help-33

[lean falcon]

let me show you everything

this is the question i need to answer

and this is all my work

[lean falcon]

ohh is that 50x??

🤦‍♂️

one sec

[cunning fiber]

@lean falcon are you done

[lean falcon]

im fixing everything

ill send a pic

[lean falcon]

i'll send a vid to make sure everything is good rn

just fixing it

[lean falcon]

can you go through this last part with me

[lean falcon]

how does it look btw

i still cant see the hole why is that

is my hole wrong

or vertical asymptote

[cunning fiber]

,w expand (t+4)(t-10)(t-15)

[elfin berry]

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[cunning fiber]

,w maximize (t+4)(t-10)(t-15)

[elfin berry]

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[cunning fiber]

Check a and b yourself against this

[cunning fiber]

Zoom in and don’t highlight every point

Although desmos sometimes breaks with that kinda stuff so who knows

[lean falcon]

the hole is -4 right

[cunning fiber]

,w discontinuity (t+4)(t-10)(t-15)/(t^2-3t-28)

[elfin berry]

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[lean falcon]

okay okay

[cunning fiber]

I’d write the hole as a coordinate point but do what you want ig

[lean falcon]

i think my teacher wants it as hole i would write it like that

[cunning fiber]

k

[lean falcon]

i dont get it i'm zooming in i just see these points

i'm following a similiar vid my teacher gave me with a diff problem and for her it shows

[cunning fiber]

The grid lines are in increments of 5

You’re looking for behavior at 4

[lean falcon]

the thing is the graph counts as some grade points so i have to find it

[cunning fiber]

.

[lean falcon]

how do i change that

[cunning fiber]

I meant to zoom in more

Maybe you can go into settings if you want to change the scale

Idk

[lean falcon]

these are the only options i think

[cunning fiber]

.

[lean falcon]

on which side

[cunning fiber]

The teacher’s graph is way more zoomed in relative to how zoomed in you are rn

[cunning fiber]

Dude just zoom in on the interval from [-5,0]

Or closer to 4 if you want to be specific

[lean falcon]

lemme take a vid

[lean falcon]

look how i'm zooming

@cunning fiber are you still here

[lean falcon]

ohh got it

but why do i have to hold and press

[cunning fiber]

floating point errors

[lean falcon]

it doesnt have a button?

[lean falcon]

i see ok

okay perfect

let me write it down

@cunning fiber everything finally looks good ty bro ur a lifesaver 🐐

[marsh citrus]

@lean falcon Has your question been resolved?

[marsh citrus]

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[sleek crater]

What do they do to go from the top highlighted to the bottom?

[hazy lion]

looks like they integrated

using the power rule

[sleek crater]

what happened to the 4/3 and 1/9?

[hazy lion]

you know the power rule?

[sleek crater]

oh shit

[hazy lion]

theres some introduced constants

they interact with the coefficients

[sleek crater]

so you multiply y^1/3 * 1

[hazy lion]

well, constants isnt clear in this scenario

[sleek crater]

and divide by the result

[hazy lion]

lets look at one term

[sleek crater]

I forgot about the divide part

[hazy lion]

$\int\frac 43 y^{1/3}\dd y$

[elfin berry]

jan Niku

[sleek crater]

y^4/3

divide by 4/3

[hazy lion]

this is $\frac 4 3 \cdot \frac 34 y^{4/3}$

[elfin berry]

jan Niku

[hazy lion]

well plus C but blah blah

hopefully that tracks

you can double check the rest of the terms but this is what is occuring

[sleek crater]

ok gotcha that tracks I forgot about dividing

Thanks

[hazy lion]

[sleek crater]

.close

[marsh citrus]

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[marsh citrus]

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[open tartan]

I don’t know how to do the last ones

[still temple]

You have the function what is the y value when x = -5...

For example, at x = -5, what is your y-axis value?

[open tartan]

0

It’s the last 2 I don’t get

[dry fjord]

think of f(4) like a number rather than something else

f(4) is = to what in this case?

i couldnt tell you tbh, i cant read the graph with your photo well enough XD

but my point being, you plug in numbers for each function there

so it will be 3(a) + b - 1/2(c) = ?
where a, b, and c are the values of f(4),f(0),f(-3)

and for the final question you work backwards, f(x) = 3, you want to look at which x value turns into a y value that equals 3

and that is your answer

[marsh citrus]

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[open tartan]

Thanks

[marsh citrus]

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[night lion]

am i tripping or is this genuinely wrong

[night lion]

slope is negative for obtuse angle right

[marsh citrus]

@night lion Has your question been resolved?

[cloud field]

yeah why

[night lion]

they took positive

tf

[cloud field]

well tan(angle) = m

your m is -k

if angle is obtuse, then tan(angle) is -(some number)

-(some number) = -k

then just remove the - signs

@night lion

[night lion]

umm wdym

[cloud field]

so the line makes an angle with the x axis

[night lion]

yes

[cloud field]

if you take the tan of that angle

you get its slope

[night lion]

of 150 degrees

[cloud field]

right?

[night lion]

tan 150 = -1/sqrt3

im not sure why u think i dont understand slope

i was asking if they did the math wrong

[cloud field]

no they didnt

[night lion]

or am i missing something mundane

[cloud field]

the equation for zero order is A = -kt + A0

you know this right?

[night lion]

ohhhh yea

fuck i assumed

k is positive and making it negative

🤦🏻‍♂️ chemistry

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[marsh citrus]

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[night lion]

thanks

[marsh citrus]

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[meager badger]

How do i do this

[meager badger]

Would this equation wor k

[marsh citrus]

@meager badger Has your question been resolved?

[meager badger]

<@&286206848099549185>

[marsh citrus]

@meager badger Has your question been resolved?

[vestal forge]

you can draw it out to visualise

the area in red is what the whole expression represents. The area in black is what the second integral represent

[meager badger]

how do you do that

[vestal forge]

do what exactly?

[meager badger]

Shade in the area

[vestal forge]

as in how I do it technically? or how I know which area to shade in?

[meager badger]

technically

[vestal forge]

you can ask desmos to graph an inequality for you, and it will shade in the region containing points (x, y) that satisfy the said inequality

in the case above, if you graph (x-2)^2 + y^2<=4, it will shade the red disk along with its border.
Equivalently you could graph r<=4cos(theta) and this will give the same result

[meager badger]

ahh makes sense

thanks so mcuh

[marsh citrus]

@meager badger Has your question been resolved?

