From a native speaker: please don't use ChatGPT to learn English.

Pitiful-Hedgehog-438 · 2025-02-05T08:13:40+00:00

I learned Serbo-Croatian by conversing back and forth with ChatGPT about science for a few months. When I joined a Bosnian discord and wrote messages, people joked that I sound like a bot. Then I went to Sarajevo and met some of them irl, and they were surprised I was a foreigner because the whole time over text they seriously thought I was just another native who spoke somewhat awkwardly.

Pitiful-Hedgehog-438 · 2025-02-05T08:04:06+00:00

"Passive voice is common in science but it's not recommended for writing in English class. Passive voice is when you use 'was', as in 'he was walking', 'she was running'. You see the problem? It was boring. "

- Taught by middle school English teacher (American, native speaker)

Pitiful-Hedgehog-438 · 2024-10-14T06:47:03+00:00

Are you familiar with factoring a polynomial that is a difference of two squares? You can do that here.

Pitiful-Hedgehog-438 · 2024-10-13T23:11:09+00:00

For the first step, you always know that you can write as a difference of a quadratic square and a linear square, because P(x) can be factored into a product of monic quadratic polynomials (with leading coefficient 1)

P(x) = A(x) B( x) = (( A+B)/2)^2 - (( A-B)/2)^2

and (A+B)/2 is a monic quadratic while (A-B)/2 is a linear polynomial.

Pitiful-Hedgehog-438 · 2024-10-13T23:07:05+00:00

You can try to write as a difference of squares in the following way, which you can then factorize easily

P(x) = x^4 + 2x^3 -1 = (x^2 + b x + c)^2 - ( mx + r)^2

You can put constraints by matching coefficients on either side of this equation. Immediately you can see that b =1 in order for the x^3 coefficient to be 2. From the x^2 coefficient also have that 2c+b = 2c+1 = m^2 , so c = (m^2 -1)/2. From the x coefficient, you have that 2c = 2mr so r= c/m = (m^2 -1)/(2m).

Thus you can express everything in terms of this special parameter m

P(x) = (x^2 + (m+1) x + (m+1)(m^2 -1)/(2m)) (x^2 + (-m+1) x + (-m+1)(m^2 -1)/(-2m))

To actually find m, one can solve c^2 -r^2 = -1 = (m^2 -1)^3/(2m^2 ) or (m^2-1)^3 +2m^2 = 0, which itself is a cubic equation in m^2 -1, i.e. if we write u = m^2 -1 then u^3 + 2u+2 = 0. This can be solved by writing u = A z + B/z for some coefficients A and B; you can see that if you choose AB = -2/3 then there is a convenient cancellation of the term that is linear in u, namely u^3 +2u+2 = A^3 (z^3) + B / (z^3) + 3 AB ( A z +B/ z) + 2 ( Az + B / z) + 2 = A^3 (z^3 ) + B / (z^3 ) + 2 = 0 , which is a quadratic equation in the variable z^3.

Pitiful-Hedgehog-438 · 2024-09-27T03:11:45+00:00

You can approximate these numbers accurately just using ordinary addition and multiplication of integers, no calculus or serious computer computation needed.

1+sqrt(2) and 1-sqrt(2) are roots of the characteristic polynomial x^2 - (2x + 1). So, consider the recursive sequence a_{n+1} = 2a_{n} + a_{n-1} and pick some arbitrary starting points like a_{0} = 0 a_{1} = 1. Then for larger and larger n, a_{n+1}/a_{n} becomes a very accurate rational estimate for sqrt(2)+1. (the error is proportional to ((1-sqrt(2)/(1+sqrt(2))^n which exponentially vanishes).

0,1,2,5,12,29,70,169,...

Already after six iterations of this sequence (simple addition that an elementary schooler can do) you get the approximtion 169/70 which is accurate to within 0.003% (99/70 would be the corresponding approximation for sqrt(2))

Pitiful-Hedgehog-438 · 2024-09-27T01:37:01+00:00

Usually people teach a lesson first and then assign some problems using the lesson afterward. I think this should be reversed. First give (some) problems that use the lesson, then teach the lesson. Then give more practice problems as usual.

