help-0

[alpine sable]

Got it

[pale kestrel]

so we will say k are vertical

Next we need to choose which k are vertical going away

[alpine sable]

And n-k is horizontal

[pale kestrel]

and then choose which k are vertical

coming back

so we will then have

n choose k squared

and sum this over k

[alpine sable]

Ok but why n choose K squared

[pale kestrel]

?

we have to make a choice twice

in your going away steps

choose your k vertical moves

[alpine sable]

Oh ok

shuri is this the same question as the one you had in the morning?

[pale kestrel]

in your returning steps

[alpine sable]

I see

[pale kestrel]

choose your vertical move

[alpine sable]

That is a good question

[pale kestrel]

when i cant remember

[alpine sable]

I understand now

I got a fun problem for u guys

If u wanna try

[pale kestrel]

go for it

[alpine sable]

U have a 29x29 chess board

Pick the diagonals

Btw no.s from 1-29 are written on the chess board boxes

Each number appearing 29 times

[alpine sable]

oh, i mean 3 hours ago 😅 time zones

2 i guess

[pale kestrel]

yes

[alpine sable]

Now above the largest diagonal, all the squares have numbers written into it which sum up to 3 times the sum of numbers below the diagonals

What is the number in the centre square of the chess board

[pale kestrel]

so i will chop the board in 2

the sum of all in the lower left triangle

the sum of all in the upper right triangle

one is 3x the other

and I need the centre number?

[alpine sable]

Yes

[pale kestrel]

the main diagonal sum is irrelevant?

[alpine sable]

Number written on the centre of the board

[pale kestrel]

its excluded i mean?

the main diagonal sum

[alpine sable]

Everything above the diagonal squares and below

Not including the diagonal

[pale kestrel]

you are not including the main diagonal

in either sum

[alpine sable]

Yes

[pale kestrel]

ok

i will try

btw

$$\sum_{k=0}^n \binom{n}{k}^2=\binom{2n}{n}$$

[alpine sable]

Oki

[ocean seal]

Shuri2060

[pale kestrel]

^ this was used in the final step

[alpine sable]

God dayum

[hasty elk]

Does that not imply the main diagonal numbers are all the same

[pale kestrel]

this identity

[alpine sable]

Dayum

Why tho

[hasty elk]

If you only have information regarding the upper right and lower left not including the main diagonal, and from that information can deduce the central square value, it follows that whatever the order of the main diagonal the value must remain unchanged

[pale kestrel]

ah smart

[hasty elk]

It's a bit of a metagaming strategy i admit

[alpine sable]

Hehe

Noice

[pale kestrel]

sudoku strat

[alpine sable]

So , what’s the number on the diagonals

[hasty elk]

Shuri can jab at it, i don't have paper with me

[pale kestrel]

neither do i

i think x is 3 mod 4

29 | S (lesser sum)

maybe parity helps? hmm

[hasty elk]

Or well

Both triangles contain 29 x 14 squares

So all we gotta do is pick two 14-element subsets of {1,...,29}

It doesn't have to be, but i think this is a promising way given that there's no other information

[alpine sable]

Do u guys want a hint

[pale kestrel]

no

[alpine sable]

Ok

[hasty elk]

It's 15

Because 1 + ... + 14 is 105

And 16 + ... + 29 = 315

Wao such nice numbers

[pale kestrel]

14.15/2

(29.30-15.16)/2

[alpine sable]

GG

[pale kestrel]

14 : 29.2-16

i cant see why

does this generalise hmm

[hasty elk]

Probably not, because you need a very nice sweet spot

[alpine sable]

very

[hasty elk]

The solution implies you have to chuck all the top numbers in one corner

And all the bottom numbers in the other corner

You need a very fine sweet spot to do this

[pale kestrel]

n(n+1)/2 : ((2n+1)(2n+2)-(n+1)(n+2))/2

n(n+1) : 3n^2+3n

no lol its true forall n

[hasty elk]

Is it?

[pale kestrel]

check my math but

[hasty elk]

Oooh

[pale kestrel]

i think yes

[hasty elk]

Learn something new everyday

[pale kestrel]

ikr

i swear visually

[alpine sable]

I used induction to solve this question

[pale kestrel]

you are just doing 3 triangles

vs 1 triangle

similar triangles

u can probably show via picture

[hasty elk]

I should totally set this for some competition

[pale kestrel]

maybe not

[hasty elk]

See what happens

Meanwhile i used basic olympiad gaming to solve it

[pale kestrel]

yh i was gonna do same method

but i kinda cant add

like best first guess hahaha

[hasty elk]

Ah, i had experience practicing addition

So what happened to the walk question anyway?

[pale kestrel]

you made the most progress by far

i see quite quickly that same argument falls apart in higher dimensions

[hasty elk]

Yeah

[pale kestrel]

Using the same format will fail to return to origin

[alpine sable]

How I solved it was using induction

higher dimensions is very hard to have a intuitive explanation for the paths

I took a 3x3 chess board

All 3’s in top of diagonal

All 1’s below

U get 2’s in the diagonal

Verify for 5x5

Hence the formula is (n+1)/2

Where n is the No of side is chess board

[hasty elk]

That doesn't sound like induction

[alpine sable]

Oh

[hasty elk]

That sounds like pattern

[alpine sable]

Anywho I solved it

[pale kestrel]

11001010
00101101

If half of each row is 0's. Then each column can be paired with its binary complement.