[marsh citrus]

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[open seal]

thats og question

heres my work

[trim quest]

you accidentally turned -3 into -3x

[open seal]

where

ok i see it

.close

[marsh citrus]

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[open seal]

.reopen

[marsh citrus]

✅

[open seal]

so I get -2h^2 -3hx all over h , im not sure where to go next

[dark leaf]

factor out h

and cancel it

[open seal]

so its just -2h-3x

[dark leaf]

yes

now put h = 0 and enjoy

$\lim_{h\rightarrow 0} \left(-2h-3x\right)=-3x$

[crystal lintel]

hmm

the answer shouldn't be that though

[open seal]

says its wrong

[trim quest]

-2h^2-4hx

[dark leaf]

the derivative should be -4

[dark leaf]

yes

[open seal]

idk what yall are on about

[dark leaf]

?

[trim quest]

you just made a little mistake here. You said -3hx but I think you should have -4hx

[open seal]

bruhh im such a noob

.close

[marsh citrus]

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[proud basin]

reel

[open seal]

idk why i closed it it's gunna reopen cuz im gunna need help with the next one (likely)

[wise musk]

on the 4th line where did 3x come from?
i think it should be 3
and the 3s cancel out each other

[dark leaf]

$f(x+h) - f(x) = (3 - 2x^2 - 4xh - 2h^2) - (3 - 2x^2) = -4xh - 2h^2$

[open seal]

guys i finished it im all good

[open seal]

yeah thanks!!

[dark leaf]

[marsh citrus]

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[velvet cairn]

In how many ways can you arrange 3 men and 3 women in a circular table such that men and women are alternate

[slender socket]

We're going to need some more context, there's a lot of ways you could interpret that question

[velvet cairn]

Like ?

[slender socket]

Do we care that a man is in a position or do we care about which specific man is in a position? Do we care about positions relative to each other or do we care about positions relative to the table? How many positions are at the table?

[velvet cairn]

The table is purely symmetry
A perfect circular transperant glass and there are 6 places

[slender socket]

So if there was just one person at the table would there be one way to arrange them or 6?

[velvet cairn]

1

Their order with respect to each other matters only
Not wrt the table

[slender socket]

And if we swap 2 men does that count as a new arrangement?

[velvet cairn]

Yes except the diametrically opposite case

[lucid turret]

if the men are congruent then only 1

and the positions also, if, don't matter

[velvet cairn]

Men and women are definately different

How can all the men be assumed to be identical

They are all distinct

[lucid turret]

m1, m2, m3
w1, w2, w3

[velvet cairn]

Yep 👍

[lucid turret]

I think they are just like apples

3 men
3 women
3men = men + men + men

[velvet cairn]

No
Say the men are Adam, Ben and mark
And the women are Stacy, mary and jane

[lucid turret]

oh you're the OP

[slender socket]

Well now that we have all that we can do some combinatorics: Let's say that the table always starts at a given man so we have a fixed starting point since rotation around the table doesn't matter. From there we have 3 women who can be positioned next to him, then 2 men next to her, etc.. Do you see where we can go from there?

[lucid turret]

let's consider point C as the base

[eternal grail]

all arrangement - women always sit next to each other and men always sit next to each other
(6-1)! - (2-1)!x3!x3!

-1 because it is a circle

[slender socket]

Firstly, wrong. Secondly, don't just give the answer.

[eternal grail]

oh my bad, i forget to consider some other possible cases

ok i find the exact problem in my textbook so i think it should be correct
let three man sit first in a circle, so is (3-1)! of ways, then let woman sit between every two mens, since theres three slot to be filled the circular property of circle is gone, so the combination is 3!
so should be (3-1)!x3!

[lucid turret]

Big case 1:
_m1 comes the first(C), then w1 and then m2 and then w2

m1 is at C, m2 at E and m3 at G
now rotate
number of possible cases = 6

Bigger Case2: m1 comes the first(C) but then comes w1 and then m3 and then w2
Here also cases = 6

further continuing, number of possible bigger cases can be calculated as:

m1,w1,m2,w2,m3,w3
m1,w1,m3,w2,m2,w3
m1,w2,m2,w1,m3,w3
m1,w2,m2,w3,m3,w1
m1,w3,m3,w1,m1,w1

• one more case of m1, w3

now m2 or m3 can come it w_x's left
when m2 is at the left of w_x, bigger case = 6
similarly for m3 it is 6

so every women and every man has experienced every women and every other man.

Total number of cases = 6 * 3* 6

= 108

point out the mistakes

[slender socket]

Firstly, wrong. Secondly, don't just give the answer.

[lucid turret]

ik it's wrong

[slender socket]

... Then why give that answer?

[lucid turret]

i said point out the mistakes

[slender socket]

I can't do that because I can't make sense if what you're even saying

[eternal grail]

i mean usually these permutation problem dont need to list out all the cases so detailed

[lucid turret]

yes

[slender socket]

Anyway, @velvet cairn, refer the the message I'm replying to for how to go about solving this

[marsh citrus]

@velvet cairn Has your question been resolved?

[lucid turret]

is the answer 72?

12* 6(for positions)

[slender socket]

No

72 would be correct if arrangments were relative to the table, but they're not, as discussed above

[lucid turret]

ohok

[velvet cairn]

So we have 3 vacant spaces for 3 women and 2 vacant spaces for 2 men
So it should be 3!×2! ryt ?

[sacred idol]

the pattern is always MWMWMW

now ignore the circle scenario

actually ill rephrase the last sentence

dont worry about rotating the table for now

there are 6 people in the table

3 men and 3 women alternating

now, do you know how many ways you can permute n different objects in an arrangement?

its a simple term

[velvet cairn]

n !

[sacred idol]

yep

so how many ways can we permute the men and women?

each separately

among themselves

[velvet cairn]

3!×3!

[sacred idol]

yes

but we have a problem

we can rotate the table

[velvet cairn]

Mhm

[sacred idol]

and if we turn it twice we may get the same permutation

this means there are three ways we can get the same permutation from this

[velvet cairn]

I get it

[sacred idol]

no rotation -> MWMWMW
once -> WMWMWM
twice -> MWMWMW

[velvet cairn]

Each linear permutation gives arise to 3 circular permutations

[sacred idol]

yes

im kind of confused of the answer though

seems too small

you should get 6*6/3 then

which is 12

ok never mind

its 12 yeah

another way people approach this problem is to start with the rotating problem first

but personally i find it hard to comprehend

[marsh citrus]

@velvet cairn Has your question been resolved?

