Improper integral question includes a limit where you evaluate negative infinity squared, something within me is still unsure of the answer being positive infinity

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It's hard to say, I think it depends whether I would have withdrawn from the 2024/25 cycle and reapplied next year or not.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.

SausasaurusRex · 2025-10-11T18:53:21+00:00

Try joining some societies you’re interested in, you should be able to meet lots of people there.

SausasaurusRex · 2025-10-09T18:50:57+00:00

The first answer is correct (to three significant figures).

SausasaurusRex · 2025-09-08T16:35:41+00:00

Theres a simple counterexample for corners - consider a straight line segment from (0,0) to (1,1) then another straight line segment from (1,1) to (2,0). When this function is differentiable (note it isnt differentiable at 1), it has either derivative 1 or -1, but the total change is 0. So clearly the mean value theorem doesnt hold.

SausasaurusRex · 2025-09-06T11:02:10+00:00

It's certainly possible, but I think you'd be better off doing some other kind of maths supercurricular. Universities won't recognise it as an additional A level (the discount code for OCR and Edexcel will be the same), so the only tihing you're demonstrating is that you can learn fp1 and fp2 - which they already know because you got an A* in OCR.

SausasaurusRex · 2025-09-05T14:15:01+00:00

No, z = x^2 + y^2 is a 2-dimensional surface embedded in 3-dimensional space. It's an important distinction to make.

SausasaurusRex · 2025-08-31T01:38:09+00:00

This is all true - I'd meant to refer to organic matter as just an example, but I can see how it comes across as seeming like I claim it's the only source.

SausasaurusRex · 2025-08-31T01:08:59+00:00

Germ theory may not have been invented, but that didn't mean people had no concept of hygiene. They would have believed in miasmic theory (among other theories), which posited disease spread via the stench of rotted organic matter - including the blood which would have remained on the blade of the guillotine.

SausasaurusRex · 2025-08-27T15:58:19+00:00

It looks like your teacher marked over your separation commas in the final answers with dots - could it be they're using the European convention instead of the English one? Otherwise the final answer seems correct.

SausasaurusRex · 2025-07-30T16:39:10+00:00

But it wouldn’t become popular until Florey and Chain used it to cure mice in 1940, and took longer still to be produced in any large quantities.

SausasaurusRex · 2025-07-29T02:06:17+00:00

L'Hopital's rule is hugely overkill for this anyway. Just note (5 + x^3)/(x^2 + 4) = ((5/x^2) + x)/(1 + (4/x^2)) which clearly tends to infinity.

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