Adding rows makes this no longer true. That is quite a special property for 2 rows (and obviously 1 row)

[alpine sable]

Come on cut me some slack , I solved it while I was 15

That’s the best I could do

I think this was from tournament of towns

[hasty elk]

Yeah, and i don't think i have a ready way to generalise it also

[pale kestrel]

what about changing the problem to fit the solution lol

hahaha

[hasty elk]

In the first place you have to add more paths, soooooo

[alpine sable]

what is the problem, path from origin to origin in higher dimensions?

[pale kestrel]

it is

[hasty elk]

Yep

We have a solution in a sum of multinomial coefficients and that's the best so far

[alpine sable]

and the solution generalizes to higher dimensions?

[pale kestrel]

The problem is that this formula does not apply for higher dimensions by changing exponent. It only applies to 1 and 2

The multinomial sum is best we can do for general dimensions

[alpine sable]

yeah, someone gave a counter example for 3 dimensions in this thread iirc

[hasty elk]

.

[pale kestrel]

.close

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[gusty cove]

Why is the K value 10 here

the book doesnt explain how that result was achived

[hasty elk]

cause you want the probability X is between 40 and 60

i.e. it differs from the mean, 50, by 10

[gusty cove]

what if it were 30 and 60?

[hasty elk]

then you need to split into two cases, one dealing with 40 to 60, and one dealing with 30 to 40

[gusty cove]

hmmmm

thanks

the book doesnt go into detail

it seems that here it got a different result

[hasty elk]

yes, Chebyshev (and Markov) inequalities are only upper bound estimates of the probability

it doesn't tell you the actual probability

it just guarantees it doesn't exceed a certain value

[gusty cove]

I mostly want to know how k = 4

[hasty elk]

it looks like a different example

i don't know, because i don't have the question

[gusty cove]

that is the entire question

[hasty elk]

if that's the entire question, they just picked a value for k to showcase the difference between Chebyshev and the actual probability

it doesn't mean anything

[vale wigeon]

they wanted to know a bound on P(|X-5|>4)

nothing else to it

[gusty cove]

Ok thanks

i thought you were suppose to derive the value

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[cunning trout]

How do I do these questions?

[gray isle]

apply the (piecewise) definition of the absolute value

[cunning trout]

Okay

I actually don’t know the piecewise definition.

[gray isle]

$|\text{this}| = \begin{cases} \text{this}\ &\text{if } \text{this} > 0 \
-\text{this} \ &\text{if\ } \text{this}\leq 0 \end{cases}$

[ocean seal]

ℝamonov

[cunning trout]

I don’t understand how I apply this

Oh OkY I understand thanks

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[ocean seal]

yayy

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[red ledge]

Hello

[red ledge]

What does set mean?

How do I know about them...?

Shouldn't he say a relation between 2 variables

Because set is a collection of things

And Y and X is just a a variable

Like variable means something keeps changing

Right as well?

I wanna understand what is a set and why he used it with X and Y

Can a variable which is singlel be a set?

[pliant dune]

A set is like you say a collection of things

This definition thinks of functions as a way to pair items from the first set to some item of the second set

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@red ledge Has your question been resolved?

[lone heart]

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[glad sluice]

In how many ways can we choose a set of k numbers out of n consecutive numbers such that no two numbers in the set are consecutive?

[lone heart]

@glad sluice Has your question been resolved?

[covert agate]

<@&286206848099549185>

[strong furnace]

look up stars and bars

[glad sluice]

how would that be used here?

[strong furnace]

ok so first I was writing it up and then I realized it is a visual tool so I should probably write it up so I am writing it up and I have made a baka mistake but I don't want to rewrite it so I will just explain it ok?

[glad sluice]

lmao ok

[strong furnace]

zzz

@glad sluice finally

[glad sluice]

oh nice

I'll read through it one second

[strong furnace]

ok so those bars represent the k bars

you can read as I type right?

[glad sluice]

yessir

[strong furnace]

and the dashes are the remaining places for other numbers to go

[glad sluice]

oki

[strong furnace]

so now for consecutive it should be easy to see that the edge dash need not be filled but every other dash must have atleast one number in it right?

[glad sluice]

okay.... wait

[strong furnace]

otherwise there is a possibility of consecutive numbers while on the edge you can have say {1,2,3..n} are chosen {1,3,...}

[glad sluice]

what do the x1, x2, x3..., x_k+1 represent?

