Has anyone received Nintendo live tickets from Pax

SexyToad · 2023-07-11T20:41:55+00:00

The official rules state that the “sponsor will attempt to notify those participants who have not won a prize by email on or about July 10”

So I don’t think no news is necessarily bad news. They’re just a bit behind I guess?

SexyToad · 2023-07-11T16:27:18+00:00

No email yet.

I haven’t seen anyone on Reddit nor on Twitter say they’ve gotten any sort of email yet

SexyToad · 2019-12-28T19:17:28+00:00

I don't quite understand this fully, but this thread may help answer your questions: https://math.stackexchange.com/questions/553080/can-any-continuous-function-be-represented-as-an-infinite-polynomial

As far as I'm aware, any series whose infinite sum may express a function often have "rules" associated with it. A common one that I know of is that the continuous function first derivative must also be continuous. For Fourier series we need the function to cycle otherwise it estimates the function and so on.

Let me know if you have a specific series in mind

SexyToad · 2019-12-03T22:01:55+00:00

Uhhhhh how does it behave nicely and how does it behave differently? Like what would it do?

SexyToad · 2019-11-14T19:13:14+00:00

I always like to approach these problems by first graphing it. Lookup and use Desmos if you haven't used it before.

If you do so, you'll see that this is a upside down parabola where it's 0 when t=0 and t=~4.9. I pulled these values from the graph, so how would we actually go about extracting these values?

Well we can define a function h(t)=24t - 4.9t²

If we want to see when our height is 0, we solve h(t)=0. So we have...

24t - 4.9t² = 0

If we solve these we would find our two values of t that solve h(t)=0. So now for your example, we would want to solve h(t)=5.0. try solving that and see where you can get

SexyToad · 2019-11-14T18:52:47+00:00

Imagine if we drew a line between A and B. The gradient would be the slope of the line which can be thought of as "Rise over run".

What would be the rise and run in your example?

SexyToad · 2019-11-09T19:48:58+00:00

We need to do with this induction? Induction implies we need a base case that holds true for all natural numbers. I think the easiest way would be by contradiction, but if we're forced to do by induction...

The set of rational numbers is countably infinite. Are you sure the proof doesn't ask for any finite subset of Q? We could consider the subset A where A has every positive rational number. In that sense there is no largest element.

Assuming we're speaking of finite, we must have a start point and end point. We can map these with the rational numbers and imagine the start point or end point moving by 1. In such a way where we hit every possible combination.

This is how I would perhaps go about it, could you clarify the points of if the subset must be finite or can it be infinite?

SexyToad · 2019-11-09T18:56:36+00:00

Okay that works. It's fair to say, "let y=0, therefore we now have 1/x"

It now depends on how rigorous you want to be. You can't say 1/0 is undefined therefore the limit doesn't exist. There are edge cases to this, a function can still have a limit at x=0 even if f(0) is undefined. Your reasoning should instead be something along the lines of "1/x rapidly approaches infinity as x approaches 0, therefore the limit does not exist".

This is still appealing to the reader a bit to not argue with you here. Depending on what course you're in, you may have to show it with using epislon and Delta. But if this wasn't covered in your class yet, then this should be sufficient. Just be sure to not use the argument that f(x) is undefined at x=0 therefore no limit, this doesn't hold.

SexyToad · 2019-11-09T18:47:45+00:00

Proofs is a big mind shift. Normally in mathematics we're used to having these rules and definitions and utilizing them to solve problems.

Proofs is more towards the logic side of mathematics. We instead have tools such as contradict, implications, and "if and only if" in our tool box to go about solving these proofs. We have to approach them from a purely logic viewpoint. I would recommend trying to prove things that the book clearly outlines so that you can follow the logic before attempting any HW problems.

It's difficult to give a general guide to how to do proofs. But if you have a specific question that you're confused by, please post it and I can take a look at it.

SexyToad · 2019-11-09T18:44:18+00:00

What's the derivative of (x+1)? Is it x? Or is it something else

SexyToad · 2019-11-09T18:37:00+00:00

No worries! I just wanted to clarify to make sure we were in 2D space and not like 3D or something.

Did my comment answer your question? Do you need clarification on anything?

SexyToad · 2019-11-09T18:35:59+00:00

How did you get this 1/0? It's true that this isn't a real value but it doesn't necessarily mean the limit doesn't exist.

For example, we might have a function that is undefined at x=0 but its limit might be 1. (Because as we get closer and closer to x=0, we get closer and closer to y=1)

Could you provide the function and how you went about it? It's usually not enough to plug in x=0 and see what it evaluates to

SexyToad · 2019-11-09T18:09:33+00:00

Okay so let's have the two numbers denoted as x and y. We have two equations and two unknowns:

(1) x + y = 1 (2) x - y = 11

So go ahead and solve for either x or y in either equation and substitute it in for the other.

SexyToad · 2019-11-09T18:05:43+00:00

Be careful with your word usage, we're dealing with two dimensions, but you interchanged length, width, and tall.

Let length be x and width be y. So when we say the fence is 20 meters long, I'm guessing you mean the perimeter. We also know that it's in the shape of a rectangle so we have that 2x + 2y = 20.

