help-13

[zealous mango]

oh nice

anyways

[upper laurel]

this rule is forgetting something, but youre on the right track

keep trying to type it

dont forget that you get points from splitting itself instead of just the points of each piece

[zealous mango]

wait wdym

i just randomly tried smth

lol

wait i thiknk ik

pn=pn-k+pk+n

right?

[upper laurel]

pn = p(n-k) + pk + (youre forgetting something)

[zealous mango]

n?

[upper laurel]

when you split a piece into k and n-k

[zealous mango]

k

k

k*1

[upper laurel]

do you really only get n points?

[zealous mango]

wait its kn

k*n

mb

[upper laurel]

its not k * n points

what is it specifically

[zealous mango]

k*(n-k)

[upper laurel]

yep

,,p_n=p_{n-k}+p_k+k(n-k)

[zealous mango]

smh my brain so slow rn

[upper laurel]

dw about it

[wraith dagger]

mtt

[upper laurel]

now this we know must be true, so by itself it cant be our inductive hypothesis

we also have a claim that $p_n=\frac{n(n-1)}2$

[wraith dagger]

mtt

[zealous mango]

so is it p(n-k)+pk+k(n-k)=n(n-1)/2

[upper laurel]

keep in mind thats assuming this

remember, we have to prove it, not assume it

look on the other side

you know, the proof is only for a specific n

technically, we dont have a formula for pk and p(n-k), since theyre both not pn

would you like to assume something about pk and p(n-k) though?

something that should make sense

[zealous mango]

wdym

[upper laurel]

do you think this is true for all n?

[zealous mango]

not all

its false for 0?

or 1

[upper laurel]

0 isnt a valid number of pieces spectre

get that outta here

[zealous mango]

wait no

oops

im tihnking completely conceptually

[upper laurel]

and even then 0 has 0 * (0 - 1) / 2 = 0 points anyway

[zealous mango]

yes

1

[upper laurel]

lets be bold about this

also not true

for 1 piece, you never got to split it

so you have 0 points

makes sense

[zealous mango]

so is it just non integers

[upper laurel]

non-integers arent valid either

[zealous mango]

huh

[upper laurel]

we are working with a non-negative number of integer squares for tony's chocolate bar

if you think theres a counterexample, youll need to name a non-negative integer

[zealous mango]

so its true?

[upper laurel]

it might be, you should try assuming it

[zealous mango]

so...

theres contradiction involved?

uhhh

[upper laurel]

there'd only be contradiction involved if we find one

we havent found one yet

we have this for instance

[zealous mango]

how do we prove tho

[upper laurel]

this is the formula that keeps track of the points after a chocolate bar is split, right

[zealous mango]

ye

[upper laurel]

its saying its the same number of points as pn, regardless of the value of k, right

[zealous mango]

yes

[upper laurel]

now we can combine this with the other assumption that maybe this formula works for all n

[zealous mango]

wait so we can base off as any value of k

[upper laurel]

I dont know what you exactly mean by that

[zealous mango]

nvm

[zealous mango]

so pn-k+pk+k(n-k)=n(n-1)/2
then algebraic manipulation?

[upper laurel]

again, no

lets remember how a proof works ok?

[zealous mango]

ok

[upper laurel]

a proof has a beginning and an end

what do you begin with in a proof?

[zealous mango]

assumption

claim

[upper laurel]

and what do you end in a proof?

[zealous mango]

answer?

[upper laurel]

wdym

[zealous mango]

qed

or do you mean last step

[upper laurel]

sure we can go with QED whatever

the last step would be the claim, because we proved it

[zealous mango]

yes

[upper laurel]

then you can put the QED thing at the end

now what is the claim?

[zealous mango]

ok

that pn=(n(n-1))/2

[upper laurel]

for a given n or for all n?

[zealous mango]

all

[upper laurel]

alr

now to do this, we'll need induction

induction lets us prove less than before and still get the same result

in this case, the claim we'll try to prove with induction is that pn = n (n - 1) / 2 only works for a specific n

[zealous mango]

wait why

[upper laurel]

thats a good question actually, let me be extra sure on that because that might be wrong

[zealous mango]

kk

[upper laurel]

oh alr this is how we phrase it

yea the claim is to prove it for all n

we'll prove the claim for all n using induction, which will assume a smaller version of this claim

[zealous mango]

yes

[upper laurel]

now for this smaller version, we'll use m

0 < m < n

[upper laurel]

also keep this formula in mind

since k and n - k can be any number 0 < k < n and 0 < n - k < n

just like the m here

[zealous mango]

ok

[upper laurel]

knowing we're doing induction, what assumption should we make?

[zealous mango]

wym

[upper laurel]

we will use this assumption on the formula and hopefully prove that it is true

whoops, Im asking you for the assumption we need

the inductive hypothesis

[zealous mango]

oh

isn't it just that pn=(n(n-1))/2

[upper laurel]

do I need to repeat myself

[upper laurel]

why wouldnt this work

[zealous mango]

its an assumption?

[upper laurel]

nope

pn = n(n - 1)/2

didnt we say this was something else earlier in the proof?

what was it called?

[zealous mango]

triangle number

[upper laurel]

nope

starts with c

5 letters...

