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SuicideFuel Math thread problem (official)

Ahnfeltia

Ahnfeltia

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Correct :feelsokman:.

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.
 
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IronsideCel

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@trying to ascend What is 2+2?



it's over
 
L

Lonelyus

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@TouhouMathcel
 
im done

im done

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

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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

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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

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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

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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!
 
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Ahnfeltia

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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.
 
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NirvanaFan1988

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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

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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
 
Logic55

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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
 
M

Med Amine

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M

Med Amine

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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

trying to ascend

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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
 
trying to ascend

trying to ascend

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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

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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

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why is this still pinned?
 
W

WojakSuffer

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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
 
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Puppeter

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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

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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

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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

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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

trying to ascend

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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

incelpardo99

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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
 
incelpardo99

incelpardo99

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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
 
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trying to ascend

trying to ascend

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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

incelpardo99

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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

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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

trying to ascend

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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

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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

incelpardo99

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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

trying to ascend

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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

incelpardo99

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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

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

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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.
 

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