help-26

[lusty marsh]

here

lmk if u have anyy doubts

[glass harbor]

so you derive x^2 into 2 after multiplying x^2 on the tops?

[lusty marsh]

huh

[glass harbor]

wait but where did the 2 over x^2 go?

[lusty marsh]

in the 5/x part u mean?

[glass harbor]

nono like

you have x^2(2/x^2...)

oh wait x^2 derived isn't even 2, hm

why did the x^2 and 2 flip places?

(not doubting you, just confused)

[lusty marsh]

cz a* b=b* a
so I can write 2* x=x *2

[lusty marsh]

the x^2 on top and bottom get cancelled out

[glass harbor]

that's a thing ?? @~@

[lusty marsh]

yepp

similarly we can write $5x^2=x^25$

[thorny flame]

swerriee

[glass harbor]

incheresting, I would never have gotten that 😭

oki lemme attempt to write this out again

oki got it! :D Thank you <3

have a great night :D

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[terse lotus]

Snagging this channel to ask a more general version of the question I asked in #help-27 . When calculating an eigenvector, for example - for the matrx ([-7,-7],[-5,-5]), you end up with something like -7a -7b = 0. You can choose an eigenvector as (1,-1), and I get that - but why not (2,-2)? And is that equivalent? When you're doing calculations with that eigenvector, how does the scaling work?

[topaz sinew]

Please don't occupy multiple help channels.

[terse lotus]

🫠

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[glass harbor]

Hi :D I would close both channels and start anew in a new one, it's a lot more likely to get a response in my experience

[terse lotus]

Ah okay thank you!

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[urban gust]

Can someone help me with finding the increasing and decreasing intervals

[topaz sinew]

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[sharp vigil]

Hello

[urban gust]

Hi

[sharp vigil]

First calculate those points

When the function is going down

It's decreasing

When function going up It's increasing

[urban gust]

Yes

[sharp vigil]

Just state the regions in which these things are happening

That's it right?

What else is their to know

[urban gust]

(+ inf, -7)?

[sharp vigil]

Nope

You see

The function is decreasing even after that

From x belongs to - infinity to 0

It's decreasing

The interval is of x

Not y

Then from x = 0 to 8/3 it's increasing

Then from x=8/3 to infinty

It's decreasing again

You can do the same for the last question

Hope this helped

[urban gust]

Ty

[urban gust]

Would it be decreasing: (+inf, -7), (8/3, -inf) and increasing is (-7, 8/3)

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[little pond]

But if the x^2 - x case directly implied that x is an integer already, then what would the point have been of them telling you that x^n - x is also an integer?

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[little pond]

Here’s what I have so far:

if $x^2 - x = k$, and k is an integer, then $x = \frac{1 \pm \sqrt{1 + 4k}}{2}$

That means that the fact that $x^n - x$ is an integer must imply that $1 + 4k$ is a perfect square, but that’s where I got stuck. How can I show that? I tried doing it out for n = 3 and n = 4, where I realised that the factorisation is $x(x-1)$(some other stuff), and since $x(x-1)$ is an integer, the rest of the stuff must be an integer too, but I wasn’t able to prove that - it’s definitely not true for n = 3 or n = 4, but I’m not sure how to deal with the n in the power.

[thorny flame]

Fear na bPónairí

[keen plinth]

I would recommend prove by contrapositive, if x is not an integer, then for all $n\ge 3$, $x^n-x$ is not integer

[thorny flame]

ManMan

[topaz sinew]

@little pond Has your question been resolved?

[little pond]

But what about x^2 - x?

[keen plinth]

similarly, x^2-x is not integer either

[topaz sinew]

@little pond Has your question been resolved?

[little pond]

I’ll give that a go after, but I wanna try to finish my first idea if possible

[topaz sinew]

@little pond Has your question been resolved?

[topaz sinew]

@little pond Has your question been resolved?

[jaunty leaf]

Thing is, since $x(x-1)$ is integer and $x(x-1)$(some other stuff) is integer, we conclude that (some other stuff) is rational. On the other hand, (some other stuff) = $x^{n-2} - x^{n-3} + ... \pm 1$. Substitute $\frac{1\pm\sqrt{1+4k}}{2}$ here and you'll get (something rational) + (something rational)$\sqrt{1+4k} \in \mathbb{Q}$, which implies that $\sqrt{1+4k} \in \mathbb{Q}$ for some integer $k$. Hence $\sqrt{1+4k} \in \mathbb{Z}$ and notably it's odd. Now it's not hard to see that $x$ is integer

[thorny flame]

EQUENOS

[topaz sinew]

@little pond Has your question been resolved?

[little pond]

Is it formal enough to say “substitute and you’ll see…”? Because of course you can’t really sub it in because of the powers of n

[jaunty leaf]

You should prove that the coefficient of sqrt(1+4k) is not zero

[little pond]

How can I do that when there are possibly infinitely many terms?

And it didn’t specify that n is an integer so technically this might not prove it, so how do I get aroun that?

[jaunty leaf]

n is definitely an integer. Otherwise the statement isn't true

For example consider x = (1+sqrt(5))/2. Then x^2 - x = 1, and we can find some n >= 3 such that x^n - x is also an integer

Solve x^n - x = 4 relative to n. It yields n = ln(x + 4)/ln(x) ≈ 3.587

[jaunty leaf]

The number of terms is finite but yes, n is arbitrary. But that doesn't mean that the task is impossible

Oh, and actually I was mistaken and that long sum should've been
$$x^{n-2} + x^{n-3} + ... + 1$$

[thorny flame]

EQUENOS

[jaunty leaf]

Actually I think it's better to express it as $\frac{x^{n-1}-1}{x-1}$

[thorny flame]

EQUENOS

[jaunty leaf]

Good luck

[little pond]

So you’re saying if I sub my expression for x into here, I should be able to deduce that x is an integer? (Or in other words, 1 + 4n is a perfect square?)