[zealous mango]

ermm

[upper laurel]

there arent very many 5-letters words that start with c and that we've just been recently talking about

try one at least

[zealous mango]

chain

[upper laurel]

spectre

this is our claim

do you know what a claim is

[zealous mango]

yes

mbmb

[upper laurel]

no you dont

youve made the same mistake four times now

when you prove something, you cant just assume the claim is true

because then your proof would already be over

we're trying to prove the claim is true

[zealous mango]

isnt that just the inductive hypothesis

[upper laurel]

nope

lets do this one step at a time

[zealous mango]

i mean part of it

[upper laurel]

its certainly part of it, but youll need to make the claim weaker

[zealous mango]

ok

[upper laurel]

so far the claim is that this works for all n

since thats what we're trying to prove, assuming that its already true isnt going to rigorously do us any favors

think about the m I told you earlier

shouldnt that be part of this hypothesis?

I told it to you

[zealous mango]

sure?

yes?

[upper laurel]

we'll try it

m here is like n, but less varying

m is an integer with 0 < m < n

[zealous mango]

ok

[upper laurel]

given the claim we're proving is about pn

what would the inductive hypothesis try and be instead?

[zealous mango]

m?

[upper laurel]

thats correct, but give more detail

[zealous mango]

the points from m

[upper laurel]

the inductive hypothesis is about pm, for any 0 < m < n

what is the hypothesis assuming for us

[zealous mango]

pm=m(m-1)/2

[upper laurel]

thats exactly it

the claim is pn = n(n - 1)/2 for all n > 0

[zealous mango]

oh.

[upper laurel]

the hypothesis is pm = m(m - 1)/2 for all 0 < m < n

[upper laurel]

think about it, if youre just going to assume this is already true, then your proof would be already done

so we cant just start there, you still dont believe that this is true

[upper laurel]

you couldve just started here

for our n, maybe its just only true for all m less than n

[zealous mango]

yes

[upper laurel]

so with the correct assumptions, lets look at the formula again

we have the assumption that pm = m(m - 1)/2 if m < n

[zealous mango]

ok

[upper laurel]

we need to prove that pn is n(n - 1)/2

[zealous mango]

how

[upper laurel]

are you just going to have me do every step of the proof for you

we've gotten through the hard part already, which was choosing the correct assumption

now we're in the part where we use what we have

[zealous mango]

so

we substitute

right?

n*(n-1)/2=(n-k)x(n-k-1)+k(k-1)+k(n-k)

[upper laurel]

youve made the same mistake again

we need to prove that pn is the same thing as n(n - 1)/2

that means you dont get to write it in the equation yet

do you know why this is actually a problem?

[zealous mango]

oh

right

because you dont know its true yet

[upper laurel]

man youre going to be like a conspiracy theorist at this point

[upper laurel]

youve also forgotten about the /2s

[zealous mango]

but still

i thought you had to use the hypothesis in the inductive step

or was i taught wrong

[upper laurel]

spectre

m and n are different letters

right

[zealous mango]

ye

[upper laurel]

on the right, youre using pk and p(n-k)

when you substitute in the formula, which claim are you using

pm?

or pn?

[zealous mango]

pm

[upper laurel]

on the left, youre using pn

when you substitute in the formula,

is it pm or pn youre using?

[zealous mango]

pn

[upper laurel]

and I just didnt want you to use pn

[uncut blaze]

can i use this channel i have a small question

[upper laurel]

the inductive hypothesis of $p_m=\frac{m(m-1)}2$ for all $m<n$ is still OK

[wraith dagger]

mtt

[upper laurel]

we didnt say the hypothesis was wrong

[upper laurel]

this channel is for spectre, go ask the question in an open one like #help-19 then someone can go help you there

[uncut blaze]

sorry my bad i already find a channel..

[upper laurel]

thats alr, I just named an example channel to use

@zealous mango back to this

[zealous mango]

P(n)=k(n−k)+k(k−1)/2​+(n−k)(n−k−1)​/2
why can't we just do that

[upper laurel]

youve typed it wrong again

[zealous mango]

oops

[upper laurel]

wait what is that

−

you cant type that with a normal keyboard

where did you get this from

[upper laurel]

that

wheres that image from

[zealous mango]

(chatgpt)

[upper laurel]

alr

[zealous mango]

idk how to write it so like i jsutaasked

[upper laurel]

what happened to the desmos latex thing from earlier?

[zealous mango]

um

idk how to write subscript but... i kinda forgot

[upper laurel]

bruh

do you know how to write them in google docs?

[zealous mango]

no. i dont usually write math online

usually i jsut do it on paper

[upper laurel]

google docs isnt for math bro

for subscript press Ctrl + _

[zealous mango]

gtg

im dum sry

.close

[cedar kiln]

Channel closed

Closed by @zealous mango

Use .reopen if this was a mistake.

[upper laurel]

bruh

so youre giving up

[upper laurel]

what you do afterwards is simplify the right side

then you get pn = n(n-1)/2

and thats how you prove it

we also didnt explain where the inductive hypothesis actually comes from, which wouldve made this make more sense

[upper laurel]

theres also why this works for induction

we didnt get to that either, this is only really proving it for a specific n instead of a general n

as soon as you get here, you then have to say something to complete the proof

you need to know where this comes from to understand the proof

[cedar kiln]

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[warm flume]

https://youtu.be/tuVd355R-OQ
6:24
what is the mental math here? Why did he divide instead of multiplying and what's this process called

[upper laurel]

,,\frac56\cdot\frac85=\frac{5\cdot8}{6\cdot5}=\frac{{\color{yellow}5\cdot}8}{{\color{yellow}5\cdot}6}=\frac86

[wraith dagger]

mtt

[raven horizon]

test

[upper laurel]

in general, youre multiplying by 5 on the top and on the bottom, so you can reduce the fraction of this factor of 5

[cedar kiln]

@warm flume Has your question been resolved?

