#help confused

243 messages · Page 1 of 1 (latest)

sterile hemlock
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Here’s the problem. It’s picture.

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rustic aspenBOT
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sterile hemlock
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Its number 8 btw

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i noticed that thats the formula for the double tangent

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so i got a closed form of:

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$s_n = tan (4 * 2^n)$. However, I was unable to find the periodicity of this function

toxic fossilBOT
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!𒐪cut rizzly bear𒐪!

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bronze igloo
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do you know the periodicity of tan

sterile hemlock
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no

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this is a mth competition problem

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or its homework

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but its for math competition

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well

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i know that tan (theta) = tan (360k + theta)

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for any integer k

timber gull
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bro doesn't know shit but won't look it up

sterile hemlock
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?

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so like every 180?

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wait im special

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that aint the tan function

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thats just a recursive

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🫢

faint dragon
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Should be tan(4°*2^n), not tan(4*2^n). We have:
tan(4°*2^n) = tan((π/45)2^n)
So, for some integer m:
(π/45)2^(n + k) = (π/45)2^n + πm
2^(n + k) = 2^n + 45m
Not sure how to continue from here, though.

sterile hemlock
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wait

timber gull
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here watch this

sterile hemlock
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wait no

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im right

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about the double tangent

sterile hemlock
timber gull
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here

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degree i think

faint dragon
sterile hemlock
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well thats what i meant

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how is 4 * 2^n not what us aid

faint dragon
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Then you should've specified it.

sterile hemlock
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oh my bad

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but what did u think it was

faint dragon
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4° is not 4, it's π/45.

timber gull
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just use the formula for tan(2A)

sterile hemlock
timber gull
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then?

sterile hemlock
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so i got the closed form of s_n

timber gull
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is that even periodic

sterile hemlock
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idk

timber gull
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hmm okay let's see

faint dragon
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Look what I wrote above.

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It just seems to require some number theory.

sterile hemlock
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um

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can u pls not use radians

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i am familiar but i don't like it

faint dragon
sterile hemlock
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i know

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but since u don't use latex

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its harderto understand

timber gull
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s1 : tan(8)
s2 : tan(16)
s3 : tan(32)
s4 : tan(64)
s5 : tan(128)
s6 : tan(256)=tan(256-180)=tan(76)
s7 : tan(152)
s8 : tan(334)=tan(334-180)=tan(154)
a9 : tan(308)=tan(308-180)=tan(128)
and we repeat so the period is 5->9

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unless I messed up the calculations

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ah crap

sterile hemlock
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you didn't

timber gull
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i did mess it up

sterile hemlock
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what

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where

timber gull
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fixed it

sterile hemlock
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what did u fix

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or change

timber gull
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s7

sterile hemlock
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what was wrong

timber gull
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i wrote 162 there

sterile hemlock
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o

timber gull
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so starting from s5 the sequence is 4-periodic

sterile hemlock
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but wouldn't you need to change the rest

timber gull
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what

sterile hemlock
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2 * 152 is not 334

timber gull
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oh right

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dude why should i do this

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you have a Calculator

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you do it

sterile hemlock
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bro

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i can't

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im not typing each individual

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also it should be assumed that no caluclator

timber gull
sterile hemlock
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@faint dragon any ideas

timber gull
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it's 12 periodic

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just write it all out

sterile hemlock
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o

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tysm

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now i need help on q 5

timber gull
sterile hemlock
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what is htis

timber gull
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use inequalities

sterile hemlock
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oh

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wdym by that

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can u specify

timber gull
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one of my favorites ever

sterile hemlock
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but they need to hbe equal

timber gull
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okay

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first let's do it with ceil(x+y)=ceil(x)+ceil(y)

sterile hemlock
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ok

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how should i approach it

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o

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i have idea

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so ceil(x) + ceil(y) <= x + y + 2

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right

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then

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ceil(x+y) <= x+y+1 right

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and each is greater than or equal to x+y

timber gull
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wat

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split each x, y into n+f with f in [0,1)

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n an integer

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do casework

sterile hemlock
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wait actually stricly less than

timber gull
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wait I'll close my laptop

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okay

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okay so first

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nvm than

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okay

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so is the area of $\qty{(x, y) \in \bR:x^2+y^2<1}$

toxic fossilBOT
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Coffey, Slavic Taiga

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timber gull
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the same as the case if we have non strict inequality

