help-4

[stark wedge]

agree so far?

[prisma flint]

yes, thank you

[stark wedge]

oh, did you understand it fully now? or should i continue

[prisma flint]

please continue

[stark wedge]

next is to divide both sides by 104 and get $$z^2 = \frac{5}{104} z + \frac{105}{104}$$

[rocky lotus]

Ann

[prisma flint]

a got it
thank you very much!

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[normal geode]

Im struggling with the following question : they give this series, (12) + (56) + (910) + (1314) + ... + (81*82) , then they ask to write it in sigma notation, but its not neccessary to calculate the value of this series. it counts 4 marks. i dont understand sigma notation well, and i dont know how to solve it in a simplistic manner

[normal geode]

its pre uni mathematics

[river shale]

$1\times 2 + 5\times 6+ 9\times 10+ 13\times 14 +...+ 81\times 82$

[rocky lotus]

🌙 ЅκψΑиdΝιɡħτ

[normal geode]

yes

sorry😅

[humble otter]

Do you notice a pattern?

[normal geode]

yes

[humble otter]

What is it?

[normal geode]

the even numbers forms a pattern\

i think its, for the even patterns, a arithmatic sequence

as well as the uneven ones

+4

thus its

give me a min lol

[river shale]

,rccw

[normal geode]

meaning Tn=a+(n-1)d

[humble otter]

Let's look at them term by term.
Look at the first term 1×2 then the next term 5×6.
Next term 9×10.
Can you figure out the nth term?

[normal geode]

if i split the rows its +4, but if i calculate the brackets its

if im not mistaken, a quadratic sequence😅

[humble otter]

Maybe

Can you come up with an expression for the nth term for 1+5+9+...?

[normal geode]

yes, its Tn = 1 + (n-1)4 for the uneven ones, and Tn = 2 + (n-1)4

for the even ones

[humble otter]

Simplify them

[normal geode]

👍

[humble otter]

The final sequence is just the product of the nth terms of the two sequences

[normal geode]

so when i find the product, i build the sigma notation based on that?

[humble otter]

What is the product of their nth terms ?

[normal geode]

just to make sure, enlish is not my first language, product means +?

[humble otter]

No. Product means ×

[normal geode]

ok cool thanks

2 secs

i got -4n+6

[humble otter]

What? The product should be quadratic

[normal geode]

i did this (4n-2)(4n-3)

oh wait

i forgot to add the ^2

its

16n^2-20n+6

[humble otter]

Correct

What did you just do?

I mean, realise what you just did

[normal geode]

i multiplied the brackets out and forgot my exponential laws folr a sec😂

[humble otter]

You took apart the original sequence into two parts. You realised that each term has two parts, each from two different sequences

[normal geode]

ohh you mean that

[humble otter]

1st sequence: 1,5,9...
2nd sequence: 2,6,10...

[normal geode]

yes, i took apart the sequence, found the product of the twoe sequences to find the n term of the combined sequence

[humble otter]

That's what you do with such type of sequences.

[normal geode]

may i ask why we had to find the product? is it because the combined sequences are being multiplied with each other?

[humble otter]

Yes

[normal geode]

ohhhh

[humble otter]

If the terms were like:
$\frac{1}{2}+\frac{5}{6}+...$

[rocky lotus]

PRAKHAR

[normal geode]

then i would divide it

[humble otter]

The nth term would be like:
$\frac{4n-3}{4n-2}$

[rocky lotus]

PRAKHAR

[normal geode]

ohhh

\ok

thank you!

[humble otter]

Also, for example,

If the terms were like:
$\frac{1}{\sqrt{2}}+\frac{5}{\sqrt{6}}+...$

[rocky lotus]

PRAKHAR

[humble otter]

The nth term would be like:
$\frac{4n-3}{\sqrt{4n-2}}$

[rocky lotus]

PRAKHAR

[humble otter]

Do you understand this concept better now?

[normal geode]

yes, i do, thank you for helping!!!

[humble otter]

Sometimes, they just give the simplified terms. You may have to convert the terms to some kind of form to sum it.

For example, the might just give: 2 + 30 + 90 + ...

[normal geode]

how do you mean simplified terms? like the product of each term, then i have to ''break'' it apart into numbers that makes a sequence?

[humble otter]

Factorising each term we can get:
1×2 + 5×6 + 9×10 +...

[normal geode]

ohhhh

and if i just get it in sigma notation, i write it out?

[humble otter]

If you get sigma notation, you're lucky and half the job is already done

[normal geode]

lol

[humble otter]

You just need to sum it

[normal geode]

i can manage that lol

[humble otter]

If you happen to get a sequence, you need to convert that into sigma notation then sum

[normal geode]

in this question they ask us todo that

[humble otter]

Good luck 👍

[normal geode]

thank you for you help, have a good day/night!!