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

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[dense shard]

Can someone help me with my performance task? (I can solve it, I just need help with creating the problem)

[slender ginkgo]

could u post the problem here

[frosty coral]

Which problem?

[dense shard]

So we’re (me and my partner) creating a problem formulation on quadratic functions. But I cant make it in to a sentence

In a basketball game, Jonathan throws the ball towards Jaidon by bouncing it. What is the minimum height of the ball when it bounced represented in y=x2-8x+24

Its maximum*

[opal hinge]

It's quite unreal how acceleration is 1 here

[dense shard]

so what were tryna find is the graph of the parabola but its going upward so thats not how its supposed to bounce and were tryna find what were supposed to find like max height of ball when it was bouncing from J1 to j2

[opal hinge]

And also the leading coefficient have to be negative to have maximum

[dense shard]

I dont understand🥲

So x2 has to be -x2?

[frosty coral]

the parabola from the expresison you gave us

[opal hinge]

Well yeah

[dense shard]

Oh i forgot how to solve when ax2 is negative

[opal hinge]

It's just the same

[frosty coral]

yes

[dense shard]

OHH

oki ill come back

[opal hinge]

Maximums when x=-b/2a

Also it should be -4.9x^2 since gravity cause object to accelerate at 9.8 m/s^2 downward

[dense shard]

Im confused

[frosty coral]

Why are you multiplying the parenthesis by 24?

on the second line

[opal hinge]

when find maximum we complete the square like this P(x)=-(ax+b)^2+c

so that maximum is c

[dense shard]

Tying to find k

[dense shard]

Oh so I have to put the negative of x2 outside?

[opal hinge]

yess

(ax+b)^2>=0 so -(ax+b)^2=<0

when want that "=<" sign so we can get maximum

[dense shard]

So its gonna be like this?

[opal hinge]

,rccw

[dense shard]

Whats that app?

[frosty coral]

google

[dense shard]

What website?

[frosty coral]

I just typed the equation on google

[opal hinge]

y=-x2-8x+24 this is the equation right?

[dense shard]

Yepp

[opal hinge]

you can try desmos

[frosty coral]

it plots the curve

[opal hinge]

so y=-(x^2+8x)+24

[opal hinge]

Idk why you multiple by 24 here

[frosty coral]

that's what I've said

[dense shard]

Why i remove 24 in the parenthesis?

[frosty coral]

Look, you should have -x^2-8x-16+b

where -16+b =24

[dense shard]

I dont understand🥲

[frosty coral]

so b would be 40

[frosty coral]

What aren't you understanding?

[opal hinge]

,tex .cts

what

[dense shard]

Whats your question on 24?

[frosty coral]

y=-x^2-8x+24=-x^2-8x-16+40=-(x+4)^2+40

[dense shard]

Wait where did 40 come from

[frosty coral]

from completing the square

completing the square is the name of the method

[opal hinge]

how can I print the CTS formulas here

,tex .cts

[wraith dagger]

Alexis_Fx

[dense shard]

I forgot that method

This is what desmos got and It’s right ball bounces like that right?

But its too narrow (?) or something

[upper laurel]

for reference, balls usually bounce like this

smaller and smaller parabolas

[opal hinge]

first of all, you need negative leading coefficient to make it look like a bouncing ball

[dense shard]

Okay

[opal hinge]

And also, the acceleration usually 9.8 or 10 m/s^2

[dense shard]

Oh wait im dumb

[frosty coral]

this should make it obvious

[dense shard]

Wait

[frosty coral]

@dense shard , do yo understand now?

[dense shard]

Im trying to

[frosty coral]

I added and subtracted 16

in order to complete the square

[frosty coral]

It doesn't change the equation

[frosty coral]

if you multiply the expression by 24, you'll get -24x^2-192+384-16

[dense shard]

So do I need to change the equation so it would look like that?

[frosty coral]

First, you need to change the sign of the squared term

and not multiply by 24

[upper laurel]

you dont necessarily have to, thats just a realism kind of issue

[dense shard]

Wait can I change my problem to a paper airplane?

Like im going to throw a paper airplane

It goes up then goes down

This is what ure doing too right Nilson?

These are the steps Ive been following

[frosty coral]

yes, but with (x+h)^2

[dense shard]

Whats that

[frosty coral]

in your slides, y=a(x-h)^2+k

I'm using y=a(x+h)^2+k

[dense shard]

Yep

But the only difference is + and -

[frosty coral]

both methods will solve your problem

[dense shard]

Oh

How do I solve it using the method in the slides?

[frosty coral]

you need to expand the expression

that's in the parenthesis

[dense shard]

How

[dense shard]

Okay

[frosty coral]

then, a=-1, -2h=-8 and h^2+k=24

[dense shard]

Wait wait

The bouncing thingy I meant on my problem is like

Jonathan throws the ball on the ground and it bounces to Jaidon

Like in real basketball

[frosty coral]

yeah

[dense shard]

So this is right?

[frosty coral]

no

[dense shard]

Why

[frosty coral]

It will bouce like in this graph

[dense shard]

Wait the graph i was using is positive ax2

WAIT WHAT

[frosty coral]

it doesn't represent reallity of a bouncing ball

[dense shard]

How

[frosty coral]

because balls don't bounce like that

[dense shard]

Wait let me show u vid

[frosty coral]

Which video?