Since the other part is also an integer?

[jaunty leaf]

Yes, but again, only if you prove that the coefficient of sqrt(1+4k) is not zero

[little pond]

Ahhh yes

That’s a pretty clever proof then

Well assuming I can prove that 😅

[little pond]

It’s pretty late for me rn, so I’ll probably just try it tomorrow, but would you mind if I shoot you a DM if I have any problems? Just so I can close this channel

[jaunty leaf]

Yes, of course. Feel free to DM

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[robust stone]

im just stuck on how they got x=cot5y (this is the mark scheme)

oh wait no lol i got it

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[coral fog]

If I have E[X^2] is there a way to find E[X^4] quickly?
without doing integrals all over again

[coral fog]

It's about probability and statistics

E[X] is the mean

[pearl peak]

Depends on the integral but in general I don’t think so

[coral fog]

Like it's so demanding

it's about question 3

where Var(3X^2+2) = 9Var(X^2)

[ionic oar]

I mean

you could do the

Expectation version of the formula

[ \text{Var} (X^2) = E(X^2)^2 - E(X^4) ]

[thorny flame]

neon

[ionic oar]

and those are pretty simple to solve with an integral

[coral fog]

Correct

My issue is finding E[X^4]

[ionic oar]

doubly easy if you know the gamma function

right

[coral fog]

if there is a shorter way to find E[X^4]

[thorny flame]

neon

[ionic oar]

do you know about this?

[coral fog]

nope

Is this calc 4?

[ionic oar]

you can prove it pretty easily with integration by parts

[ionic oar]

ehh idk not really

[coral fog]

I mean It's a probability course

[ionic oar]

it's fairly easy to prove

[coral fog]

I don't think they'll ask for proofs about integrals

[ionic oar]

yeah alright

[coral fog]

thanks 🙂

[ionic oar]

Have you corm across the gamma distribution.

in your course yet

[coral fog]

I'm still in binomial distribution

and poisson

[ionic oar]

you haven't gotten to continuous distributions yet?

is it a part of the course?

[coral fog]

This is a continious distribution correct?

[ionic oar]

yes

I mean like well known ones

normal distribution and stuff

[coral fog]

I don't think so

[ionic oar]

okay no problem then

[ionic oar]

this is useful to remember though

it will be of use to you here for instance

[coral fog]

I'll stick to solving the integral

can I split e^(-x/4)

[thorny flame]

neon

[ionic oar]

wdym by "split"

[coral fog]

It should look like this form

right?

[ionic oar]

yes

so just substitute x/4

[coral fog]

I did not understand

can you elaborate

[ionic oar]

x/4 = u

[thorny flame]

neon

[ionic oar]

do you understand this?

[coral fog]

Yes got it

thanks

[ionic oar]

so that integral is just 4!

[coral fog]

correct

[ionic oar]

,calc 4^4*4!

[thorny flame]

Result:

6144

[ionic oar]

there you go

[coral fog]

ty

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[distant totem]

If i understood correct, f'(x) is the growth factor of a given point x when delta x goes towards 0, but at the same time is f'(x) the slope of the tangent?

[shut night]

f'(x) is the growth of the function f(x) at a given x. the tangent is the same thing

[distant totem]

so what really is the difference then between a tangent and secant

[fading owl]

there is no secant in a graph

[shut night]

secant is the line through two points, and when you move the second point really close to the first one it becomes the tangent is probably the best intuition

[distant totem]

so a secant is basically something we can draw to help us find the slope of a tangent?

[neon iron]

Yeah pretty much

[shut night]

well

[neon iron]

Formally the derivative of a function is defined as the instantaneous rate of change

At a point

[shut night]

if you find the slope of the secant it gives you the average rate of change

and from that the idea is, but what if i want to know the rate of change at a given point

but if i tried to divide y_1-y_0 / (x_1 - x_0)

and i took some point

i would be dividing by 0 and thats not possible

[distant totem]

so when the question asks "Find the momentary growth factor when x=2". They're really asking for what the slope of the tanget is when x=a, and we can draw a secant through 2 points to find the average rate of change by doing a = f(x2) - f(x1) / delta x

[shut night]

so the idea is we move the second point really close to the first point

Yes and no

you want the rate of change at x=2

if you took a second point f(a) really far away from the first point

youd get the average between the two points, but thats not the rate of change at the first point

So you take the limit as a gets really close to x

which is the derivative of x or the tangent

[neon iron]

3Blue1Brown has an amazing video on what @shut night with a visualisation

https://youtu.be/9vKqVkMQHKk?si=-n-gaTfX9PUOkGwK

do check it out if you wish

[distant totem]

Ah alright, so to summarize a bit.

We can use the equation for f'(x) to find the rate of change at a given point x = a, when lim delta x -> 0. However, we can also find the rate of change of a secant by basically drawing it through 2 points in the graph and use delta y / delta x

[distant totem]

I'll definitely check it out, thanks!

[shut night]

Yeah just remember the secant is the average rate of change between two points, and the tangent is the rate of change at a given point

[distant totem]

Yeah alright, thank you so much! Looked really like they were the same in the book, but I'll watch the video to maybe make it easier by visualizing it 🙂

[fading owl]

we concur on what we watch

3b1b is an amazing channel

[shut night]

But the idea is you take a secant and move the second point reallllyyyy close to the first one (the limit) and then two points super close to each other must have the slope of the given point!

