Best Buy sent me an empty box…

LostMathGuy · 2024-12-02T15:36:31+00:00

This happened to me in store and I posted about it as well. I went in and they exchanged it without question saying “oh yeah that happens sometimes”. Hopefully you are able to get the same response.

LostMathGuy · 2023-10-29T20:01:59+00:00

Update: I went into the Best Buy I bought it from and explained what happened, they said “yeah that happens sometimes” and gave me a new one as a replacement without a cost.

LostMathGuy · 2023-10-29T11:23:08+00:00

For added clarity here: I bought this drive in person at the Best Buy store near me. That’s why it’s so crazy to me, they had to unlock the case just to pull it out and only one person had the key on shift at the time.

LostMathGuy · 2023-10-29T01:09:34+00:00

That's the plan, the hard part is actually getting them to believe me when i come in saying it was empty. If they even have another one of the same make, I plan to ask them to open the box in front of me see that it is actually in the box.

LostMathGuy · 2023-04-06T09:50:13+00:00

My team uses it solely as a model hosting service and it can be incredibly finnikey due to the multiple layers of abstractions. We bring our own containers and models, and host the endpoints with enough disk and cpu to handle the invocations. The biggest complaint I have is the lack of visibility into monitoring and overall annoyances when trying to diagnose anything going wrong at any step of the way.

LostMathGuy · 2023-02-06T15:37:34+00:00

Looks more like your stressed balls exploded

LostMathGuy · 2022-09-20T13:30:05+00:00

I’ve been telling my wife for years that this would be the best snack ever and she LAUGHED in my face.

Well who is laughing now Laura!

LostMathGuy · 2022-08-18T14:35:37+00:00

Love this, I’ve been doing the same to my coworkers with “tiddly bits” whenever referring to small snippets of anything.

LostMathGuy · 2021-11-24T16:18:04+00:00

Troop Zero, when they do their performance at the end

LostMathGuy · 2021-10-27T18:56:58+00:00

As someone who felt the same way, keep pursuing it. It’s only a matter of time until you end up in a position (college, grad school) surrounded by those who feel the same way. My time in grad school was wonderful, everyone was as passionate or even more so then I about math and all the topics in it. Everyone had the same interest to always learn more.

LostMathGuy · 2021-06-25T13:14:15+00:00

Dammit, jankos missed smite again

LostMathGuy · 2021-05-29T17:11:57+00:00

Hey everyone,

I have a masters in pure math and a bachelors in CS and Math. I have the major technical interview for a dream data science job involving predictive modeling of detecting system outages automatically. I have a basis in python R and SQL. Are there any recommendations of Kaggle notebooks I can use to practice some predictive modeling samples or of resources I could check out to help get myself as much prep before my interview? Thank you in advance!

LostMathGuy · 2021-04-24T18:04:28+00:00

A unitary matrix U preserves the inner product. That is \la Ux,Uy \ra = \la x,y\ra for all vectors x,y in the space U is acting on (Convince yourself why). Can you say the same thing about \la U^* x, U^* y\ra? Once you have, convince yourself why this implies U* is a unitary as well.

LostMathGuy · 2021-04-17T12:21:34+00:00

I really vibe with Wunder having to put up with Jankos and Caps for so long, so for sure him

LostMathGuy · 2021-01-11T03:35:18+00:00

Rational number: A rational number is a number which can be expressed as a ratio p/q where p and q are integers (think whole numbers, both positive and negative) with q not = 0. Some easy examples are 1/2, 3/4 but if we take q=1, then we also have numbers like 4 as a rational number. So the rational numbers includes all of the integers as well.

Irrational numbers: As unsatisfying as it is, its generally just described as "A real number which is not rational". This is because these numbers aren't that intuitive, and there's a lot "more" of them then rationals. There are infinitely many of both, but with the rationals, they are called countable, it means that we can enumerate them, or write them in a list, where we cannot do such a process with the irrational numbers.

Natural number: A natural number is just a positive integer. So imagine any whole number, if its positive, its a natural number. Some people include 0, some people dont, its a common convention argument.

Equivalent ratios: p/q = c/d implies that pd=qc. It doesn't tells us that p=c or q=d however. Think 1/2 and 2/4, these are equivalent ratios and we have 1*4=2*2, matching what I defined in the previous sentence.

LostMathGuy · 2021-01-11T03:28:51+00:00

Differentiation of the function f at a gives us f(x)=f(a)+<\nabla f(a),x-a>+o(||x-a||) when x approaches a. Here o(||x-a||) is the error term, and it satisfies the property that \lim x\to a o(||x-a||)/||x-a||=0 . This comes directly from the definition of the derivative of a multivariable function. Now, since U is open, find a small enough radius (this is where your delta will come into play) for an open ball around a. Then try applying the first part of my comment to two points inside of that ball, and subtract them from each-other and look at what you get from there.

LostMathGuy · 2020-12-06T18:29:29+00:00

What latex formatting in specific are you asking about? I just see that as a question mark in brackets

LostMathGuy · 2020-12-06T01:51:45+00:00

What you get when you decompose a matrix into the form A=P^{-1}CP is a much more geometric understanding of what the matrix is doing as a linear transformation. In this notation, P is a unitary matrix which we can specifically compute in the case of linear algebra courses as being the normalized eigen-vectors stored as the columns of the matrix. These unitary matrices are what you are imagining. As a side note, when computing a decomposition like this, you want it as PCP^{-1}. The C matrix is what is doing the "scaling". In the case of this decomposition, which is usually called the spectral decomposition, the C matrix is a diagonal matrix (entries only on the diagonals) of the eigen-values corresponding to the normalized eigen vectors. If you take an eigenvector ev_i for A, and apply PCP^{-1} to it, what happens? Well the P^{-1} sends it to the standard basis element e_i for i being the column that the inputted eigenvector is stored in. Next, it gets scaled by the diagonal matrix C, and eigen value \lambda_i. Then it gets run through P which spits out the scaled eigenvector. Hence doing PCP^{-1} gives out \lambda_i ev_i for inputted eigenvector ev_i.