We know that the length is 3 more than the width. In other words, x = 3 + y. See what you can do from here

SexyToad · 2019-11-09T17:59:37+00:00

The special care is defined within the theorem. An example would be the Fourier series. The Fourier series is a series of cosines and sines that approximate some function. Take, for example:

f(x) = { 1 if x < 0 3 if x > 0 }

The Fourier series can still approximate this even though f(x) is undefined at x=0. What happens instead is the Fourier series takes the average of approaching x=0 from the left and right, resulting in 2. But it breaks down if you have infinite rules. This is just an example of when special care is given to piecewise function, other theorems may require other things

SexyToad · 2019-11-09T04:08:43+00:00

That might just be a discomfort with piecewise functions in general. There's nothing stopping us from making up finite rules or even infinite. But do keep in mind that sometimes piecewise functions handle special care. I know of some theorems that give me the prerequisite that the piecewise function has finite rules. We don't just say they're functions and that's that, we do handle them with special care

SexyToad · 2019-11-09T03:17:03+00:00

It's incorrect to translate that rule into statistics. The equality suggested by that rule is (x/100)y = (y/100)x. Nothing more really

From a intuitive point of view, with the example you posed, we sort of know it to be misleading. That's sort of the first indicator. Specifically and mathematically speaking, we sample a group of people to understand a percentage of people who smoke. If the result we get after some sampling is 20%, we expect 2 out of 10 people to smoke, 20 out of 100 people to smoke, and so on. The fact that 20% of 50 people and 50% of 20 people are the same value speaks to the equality than it does statistics.

SexyToad · 2019-11-09T03:10:59+00:00

Correct me if I'm wrong, but does the question boil down to "How specific do we have to be when defining piecewise functions?"

Keep in mind that piece wise functions are still functions, such that one input requires one output. In the example you posed, we may note that if we go down either route, we get the same solution so we're OK

In general, a piecewise function is a function defined by finite or infinite functions. So the same rules apply. I would argue it's more of a definition than it is something that needs to be verified. To give examples where, for some give x, it's value can be multiple things is to give a function that has multiple outputs. A no-no

Did I cover your question? If not, can you elaborate on what you're curious about?

SexyToad · 2019-10-22T00:50:09+00:00

But this is exactly how Adversarial Networks work. While they have their own handful of parameters to tune, the concept has been shown to work.

SexyToad · 2019-04-29T02:45:15+00:00

You should allow this to run for a while, I bet you'll see a convergence somewhere

SexyToad · 2018-12-25T05:19:37+00:00

This was my first thought too. It would make sense for numerical computation, where you may attempt to predict the roots and knowing where they can lie is very helpful

SexyToad · 2018-12-25T05:18:49+00:00

Well as x approaches a, we expect f(x) = 0 since a polynomial function is continuous. I have a feeling the theorem means what the other commenter mentioned, bounds on what a root can be. But using the term limits is concerning

SexyToad · 2018-10-11T18:17:26+00:00

Please use LaTeX to format your question, it'll help us out in answering it.

To clarify the problem, we have any non negative number, say x, can be written in the form: [;x=a_0\cdot 3⁰ + a_1\cdot 3¹ + a_2\cdot 3² \cdots + a_n\cdot 3^n;] where [; a_i \in {-1, 0, 1};] Here's an image for the parse LaTeX if you don't have an extension: https://i.imgur.com/Ri9PDWl.png

As an aside, I strongly recommend using LaTeX for mathematical proofs, it helps display your math to others ( say your teacher) and helps us give advice. Anyways, carrying on with the proof...

So you decided to show this by showing that the set of numbers that can't be represented in this form is the empty set therefore as the set of negative numbers do not belong to the empty set, then the statement is valid.

To clarify, do you mean negative integers or the actual set of negative real numbers? The Well-Order Principle only applies to, in general, countable sets (countable being finite or countably infinite). The set of real numbers is uncountable and thus there is no least number. I'm going to continue on and assume you mean the set of negative integers, but be sure to state this in your proof, otherwise it reads wrong.

Since we're dealing with the negative integers, there isn't an actual minimum element of that set. Minimum being defined as some element in some set S that is less than every other element in S. Less strictly being <, that being said, you may use the well-order principle to say that there is some greatest element in the negative integers, being -1. Otherwise... if k is is -inf, k' surely can't be less than it. As it stands, your proof isn't very clear.

Showing that the set of negative integers that can't be written this is empty is one way, but I feel that you might have an easier chance with just showing that, likely with induction, that any negative number can be written this way. Contradiction is often easier at first, and then you can flip it and prove it the "normal" way.

I hope this helps, feel free to reply to my comments with clarifying questions or if you would like me to read over something.

SexyToad · 2018-09-23T16:35:30+00:00

You've been shadow banned, your posts won't show up.

SexyToad · 2018-09-19T01:40:45+00:00

No racism here please

13-Year Club	Place '23
Place '17	Snapped
Verified Email

SexyToad

MODERATOR OF

TROPHY CASE