[strong furnace]

yeah I was gonna come to that

they are the "amount of numbers in that dash"

[glad sluice]

ah

lemme try to understand this gimme a moment

[strong furnace]

sure ping me

[glad sluice]

@strong furnace i dont get why we're doing this though
like why do the amount of numbers in each dash even correlate to consecutive numbers

and why are the bars even there

[strong furnace]

well the one big reason is because this is a popular way of doing it and it is taught

but while it visually helps you don't need bars

however this gives you a more precise way of writing a set of numbers with no consecutive two numbers

so while bars are not as important

x1,x2... are pretty important

they are pretty much saying between two consecutive numbers

there is at least one other number

[glad sluice]

OHHHH

so the bars represent the numbers??

like the consecutive n numbers

[strong furnace]

I am not having the best day

[glad sluice]

I'm sorry for bothering you

[strong furnace]

this was supposed to read bars represent the k numbers

yes

bars represent bars 🧠

[glad sluice]

xddd

[strong furnace]

anyway yes the bars represent our choice yes

[glad sluice]

alrighttt this makes much more sense

now why do x1, x2, ..., xk+1 add upto n*k?

[strong furnace]

n-k

because we chose k numbers leaving us with n-k

[glad sluice]

oh that's n-k

right makes sense

ok I can proceed from here on

thanks a lot!!! have a good day!

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[storm tundra]

ive already done the first part to get y =(x+4x)^2 + 9 but forgot how to do the next bit

would the turning point be (-4, 9)?

[pale kestrel]

when does the right hand side 'turn'

when is it going to stop increasing, and start decreasing (or vice versa)

[glass lichen]

this is wrong

[gray isle]

you didn't complete the square properly

can you show your work

[storm tundra]

#4

[pale kestrel]

The 2nd line is wrong

[storm tundra]

is it -?

[gray isle]

there are multiple things wrong with the second line

[pale kestrel]

u should complete the square step by step

x + 4x

is 5x

this squared

is 25x^2

[gray isle]

redo the whole thing, showing all steps (and not taking supposed "shortcut methods")

[pale kestrel]

so apparently we have 25x^2 + 9

[storm tundra]

no that seems wrong

uhh lemme show you the thing

i think its different methods but im not supposed to change the brackets just half of the x^2 and the x

[gray isle]

different methods if valid and applied properly will lead to correct results

[pale kestrel]

The best way to avoid mistakes is to add 0

u want

(x + b/2)^2

that is $$x^2 + bx + \frac{b^2}{4}$$

[ocean seal]

Shuri2060

[pale kestrel]

^ u need to convert this into the squared term

So if I have

$y = x^2 + 2x + 5$

[ocean seal]

Shuri2060

[pale kestrel]

I will need to take out a x^2 + 2x + 1.

So i write the original thing like this

$y = x^2 + 2x + 1 - 1 + 5$

[ocean seal]

Shuri2060

[storm tundra]

for th keywords "in the form y = (x-a)^2 +b we were told to do this

say we had x^2 + 4x -3
make x^2 just x, half the 4x to just get 2x and that put that in a bracket thats squared and add the last value of -3 so it would be
(x+2x)^2 -3
than square second number in bracket and add outside so
(x+2x)^2 +4 -3
then collect terms
(x+2x)^2 +1

[pale kestrel]

that skips some steps

[pale kestrel]

I think writing this step

[storm tundra]

did i forget some steps

[pale kestrel]

Avoids mistakes

[storm tundra]

what steps have i missed

[pale kestrel]

1. Add '0' to the original equation
2. Spot the square

Is what I would show

Your first step (after figuring out what b is) is to add 0

[gray isle]

you're not applying the method properly

in more ways than one

[storm tundra]

what am i doing wrong then, skipping a few steps, i cant remember getting taught the add 0 step

[gray isle]

you're misremembering the exact process

[pale kestrel]

$$y = x^2 + 2x + 5$$

$$y = x^2 + 2x + 1 - 1 + 5$$

$$y = (x^2 + 2x + 1) - 1 + 5$$

$$y = (x+1)^2 - 1 + 5$$

[gray isle]

$\cts$

[ocean seal]

Shuri2060

ℝamonov

[pale kestrel]

If you write the working like this, the whole process becomes easier to understand

That aside, you're factoring incorrectly

yh

[storm tundra]

That’s the method I was using

oooo so i just found that method

[gray isle]

argh

[storm tundra]

the second number squared has to be negative

[gray isle]

that page is plagued with mistakes

[storm tundra]

that page is exactly how we were taught to do it

that method at the top is directly copied from my teachers notes

[gray isle]

you should actual understand the basic principles of completing the square outlined above

can't read the second point

add on half of the (x term)^2 inside the brackets?

that's wrong

[pale kestrel]

Remembering a method without understand will make mistakes more likely

[gray isle]

and is probably why you're messing all this up

[storm tundra]

2. Add on (half of the x term)^2 inside the brackets
3. Subtract the same number outside the brackets
4. Write the extression in complete square form.```

[gray isle]

yeh. #2 is wrong

[storm tundra]

sorry i know you guys are saying its wrong and im trying to understand but ive been doing it that way the past few months and getting marked as its right which is why its hard for me to understand what im doing wrong

have i maybe copyed #2 wrong

[gray isle]

probably

[storm tundra]

what should it be changed to

[gray isle]

+(b/2)^2

add the square of half the coefficient of the x term

[storm tundra]

i thyink i could havei wrote the method down wrong but then just learned how to do it when he was going over examples so now when i forgot and am going over the method thats wrong its all messing it up

[gray isle]

the combination of step 2 and 3 is
+(b/2)^2 - (b/2)^2

(effectively adding 0)
outlined in the guide i made above

[storm tundra]

so that method there is wrong

[gray isle]

yes, the description is wrong

[storm tundra]

and those two examples are wrong

[gray isle]

despite the description being wrong,
the actual work is correct

[storm tundra]

oh ffs 😂

this is the most confused ive ever been

bu t is that not EXACTLY what ive done here

[gray isle]

is wrong

[storm tundra]

but its the exats same working...

