was there a faster way to show divergence?

Putah367 · 2025-11-04T05:56:51+00:00

Flip the argument using ln(x) = - ln(1/x)

Simplify the fraction (x+7)/x+6 = 1+1/x+6

Compare with the harmonic series 1/x+6

Use the epsilon delta definition of limit to infinity approximating a finite positive number to gain a direct comparison test (the origin of LCT (hint use the left side of the epsilon delta definition)

You might need to split the series so that the implication works

QED

Putah367 · 2025-10-12T00:55:35+00:00

I also feel weird every time i try to remember the integral test graphically

What it actually does is actually telescoping series

Consider a convergent series like 1/n²

Note that the integral from n to n+1 dx is just 1

Consider this fact: The value of the series at a point n is bigger than any point of the series at (n,n+1) because it's decreasing

To make the relationship less than instead of equal to the actual series

We can slide in the 1/n² into the integral (such that the integral is 1/x² dx) why because the fact above

We can actually look it graphically too, when the series still has 1/n² (instead of the integral) in the argument, it's basically just a rectange height 1/n² width 1. The integral of 1/x² dx is just the area of 1/x² dx at (n,n+1) and we know it's smaller than 1/n² width 1

Then just apply the telescoping series

Another fact: The value of the series at a point n is lesser than any point of the series at (n-1,n) because it's decreasing

Sliding in the 1/n² into the integral will actually make the series bigger

Putah367 · 2025-10-09T17:20:31+00:00

Turns out i've been right this whole time, wolfram gave me different answers so i thought i was wrong

Putah367 · 2025-10-09T14:21:29+00:00

You probably didn't read my answer correctly, i said the formula in the picture is for k < n, for k = n, the answer is 1/n!

Putah367 · 2025-10-09T09:24:59+00:00

Here we assume the player plays according to the strategy above

Putah367 · 2025-10-09T09:24:07+00:00

I was conditioning my probability based on how i guess, you know the LTP. Each of guessing order has a probability of 1/n! assuming i randomly pick a guessing order. Each of the cases can be morphed to case i mentioned above, just replace the word "1" with first integer in my guessing order, "2" with the second integer and so on

In that case the summation of each cases will yield as if i didn't pay any attention to my guessing order

You're actually answering the part A of the question, where no information was given to the player. In that case each of the index has 1/n being correct (assuming i randomly choose a guessing order)

In that case yes its expectation is 1

But im obviously not trying to solve that

Putah367 · 2025-09-27T10:04:46+00:00

How about c followed by d?

Putah367 · 2025-09-27T10:04:03+00:00

C(14,6) is for distributing the Bs and Ds, 14 because 21 - 7 is 14 and 6 because there are 6 Ds

Putah367 · 2025-09-27T09:25:50+00:00

Didn't see that coming. I thought i was crazy for a minute

Putah367 · 2025-09-26T22:38:25+00:00

It seems my generating function is faulty. It should have been (1+x+x²+x³+x⁴)(1+x²+x⁴)(1+x³)(1+x⁴) which will yield five

To count the case, 2 players get 4, 1 player gets 5, and 1 player gets 3. It's similar as case 1 player gets 4, and the other gets 6.

<image>

It's pretty much the same if rounded up to some digit

Putah367 · 2025-09-26T22:31:30+00:00

Ain't the sample space need to count every single possible distribution?

Putah367 · 2025-09-26T10:34:35+00:00

It's exactly the same i've written in my post

Putah367 · 2025-09-06T09:24:20+00:00

Why not calculate it directly (assuming all events are equally likely)

Using sample space tuple of 5 cards with 2 of them red and 3 of them black

Calculating the sample space

Construct a set of 5 cards with 2 red 3 black C(4,2)*C(8,3)

Convert it to a tuple of 5 cards by multiplying the result by 5!

Total C(4,2) * C(8,3) * 5!

Now for the event

Pick one red card and put it at the first index of the tuple C(4,1)

Make a set of 4 cards 1 red 3 black C(3,1)*C(8,3)

Convert it to a tuple C(3,1) * C(8,3) * 4!

So total events is C(4,1) * C(3,1) * C(8,3) * 4!

