help-13

[gloomy pewter]

Like this

[flint cape]

$$F(a; b; c) = {\underbrace{a^{a^{a^{\dots}}}}_{\mathrm{c,\text{times}}}}^{^{^{^{b}}}}$$

[wraith dagger]

Waes (Wires)

[gloomy pewter]

No, unfortunately this is not true

[flint cape]

Again, though, point of this?

@gloomy pewter Yes I did point that in the next damn message

[gloomy pewter]

Let me remind you, I need a recurrent function, where as the number increases, the result becomes more accurate

[flint cape]

"Accurate"?

That assumes this has a limit

[gloomy pewter]

And the result should tend to the result of tetration with three selected arguments, and they can be anything, that is, any complex numbers

[flint cape]

Or do you mean some sort of sequence of terms, whose limit is this?

[gloomy pewter]

For example:
lim (d→+∞) F(a; b; c; d) = a^a^a^...("b" times)...^c

[flint cape]

Surely that's governed by F(a, b, c+1) = a ^ F(a, b, c)?

[gloomy pewter]

I just need to be able to plug in any number other than integers and get an analytically correct result

[flint cape]

okay...

But exponentiation is already a defined thing even in C?

...You did check that exponentiation already has a meaning in C, right?

[gloomy pewter]

Listen: it has long been known in mathematics that the exponent can be any complex number, that is, a + ib, and the result will be evaluated analytically, since the exponentiation is a multiple multiplication. Although personally I still don't quite understand how the indicator can be expressed by imaginary numbers, since I don't understand how the number should behave.

But now we come to the situation where we need to exponentiate repeatedly. If we use an exponent repeatedly, then this is a recursion of the exponent with some number of uses of the exponent, this is a nested power tower, since the top of this tower can be any number, and the tower itself is a series of instances of the base of the exponent

[flint cape]

Okay...

Where're you going with this?

[gloomy pewter]

If the top of the nested power tower is equal to one, this is already tetration, but it only works with natural numbers and zero.

[flint cape]

We can evaluate exponents even if both the base and the power are complex

[gloomy pewter]

I need to have a formula that respects the behavior of a function like tetration and can take any number as its arguments

[gloomy pewter]

Well, yes, that's what I said, but there are questions about tetration

[flint cape]

You're wondering if a and b are complex, how to define a^^b?

[tawny sage]

there are ways people have tried to expand tetration to complex heights, it's talked about in that wikipedia article linked above

[gloomy pewter]

Wikipedia doesn't help me because it provides some data, of which linear and quadratic approximation

[gloomy pewter]

Yes, I need to understand what the general definition it

[flint cape]

So

https://en.wikipedia.org/wiki/Tetration#Complex_bases is about a being complex

And https://en.wikipedia.org/wiki/Tetration#Complex_heights is about b being complex (i.e. complex heights)

And https://googology.fandom.com/wiki/Tetration#Generalization already suggests there's no consensus on what to label non-integer towers, let alone complex ones

[gloomy pewter]

The Schroeder function is used, this is the closest definition to tetration with any index

This method of tetration is by the way taken from William Paulsen and Samuel Cowgill

I have more questions about how to make it possible to flexibly choose the base and tetration index

[tawny sage]

Considering the paper finding a way to evaluate a holomorphic extension came out only this year, I'm guessing it's going to be well above all of our heads at this point

[tawny sage]

I'd recommend trying to just work through these sorts of issues before you look more into that

also, you have a habit of using lots of non-standard terminology. That's gonna make it harder for people to help you in the long run

[flint cape]

It's the mathematical equivalent of

https://tenor.com/view/princess-bride-you-keep-using-that-word-i-dont-think-it-means-what-you-think-it-means-gif-15708582

[for the record, by the way - you don't "need" to understand this, you "want" to understand it]

[there is a difference there]

[gloomy pewter]

I'm sorry about that

I understand that such a term as iterativity does not yet exist in the regulations, but I have already explained to you what it means

[flint cape]

Correction: you've explained what you mean by it

That isn't the same as "what it means" (because that implies there is a standard definition, or at least some definition that has some consensus)

[tawny sage]

it's not a big problem it's just gonna lead to less people talking math with you, and I don't want that. It also leads to lots of conversations trying to figure out exactly what you mean, rather than talking about the math itself. I think we do know what you mean by "iterativity" now, but the conversation could have gotten to the math parts a lot faster if we didn't have to figure that out first.

[gloomy pewter]

The iterability of a function is the argument of the function that indicates how many times it has been applied to the argument/s in brackets.

$$sin^(-1)(x) = arcsin(x)$$

$$sin^{-1} (x) = arcsin(x)$$

The number -1 is the iterativity value, and it is negative, that is, this is the function that if applied together with a function that is equal in absolute by iterativity, then we will get the argument back

$$\sin^{-1} (x) = \arcsin(x)$$

[wraith dagger]

Pentalogue

[gloomy pewter]

That is, the iterativity of a function is some kind of material value of the function

$$\arcsin^{-1} (x) = \sin(x)$$

[wraith dagger]

Pentalogue

[tawny sage]

again, no idea what "material value" would mean

it's not intrinsic to the function, if that's what you mean

it relies on some other base function

[gloomy pewter]

$$\sin^{1} (x) = sin(x)$$

$$\sin^{1} (x) = \sin(x)$$

[wraith dagger]

Pentalogue

[gloomy pewter]

$$\sin^{2} (x) = \sin(\sin(x))$$

[wraith dagger]

Pentalogue

[gloomy pewter]

$$\sin(x)^{2} = \sin(x)×\sin(x)$$

$$\sin(x)^{2} = \sin(x)\sin(x)$$

[wraith dagger]

Pentalogue

[gloomy pewter]

$$\sin^{i} (x) = ???$$

[wraith dagger]

Pentalogue

[gloomy pewter]

$$\exp_{a}(x) = a^{x} $$

[wraith dagger]

Pentalogue

[gloomy pewter]

$$\exp_{a}^{-1}(x) = /log_{a}{x} $$

$$\exp_{a}^{-1}(x) = \log_{a}{x} $$

[wraith dagger]

Pentalogue

[gloomy pewter]

$$\exp_{a}^{2}(x) = \exp_{a}{\exp_{a}{x}}$$

$$\exp_{a}^{2}(x) = \exp_{a}{(\exp_{a}{x})}$$

[wraith dagger]

Pentalogue

[gloomy pewter]

$$\exp_{a}^{2}(1) = \tetration(a, 2)$$

$$\exp_{a}^{2}(1) = {tetration}(a, 2)$$

$$\exp_{a}^{2}(1) = tetration(a, 2)$$

[wraith dagger]

Pentalogue

[gloomy pewter]

How to make a non-italic font?