[distant totem]

I usually watch, the Chemistry Tutor as he also explains physics

[neon iron]

Agreed. This legend literally made an open source programme (which a lot of other maths channels use as well) for creating his maths animations

Grant Sanderson is really god tier

[fading owl]

yep he deserves an award

no scrap that

[neon iron]

Yeah he’s amazing as well

[fading owl]

five awards

[distant totem]

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[rare atlas]

'

[opal vault]

!status

[topaz sinew]

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

[rare atlas]

2

[topaz sinew]

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

[rare atlas]

when taking derivative , there is always one that does not change, (x-1)^2024
and there is one at the final 2024! but i can not find the middle one

[opal vault]

did you use leibniz formula of derivation?

[rare atlas]

dont know what is that

[opal vault]

f = g*h, all of which being functions

$f^{(n)}(x) = \sum_{k=0}^n \binom{n}{k}g^{(k)}(x)h^{(n-k)}(x)$

[thorny flame]

rafilou is not not born in 2003

[rare atlas]

no, i did not use

this

[opal vault]

you should, it helps a lot

[rare atlas]

ok i am trying

what is g^(0)?

[fair thorn]

just g

the 0th derivative of g(x) is defined to be g(x)

[rare atlas]

ok ty

with this formula, I find answer 0?

is it correct?

@opal vault ?

@fair thorn ?

<@&286206848099549185>

[somber geyser]

hai

[fair thorn]

sorry i had to go smwh

show your work

[rare atlas]

.

guys, help please

<@&286206848099549185>

[fair thorn]

like i said

show your work

[glad folio]

<@&286206848099549185>

[fair thorn]

@glad folio

why tf are you pinging helpers

a. in someone else's channel

b. within 2 minutes of the last ping

[topaz sinew]

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.

[rare atlas]

I said g(x)= x-1
h(x)= x+1

and used the formula given

it givess 0

[fair thorn]

show how you used it, maybe?

[rare atlas]

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[opal vault]

incorrect answer

[rare atlas]

why

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[vapid seal]

For question f are they orthonormal?

[vapid seal]

I found the dot product of the first vector and second = 0.000448

[vernal vale]

bad sign

[vapid seal]

Shouldn't we find if they are perpendicular?

[vernal vale]

yea they have to be an orthogonal set before they can be orthonormal for sure

[vapid seal]

The books says they are orthonormal, but I don't understand how

[eager compass]

orthonormal means each vector has length 1 and is orthogonal to each other

[vapid seal]

If the dot product was close to 0 can we say that they are perpendicular?

[eager compass]

no

for which one did you found 0.000448?

[vernal vale]

,w Dot[{0.921, 0.294, -0.254}, {-0.254, 0.951, 0.178}]

[thorny flame]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by janniku

[vernal vale]

not orthogonal

[eager compass]

close to 0 is not 0

[vapid seal]

ok

thanks for the help

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[mossy nest]

can someone help me with my homework?

this is my solution but im not 100% sure

also not sure with the derivatives

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@mossy nest Has your question been resolved?

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@mossy nest Has your question been resolved?

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@mossy nest Has your question been resolved?

[stark crater]

<@&286206848099549185>

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[merry bridge]

I'm supposed to find the derivative of this parametric equation, and I did. (sin(t)(2cos(t)-1))/(cos(t)-cos(2t))
But I'm supposed to simplify it further using the double angle identities. The problem is, when I do, I get left with a really ugly expression, and my prof said it's supposed to look "better"?

[merry bridge]

I simplified it to (sin(t)(2cos(t)-1))/((-2cos(t)-1)(cos(t)-1))

let me actually put this into neat formatting because it's unreadable inline

[agile harness]

$\frac{dy}{dx} = \frac{-2\sin(t) + 2\sin(2t)}{2\cos(t) - 2\cos(2t)}$

[thorny flame]

knief

[merry bridge]

yeah

because double angle on the sin 2t gives 2sin(t)cos(t)

but then

[agile harness]

yea

and he wants you to expand double angle

[merry bridge]

right and I did lol but this isn't neat

[agile harness]

is this your answer

[merry bridge]

he said that there's a way to simplify it further

[agile harness]

and what have you tried for that

if it helps, think of cos(t) as x

or u

u - 2u^2 + 1

[merry bridge]

we're not supposed to use half angle identity so I can't make it cot(3t/2)

i see no other way to simplify it

[agile harness]

how did he want it then?

[merry bridge]

that's the issue, he didn't really specify lol. We're supposed to take the limit for when t approaches 0, but like

im not sure how to take the limit of it if it weren't in this quotient form... so i guess this is the best i can do lol

nvm, sorry for wasting your time lol

[agile harness]

all good

[merry bridge]

!close

.close

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[royal sky]

why does the line here must be a homogenous equation

[sweet shard]

As opposed to what?

[royal sky]

just that, like why cant the line be (x,y) vector here why does it have to be a homogenous equation

[sweet shard]

so you mean c=0?

[royal sky]

no like ax+by+c=0

[sweet shard]

Yes that's the line in the question

What alternative line do you propose?

[royal sky]

nvm i realized

it was infinte

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[neon iron]

<@&286206848099549185>

[thorny remnant]

!15mins

[topaz sinew]

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.

[smoky sparrow]

angle in the centre is double angle on the circumference

[neon iron]

Yeah

[smoky sparrow]

great so you need the non-reflex angle of m(AB)

what must that be then?

[neon iron]

Then the answer is 50

degrees?

[smoky sparrow]

yep!

