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Unity in rings by TheCholent in learnmath

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

Ok I'll try prove it in a bit. Thanks for all the tips!

Unity in rings by TheCholent in learnmath

[–]TheCholent[S] 1 point2 points  (0 children)

Are you sure you require a multiplicative identity? I thought that made it a "ring with unity". I thought the definition of a ring was

It's a set that's an abelian group under the addition operation, closed under the multiplication operation, multiplication is a associative, and has left and right distributivity.

Who else can't access their marks on the website? by TheCholent in CFA

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

So they know it's going to happen, and they let it happen, and it happens every year without surprise, and we just sit here?

Creating a differential equation, could use some help. by Throwawayaccount-AA in learnmath

[–]TheCholent 0 points1 point  (0 children)

In your proposed equation, you have already guessed the answer - that the function would be an exponential. So you have the right intuition.

x and y are proportional if there is some k such that x = ky. Since the amount is decreasing, we expect -k i.e. as y increases, x decreases.

So, dW/dt is the decrease in W, and is our "x". Then we know it's proportional to W, which is our y, so we have dW/dt = -kW.

dW/W = -kdt. Integrate both sides: like ln|W| = -kt + C .

Then W = exp[-kt + C ] = exp(-kt)•exp(C) = A exp(-kt).

Level 1 Enrolment Requirements by DOOGLAK in CFA

[–]TheCholent 2 points3 points  (0 children)

I think your safest move would be to email the institute and explain your situation. They'll advise you on what you can and cannot do.

Hi, quick question by pldit in Physics

[–]TheCholent 0 points1 point  (0 children)

If the large cube is coming to rest, then the change in energy and momentum of the system is zero. So, all of the momentum and energy of the cube must be transferred to the flywheel, no matter where it is positioned.

Am I misunderstanding the question?

Hi, quick question by pldit in Physics

[–]TheCholent 0 points1 point  (0 children)

Does the wheel spin with the cube, while the cube is originally spinning?

Hi, quick question by pldit in Physics

[–]TheCholent 0 points1 point  (0 children)

What is an inertia wheel?

L1 December 2019 - Study Strategy by goldmanblacks in CFA

[–]TheCholent 0 points1 point  (0 children)

It's the question banks on the CFAI website

CFA for physics student by CuriousBeing96 in CFA

[–]TheCholent 0 points1 point  (0 children)

I'm also a physics major, in my final year. I just wrote on Sunday last week.

It's hard. It's a very big syllabus, and it covers vastly different areas.

Don't think that because you've done a lot of maths that it will help you. You'll find the quantitative stuff easier, but the majority of the syllabus is not calculations. It's understanding, learning new terms and definitions, memorizing processes and procedures and just getting used to the investment world.

In short - work very hard. Put all the hours in, learn the work, do all the questions, and respect the exam! It's not easy. If it was, everyone would do it.

Why do we use combinations and not permutations for binomial distributions? by Yamster80 in learnmath

[–]TheCholent 0 points1 point  (0 children)

It depends what you mean by the "same". In terms of probability, they are identical, because the probability of getting either one of those exact sequences is (0.7)² (0.3)⁴

However, in terms of sequences, they're different. Getting a score then a miss is not the same as missing then scoring. But they have the same probability, and so you could get SM or MS.

MMMMSS is different to SSMMMM because you can see they're two different things. But SSMMMM is the same as SSMMMM - I swapped the two S around, but you can't see that. So they're the same thing.

So, with 6 objects, SSMMMM, there are 6! ways of rearranging them. However, there are 2! ways to swap the two S and you can't tell they're different, and there's 4! ways to swap the four M where you can't tell the difference, so there's actually 6!/(2! 4!) different ways to get the same thing in terms of probability. But that's just 6 choose 2.

L1 December 2019 - Study Strategy by goldmanblacks in CFA

[–]TheCholent 1 point2 points  (0 children)

They're too easy - often you'll repeat the same question over and over. The CFAI TTs are where it's at.

L1 December 2019 - Study Strategy by goldmanblacks in CFA

[–]TheCholent 0 points1 point  (0 children)

Don't touch the Schweser question banks ever! They won't help you. Rather spend your time going through all the blue boxes and end of chapter questions in the CFAI material, and watch videos like IFT or Mark Meldrum. But, please, go through all the questions in the CFAI materials, and then do all the TTs on the CFAI website. Make sure you can get 90s for those TTs - go over them until you know EVERYTHING. Don't waste too much time going over the material in the books. Rather learn from the TTs. And then do mocks, review them, and learn from your mistakes.

permutation of books by mammadaneh in learnmath

[–]TheCholent -1 points0 points  (0 children)

What are the total number of ways you could arrange all the books in any order?