<@&286206848099549185>

[calm sierra]

$\mathrm{tetration}(a,2)$

[wraith dagger]

ロケット・ジャンプ

[gloomy pewter]

Thanks a lot

$$\exp_{a}^{2}(1) = \mathrm{tetration}(a, 2)$$

[wraith dagger]

Pentalogue

[gloomy pewter]

$$\exp_{a}^{b}(1) = \mathrm{tetration}(a, b)$$

[wraith dagger]

Pentalogue

[gloomy pewter]

So what should I do next?

$$\exp_{a}^{b}(1) = \mathrm{tetration}(a, b)$$
Where $a$ and $b$ can be any complex numbers

[wraith dagger]

Pentalogue

[gloomy pewter]

How to remove indentation in the first line?

[calm sierra]

you should do $$ instead

$\exp_{a}^{b}(1) = \mathrm{tetration}(a, b)$ where $a$ and $b$ can be any complex numbers

[wraith dagger]

ロケット・ジャンプ

[gloomy pewter]

$\exp_{a}^{b}(1) = \mathrm{tetration}(a, b)$
Where $a$ and $b$ can be any complex numbers

[wraith dagger]

Pentalogue

[calm sierra]

wdym hyphenate

[gloomy pewter]

How to make a line break and prevent indentation on the first line?

[gloomy pewter]

I made a typo, don't pay attention

[calm sierra]

\\

[gloomy pewter]

$$\exp_{a}^{b}(1) = \mathrm{tetration}(a, b)$$
\where $a$ and $b$ can be any complex numbers

[calm sierra]

$\exp_{a}^{b}(1) = \mathrm{tetration}(a, b)$ \
Where $a$ and $b$ can be any complex numbers

[wraith dagger]

ロケット・ジャンプ

$\exp_{a}^{b}(1) = \mathrm{tetration}(a, b)$ \\
Where $a$ and $b$ can be any complex numbers

[gloomy pewter]

Thanks a lot again

$\exp_{a}^{2}(1) = \mathrm{tetration}(a, 2) = a^{a}$

[wraith dagger]

Pentalogue

[gloomy pewter]

$\exp_{a}^{1.5}(1) = \mathrm{tetration}(a, 1.5) = a^{\superscript{0.5}a}$

$\exp_{a}^{1.5}(1) = \mathrm{tetration}(a, 1.5) = a^{^{0.5}a}$

[gloomy pewter]

How can you implement the exponent in general? And is it possible to make the exponent not on the right, but on the left?

$\exp_{a}^{3}(1) = \mathrm{tetration}(a, 3) = a^{a^{a}}$

[wraith dagger]

Pentalogue

[gloomy pewter]

$\exp_{a}^{∞}(1) = \mathrm{tetration}(a, ∞) = a^{a^{a^{a^{a^{a^{a^{a^{a^{\dots}}}}}}}}}$

$\exp_{a}^{\infty}(1) = \mathrm{tetration}(a, \infty) = a^{a^{a^{a^{a^{a^{a^{a^{a^{\dots}}}}}}}}}$

[wraith dagger]

Pentalogue

[gloomy pewter]

I came up with my own notation to denote iterativity

$\mathrm{iter}(F(x); n) = F(F(F(\dots(F(x)\dots)))$

$\mathrm{iter}(\mathrm{F}(x); n) = \mathrm{F}(\mathrm{F}(\mathrm{F}(\dots\mathrm{F}(x)\dots)))$

[wraith dagger]

Pentalogue

[gloomy pewter]

$\mathrm{iter}(\mathrm{F}(x); n) = {\underbrace{\mathrm{F}(\mathrm{F}(\mathrm{F}(\dots(\mathrm{F}(x))\dots)))}_{n}$

[wraith dagger]

