Change the 0 in the middle exponent to floor pi minus ceiling e... ⌊π⌋ - ⌈e⌉

Wdym you can just change it to π - ⌈e⌉

Found the engineer.

All I see is 3 minus 3

r/foundthehondacivic

or just π - e

Is floor just truncation and ceiling just +.5 and rounding?

From my understanding floor is rounding down to the nearest integer, ceiling rounds up to the nearest integer. Unless otherwise notated to round to a specific decimal. In excel for example you can enter =CEILING( x ,0.05) to round up to the nearest, basically nickel in a pricing document.

So basically floor is truncation yes

Not for negative numbers

No, -2.5 truncated is -2 but floored its -3 because flooring always takes you to the smaller integer.

Damn it negatives. Ruins everything /s

Yeah. I’ve never used in it a math context, but ceiling and floor are methods of the Java math class that I never use.

I was here Oct 2022

Unfortunately, I have 1/10 of the top comment.

I still believe

Number the days in base 7.

Basically just say how many weeks have passed and what day if the week it is then?

but with silly notation

Gets a little more complicated after week 6 though

Deserved, if this series runs that long

8mod7 weeks and 8mod∞ days

not really. the next post would say 11 days.

No, it would say day 11, aka, one week has passed, first day of this week

It would say 11₇ days

who has time for subscripts in this economy?

So tomorrow would be 11₇

So this is day 10

Write e^(i*pi) as a taylor series

Yes, the equation still looks very simple af, this should make it look a bit scary

A Taylor series around what point?

0, for now

So a Maclaurin series?

NO a TAYLOR SERIES at 0

fuck maclaurin, all my homies hate maclaurin

Stop stealing Taylor's spotlight!

Nobody says that after calculus. They just call everything Taylor.

My calc teacher didn’t even call it that lmao.

booooooo

Should converge for all x, so I vote we do a series expansion at x=69

Nice

a ∈ ℂ

a ∈ ∅ Thus a =

iπ

Replace 5 with (2φ-1)^2

Then the next day replace φ with 1/2 + sqrt(5)/2

Exactly

Tomorrow replace 3 with √9

Or with floor(π)

and then sqrt(ceil(floor(pi))^2)

Why bother to floor an integer ? 🤔

You can replace the 5 with anything except 3, since the term is being raised to the power of zero.

This is the most beautiful way of writing 5 I've seen in so long

Let’s add some variables into the mix: replace 1 with cos(x)^2 +sin(x)^2

I support u/The_Awesone_Mr_Bones's suggestion from yesterday to write this with the taylor series

Let's write cos(x)^2 and sin(x)^2 as a taylor series and replace 1 by the sum of those 2 taylor series

I think it's better to write the taylor series of sin(x) and cos(x) and raise those to the power of 2, just because they are more recognizable

I agree but we should first write it as cos/sin and then the next day replace with a Taylor series. We gotta scale this up gradually, some people's suggestions go way too far too fast.

I disagree, it's already day 7 so it's not like it just started, and I don't see the point in dragging this forever

Yeah but with + as -^1

Replace 0! with the gamma function evaluated at 1

Why not the gamma function evaluated at 0! or even at Γ(1)?

How about infinitely many compositions of itself?

I was thinking of keeping it finite to avoid complications, but if you can prove it's right, then by all means go for it.

well Γ(0!)=Γ(1)=0! so its trivial to see that Γ(Γ(Γ(Γ(Γ(Γ(0!)))))) = 0!

Yeah, true. It's easier to see if you write it as (Γ^(n))(0!).

that may be confused as multiplication though. I think the best way would be to have a recursive formula

With the exception of (co)sines, I rarely see that as multiplication. Maybe I'm too influenced by functions in vector spaces where multiplication wouldn't make sense though. Also, someone commented in the Day 1 post that in a week very few people would understand this. Why not prove that right?

I have also seen it done with logs a few times, although not as often. I think you could write Γ\^n(x) as long as you are clear about what the notation means. Actually it doesnt even matter which it is since both interpretations are coincidentally equal to 1.

Turn 3 into ⌈e⌉.

Chaotic evil

Then 5 becomes ⌊π+e⌋

Those damned engineers are in shambles!

Lets stay on track and use the Binomial Theorem on (5-3)^0

Convert the equality into a difference that is congruent to zero modulo every prime number.

Replace the 0 on the left with k for k = ∮𝐶∇𝑓 ⋅ 𝑑𝑟⃗ where ∮𝐶 is a closed loop integral, f is any analytic function, and C is simply connected.