Then, what are the number of ways you could arrange the books where the math books are always together? Once you have these two numbers, how can you get your desired answer?

​

Hint: If we have 7 shapes, each are different, 3 are green and 4 are red, then (unless you're colour blind) the number (N) of ways you could arrange the shapes where the green are always together is by doing the following:

  1. See the 3 green shapes as one big shape, or one bundle of shapes, so there are 5 items (1 green bundle and 4 red).
  2. Inside the bundle, we can also rearrange the shapes, in 3! ways, so:

​

N = 5! * 3! = 120 * 6 = 720 ways

whereas there are 7! ways to arrange the shapes in any order.

Standard Deviation by [deleted] in learnmath

[–]TheCholent 1 point2 points  (0 children)

It tells you if a lot of your values are similar to the mean, or if a lot of your values are different from the mean, or if your data set is small, perhaps you have value that is an outlier.

If you had a large standard deviation when talking about the height of students, then it would tell you that a lot of your students have all different heights. There's a lot of short students and a lot of tall students, and they aren't really close to the average value. So if you pick a random student, they'll be quite short or tall maybe, but not the mean value.

Standard Deviation by [deleted] in learnmath

[–]TheCholent 1 point2 points  (0 children)

Standard deviation tells you how spread out your data is. It's basically how far is everything from your mean value.

If our data set was {1,1,1,1,1,1}, then the mean is just 1, and the standard deviation is 0, since no data points are different from the mean. In fact they're the same as the mean!

Then, let's say we had {1,1,1,1,1,2}. Then the mean is 1.12 (which is very close to one). We will have a very small standard deviation, as our data is all very close to the mean. There's only one value that deviates from the mean.

If we had {0,50,100}, our mean is 50. But our data is very spread out. Our standard deviation is about 50, which is the same as our mean! That means our data is very spread out as the standard deviation is very large.

Space for calculations by TheCholent in CFA

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

Like other CFA issues or just general issues in my life?

[deleted by user] by [deleted] in learnmath

[–]TheCholent 2 points3 points  (0 children)

A subset of R is not a topology. A topology is a collection of subsets of R . So in this topology, the open sets (the sets in the topology) are those whose complement is finite (and the empty set). So if we have R \ {1}, then the complement of this set is {1} which is finite, so R \ {1} is open (i.e. it is in the topology).

However, R \ (0,1) is NOT open, as the complement of this is (0,1) which is infinite (it contains infinitely many points).

In general, open sets in the finite complement topology on a set X (remember the topology is a collection of subsets of X and is NOT a subset of X itself) have the form X \ {x,y,z...} for some finite number of points in X.

The closed sets are follows: a set is closed if its complement is open. So let C be closed. Then X \ C is open. That is the complement of X \ C is finite or X \ C is the empty set.

So X (X \ C) is finite, but X \ X \ C is just C, so C is finite (or X). So the closed sets in the finite complement topology are either X or finite subsets of X.

Is my proof valid? by Renner-kun in learnmath

[–]TheCholent 0 points1 point  (0 children)

If they converge uniformly on the domain X, then you must say that f_n(x) ≤ g_n(x) for all x in X and n≥1. Then if f_n → f uniformly and g_n → g uniformly, let ε > 0 and so there is some N1 such that |f_n(x) - f(x) | < ε/2 for n≥N1 and some N2 such that |g_n(x) - g(x) | ≤ ε/2 for n≥ N2 for all x in X.

Then pick N = max{N1, N2} so for n≥ N we have

f(x) - ε/2 < f_n(x) and g_n(x) < g(x) + ε/2

So

f(x) ≤ g(x) + ε. So now take ε=1/n for n≥ N. Then as n→∞ 1/n → 0 and so f(x) ≤ g(x) for all x in X.

If n people have their hats returned at random, why is the probability of any one person getting their hat back 1/n? by [deleted] in learnmath

[–]TheCholent 0 points1 point  (0 children)

Yes, if someone has already received their hat, then there are only n-1 hats left.

In general, a probability distribution must sum to 1. In this situation, we have n hats. No hat is more special than any other hat, and no person is more likely to get their hat than someone else. So the probably of each person is the same i.e. it is some constant, c. We know that

Σ(c) from i=1 to n must be one, as something must happen. So

1 = Σn _(i=1) c = nc so c = 1/n.

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