Pentalogue

$\mathrm{iter}(\mathrm{F}(x); n) = {\underbrace{\mathrm{F}(\mathrm{F}(\mathrm{F}(\dots(\mathrm{F}(x))\dots)))}_{n}$
```Compilation error:```! Missing } inserted.
<inserted text> 
                }
l.49 ...thrm{F}(\dots(\mathrm{F}(x))\dots)))}_{n}$
                                                  
I've inserted something that you may have forgotten.
(See the <inserted text> above.)
With luck, this will get me unwedged. But if you
really didn't forget anything, try typing `2' now; then
my insertion and my current dilemma will both disappear.```

[gloomy pewter]

$\mathrm{iter}(\mathrm{F}(x); n) = \underbrace{\mathrm{F}(\mathrm{F}(\mathrm{F}(\dots\mathrm{F}(x)\dots)))_}$

$\mathrm{iter}(\mathrm{F}(x); n) = \underbrace{{\mathrm{F}(\mathrm{F}(\mathrm{F}(\dots\mathrm{F}(x)\dots)))}_{\mathrm{F}()} n \mathrm{times}}$

$\mathrm{iter}(\mathrm{F}(x); n) = \underbrace{{\mathrm{F}(\mathrm{F}(\mathrm{F}(\dots(\mathrm{F}(x))\dots)))_\mathrm{F}()} n \mathrm{times}}$

$\mathrm{iter}(\mathrm{F}(x); n) = \underbrace{\mathrm{F}(\mathrm{F}(\mathrm{F}(\dots(\mathrm{F}(x))\dots)))}_{\mathrm{F}() n \mathrm{times}}$

[wraith dagger]

Pentalogue

[gloomy pewter]

How to make a space between words?

[supple grove]

damn

[gloomy pewter]

No

[wraith dagger]

DaveyLovesSocks

[supple grove]

?

[gloomy pewter]

«F() n times»

[supple grove]

but why

[gloomy pewter]

This is a notation designation

[supple grove]

ok...

[gloomy pewter]

$\mathrm{iter}(\mathrm{F}(x); n) = \underbrace{\mathrm{F}(\mathrm{F}(\mathrm{F}(\dots(\mathrm{F}(x))\dots)))}_{\mathrm{F}(){n} \mathrm{times}}$

[wraith dagger]

Pentalogue

[gloomy pewter]

$\mathrm{iter}(\mathrm{F}(x); n) = \underbrace{\mathrm{F}(\mathrm{F}(\mathrm{F}(\dots(\mathrm{F}(x))\dots)))}_{\mathrm{F}(){ n} \mathrm{ times}}$

$\mathrm{iter}(\mathrm{F}(x); n) = \underbrace{\mathrm{F}(\mathrm{F}(\mathrm{F}(\dots(\mathrm{F}(x))\dots)))}_{\mathrm{F}(),{n} \text{ times}}$

$\mathrm{iter}(\mathrm{F}(x); n) = \underbrace{\mathrm{F}(\mathrm{F}(\mathrm{F}(\dots(\mathrm{F}(x))\dots)))}_{\mathrm{F}() n \text{ times}}$

$\mathrm{iter}(\mathrm{F}(x); n) = \underbrace{\mathrm{F}(\mathrm{F}(\mathrm{F}(\dots(\mathrm{F}(x))\dots)))}_{\mathrm{F}(){ n}\text{ times}}$

$\mathrm{iter}(\mathrm{F}(x); n) = \underbrace{\mathrm{F}(\mathrm{F}(\mathrm{F}(\dots(\mathrm{F}(x))\dots)))}_{\mathrm{F}()\text{ }n\text{ times}}$

Here, perfect

Expression $\mathrm{iter}(\mathrm{F}(x), n) = \underbrace{\mathrm{F}(\mathrm{F}(\mathrm{F}(\dots(\mathrm{F}(x))\dots)))}_{\mathrm{F}()\text{ }n\text{ times}}$, \
where $\mathrm{F}(x)$ – some function, and $n$ – number of iterations, which can be any complex number

[supple grove]

you can just edit your message by the way

you dont need to post a new one every time

[gloomy pewter]

I didn't know that I could edit the message and the picture would change after that

<@&286206848099549185>

[split ice]

so what is the question?

[wraith dagger]

Pentalogue

[supple grove]

surely if n can be complex this definition fails

[flint cape]

cf. how you define addition (by adding b multiples of 1 to a), multiplication (by adding b copies of a) and exponentiation (by multiplying b copies of a)

[gloomy pewter]

The question is, what recurrent function approximates the results of tetration with any base and index, where the higher the value of the recurrent term, the closer the result to the value of tetration

[flint cape]

These definitions already fall apart at the first hurdle when "b is not an integer"

[gloomy pewter]

But later I gave a clarification: what is the general definition of an exponentiation where the value of its iteration is not expressed as an integer?

[flint cape]

Similarly, we can't just use the same definition of tetration as repeated exponentiation directly, if the power is not an integer, forget it being complex

Generally we take $a^b = \exp (b \log a)$ when $a,b \in \bC$

[wraith dagger]

Waes (Wires)

[flint cape]

(where that should actually be __L__og, i.e. the complex logarithm)

You can see that this is already wildly different from "iterative multiplication"

[gloomy pewter]

Expression $\mathrm{iter}(\mathrm{F}(1), 3) = \mathrm{tetration}_{a}(3)$, \
where $\mathrm{F}(x) = a^{x}$, and $n$ – number of iterations

[flint cape]

Again, what does it mean to do something a complex number of times?

Heck, what would it mean to do something a non-integer number of times?

You're tripping over the first three hurdles while claiming you can jump over the 4th, 5th and 6th in one go

[gloomy pewter]

Well, I also want to find out what the hell this means

[flint cape]

Slight clarification - this doesn't mean anything. If we want to make this work first for non-integers, we have to redefine it in a way that agrees with what we already got without having that integer limitation

[gloomy pewter]

You understand what my iter() function notation means, right?

[flint cape]

I don't need to make sense of the function to understand the theory you're trying to gun for

But the premise is still flawed

[gloomy pewter]

Damn, it's very difficult

[flint cape]

As I already mentioned here, there's not even a consensus on how to alter the definition so that non-integer tetration can make sense

yeah no shit

[wraith dagger]

Pentalogue

[flint cape]

This sits WELL outside school maths

And that complex-tetrand paper that's mentioned in the Wikipedia article, that was written THIS YEAR

[placid turtle]

i assume n gotta be an integer right

[gloomy pewter]

Yes, tetration is too fresh for our brains to understand

[flint cape]

So we're talking the boundary of the field itself

[flint cape]

You're right - this is the ONLY way this definition makes sense; the only problem is, Pentalogue's trying to extend this into n being complex (or somewhere along those lines), but under a flawed definition

[gloomy pewter]

It must be chosen without restrictions, that is, it can be any complex number, but due to the limitation the number n is always a natural number or sometimes an integer

[flint cape]

Right, so the definition ITSELF is restrictive

You need a new definition; this one will not do

[gloomy pewter]

Now I'll start over

[flint cape]

Recall this

These definitions already fall apart when b is, say, a fraction

Or negative

Or irrational

Forget even whether b can be complex

[gloomy pewter]

I found a recurrent function that with increasing step gives the value of the square superroot

[flint cape]

These definitions don't hold for most real numbers b

[gloomy pewter]

How to make a fraction using TeXit markup?

[supple grove]

$\frac{a}{b}$

[flint cape]

$\frac{a}{b}$

[wraith dagger]

DaveyLovesSocks

[gloomy pewter]

Thanks

$\mathrm{F}(n) = \mathrm{F}(n-1) - \frac{\mathrm{F}(n-1)\log(\mathrm{F}(n-1))-\log(a)}{1+\log(\mathrm{F}(n-1))}$, \ where the number $a$ is the number under the superroot

[wraith dagger]

Pentalogue

[gloomy pewter]

I recommend you take a look at this

I also need a recurrent function, but one that gets closer to the results of tetration

[flint cape]

?

I'm referring to b, the number in the definitions I listed

I'm not calling you a bitch

[gloomy pewter]

Are you familiar with the recurrent approximation to the super square root?

[flint cape]

Or are you asking me to watch "It", the horror film?

[gloomy pewter]

What?

[flint cape]

What does this mean here?

Because, seeing as no film was mentioned, this comes off as rather antagonistic

[gloomy pewter]

Do you know what a superroot is?

[gloomy pewter]

What do films have to do with it?

[flint cape]

I'm assuming you mean the inverse of a tetration

[gloomy pewter]

No, just superroot