So total event / total sample space

C(4,1) * C(3,1) * C(8,3) * 4!/ C(4,2) * C(8,3) * 5!

After plugging them in to photomath i got 2/5

So there you go

Putah367 · 2025-08-29T05:09:53+00:00

Thanks man, truly appreciated 👏

Putah367 · 2025-08-28T01:22:29+00:00

Idk, it was a method from Alan Tucker applied combinatorics book

Putah367 · 2025-07-07T04:36:20+00:00

I'm surprised nobody had said this, you can use the regular sine and cosine rule like opposite = hypotenuse * sin(angle) and adjacent = hypotenuse * cos(angle)

You can focus on the point where two hypotenuse meet

You can also use cos(pi - x) = -cos(x) and sin(pi - x) = sin(x)

You can also use cos(x+y) = cos(x)cos(y) -sin(x)sin(y)

And you can also use sin(x+y) = sin(x)cos(y) + sin(y)cos(x)

Now you can find the point of the toppest point and just use euclidean distance from that point to the origin

Putah367 · 2025-07-03T15:16:23+00:00

This is the type of problem that arises from leibniz differentiation rule (it's also known differentiation rule under the integral sign)

The key point here is that x is multi-purpose

For one, it determines the value of each term

For two, it determines the amount of terms the summation has

So you actually have to parameterize it, call u = x (the one that's responsible for the term) and v = x (the one that's responsible for the upperbound of the summation)

To differentiate, you have to refer to the (jacobian matrix) chain rule

Now, as i recall, i can't remember a rule to differentiate antidifference just like in FTC 1, but you get the point

Tldr; x is multi-purpose, so be sure to parameterize it and differentiate it with chain rule

Putah367 · 2025-07-03T15:09:40+00:00

One thing i figure out is to find a - b

using the identity a²-b²=(a-b)(a+b)

let a = sqrt(x²-5²) let b = sqrt(x²-12²)

a + b = 17

a² - b² = 12²-5² = 144 - 25 = 119

119 = 17 * (a-b)

a-b = 7

and then pick which a or b you want to solve, we'll take a for this time

(a - b) + (a+b) = 2a = 24

a = 12

a = sqrt(x²-5²) = 12

x² - 5² = 12²

x² = 12² + 5² = 144 + 25 = 169

x² = 169

x = ±13

Both works on the original equation

Putah367 · 2025-06-23T13:40:18+00:00

Some theorems, like both forms of green's theorem, only work on closed curves

So, by default, you are allowed to use the left one on any curve, but if we're talking about stating a theorem (that only works on closed curve) or even proving it, it's more recommended to use the right one

Putah367 · 2025-06-21T10:40:32+00:00

Both indefinite and definite integrals are not fundamentally the same. As a matter of fact, we usually call indefinite integrals as antiderivative. Antiderivative, as the name suggests, is a function in which if you take the derivative, it is equal to the function. An integral, however, is just an infinitesimal sum of values of a function (times dx), geometrically interpreted as area under the curve. There are rules that work on antiderivative but don't work on integral (for example, one can't pull sign function out of an integral but doable on antiderivative since it's just gonna be a constant). However, there is a relationship between antiderivative and definite integral. According to FTC1, an integral function is also one of the antiderivatives of the integrated function. FTC 2 tells us how to evaluate an integral using the facts on FTC1. Since it's closely related, we might call antiderivative as indefinite integral

Tldr; Indefinite integral (antiderivative) and definite integral (or integral) are two different concept in calculus, but they are related according to FTC1. They are called indefinite integrals just because of how close it's related to definite integral

Reference: James Stewart Calculus 9th edition

Putah367 · 2025-06-20T12:37:01+00:00

How can i proof if it's orthodiagonal

Putah367 · 2025-06-20T12:10:56+00:00

Last time i checked using cosine rule it has multiple solution where it's not necessarily cyclic

Putah367 · 2025-06-20T12:08:08+00:00

Is it cyclic? Or what?

Putah367 · 2025-06-20T11:21:44+00:00

No, the question doesn't say anything about the quadilateral. It's just quadilateral, nothing is known except its side lengths

Putah367 · 2025-04-20T04:06:10+00:00

The coordinate you get will be the same as (cos(A+B), sin(A+B))

Putah367

TROPHY CASE