[marsh citrus]

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[night mica]

what a shining keyboard

[still bane]

Bruh

Anyone ?

[marsh citrus]

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[marsh citrus]

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[cunning sapphire]

I'm honestly very confused as where to even start

[frail orbit]

what do vertical asymptote and horizontal asymptote mean?

[cunning sapphire]

that the graph never touches those points, only approaches them

[frail orbit]

i mean more in a limit sense, or what it tells you about a functions domain/range and such

like what does x=-9 being a VA actually tell you about (x+p)/(qx+r)

[cunning sapphire]

that the range is infinite?

[frail orbit]

how do you normally identify vertical asymptotes on rational functions?

[cunning sapphire]

equal the denominator to 0

[frail orbit]

that works here too

[cunning sapphire]

qx+r will eventually give me -9

[frail orbit]

no

[cunning sapphire]

okay

uh

no clue then

[frail orbit]

remember that it says vertical asymptote at x=-9

[cunning sapphire]

oh

im dumb okay

so q(-9)+r=0

[frail orbit]

indeed

[cunning sapphire]

can it be stated that q+r=0 then? if i move the -9 dividing to the other side

[frail orbit]

no

[cunning sapphire]

okay I won't do that then

[frail orbit]

you have to divide all terms by -9 to each side

you didn't divide r by -9 in that, but yes. I wouldn't focus on doing that yet

now

[cunning sapphire]

the horizontal asymptote is what trips me up

cuz how is 1/q gonna give me -3

[frail orbit]

are you confused why that's true?

or

[cunning sapphire]

yea

like

how is a fraction smaller than 1 gonna give -3

[frail orbit]

,calc 1/(0.5)

[elfin berry]

Result:

2

[cunning sapphire]

ahhhh

,calc 1/(-0.25)

[elfin berry]

Result:

-4

[cunning sapphire]

okay!!!

[frail orbit]

q is definitely smaller than 1

[cunning sapphire]

gotcha

[frail orbit]

in general 1/(1/x)=x (given x not equal to 0)

[cunning sapphire]

so q is -1/3

[frail orbit]

yep

[cunning sapphire]

,calc (-1/3)*(-9)

[elfin berry]

Result:

3

[frail orbit]

and this equation allows you to solve for r

[cunning sapphire]

so r=-3

[frail orbit]

yep

now all we need is p, and we use the information we haven't used yet

[cunning sapphire]

and to solve for p i can just substitute x and y with the coordinates

[frail orbit]

yes

[cunning sapphire]

okay!!!! gotcha

ill do that rq

so p=-2

awesome!

thanks so much

.close

[marsh citrus]

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[marsh citrus]

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[graceful nacelle]

i have

[graceful nacelle]

some money

that will grow 50% per week

it is $150

i want to calculate how much money i have made in the last month after x amount of weeks

my total amount after x weeks would be 150*1.5^x

so for the last month it would be something with a sigma right

difference of 2 sigmas

something like this?

hmm idk

[dark leaf]

compound or simple?

[graceful nacelle]

compound

[marsh citrus]

@graceful nacelle Has your question been resolved?

[marsh citrus]

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[sage fable]

Hello! I would like some hints on how to solve the (C) part, this question is part of a self-assessment in my uni! Thanks

[sage fable]

I was able to solve (a), (b), (d) with sandwich theorem, so i believe (c) can also be solved using sandwich theorem

But i cant find convergent sequences to use as bounds for the question to get a convergence point

<@&286206848099549185>

[granite wren]

Use sandwhich properly!!

[sage fable]

yeah to use it i need the bounds

but im struggling to figure out the bounds

[granite wren]

try it again

Bye

[sage fable]

what 😭

ok

[mellow galleon]

welp, i dont take uni math, but if i had to gamble a guess, i would try to show that its always > 0 and thats its monotonously decreasing (is that even the right way to use that word...)

dont mind me if it doesnt work 😛 just here to learn as well

[sage fable]

thanks, ill try that

yeah i did that, is that enough to prove it converges?

also this way i cant find the point of convergence tho right?

[mellow galleon]

i mean ill think thats enough (dont trust me on this rly im just a high schooler)

as for what it converges too, thats a good point i didnt think of thta, i only thought you needed to know whether it converges

[mellow crag]

that would work yeah

[mellow crag]

that is enough to prove it converges

but it doesn't tell u about the point of convergence

[sage fable]

what should i do for the point of convergence?

[mellow crag]

ngl normally when questions say "investigate the convergence/divergence of X" i feel like it usually only means determine whether or not they converge but it'll depend on who wrote the q

[mellow crag]

i'll work on it first and lyk but i think u should probably take a guess as to what it converges and see if u can prove that

oh wait no i remember how to do this question now

[sage fable]

alright, and yeah this question does only ask to check if it converges, but the prof said if u find the point of convergence u get more marks

[mellow crag]

yeah there is a way you can 'sandwich' it it's not really a sandwich, ignore me

[mellow crag]

||what's kinda annoying is that we have (n+1)^a - n^a, which intuitively should be really close to 0, but it's annoying that we don't have something obvious like 1/n - think about how you can manipulate the expression to get x_n = 1/stuff||

[sage fable]

lemme try then

[wanton willow]

by sandwhich theoreom do yall mean the squeeze thereom?

[grave dove]

you showed that its > 0 for all n, and you showed that it is decreasing for all n

[wanton willow]

im assuming so since they both seem to imply the same thing ig

[mellow crag]

yh

[wanton willow]

ok

[grave dove]

Now ask yourself, at what value does the sequence start?

x0 = ?

[sage fable]

i think n belongs to N

so gotta start with 1

i mean in class we always started with n = 1

[grave dove]

suppose it starts from 0

x0 will be 1

all next terms will be less than 1

but must remain > 0

so it converges to 0

[mellow crag]

that doesn't imply it converges to 0?

[wanton willow]

w h a t

[mellow crag]

i.e 1/(n+2) + 0.0005 doesn't converge to 0

[mellow crag]

but satisfies everything u've said

[wanton willow]

thats a very weird assumption tbh

well it also remains greater than -5 why doesnt it converge to -5 in that case?