[flint cape]

Because "to watch" is a verb for "to watch media, e.g. videos", or "to spectate (a sports match)"

Outside of that, "Watch it" has its own set meaning: https://en.wiktionary.org/wiki/watch_it

[gloomy pewter]

I understand that you decided to find fault with the verb "to watch" that I wrote, okay

I get that, what you want to tell me

[gloomy pewter]

In general, I wanted to say that this is a recurrent formula, where the larger the number n, the closer the value of F(n) is to the square superroot of the number a

[gloomy pewter]

https://www.desmos.com/calculator/hc2s5geckj

By analogy with this formula, I need a formula that gives a recurrent approximation to the values of tetration

[gloomy pewter]

There may have been a misunderstanding, but let's get down to business, I really need help

<@&286206848099549185>

.close

[cedar kiln]

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[cedar kiln]

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[pseudo merlin]

can i pls have help witb this

i just started with dm/dt = km

[storm zealot]

I believe that this is a first order half life reaction given by the formula

Do you know how to derive this formula from the first order kinetics formula? @pseudo merlin

[tropic oxide]

-km

cause it's decaying hence getting smaller hence its rate of change is negative

[tropic oxide]

kinetics??

[storm zealot]

https://www.biologyonline.com/dictionary/first-order-kinetics#:~:text=First-order kinetics is when,concentration of a single reactant.

[cedar kiln]

@pseudo merlin Has your question been resolved?

[pseudo merlin]

oh

how do i find it when its halved

dm/dt = km/2?

[tropic oxide]

well first solve for m as a function of t

[pseudo merlin]

oh

ok let me try

[tropic oxide]

afterwards, take $m(0)=1$ and then, to find the \textbf{time at which} the mass is halved, you solve for $t$ (time) in $m(t)=\frac12$.

[wraith dagger]

Ann

[pseudo merlin]

wait i forgot + c

[tropic oxide]

you need to solve for m anyway and also you can and should assume m>0.

[pseudo merlin]

oh ok that makes sense

[storm zealot]

You are supposed to find when t is halved, not t

[tropic oxide]

when m is halved, you mean.

your correction had a bit of a fatal typo in it

[storm zealot]

Yes, and also I actually meant t_1/2

[pseudo merlin]

wait im kind of confused at the m(0) = 1 part

why?

[tropic oxide]

unironically: for convenience

you can pretend that we're looking at like, 1 kilogram of your radioactive substance or something

but really the 1 just stands for the starting amount

so that half of it can be just 1/2

[pseudo merlin]

ooohh ok ok

[cedar kiln]

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[storm zealot]

3 minutes timeout lol

[cedar kiln]

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[pseudo merlin]

A

[pseudo merlin]

Im still kinda stuck

[tropic oxide]

yeah cause you're doing a bit of cart-before-horsing

$m = e^{-kt+c}$, work out $c$ such that $m=1$ when $t=0$ first and foremost!

[wraith dagger]

Ann

[storm zealot]

First you use distributive property

Then to find initial mass we consider the time was 0

[tropic oxide]

i mean i literally told him to take m(0)=1...

what distributive property are you talking about

i think you may be calling something by the wrong name

[storm zealot]

By We have e^-kt + C

Can we write this as e^C e^-kt for convenience?

[tropic oxide]

no tf we can't

$a^{m+n} \neq a^m + a^n$ mate

[wraith dagger]

Ann

[storm zealot]

Sorry

[tropic oxide]

we can write $e^{-kt+c}$ as $e^c e^{-kt}$ yes, but that's not a distribution

[wraith dagger]

Ann

[tropic oxide]

it's an exponent law you would apply

[pseudo merlin]

Ohh omg i of it

Got it

[tropic oxide]

also e^(-kt+c) please not e^-kt + c

[pseudo merlin]

I was so confused cause i got -ln(1/2)/k

[storm zealot]

Yes exponential law

[pseudo merlin]

but its the same as ln(2)/k

Lol

[cedar kiln]

@pseudo merlin Has your question been resolved?

[cedar kiln]

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[manic kelp]

im not rly sure what im meant to find

[austere ember]

do you know what expectation / expected value is?

[manic kelp]

the value that is most likely?

[austere ember]

yeah sort of, you could think of it like the average value?

not necessarily the most likely

[manic kelp]

alr

[austere ember]

for example, let's say we were flipping a coin, and if it's heads you give me a dollar, but if it's tails i give you a dollar

[manic kelp]

the expectation is 1/2?

[austere ember]

there's a 1/2 probability of you netting -1, and a 1/2 probability of you making 1

[manic kelp]

is that how that works

[austere ember]

not quite, those are the raw probabilities

[manic kelp]

ah

[austere ember]

so if it's equally likely for you to make a dollar or lose a dollar

[manic kelp]

yes

[austere ember]

what do you think the expected value is for flipping the coin?

in other words, if this was a game, how much would you pay to play it?

[austere ember]

maybe this is a worse way to think about it so ignore it

but it's equally likely for you to gain or lose the same amount

[manic kelp]

is it 0..
im thinking cause theres a 50/50 i get a dollar, and 50/50 i lose

[austere ember]

yes

[manic kelp]

alr sweet

[austere ember]

expected value is kind of like the "average" outcome

but that average is weighted by the probabilities

in this case it's just the average of -1 and 1 because it's 50/50

but let's say instead of being 50/50, it was a biased coin so it had a 90% chance of landing on heads

do you see why the "average" outcome is no longer $0?

[manic kelp]

yea i think

well yes i see it because it is no longer 50/50

but how i would write the answer is a different story

[austere ember]

by the rules of the game, it's now much more likely for you to lose a dollar than to make a dollar, so the average gain/loss has shifted from 0

[manic kelp]

yes got that

[austere ember]

this idea of a weighted average is what motivates this formula

[manic kelp]

oh wtf

[austere ember]

do you kinda understand what this says or no

notation can be scary

[manic kelp]

no XD

[austere ember]

no worries, it's a little complicated looking

[manic kelp]

never used that symbol in my life

[austere ember]

that's a sigma, it just means a sum

[manic kelp]

alr

[austere ember]

so let's say there were n possible outcomes which are the outcomes x_1, ..., x_n

and you have the probabilities for each of these outcomes, P(x_1), ..., P(x_n)

what you do to calculate the expected value is multiply each outcome by its probability, and then sum these all up

so x_1 * P(x_1) + ... + x_n * P(x_n)

[manic kelp]

that seems like a very confusing formula to do just that

[austere ember]

consider the really simple case from earlier, the 50-50 coin

we agree the expected value is 0

there are two outcomes there, either +1 or -1, agreed?

[manic kelp]

yes

[austere ember]

and their probabilities are each 1/2

using this formula, you'd find that the expectation is (1) * (1/2) + (-1) * (1/2)

which gives 0

in the weighted coin example, the expectation would've been (-1) * (9/10) + (1) * (1/10)

so what the formula is doing is assigning a higher "weight" to outcomes with higher probability

does that kinda make sense?

you can kind of think about it as a game of tug of war. there's a person pulling the rope towards each outcome, and their strength is kind of proportional to the probability of that outcome. so the person with the more likely outcome pulls harder. the expectation is kind of like where the center of the rope ends up