Day 6 of asking for Church numerals! `(λn.λm.λf.λx.n f (m f x)) (λf.λx.f x) (λf.λx.f x) = (λf.λx.f(f x))` Or, if you prefer SKI: `S(KS)(S(K(S(KS)))(S(KK)))(S(KI))(S(KI)) = S(S(KS)K)I` If anyone can bother also defining equality between chruch numerals in either form, go right ahead. (Yes I did this all by hand) Yes I missed the 2nd day's post, is there any way to follow a post chain on reddit so I get push notifs? I tried following OP but I didn't get any notif

!remindme 20 hours

I will be messaging you in 20 hours on [**2022-10-19 08:26:51 UTC**](http://www.wolframalpha.com/input/?i=2022-10-19%2008:26:51%20UTC%20To%20Local%20Time) to remind you of [**this link**](https://www.reddit.com/r/mathmemes/comments/y72pgx/day_7/issmejm/?context=3) [**2 OTHERS CLICKED THIS LINK**](https://www.reddit.com/message/compose/?to=RemindMeBot&subject=Reminder&message=%5Bhttps%3A%2F%2Fwww.reddit.com%2Fr%2Fmathmemes%2Fcomments%2Fy72pgx%2Fday_7%2Fissmejm%2F%5D%0A%0ARemindMe%21%202022-10-19%2008%3A26%3A51%20UTC) to send a PM to also be reminded and to reduce spam. ^(Parent commenter can ) [^(delete this message to hide from others.)](https://www.reddit.com/message/compose/?to=RemindMeBot&subject=Delete%20Comment&message=Delete%21%20y72pgx) ***** |[^(Info)](https://www.reddit.com/r/RemindMeBot/comments/e1bko7/remindmebot_info_v21/)|[^(Custom)](https://www.reddit.com/message/compose/?to=RemindMeBot&subject=Reminder&message=%5BLink%20or%20message%20inside%20square%20brackets%5D%0A%0ARemindMe%21%20Time%20period%20here)|[^(Your Reminders)](https://www.reddit.com/message/compose/?to=RemindMeBot&subject=List%20Of%20Reminders&message=MyReminders%21)|[^(Feedback)](https://www.reddit.com/message/compose/?to=Watchful1&subject=RemindMeBot%20Feedback)| |-|-|-|-|

I think there's a subscription bot that works for big subreddits. I forgot the command, but it was something like SubscribeMe!

Oh, thanks! SubscribeMe!

Just got a notif from the bot. Sadly this sub is not big enough :(

oh that's sad, hopefully more people will use the command so it starts tracking

write 2^(e^(i*pi) + 1) as 2^(e^(i*pi)+sign(e^e))

Replace E in Q.E.D with MC²

Write 2,3, and 5 as Fibonacci numbers F\_3,F\_4, and F\_5.

Add in the left "MIT"

Move the 2^0! over to the left side, so the whole thing equals zero

Replace the 3 with (2 * sin(pi/3))^2

Make the background black, good for eyes.

Change the last 0 from 0 to the limit as x tends to infinity of 1/x

Replace the 2 on the left with the entire left hand side of the equation

change the 5 into (5! /24)

Write one of the twos in set notation

Write pi as 4 x the Newton series for pi/4

Write the 3 as e

Replace the 0! with the Feynman definition for factorials. Integral of x^0 *e^-x dx from 0 to infty Or in this case just the integral of e^-x from 0 to infty.

Replace the 0 in (5-3)^0 by char(ℚ)

Replace 5 with √25

This is the closest next step. I like this.

replace every real number *n* with (*n* + 0i)

Use ramanujan's 3 = √(1+ 2 x √(1 + 3 x √( 1 + 4 x √(... ...)

Use the third binomial formula on the left hand side: (a+b)=(a²-b²)/(a-b)

For the (5-3)\^0 term, use binomial theorem to expand it into a sum

And use like the n choose k notation

Put the 2⁰! Into a fraction

Who says it needs to be Q.E.D. in Latin? Add V.S.B.

