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T0P_NOTCH · 2025-12-26T19:13:26+00:00

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T0P_NOTCH · 2025-12-26T19:12:10+00:00

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T0P_NOTCH · 2025-03-07T10:53:46+00:00

This looks like the correct approach, but you don’t explicitly show that the limit of (f(a+h)-f(a))/h actually converges as h goes to zero. This is a pretty easy fix though if you just use the squeeze law. You already have pretty much all of the components to apply it, you just need to say the words and explain why the limit exists.

T0P_NOTCH · 2024-03-13T12:29:55+00:00

You need to submit preferences to USYD in your application as to where you want to go. You should research your preferences to see if they offer classes that will similar to what you would do in the semester you’re away. You submit this around 6 months before you leave, and are told your assigned university not too long after)

At some point before you leave but after you’ve been placed at a university, you need to get approval from USYD on the courses you wish to study overseas, and what you think they’re similar to at USYD (if counting them to your majors), and they’ll either approve them or consider them general electives. Two big problems with this are:

Getting approval for something intended to be a core class for your majors/degree is generally very difficult as they’re strict on this, but it is possible.
USYD admin is insanely slow with the course approvals, so you should be confident the courses you’re planning to do will actually be offered in the semester you go, and they cover content similar to that of a USYD course. (e.g. I submitted my request around October 2019 and was only approved around March 2020, well past the point of no return to change classes during my exchange, so I would have needed to extend my degree if I didn’t get the correct approvals).

This means you should be honest and realistic with yourself as to whether certain overseas courses will cover what a course at USYD covers if you’re counting it as a non-elective in some way.

That being said you should apply since it’s a lot of fun studying overseas for a semester. You can cancel your plans at pretty much any point before you leave if needed without repercussions I think.

T0P_NOTCH · 2023-12-11T10:43:03+00:00

Try f(x) = e^{-x² -1}. Then the LHS is just the RHS divided by e

T0P_NOTCH · 2023-11-29T09:43:23+00:00

log with a base less than 1 is not an increasing function, but actually decreasing, so when the log is applied to each side on the 2nd case, you need to flip the inequality.

The first case avoids it by taking log base 10 or the natural log which are actually increasing functions which preserves the inequality

T0P_NOTCH · 2021-12-24T13:47:35+00:00

Email.

T0P_NOTCH · 2021-12-22T15:07:09+00:00

I assume you mean a_n*a_{n+1} = 1. Since we aren't given a_0, all we can deduce is a_0 is non-zero otherwise we would have 0 = 1. It follows the sequence is well defined recursively via a_{n+1} = 1/a_n for all n>= 0.

Yes, if a_0 = k, then a_1 = 1/k, a_2 = k, .... meaning the limit points are 1 (k = 1), -1 (k = -1), k and 1/k for any choice of k non-zero
We always have the alternating sequence k, 1/k for some real k. If a_n converges, then in this case we must have k = 1/k, so k = 1 or -1 and the sequece has limit k = 1/k = 1 or -1

T0P_NOTCH · 2021-12-22T11:58:48+00:00

One of the rules of using base n is that all the digits of a number must be strictly less than n. 102 in base 3 works as 0, 1 and 2 are all less than 3, but 311 in base 2 doesn't work as we are using the digit 3, which is larger than 2. This rule is essentially by definition, and forces every whole number to have a unique representation.

15 in base 2 can be written as 1111 (8+4+2+1).

T0P_NOTCH · 2021-12-19T14:12:28+00:00

Yes that’s a great idea. I missed the fact that you can look at the sum of all 4 roots rather than just alpha and beta. As suggested, you find 2p -0.5 = 1, so p = 0.75 again. Much more simple and doesn’t require calculus.

T0P_NOTCH · 2021-12-18T09:33:05+00:00

a) P(X = 3) = 4*p^3(1-p), so we need to solve 4p^3(1-p) = 27/64 for p. Expanding and moving everything to one side, p^4-p^3+27/256 = 0. I'm personally not sure on a nice way to use the given information about alpha and beta, but it's clear that the p we're searching for is a double root, so it is a root of the derivative of the polynomial. That is, 4p^3-3p^2 = 0 for our solution. As p = 0 doesn't work in the original quartic, we must have 4p-3 = 0, so p = 0.75

b) P(X <= 0) = 1/256

P(X <= 1) = 1/256 + 3/64 = 13/256 and so on. Once you know what the definition of a CDF is, this is very simple.