[neon iron]

Ok thank you

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[forest plank]

Someone please explain this statement in the solution

[alpine dirge]

which book is it? i really like these questions

[crimson wasp]

u live in India

[smoky sparrow]

suppose that g is not 1-to-1, that there exist natural a, b such that g(a) = g(b)
then f(g(a)) = f(g(b)) so that would imply that h(n) = n is also not 1-to-1

[alpine dirge]

thankss

[crimson wasp]

u do live in India

which state

[smoky sparrow]

f must be surjective also, shouldn't be too hard to see what happens if f is not surjective

[forest plank]

Oooooh got it

Thanks

[crimson wasp]

where u live

[smoky sparrow]

no worries!

[forest plank]

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[crimson wasp]

i livw in india too

[forest plank]

F is surjective because g is surjective correct?

[smoky sparrow]

g doesn't actually need to be surjective

[forest plank]

.reopen

[topaz sinew]

✅

[smoky sparrow]

wait

yeah you could have f(x) = tan(x) but g(x) = x for -pi/2 < x < pi/2

for instance

[obtuse garden]

writing rmo tomorrow?

[forest plank]

This ones a polynomial tho

[forest plank]

Ya

[smoky sparrow]

it doesn't matter
we just need to find what conditions $f \circ g$ being 1-to-1 imposes on $f, g$

[thorny flame]

south's secret twin brother

[smoky sparrow]

also tan x is not a polynomial but

[obtuse garden]

ooh noice which grade

[smoky sparrow]

and this is abstract I know

[forest plank]

ya

[forest plank]

message me in dm i dont want to clutter this chat

[forest plank]

since f o g of n = n is injective, g(n) is also injective, but how do we know that f is onto?

[smoky sparrow]

well if f is not onto, then there must be some k such that f(n) = k has no solutions

so n can be anything, and g(n) can be anything, so we can just replace $n \mapsto g(n)$

[thorny flame]

south's secret twin brother

[forest plank]

right got it

and since g(n) = f(n) + 2, that means that f is also injective and g is also surjective correct?

ok final question

what does it mean that f and g are inverse of each other

nvm got that too. becuase f returns the preimage of n on g right?

thanks

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[cedar spindle]

.open

.reopen

Solve for x, y
x/4 = y/5 and x + y = 54

[cedar spindle]

i know the answer but dunno how to solve 😔

<@&286206848099549185>

nobody replies?

ig imma close

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[regal cave]

hey

do you still need help

@cedar spindle

so you can write it as 5x = 4y

and then you can solve it via substitution

x = 4y/5

4y/5 + y = 54
9y/5 = 54
9y = 260
y = 260/9

and then you can solve for x

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[broken stone]

If you’re dividing 1/2x^2 by 1/2x, wouldn’t it be 1/4x instead of x? Watching a tutorial and apparently it’s x - 12, not 1/4x - 12

[heady bay]

Two ships leave from the same port. One ship travels on a bearing of 157° at 20 knots. The second ship travels on a bearing of 247° at 35 knots. (1 knot is a speed of 1 nautical mile per hour).

a. How far apart are the ships after 8 hours?

b. Calculate the bearing of the second ship from the first?

I cant answer question b

<@&286206848099549185>

[neon iron]

! occupied

[topaz sinew]

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).

@broken stone Has your question been resolved?

[broken stone]

^

[worn trout]

how did 5x become 6x on the second line?

oh nvm im blind

[worn trout]

so ur asking what happens if you divide x^2/2 by x/2? or 1/2*x^2 by 1/2**x

[broken stone]

Id guess the 2nd, I’m dividing everything in the 2nd line (except 0) by the GCF, which is 1/2x

[worn trout]

Yes and u did it correctly, I don't see the problem here

[broken stone]

My original answer was 1/4x - 3, I just corrected it according to a video guide

[worn trout]

okay so (1/2)*x^2 divided by (1/2)x is just x

Its pretty clear to see

this is mad confusing, let me work it out and ill send a pic

[broken stone]

Subbed in 2 for x and see it now, just saw dividing half by a half and thought “half of half is a 4th”

[worn trout]

just work it out like that

so much less confusing

[topaz sinew]

@broken stone Has your question been resolved?

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[lusty cedar]

Im so lost here

I got it wrong

for 6b

I used the addition formula

to get 5 root 3/2 cos x - 5/2 sinx = 4/5 sinx

and then just solve from there

Idk why its wrong

😭

[eager compass]

Can you show us your work?

[subtle dust]

solve cos(x-30) further

[subtle dust]

yo mb didint notice u opened it

but anyways

u opened it wrong

[agile harness]

$\cos(u-v) = \cos(u)\cos(v) + \sin(u)\sin(v)$

[thorny flame]

knief

[lusty cedar]

no because sin (-30) returns -1/2

[agile harness]

💀💀💀💀

🙏🏻🙏🏻🙏🏻

[lusty cedar]

wait what

[agile harness]

brother

[lusty cedar]

what have i done wrong

😭

[agile harness]

you’re not subtracting -30

youre subtracting 30

v IS 30

[agile harness]

^

[agile harness]

the audacity lol

[lusty cedar]

OHHHHHHHH

yeah my b hahaha

thank you

[agile harness]

you’re welcome

[lusty cedar]

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[faint otter]

how is this wrong

[neon venture]

huh (-2, oo) should be enough tho

[loud oasis]

i think you can combine your answers for the increasing part

[faint otter]

yeah i think im a bit slow

i didnt even notice i could combine them 😭 thank u guys

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[iron blaze]

If xy is a perfect square

[iron blaze]

and yz is a perfect square

prove that xz is a perfect square

So what i figured

is let xy = i^2

yz = j^2

therefore xy^2z = i^2j^2

and xz = i^2*j^2/y^2

so I have to prove that i * j /y is an integer

so if y is a prime number

than this is obvious

however if y is a composite number, how do i go about proving this?

yo

[outer salmon]

Since the square root of any natural number that is not a perfect square is irrational, xz must be a perfect square here

[iron blaze]

Oh i get it

[outer salmon]

(Assuming x and y are natural numbers?)