[sage fable]

right

thats true

[wanton willow]

you cant base convergence on the minimum value for it

[grave dove]

its like 1/x function

decresisng but >0

[mellow crag]

^

1/(n+2) + 0.00005 is less than 1 for all n>=1, and it's greater than 0 but it doesn't converge to 0

[wanton willow]

yeah but the decreasing can be logarithmic
so it becomes slower and slower and approaches a point still greater than 0, like 0.0004 or something

[grave dove]

bruh were talking limits here...

[mellow crag]

while it is true that the minimum value of the sequence is what it converges to, we don't know that 0 is the minimum value

[wanton willow]

and?

[wanton willow]

exactly yeah

[grave dove]

covergence point is the limit of xn when n tends to infinity, right

[wanton willow]

thats what i meant mb

[sage fable]

i got it

[grave dove]

so just say 0

[wanton willow]

but we dont know definitively that x = 0 is the minimum value

nor have we PROVEN that the function is always decreasing

even if its obvious we've still gotta prove it

[sage fable]

so basically, the lower bound is 0, the upper bound can be the starting point of the sequence which is always greater than any other element of the sequence because its monotonously decreasing

for alpha -> 0, the first element of sequence also tends to 0

does that work

[mellow crag]

no

what if the answer depended on alpha?

if u can prove that for all alpha, we have the same limit then yeah ur thing should work

[sage fable]

no that wont work then

[mellow crag]

but if the answer were i.e. alpha^4 then no it wouldn't

[sage fable]

hmmm

[mellow crag]

btw this is an analysis question, not really a calculus question

[wanton willow]

yeah

[mellow crag]

in analysis if a question asks for a limit l, you need to prove it rigorously

(level of rigor depends on how far into analysis u r)

[wanton willow]

btw are yall SURE its monotonely decreasing?

even if it seems obvious shouldnt we prove it rigoriously first before moving on?

[sage fable]

i proved using its first derivative

the first derivative of x_n is always less than 0

which means x_n is decreasing

i hope thats what it means, otherwise i will fail calc

[wanton willow]

for some reason rigorous limits scare me to be honest so ive never really learned the rigorous epsilon-delta definition of limits
(this is only possible cuz i self-learned calculus ofc, i didnt skip a part of cirriculum in a school course just cuz i felt like it)

[sage fable]

we are doing epsilon delta definitions

[wanton willow]

yeah

youre right i think dw

[sage fable]

um so

[mellow crag]

^

that's the best way to go about actually finding the limit

[sage fable]

i am still thinking about that yea

i got it to 0 x oo form

[wanton willow]

damn i hate stuff that is so obvious un-rigorously but is hard to prove rigorously

[sage fable]

wait what is oo^a

[wanton willow]

can you play with it abit and use lhopitals rule?

as in turn it into a 0/0 or infinity/infinity fraction

since in most cases you can

[mellow crag]

0 x infinity?

[sage fable]

yes

[mellow crag]

what did u do?

[sage fable]

let me send a screenshot

[wanton willow]

depends on a
0<a then infinity
a = 0 indeterminate form
a<0 then zero

(correct me if im wrong please i havent dealt with limits in a while)

[sage fable]

aaaaaa i feel like it needs a bit more playing with to apply l'hopital's rule

[wanton willow]

well to use lhopitals rule you can just do whatever that second part is divided by 1/n^a

its the same as multiplying by n^a except now its in a fraction form

[mellow crag]

||normally when we have a fraction like 1/(sqrt(3)+sqrt(2)), we'd multiply top and bottom by sqrt(3)-sqrt(2) to rationalise the denominator - can you do this in reverse?||

[wanton willow]

and the denominator is zero and the numerator is also zero i think

so youre free to use lhopitals rule once you do that im pretty sure

[sage fable]

ok

yeah no applying lhopital here wouldnt work

huge rabbit hole

[sage fable]

checking this now

i did that

but lhopital on it doesnt work either

[mellow crag]

you should get $x_n = \frac{1}{(n+1)^{\alpha}+n^{\alpha}}$

[elfin berry]

LY

[mellow crag]

which obviously converges to 0

[sage fable]

the numerator didnt come 1

[mellow crag]

wait does it not

[wanton willow]

did you try using algebraic manipulation after lhopitals rule?

[sage fable]

$x_n = \frac{(n+1)^{2\alpha}-n^{2\alpha}}{(n+1)^{\alpha}+n^{\alpha}}$

[mellow crag]

oh yeah you don't

[wanton willow]

cuz i did and i got (n+1)^(a-1)

[wanton willow]

and you can just let u = n+1 and u approach infinity

[mellow crag]

but it doesn't really matter cus obviously the numberator < 1

[sage fable]

ugh

cant use latex bot

[wanton willow]

o it becomes u^(a-1)

[elfin berry]

Osiris

[sage fable]

yes this is what i get

[mellow crag]

note that numerator <= 1 by either using the fact that u've already shown that x_n is decreasing

or like just algebra manipulation

[sage fable]

right

[wanton willow]

we can let a-1 = b and since the range for a is greater than 0 and less than one, the range for be b is greater than -1 and less than 0
and we know for that range the limit as n approaches infinity of n^a is zero

[mellow galleon]

uhhhhhhhh.......

isnt 2 * alpha now larger than 1 tho

[wanton willow]

guys are yall listening to me??

[sage fable]

oh yeah hes right

yes m3nny im listening to u as well

illl try some algebraic manipulation

[mellow crag]

good point!

[wanton willow]

yeah then try to get it to the form (N+1)^(a-1) and after that you should be able to easily prove it approaches zero by just substituting some variables and using the range we're given

[sage fable]

how did u get this? i am not getting that on manipulation after l'hopitls

[wanton willow]

ok so after applying lhopitals rule you get
(a(1+1/n)^a-1 x -1/n^2)/(-an^(-a-1))
divide the top and bottom by negative a to get
(1+1/n)^a-1 x 1/n^2) / (1/n^a+1)
then multiply the top and bottom by n^2 to get
(1+1/n)^a-1 / (1/n^(a-1))
then simplify the fraction by multipling top and bottom by n^(a-1) to get
n^(a-1) x (1+1/n)^(a-1)
factor out exponentiation to a-1
(n x (1+1/n))^(a-1
distribute the n inside the brackets
(n+1)^(a-1)

lemme know if i went wrong anywhere

[mellow crag]

||if alpha <= 1/2 then the method above works, if alpha > 1/2 then multiply our original expression by (n+1)^(1-a) + n^(1-a) instead||

[sage fable]

wait i think m3nny might be going in the right directon

lemme just write that down and see if its correct

WAIT

thats right

i got the first step wrong initially where the denominator is 1/-an^(-a-1) i wrote it as 1/-an^(a-1)

and yeah thats proof enough

wow

[mellow galleon]

wait... can i just like offer up an idea? not sure if its rigorous enough

[wanton willow]

yay!!