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[foggy lantern]

if a function on the graph looks like this is it odd or neither

[hidden terrace]

Neither

[foggy lantern]

is it because it's not touching the origin

[hidden terrace]

Yes

[stark wedge]

one handy property of odd functions that you can easily check at a glance is: any odd function that's defined at x=0 passes thru the origin

therefore if your function goes through (0, anything else) then there's no way for it to ever be odd

[foggy lantern]

okok

tysm

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[odd sequoia]

can you guys help me?
or how can i make my own help channel?

[foggy lantern]

under the category math help (available) just choose an available text channel with your question

[odd sequoia]

i am from korea and i am trying to ask for help and i used gemini to translate the words to english

the three following conditions are 가,나,다 in korean

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[sterile knoll]

Am i doing it right?

[sterile knoll]

<@&286206848099549185>

[honest stone]

!15min

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[dire cloud]

It does seem right

[sterile knoll]

Wait

One more

[pliant nova]

can you elaborate on this step?

idk how you got there

[sterile knoll]

Erm make Common denominator

[agile ginkgo]

Erm

No

If it was common denominator then the bottom is multiplied

[sterile knoll]

Oh

Yes

But the final answer still rigth?

[pliant nova]

ermm well a^2 + b^2 isnt (a + b)^2

it's (a + b)^2 - 2ab

[sterile knoll]

Where

Which step

Oh

What about this time @pliant nova

[pliant nova]

uh

[sterile knoll]

兄弟 幫幫我

Is that mean im right?

[pliant nova]

when i said that i meant the top not the bottom

bottoms fine thats just product of roots squared

[sterile knoll]

Oh

Damit

Let me fix it

The answer is 8

?

@pliant nova

I fixed that

[pliant nova]

idk

[sterile knoll]

What about the 30b

Wait

[pliant nova]

it had the same issue

[sterile knoll]

It wasn't ^2

Erm

N m

Nvm

I got -37/2

Is that right

[pliant nova]

no idea

[pliant nova]

36 - (-4) = 32?

[zealous pendant]

Yeah it should be 10

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[worthy lily]

wrote up this diff eq equatiom, i wanted to make sure i did everything right

[craggy girder]

I can't read what you have written

[worthy lily]

phooey

uhhh

working a different equation rn i'll post that up then i guess

heres this one, class starting in a bit so i cant finish it, but i hope im kn the right path

for reference, this is the original problem

just realized i fucked up

-1/x should be - (y+1)/x

[untold verge]

@worthy lily is your doubt resolved?

[worthy lily]

honestly idk, it feels like the question is missing something

[untold verge]

i am getting if to be 1/rootx

[worthy lily]

can you explain how you’re getting that

[worthy lily]

im getting the integrating dacgor as this

[untold verge]

why would if be a function of y?

Pls adjust for handwriting

@worthy lily

[worthy lily]

i’ll take a look into it after class

[untold verge]

okay

[worthy lily]

screw it i’ll take a look now

oh

huh

i’ve never though of it like that

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@worthy lily Has your question been resolved?

[worthy lily]

i guess?

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[polar surge]

this solution only exists if them two terms in the brackets are greater than 0

and my notes state that this is only the case if a12a21 < 1

but how is this condition sufficient? could one of a12 or a21 also be "large" negative number and cause a term to be negative?

[drifting hornet]

i feel like we might be missing some context

whats meant by solution?

what are we solving?

[polar surge]

i am not providing too much context as the question is massive and i think may cause some confusion. this is a steady state solution from a coupled ODE system.

all im asking is, ive been told that this solution only exists if both terms in that solution (ie x1 x2) are positive, and this is only the case if a12a21 < 1. But i dont understand how the condition a12a21 < 1 is sufficient enough to ensure that both terms in the solution are positive

[drifting hornet]

its not sufficient

[polar surge]

ie if a12a21 is < 1, how does that suggest that this term is > 0

[drifting hornet]

if a12 was e.g. -2 and a21 was 2, it would be negative

[polar surge]

indeed i thought the same

[drifting hornet]

The real conditions should be
a12a21 < 1, a12 > -1, a21 >-1
or
a12a21 > 1, a12 < -1, a21 < -1
i think

so unless there is more context, which there might be, its wrong

there could be something that guarantees e.g. that a12 and a21 are both positive

[polar surge]

i am finding the steady states of this systen. u1 and u2 are the population of two species

u cant have negative population

hence u1 and u2 must be greater than 0

[drifting hornet]

what are a12 and a21?

[polar surge]

just constants

[drifting hornet]

well, hard to say then

sorry, idk what went wrong. Either your source is wrong or you missed some important constraints

[polar surge]

yh np i will check with my professor

thanks for the help

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[dull cipher]

how do i find a with just this

[dull cipher]

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[wicked ether]

I realised that I could somehow relate the angle theta to the two lines to use the slope formula, but no clue on how to proceed from here.