[dense shard]

Bounce like this

[frosty coral]

it's bouncing like this in the first half

till the middle point

than it bounces like that in the other half

[opal hinge]

keep in mind that ball is affected by air resistance , friction and it also rotate so I would lose a lot of energy

so it won't be a perfect parabola

[dense shard]

So the problem is..?

[opal hinge]

if we do with no air resistance , friction, and rotation, It's a parabola like -4.9x^2 + v_0x + xo

[dense shard]

We don’t need to make it like too realistic

Just a problem where we can see parabolas then find max/min value by using h= -b/2a or k=4ac-b2/4a

[opal hinge]

then any parabola with leading coefficient negative is fine

make sure that

maximum happen when x>0

[dense shard]

Yep

[opal hinge]

since time can't be negative

and also maximun value >0

[dense shard]

Yea

So do I need to change anything with my problem?

[dense shard]

This

[opal hinge]

-x^2+8x+24?

[dense shard]

But the people are supposed to be bouncing like this

[dense shard]

The ball is*

[opal hinge]

the ball bounce twice in the video

you only have one parabola

I mean

[dense shard]

Yea but in a real basketball game you would only need to pass it once

I only showed it 2 times because it wouldnt cut

[opal hinge]

in your equation the ball bounce like the red curve

here

What do you want? it bounce again when it hit the ground ?

[dense shard]

Yes

[dense shard]

If its like that it would be an overhead pass

[opal hinge]

well you would need 1 more parabola for that

[dense shard]

Oh

So I should just make the problem to an overhead pass instead of a bounce pass?

[opal hinge]

yeah

you would need 2 parabola like this

[dense shard]

the function Ill use is -x2 + 8x + 24 to make this parabola and ill change the problem from a bounce pass to an overhead pass

Okay

So no more problems after that or do I need to make my problem clearer so my teacher would understand it

[opal hinge]

It's fine ig

[dense shard]

I need to remove the negative from x2 right?

[opal hinge]

it's kinda unreal when accelaration here is 2 m/s^2

[dense shard]

Remove from parenthesis

Oh

[opal hinge]

y= -x2 + 8x + 24

[dense shard]

Wait whered u get 2

[opal hinge]

this is okay

[opal hinge]

well, if you have learnt projectile motion

[dense shard]

I havent

[opal hinge]

$y=\frac{a\cdot t^2}{2} + v_0\cdot t + y_0$

[wraith dagger]

Alexis_Fx

[opal hinge]

this's formula for any object get projected

a is accelaration ofc

[dense shard]

Ohh

[opal hinge]

a usually -9.8 m/s^2

it's caused by gravity

[dense shard]

Wait my given is -x2 - 8x + 24

[dense shard]

Yess

[opal hinge]

that's why I called it unreal

But your teacher won't care i think

[dense shard]

Ye shes more of a teacher for math

[opal hinge]

that's fine, 1 problem tho

[dense shard]

Ye

[opal hinge]

when it's on the ground in the first time, x<0 here

Idk if your teacher would accept that

[dense shard]

Why would it be in the ground tho

[opal hinge]

you can't say the ball is throwed from the ground at -2 second

because y=0

[dense shard]

It would be in the player’s hand (behind his head) then throws to the orher player

[opal hinge]

you should say that

in your problem

[dense shard]

In the problem?

Alright

So no more problems after that ?

[opal hinge]

well yeah

I think that's it

[dense shard]

oki thankss

Wait how do i solve it…

[opal hinge]

completing the square

[dense shard]

How do i do that

[opal hinge]

you should watch youtube at this point

It will explain it better than me

[cedar kiln]

@dense shard Has your question been resolved?

[cedar kiln]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[cedar kiln]

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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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• Be polite and have a nice day!

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[proper spire]

I want help regarding this problem.. It seems a bit complicated..

[slender ginkgo]

interesting

[zealous mango]

that is...

[slender ginkgo]

${a_{31} = a_{25} + 6d}$

wait gimme a sec

[proper spire]

5d??

[wraith dagger]

k

[tropic oxide]

a_25 and a_31 are six steps apart, not five

[slender ginkgo]

${f(a_{31}) = f(a_{25} + 6d)}$

[tropic oxide]

k is correct here

[wraith dagger]

k

[tropic oxide]

first off though

@proper spire do you understand the meaning of f(x+y) = f(x)f(y)

[proper spire]

Yes it's a the func when you take the variable (x+y).. And the func will be equal to the multiplication of the output of the functions which have x and y as variables

[tropic oxide]

ok thats the literal meaning.

let me say this differently

there's a special name for the class of functions which satisfy this.

do you know it?

[proper spire]

Nope pls educate me

[slender ginkgo]

think of the exponent rules

[proper spire]

Wait.. Functions can work like exponents??? I don't think so..

[slender ginkgo]

${e^{x+y} = e^x \cdot e^y}$

[wraith dagger]

k

[tropic oxide]

try $f(x) = 10^x$, you'll find that this function satisfies $f(x+y)=f(x)f(y)$.

[wraith dagger]

Ann

[tropic oxide]

and in fact nothing special about the number 10

all exponential functions of the form f(x) = a^x will satisfy this

[slender ginkgo]

i think we can use f(x+y) = f(x)f(y) directly here tho

to get f(d)

[raven shard]

i wouldn't do this question with this in mind

$a_i = a_1 + (i - 1)d \Longrightarrow f(a_i) = f(a_1 + (i - 1)d) = f(a_1)f(d)^{i - 1}$

[wraith dagger]

Mqnic_

[raven shard]

is all you need

[proper spire]

So uhh whats y in this case..??