[sage fable]

yessir

[wanton willow]

thank god for lhopitals rule

truly one of the best things to happen to mathematics

[mellow crag]

are u allowed l'hopital's rule?

[sage fable]

now if someone can explain why we applied l'hopitals to x_n

[mellow galleon]

what if we notice that its like an integral of alpha * (n+1)^(alpha-1) from the interval of n to n+1

[sage fable]

oh shit now that u remind me.... we arent LMAO

[mellow galleon]

then like since (n+1)^(alpha-1) is convergent like said then everything just goes to 0?

[sage fable]

nooooooooo i just got reminded about this fact

[wanton willow]

i dont see any reason for it not to be allowed other than cases where they havent learned it yet or its used in a way thats self-dependent like in sinx/x

[mellow crag]

in general, if u've just done epsilon delta definitions, you probably haven't proved l'hopital's yet

[mellow crag]

which is why i thought that Osiris might not have been allowed it

[wanton willow]

oh yeah fair enough then

wait i just realized ive never actually seen the rigorous proof for lhopitals rule i only took it as fact and just kinda used intution

[sage fable]

that is correct but still relies on the fact that we need the expression (n+1)^(a-1) which is derivable only from l'hopitals it seems

[wanton willow]

i should probably check that out later

[mellow galleon]

ah ic thanks

[wanton willow]

especially cuz i wanna take real analysis and im sure you need to prove limits more rigorously for that

anyways damn it sucks you cant use lhopitals rule

[sage fable]

ok so far this is what we have established

1. x_n > 0 for all natural n but its monotonously decreasing so it converges
2. if we find an upper bound that also converges to 0 i can sandwich proof it to converge to 0

[mellow crag]

yeah actually that works lol

[wanton willow]

i kinda get it cuz its like banning calculators to make students comfortable with using other methods but still its so inconvenient

[sage fable]

we are allowed to use l'hopitals on bounds if we use sandwich theorem

[mellow crag]

it's not really to get u to use other methods, it's just like if u don't know how to prove l'hopital's rule then using l'hopital's rule ends up being like 'the same' as going "this thing should tend to 0"

[sage fable]

prof said u just cant straight away prove it by applying limit, cuz thats unreal

i literally forgot about this until u mentioned it just now, i would have failed calculus omg

[mellow crag]

obviously u need to have done integrals first but if u've done integrals in analysis then it does work lmao

[wanton willow]

well most students dont learn how to prove the quadratic formula but theyre still allowed to use it

ik theres completing the square but i dont think thats taught before the quadratic formula though

[sage fable]

nono we learned proof of quadratic formula before applying\

using completing the square yes

[mellow galleon]

was just a wild thought since ive never touched analysis, might look into it after this tho this seems cool

[wanton willow]

oh my bad then
We still havent took quadratics at school yet so i dont know what order theyre typically taught

[sage fable]

thats HS math tho, rn i just need an upper bound 😭

that converges to 0

[mellow crag]

also it's a bit different cus it's school rather than uni but like if ur taking a pure maths course, u try to learn the proofs of everything before u can use them

[wanton willow]

oh right this is an anlalysis not a calculus 2 class

makes sense then yeah

[mellow crag]

i mean if ur not allowed to use l'hopital's then i don't think ur allowed to use l'hopitals on ur sandwich expressions lol

[wanton willow]

maybe they just want students to use the sandwhich method somewhere even if they supplement it with other stuff?

[mellow crag]

this works but idk if there's a cleaner way to do it

[sage fable]

we can use anything to get the limits in our bounds of sanwich actually

[mellow crag]

oh interesting

well hf!

[wanton willow]

ok so we need an expression greater than (n+1)^a - n^a when 0 < a < 1 that converges to zero

hmm

[sage fable]

gonna try this

nope too messy

its supposed to be simple

cuz we have time limits and shi in our exams

[wanton willow]

btw isnt saying n^a when 0 < a < 1 the same as saying n^(1/a) for any a > 1?

wait actually that doesnt account for alot of cases nvm

[sage fable]

yeah..

tho i feel like if we cant get the point in this question, maybe its not meant to be

after all the main question was to just investigate

with that said can i also get some help with (D) and (E) c:

[wanton willow]

well for d i think it has to converge to sqrt2 - 1

since the limit as n approaches infinity of 2^(1/(2n+1)) is the same as the limit as m approaches 0 of 2^m which is just 1

so sqrt2 - 1

and im not sure how to approach e yet tbh

factorials, especially double factorials, are pretty awkward to handle

[sage fable]

oh okay

thats fine, i think i understand what i need to do now

thanks guys, i learned a lot @mellow crag @wanton willow @grave dove @mellow galleon (sorry for pings just thanking)

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[graceful nacelle]

I have 150 dollars and it grows 50% every month. i want to know how much my amount has increased in the last month after x months

[graceful nacelle]

how do i do that

would it be the balance after x weeks minus the balance after x-4 weeks

[tall pewter]

WHERE DID YOU GET 4 FROM

OHH NVM I SEE YOU TALKING WEEKS

[graceful nacelle]

like this right

[tall pewter]

IS THE TIME IN WEEKS OR MONTHS?

[graceful nacelle]

weeks

[tall pewter]

IT SAYS IT GROWS 50% EVERY MONTH THOUGH

[graceful nacelle]

oh fuck mb

meant weeks

50% every week

this would be the dollar amount and the increase over the last month correct?

