SuicideFuel Math thread problem (official)

[Ahnfeltia]

Correct .

Did you use the inclusion-exclusion principle?

Sure, but brute forcing the problem ain't worth any credit. Edit: just now seeing your edit. No, I had the computer just calculate all the possibilities (this is called brute forcing) because I couldn't figure out how to make inclusion-exclusion work.

 

[IronsideCel]

@trying to ascend What is 2+2?



it's over

 

[Lonelyus]

@TouhouMathcel

 

[im done]

Find the generating function for the sequence
a(n)-(n+1)a(n-1) =0
Where a(0)=1

 

[Ahnfeltia]

Find the generating function for the sequence
a(n)-(n+1)a(n-1) =0
Where a(0)=1

Let G(x) be the ordinary (or did you mean exponential?) generating function of a(n) -- i.e., G(x) = sum a(n)x^n (from n = 0 to oo). Then

G(x) = 1 + sum a(n+1)x^(n+1) = 1 + sum (n+2)a(n)x^(n+1) = 1 + d/dx sum a(n)x^(n+2) = 1 + d/dx [ x^2G(x) ] = 1 + 2xG(x) + x^2G'(x)

so we have to solve the first-order linear ordinary differential equation y = 1 + 2xy + x^2y' with y(0) = 1. While doing so is standard (integrating factor) the solution is ugly, involving an exponential integral. The exponential generating function seems as tho it'll be even uglier, however. Am I missing something?

 

[Ahnfeltia]

Let G(x) be the ordinary (or did you mean exponential?) generating function of a(n) -- i.e., G(x) = sum a(n)x^n (from n = 0 to oo). Then

G(x) = 1 + sum a(n+1)x^(n+1) = 1 + sum (n+2)a(n)x^(n+1) = 1 + d/dx sum a(n)x^(n+2) = 1 + d/dx [ x^2G(x) ] = 1 + 2xG(x) + x^2G'(x)

so we have to solve the first-order linear ordinary differential equation y = 1 + 2xy + x^2y' with y(0) = 1. While doing so is standard (integrating factor) the solution is ugly, involving an exponential integral. The exponential generating function seems as tho it'll be even uglier, however. Am I missing something?

Upon further inspection, the initial condition y(0) = 1 is superfluous. It's actually very easy to solve the recursion directly -- i.e., a(n) = (n+1)!
As such, it's not at all surprising that the ordinary generating function is ugly, given that it's radius of convergence at 0 has to be 0.

 

[Ahnfeltia]

The exponential generating function seems as tho it'll be even uglier, however. Am I missing something?

I was wrong about the exponential generating function. It'll be G(x) = sum (n+1)x^n = d/dx sum x^(n+1) = d/dx [ x/(1-x) ] = 1/(1-x)^2.

 

[im done]

I was wrong about the exponential generating function. It'll be G(x) = sum (n+1)x^n = d/dx sum x^(n+1) = d/dx [ x/(1-x) ] = 1/(1-x)^2.

Based
This was not the original problem i was working on i think i made a mistake while defining recursive relation the problem was if you begin a 4*4 matrix with all entries 1 and then multiply it by a 4*4 matrix with all entries 2 and continue doing it till n matrices the base case a1 is entry of first matrix and a2 is entry of (1)4*4 x (2)4*4 and then next is a3 is (1)4*4x(2)4*4x(3)4*4 and so a4...like this where ($)4*4 represents all matrix entries are $
I think the nth product will be 2^nxn! But i think there is no generating function for an=2^nxn!

 

[Ahnfeltia]

Based
This was not the original problem i was working on i think i made a mistake while defining recursive relation the problem was if you begin a 4*4 matrix with all entries 1 and then multiply it by a 4*4 matrix with all entries 2 and continue doing it till n matrices the base case a1 is entry of first matrix and a2 is entry of (1)4*4 x (2)4*4 and then next is a3 is (1)4*4x(2)4*4x(3)4*4 and so a4...like this where ($)4*4 represents all matrix entries are $
I think the nth product will be 2^nxn! But i think there is no generating function for an=2^nxn!

You're close. Your a(n) = 4^(n-1)n! as can be easily checked via eigendecomposition. As for there not being a generating function, there's indeed no sensible ordinary generating function, but once again there is an exponential generating function that's nice. Indeed, the exponential generating function works out to

G(x) = sum a(n)x^n/n! = sum 4^(n-1)x^n = ¼ sum (4x)^n

which is just a good old geometric series with radius of convergence ¼.