[slender ginkgo]

look at the rhs of this

we also know that ${f(a_{31}) = 64f(a_{25})}$

[wraith dagger]

k

[slender ginkgo]

can u find f(d)

[slender ginkgo]

once we find f(d), this will be useful in finding the gp sum to get f(a_1)

[proper spire]

Now can jus substitute f(a_31) for 64f(a_25)

[slender ginkgo]

yes

so

${64f(a_{25}) = f(a_{25})f(6d)}$

[wraith dagger]

k

[slender ginkgo]

Notice when ${n \in \mathbb{N}}$,
[ f(n \cdot d) = f(\underbrace{d + d+ \dotsm + d}{n \text{ times}}) = \underbrace{f(d)f(d)\dotsm f(d)}{n\text{ tunes}} = [f(d)]^n]
As Mqnic has stated.

[wraith dagger]

k

[slender ginkgo]

so

[slender ginkgo]

working from this

can u find f(d)

[proper spire]

f(d)^6 = 64

???

[slender ginkgo]

so whats f(d)

remember

[proper spire]

It's 2

[slender ginkgo]

cool

so

lets look at ${\sum_{i=1}^{50} f(a_i)}$ now

[wraith dagger]

k

[slender ginkgo]

from the formula for ap

we know that ${a_i = a_1 + (i-1)d}$. So ${f(a_i) = f(a_1 + (i-1)d)}$

[proper spire]

Yes yes..

[wraith dagger]

k

[slender ginkgo]

can u see how the sum is a geometric sum now

[proper spire]

How did arithmetic turn to geometric???

[slender ginkgo]

can u simplify the rhs

[slender ginkgo]

^

[proper spire]

Yes correct.. It's a new property for me so it kinda didn't click me

[slender ginkgo]

ok

so can u see that

[ \sum_{i=1}^{50} f(a_1) = \sum_{i=1}^{50} f(a_1)(d)^{i-1}]

[wraith dagger]

k

[proper spire]

F(d) is 6 so wait 6^i-1

[slender ginkgo]

howis f(d) 6??

[proper spire]

Sorry 2

2^i-1

I'm srry

[slender ginkgo]

cool

so it becomes

[proper spire]

1 = 2^i-1

I think

[slender ginkgo]

[ \frac{f(a_1)(1-2^{50})}{1-2} = 3(2^{25}+1)]

[wraith dagger]

k

[proper spire]

How'd u get the RHS

I get how u got the LHS

[slender ginkgo]

lhs by this

rhs is given

[proper spire]

Yeee

Okayyy

Thanks for helpp

[slender ginkgo]

u could do the rest, ya?

[proper spire]

Yeee

I'm closing this now

It's .close ryt?

.close

[cedar kiln]

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[cedar kiln]

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[prisma pollen]

shouldnt we use the max factors for gcd and min factors for lcm?

[tropic oxide]

no, why would we do that

[crimson delta]

it might help if you did an example

[prisma pollen]

say that we have 2 numbers m= 2^2 * 3 * 7 and n = 2^3 * 3^2 * 11\

lets say we want to find the gcd

so for each factors present in both numbers wont we take the factor which has the least power?

[crimson delta]

which is what the formula is saying

[tropic oxide]

yes that's what min() means and that's what is said

[prisma pollen]

wtf

shouldnt it be called max()?

cuz we're taking the bigger power

[tropic oxide]

no

[tropic oxide]

we are taking the smaller

and you yourself said least

[prisma pollen]

nvm i got confused because gcd has "greater" in it

.close

[cedar kiln]

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[gloomy pewter]

What is X equal to if X = cos(X)?

[wicked mantle]

,w sketch cos(x) - x = 0

[wraith dagger]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by skyandnight

[tranquil oracle]

We can probably show you how to calculate decimal digits of this number, but I don't think there's any closed form.

[opal pike]

any math exam tips please

I am getting low scores in math recently

[wicked mantle]

Plz claim ur own channel

#❓how-to-get-help

[gloomy pewter]

Why doesn't it exist?

[wicked mantle]

It does exist

[wicked mantle]

Look at this sketch

[tranquil oracle]

I think Pentalogue was asking why a closed form might not exist

[wicked mantle]

Oh

Mb

[gloomy pewter]

$x = \cos(x)$
$x = \arccos(x)$

[wraith dagger]

Pentalogue

[tranquil oracle]

You can consider that there are only countably infinite many "closed forms" but uncountably many reals, so it motivates the idea that most real numbers don't have closed forms.

[gloomy pewter]

No, I asked what X is if X = cos(X), I don't need a decimal approximation, but a formal form of the number

[tranquil oracle]

You already have one form of it - "(real) fixed point of cos"

[gloomy pewter]

You got it right, I need a fixed point of the cosine

Probably because the function is periodic, there are problems with finding a fixed point of the cosine

[flint cape]

Do you know what we mean by "closed form"? /gen

[crimson delta]

the issue is not with being periodic

[gloomy pewter]

No, explain please

[flint cape]

[then admittedly you can't say "no" necessarily to this if you didn't know lol]

[gloomy pewter]

Okay

[flint cape]

An expression is in closed form if it's formed only using "basic" functions and function compositions

Typically, we mean these to be finitely many

By "basic", we generally allow basic arithmetic, integer powers, nth roots, exponential functions, logs and trig

[gloomy pewter]

Okay, I get it, next

[flint cape]

Now, technically, there is a closed form in this specific instance: the solution to x = cos(x) is called the Dottie Number, and according to https://mathworld.wolfram.com/DottieNumber.html it is given as

But this is making use of another function $I_z^{-1}(a,b)$, which is defined using a bunch of integrals

[wraith dagger]

Waes (Wires)

[gloomy pewter]

Beta function?