[tall pewter]

SEEMS So

[graceful nacelle]

great

thank you

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[still temple]

Don’t have answer key

But can someone solve and explain

I don’t get this problem

[sage fable]

@still temple the question says that richard has 19 different football player cards already, he is missing a single specific one. now he buys 5 bars that can contain any card from the 20 cards in the pool. u need to find in part 1 if after buying the 5 bars he was unable to complete his collection

theres 19/20 chance that each chocolate bar he buys does NOT have the card he needs

so in part (i) are u able to calculate his probability of not completing his collection now? remember he buys 5 bars

[marsh citrus]

@still temple Has your question been resolved?

[marsh citrus]

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[harsh pumice]

Hey. Not sure if there will be anyone to help with this, but just in case I'm having an issue with this graph.
This is the graph I have entered its (0,5)

[harsh pumice]

And this is the function instruction

[harsh pumice]

The issue is I know this is wrong because when I input in desmos its (0,5.5) but with this graph I can't (or dont know how to) make it that precise. I'm not sure what to do.

<@&286206848099549185>

[pale bough]

Did you try any other graphing tools?

[harsh pumice]

no this is the platform I have to submit it on

[marsh citrus]

@harsh pumice Has your question been resolved?

[marsh citrus]

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[grand estuary]

like, help me understand it so i can actually solve it. left column is set-builder notation, middle is interval notation, and right is the line graph

[grand estuary]

<@&286206848099549185>

[sage fable]

when ur talking about intervals
if you take a<b<c
this means b belongs somewhere between a and c, not including the bounds a and c

which means at a and c the line graph will be open (unfilled dot)

whenever it can also be equal to bounds, the graph of closed at that point

[marsh citrus]

@grand estuary Has your question been resolved?

[grand estuary]

what about set-builder notations?

[sage fable]

is the same there as well

just uses actual words instead of symbols like union or intersection

[grand estuary]

heres the things, i still dont understand it 😭 as this part of the subject is still entirely new to me

[grand estuary]

what would that look like?

[grand estuary]

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[wooden ridge]

How can I express this vector in the specified form?

[wooden ridge]

does each point correspond to the different v's

[hollow oracle]

you can think of it like that

[vestal sandal]

i, j and k are unitary vectors. think in 3 dimensions: if you have a vector V of the above form (V1 i + V2 j + V3 k) then you move V1 units on the x axis, V2 units on the y axis and V3 units on the z axis

so, look at your two points and think how much you "moved"

for the x coordinate u went from -4 to -4...so how much did you move?

[wooden ridge]

it did not move at all

[vestal sandal]

exactly

so i's coefficient will be0

what about the y axis, from 2 to -2

[wooden ridge]

4

[vestal sandal]

mhm but in what direction?

that also matters

you went from +2 to -2, so you actually moved "-4" steps

[wooden ridge]

-4 because it went down from 2 to -2

yes

[vestal sandal]

and what about the z axis?

[wooden ridge]

+3

[vestal sandal]

right

a more "mathematical" way to think about it is that if P1 has coords (x1,y1,z1) and P2 has coords (x2,y2,z2) then the vector P1P2 = V1 i + V2 j + V3 k will have coefficients:
V1= x2-x1
V2= y2-y1
V3= z2-z1

makes sense right?

since the difference of the two coordinates is the distance we travelled, basically

[wooden ridge]

yes

so it would be 0i+(-4j)+3k

[vestal sandal]

yes thats right

[wooden ridge]

now how can i determine length

i overthougt the over part

[vestal sandal]

length?

you mean the magnitude of the vector?

[wooden ridge]

never mind different question

[hollow oracle]

same thing

no need to get too technical

[vestal sandal]

oh well

[vestal sandal]

we use the distance formula/ the pythagorean theorem

they're both the same thing

|V| = sqrt(v1^2+v2^2+v3^2)

[wooden ridge]

oh okay makes sense i thought that was what it was

thanks for you help

you explained it really well

[vestal sandal]

no worries, have a good one

[wooden ridge]

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[still temple]

So confused, how is this not 3

[wary kite]

looks more like 3.5 to me

maybe 3.25

[hollow oracle]

i hate this problem lmao

[wary kite]

yea it’s more of an eyesight test lol

[still temple]

It was js that I couldnt tell it was 3.5 sorry dumb question (im so cooked all of this is due in a hour and a half)]

[wary kite]

so it was 3.5?

[still temple]

Yes

[lilac siren]

does this question literally have no context

just guess the number

[still temple]

literally none

[lilac siren]

nice

[still temple]

literally was a guess

What did you use to draw the lines?

Alright thanks man I appreciate all the help

Do you guys js do this for fun or is it to help you orr

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[pale swan]

solve for SRA

<@&286206848099549185>

[quiet anvil]

Are the following lines: VM, UI, SD, AP?

Also

!15min

[marsh citrus]

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.

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[marsh citrus]

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[lyric kelp]

In a system of linear equation,does a homogenous eqution ALWAYS go through the origin?

[lyric kelp]

And why

[crystal wraith]

Because if you sub in zero for x and y the equation is satisfied. (0,0) are the coordinates of the origin.

[lyric kelp]

oh yeah im kidna dumb lol ty

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[small stream]

Might be a silly question, Can we find a one-sided derivative without the definition method? If so, what conditions should the function meet? like this function, is there any other way except using definition to find the one side derivative of f at 1? The following is my take

[devout mauve]

well for the one side you can just differentiate as usual. i.e. power rule

you dont know the differentiation rules yet?

like x^n -> nx^(n-1) ?

[small stream]

I know it, I'm just not famaliar with the terminology in English

[small stream]

so what conditions should the function meet if we want to do this way?

[devout mauve]

both the point and the side you come from follow the same "rule"

i.e. 2/3 x^3 here

[small stream]

yes I mean, what conditions should the function meet if we want to do this way?The way that just differentiate it. Like, follow what rules?

[devout mauve]

the function at x=1 and for x>1 follows the same def here. so you can just differentiate that

thats not true for x=1 and x<1, so you cant do it for the left side limit

btw sidenote, the function isnt even continuous at x=1 from the left, so you really dont have to worry about the derivative

[devout mauve]

well if your assumption already says that f is differentiable there...

the point is, the limit you wrote down doesnt even know that on the left of x0 the function behaves badly

if you tried to take the derivative from the right for the function 2/3x^3 (everywhere) at x=1, you would compute the same limit

[devout hill]

I think what the other helper is trying to get at is: the function x^n has a known derivative, it is equal to nx^{n-1}.

Now you function coincides with a nice function on [1, 1.5): 2/3 x^3.