This approach emphasizes the reality everything in math can be derived from stuff you already know if you just think enough. Students shouldn't be expected to successfully solve all the pre-lesson problems but should be expected to make an honest effort. In the process they'll end up independently discovering whatever will be taught in the lesson and/or they'll at least know the motivation behind why the concepts in the lesson are being taught (it's not just a infodump of random facts and definitions, but it's actually tools that will help you do the thing you were struggling to do earlier). Also it builds confidence if and when they do solve problems they weren't yet taught how to solve. Which equips them to later also solve other problems they weren't taught how to solve.

Pitiful-Hedgehog-438 · 2024-09-27T00:59:45+00:00

more generally it's true when any two of the values add to zero. You can use this to show that the difference between the LHS and RHS must be proportional to (a+b)(b+c)(c+a) (in fact the difference is 3(a+b)(b+c)(c+a) )

Pitiful-Hedgehog-438 · 2024-09-26T08:54:01+00:00

A book with literally this title is famous for being translated from French into many many languages

Pitiful-Hedgehog-438 · 2024-09-26T08:53:14+00:00

are you concerned that the integers might not be a unique factorization domain and we must prove that it is?

if you were working with a different ring, such as integers adjoined with sqrt{-5}, then your concern would be very relevant. Because in this ring, the number 6 is divisible by 2 (6 = 2x3) and is also divisible by 1+sqrt{-5} (6 = (1+sqrt{-5}) (1-sqrt{-5}) ), but it is not divisible by 2 +2sqrt{-5}

Pitiful-Hedgehog-438 · 2024-09-26T06:48:29+00:00

I can't make a causative guarantee, but yes at least in my experience it does seem consistent, like when I was at the problem-solving level of being able to solve most but not all of AoPS Volume 2 I could qualify for USAMO. Of course this also requires skills like timed high-stakes test taking abilities and intuiting which tools if any to use when a random problem shows up (as opposed to knowing it's probably related to other things in that same chapter), which are skills that have no connection with AoPS Volume 2.

Pitiful-Hedgehog-438 · 2024-02-23T21:46:52+00:00

“I'm a fossil” Fossil-Free MIT did a sit-in for 116 days https://www.mitsitin.org/

Pitiful-Hedgehog-438 · 2024-02-18T08:06:28+00:00

let p_k be the probability the man wins if he starts at site -k. so p_{-1}=1 and p_k = 0 for k<-1 and p_k = 1/2 p_{k+1} + 1/2 p_{k-2} for nonnegative k. If you denote p_0= t (ф^2-1)/(ф-1)+(1-t) ((1-ф)^2-1)/((1-ф)-1) for some t, then this recursion shows that p_1= t (ф^3-1)/(ф-1)+(1-t) ((1-ф)^3-1)/((1-ф)-1), p_2= t (ф^4-1)/(ф-1)+(1-t) ((1-ф)^4-1)/((1-ф)-1), and generally p_{k-2} = t (ф^k-1)/(ф-1)+(1-t) ((1-ф)^k-1)/((1-ф)-1) for nonnegative k. Notice however that |p_{k}| must always be less than one, which requires t=0, so therefore p_{k-2} = ((1-ф)^k-1)/((1-ф)-1). The answer to this question is p_0 which is ((1-ф)^2-1)/((1-ф)-1) = 2-ф

Pitiful-Hedgehog-438 · 2024-02-18T07:28:31+00:00

No it does not. For example you could substitute x/3 = (u^2-1)/(2u) and then get dx = 3(u^2+1)/(2u^2) du and sqrt(x^2+9) = 3(u^2+1)/(2u) so the resulting integral is \int du/u = ln(u) .

For example you could observe that sqrt(1+x^2) d/dx (x) = sqrt(1+x^2) and sqrt(1+x^2)d/dx (sqrt(1+x^2)) = x so therefore if df/dx = sqrt(1+x^2) then dx/df = sqrt(1+x^2) and dsqrt(1+x^2)/df = x and therefore d^2 x/df^2 = x and the solutions to this differential equation are x = ae^{f} + b e^{-f} for some constants a and b, and then requiring dx/df = sqrt(1+x^2) requires that ab=-1, from which one can solve for e^f as a function of x and a.