[iron blaze]

ya and xz isnt a fraction

its an integer

yes

x y and z are

natural numbers

not integers

natural numbers

ya

aight thanks

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[copper fossil]

can someone explain what is going on in this problem lol

[copper fossil]

This section is taking forever and the solutions given skip tons of steps

Specifically the part where it says "From the volume equation. h = ..."

Why is there a 4 in the numerator all of the sudden? don't I plug r and V into h to find h?

Here's the full problem:

I've already spent way more time on this than I can afford so I need to move on

[topaz sinew]

@copper fossil Has your question been resolved?

[acoustic pecan]

C=2pi r h + 2pi r^2

V=pi r^2 h + 2/3 pi r^3

V/(pi r^2)=h+2/3 r, h=V/(pi r^2) - 2r/3)

C=2pi r[V/(pi r^2) - 2r/3]+2pi r^2 = 2V/r - 4/3 pi r^2 +2pi r^2 = 2V/r +8/3 pi r^2

dC/dr=-2V/r^2+16/3 pi r =0

v/r^2= 8/3 pi r V=8/3 pi r^3, then r=...

subbing it into the h equation
h=V/pi * 1/(3V/8pi)^2/3 -...

=V/pi * (8pi/3V)^2/3 -... 8^(2/3)=4, is where the 4 is coming from

just follow the calculations

[copper fossil]

So I don't plug V = 8/3* pi * r^3 into h?

[acoustic pecan]

no youre plugging r=(3V/8pi)^1/3 into h

[copper fossil]

Is V not plugged in because the problem said volume is fixed?

[acoustic pecan]

yeah, since we know V is a fixed value it just makes it easier to work with than r

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[acoustic pecan]

r and h are our only changing things

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[boreal surge]

oops

[strong sable]

sry lol

[boreal surge]

nah i was just bad timing lol

I was wondering if I used the right formula for this

i used the 1-PropZtest formula for this

<@&286206848099549185>

[short mesa]

You have to transform it into the sampling distribution

but a 1-propZtest is right

[boreal surge]

I got 0.1935

do you want me to show my work?

[short mesa]

yes that would be good

[boreal surge]

Little messy

[short mesa]

looks correct to me

[boreal surge]

okay thank you

[short mesa]

if i’m wrong im sorry; it’s been a bit since i’ve done statistics but the first formula is right and im assuming you used a stats list for the second portion

[boreal surge]

This is all I did

[short mesa]

yep

looks like what i would do

do you know if it is correct? if so; you did it correctly

i don’t have access to a calculator rn

[boreal surge]

the answer choices were 0.1835 0.1822 0.1935 and 0.1922

the professor said to choose the closest answer

so 0.1935 is what I chose

[short mesa]

well you got a direct answer and your formulae are correct

[boreal surge]

Okay thank you

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[teal umbra]

Given triangle ABC inscribed in (O). Any circle through B, C intersects AC, AB times as at E, F. Prove that AO is perpendicular to EF

[teal umbra]

help me pls

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@teal umbra Has your question been resolved?

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[strange mortar]

Im unsure what to do for this. why would i break it up into 30 sets based on the odds?

[strange mortar]

this is a pigeonhole principle proof, but i dont really know where to start

i thought of the subset of primes or something but idk if that helps

and im guessing you cant really use that anyway

[topaz sinew]

@strange mortar Has your question been resolved?

[vernal ledge]

okk

so i tried searching up on this question

and i think i get it

so as the hint says you indeed have 30 odd numbers

but, the even numbers are also just odd numbers multiplied by a power of 2

say we have 5

and another even number 20

but then 20 is 2^2 * 5

and the even number which divides 5 must obviously be greater than 5

[strange mortar]

not necessarily

[vernal ledge]

example?

[strange mortar]

but like i think ive seen what ur talking ab

2*2=4

[vernal ledge]

like take a prime factorization of an even number

and remove ALL the 2 powers

so 4 = 2^2 * 1

so if you have an even number

you can remove all the two powers and map it to an odd number

which will divide the even number

[strange mortar]

im guessing ur reading off the same thing im seeing

but i just dont know what the number pairing hint means

or i meant the 30 sets thing

[vernal ledge]

so say you take a set containing all the 30 odd numbers

then no matter what single even number you take

youll have at least one odd number mapping to it

say we take 52

which is 2^2 * 13

but then we already have 13 in our set

[strange mortar]

ill just try figuring this out

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[jolly vale]

How do we solve this? This chosen answer is wrong btw

[loud oasis]

we would like to apply the intermediate value theorem

[jolly vale]

Yes

D can't be the answer bc the product would have to be negative to have a solution in the interval?

[loud oasis]

we can see that all of the solution choices involve 0, so we need to find a function which, given this information, is guaranteed to equal 0 in this interval

[jolly vale]

Such as?

[loud oasis]

well let's consider the statement of the intermediate value theorem

can we find a function, using the given information, so that 0 is between f(a) and f(b)?

[jolly vale]

Any function?

[loud oasis]

well we would like for it to be related to f(x) and g(x)

using information from the problem statement

[jolly vale]

Sorry cant think of anything

[loud oasis]

what information are we given in the original problem statement, about f(x) and g(x)?

[jolly vale]

both continuous on the interval could polynomials I think

[loud oasis]

let's not assume they're any particular class of funciton

is there anything else that we are explicitly given about them?