2/3x^3 has the nice derivative 2x^2, which, evaluated at 1, is 2.

Since your function coincides with 2/3 x^3 on [1,1.5), the right-sided derivative is equal to 2x^2 evaluated at 1, that is 2.

these rules: "x^n has derivative nx^{n-1}" are basically derived from definition.

They are there cuz its kinda tedious to do def over and over again whenever you encouter functions like this

[small stream]

I guess I didnt express myself correctly. The pic I made is, I mean I made a summary like, what conditions should the function meet if we want to do this way. so I'm trying to find the conditions fit in general not only in the example I posted above

My point is that if these two parts are good, or just open interval is enough

[devout hill]

lemme give an example:
If you know what $g'(1)$ is, and $f(x)=g(x)$ on $[1,1+h)$, then $f'_{+}(1)=g'(1)$

[elfin berry]

qwertytrewq

[devout hill]

in our case g(x)=2/3 x^3

g'(x)=2x^2 so g'(1)=2

and f coincides with g on [1,1+h)

so $f'_+(1)=g'(1)$

[elfin berry]

qwertytrewq

[devout hill]

so its not abt whether the derivative exists or not, it more about: "oh we know this nice function's derivative, it must coincide with our uglier function"

In general: if $g'(a)$ exists and $f(x)=g(x)$ on $[a,a+h)$ for some $h$, then $f'_+(a)=g'(a)$

[elfin berry]

qwertytrewq

[small stream]

Thank you, but what about this doubt?

@devout hill where did you go I need you

[devout hill]

i am pretty sure this is wrong. lemme find an counterexample

[small stream]

you mean we cannot make it diff at an open interval?

[devout hill]

take $f(x)=\left{\begin{array}{cc}x^2\sin\left(\frac{1}{x}\right)&,x\neq 0\ 0&,x=0\end{array}\right}.$

[elfin berry]

qwertytrewq

[devout hill]

wait nvm

you also asserted limf'(x) exists

[small stream]

sure cuz we want that side of f'(x0) exist, ofc we need to assure limf'(x) existing

[devout hill]

nvm ur statement should be correct through mean value theorm

but it seems somewhat irrelevant to what method we used to get the answer

[small stream]

wait how thru mvt I just feel it might be right but idk how to prove it

[devout hill]

$\frac{f(1+h)-f(1)}{h}= f'(c_h)$

[elfin berry]

qwertytrewq

[devout hill]

for some $c_h\in (1,1+h)$

[elfin berry]

qwertytrewq

[small stream]

It is relevant, the right side of the original function fits for this statement, continuous, diff, and lim exits, then we can just take the derivative cuz we know it will exist

[devout hill]

as $h$ gets small close to $1^+$, $c_h$ also gets close to $1^+$

[elfin berry]

qwertytrewq

[devout hill]

so the limit equals to f'_+(x_0) exists and must be A

[devout hill]

in fact the derivative will be equal to the limit of the derivative here

[devout hill]

yeah, however,

[small stream]

instead, the left side will fail this statement: continuous × , so we will know that left side f' doesnt exist

[devout hill]

this statement right here alr asserts that f'_+(1) exists

so it sort of feels redundant to apply the other lemma we proved

[devout hill]

because you don't need to check for limits of derivatives here (this works if the limit of derivative doesnt exists)

heck, you don't even need g to be differentiable anywhere else but 1

[small stream]

and f doesnt have to be diff at 1

[small stream]

so here it should be open interval !

fair enough now

Thank you

.close

[marsh citrus]

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[small stream]

Let $ a_n = \int_{0}^{1} x^{n} \sqrt{1 - x^{2}} , dx $ for $n = 0, 1, 2, \ldots$.\
(1) Prove that the sequence ${ a_n }$ is monotonically decreasing, and that $a_n = \frac{n - 1}{n + 2} a_{n - 2} $ for $ n = 2, 3, \ldots $.\
(2) Find $\lim_{n \to \infty} \frac{a_n}{a_{n - 1}}$

[red nimbus]

trig sub maybe

[limber hearth]

try to calculate an+1 - an

And get a negative result

[elfin berry]

miyo

[red nimbus]

,, \int_0^1 x^n\sqrt{1-x^2}(1 -x) : \dd x < 0

[elfin berry]

bacc

[small stream]

seems not very nice to handle with

[limber hearth]

Yes with x between 0 and 1 this is negative, so its monotonically deacreasing

[red nimbus]

Ok but how do we prove this

[limber hearth]

The value of a_n ?

[red nimbus]

,, \int_0^1 x^n\sqrt{1+x}(1 -x)^{3/2} : \dd x < 0

[elfin berry]

bacc

[red nimbus]

I meant to show that this holds

[red nimbus]

When does any task you post seem nice to handle 😭 but they are very interesting

[limber hearth]

It is x-1 not 1-x

[limber hearth]

So you can't get this

[red nimbus]

how

[limber hearth]

See

X^n+1 - x^n = x^n(x-1)

[red nimbus]

Ok I see

I need to switch the < to >

I did in my head

a(n) - a(n+1) > 0

arghh

[still temple]

$0 \leq x \leq 1$

$\implies 0 \leq x^n \leq 1$, $n > 0$

Also, $-1 \leq x-1 \leq 0$

$\implies 0 \leq x^n\sqrt{1-x^2} \leq \sqrt{1-x^2} \leq 1$

$\implies 0 \geq (x-1)x^n\sqrt{1-x^2} \geq x-1 \geq -1$

$I \leq 0$ where $I$ is your integral

[limber hearth]

Thats the wording indeed

[elfin berry]

Miles

[marsh citrus]

@small stream Has your question been resolved?

[red nimbus]

it looks good what miles cooked

[small stream]

wait but how to show that $a_n = \frac{n - 1}{n + 2} a_{n - 2}$

[elfin berry]

miyo

[red nimbus]

I was thinking maybe you can deduce that with the reduction formula

If we use a trig sub there is reduction formula for sin(x)^n

[devout hill]

I haven't wrote it in detail. but my hint would be (rather not obvious): the derivative of $(1-x^2)^\frac{3}{2}$ is? Try Integration by parts.