[manic kelp]

one sec, sorry I don't think I fully understood
we have this formula, which tells us that the expected value is the sum of (what exactly?) times the amount of outcomes? times the probability of those outcomes?

[austere ember]

each x_i is an outcome

so in the coin example it's the money

+1 and -1

and yes, the P(x_i) means the probability of that outcome

so for each outcome, you take it and multiply it by its probability

and then add all of those results up

i tried kinda giving the intuition for expectation as a weighted average but it might also help just memorizing the formula, it should've been taught if they're asking this kind of question

[manic kelp]

oof sorry i might be too slow with this but ur explanation has helped, i just need the example and formula separately at first

[austere ember]

no it's no worries, it can be tough if you're seeing it for the first time

[manic kelp]

(im doing a uni entrance exam to cover a year of education i didnt have so im learning all this on my own)

[austere ember]

the reason you multiply an outcome by its probability is because it matters how likely it is. it's not exactly an average if you consider each outcome to be equally likely

[manic kelp]

are you able to vc to explain this to me if not totally okay ill try to reread all this

[austere ember]

sure

[cedar kiln]

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[steady stag]

how do i solve for the three variables in this system of nonlinear equations

$\begin{cases}
d + a^3 - 2 a b - \frac{1}{d} = 0, \
a d + \left(a^2 b - b^2\right) + \frac{a}{d} = 1, \
b d - \frac{a^2 - b}{d} = 1.
\end{cases}$

[wraith dagger]

∲ビジョン∲

[tropic oxide]

that looks a bit weird

@steady stag was this system given to you as is or did you get it from somewhere else

[steady stag]

it was given to me like this

well it was just a nonlinear system of three equations

[tropic oxide]

hm so it did not arise from a different problem

but ok guess i'll try poking around with it

[steady stag]

the only issue was when i solved and substituted independent variables, the degrees rack up quickly

[tropic oxide]

were you just given this system and absolutely zero other context or other info

like just something to grab onto

cause ngl i dont see any way to progress at all lol

[sullen cipher]

Wanted to suggest cramer's rule but it still seems complex

[tropic oxide]

this is explicitly stated as a nonlinear system

[sullen cipher]

Oo

[steady stag]

just a word problem

but its what gave the equation

[tropic oxide]

share the word problem!

[steady stag]

Three friends, Alice (a), Bob (b), and Dave (d) are working together on a project where their productivity follows some teamwork rules. Dave’s total work done plus Alice’s work cubed equals twice the product of Alice and Bob’s work plus the reciprocal of Dave’s work . The combined effort of Alice times Dave plus the difference between Alice squared times Bob and Bob squared plus Alice divided by Dave equals one. Finally, Bob’s effort times Dave equals one plus the quantity. Find the productivity levels a, b, d that satisfy these teamwork conditions.

[tropic oxide]

bruh that's contrived as hell

[steady stag]

i know

[cedar kiln]

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[clever hazel]

heko

[clever hazel]

kinda new to this chain rule thing

but so for the actual answer here

they somehow disregard the negative

but if i just use chain rule that i just learnt from chatgpt

it accounts the negative

the question is regarding

maximum

would appreciate help

[proven summit]

!nogpt

[cedar kiln]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[clever hazel]

well

u can see the workking out

it looks accurate to me

so like

erm

hold on

we gotta derive this

so u make the denomintor =u

[blazing dune]

The answer seems correct

[clever hazel]

so u do the derivative of the outer x dervative of the inner

[clever hazel]

yh that answer is correct

its from a video

i didnt get that

i got p=1/2

not -1/2

[proven summit]

you got 12p + 6 = 0 in your working out

[blazing dune]

Then show your steps

[proven summit]

i'm guessing you forgot the minus sign for -6 when you subtract to both sides by 6

[clever hazel]

this is my working out

like

the final answer

here

ok

i set it =0

so then

multiply by denominator

oh WAIT

im

so dumb

nvm guys

wasted ur time

[proven summit]

it's alright

[clever hazel]

bracket error in my working out

i may have taken a shower in between solving the problem

my bad guys

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[cedar kiln]

Channel closed

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as this action is irreversible, and this channel may abruptly lock.

[severe oasis]

.close

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[civic coral]

For the C sign, it stands for combination.

I would like someone to check the result, I got 11/24 which is different from the answer provided by the book

[opal hinge]

what's the question?

[proven summit]

!original

[cedar kiln]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

@civic coral Has your question been resolved?

[civic coral]

It doesn’t matter

[civic coral]

This is off topic

[opal hinge]

you mean you just have to calculate all this?

there is a device called "calculator", ever heard of it?

[silk gust]

are you 100% sure that the expression in and of itself is correct, as well?

[upper ruin]

!original

[cedar kiln]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[civic coral]

I cannot put combination on it

[opal hinge]

mine can, get a better one

[flint cape]

1. What calculator do you even have?

2. You do know there's a formula for nCr, right?

Specifically: $nCr = \frac{n!}{r! (n-r)!}$

[wraith dagger]

Waes (Wires)

[cedar kiln]

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[shell kraken]

don't understand this

[shell kraken]

im not understanding the concept if its multiple periods

I thought that because 6 months, that means 2 compounding periods in a year, so it should be m/2. It wasn't in the options so i just chose randomly

[flint cape]

(Would it matter how long a year is? The question's relevant to months)

[shell kraken]

oh yea, you're right

[flint cape]

If the result is that you multiply by 0.8 after 6 months, then

per month, you need some term that, when raised to the 6th power, is 0.8

i.e. 0.8 ^ (1/6)

[shell kraken]

OHHHH RIGHT

[flint cape]

You then raise that to the power m to determine how much you multiply after m months, which gets you expression C

[shell kraken]

AND BECAUSE OF EXPONENT LAWS THEY ADD UP WHEN MULTIPLIED

thats a brain freeze

thanks a lot

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[wicked panther]

I'm looking for a logical order of math expressions, symbols, or terms, I'm designing a memory palace and want to do it attached to as many things possible so that I can recall instantaneously a math operation and be able to apply it. Any lists or places I can look into? (sorry if it was a bit vague)

[rare schooner]

what level of math are youlooking for

[wicked panther]

i want to be a physicist, but from beginner to beyond expert, I want to become proficient at math and will dedicate the time to it, anything would be really appreciated

[rare schooner]

ah great!

I see that you're an undergraduate?

[wicked panther]

Yeah

[rare schooner]

So I suppose you're already proficient?

This is quite nice. I'm doing my MPhys right now on astronomy

[wicked panther]