[jolly vale]

f(a)>g(a)

[loud oasis]

could we rearrange that inequality so that one side is 0?

[jolly vale]

f(b)<g(b)

[loud oasis]

yes. so could we do this with both inequalities

[jolly vale]

f(a)-g(a)>0

[loud oasis]

yes

what about the other one?

[jolly vale]

f(b)-g(b)<0

[loud oasis]

yes. can we combine those two inequalities now?

[jolly vale]

How?

[loud oasis]

... < 0 < ...

[jolly vale]

Oh okay
f(b)-g(b)<0<f(a)-g(a)

[loud oasis]

yes

so can we use the intermediate value theorem here?

[jolly vale]

Idk the only way I've used ivt was when both a graph and points were given

[loud oasis]

the intermediate value theorem says, that if we have some function $h(x)$ which is continuous on $[a,b]$, and we have a function value $y_0$ where either $h(a) < y_0 < h(b)$ or $h(b) < y_0 < h(a)$, we are guaranteed an input value $c$ in the interval $(a,b)$ such that $h(c) = y_0$

[thorny flame]

cloud

[jolly vale]

Is it similar squeeze theorm where if h(a)=h(b) then it's y

So answer c?

[loud oasis]

the y_0 we are interested in this case is 0

so do we have any functions $h(x)$ where we have $h(a) < 0 < h(b)$ or $h(b) < 0 < h(a)$?

[thorny flame]

cloud

[loud oasis]

(thinking about the inequality we just made)

[topaz sinew]

@jolly vale Has your question been resolved?

[jolly vale]

I'm not sure I understand what you are talking about, sorry got busy

[loud oasis]

remember that sums, differences, products etc of functions are also functions

[loud oasis]

we would like to use this inequality

[topaz sinew]

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[jolly vale]

Yeah h(a)and h(b)

[topaz sinew]

@jolly vale Has your question been resolved?

[smoky sparrow]

f(a) - g(a) > 0 and f(b) - g(b) < 0

now if you let h(x) = f(x) - g(x), what can you conclude

[jolly vale]

That's the answer is c cuz f(x)=g(x) when x=0

[smoky sparrow]

yeah

IVT implies that f(x) - g(x) = 0 has at least one sol

cause f(x) - g(x) is continuous

[jolly vale]

Thanks

[smoky sparrow]

cool no worries

[jolly vale]

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[smoky sparrow]

as helpers we do our best to explain
I've definitely been in situations where I wasn't explaining very well

also I see you've done the work above

[topaz sinew]

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[sweet trout]

Hi guys may I know how to solve this question?

[verbal crater]

what does diagram 13.1 look like

[sweet trout]

is it blurry

[verbal crater]

no its good

[verbal crater]

TS and PQ are not the same no?

[sweet trout]

uh ignore what i drew

i realised that after i took the picture

🗿

[verbal crater]

hmm then i guess heron's formula is your best bet

[sweet trout]

really?

[verbal crater]

you can find TQ and TS by using papa pythagoras

[sweet trout]

hmm

ill try

my other question isnt loading 🗿

[sweet trout]

its not 90 degrees though

[verbal crater]

is it not?

[sweet trout]

doesnt seem like it

[verbal crater]

T is vertically above P though

[sweet trout]

bro my other q isnt sending

.

[sweet trout]

fair enough....

ok wow

[verbal crater]

classic tm speed

or whatever isp u using lol

[sweet trout]

tm?

[verbal crater]

telekom malaysia

[sweet trout]

how tf do u know

no way

🗿

im bad at ambigious

shiorin left the chat pov

[verbal crater]

nah lol im thinking

[sweet trout]

this is the one time i got confused in ambigious

usually is i can imagine a isosceles triangle

[verbal crater]

i used to just randomly draw it lol

[sweet trout]

u graduated?

[verbal crater]

ye

[sweet trout]

damn

hard to comeby a senior in the same country

• holofan and a hoshiyomi

💀

[verbal crater]

lmao

hmm

maybe something like this

[sweet trout]

heh

how does that work

[verbal crater]

wait nvm that doesnt work lol

[sweet trout]

ya thats what i was thinking as well

🗿

[verbal crater]

oh i got it maybe

you gotta project a line down to PS such that Q'S' is the same as QS

[sweet trout]

i beg ur pardon?

u mean reflect on PQ?

[verbal crater]

something like this

[sweet trout]

i c...

it didnt specify PS has to be same length

wow

[verbal crater]

ye only PQ and QS has to be the same

[sweet trout]

now that u drew it out

i see the isosceles triangle now

[topaz sinew]

@sweet trout Has your question been resolved?

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[crisp night]

How do I find the sum of two vectorial spaces? Exercise number 3. The vector that generates W are v1(1,1,0,0); v2(-2,0,1,-1)

[desert peak]

Posso scriverti in italiano?