[gray isle]

the x doesn't belong next to the 4

[storm tundra]

that is true

[gray isle]

hence why the description is wrong

[storm tundra]

scribbled out

[gray isle]

should be half of the coefficient of the x term

[storm tundra]

okay even though thats messy thats the exact same working as #help-0 message

[gray isle]

yes

[storm tundra]

so that's right?

[gray isle]

yes

[storm tundra]

I feel as though this is where qualification knowledge of certain countries qualifications and methods would be of benefit as if someone that is from Scotland saw that they'd say "oh if your doing nat 5 then you need to subtract the squared number not add it and you shouldn't add the x inside the bracket" although for what i wrote with the x's what you guys were saying was right to deal with that

I came with a different problem so got a different method/solution so ty peeps for dealing with me xD

[gray isle]

the method is supposed to be the same, its just that one was poorly described

and skipped half a step

[storm tundra]

lmao thanks anyways troops

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[chrome crane]

Hi, i've got this proof set that I' struggling with:

𝐵={𝑥∈ℝ:∃𝑦∈[0,1] s.t. 𝑥<𝑦}

I read it as "the set of all real X for some Y between [0,1] such that x < y." but i still cant quiet understand it. Would it be (-inf, 0) since x has to be lower than y and x cant be inside [0,1]?

[abstract fractal]

Why can't x be in [0,1]?

[chrome crane]

because y is [0,1] and x has to be lower than y

i mean i think right? isnt that what this 𝐵={𝑥∈ℝ:∃𝑦∈[0,1] s.t. 𝑥<𝑦} says?

[abstract fractal]

Why does that exclude x from being in [0,1]? Let's say x is 0.5. Does there not exist a y in [0,1] such that 0.5<y?

[chrome crane]

ohhh i see. because ∃ means 'for some' which means it isn't every y in [0,1] correct?

[abstract fractal]

Yes

It's just saying there exists a y, not that it's every y

[chrome crane]

so would it be (-inf, 1)

because the ) would be excluding at least one y

[abstract fractal]

Yes

[chrome crane]

Thank you so much for your help sir!

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[hot mist]

Is the notation 6 | x, standard notation to signify that x is divisible by 6?

[hot mist]

Would most math markers recognize it?

[echo socket]

I think so

[hot mist]

ok great thanks a lot, I will use it from now on so that I skip writing a few words

thanks for the help

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[deep radish]

Could someone explain this to me please I dont understand the part I circled in red

[alpine sable]

Oh they just took 3 common

From both the terms

4(3m) and 3

[deep radish]

oops sorry I meant to put red on here I dont understand where this red part comes from

[lone heart]

@deep radish Has your question been resolved?

[alpine sable]

Oh

I write -1 as -4 + 3

Then u take 4 common

From 4^k.4 and -4

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[deep radish]

ohh ok thanks also one more thing im confused about is what hte 4^k means/comes from

[alpine sable]

From here

U need to prove 4^n - 1 always divisible by 3 for n belong to natural numbers

Put n = k+1

[lone heart]

@deep radish Has your question been resolved?

[lone heart]

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[true pulsar]

Use sin2x = 2sinxcosx

[dim vine]

Here I substituted u=cosx

Oh no... I just saw the 2x

Sorry

thanks

.close

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[shrewd smelt]

sorry but could someone help me on this?

[gleaming night]

Yo, can anyone please explain why the total amount of values in the progression when using the first and last term is the same when using the third and third to last term.

[shrewd smelt]

help-9,12,20 are open

[gleaming night]

sorry, thanks

[lone heart]

@shrewd smelt Has your question been resolved?

[true pulsar]

Well just look at the options and think about what happens when you assume them to be true

Take option A, where x = 10

So the box is then 10 inches tall right

But what about the other dimensions

If x = 10, how much width will the box have?

[shrewd smelt]

0?

[true pulsar]

Precisely, because you obviously saw 10 + 10 = 20, so you cut away the whole side

And if width is 0

What's the volume?

[true pulsar]

Exactly, and it's hard to have volume smaller than 0

[shrewd smelt]

okay thank you

.close

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[true pulsar]

No problem

[lone heart]

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[covert agate]

hi

[covert agate]

let f = x^n sin(1/x) for x ≠ 0 and 0 for x = 0

for what values of n is f everywhere differentiable?

[pale kestrel]

$$f(x) = \begin{cases}x^n\sin(\frac{1}{x}) & x\neq 0\end \ 0 & x=0 \end{cases}$$

zzz

[covert agate]

lol

[ocean seal]

Shuri2060
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[pale kestrel]

uh i dont even know

$$f(x) = \begin{cases}x^n\sin\left(\frac{1}{x}\right) & x\neq 0 \ 0 & x=0 \end{cases}$$

no idea what the heck is wrong here

#latex-help

nvm i see

[ocean seal]

Shuri2060

[pale kestrel]

@covert agate Right ok, so this?