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[cerulean cairn]

I need help answering this problem

[cerulean cairn]

Especially the second one

I found that b is included in x and X is included in the complementary of A which is equivalent to A inter B is equivalent to the empty set

In the second problem I verified that X is the solution of X = (\overline{A} \cap Z) \cup (B \cup \overline{Z}), \quad Z \in \mathcal{P}(E). Using the first equation but I can’t demonstrate the converse

Can someone give me a lead

Umm hello

<@&286206848099549185>

Ok im being left on read

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@cerulean cairn Has your question been resolved?

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@cerulean cairn Has your question been resolved?

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[solid iron]

I'm on part (b).

[solid iron]

My prof says were allowed to use a corollary, cor. 1.3.6, which says that a set is closed iff it contains it's limit points.

for the first case, where you show S is closed -> S = S bar, my idea is to pick any element x in S and break into cases, showing x is either in S or dS

my book is using dS to denote the set of boundary points

i missed something else in the book

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[sweet sinew]

how are these two equal?

[frozen ledge]

first off, do you agree with the right sided limit?

[sweet sinew]

yea

[frozen ledge]

ok now for the limit to exist we need the right side limit to be equal to the left side limit

[sweet sinew]

yeah

im just confused how the two highlighted expressions are equal

[frozen ledge]

you mean how to factor it?

[sweet sinew]

yeah sure

[frozen ledge]

oh i see

[steady charm]

they don't appear to be equal

[sweet sinew]

shi that what im saying bruh

[frozen ledge]

yea they are not

[sweet sinew]

the answers wrong right

if a = 2 then the lim from the left side as x -> 1 would be 0

not 1/2

[steady charm]

yeah the answer doe not appear to be correct

[frozen ledge]

yea whoever wrote this just made a factoring mistake and thought it was -a instead of + a i guess

[sweet sinew]

what would the answer be in this case then

for a

[frozen ledge]

well first off if we need it to be equal to 1/2 then we must have a 0/0 situation so the denominator has a root of x = 1

[sweet sinew]

then 2a = 0

[frozen ledge]

which would force a = 0 but if a = 0 then you have (x^2 - 1)/(x^2 - x) and the limit as x goes to 1 of this would be 2 not 1/2

[steady charm]

so the problem is unsolvable unless we go back and change it to actually be x^2 + (a - 1)x - a, which is maybe the problem the question maker had in mind (had the factorization already and reverse engineered the problem)

[sweet sinew]

okay that makes sense

also could i get help with this

[frozen ledge]

sure

what have you tried

[sweet sinew]

idk tbh ik that the lim x -> 1 is 2

[frozen ledge]

well of f

[sweet sinew]

yeah

[frozen ledge]

in these scenarios its best to consider x < 1 and x > 1 separately

for instance when x < 1 what does f approach and is it from above or below

[fresh drift]

Daddy knief

Ye knief is right

[frozen ledge]

word

[sweet sinew]

it approaches 2

[frozen ledge]

from above or below?

[sweet sinew]

wdym isnt it from left

[frozen ledge]

yea but i mean when x < 1 are we > 2 or < 2

[sweet sinew]

ooh

less then 2

[frozen ledge]

right so what does f approach when we are a bit less than 2

[sweet sinew]

-1

[frozen ledge]

now do the same thing but for x > 1

and see if they agree

[sweet sinew]

ooh ok i see

alr thx

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[pine prairie]

is there a clever way to inspect the antiderivative of "nice" polynomials?

[neat locust]

No

What is "nice"?

[pine prairie]

factorisable, known roots etc

symmetric, odd/even

[neat locust]

Standard form or factored?

[pine prairie]

low order

factored forms

or can easily be factorised

like even for linear functions

like trying to graph the antiderivative of this

[steady charm]

well we know it should be increasing where the function is positive, have turning points at the roots, and be decreasing where the function is negative

[bold dagger]

you can probably know a few things from this like the extrema and where the function is increasing/decreasing

[steady charm]

you'd need some initial condition to actually have values ofc, otherwise it's only unique up to vertical shift

[pine prairie]

mhmm

it gives a point to pass through

but it doesn't seem realistic to find visually the roots and intercepts

[steady charm]

yeah you can get qualitative up-and-down behavior easily but not quantitative root/intercept bevavior

[bold dagger]

i mean if you have a graph and need to figure things about the antiderivative from it then the "special points" should have nice coordinates that you can get from the graph

[pine prairie]

i guess it looks like this?

but you can't get much more than this

[bold dagger]

sure

[pine prairie]

it might even cross the x axis

but i couldn't tell unless i tried to solve this analytically?