[crisp night]

I found the intersection of the two spaces by a system of the Cartesian equations of the two vector spaces

Va bene

Anzi molto meglio

[desert peak]

Ok, l'intersezione te la ottieni come intersezione delle equazioni cartesiane dei sottospazi

[desert peak]

L'altro spazio mi pare che si ottenga (ho fatto geometria 1 sei anni fa, quindi verifica) come span dei vettori che generano i due spazi

[crisp night]

Si però essendo una base devo rimuovere i vettori che possono essere scritti come combinazione lineare

[desert peak]

Questo è vero

[crisp night]

Applicando il teorema della dimensione di uno spazio vettoriale la dim di U+W deve per forza essere 3

Quindi ho pensato di togliere uno dei vettori che mi dava lintersezione se sommato

Che sicuramente può essere una soluzione ma cercavo qualcosa di più generale

[desert peak]

Mi sa che ti tocca fare questo oppure provare quello che dicevo tu. Se metti i 4 vettori assieme, sei sicuro che i due vettori di U (o di V se parti da V) stanno nella base di U+V

Bisogna vedere quale degli altri due che rimangono è indipendente o meno con gli altri

[crisp night]

Conviene procedere con un sistema a questo punto riscrivo tutti e quattro i vettori e trovo i parametri liberi che generano lo spazio U+W

[desert peak]

Anche questo si può fare

[crisp night]

Brutta situazione però

Ponendoli a sistema mi da che nessuno è linearmente indipendente

Sia ad occhio sia utilizzando Sympy

Però per il teorema sulla dimensione deve essere per forza 3 e quindi uno va scritto come combinazione lineare per forza

Io concluderei togliendo uno dei vettori che sommati mi da il vettore base di U intersecato W a questo punto essendo la somma definita seguendo il teorema vettori di U vettori W - vettori dell Intersezione

[desert peak]

Io ho fatto questo

Ho messo riga per riga i vettori e poi ho ridotto per riga la matrice. Siccome l'ultima riga è nulla, allora il quarto è c.l. degli altri 3

Questo funziona per le regole della c.l.: quando hai uno spazio che è generato da una combinazione lineare di vettori, tu puoi cambiare uno dei vettori con lo stesso vettore a cui sommi o sottrai una combinazione lineare dei rimanenti

[topaz sinew]

@crisp night Has your question been resolved?

[crisp night]

Corretto

Grazie mille

Il risultato è analogo a quello che ho trovato io

Questo metodo è molto molto più generale grazie mille

.close

[topaz sinew]

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[topaz sinew]

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[burnt fjord]

Show that B=5^n×12−12^n×5 can be divide by 7.

[burnt fjord]

can you help me resolve this with congruence ?

[smoky sparrow]

oh that is just congruent to 5^n * 5 - 5^n * 5

12 = 5 (mod 7)

[burnt fjord]

can you explain a little more please i dont really understand

[smoky sparrow]

https://www.youtube.com/watch?v=6dZLq77gSGU

[golden blade]

You can subtract 7 from 12

[smoky sparrow]

you can just replace the 12 with 5 wherever 12 appears

[burnt fjord]

just because 12 = 1*7 + 5 ?

[smoky sparrow]

yes

[burnt fjord]

so in every congruence expression you can just simplify numbers with no others impacts on other ?

[smoky sparrow]

yeah

[burnt fjord]

ok i didnt know that thanks 👍

and so 5^n5-5^n5 is just 0 so ≡ 0 ?

[smoky sparrow]

https://math.stackexchange.com/questions/1070722/find-the-least-positive-integer-x-that-satisfies-the-congruence-90x-equiv-41?rq=1

here's another example

[smoky sparrow]

yes

[burnt fjord]

ok thanks 👍👍👍

.close

[topaz sinew]

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[burnt fjord]

hey excuse me i dont understand this reply in the link you gave me

why 90x is equivalent to 15x

[topaz sinew]

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[pearl fog]

find all integers n such that x^3+y^3+z^3-3xyz=n has a positive integer solution (x,y,z)

[pearl fog]

so like if you factor tou can get

(x+y+z)(x^2+y^2+z^2-xy-xz-yz)=n

but idk what next

ig n is positive

idk what next

wait gold on

doesent this like make no sense

x^3+y^3+z^3-3xyz has like infinite ranges

[prisma mesa]

it does, the question is what exactly is its range

if x, y, z in N

[pearl fog]

yeah but im pretty sure there is infinite n

[prisma mesa]

yeah, there 100% is

[pearl fog]

wtf

.close ok

[topaz sinew]

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[topaz sinew]

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[left egret]

I type

[left egret]

Oh ok

I need help

With a math problem

[topaz sinew]

No need to ask “Can I ask…?” or “Does anyone know about…?”—it’s faster for everyone if you just ask your question! See https://dontasktoask.com/

[left egret]

Look this is the math problem if someone knows please help me

I did the basic stuff already but don't know what's next

[finite moth]

what’s the length of the blue line?

[left egret]

20

r+r

[finite moth]

what’s the length of each side of the square

(use x for the little circle radius)

[left egret]

10?

no wait

I separate the square into a triangle

[river marsh]

remember its the 45 45 90 triangle

[left egret]

wdym

like angle?

right angle yeah

[finite moth]

the red is the radius of the large circle

the green is the radius of the small circle

what should the length of the side be?

[left egret]

so i use pythagorem theorem to find the hypothenius?

[finite moth]

yes

and then solve for x

[left egret]

once i have the answer for the hypothenius what should I do next?

[finite moth]

if we use s for the side of the square

we know that s^2 + s^2 = 20^2

we also know that s = 10 + x

which means that (10+x)^2 + (10+x)^2 = 20^2

now you can expand and then solve for x

[left egret]

would it be 10√2 − 10?

[river marsh]

Correct

[left egret]

aah thank you very much

[topaz sinew]

@left egret Has your question been resolved?

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[keen plinth]

so i’m really confused about how an isolating point could be continuous, mainly the difference of continuity of function and point

[keen plinth]

so is my drawing of understanding p, E and f(x) correct, according to the statement?

[pseudo jetty]

what do the arcs represent?

[keen plinth]

upper arc represents the function f(x), lower arc represents the domain

[pseudo jetty]

you've draw p somewhere floating in midair, why?

p is a point of the domain E

it should be sitting on the horizontal axis

[keen plinth]

yes, sorry it means the x-coordinate of that point

[pseudo jetty]

is p isolated in your drawing?

