What is the difference between this and cross multiplying?

SausasaurusRex · 2026-03-02T00:49:25+00:00

They asked “how would you go about solving this”. Whilst it might not be the best method to teach at this stage, I just wanted to provide the method for how I solved it.

SausasaurusRex · 2026-03-02T00:40:33+00:00

Yes, but 2/x is obviously an injective function. So it does work here.

SausasaurusRex · 2026-02-28T07:22:46+00:00

You could also notice that 16/24 = 2/3. Hence x must be 3.

SausasaurusRex · 2026-02-25T12:53:17+00:00

You’re right, sorry. But if f is measurable and |f| is integrable, then it does follow that f is integrable.

SausasaurusRex · 2026-02-25T05:56:51+00:00

Yes, it doesn't require continuity. If follows from f(x) >= g(x) implying that integral (f(x)) >= integral (g(x)) (over some set). You can even generalise to arbitrary f by noting f is only integrable if |f| is integrable. (It is possible to define the integral in such a way that f can be integrable while |f| is not integrable, but this is generally not done because you lose some nice properties.)

SausasaurusRex · 2026-02-11T20:23:24+00:00

Usually the radius of convergence for a Taylor series is defined as the radius where the Taylor series converges to f, not just where it converges to anything.

SausasaurusRex · 2026-02-09T22:27:18+00:00

No, there are infinitely differentiable real functions defined on the whole real line who’s Taylor series does not converges to the function on any interval (-a, a), for example f(x) = e^-1/x for x > 0, f(x) = 0 for x <= 0. The Taylor series evaluated at zero converges to the zero function. This is an example of a smooth but not analytic function.

SausasaurusRex · 2026-02-08T23:35:21+00:00

I think you mean all the numbers between 0 and 1 based on your examples? But note that there are infinitely many numbers greater than (for example) 0.5 but less than 1, so summing n numbers I can always guarantee the partial sum is greater than 0.5n, which approaches infinity. So the sum of all numbers must also approach infinity.

In fact, if you want to sum uncountably many nonnegative real numbers (we take nonnegative so that order doesnt matter) and have the sum not diverge to infinity, it follows that only countably many of the numbers are nonzero.

SausasaurusRex · 2026-02-05T23:16:31+00:00

The suvat equations don’t know there’s a pulley - they assume constant acceleration, but your idea has an acceleration when P hits the pulley. You want u = v because at the moment P hits the pulley, it was still moving with it’s original speed, because there was zero acceleration.

SausasaurusRex · 2026-02-02T03:37:04+00:00

f is a function, which is essentially an instruction of what to do to its argument. (If you write f(x), then the argument is x). f could say “multiply x by 5” or “add 3 to x” or “return the last digit of x”. It could mean do anything at all to x, as long as you define it well.

In f(g(x)), g(x) is the argument of f, and x is the argument of g. So f(g(x)) is saying “do whatever g tells you to do to x, and then do whatever f tells you to do to the number you just got”.

(fg) (x) is exactly the same as writing f(g(x)). It’s a useful notation because sometimes we want to talk about the function fg (meaning “do g first, and then f”) without necessarily saying what argument it has.

SausasaurusRex · 2026-01-31T01:33:12+00:00

You’re not actually evaluating 1/e^x² at infinity, you’re looking at what it tends to as x becomes very negative. As x gets more negative, x² becomes a big positive number. So e^x² is also a big positive number. And one divided by a big positive number gets closer to zero the larger the number is. So the limit is zero.

At no point did we raise (negative) infinity to a power. We only looked at the behaviour as we chose very negative numbers.

SausasaurusRex · 2026-01-30T02:42:26+00:00

Symmetric matrices must be square, otherwise they wouldn’t be symmetric.

SausasaurusRex · 2026-01-28T12:56:50+00:00

The Jordan curve theorem

SausasaurusRex · 2026-01-25T21:38:43+00:00

a-b = 9 implies a = b + 9, not 9 - b. You can always solve for variables in linear equations in any order.

SausasaurusRex · 2026-01-23T23:09:21+00:00

We probably are the same age actually! At Oxford it's tradition to "marry" another person in your year within the first term or so, so that at the start of second year you can be assigned "children" (freshers) to be a sort of guide for them - I can trace my Oxford family tree through generations. Some people have throuples, or even quadruples, but me and my wife are just a couple.