[covert agate]

yea

[pale kestrel]

k, so what u done

[covert agate]

n = 2 is a solution

[pale kestrel]

huh?

oh uh sure

[covert agate]

i also got lim_(x → 0) x^(n - 1) sin(1/x) = 0

no idea what to do

[pale kestrel]

alright first

For all n, this function is differentiable everywhere except 0

agreed?

[covert agate]

uhh sure

[pale kestrel]

so u only need to check at 0 with the defn

[pale kestrel]

like this

[covert agate]

i used alternate form

[pale kestrel]

and then solve for n

[covert agate]

formal form is bad

[pale kestrel]

?

[covert agate]

how

[pale kestrel]

idk, let me try

[covert agate]

epsilon delta???

[pale kestrel]

no

$$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}{h}$$

[ocean seal]

Shuri2060

[pale kestrel]

u need this thing to exist at x=0

[covert agate]

bruh

why not use alternative form

[pale kestrel]

what?

[covert agate]

much neater

[pale kestrel]

no idea what that is

This is just defn of f'x

[covert agate]

lim_(x → a) (f(x) - f(a))/(x - a)

= f’(a)

[pale kestrel]

it doesnt make a diff

both notations means the same thing

$$f'(0)=\lim_{h\to0}\frac{f(h)-f(0)}{h}$$

[ocean seal]

Shuri2060

[covert agate]

yea we know f(0)

so we get sth much simpler

[pale kestrel]

$$f'(0)=\lim_{x\to0}\frac{x^n\sin\left(\frac{1}{x}\right)-0}{x}$$

[ocean seal]

Shuri2060

[pale kestrel]

$$f'(0)=\lim_{x\to0}x^{n-1}\sin\left(\frac{1}{x}\right)$$

k, so thats what u got

[ocean seal]

Shuri2060

[pale kestrel]

Then you have to make this limit exist

Can you use lhopital, just checking?

[covert agate]

no

[pale kestrel]

k. dont think u need it anyways

[covert agate]

using special limits somehow…?

[pale kestrel]

From this step, I might try a substitution

y - 1/x

but hmm

perhaps consider the behaviour of sin x

$$f'(0)=\lim_{x\to0}x^{n-1}\lim_{x\to0}\sin\left(\frac{1}{x}\right)$$

[covert agate]

im thinking of parity

[ocean seal]

Shuri2060

[covert agate]

but doesnt seem helpful

[pale kestrel]

Consider this rearrangements of limits.

IF the limit of the left term exists

THEN the entire limit cannot exist.

[covert agate]

$$f'(0)=\lim_{x\to0}x^{n-2}\lim_{x\to0}x\sin\left(\frac{1}{x}\right)$$

[ocean seal]

Chromium

[covert agate]

doesnt this work better

[pale kestrel]

I don't think so?

Or at least, that is not my approach

[pale kestrel]

.

I am saying this

[covert agate]

huh why

[pale kestrel]

Suppose the entire limit AND the limit of the left term exists

THEN the limit of the right term also has to exist?

if I'm not mistaken

Maybe I need to fine tune this argument

[covert agate]

ok thats wacky asf lol

[pale kestrel]

hmmm

I basically want to make some argument to show L = 0

or something, I think

[covert agate]

i think youre abusing the well-defined-ness of limits

which is wacky

[pale kestrel]

Ok that argument doesn't quite work

I'm not abusing anything

I am arguing via contradiction essentially

I need to split that argument into cases, though

I will call this L = AB

Suppose A exists and A ≠ 0

Then the entire limit cannot exist.

Agreed?

[covert agate]

hmm

wait

you’re abusing the fact that RHS isn’t well defined and thus doesn’t exist lol

[pale kestrel]

I'm not

[covert agate]

i mean it works i guess..?

[pale kestrel]

this isn't abusing

I am basically saying suppose A exists and isn't 0. Then divide both sides by A

[covert agate]

ok

[pale kestrel]

Then we have a limit that exists on the left, but not on the right

This is a contradiction

Hence, what I assumed isn't possible.

This is a perfectly valid argument

[covert agate]

seems wild

[pale kestrel]

Suppose A exists and A = 0

[covert agate]

wait

[pale kestrel]

Basically there are 3 cases to assume on A

I think you might get the answer if you consider all of them

[covert agate]

isn’t lim(fg) = lim f lim g iff lim f AND lim g are well defined?

thats by property of limits

in your case lim g isn’t

[pale kestrel]

let me think

[covert agate]

so you can’t just split it into those

or else lim (x/x) = lim x lim 1/x

shit goes down

[pale kestrel]

I agree with this statement.

Perhaps I need to make a more precise argument

Let

$$L=\lim_{x\to0}x^{n-1}\sin\left(\frac{1}{x}\right)$$

[ocean seal]

Shuri2060

[pale kestrel]

We will assume L exists.