[flint cape]

Generally we don't say "we can write it using some integral" to mean it's closed, so I disagree with how the above link's presented it

cf. the Gamma function

[fossil dawn]

integration is not closed imo

see: the error function too

[flint cape]

In general no, you're right

[gloomy pewter]

How to implement this function?

[flint cape]

My guess is by this they mean that the relevant integrand (the thing being integrated) is in closed form; this was on Wikipedia for instance (where D is the Dottie Number)

[flint cape]

This function's an integral-based function. i.e. it uses integrals

It's unclear what you mean by "implement" here

[gloomy pewter]

To implement is to recreate (perform calculations in graphical form)

[crimson delta]

if you want to find a decimal approximation of the number there are much easier ways than implementing that formula

[flint cape]

The obvious choice is brute iteration

$x_{n+1} = \cos x_n$

[wraith dagger]

Waes (Wires)

[gloomy pewter]

Yes, I understand, but in general I need both a decimal approximation and a calculation formula

[flint cape]

And just spam that shit

[gloomy pewter]

Interesting...

[crimson delta]

who cares about a formula if you have an algorithm to compute it to arbitrary precision

[tranquil oracle]

a better choice is probably newton iteration

[tranquil oracle]

the convergence of this iteration is slow

[flint cape]

-# "want" (хочу́) not "need" (нужда́юсь), we've discussed this 👍

[flint cape]

yes

You see how it's converging there?

Like, from this point, I can gather that it's approx. 0.74 to 2dp

[gloomy pewter]

Yes, I see that the members of the sequence converge with the root of the equation X = cos(X)

[gloomy pewter]

I didn't know that a recursive formula would work for approximation

[flint cape]

This might be a lost-in-translation thing; I think he did mean by "calculation formula" this algorithm (or similar algorithms)

[tranquil oracle]

another approach is substituting the terms into one of these "sequence acceleration techniques"
https://en.wikipedia.org/wiki/Aitken's_delta-squared_process

[gloomy pewter]

Do you know Russian language?

[flint cape]

-# those have been banned since 2067, what are you doing with them in Night City!? /s

[flint cape]

Not substantially

But I can tell there are language barriers in place here

[gloomy pewter]

Well, that's clear

[gloomy pewter]

Yes, there are language barriers!

Thanks for the answer! I am satisfied with it.

I am satisfied with the solution to the problem

[tranquil oracle]

.close if you think it's solved

[gloomy pewter]

And using iterations, can we find a fixed point of any function?

[tranquil oracle]

not necessarily

[wicked mantle]

[tranquil oracle]

consider f(x)=2x+3

[flint cape]

Not generally; you can at this point probably Google this to see what's going on

Your search term being "fixed point iteration method"

[gloomy pewter]

Yes, but I would like to get a reliable answer from helpers

[flint cape]

Generally solutions, if they can be found, fall into one of two camps:

They either spiral inwards like a "cobweb", or they step into a solution like a "staircase"

[gloomy pewter]

Okay, I get it

[flint cape]

You can imagine though that if the iteration method for instance steps outwards, then it's not going to get to the fixed point

[gloomy pewter]

And the fixed point must always be a real number?

[flint cape]

I mean if you can explain why you were thinking it wouldn't be...?

[gloomy pewter]

What do you mean? That is, if the step turns out to be further than necessary?

[gloomy pewter]

Anyway, thank you a lot for your answer

[flint cape]

Here's an example of what I meant

You can see that x0, x1, x2, x3, ... is approaching the fixed point there

But you can probably see that if I extend this graph to the left, there should likely be another fixed point there

But if I pick a starting point (x0) anywhere after that fixed point, I'm going to climb away from that FP

[gloomy pewter]

From the graph, we can assume that the fixed point will be on the left

[flint cape]

That's what I said here

[there's already one fixed point up top, which this "staircase" is hunting]

[gloomy pewter]

Yes, and with each step there is a greater and greater approach to this fixed point

[flint cape]

So the main crux of the method is that your starting point determines which fixed point, if at all, you're going to converge to

https://www.desmos.com/calculator/223ygogmaj Have a play with this graph, what x_1 (the starting point) and f(x) are

Here's a good example where the starting point doesn't lead to a fp

In fact if it converges at all, this method only converges to the fp at the origin

It can't ever locate the fp at (1,3)

[That is, unless you happen to start with x = 1 or x = -1; but at that point you would have likely already known there to be an fp right there before even testing this

This one is an example of there being only one fp yet being incalculable anywhere via this method

You could probably deduce that -1 leads to a positive infinity and -2 leads to a negative infinity, so the fp must be somewhere between -2 and -1

[cedar kiln]

@gloomy pewter Has your question been resolved?

[cedar kiln]

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[silk pilot]

|x-1| + |x-a| = 1 - a
Find a, such that the set of solutions contains 3 whole numbers.

[opal hinge]

exactly 3?

or at least 3

[silk pilot]

the set x must have 3 numbers wihtout a floating point

[paper edge]

That doesn't answer the question he asked

Does that mean exactly 3 or ≥3?

[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.

