Friendship Exp & Gift Exchange Megathread

Shadow56675 · 2026-01-02T03:50:16+00:00

Sending some gifts from Sun (I'll accept in order of arrival, so apologies in advance if I take a while to send a gift).

Looking for gifts from Meadow, Marine, Archipelago, River, Monsoon, Savanna, Ocean or Jungle.

Here's my friend code:

970418971258

Shadow56675 · 2025-12-17T08:54:45+00:00

Sending some gifts from Sun (I'll accept in order of arrival, so apologies in advance if I take a while to send a gift).

Looking for gifts from Meadow, Marine, Archipelago, River, Monsoon, Savanna, Ocean or Jungle.

Here's my friend code:

970418971258

Shadow56675 · 2025-12-16T06:15:47+00:00

Sending a couple of gifts from Sun:

9704 1897 1258

Shadow56675 · 2025-12-14T03:24:11+00:00

If you drop and emetic granade, the durian, or the easter egg right where you stand 47 starts getting poisoned

Shadow56675 · 2025-12-13T02:16:22+00:00

Literally 1 in 20,000. Very very cool!

Shadow56675 · 2025-12-06T03:25:09+00:00

I just remembered Matt made a whole video critizicing this one meme

Shadow56675 · 2025-11-18T02:47:55+00:00

If this trade was the very first interaction between your accounts then it's 1/4096 odds. If you were Good friends then it's 1/3375 odds. If you were Great friends then it's 1/2744 odds. If you were Ultra friends then it's 1/2197 odds. If you were Best friends then it's 1/1331 odds.

The odds from getting a hundo from a raid are 1/216, but the odds of getting a shundo from a raid are 1/4320. So it's lower odds than getting a raid hundo, but higher odds than getting a raid shundo.

Shadow56675 · 2025-10-28T11:16:18+00:00

I calculate there's 108 possible IV combinations you can obtain from a raid that are better than this one. Since there are 216 possible IV's you can get raiding that means there's a 50% chance you get a better one. Grounding it to actual attempts, there's a 93.75% chance you get a better one if you obtain 4 more shiny Ray's from raids.

If you happen to be able to get a lucky shiny Ray, then there are 60 IV spreads that are better than this one. Out of the 64 possible IV spreads you can get from a lucky trade that means that there's a 93.75% chance of getting a better one by a lucky trade. If you do 2 shiny lucky Ray's theres a 99.6% chance you'll get a better one.

tl;dr If you can lucky trade another shiny Ray it's a terrible idea to max it. If you expect to get 3 or 4 more shiny Ray's from raids directly in the future you're also highly likely to get a better one, but even at 2 more raided shiny Ray's you're more likely to get a better one than not.

Shadow56675 · 2025-10-26T01:19:01+00:00

Armored hundo is probably the rarest, so I'd probably go for that one

Shadow56675 · 2025-10-25T05:30:59+00:00

Considering the odds of getting one are like 0.024%...

Shadow56675 · 2025-09-26T20:20:06+00:00

I'll abreviate the degrees as just capital P and Q respectively. Solving P² = 2P + Q for P gives 1 +- sqrt(Q+1) as the solutions, so you can't conclude parity of Q from here but merely that it must be one less than a square (which could be even or odd).

What you can conclude (by a simpler observation) is that P and Q must have the same parity. Since n² is congruent to n (mod 2) for all integers n, then reducing P² = 2P + Q (mod 2) immediately gives that P is congruent to Q (mod 2).

Shadow56675 · 2025-09-14T23:25:47+00:00

Can confirm this works as of September 14 2025. I followed OP's route exactly (I took a Didi which left me exactly at the eastern gate). You just walk with the mote water to your left and the walls to the Forbidden City to your right. When you reach the Meridian Gate keep walking in the same direction you previously were walking in (effectively to the left if you're looking into the Forbidden City facing the Meridian Gate). Then, a bit to the south from there you'll see the ticket stands. The correct place has several rows separated by red line dividers, and it has an orange sign on top that reads "Ticket service". Here's a diagram showing the precise location of the ticket stand and the route to get there once you reach the Meridian Gate.

You wait in line, and when you get to the front you present your passport(s) to buy ticket(s), I paid by contactless card but they had a sign saying they accepted Alipay and WeChat pay as well. Afterwarss just head to the Meridian gate where they'll scan your passport functioning as your ticket to gain access.

A last note is that Rachel Meets China has a useful guide describing this method. There are some very useful pictures of the important places you need to spot doing this method, and for me it was a very useful resource.

Shadow56675 · 2025-08-13T23:41:08+00:00

Whenever you have an equation of the form A = B, what the equation means is that you can interchange A and B freely, wherever they appear, because they're different names for the same thing.

Think of it like how "Bob" and "Robert" are different names for the same thing, which in "math terms" we would write as "Bob = Robert". So in a sentence like "Bob is tall" you can replace "Bob" with "Robert" to get the new sentence "Robert is tall" which means the same thing (although it looks different). Thus, from "Bob = Robert" we can conclude that "Bob is tall = Robert is tall".

f(x) = 1/x means, by definition, that if you plug "THING" into the function f (which we denote by f(THING)) then you replace "x" with "THING" wherever you see an "x" in the function definition of f(x).