[keen plinth]

ohhhh

so what i've drawn is an isolating f(p) but not p right?

[pseudo jetty]

well im not sure which set you consider f(p) to be part of

perhaps of the graph

then you've drawn (p, f(p)) as an isolated point of the graph of f

[keen plinth]

yes but p itself isn't isolated in E

[pseudo jetty]

it is most definitely not in your picture

[keen plinth]

so what will a correctly depicted graph looks like

[pseudo jetty]

every point of Z as a subset of R is isolated

try drawing a function defined on Z

then see if its continuous

[keen plinth]

is it a must that it is only defined on Z?

[pseudo jetty]

i guess it isnt made explicit in what you sent

f should be a function E -> some other set

presumably R

f : E -> R

[keen plinth]

so having an intuitive thought on Z only tho

[pseudo jetty]

Z is just a convenient example where every point is isolated

feel free to use other sets

,, (-\infty, -1) \union \set 0 \union (1, \infty)

[pseudo jetty]

here 0 is isolated

,, \set {\tf1n \where n \in \Z \setminus \set 0}

[pseudo jetty]

here again every point is isolated

[keen plinth]

I still don't get how it can be continuous tho, what will be the visulization of the graph, the continuity at a limit point statement is only true if it is a limit point?

[pseudo jetty]

its just requiring the limit to equal the value of the function

neighbourhoods of an isolated point eventually only contain the isolated point itself

so you are taking the limit over a single point

[keen plinth]

so are we using the epsilon delta definition here

[pseudo jetty]

sure why not

$L$ is a limit of $f$ at $x$ if for all $\epsilon > 0$, there exists $\delta > 0$ such that if $\abs {x - a} < \delta$ then $\abs {f(x) - L} < \epsilon$

[pseudo jetty]

if $p$ is an isolated point of $E$, then there exists $\eta > 0$ such that $\set {x \where \abs {x - p} < \eta} = \set p$

[keen plinth]

Ohh wait, so as we have the value of f(x) defined here as p, any $\epsilon > 0$ will be automatically true for this statement

[thorny flame]

ManMan

[pseudo jetty]

then taking $\delta = \eta$, you yield $\abs {x - p} < \eta$ if and only if $x = p$, whence $\abs {f(p) - f(p)} < \epsilon$ holds trivially for all $\epsilon > 0$

[keen plinth]

wait, lemme digest

[keen plinth]

So everything is based on definition

analysis course😭

[pseudo jetty]

what isn't based on definitions

all of mathematics is based on definitions

[keen plinth]

yeh, it is challenging but interesting

through out those definitions, a lot of misconceptions in calculus are cleaned

[pseudo jetty]

im sure you can get a nice picture of this

[keen plinth]

thx for your help

.close

[topaz sinew]

Channel closed

Closed by @keen plinth

Use .reopen if this was a mistake.

[pseudo jetty]

here is a picture btw

0 on the x-axis is isolated

so whatever point you put as the blue one, the function will be continuous

[keen plinth]

I was confused about HOW it looks so discontinuous, but then realise that the requirement of continuity in the statement only holds for domain E, so it is still continuous in the domain E

[topaz sinew]

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[novel solar]

a couple decide to start having children and keep having children until they have the same number of boys and girls. how many children do they expect to have?

[neon iron]

Is this real

[fair thorn]

yes

[novel solar]

lol

[neon iron]

So isn't each birth a 50/50 chance

💀💀

[novel solar]

yeah like the prob of either child is 50/50

we're trying to find the expected number of children

[fair thorn]

i think it's fairly obvious?
2?

[novel solar]

so like the expected number of coin flips if we want to stop at an equal number of heads and tails with a fair coin

n

[fair thorn]

actually yeah interesting question

[novel solar]

why'd you say 2? the value is 2 if you have one boy and one girl or one girl and one boy. that happens with probability 1/4 each so 1/2 total

[neon iron]

33

3*

[novel solar]

so it's 2 only 1/2 of the time

[neon iron]

I think it's 3 man

[novel solar]

how'd you get that

I wrote code to solve for this and I got

[fair thorn]

some power of two no doubt

[neon iron]

Okay lile

[hearty turret]

define X_n = difference between #boys - #girls

[neon iron]

Don't u need atleast one gender to even start counting

😭

[hearty turret]

then X_n is a simple symmetric random walk starting at 0

define T = inf{n >= 1 s.t. X_n = 0}

we want to compute E(T)

it isn't too hard to show E(T) is infinite

[fair thorn]

LY what year of undergrad ru in

[hearty turret]

so tell that couple to better start getting busy!

[hearty turret]

2nd

[neon iron]

😭😭

[novel solar]

infinity doesn't sound right...

like

[novel solar]

first of all i don't really understand this my bad

[hearty turret]

if u were wondering abt the courses i've done/doing,

• numbers&sets (introductory maths course), differential equations, vectors&matrices, groups, vector calculus, dynamics&relativity, probability, analysis, groups rings & modules
• analysis&topology, markov chains, quantum mechanics, linear algebra, methods, galois theory & prob&measure (unofficially)

[hearty turret]

yeah probability is weird

[neon iron]

Well if u think abt it

[novel solar]

wow i ran a simulation with 1000 trials and it gave me 25k as the answer

a lot larger than i thought

i thought it'd be between 2 and 6 dang

[neon iron]

I think it's bc achieving the same amount of boys and girls

IS infinite

[neon iron]

💀💀💀

[hearty turret]

yeah basically the answer is infinity, so if you keep running more and more trials, you should find that the answer just keeps going up and up