 

[NirvanaFan1988]

Is there a correlation between being good at math and being unattractive to foids

Lol like an inverse relationship, if you are good at math chances are you are a professional or are studying in STEM and you are basically a betabuxxx bitch which makes you a volcel for not trying hard enough in your academic/professional area and rotting here instead

 

[Ahnfeltia]

Lol like an inverse relationship, if you are good at math chances are you are a professional or are studying in STEM and you are basically a betabuxxx bitch which makes you a volcel for not trying hard enough in your academic/professional area and rotting here instead

Just be a betabuxx theory

 

[Linesnap99]

-(Lim x--->0 (| 2x - 1| - | 2x + 1|)/x)

when multiplied by 0 the answer is 0

 

[Linesnap99]

(144^x + 324^x)/64^x + 729^x = 6/7

multiply with 0 you get 0

 

[Linesnap99]

In the following expansion, what's the sum of the coefficient of all x to the power of a multiple of 3?

(1 + x² - x³ + x^4)^10

0 when you multiply with 0

 

[Linesnap99]

Find the generating function for the sequence
a(n)-(n+1)a(n-1) =0
Where a(0)=1

0 because a(n)-(n+1)a(n-1) =0 * 0

 

[Epedaphic]

3+2

 

[Logic55]

agepill-teenlovepill=

Even university girls that choose to remain single and independent still have sex with Chads behind closed doors. Meanwhile, nerds at university have no social life and are rotting in loneliness

 

[Linesnap99]

3+2

=2+3

 

[Epedaphic]

=2+3

 

[Med Amine]

 

[Med Amine]

Lol like an inverse relationship, if you are good at math chances are you are a professional or are studying in STEM and you are basically a betabuxxx bitch which makes you a volcel for not trying hard enough in your academic/professional area and rotting here instead

you can be bad at math and still be good at STEM if you're a some surgeon or doctor
betabuxxing is based, i wanna betabuxx a fat piss haired woman.

 

[trying to ascend]

you can be bad at math and still be good at STEM if you're a some surgeon or doctor
betabuxxing is based, i wanna betabuxx a fat piss haired woman.

That's medicine, not STEM

 

[Med Amine]

That's medicine, not STEM

medicine is a part of science

 

[trying to ascend]

medicine is a part of science

So are geography and sociology, they are social sciences.

But the S in STEM doesn't refer to all sciences

 

[turbosperg]

agepill-teenlovepill=

Over

---

If an oven can bake one cake every hour,

how many cakes can 40 ovens bake in 3 years?

---

 

[Ahnfeltia]

View attachment 844872

 

[Ahnfeltia]

If an oven can bake one cake every hour,

how many cakes can 40 ovens bake in 3 years?

Is that a Holocaust reference or am I way too far gone?

 

[turbosperg]

Is that a Holocaust reference or am I way too far gone?

If you forget the cake in the oven for too long it will surely become holocausted down to ash.

 

[CruxGammata]

why is this still pinned?

 

[trying to ascend]

why is this still pinned?

Because it's the HIGHEST IQ thread on the site

 

[belysning]

when multiplied by 0 the answer is 0

Duh

 

[WojakSuffer]

Chad and Pepe are playing with a pile of black, red and green stones.
When the game starts, there are k many black pieces (stone piles) on the game board
k could be any positive integer.

Players take turns. Chad took the first move. There are 2 options for a move:
(1) Place a red or green stone (from an infinite supply) on top of an existing stone pile. At the end of the round, there may not be two tiles of the same color in the stack.
A stack can contain up to three stones.
or
(2) Take out 2 stacks, but the top colors of the two stacks must be the same. Stack height doesn't matter.

If you can't take any action, you lose.
Determine whether either player can use a strategy to guarantee victory, and if so, explain how.
k=6

 

[Puppeter]

Chad and Pepe are playing with a pile of black, red and green stones.
When the game starts, there are many black pieces on the game board
could be any positive integer.

Players take turns. Chad took the first move. There are 2 options for a move:
(1) Place a red or green stone (from an infinite supply) on top of an existing stone pile. At the end of the round, there may not be two tiles of the same color in the stack.
A stack can contain up to three stones.
or
(2) Take out 2 stacks, but the top colors of the two stacks must be the same. Stack height doesn't matter.