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@sterile hemlock

faint dragon
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Oh, hold on.
2^(n + k) = 2^n + 45m
2^n (2^k + 1) = 45m
If the left side is even, then the right side has the largest odd factor 45, which isn't 1 above a power of 2. So, n = 0 and m = 2m' + 1 is odd:
2^k + 1 = 45(2m' + 1)
2^k + 1 = 90m' + 45
2^k = 90m' + 44
From here... Well, we probably need number theory. I tried brute force, but couldn't find a solution.

timber gull
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via brute force

timber gull
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Ike

faint dragon
timber gull
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$\sum_{1\leq i\leq k} \ceil{x_i} = \ceil{\sum_{1\leq i\leq k}x_i}$

toxic fossilBOT
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Coffey, Slavic Taiga

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sterile hemlock
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ermm

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sure i guess

timber gull
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preety trivial but still

sterile hemlock
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if its simpler

timber gull
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okay

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lemme turn off my notebook wait

sterile hemlock
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ok

timber gull
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did you know that the intersection of countably many dense subsets in a complete space is non empty and even dense in the space this is known as baire theorem

sterile hemlock
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i have no idea what ur saying

timber gull
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okay ill stop

sterile hemlock
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ok so how to do the prolem

timber gull
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gimme 45 seconds

sterile hemlock
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k

timber gull
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do yk what linearity is

sterile hemlock
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no

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how about we not gereralize it

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and just do the problem

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it seems we are getting too out of hand

timber gull
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wii

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okay

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so

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let's start with

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What's the probability of having $\ceil{x+y}=\ceil{x}+\ceil{y}$ Where $x, y\in [0,1]$

toxic fossilBOT
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Coffey, Slavic Taiga

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sterile hemlock
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idk

timber gull
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use your brain

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betterSkull bro

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it's literally just the probability of $x+y>1$

toxic fossilBOT
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Coffey, Slavic Taiga

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sterile hemlock
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what

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how

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ok well in that case its 1/2

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but how

timber gull
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because then ceil(x+y) is 2

sterile hemlock
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oh

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yes

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very clever

timber gull
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while ceil(x)+ceil(y) is always 2

sterile hemlock
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ok 1/2

timber gull
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okay

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now we see that

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in any square in R2

sterile hemlock
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R2?

timber gull
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nvm I'll write it

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The above results in a probability of $0.5$ in any square $\$ $[m, m+1]\times [n, n+1]$ where $m, n\in\bZ$

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okay?

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yeah

sterile hemlock
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yes i ager

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cuz thats just a 1x1

toxic fossilBOT
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Coffey, Slavic Taiga

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timber gull
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yes

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precisely

sterile hemlock
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wait

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how do we know it will always be 1/2

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even if the square is anywhere

timber gull
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so can we say that for any rectangle $[a, b] \times [c, d]$ with $a, b, c, d\in \bZ$ such that $a<b, c<d$ it holds

toxic fossilBOT
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Coffey, Slavic Taiga

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timber gull
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any real can be split into n+f where n is an integer and f in [0,1)

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so if we move the square up or down or in any direction by 1

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we just add/subtract an 1 from both sides of the equation

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resulting in the same equation

sterile hemlock
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but if you move it

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it might be difff

timber gull
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we move it by an integer

sterile hemlock
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for example

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if the square is at 2,2

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the line does not even intersec it

timber gull
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2,2?

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tfym 2, 2

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at?

sterile hemlock
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the bottom left vertex it s2,2

timber gull
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how does 2,2 describe a square

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okay

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sure

sterile hemlock
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side length 1

timber gull
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then all the numbers we work with can be broken into 2+f

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done

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bro dissappear

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anyways this is the trivial idea

sterile hemlock
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oh

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ok ty

timber gull
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you can generalise it to any dimensions

sterile hemlock
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yes

timber gull
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the probability is

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$\text{vol}\qty(\qty{(x_1,\ldots, x_n) \in \qty(0,1)^n : \Sigma_{i=1}^nx_i>1})$

toxic fossilBOT
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Coffey, Slavic Taiga

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sterile hemlock
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??????????

timber gull
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for n variables

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which is still 1/2 don't get me wrong

sterile hemlock
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wdym which is still 1

timber gull
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basically

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the probability is the same if we have x+y+z

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or x+y+z+w...

sterile hemlock
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oh

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how'

timber gull
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okay let's talk 3D

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what's the area of the 1 side length cube

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draw the plane x+y+z>1

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,w graph x+y+z=1

toxic fossilBOT
timber gull
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bad graph

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either way thus plane splits the square in 2

sterile hemlock
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gtg sry

timber gull
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.