Let
$$A=\lim_{x\to0}x^{n-1}$$

[ocean seal]

Shuri2060

[pale kestrel]

Case 1: Assume A exists and A ≠ 0

Then L/A is well defined

$$\frac{L}{A}=\frac{\lim_{x\to0}x^{n-1}\sin\left(\frac{1}{x}\right)}{\lim_{x\to0}x^{n-1}}$$

[ocean seal]

Shuri2060

[pale kestrel]

Hence we can combine the limits

$$\frac{L}{A}=\lim_{x\to0}\frac{x^{n-1}\sin\left(\frac{1}{x}\right)}{x^{n-1}}$$

[ocean seal]

Shuri2060

[pale kestrel]

$$=\lim_{x\to0}\sin\left(\frac{1}{x}\right)$$

[ocean seal]

Shuri2060

[pale kestrel]

And this leads to a contradiction

Is that better?

[covert agate]

i guess

yea

[pale kestrel]

So if L exists, A must be 0 or undefined

Then you check either case

and figure out n?

$$0=\lim_{x\to0}x^{n-1}$$
$$DNE=\lim_{x\to0}x^{n-1}$$

[ocean seal]

Shuri2060

[covert agate]

and how does that aid us in finding n lol

[pale kestrel]

consider either case

uhh which is easier

DNE?

wait does this not help 🤔

Uhhh you should be able to make argument

that A = DNE means L cannot exist

If you think about the original thing

[covert agate]

x^(n - 1) can never be 0 lol

exponent rules

[pale kestrel]

can

you are taking x to 0

[covert agate]

wait fuck

cannot think at 2 am lol

[pale kestrel]

Yeah I think this argument is almost there

just got to rule out DNE case

Because in the end, A = 0 is a must basically

[covert agate]

n - 1 > 0?

then n > 1

[pale kestrel]

Yeah, that is the final answer

So why is DNE case not possible hmmm

[covert agate]

what a strange question

[pale kestrel]

if its DNE you kinda have to have

L = infinity * (something that alternates between -1 and 1)

but uhhh

not sure how to make that argument rigorously

[covert agate]

if dne, again we can’t simply split into two expressions lol

[pale kestrel]

Have you done analysis?

[covert agate]

no

[pale kestrel]

uhhh

have you heard of

if a sequence converges

[covert agate]

no

[pale kestrel]

all subsequences converge

rip

[covert agate]

ive only heard of convergence in school physics

[pale kestrel]

where is this Q from

[covert agate]

friend

hes much better lol

[pale kestrel]

well it doesnt need to be super rigorous

[covert agate]

he told me to give it a try

[pale kestrel]

a graphical proof will prove DNE case is impossible

only way x^{n-1} DNE is if it goes to infinity as x goes to 0

I believe?

Then that has to be multiplied by a sin(1/x) that is varying between -1 and 1

So the final limit cannot exist

because the entire thing will vary wildly between -x^{n-1} and +x^{n-1}

So you can at least convince yourself that case is impossible and get the correct final answer

[covert agate]

uhh i suppose there’s a more rigorous approach than considering whether lim x^(n - 1) exists lol

[pale kestrel]

hmm

I can think of a rigorous approach using subsequences

[covert agate]

i know no analysis ;-;

[pale kestrel]

The concept being used isnt too hard

if I take any function and consider its behaviour as x goes to infinity

I can consider a discrete subset (that is unbounded)

so like x in {1, 2, 3, 4, 5, 6, ...}

And if I consider {f(1), f(2), f(3), f(4), ...}

this must also converge if the original limit exists

[covert agate]

friend’s approach

works?

[pale kestrel]

I'm not sure about that last 3 lines

in fact M >= 2 isnt the answer?

[covert agate]

bruh this checks for limits

[pale kestrel]

Yeah, whats wrong with this?

I agree visually the answer must be n < 1

check something like desmos

nvm thats hard to see

[covert agate]

??

[pale kestrel]

Does that converge or not?

[covert agate]

not…sure?

[pale kestrel]

Check math maybe

[covert agate]

or just change 1/x to1/(999x)

[pale kestrel]

wait what am i doing

i should check the original function

maybe

Huh yeah, I don't think this is differentiable at 0...

🤔

or maybe it is

Well apparently.

not obvious from desmos graph ig

So yh answer should be n > 1

[pale kestrel]

I think from the rearrangement, they can conclude m >= 2 means g_m(x) is differentiable.

However, they haven't shown what happens for m < 2

[lone heart]

@covert agate Has your question been resolved?

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[steep flax]

i need some help with how to reduce this

[remote heron]

well what cancels?

[steep flax]

oh

[remote heron]

what can we find in both the top and the bottom

thats the stuff that cancels

[steep flax]

i guess its reduced to 5*b^3

[remote heron]

yea that sounds reasonable

[steep flax]

alright cool

[wary stream]

Don't open multiple channels for the same question

[steep flax]

sry

[lone heart]

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[quiet maple]

Whats the gradient of AC, and why

[quiet maple]

Or more like, why is it -2

Thanks 🫂

.close

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[wind tartan]

Hi there, I'm wondering how you get from:

[wind tartan]

$\lim_{h\to0}\frac{f(a+h)+f(a-h)-2f(a)}{h^{2}}$

To

[ocean seal]

omega

[wind tartan]

Vía L'hopital's rule

[glass lichen]

to what...?