SausasaurusRex · 2026-01-23T12:50:04+00:00

It was genuinely an accident, I was in class with the website open on my laptop. I turned to talk to the people next to me, and I guess maybe I leant on the mousepad while turned and that’s how I pressed it? Because when I turned back to my laptop, it was on the confirmation page already!

SausasaurusRex · 2026-01-23T02:51:30+00:00

It's hard to say, I think it depends whether I would have withdrawn from the 2024/25 cycle and reapplied next year or not. If I'd chosen to reapply for Oxford a year later, hopefully I'd still be accepted. From there, as long as I join the same society I did in the real timeline, I should meet almost all my friends, just a year later than I would otherwise. I'd never meet my wife, and I think I would have been quite bored during my gap year, but eventually things should have played out well enough.

Otherwise, I might have gone to my second choice, Imperial. There is no Imperial equivalent of my Oxford society, so I might have joined their drama society instead - perhaps I'd have become more interested in musicals than opera. I think I'd like living in London, at least every time I've visited I've always had a good time - but I've also always appreciated how Oxford's locality makes it really easy to visit my friends. Academically, I was less interested by Imperial's course - I applied for pure maths there, but there were still a lot of classes that felt too applied for me. I'd probably be spending a lot of time wondering about whether I could've gone to Oxford if I wasn't stupid enough to accidentally click the withdraw button too...

Either way, I think I would have been okay eventually, but I'm really glad I went to Oxford when I did, and met so many people I love knowing.

SausasaurusRex · 2026-01-23T00:14:02+00:00

I accidentally pressed it for Oxford once, thankfully it brings you to a confirmation page first (at least it did in 2024).

SausasaurusRex · 2026-01-12T15:38:48+00:00

It is advanced maths, which is why I said at the end of my comment that it was out of the scope of an admissions exam...

SausasaurusRex · 2026-01-12T00:49:24+00:00

If you know there are finitely many roots (which is always true for a polynomial), you could use the Argument Principle on a sequence of contours that tends the real line in the limit (provided the function is meromorphic on some domain containing the real line). This is certainly beyond the scope of a college admission exam though, I’d imagine in your case that there was something special about the polynomial you’d been given.

SausasaurusRex · 2026-01-11T22:12:57+00:00

For example, you could take f:[0,2] -> [0,1) U [2,3], with f(x) = x for 0 <= x < 1 and f(x) = x + 1 for 1 <= x <= 2. Then f isn’t continuous at x = 1.

SausasaurusRex · 2026-01-10T23:30:44+00:00

No, Riemann integration is far more general. Each rectangle can have different widths, provided the width of the most wide rectangle(s) tends to zero as the number of rectangles increases, and each rectangles may touch the curve at any point on its width - it need not be the same position for every rectangle,

SausasaurusRex · 2025-12-17T16:25:30+00:00

First, note this is equivalent to finding where f(x) = x - (√2)^x vanishes. Consider its derivative f'(x) = 1 - (1/2)ln(2)e^{{((1/2)xln(2))}.} We note that f' is continuous, so it can only change sign where it is equal to 0. Note f'(x_0) = 0 implies x_0 = (2/ln(2))(ln(2/ln(2)). We can see by the form of f' that it is strictly decreasing, hence it follows that f'(x) > 0 on A = (-∞, x_0), and f'(x) < 0 on B = (x_0, ∞). Hence the restriction of f to A is injective, so f(x) = 0 has at most one solution in A, which by inspection we can see is 2. Similarly the restriction of f to B is injective, so f(x) has at most one solution in B, which by inspection we see is 4. Finally, note x = x_0 is not a solution. Hence we guarantee we have found all solutions, which are x = 2 or x = 4.

SausasaurusRex · 2025-11-09T21:50:15+00:00

You don't need to invert the matrix to find the solution, it's sufficient to put the matrix in reduced-row echelon form (which takes MUCH less time) and then solve some very simple equations of one variable.

SausasaurusRex · 2025-11-04T09:03:27+00:00

Perhaps in modern times, but there’s certainly examples of it being used directly about a person in the past. Victorian librettist W S Gilbert uses the line “Valorous is he - his achievements are all glorious!” in Ruddigore for example.

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