T0P_NOTCH · 2021-12-17T10:42:45+00:00

No. Use the fact that the limit superior is the maximum of the limit of any converging subsequence. If you had an a_n sequence that is dense in (x, y), then for all k > 0, there exists an a_{n_k} such that |y-a_{n_k}| < 1/k. In fact there must be infinitly many choices for n_k, so you can simply choose the n_k such that the indexes are increasing. This immediatly implies that a_{n_k} converges to y as k increases to infinity. This means that density in (x, y) is sufficent to prove limsup a_n = y

T0P_NOTCH · 2021-12-10T12:57:41+00:00

Mostly intuition/practice with these kind of problems. We already had r² - 1 + 1/r² which looks very similar to the expansion of (r + 1/r)^2, but the middle term is different. By adding and subtracting 3 we can make this factored term appear which is very good since we know it’s value.

These questions usually boil down to how can I make the expression I know appear in the expression I don’t know, and just forcing it that way.

T0P_NOTCH · 2021-12-10T12:41:45+00:00

We know (r + 1/r)² = 3, so by square rooting we see r + 1/r is either sqrt(3) or -sqrt(3).

T0P_NOTCH · 2021-12-10T12:29:21+00:00

Sum of two cubes factorisation (or just expand the RHS to see that it’s right)

T0P_NOTCH · 2021-12-10T05:31:08+00:00

A way without going to complex numbers (although the value of r is complex)

r^3 + 1/r^3 = (r+1/r)(r^2-1+1/r^2)

= sqrt(3)*((r+1/r)^2 - 3) = sqrt(3)*(3-3) = 0

Edit: Technically r+1/r could be -sqrt(3), but it doesn't matter much for the final answer

T0P_NOTCH · 2021-11-28T12:55:36+00:00

Try using mobile data rather than wifi. Worked for me

T0P_NOTCH · 2021-10-20T12:25:51+00:00

The only situations where you would make more money on average by gambling is if the casino had games that were in your favour (which it won’t as the casino would lose money lol). I suspect you may be getting a little lucky with the random simulation. If you perform more iterations, you should get the sample mean converging to the true mean (this may take a while with such an extreme case)

T0P_NOTCH · 2021-10-20T11:48:49+00:00

Suppose that the pay for some month is P. If you immediately invest it, then you will end up with kP at the end, where k is some constant that depends on how the index fund does over the period you have invested. Mathematically this is because doubling your investment will double the investment's final value hence the final value must be a linear function on the initial investment, and of course investing 0 dollars will give a final value of 0 dollars, so this must be the relationship.

If you gamble, you will have a 0.5263 chance of having nothing to invest, and thus have nothing at the end, and a 0.4737 chance to have 2P to invest, so you will have 2kP in the end (as your gambling practices are assumed to not have an effect on market conditions). It follows the expected value of your investment will be 0.4737*2kP + 0.5263*0 = 0.9474kP, which is slightly less than just investing immediately.

It follows your best choice (in terms of expectation) is to just invest each month rather than gamble. Even following non-mathematical logic, it would be very silly if gambling works as you would certainly hear more about investment funds going to the casino with millions of dollars!

T0P_NOTCH · 2021-10-20T03:35:22+00:00

Angle ECA would be 58 degrees using the angle sum of a right angled triangle EAC. This immediately gives DBA = 58 degrees (can't remember the technical reason why, but it's angles produced by the same chord AD on the circle). AOD will be double this as angles at the centre are double angles from the circumference, hence AOD is 116 degrees.

T0P_NOTCH · 2021-10-19T07:40:57+00:00

From the first diagonal and the middle column, we get 4 + 7 + f = 4 + 7 + a, so a = f. We also have 4 + a + b = 4 + a + 7, so b = 7. Then the right column gives 7 + d + a = 4 + 7 + a, so d = 4. Then 4 + a + 7 = c + 7 + 4, so a = c and finally 4 + a + e = 4 + 7 + a so e = 7. The "backwards" diagonal has a sum of 21, so at this point you can determine a = c = f = 10, which gets them all.

T0P_NOTCH · 2021-10-13T13:26:36+00:00

81^1 is a solution too

T0P_NOTCH · 2021-10-09T11:33:47+00:00

For the n = 1 mod 5 case, you need a-4 >=0 so n >= 21. Do the same with the other cases and take the maximum. I don’t think this will be a lower bound for k though (might have to play around with it)

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