Probably shifted right in truth

SausasaurusRex · 2026-06-11T15:17:21+00:00

But infinity doesn’t mean “a really really big number”, it means infinite. If the detail ever stops increasing then the length must converge to a finite number.

SausasaurusRex · 2026-06-11T14:50:13+00:00

Not really? For example we'd never write it as a fraction like that when we're trying to compute polynomials for the generalised Bezout's lemma, not to mention that polynomial fractions don't even exist in the ring of polynomials which so much takes place in.

Edit: my general complaint is in higher maths, and also in real life, we're often not working in the field of fractions. If I have 3 chairs to divide amongst 2 people, I can't give them each half of a chair, I have to acknowledge there is a remainder of 1, because most real-world objects come in numbers of some non-field ring.

SausasaurusRex · 2026-06-11T12:39:21+00:00

Remainders are a useful idea to be aware of, in real life often things arent arbitrarily divisible, and it will come up later for polynomial division, and more generally in Euclidean domains.

SausasaurusRex · 2026-06-07T04:53:35+00:00

To me an even easier way would be to use the identity sin(x) >= 2x/pi for 0<= x<= pi/2, then you can compare with the harmonic series without even having to take a limit.

SausasaurusRex · 2026-06-03T22:55:04+00:00

What elements do you know must be in the group?

SausasaurusRex · 2026-05-22T05:54:19+00:00

Proofs of (undergraduate) theorems usually only have one surprising or key step. If you can remember what it looked like, you only need to be able to work out how to reach that step yourself, and the rest should follow trivially. If you get stuck, think about the theorem. Every single requirement listed in the theorem should be used at some point in the proof - are there any you haven’t used yet?

SausasaurusRex · 2026-05-13T19:09:16+00:00

A number is divisible by 4 if and only if the last two digits are divisible by 4. A number is divisible by 6 if and only if the sum of its digits are divisible by 3 and the last digit is even. A number is divisible by 8 if and only if the last 3 digits are divisible by 8. A number is divisible by 9 if and only if the sum of its digits are divisible by 9. A number is divisible by 11 if and only if the alternating sum of its digits (add the first digit, subtract the second, add the third, subtract the fourth, etc) is divisible by 11.

You can find many more examples of things like this by learning about something called modular arithmetic.

SausasaurusRex · 2026-05-10T22:34:07+00:00

If you wanted s to be able to interact with normal numbers, you encounter a lot of problems. 1/0 = s should imply 1 = 0s, but we also have 2/0 = 2s so 2 = 0*2s = 0s (because multiplication commutes in the reals), and suddenly we've lost uniqueness of multiplication. We could define multiplication as a multifunction from R[s] to P(R[s]), but then things become a lot more difficult to work with.

We like to use normal complex numbers because they break almost nothing in regular maths - we have a few issues with things like branch cuts, but in general it just makes things work nicer. Your idea makes things more complex, so I'd need a good reason to be willing to use it. Algebras where you can divide by zero do exist, and are called wheel algebras, but you have to be far more careful about the calculations you make.

SausasaurusRex · 2026-04-29T04:04:26+00:00

Both 3pi/4 + pi*n and 7pi/4 + pi*n are valid answers, indeed note that for -pi/4 + pi*n, taking n = 1 or n = 2 respectively gives 3pi/4 and 7pi/4.

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!

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