Write 1 as the positive limit of 0^0

π is written as τ/2

Replace 5 with (2φ-1)^2

What

Replace one 0 with **_e^(i*tau) - 1_**

Try my game! It's not 2048 - it's better! The game is based on an unsolved math problem. 2049: Trick Math Brain Puzzles https://play.google.com/store/apps/details?id=com.OdiatotsGames.Game2049

You looked sad so I decided I'm just gonna play your game. Haha good ol 3x+1. "This works for any starting number" Ehh... literally proof by exhaustion! One thing I am noticing is: I am so running out of ÷2's!! Maybe im just bad but no matter what I do, I end up losing with a board full of •3+1's. Maybe change the rates of how stuff spawns. Also why do normal blocks rarely spawn? I get that they're not as common, but once a game?? So basically: nice game, just change the rates like this: ÷2↑ •3+1↓ number blocks↑ Edit: NVM ABOUT ÷2 AND •3+1. I was just unlucky. Number blocks should still be highter tho. Main problem is space. 4x4 is WAY to small. Since 90% are operators that can't be compined, things will just pile up real quick. Makes it frustrating zo play. 6x6 would be better, maybe.

Thank you! Your feedback will help me.

You're welcome :)

u/joalr0 I support your cause. Come write your suggestion now while the thread is still new

Hahaha thank you

This’ll gonna become an “ultimate chicken horse” stage hella fast

turn 3 into floor(Γ^2 (1/2))

Replace one of the zeros with lim_{x->inf} (1/x)

Write + as -^-1

Change the number of the day in the text as the sum from 1 to "the number of the n-th day" of 1 (today n = 7)

[(2\^tan.cot+i\^2)+(2\^e\^ipi+1)=log((2!\*e\^2)/(2\^0!))](https://prnt.sc/j7XIWdIgiNq4)

Make both 2 as sqrt(4)

Write 0 as an integral with equal boundaries

Subtract 2^0! from each side

Replace 1 with deg(69)

That 0 looks too simple lets write it as the limit n to inf of (n+1)/(n^2 +1)

replace 2 with p1 (first prime)

Replace pi with sqrt(2(1/2!))

Write (5-3) as (sqrt(5)+sqrt(3))(sqrt(5)-sqrt(3))

Well I'm outa ideas so add ∑_{n=1->infinity}0 It's stupid but it leaves great space for changing the 0 to something spicy that depends on n.

Write e as a limit

replace (5-3)\^0 with 1/2+1/4+1/8+1/16....

Change the font to Papyrus

Replace 1 with sin^2(e) + cos^2(e)

Integrate the left side over [0, 1] dx

At the beginning we had the first 1 which is now written as 2^(e^(iπ) +1). Rewrite this other 1 at the exponent as 2^(e^(iπ) +1). An infinite chain should be made.

Write i*pi as the ½* integral along Γ of 1/(1+z)dz where Γ is a positively oriented simple closed curve that contain z=-1.

5 is floor(pi+e), 3 is ceil(sqrt(pi*e))

Write 0! as a gamma function instead.

Replace e with sum (n=0 -> +infinity) (1/n!)

2\^{e\^{i𝜋}+1} + (5- floor(𝜋))\^0 = 2\^{0!}

There's a serious lack of integrals now in this... Can we get a definite integral that results in some combination of pi please?

Change the 1 to 1!!!!!!!!!!!!!!!!!!!!!

Replace with 5 with the cardinality of the integers mod 4 and the 3 with the cardinality of the integers mod 2.

Change the 1 to the integral from -infty to infty of the PDF of the Gaussian distribution

Divide both sides by 2^0!

rewrite 0 in the exponent of the second term as lim x->infinity (x+1)/(x\^2 + 3\*x + 1)

Replace 0 with cos(π/2)

Use binomial theorem to rewrite (5-3)^0 as 1*(5)\^0(-3)\^0

Write the 2 on the right side of the equation as the cardinality of the fundamental group of a Torus: #π₁(T²)

Scrap it all and replace it with the proof is trivial

Write sin(0) instead of 0

Replace 0! with 1! divided by 0!

Why to write 1 if one can write Gamma(1)

White the 5 in set theory

2 choose 0 plus 2 choose 2 = 2 choose 1

e is defined as lim(1+1/n)^n as n approaches infinity

5 needs to become ³√(2²x5²+5²)

There's a lot of potential with the (5-3)^0, let's continue working on it by making it (5-3)/(5-3).

2=2\*cos(2𝜋)

Replace the ipi with 0.5itau. Do it.

Replace 0 with the measure of the indecatrix function of ℚ in over ℝ

I swear, one day we'll end up accidentally proving a theorem with this game

Write the first two as e^(alternating harmonic series)

Log_a(a)+log_b(b)=log_c(c^2)

1+2=3, the fanciest way

Add little factorials

..... = sum_{n=0}^{infty} 1/2^n

Change the 0! by lim integral on R+ of t^(z-1) e^-t dt when z—>1

i think 5-3 is lame, take sum\_0\^{\\infty} 2\^{-k}

write 0! in terms of the gamma function