[wind tartan]

the latex isn't rendering

or the message didn't send

so sorry one second

$\lim_{h\to0}\frac{f'(a+h)-f'(a)}{h}$

[ocean seal]

omega

[wind tartan]

rather going from the limit directly above

to the one that I sent before that

L'opitals rule was mentioned

but it doesn't make complete sense to me

unless you're for some reason integrating instead of taking the derivative

as stated in this stack exchange post: https://math.stackexchange.com/questions/1809060/proof-of-the-second-symmetric-derivative

the most upvoted answer

I don't understand the intermediate step between the first derivative representation and the second

@glass lichen

nevermind

I have found a sufficient explanation

.close

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[thick flame]

I asked: Is there a way for me to find a point some number of units along a function? (ex: f(x) = sin(x), and I want to find the point when the length of the function up to that point from 0 is 17)
and got this answer, I'm looking for a generalized solution and I don't think I have the mathematical knowledge to solve for a on my own.

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@thick flame Has your question been resolved?

[lone heart]

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[alpine sable]

if a sequence converge to 0 that means that the limit of the sequence it's 0?

[glass lichen]

yes

[alpine sable]

thank you!!!!!!!!!

.close

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[hallow stone]

can someone explain this?

[hallow stone]

nevermind

.close

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[signal falcon]

Hey

[signal falcon]

I have this exercise

I wanted to model the probability using indicator variables (the draws)

and using linearity of expectation

but looking at the b)

I dont really know how to follow up

As the Excepted value of the sum of indicator variables would give me the expected number of successes (here diamond drawed) in those 4 draws

but the question is the expected number of turns to draw 2

but if I do E(X) where X is the sum of those 4 indicator variables, I end up with 2

So if the expected number of diamonds after 4 draws is 2, I felt that naturally the answer of the expected number of draws to get 2 diamonds is 4

But I feel like something ain't right

[wintry lotus]

I don't think that is right

[signal falcon]

I tried to brute-force the expected value, by drawing the complete tree and computing the mean length is the end nodes (end node is where we get finally 2 diamonds)

and the result I got was 3 and 1/3

[wintry lotus]

That is what I got too

[signal falcon]

I can send a picture of that tree if wanted, that result makes a little bit more sense than 4, but I still feel there has to be more intuitive solution than brute-force

[signal falcon]

Did you also calculated the mean in "raw" way, or is there a more straightforward way?

[wintry lotus]

Raw, haha. I am trying to think of the programmatic way

[signal falcon]

I was expecting some linearity of expectation in this exercise, but I can't really find a way to apply it 😄

Also the c)

I am not sure how to follow, but I had idea

with the linearity of expectation

because there are 2 cases, we are waiting to get first diamond

and we are waiting to get second diamond

If we model each separately as a geometric distribution

and get expected value of each

we can then sum the expected value, to get the final expectation

and expectation of geo distribution already represents the number of trials

This would result E(X) = E(g1) + E(g2) = 2 + 3 = 5

expected number of throws is larger, which is intuitive as the probability gets worse a bit

[wintry lotus]

I think you need to mulitply those together?

[signal falcon]

Multiply expectations?

[wintry lotus]

thinking

[signal falcon]

I don't think so, because if you think about this way

probability on drawing the first diamond is 1/2

so you do that until you get it, the expected number of draws is 1/p = 2

then you can start the second drawing

but the p = 1/3 now

i find it similar to the coupon collector problem

[wintry lotus]

That does make more sense. I came up with 6 instead, but adding and 5 is much more reasonable

[signal falcon]

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[rich basin]

why is this the anbswer

this is why i did

I am confused on where can i tell which direction of the integration should i bound it to

[lone heart]

@rich basin Has your question been resolved?

[pale kestrel]

I would really suggest drawing a picture/visualizing what exactly you're integrating

The drawing should tell you the bounds

[lone heart]

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[rich basin]

.reopen

[lone heart]

✅

[rich basin]

Here

[pale kestrel]

when you integrate for that volume

what are you actually adding up?

For the hemisphere bowl

[rich basin]

you are jsut adding multiple tiny cylinders

with the height of it being dy

[pale kestrel]

tiny in height

https://i.imgur.com/7GZZmgt.png

👌

So for this they are stacking upwards

since u integrate along y

Now, for the limits, there are usually tricks like

chopping the sphere into halves or quarters

Since x^2+y^2 = r^2 isn't actually a function

======

[rich basin]

yeah

[pale kestrel]

but let me see how they do this. ..

[rich basin]

but that is why it is a hemisphere, so I would take the upper bound

[pale kestrel]

I would integrate from y = 0 to y = a/2

[rich basin]

would that produce the same answer?

that was what i did for my solution

[pale kestrel]

Yes, doesnt matter which way u did it

OH

[rich basin]

but then i got a different answer

[pale kestrel]

ok ok no no

I understand

The question

The radius of the bowl is a

The bowl is only filled halfway

to a/2

That explains the limits --- you integrate from a/2 to a

[rich basin]

so it's a hemispher but it is only half filled?