If you can't take any action, you lose.
Determine whether either player can use a strategy to guarantee victory, and if so, explain how.
k=6

Tldr

 

[incelpardo99]

Finally, some good fucken high IQ level thread, get them one of the Guidorizzi's books so they can whack their brains in OP

 

[incelpardo99]

Finally, some good fucken high IQ level thread, get them one of the Guidorizzi's books so they can whack their brains in OP

Just remembered this one is brazilian, might be hard to understand the question, maybe a Strang or Stewart might be better

 

[incelpardo99]

Lol like an inverse relationship, if you are good at math chances are you are a professional or are studying in STEM and you are basically a betabuxxx bitch which makes you a volcel for not trying hard enough in your academic/professional area and rotting here instead

When your realize that your qt foid STEM class coleague chooses to be declared volcel but in reality fucks a chad a day just like everyone else but you you might end up in here too

 

[trying to ascend]

When your realize that your qt foid STEM class coleague chooses to be declared volcel but in reality fucks a chad a day just like everyone else but you you might end up in here too

Consider the function f(n), such that f(0) = 0 and f(⌊n/10⌋) + n - 10⌊n/10⌋, how many algorisms does the smallest positive integer m has, such that f(m) = 2015?

 

[incelpardo99]

Consider the function f(n), such that f(0) = 0 and f(⌊n/10⌋) + n - 10⌊n/10⌋, how many algorisms does the smallest positive integer m has, such that f(m) = 2015?

Is the function f(n) defined by f(n) = f(⌊n/10⌋) + n - 10⌊n/10⌋? You didn't make that clear

 

[trying to ascend]

Is the function f(n) defined by f(n) = f(⌊n/10⌋) + n - 10⌊n/10⌋? You didn't make that clear

Yes

 

[incelpardo99]

Yes

def f(n):
if n == 0:
return 0
else:
return f(n // 10) + n - 10 * (n // 10)
m = 1
power = 10
algarism_counter = 1
while True:
if f(m) == 2015:
break
m += 1
if m == power:
power *= 10
algarism_counter += 1
print("m has:", algarism_counter)

Could only count to 9, gave up

 

[trying to ascend]

def f(n):
if n == 0:
return 0
else:
return f(n // 10) + n - 10 * (n // 10)
m = 1
power = 10
algarism_counter = 1
while True:
if f(m) == 2015:
break
m += 1
if m == power:
power *= 10
algarism_counter += 1
print("m has:", algarism_counter)

Could only count to 9, gave up

What is that? Why do you write like a computer?

 

[incelpardo99]

What is that? Why do you write like a computer?

It's code in python for calculating functions with recursive method, but it took so much heap for a long time that I could only see m reaching 9 algarisms, by the way what is the actual answer I'm curious

 

[FemoidsGTFO]

IT: Incel forums are full of toxicity and people encouraging each other to be miserable

Incel forums in reality:

Post all math related problems and solutions here. Don't cheat.

First problem: If the coefficients of x³ and x^4 in the expansion of (1+ ax+ bx² ) (1−2x) ^18 in powers of x are both zero, then (a, b) is equal to?

 

[trying to ascend]

It's code in python for calculating functions with recursive method, but it took so much heap for a long time that I could only see m reaching 9 algarisms, by the way what is the actual answer I'm curious

It's higher than 200 bhai.

Also, why do you use that? YOU CAN'T USE PYTHON AT OLYMPIAD

 

[Grim_Reaper]

def f(n):
if n == 0:
return 0
else:
return f(n // 10) + n - 10 * (n // 10)
m = 1
power = 10
algarism_counter = 1
while True:
if f(m) == 2015:
break
m += 1
if m == power:
power *= 10
algarism_counter += 1
print("m has:", algarism_counter)

Could only count to 9, gave up

I don't think it's anywhere near 9. The highest possible value for n - 10(⌊n/10⌋) is 9, so there's at least 223 algorithms. Idk how you can do this question without brute force, which is already impossible.

 

[incelpardo99]

It's higher than 200 bhai.

Also, why do you use that? YOU CAN'T USE PYTHON AT OLYMPIAD

Didn't know that, sorrie. Also, I reckon floor functions are not differentiable, so I assume you can't use derivatives on this one for m to reach 2015, how do you solve it?

 

[trying to ascend]

Didn't know that, sorrie. Also, I reckon floor functions are not differentiable, so I assume you can't use derivatives on this one for m to reach 2015, how do you solve it?

It's from a high school olympiad, you can solve it without calculus.

I haven't solved it

 

[incelpardo99]

I don't think it's anywhere near 9. The highest possible value for n - 10(⌊n/10⌋) is 9, so there's at least 223 algorithms. Idk how you can do this question without brute force, which is already impossible.

Ye, no heap anywhere near as capable of reaching this result

 

[Monke]

Apes should live comfortable life in the jungle, not this shit!!!!

 

[CountBleck]

Let V be a vector space with dim(V) = n < ∞. Suppose that L: V -> V is a linear map and that 0 is an eigenvalue of L.
Prove that L cannot be surjective.