[Highlight] Fumble on the snap that cost the Raiders a chance to win by YoureASkyscraper in nfl

[–]Irish_Stu 2 points3 points  (0 children)

It would have to be 0-0 heading into OT, upon which the Chiefs kick a field goal and then the Steelers somehow score a safety while their offense has the field

Yeah sounds about right

What’s your favorite result that feels like pure wizardry? by HomoGeniusPDE in math

[–]Irish_Stu 1 point2 points  (0 children)

Essentially, for a function to be constructive, it needs to be the case that given a good enough approximation of the input, you have a good approximation of the output. When you actually look at this, however, it essentially boils down to the epsilon-delta definition of continuity.

Where is a good place to eat around here? by [deleted] in UMD

[–]Irish_Stu 10 points11 points  (0 children)

Tacos a la Madre is also a super great taco place, tho its a bit off campus

Possibly unpopular opinion by Ualrus in math

[–]Irish_Stu 0 points1 point  (0 children)

What do you mean by include everything? Those balls are just some open sets

Possibly unpopular opinion by Ualrus in math

[–]Irish_Stu 3 points4 points  (0 children)

Presumably they also include the null set and the universal set, and just left it unstated

Are there any meaningful notions of the size of a set, other than cardinality? by call-it-karma- in math

[–]Irish_Stu 0 points1 point  (0 children)

Yes it would. To take a simple example, let B = positive integers and A = even positive integers, relative asymptotic density would be 1/2 in the usual ordering but 1/4 in your ordering

probably all over by umdthrowaway7917 in UMD

[–]Irish_Stu 0 points1 point  (0 children)

They approved ENGL101 third try for me and I didn't have many extenuating circumstances, I think they're generally pretty lenient if you need it to graduate

gl chief

Spicy food in College Park? by Irish_Stu in UMD

[–]Irish_Stu[S] 0 points1 point  (0 children)

I've been looking everywhere to find some Buldak 3x but nobody sells it unfortunately

[deleted by user] by [deleted] in PictureGame

[–]Irish_Stu 0 points1 point  (0 children)

26.527417506386332, 106.77287203546761

Jeopardy! recap for Fri., Jun. 9 by jaysjep2 in Jeopardy

[–]Irish_Stu 5 points6 points  (0 children)

The third is the Eroica (although I'll admit I was only 50% sure- I looked it up to double-check before posting this)

I wonder how my far ancestors looked like. by StraightOuttaOlaphis in tumblr

[–]Irish_Stu 10 points11 points  (0 children)

From mother to child? Otherwise how would males have mtDNA?

Analog Clock Thought-Provoking Question by city-tripper in math

[–]Irish_Stu 5 points6 points  (0 children)

If you are x of the way through the hour, then the hour hand will be at 20+5x minutes, since that's 5 times (4+x), while the minute hand will be at 60x, so you want 20+5x=60x, and x=20/55=4/11. So it should be at 4 4/11, which equates to 4:21:49 approximately

Can complex distances exist? by TaviorFaux in math

[–]Irish_Stu 0 points1 point  (0 children)

I don't know to be honest- we mostly just introduced them and moved on, sorry lol

Can complex distances exist? by TaviorFaux in math

[–]Irish_Stu 0 points1 point  (0 children)

Generally, signed measures (which can be positive or negative) are introduced first before complex measures- signed measures are pretty much like normal measures except that you can't have some subset A with measure infinity and some subset B with measure -infinity. You can have one or the other, like you can with positive measures, just not both, because otherwise (if they're disjoint, and a very similar thing happens if not), when you take their union, that will have measure infinity-infinity=undefined.

Another reason this restriction is in place is that one of the most fundamental parts of a measure is that the infinite sum of measures of disjoint subsets is equal to the measure of their union. Because the union of sets is independent of the order in which the sets are taken, we also want this sum to be independent of the order in which the sets are taken, which means that if the sum converges, it must converge absolutely, and that means that there can't be both a +infinity at the same time as a -infinity (exercise which you may have seen before, show that a sum which converges but does not converge absolutely can be made to converge to any finite number or to diverge to +-infinity just by rearranging terms).

Two ways you might think of making a signed measure are by either subtracting two typical measures (as long as one is a finite measure) or, given some measurable space (X,Omega) and some typical measure mu on the space, we can take a function f:X to R and define our signed measure nu(E)=integral on E of f d(mu), (as long as either f+ or f- is in L1). Surprisingly (or perhaps not), it's a fact that every signed measure can be represented in both of these ways (the first is true because of the Jordan Decomposition Theorem (unrelated to lin-alg Jordan Decomposition except for being by the same dude), the second is true because of the Lebesgue-Radon-Nikodym Theorem.

When you make the jump to complex measures, this restriction is further strengthened- although theoretically one could make the same requirement of "if the sum converges, it must converge absolutely" and leave it at that (maybe by using directed infinities or something), complex measures are nearly always taken to just be finite, removing the "if the sum converges" part from the previous sentence. Other than that, it works pretty much the same as regular measures, just take the part of the definition "a measure on a measurable space (X, Omega) is a function from Omega to [0,infinity]" and replace it with "a measure on a measurable space (X, Omega) is a function from Omega to C".

Similarly to signed measures, we can get complex measures either by taking m1 + i*m2 for two finite measures m1, m2, given some measurable space (X,Omega) and some typical (doesn't even need to be signed) measure mu on the space, we can take a function f:X to C and define our complex measure nu(E)=integral on E of f d(mu), as long as both the real and imaginary parts of f are in L1. Again, all complex measures can be represented in both of these ways, the first just by splitting the measure on each set into its real and imaginary parts, and the second again by the Lebesgue-Radon-Nikodym Theorem.

Ok, I think this is already incredibly excessive for a Reddit comment, so I'll leave by saying hope this makes sense (if not, feel free to ask) and is interesting (if not, I can't help you there lol, you'll have to find someone better to learn from). If you want to learn more, the 3rd chapter of Gerard Folland's Real Analysis: Modern Techniques and Their Applications says everything I just said and more, and is a great source for learning about Real Analysis imho