[pale kestrel]

^ but in 3d

like if i drew this upside down

it would be a half filled bowl

[rich basin]

okay thanks, and also do you know how to integrate irregular shapes

[pale kestrel]

depends what shape it is 🤔

[rich basin]

like a nonconvex 3d shape

[pale kestrel]

I mean I guess??

If you gave me a description of it I could work with

But they usually only ask about solids of revolution . ..

[rich basin]

forexample a trapezium

[pale kestrel]

The other thing is, there's not much point integrating non-curved objects

since geometry should get answer

[rich basin]

okay thanks

.close

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[rich basin]

can someone please explain (i)

[pale kestrel]

not integration wut

[pale kestrel]

Ok, so.

If there are exactly 7 games

You have to think about how the scores will progress

ok?

[rich basin]

yeah

[pale kestrel]

So the last game

has to be won by A

So we kinda don't care about that. It's fixed. (it's a constant times 1/2 to the final answer)

[rich basin]

okay

[pale kestrel]

So now we need to count for the 6 games

Of those 6 games, A has to win 4

So, we have 6 choose 4

as the number of possible game progressions

The probability each of these progressions happen is (1/2)^6

Eg. I tell you AABABA

The probability this exact progression happens is 0.5^6

Adding on the 7th game, that gives the final answer.

[rich basin]

Okay thanks

@pale kestrel do you know how would you do iii

this looks like we would have to use (i)

[pale kestrel]

hm

im pretty sure we need the answer to (ii)

what did u get?

[pale kestrel]

itll take me a while to think of it

[rich basin]

how can we prove (iii) by induction

[pale kestrel]

huh?

Maybe im not sure

[rich basin]

@pale kestrel do you perhaps can explain (i)

I'm more confused about the P(A wins T1 or T2)

[pale kestrel]

isn't i just basic prob

[rich basin]

why are they taking q ?

[pale kestrel]

Alright, so player A goes first.

A_n will be notation for the event that A wins within the first n turns

[rich basin]

yeah

[pale kestrel]

nvm

$$P(A_2) = p + r(q)$$

[ocean seal]

Shuri2060

[pale kestrel]

Do you agree with this?

[rich basin]

yeah

[pale kestrel]

k, so we need to rearrange into the form they want

all we have is that 1 = p + r + q

so uhh

q = 1 - p - r

p + r(1 - p - r)

uhhh

p + r - rp - r^2

huh what the heck

I'll try rearranging the answer I guess

[rich basin]

okay thanks I get it

[pale kestrel]

wait u got it?

[pale kestrel]

i might be wrong.

[rich basin]

I didn't think about when B lands on q

i thought the only way for A to win is when A lands n p

[pale kestrel]

(1 - r)(p + r) = (1 - r)(1 - q) = 1 - r - q + rq = p + rq

ok ok, we're good.

urghhh thinking about this

[lone heart]

@rich basin Has your question been resolved?

[hasty elk]

heya

what's the problem

[pale kestrel]

last

thank you 💙

[hasty elk]

it's pretty easy, no?

[rich basin]

how would you think you would do (ii)

[hasty elk]

(ii) i would just leave it as a sum

[rich basin]

Currently I have this?

[pale kestrel]

probably but words get me

[hasty elk]

right

so for (iii) the first thing you want is to prove the equivalent identity
$\binom{n}{n}\frac{1}{2^{n+1}} + \binom{n+1}{n}\frac{1}{2^{n+2}} + \dots + \binom{2n}{n}\frac{1}{2^{2n+1}} = \frac{1}{2}$

[ocean seal]

Camilleone

[hasty elk]

wait hold on

i messed up my count

brb

so for (iii) the first thing you want is to prove the equivalent identity
$\binom{n}{n}\frac{1}{2^{n+1}} + \binom{n+1}{n}\frac{1}{2^{n+2}} + \dots + \binom{2n}{n}\frac{1}{2^{2n+1}} = \frac{1}{2}$

[ocean seal]

Camilleone

[hasty elk]

here, i think this is the right one

each term is the probability that A wins in exactly k games, for n+1 <= k <= 2n+1

now, this covers all the ways A can win the prize

and the overall probability that A wins the prize is 1/2

so the identity is proved

that is, the sum is equivalent to $$\sum_{k=n+1}^{2n+1} \mathbb{P}(A \text{ wins in exactly } k \text{ games}) = \mathbb{P}(A\text{ wins})$$

[ocean seal]

Camilleone

[rich basin]

where did they get the GP from?

How is that part of the GP?

the denominator

[hasty elk]

wait what

where did that come from

i assume this is another question, so @pale kestrel have fun

[rich basin]

@hasty elk how did they derive the GP?

[hasty elk]

how did who derive what GP?

[hasty elk]

i only know this question

[hasty elk]

and this looks like a completely different question i haven't seen

[pale kestrel]

what

whats GP

geometric progression

[hasty elk]

also, (iii) is solved

[pale kestrel]

wait what question is this

@rich basin is this the same game or what

[hasty elk]

suddenly an extra p and r appears!

[pale kestrel]

nvm

[rich basin]

yeah it's the pie chart question

[rich basin]

How did they derive the GP for this?