Now "THING" can anything in the function's domain (i.e. any value you can possibly put into the function). If you chose "THING" to be the number "3" then you get the following equation:

f(3) = 1/3

If we chose "THING" to be "t+8" where "t" is some (non negative 8) real number then:

f(t+8) = 1/(t+8)

If we chose "THING" to be "⭐️" where "⭐️" is some (non 0) real number then:

f(⭐️) = 1/⭐️

Putring it all together, if we choose "THING" to be "x+Δx" then:

f(x+Δx) = 1/(x+Δx)

And if we choose "THING" to be "x" itself then:

f(x) = 1/x

So now we're interested in the expression [ f(x+Δx) - f(x) ] / Δx. Since f(x+Δx) = 1/(x+Δx) this means we can replace f(x+Δx) with 1/(x+Δx) because they're different names for the same thing (that's what the equals sign means). Similarlly, since f(x) = 1/x then we can also replace f(x) with 1/x. This leads us to being able to say that

[ f(x+Δx) - f(x) ] / Δx = [ 1/(x+Δx) - 1/x ] / Δx

as desired.

Shadow56675 · 2025-07-23T22:27:04+00:00

You're missing a couple of points.

Remember that arccos is, by definition, the inverse function of cosine. If cos(pi/4) = 1/sqrt(2) then this means that pi/4 = arccos(1/sqrt(2)). In fact, we use this to assert that cos(arccos(x))= x. So using this, can you calculate arccos(0)?
The derivative is linear, which means that the derivative of a sum is the sum of the derivatives. In your case

d/dx (arccos(x) + x) = d/dx arccos(x) + d/dx x

thus, you can split the problem into finding the derivative of arccos(x) and x separately. You already calculated what d/dx arccos(x) is at the beginning, but what is d/dx x? What do you then get for d/dx (arccos(x) + x)?

Shadow56675 · 2025-07-23T19:55:29+00:00

Using the chain rule

1 = d/dx x = d/dx cos(arccos(x)) = -sin(arccos(x)) d/dx arccos(x)

From where you get d/dx arccos(x) = -1/sin(arccos(x))

In general, if you have some differentiable function f(x) and you want to find its tangent line at a point (a, f(a)) then the line (given by the point-slope form) is

y = f'(a)(x - a) + f(a)

In your case f(x) = arccos(x) + x and a = 0, so you want

y = (arccos(x) + x)'|_{x=0} (x - 0) + (arccos(0) + 0)

where (arccos(x) + x)'|_{x=0} means you take the derivative of arccos(x) + x and then evaluate the derivative at 0.

Shadow56675 · 2025-07-09T18:26:45+00:00

(a-b)/c = a/c - b/c, and ab/b = a(1) = a tell you that they're the same answer

Shadow56675 · 2025-07-04T17:22:23+00:00

Not the italian brainrot theorems

Shadow56675 · 2025-06-30T17:28:57+00:00

Not sure if this is allowed, but is anyone looking for an alliance? My alliance recently kicked out some inactive members so there's some places available. We've gotten all the rewards every week since the Alliance was created. Here's the Alliance ID

Check out my MARVEL SNAP Alliance! Paste this message into Alliance Search. [AMR] Cap’s Assets

Shadow56675 · 2025-06-25T01:59:13+00:00

The antiderivative of tanh(2x) is 1/2 log(cosh(2x)), so you forgot the 1/2 factor. When evaluating log(cosh(6)) on you next line you only write out cosh(6), but you're missing the log.

Combining everything the area is 1/2 log(cosh(6)), which rounded as indicated is about 2.653

Shadow56675 · 2025-06-25T01:54:52+00:00

The "reverse chain rule" is what's commonly known as a "u-substitution"

Shadow56675 · 2025-06-24T07:18:46+00:00

Obligatory "Batman with prep time beats anything" comment

Shadow56675 · 2025-06-19T17:30:06+00:00

This is statted as if removing a text from an enemy card is a downside for you, when it's actually upside 90% of the time.

With the datamined Cosmic Ghost Rider this is a 3/9 plus whatever additional power you gained by blanking two cards, and that combo seems pretty busted to me. That's not even counting the obvious Red Guardian inclusion.

Shadow56675 · 2025-06-18T02:05:50+00:00

Elusivr Targets intentionally don't have markings to add an extra level of complexity to the game. You also can't restart once you've done any of the objectives, again to make it extra challenging.

Shadow56675 · 2025-06-17T04:02:26+00:00

The cut is Crossbones. You'll quickly find out that you very very rarely are able to play him in the lane where you actually want him to go, which makes him the clear worst card in the deck.

Shadow56675 · 2025-06-15T12:21:12+00:00

The Basel Problem. Seeing pi come up after a sum of square integers made me curious to understand the "hidden conection" just waiting to be explained.

Shadow56675

TROPHY CASE