I have a highschool level, currently at a rly busy military school adjacent to the difficulty of a "nuke" school, but i have goals, and I also want to get 100s on every test forever, I'm tired of having bad memory and falling behind on encoding and such

[rare schooner]

I see

So what field are you looking at?

Trigs?

pre-calc?

calc?

[wicked panther]

from basic, to trig, calculus geometry, to as highly complex as it can possibly get, I will dedicate a lot of time to studying these things, however up to SATs are my priority and above at the moment

[rare schooner]

Right

so you're at pre-university at the moment?

[wicked panther]

something like that

[rare schooner]

right

I can recommend a few playlists if youd like

[wicked panther]

Yes please

[rare schooner]

https://youtu.be/MHeirBPOI6w?si=t0LtbgyI30_eMC80

[tardy parcel]

I'm still on quadratic..

[wicked panther]

what's that?

[rare schooner]

who are you?

[tardy parcel]

Help y'all, i hate this🙏🙏

[rare schooner]

open a new help channel

[wicked panther]

that looks so fun

[tardy parcel]

It doesn't 😭

[rare schooner]

if you dont know that you're most likely below 14 lmao

[wicked panther]

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[eternal fern]

Hi teachers,
may i generally understand the "range" of group, ring, field, ordered field by using terminologies of subset, like:

[eternal fern]

To make it easier to remember the definitions of these terms

[golden raven]

uh what do you mean by the "range" @eternal fern

your notation:

can be interpreted in multiple ways

i.e. a ring is a group with another operation with additional conditions

hence i would consider rings to be a subset of groups, rather than the other way around

but its not clear per se

[eternal fern]

I've recently started studying analysis 1, and found there's something in common between these four definitions.

[eternal fern]

i mean, in similar way like this

if a ring have an inverse, then field

[golden raven]

yeah

this is the best way to properly understand such concepts

[eternal fern]

if field contains

1. linearity 2. compatibility of plus and multiplication
   then ordered

[golden raven]

uhhh

what does linearity mean in the context of a field

[eternal fern]

it may have different name, but generally (x <= y or y <= x) always true

said the professor

something like this

[eternal fern]

anyway, thank you Master Arnavutköy for giving me answers

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[raven shard]

you cannot compare groups and rings at all like this

[raven shard]

a ring is not a group, a group is not a ring

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[eternal fern]

is there any methods that i could use to describe these kind of properties more precisely?

i.e. a ring is a group with another operation with additional conditions

[austere hull]

.u can start chatting like this when channel is already closed 🙂

it wont open the channel

[raven shard]

because i closed it

[raven shard]

notice the channel was open here

[raven shard]

you can think of it as, that a ring has an underlying group structure (via the addition operation)

[austere hull]

yea i know but im just saying when u type something in a closed channel u can always start with period

[raven shard]

you say a ring “is” a group with […] but note the “is” is very loose because a ring is not a group at all

[eternal fern]

hmm

[austere hull]

even if its open u can still chat without opening the channel

anyway moving on

[eternal fern]

I see, so structures are similar, but there's no any relations between them at all

[raven shard]

sure

though you can note that there are certain ways to translate information from the language of rings into the language of groups and vice versa

one very obvious way in one direction is to take a ring and “forget” everything to do with multiplication so you are left with a group under the original addition operation

but sure saying “there’s no relations between them” is fine

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[eternal fern]

thank you master mqnic_

[eternal fern]

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[dire geode]

you've been promoted to master

[raven shard]

honored, really

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[south moss]

could someone help me solve this inequality?
determine all real x numbers for which the inequality (x^2-x-12)/(3^x-9) greater or equal to 0

[south moss]

i attempted to break it into a table and the values i got were {3} and [4, +inf)

book answer says [0,2) reunited with [4,+inf)

[dusk goblet]

how did you get {3}

[south moss]

in 3 the function equals to 0

[dusk goblet]

$\frac{(x - 4)(x + 3)}{3^x - 9}$

[wraith dagger]

knief

[dusk goblet]

nah

[south moss]

oh wait

okay something went wrong in there

[dusk goblet]

is it x^2 + x - 12

[south moss]

minus x

not plus

[dusk goblet]

then 3 is not a root

[south moss]

wait it was -3

it might still be wrong, ill recheck

[dusk goblet]

i mean you're both wrong

[south moss]

doesnt it go to 0 in x=-3?

[dusk goblet]

yea

[south moss]

so doesnt -3 fulfill the f(x)>/= to 0?

[dusk goblet]

yep

but your answer is still wrong

[wicked mantle]

[dusk goblet]

when are the numerator and denominator both negative?

[south moss]

(-3, 2) ?

[dusk goblet]

yea

so whats the answer

[south moss]

ohh

i'm dumb -/- gave me -

so it's (-3,2) U [4,+inf)?

[dusk goblet]

nope

[south moss]

ill try to redo the table

[dusk goblet]

lol

you're right there

[dusk goblet]

^

[south moss]

-3 with [

[dusk goblet]

yes

[south moss]

fuck

thanks

[dusk goblet]

you're welcome

[south moss]

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[south moss]

hiya. i need to calculate a limit where x goes towards 1 of [3f(x)-6 ]/[f^2(x)-4] , where f(1) = 2, f is a continuous function defined on R with values in R and i don't know where to start.

[south moss]

limit goes to 0/0, but i can't use l'hopital because they're all constants before 0/0

[kindred elm]

simplify the equation

use a2-b2

[south moss]

by f(x)?

[kindred elm]

wym

take f(x) as some random variable

[south moss]

oh i thought you meant something else

[calm sierra]

do a little algebra

[south moss]

i think i have to do something with lateral limits

but i'm not sure where that will get me

[kindred elm]

eh

do algebra

[calm sierra]

try a little algebra

[south moss]

i can simplify it to 3(f(x)-2) / f^2(x)-4

[calm sierra]

what about f^2(x)-4

[south moss]

are you talking about simplifying the top part to that or are you trying to imply something else i can do with the bottom part?

or can i move the exponent to encompass the entire function of x

wait

[calm sierra]

rewrite f^2(x)-4

[south moss]

newton's binomial?

or the

[kindred elm]

we are trying to cancel the part that is causing the 0/0 form

[kindred elm]

u r overcomplicating it

use simple algebra

[south moss]

fx -2 times x2 + 2

and i simplify by the top

oh i found it

[kindred elm]

ye so the expression becomes?

[south moss]

3/fx +2 where fx is two

3/4

thanks

[calm sierra]

np

[south moss]

i got fucked up whether f^2(x) was equal to f(x)^2

[calm sierra]

it is

[south moss]

thanks again

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[calm sierra]

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[true sapphire]

is it ok to multiply like this\

[true sapphire]

ior==or do i m,multilpy both of them by 2

[austere ember]

multiply like what?

[true sapphire]

liek first by 1, and second by 2

[austere ember]

yes

that's allowed

you can multiply any equation by any nonzero number

[true sapphire]

thank you

i shall clos