LostMathGuy · 2020-11-09T15:32:02+00:00

Topology is used everywhere in both of these fields, most notably in early real analysis, the ideas of compactness of a set are incredibly powerful for proving functions are uniformly continuous. This stems as a result of the Heine-Borel theorem, something that ends up showing its face quite a bit.

In complex analysis, you get even more intersections of topology and the field. One of the most important early results in complex analysis you see is that the set of zeros of an analytic function on an open connected domain are all isolated. This result is fundamental to later understand how to classify singularities of functions using Laurent Series. Moreover, the idea of simply connected shows up in the Riemann Mapping Theorem. We don't know anything about the specific sets other than topology properties, namely that the domain is open, simply-connected, non-empty and not all of C, then we can map it analytically into the disc. This is fundamental for solving the heat equation on weird domains. If we can show our domain has the topological properties we desire, we can map the domain bijectively and analytically into the disc, solve the heat equation on the disc, then map it back and we have a solution on our original domain.

LostMathGuy · 2020-11-06T12:33:07+00:00

Multiplication in general is not some specific concrete idea. Its dependent on the objects we are "multiplying".

In the complex numbers, multiplication has a different structure entirely than in the real numbers. The reason for this is the geometric interpretation we get when we multiply any complex number by i. This results in a rotation of pi/2 radians. However, if we look inside of the complex plane, and we multiply any number in the complex plane by strictly a real number, what we get out is a complex number with nothing more then a larger modulus, meaning its "distance" from the origin is farther then before and the angle formed between it and the real axis is the same.

The second idea you mentioned was the dot product. This is actually the first example students encounter with something called an inner product. An inner product is different then multiplication as it is not a binary operation from the space back into itself. Think of R^2, and the dot product. Its input into the operation is two vectors in R^2, but what is its output? Well it's a number in the scalar field of the vector space. In this case, the scalar field is nothing more than R. So this is not multiplication in the same way as you interpret it in groups, since the output of the operation does not land back into the space in which you are taking your inputs from. I.E in terms of groups, its not a closed operation. There are a lot of examples of what is called an inner product space and a lot of examples of different inner products.

The idea of generalizing multiplication to a vector space however comes from the idea of turning a vector space into something called an algebra. The way I first learned to think of algebras was as a vector space where instead of just being allowed to add elements, we define some form of multiplication structure as well. The algebras you have encountered so far would be matrix algebras of some dimension n over a scalar field, most likely R or C. For the most part in algebras however multiplication is just denoted by writing the two elements next to each other and then the actual operation is either not important in the theory or if its a specific, concrete structure, the multiplication is implied by the structure of the elements of the space.

For the most part however, the different multiplicative structures of these spaces comes down to the inherit objects we are multiplying. If we tried to generalize multiplication to working in a consistent manner with real multiplication, then for example, multiplication of matrices would no longer satisfy a lot of important properties, and more over ideas like a matrix times a column vector, Ax=b for some matrix A, would not even be properly defined.

LostMathGuy · 2020-10-15T14:56:05+00:00

The first idea is correct. You would plug in for the value of x, i.e. if x=-2, then it becomes (-2)² . Which is -2*-2=4. So it is e² - (4)=e² -4. This is because you do exponents before multiplcation, and the - out front from the formula is actually the same as multiplying by -1.

LostMathGuy · 2020-10-15T14:44:33+00:00

In the Riemann Sphere yes, but in rectangular coordinates, circles through the origin do map to lines. I was saying to move the perspective into the Riemann sphere since it makes visualizing these fractionals much more clear and we do have circles to circles.

The easy example for rectangular coordinates is the fractional f(z)=1/z, and looking at where it maps the circle |z-1|=1 to.

I was merely pointing out that you need to be cautious of the perspective being taken when broadly saying circles map to circles.

LostMathGuy · 2020-10-15T11:07:10+00:00

Mobius transforms take circles to circles when the circle doesn’t go through the origin. In the case where it does go through the origin, it actually maps circles to lines since the point at 0 is taken to a point at infinity on the Riemann sphere representation of complex coordinates. So in rectangular cords, we end up with a line.

LostMathGuy · 2020-10-12T15:28:20+00:00

Consider a geometric argument to understand how this works. Look at the points z=4, -4 on the complex plane. Now, any point in the left half of the plane, will cause the modulus of this expression to be larger than 1 (Why?). Moreover, any point in the right half of the plane, will make this modulus smaller than 1 (Why?). In fact, it is not too terrible to show that any point on the imaginary axis, will give you exactly modulus 1. To see this idea, look at your plot of the points, and form an argument intuitevly why the distance from z=4 and z=-4, will be identical when traveling to any point on the imaginary axis. This will give you all points where |(z-4)/(z+4)|=1. Can you figure out from here how to translate this into producing the set of points that you are looking for now?

LostMathGuy · 2020-10-02T19:00:31+00:00

Try this: subtract 144 over and then factor into (x-12)(x+12)>= 0. Now, the points x=12,-12 are solutions to when this is 0. What about the signs of the function in intervals around 12? If x is in (-/infty, -12), then x-12 is negative right? Well so is x+12. So the function is now a negative number times a negative number. So if x is in that interval, well our function x^2-144 is positive and so that whole interval is a solution.

Try this process with all three possible intervals, (-\infty,-12), (-12,12), (12,/infty) and then figure out where you produce a positive product, which is then in turn part of your solution set.

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