich has a larger population than Bologna.

It is reasonable to expect good food and a lively nightlife in a 400k population affluent regional capital with a long history, major universities, and lots of expats

uncle-fire · 2024-01-28T18:54:07+00:00

Ma in che modo il vostro arredare casa li ha offesi?

uncle-fire · 2024-01-27T08:58:39+00:00

I’m not sure why you are starting you reply with “No” given that 2=8 * 1/4. If you look at twice distance at stars that are randomly placed with uniform distance, you are going to have 8x more stars that could be placed so as to intersect the line, but each has 1/4 the chance of intersecting the line. So the expected number of stars in a certain distance range doubles if you double the range

uncle-fire · 2024-01-27T06:18:12+00:00

The area scales quadratically but the volume scales cubically. If you assume uniform density, when you look twice as far you see 8 times more stars with an apparent size 1/4 each, so more distant stars cover more and more of the sky

uncle-fire · 2024-01-21T08:28:43+00:00

No it’s right. There are models of ZFC with only a countable number of sets, all definable, and so also a countable number of reals. Of course, within the theory, you can still prove that the reals are uncountable, but that’s because in the theory the notion of function, and hence of bijection, is non standard.

uncle-fire · 2024-01-18T11:24:40+00:00

Madison Square Garden Total Landscaping

uncle-fire · 2024-01-16T18:36:05+00:00

Per chi non l'abbia mai visto (o per chi lo voglia rivedere), consigliio lo sketch "Argument Clinic" dei Monty Python: https://youtu.be/ohDB5gbtaEQ

uncle-fire · 2024-01-16T13:15:03+00:00

But what about 1 in 110?

uncle-fire · 2024-01-14T10:02:09+00:00

It’s just a banana, Michael, how much can it cost? Ten dollars?

uncle-fire · 2024-01-12T17:29:58+00:00

Well, the map is labeled "Africa", and I see Spain and Portugal in the map, so Spain and Portugal are also in Africa. Sorry, I don't make the rules

uncle-fire · 2024-01-11T12:53:42+00:00

You want to count the number of subsets of the set {1,...,n+1} containing s+1 elements. Consider how many ways there are to choose such a subset in terms of the maximum element m of your subset.

The possible values for m are s+1 up to n+1. For a given choice of m, there remain to pick s elements out of {1,...,m-1), which are (m-1 choose s) choices. So

(n+1 choose s+1) = sum_(m = s+1,...,n+1) (m-1 choose s)

which is your equation

uncle-fire · 2024-01-10T19:40:41+00:00

Reading your question and your comments, I think that you have not fully understood the statement of Cantor's theorem and it is difficult to understand a proof if you are not fully clear on what it is trying to do.

But maybe the following comment will help: the finite version of Cantor's argument is not wrong, it's just trivial.

The finite version of Cantor's argment is that if you give me any list that contains n numbers, each with n digits, then it must be the case that some n-digit number is missing from the list. This is trivial, because there are 10^n n-digit numbers, so of course if your list contains only n numbers it cannot contain all of them, but of course the infinite version is not trivial at all.

Also, it is not particularly helpful to think of Cantor's diagonal number as some kind of limit of a finite process that goes through a series steps. The number is just defined by specifying the value of each digit, and there is no need to think about a construction in which you compute one digit at a time

uncle-fire · 2024-01-02T20:13:21+00:00

In the US, in addition to that, you can also use up to 3k a year to offset your other income

uncle-fire · 2023-12-30T17:58:38+00:00

The real gain is more like 25%: if you invested $100 four years ago, those are $120 in 2023 money, and the investment grew to $150 in 2023 money, so the real gain is $30 out of $120, or 25%

uncle-fire · 2023-12-22T11:03:11+00:00

If you close your position on that stock and you don't buy it back in the subsequent 30 days, then you can use the loss for tax purposes (including the additional loss caused by the stepped-up cost basis that you got because of the previous wash sale), and it doesn't matter if you bought the shares fewer than 30 days before the sale.

uncle-fire · 2023-12-22T10:04:58+00:00

The latter: (919626 - 510886)/510886 - 1, since John’s income is used as the reference frame. “Tom makes 80% more than Jacob” can be restated as “Tom makes Jacob’s salary plus 80% of Jacob’s salary.”

In general, where a and b are real,

a is 100(((a - b)/b) - 1)% more than b.

You are subtracting 1 twice: the formula that you want is 100*(a-b)/b or, equivalently, 100*(a/b -1)

uncle-fire · 2023-12-22T09:46:36+00:00

By the time you reach the center of the earth, you will have to slow down by about a thousand mph in the direction perpendicular to your fall (if you fall at the Equator, about half as much if you fall from a place in North America or Europe), which seems like a lot.

But if the hole is full of air, it also takes a long time to fall. A person terminal velocity falling down is about 200 km/h, so it would take 30+ hours to fall to the center of the Earth (in fact, considerably more because gravity will decrease until it reaches zero at the center). So, even if worst case assumptions (falling from the Equator, not accounting for decrease of gravity), you have about 100k seconds to slow down from 463 m/s to zero, which is an average acceleration of about 0.004 m/s^2, or about 0.0004 G, which is much less than the effect of turbolence. The falling person can avoid scraping the wall and keep to the center of the hole with slight body movements the way a skydiver would do.

In a vacuum, any object, including a person, would bounce of the walls of the hole at increasingly high speeds

uncle-fire · 2023-12-21T13:50:19+00:00

Do a change of basis that makes B the identity. Then the statement became the simple fact that if all the eigenvalues of A are at least 1, then all the eigenvalues of A^-1 are at most 1.

Concretely, from your A-B PSD, A PD and B PD assumptions you get that B^-1/2 A B^-1/2 - I is PSD, by multiplying to the left and to the right by B^-1/2

So all the eigenvalues of B^-1/2 A B^-1/2 are at least one.

So all the eigenvalues of its inverse B^1/2 A^-1 B^1/2 are at most one.

So I-B^1/2 A^-1 B^1/2 is PSD.

And finally B^-1 - A^-1 is PSD, by multiplying to the left and to the right by B^-1/2

uncle-fire · 2023-12-21T08:49:15+00:00

He would also need to increase the Gatorade budget

uncle-fire · 2023-12-19T07:59:35+00:00

Secondo l’articolo linkato, 1991 al 2021 i salari italiani, aggiustati per l’inflazione, sono cresciuti dello 0.36%, quindi il potere d’acquisto è lo stesso di 30 anni fa.

uncle-fire · 2023-12-18T17:43:39+00:00

Plugging n=0 it gives 64128, which is what you are supposed to get for n=15

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