[silk pilot]

sorry, let me retry

I myself am confused

The set of solutions must contain 3 integers

I figured out that 1 - a >= 0 since the left-hand side is a sum of absolute values

[fossil dawn]

the question that was asked of you was: is it exactly 3, or at least 3?
from the wording it seems like you need exactly 3, but best to confirm

[silk pilot]

yes

oh my god

yes, exzcalty 3 integers

not whole numbers

[upper ruin]

Integers are whole numbers

Unless 0 is counted differently

[silk pilot]

but whole numbers are positive integers with maybe 0

[cerulean sail]

(negative integers)

[nova snow]

carrttbitttt

[upper ruin]

So whole = Z \ {0} ?

[lavish vine]

@silk pilot what does the graph look like ?

[silk pilot]

what graph

[lavish vine]

The graph of lhs

[cerulean sail]

Non-negative integers

[upper ruin]

Ohhh so it's the opposite?

[surreal cave]

whole is W = {z in Z | z >= 0 }

[silk pilot]

[upper ruin]

Never knew that difference, wow

[lavish vine]

I would recommend trying to graph the LHS for a few values of a

[silk pilot]

|x-1| + |x-a| = 1 - a
Find the value(s) of a, such that the set of solutions contains 3 integers
(the task)

[silk pilot]

with desmos?

[lavish vine]

You can use that or graph it yorself

The latter is preferred

[silk pilot]

this always confused me

"let's try some values"

what values

0 is probably the obvious choice

[lavish vine]

Like you can take a as 1

[silk pilot]

but after that?

[lavish vine]

2 3 any number

Now that you know how the graph looks like

You can see where you will get 3 solutions

[silk pilot]

so your recommendation is to graph it, and use the graph to solve it?

can the task be solved with numbers?

[lavish vine]

It can be but it can be slightly harder to see why it works

Here is the graph for some random A

For which value do we get 3 or more solutions

[silk pilot]

for a = 3

[lavish vine]

y=3 here

[silk pilot]

oh

[lavish vine]

But now we know what to do

[silk pilot]

...

[lavish vine]

Wait

[silk pilot]

oh my godc

[lavish vine]

I am back

Sorry for the delay

We know that all values of A less than 1 have 3 solutions not necessarily integers

[silk pilot]

so all values that are in (-inf; 1)?

[lavish vine]

But we need 3 whole numbers

[hollow temple]

Hmm.

[lavish vine]

This condition isn't satisfied for a few A in this interval

[silk pilot]

yeees...?

[lavish vine]

How can we ensure 3 integers values of x from this graph

[silk pilot]

what do you mean by "ensuring a value"?

[lavish vine]

I mean that we know our solutions are that horizontal line in the graph

[crimson sedge]

geometry without figure pmo

[lavish vine]

! occupied

[cedar kiln]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[silk pilot]

yes, so...?

[lavish vine]

Wait let us start from the beginning

[silk pilot]

good idea

[lavish vine]

What do we need here

[silk pilot]

this

[lavish vine]

Let us focus on a simpler goal

For which values of y does the equation of the form y= |x-a|+|x-b| have 3

Solutiona

Not necessarily integers

[hollow temple]

字母A六十四.

[lavish vine]

???

What do you mean

[silk pilot]

I'll try to graph it?

[lavish vine]

Yes you can graph it or try numerically if you want to

[silk pilot]

how would I do it numerically?

if a and b are parameters, that is set numbers, why would x be a set?

[lavish vine]

Basically you can break it into 3 piece wise functions

That should give you the answer

[silk pilot]

oh

so
y = |x - a| + |x + b|
and then we check x - a > 0, less then, and the same for x - b?

[silk pilot]

but still

[lavish vine]

,rotate

[lavish vine]

It will be this function

[silk pilot]

okay

[lavish vine]

Now assume y is less than a-b

Then can we have a solution

[silk pilot]

...

hold on

y is bound to a and b, right?

then your suggestion can be expanded to 2x-a-b < a-b and a + b - 2x < a - b?

[lavish vine]

?

The other 2 equations are greater than a-b

I would now suggest graphing it to see why this is the case

You can try it for any value of a and b

[silk pilot]

why are they greater?

I am sorry, I know that I probably won't understand this now, but still

[lavish vine]

A way to understand this can be that when we break it into piecewise functions the first function is increasing

So as x gets smaller it decreases

Now for the second interval it is constant

And the third interval it is a decreasing function

So it gets larger

Hence never goes below a-b

[silk pilot]

oh okay

you know what, I should probably take a break and try again later
thank you for your help

[lavish vine]

But this is specific to |x-a|+|x-b| since any other coefficient changes the piecewise function

[silk pilot]

.close

[cedar kiln]

Channel closed

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[cedar kiln]

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[obsidian ravine]

The figure shows a square and an inscribed circle. In the upper right corner of the square, we draw a rectangle with sides 10 and 5. We note that the rectangle lies completely outside the circle and that one corner of the rectangle lies on the circle. What is the radius of the circle?

[obsidian ravine]

I actually have no idea what to do here

[magic solar]

I’m pretty sure this is unsolvable

[obsidian ravine]

it's solvable

[opal hinge]

The key is symmetrical here, draw the same rectangles in three other corners

[obsidian ravine]

how would that help

[opal hinge]

Do you have the answer key

[obsidian ravine]

yea

the radius is 25

I dont understand how to get there

<@&286206848099549185>

[opal hinge]

Yeah, that's what i thought

Try to draw three other rectangles like I said

[tulip tinsel]

ah I get it

@obsidian ravine

do u just straight up want the answer or a hint

the hint he gave is good

[obsidian ravine]

for now a hint

if I don't manage to solve ill ping

[tulip tinsel]

yeah draw all four rectangles and see if you notice any way you narrow down on a value for r

the solution is actually pre elegant

[minor dust]

Can you find a triangle in this figure in terms of the 5x10 rectangle and the radius?