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[dusky furnace]

I've never seen the following function application notation before. The span function takes vectors as arguments like span(v1, v2, ..., vN), so I assume this notation is similar to the set builder notation, except the resulting elements are not part of a set, but instead are "splatted" into the function. Thus, instead of span({v1, v2, v3}), it is span(v1, v2, v3). Is my interpretation correct?

[violet flume]

i think theyre just collapsing set notation

are you sure this isnt a set?

[dusky furnace]

This is how the book defines the span function:

It doesn't take a set as an argument, but individual vectors.

[violet flume]

"the span of a set of vectors"

[dusky furnace]

I think they mean the list of individual vectors can be thought of as a set, not that it takes one literal set of them.

[violet flume]

i don't know why you're resisting lol

it would appear to be a set, to me

[dusky furnace]

I understand that it's a distinction without that much difference, but I'm trying to make sense of the syntax.

[violet flume]

it seems like this is helpful

because it tells you that they're just being lazy with brackets

and really they mean $\text{span} \qty( { \dots } )$

[wraith dagger]

jan Niku

[dusky furnace]

Have you personally seen the syntax of the original image? i.e. f(x|x = ...)

Or is this new notation for you as well?

[violet flume]

i think so, yea

using it here

$\text{span} ( { x_k : k=1,2,\dots} )$

[wraith dagger]

jan Niku

[violet flume]

unless this isnt what you mean

[dusky furnace]

I was just looking at this wikipedia page about sets. The function is defined there to take an actual set as argument.
https://en.wikipedia.org/wiki/Linear_span

[violet flume]

what you posted is an actual set

[dusky furnace]

In this question, I was mainly interested to know if this was a type of standard notation, to avoid extra {, }, like span(x|y), rather than span({x|y}), or if this was just this author's personal style.

[worldly chasm]

Notation is notation, and almost none of it is actually standard in any real sense. In this case the extra {}s don't really add much, so they were omitted.

You are expected to interpret it as a set inside of the parens.

[violet flume]

they're being lazy, but its clear what they mean, in my mind

if i am reading VS stuff and i see $v \in V : \sum ^j a_i v_i$ my brain completely glosses over all these symbols because that construction is so common personnaly

[wraith dagger]

jan Niku

[dusky furnace]

I hear you. But, don't forget that this type of interpretation makes alot of sense to people used to reading higher mathematics, but us muggles can struggle with the stuff. So, when I notice new notation that I've never seen before, I try to confirm my interpretation, because I've interpreted new notation wrong in the past.

[violet flume]

and even the v e V is kind of superfluous

it seems like your author is being pretty loosey goosey

so i wouldnt worry too much about their notation unless you feel like youre confused about what theyre trying to say

[dusky furnace]

I think I'm good; just wanted to double check.

Thank you

🙂

.close

[cedar kiln]

Channel closed

Closed by @dusky furnace

Use .reopen if this was a mistake.

[violet flume]

sorry if im dismissive

i didnt mean to be that way

[dusky furnace]

you're good

[violet flume]

its always good to ask

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[sonic epoch]

Hello anyone got a tip to not miss multiplying a Negative sign to a parenthesis? Ex
-(3x+5) is actually -3x-5 and not -3x+5 cuz omfg

[upper laurel]

any time you see a -(), you remember to do it, and anytime you see a - without a (), you remember not to do it

[sonic epoch]

im losing my mind that my grades are dying cuz i keep MISSING this mfs

[upper laurel]

more seriously, you can consider a () as grouping terms together, so any actions done on the parentheses should be done to each term

keep in mind if its a product of things inside the parentheses, then thats just 1 term so it doesnt count
it has to be addition/subtraction in the parentheses

[sonic epoch]

ight ty ill try that

.close

[cedar kiln]

Channel closed

Closed by @sonic epoch

Use .reopen if this was a mistake.

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[zealous mango]

hi

[zealous mango]

Tony has a 20 by 21 chocolate bar. Each time he breaks a chocolate bar
with n squares into two pieces with k and n − k squares, he gains k(n − k) points. What is the
maximum number of points he can gain?

anyone

<@&286206848099549185>

[drifting cedar]

!15min

[cedar kiln]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[surreal cave]

!15m

[cedar kiln]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[zealous mango]

ok

!15m

[cedar kiln]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[austere hull]

what have u tried

[zealous mango]

small cases

idk

[manic snow]

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[zealous mango]

2

[zealous mango]

2

um

guys?

[austere hull]

!show

[cedar kiln]

Show your work, and if possible, explain where you are stuck.

[zealous mango]

um i just did like a grid

i can't find any patterns

i think its induction?

or strong

[manic snow]

It has something to do with binomial coeffecients im pretty sure

[zealous mango]

uh

ok

wait lemme test some more

[upper laurel]

whats the highest score youve gotten for this

[zealous mango]

oh i haven't tried yet

im trying to find a induction proof

or a recurrence

idk im stil lstuck

<@&286206848099549185>

<@&286206848099549185> are you guys... here?

[upper laurel]

spectre not all questions can be immediately solved in seconds

[jagged tide]

Don't ping twice

[zealous mango]

ok

mbmb

[upper laurel]

give me some time alr

[jagged tide]

I was on it dw

[zealous mango]

ok

i thought there was no one on

lol

[fossil dawn]

give people time to read and respond pls

[zealous mango]

im new here

smh

😭

[upper laurel]

new to discord?

[zealous mango]

like new to this server

idk how it works

[solid wave]

U just ping us once and then we flood the channel

[zealous mango]

ok

[solid wave]

Wths the question

[zealous mango]

btw i just need some hint

idk ill try to solve myself

[jagged tide]

Ok so one thing to be noticed, once you've broken an n square, k=n/2 maximises the points for breaking the n square

So we are left with the task of breaking it optimally

[upper laurel]

@zealous mango you need to prove that, regardless of how you break this apart, you will always be left with the same score

[zealous mango]

yeah ive noticed that in some smaller cases

idk how to prove tho

[upper laurel]

Ive just broken the square in two and gotten 87990

Ive also just tried breaking the square one bit at a time and gotten 87990

87990 being 419 * 420 / 2, short for 1 * 419 + 1 * 418 + 1 * 417 + ... + 1 * 1

[jagged tide]

Hmm

21×20

It's 420

I see the pattern

[upper laurel]

some? was there a case where this pattern didnt work?

you should be able to just notice the pattern then prove it with strong induction

[zealous mango]

no, i meant not necessarily

yeah

[jagged tide]

There was an exception?

[zealous mango]

idk, i just meant that there might have been one

[upper laurel]

you need to rephase it as "in all the cases I tested", because "some" can mean "I didnt test one" and also "I found one where it didnt work"

[zealous mango]

oh

ok

[jagged tide]

Oh

[zealous mango]

but how do i create a proof with that

[jagged tide]

Give me a minute gang

Be patient will you