[elfin berry]

qwertytrewq

[devout hill]

hopefully it works

[small stream]

why $(1-x^2)^\frac{3}{2}$ What did you spot that triggered you this

[elfin berry]

miyo

[devout hill]

cuz when a_n-a_{n-2} gives you another 1-x^2

so you just have (1-x^2)^\frac{3}{2}

[devout hill]

i just did it, it works

thank god

ofc integration by parts is the first thing that pops up in my mind, cuz "recursive" formulae like this using integration by parts appears in deriving logarithmic integral things too

basically id expect the degree of x to raise/drop by a certain amount if i apply it correctly

🤷‍♂️ like how else are they gonna do it

[devout hill]

this is what i mean: you just apply integration by part over and over again to get k!/ln x^k (of course with some detail one needs to take in account for)

[small stream]

Jeez

ok fine lemme do it from the strach to understand these

[devout hill]

why don't u use the texit bot here

[small stream]

Jeez. Seeing this expression makes me feel like I'm about to faint

[small stream]

why should we use ibp to handle this intergration? Sorry i forgot a lot of knowledges about intergration

And why did you think of compute an-an-2, i mean, what inspires you

[devout hill]

what does integration by parts achieve? it make udv=uv-vdu

what we hope is that uv vanishes (since it is evaluated at 1 and 0). And vdu gives us something like x^{differnt power} (1-x^2)^{a value we hope is 1/2}

because if x is to a different power then this integral then relates to a_{different number}

which gives us a recurrence (which is what we want)

[devout hill]

now the choice of "u" and "v" on the other hand.... is more like "play around and find out"

[devout hill]

a lil bit of simplifying gets the job done

[small stream]

Since ibp can help establish recursive relationships between integrals, when we see recurrence in sequences and series, we can always try to use ibp?

[devout hill]

where each term is in terms of integral? probably.

[small stream]

wait a min I need to ponder for a while

[devout hill]

in this problem the functions also gives some "hint": a_n compared to a_n-2, the power on x decreases. Similarly, if you were to apply ibp with u part as x^{some power}, you also establish some equation with x^{some power-1} on the other side

these are all vague hints that tells you maybe you should at least try ibp

it might not always work tho

[small stream]

This gives me headaches, need to reconsider later, I'll repost it maybe tmr to finish it, it destroyed all my little confidence, i need to do some simple ones to regain my poor confidence

.close

[marsh citrus]

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[marsh citrus]

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[celest folio]

did i interpret this problem correctly

change in altitude is change in hypotenuse

[main idol]

Altitude is vertical

So the side opposite theta in your diagram

[celest folio]

so sin 25 = x/y

how do i make an equation out of that tho

[main idol]

y = speed of hypotenuse

Don't use y, that's confusing

Use h or something

[celest folio]

kk

so sin 25 = x/h

so is that x/dhdx

[main idol]

everything in your diagram is already speed

[celest folio]

hm so what do i do

[main idol]

Use trig to find the missing side

[celest folio]

do i take the derivative of sin 25 = x/h

kk thanks

.close

[marsh citrus]

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[marsh citrus]

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[marsh citrus]

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

@green rover Has your question been resolved?

[green rover]

1

I mean 2 💞

<@&286206848099549185>

Avengers assemble

[still temple]

wth

why avengers lol

[raw hawk]

Lol

[green rover]

Idk what u want me to say

Teen titans go!

Are you hype yet

[raw hawk]

DC

[green rover]

SOMEONE HELP ME

[raw hawk]

Ok, with everything?

[green rover]

well I got the abcdefgh because I’m just that good

But need help with 2 and 3 😢

[raw hawk]

Do you know what horizontal translation is?

[green rover]

Yes

[raw hawk]

Give an example of horizontal translation of a function f(x)

[green rover]

No

[raw hawk]

Ok

Bye

[green rover]

NOOO

COME BACK

im sorry

I will lock in

F(x)=(x)-2

😘

[marsh citrus]

@green rover Has your question been resolved?

[marsh citrus]

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[foggy furnace]

In formal logic, How would I prove or disprove
(Ex)[A(x) ^ B(x)] → (Ex)A(x) ^ (Ex)B(x)
???

[foggy furnace]

Please ping btw

[marsh citrus]

@foggy furnace Has your question been resolved?

[marsh citrus]

@foggy furnace Has your question been resolved?

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[ripe glen]

im kinda confused by the last two in this table, if p is false shouldnt q also be false?

[marsh citrus]

@ripe glen Has your question been resolved?

[sleek lake]

this is talking about the thing where this is how it works

there are 16 "things", operations on two variables
or is one of them, and is one of them

one of 16 is this thing, it exists whether you want it or not, we call it →

there's nothing to be confused about, there's no why

[sleek lake]

if the kettle is boiling, it has water
boiling → has water
it tells you that one situation is impossible, and nothing else, the other 3 could happen

[marsh citrus]

@ripe glen Has your question been resolved?

[ripe glen]

So as long as the water is boiling there is water.
And the one F on the p of q is if the water wasn't boiling right?

[sleek lake]

if there was no water in the boiling kettle

this never happens, so when you say "boiling → has water" is true that's what you're saying

so another way to say p → q is ~(p & ~q)

and another is (~p V q)

[ripe glen]

Whatabout the last one? Does it just work like multiplying negatives and two make a positive?

[sleek lake]

it doesn't "work"

it's just a language

p → q
short for "p→q is true"
short for (p and q) or (~p and q) or (~p and ~q)

"i swear it's not the second case"

[ripe glen]

So p->q is the same as (~p -> ~p)?

[sleek lake]

(~q → ~p)

no wait

yeah

(~p -> ~p) is not the same, you can't make it false, so it doesn't tell you anything new

(~p -> ~p) is the same as (g V ~g)

[ripe glen]

Then why does p: F and q:F equal T then?

[sleek lake]

there is no why

i don't understand what are you talking about

[sleek lake]

there are 16 tables you can make

they are saying, this table has a name

we call it →

that's all they are saying

there's no logic underneath

[ripe glen]

Ah

I think I get it now

[sleek lake]

because this table gives you 3 options of what pq could be

it's equivalent to saying "if p then q"

like the kettle example

i guess it's a bad example, because you would never say it in real life

like, you would as a joke, but never to inform someone

"What day is it?"
"If Bob is at home, it's tuesday"

if it happens that bob is at work, and it's wednesday, the statement wasn't a lie

[ripe glen]

So I follow the truth table, even if it doesn't make sense?

[sleek lake]

well it can't not make sense

it's a table with letters