[obsidian ravine]

i cant solve this

can u give me the answer

[tulip tinsel]

you sure

?

[obsidian ravine]

yes

[tulip tinsel]

this is a really really good hint

if you can manage to find a right angle triangle with sides that are in terms of r, you can find an equation that relates all three sides and solve for r

cannot give a better hint than that without giving the solution

[obsidian ravine]

okay i'll try

[tulip tinsel]

if not I have the solution ready for you

will send it in spoilers cause I gtg

gl

[obsidian ravine]

thank you

[tulip tinsel]

oops

there you go

hopefully you didnt see that 😂

[obsidian ravine]

i didn't look

thanks

I understand it noqw

I didnt use answer key

thanks

.close

[cedar kiln]

Channel closed

Closed by @obsidian ravine

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[opal hinge]

You should have trust me when I said draw more rectangles lol

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[slender ginkgo]

Splendid

[cedar kiln]

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[sleek drum]

hey guys i was doing exponents and i needed some help with 2^2x - 5.2^x + 4 = 0

[sleek drum]

i know i have to do mid term splitting

[rare schooner]

it's a decimal in the middle right?

[sleek drum]

yes its decimle

[rare schooner]

what exactly are you supposed to do?

[sullen cipher]

Subbing

[sleek drum]

decimal*

[green coral]

no thats multiplication no?

[sullen cipher]

Oh

[sleek drum]

exponents im finding out x

[green coral]

wat

[sleek drum]

decimal

[green coral]

i dont think so

[sullen cipher]

U sure

[rare schooner]

you cant

[green coral]

contexually

[rare schooner]

it's an expression

nevermind

im stupid

[sleek drum]

;-;

T_T

[slender ginkgo]

$2^{2x} - 5.2^x + 4 = 0$

[wraith dagger]

k

[green coral]

it is an equation but yeah

[slender ginkgo]

This?

[rare schooner]

can you send the original problem?

[sleek drum]

its just this

i need x

[rare schooner]

actually?

are you sure the dot isnt in the middle?

[boreal grove]

2^x = t

[sleek drum]

yes iam sure

[boreal grove]

Then it's a quadratic

[sleek drum]

yes im taking 2^x as y

so i did

4y - 5y + 4 = 0

[rare schooner]

you cant do shite with the middle term

[sleek drum]

but i think thats wrong

[boreal grove]

I'll assume it's (2^x)^2 too and not 2^2x

[slender ginkgo]

Is it 5.2 or $5 \cdot 2$?

[wraith dagger]

k

[rare schooner]

$2^{2x} - 5*2^x + 4 = 0$

[wraith dagger]

The golden fish

[sleek drum]

uhh there are few examplesin the book similar to this and they are done via middle term

[green coral]

same thing no?
like 2^(2x)

[sleek drum]

its 5.2 trust

[rare schooner]

i am very much inclined to believe this is the case

[slender ginkgo]

Shi

Now it’s more complicated

[rare schooner]

you have the problem?

ss

[sleek drum]

in the book its 5.2

[rare schooner]

take a photo lol

[sleek drum]

alrighty

wait lemme grab my phone

[green coral]

give us some context , i feel there is a possibility that dot is also an alias for multiplication

[rare schooner]

These kind of problems are usually quadratics

[sleek drum]

it could be but i dont have context to give

the question states "solve the following equations" and then the equation i sent

im sending the picture my phone is turning on

[green coral]

ah ic so we can just formulate an answer considering it one , and then see where it takes us ?
it would be amazing if you have the answer key so we can be sure

[sleek drum]

oh yeah i can tell this

in the examples

[green coral]

if we find out the answer didnt match we can think it as an decimal then

[sleek drum]

there are 2 possible answers

[rare schooner]

yeah

[sleek drum]

and in the book the answer for this question is

[rare schooner]

then it shouldnt be a decimal

[sleek drum]

either 0 or 2

[rare schooner]

yeah

treat it as a multiplication sign

[green coral]

bingo

[sleek drum]

how do i differenciate between a multi and decimal btw

[rare schooner]

one should be in the middle

one should be at the bottom

[sleek drum]

so the book is just dumb?

[green coral]

usually the multiplication is used in the middle , however it depends on context too

[rare schooner]

but sometimes textbooks fuck it up for some reason

[sleek drum]

because the decimal is at the base level

[rare schooner]

yeah

[sleek drum]

or dot atp

[green coral]

nah in scalar multiplication while doing vectors some books have it in bottom

[rare schooner]

anyways lets try this problem correctly then

[green coral]

^

[rare schooner]

think of the possible things you can substitute

[sleek drum]

3i

the question is 3i

sorry for sideways

[rare schooner]

huuh

does appear that it's a decimal

[sleek drum]

right

i could send a picture of an example if that would help

[sullen cipher]

Unless ques was typed wrong

[sullen cipher]

Yes maybe?

[sleek drum]

okay wait

[rare schooner]

they wrote it wrongly

[boreal grove]

It's multiplication

[rare schooner]

I solved it perfectly fine with multiplication