[zealous mango]

do i include small cases or just base cases

kk

[upper laurel]

lets consider a proof for n squares

we want to prove that the score is always (n - 1) n / 2

now first, we need an assumption

what do you think would be a good assumption here, seeing the pattern at play?

[zealous mango]

uh

[upper laurel]

it helps to go for the strongest pattern first, because as it stands that seems to be the only pattern

[zealous mango]

wait

why is the score always (n-1)n/2

[upper laurel]

spectre listen to me

we are building a proof right now

[zealous mango]

oh ok

[upper laurel]

and the key thing about an induction proof is?

[zealous mango]

inductive hypothesis

[upper laurel]

you dont need to know why at all, you just need to prove that the pattern continues

[zealous mango]

ok

[upper laurel]

so what will we be assuming to start the proof?

[zealous mango]

that the max points is 210?

[upper laurel]

what?

we're talking about a proof for n squares

[zealous mango]

oh

n(n-1)/2

[upper laurel]

thats at the end of the proof

[jagged tide]

Can induction really do it tho

[upper laurel]

yes it can

we need something else to begin the proof instead

[zealous mango]

um

base cases

[upper laurel]

no

think about what induction hypotheses usually look like

dont consider base cases for now, the key thing about strong induction is that you can almost entirely skip them

lets just focus on one key question:

lets say you have n squares

and you split one of the pieces into k squares

and the other piece is n - k squares

[zealous mango]

yes

[upper laurel]

what can we assume about these two pieces

something presumably related to n (n - 1) / 2

that will ensure our score will add up to n (n - 1) / 2 regardless of k?

[zealous mango]

uh

they add to n

idk

[upper laurel]

we already know taht, we dont need to assume that

[zealous mango]

uhhh

[upper laurel]

remember, we are assuming something

assumptions mean we need to just make a claim up

make a claim

up

think about the pattern you've been seeing for small cases

what kind of bold claim can help us out?

[zealous mango]

is there some kind of recursive function

cause like

[upper laurel]

we already know this function or whatever would presumably be recursive

the proof will ignore this

an inductive proof is already recursive, sure if you want you can try and predict how this function will act, but the proof doesnt need that kind of machinery to prove the statement

[zealous mango]

hmm

idek

[jagged tide]

@zealous mango do you have the answer key for this question?

[zealous mango]

no

[upper laurel]

lets try a few smaller cases so we're not staring at variables

[zealous mango]

ok

[jagged tide]

The source?

[zealous mango]

hw

[upper laurel]

if we have 1 square, itll always be 1 square

[zealous mango]

ye

[upper laurel]

if we have 2 squares, whats the only score you can get?

[zealous mango]

1

[upper laurel]

so 1 square is score 0
and 2 squares is score 1

3 squares is?

[zealous mango]

3

[upper laurel]

4 squares?

[zealous mango]

2x2 or 4x1

2x2 is 6 4x1 is also 6 nvm

[upper laurel]

???

you know you dont need to break squares apart in a grid right

you can see in the question itself what that looks like

[zealous mango]

oh yeah

wait is there like triangle numbers involved

[upper laurel]

I thought you said you noticed a pattern

you only seeing it now?

[zealous mango]

oh i was doint the whole grid

not just focusing on the 1x something

i saw a different pattern smh 😭

[upper laurel]

anyways its not always possible to break the squares apart into two pieces, where one square has k pieces and the other square has n - k pieces
its not guaranteed that k^2 + (n - k)^2 = n^2

however the 20 x 21 chocolate bar is presumably made out of 420 squares, so thats what it means by breaking them apart

[zealous mango]

yes

[upper laurel]

so we can assume that its just a number 420, that we're breaking apart into pieces like 419 + 1 or something like that

alr so given that
1 square is score 0
2 squares is score 1
3 squares is score 3
does it look like any way you split 4 squares will be score 6?

[zealous mango]

yes?

[upper laurel]

have you tried it?

[zealous mango]

yes

[upper laurel]

score 6 then

[zealous mango]

wdym

[upper laurel]

brb

[zealous mango]

i tried all ways from 1x1 to 4x4

kk

[upper laurel]

have you also tried 3 + 1 for 4 squares btw

[zealous mango]

yes

[upper laurel]

ok making sure, Ill brb

[zealous mango]

3+3

kk

btw the pattern i found was after breaking off a piece, you can then add the number of ways to break those pieces individually, which is recursive, so that waas pretty dumb of me, it was pretty obvious

also i suspected something to do with traingular numbers but wasn't sure lol

[upper laurel]

Im not sure what you mean by this, the problem is asking for the total point amount instead of the number of possible pieces

[zealous mango]

wait i meant the points you get from breaking them individually

oops

[upper laurel]

oh I see

yes thats important

until you tell me what the inductive hypothesis is though, we cant really see this in action

[zealous mango]

isn't it just the (a*b-1 ) triangular number?

[upper laurel]

be more specific

[zealous mango]

wait brb

[upper laurel]

alr

[upper laurel]

oh its important for the induction hypothesis that we write this recursive formula down, now that youre mentioning it

[zealous mango]

im back

uh

[zealous mango]

(a*a-1)/2

[upper laurel]

first, the parentheses should be a * (a - 1) / 2

[zealous mango]

oops mb

[upper laurel]

second, are you going to have a be the number of squares? why not n?

[zealous mango]

sure ig

(n* (n-1))/2

[upper laurel]

alr, then what will we be using to represent the number of points we get

[zealous mango]

x

[upper laurel]

like for the function n (n - 1) / 2

[zealous mango]

f(x)

no idea

[upper laurel]

choose a letter

then we'll use it

[zealous mango]

p

for points

[upper laurel]

makes sense

so what we are setting out to prove is that $p_n=\frac{n(n-1)}2$, right

[wraith dagger]

mtt

[zealous mango]

yes

[upper laurel]

now to make up the hypothesis, I'll need a more specific formula, so that we can use it

[upper laurel]

try describing this adding behavior in more detail

[zealous mango]

so more riggourously

[upper laurel]

rigorously

[zealous mango]

ok

pn=pn-k+pk

i think?

$p_n=p_n-k+p_n

$p_n=p_n-k+p_n$

[wraith dagger]

Spectre

[zealous mango]

$p_n=p_(n-k)+p_n

$p_n=p_(n-k)+p_n$

[wraith dagger]

Spectre

[zealous mango]

what the

$p_n=p_(n-k)+p_n$

[wraith dagger]

Spectre

[upper laurel]

as a reminder, what youre trying right now is "latex"

this is not a language that will automatically format the things for you

[zealous mango]

oh

ok

[upper laurel]

you will have to instruct it to place everything in the subscript

[zealous mango]

how

[upper laurel]

texit is just a discord bot to render in latex for you

but if you want to skip typing it in manually,

you can go on https://desmos.com/calculator and type it in there

then copy-paste it here

desmos handles the math you type as latex, so copy-pasting it will already give you good latex to get the bot to render

for example I type this here

then copy-paste it: \frac{n\left(n-1\right)}{2}

,,\frac{n\left(n-1\right)}{2}

[wraith dagger]

mtt

[upper laurel]

then the bot renders it

[zealous mango]

wow

[upper laurel]

to tell the bot to do this, you can type ,, before it as a bot command

[zealous mango]

\frac{n\left(n-2\right)}{123}

,,\frac{n\left(n-2\right)}{123}

[wraith dagger]

Spectre