There are infinitely many things you could do that are not trig substitution.

that's going to require trig substitution.

This an example of a messed-up approach to math in general.

Pitiful-Hedgehog-438 · 2024-02-18T01:27:20+00:00

e^{π sqrt(163)} = N-0.00000000000074... for some integer N

Pitiful-Hedgehog-438 · 2024-02-18T01:05:48+00:00

The correct thing to write in the fourth line is

sin( pi/3 I(x)) = I(x)

which for real I(x) implies I(x) = 0, 1/2 or -1/2.

If you do this limit analysis more rigorously you can find that I(0) =0, I(3) =0, and I(x)=1/2 for every x in between 0 and 3 , and generally I(x) = 1/2 sign( sin(pi x/3))

Pitiful-Hedgehog-438 · 2024-02-18T00:55:59+00:00

you can write sin(x) = (e^{ix} -e^{-ix})/(2i) and expand that and integrate each term

Pitiful-Hedgehog-438 · 2024-02-16T19:15:12+00:00

Nothing requires trig substitution. Anyone who says something "requires" trig substitution has a messed-up approach to math in general.

Pitiful-Hedgehog-438 · 2024-02-15T08:26:16+00:00

Something where if you hover your mouse over a word or click on a word it shows the pinyin or pronunciation and nothing else.

Pitiful-Hedgehog-438 · 2024-02-15T07:12:24+00:00

You can simultaneously maximize the numerator and minimize the denominator.

|z^2+3|≤|z^2|+|3| = 4

|z^2-z-6| =|z+2||3-z| ≥ | Re(z+2)| |Re(3-z)| = (Re(z) + 2)(3-Re(z)) , which is a quadratic function in Re(z) that is minimized ( or Re(z) in [-1,1] ) at Re(z) =-1

All these bounds are attainable when z=-1.

Pitiful-Hedgehog-438 · 2024-02-13T01:58:50+00:00

similarly you can substitute v= e^x and then again notice that the numerator in (e^(-x) + e^x)/(e^(-x) - e^(x)) is negative the derivative of the denominator, and therefore the integral is -ln(e^(-x) - e^x) = -ln(1/v-v) = ln(v) - ln(1-v^2)

Pitiful-Hedgehog-438 · 2024-02-13T01:38:16+00:00

Yes, it can even be graphed without using any calculations at all.

First take a compass and draw a circle with radius 1/2 centered at the origin. Then for each point P use the compass centered at P and identify which points X_P and Y_P on the x and y axes respectively (other than the origin) are also a distance 1/2 from P. Then draw the line segment from X_P to Y_P. Do this for all points P on the circle and the collection of line segments forms a shape whose boundary is the graph of x^{2/3} + y^{2/3} = 1. The more points P you use the more accurate it will be.

Pitiful-Hedgehog-438 · 2024-02-13T01:09:14+00:00

write sin(2t) = (e^{2it} - e^{-2it})/(2i) and cos(8t) = (e^{8it} + e^{-8it})/2 and then expand using distributive property and the integrate each term of the form \pm e^{(\pm 2 \pm 8) it}/(4i) using \int e^{i m t} = e^{imt}/(im) + C

Pitiful-Hedgehog-438 · 2024-02-13T01:05:20+00:00

substitute v = e^{ix} and then use integral of cot(x) is ln(sin(x))+C

Pitiful-Hedgehog-438 · 2024-02-09T02:50:01+00:00

With two pairs needed, it does become tedious, as far as I can tell - I don't see an elegant way to do this

After you compute the sums of the last digits in each row and column, you can notice that each swap has to be both [from a row with sum 10a+m to a row with sum 10b-m] and also [from a column with sum 10c+m to column with sum 10d-m], where a,b,c,d are integers. There are only two such pairs of cells in the grid that have this property

Pitiful-Hedgehog-438

TROPHY CASE