[bold dagger]

i mean you cant get much if you arent given a sufficient amount of given

like if you are just given a graph of the derivative then that would be little info no?

at least i dont see a way to get the graph of the antiderivative just from the graph of the derivative

that being said there could be a way that i am not aware of

[tawny moon]

I don't think it's possible. that's why we get a +C when integrating, no? we can draw the shape of the graph, but we'll never know where on the y-axis it is

(if only the graph of the derivative and nothing else is given)

[pine prairie]

it gives a point too

but that seems so much effort to find everything

also without the actualy shape parameter i dont think you can definitively say what scale the quartic is tho

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[celest coyote]

Confused about the derivative problem

[celest coyote]

To find max height I think I just plug in 20 sec

But for value of gs idk what to do

[hot tide]

the time at maximum height occurs when ds/dt = 0

[celest coyote]

I think I just take the derivative of the position which would give us velocity, then I would take the derivative of the velocity which would be acceleration

[celest coyote]

??

Oh wait ye

because it has to stop moving

[hot tide]

yup

[celest coyote]

but I think its is more theoretical

not like physics

because I was also assuming about that

[hot tide]

well ok what's the maximum height you get?

in terms of g_s

[celest coyote]

oh wait this is a projectile motion equation I just realized lol

[hot tide]

yeah xd

[celest coyote]

gs is the acceleration due to gravity

[hot tide]

yes

[late creek]

15 - g * 20 = 0?

[celest coyote]

Well it says it reached its max height at 20 secs

[hot tide]

indeed

[celest coyote]

so don't I just plug in 20?

[hot tide]

you'll be left with s = 15*20 - (1/2) g_s (20)^2

[celest coyote]

since we are given a position function

[hot tide]

you still have both s and g_s undetermined

in fact you don't need to find the maximum position at all

all you need to know that $\frac{ds}{dt} = 0$ occurs at $t=20$

[rocky lotus]

artemetra

[late creek]

Try thinking of it intuitively. After launch, what does gravity do to the velocity?

[celest coyote]

decreases it?

due to air resistance

and then stops

then comes back down

[late creek]

Yea. And, what happens at the maximum height to velocity?

[late creek]

Yes

[celest coyote]

it stops

[late creek]

What height does this happen?

[celest coyote]

0???

[late creek]

No....

[celest coyote]

well if we plug in zero for t

then we get

[late creek]

If it stops, then don't you think it's the maximum height because then it starts coming downwards?

[celest coyote]

15(0) - (1/2) (gs) (0)

which is 0

[late creek]

..

[celest coyote]

Ye but what is that value??

[late creek]

Ye so, it starts at 15 decreases by g constantly until it reaches 0 velocity at the maximum height.

What does the question say about time taken to attain maximum height?

[celest coyote]

20 secs

[late creek]

Yes. Can you form an equation of this?

Think about it.

[celest coyote]

I take the derivative of the position function?

[late creek]

Starts at 15, decreasing by g constantly for 20 sec until its zero

[celest coyote]

This isn't a physics question

this is calc

[late creek]

Ik..

[celest coyote]

which is why I'm confused on how to express it in calc

not physics

Oh wait

g is constant

[late creek]

..

[celest coyote]

so it wouldn't play a role right?

so the derative of the position function = 0?

im so lost 😭

[hot tide]

s is a function of t

do you agree?

[celest coyote]

ye

ds/dt

[hot tide]

the derivative of s is also a function of t

it's not just 0

[celest coyote]

velocity is 0

[hot tide]

at t=20

otherwise not

[celest coyote]

so wait

just to be clear

s prime (20) = 15 - gst = 0?

[hot tide]

yes!

[celest coyote]

ok

[hot tide]

or

replace t with 20

[celest coyote]

oops

[hot tide]

15 - g_s * 20 = 0

[celest coyote]

15-gs(20) = 0

[hot tide]

yes

[celest coyote]

so

15/-20 = gs

gs = -3/4

[hot tide]

uhh mind your sign

[celest coyote]

?

[late creek]

Noooo

[hot tide]

not -

15-g_s(20)=0 means
15 = g_s(20)

[celest coyote]

oh

ok so 3/4

[hot tide]

indeed

ms^-2

and that's your final answer

[celest coyote]

one sec

I mean it makes sense

but why couldn't I just do a double derivative of the position function?

that would also give me acceleration

[late creek]

Yes

But, you'll not have the value.

ds/dt = -gs

double*

[celest coyote]

Ah

Right

[vale dock]

@celest coyote Has your question been resolved?

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[river arrow]

To confirm, if $f(x) = e^{\ln(x^2-4)}$, then f(2) and f(-2) are both undefined?

[rocky lotus]

Plvzfq_rit

[uncut thistle]

Yep

[river arrow]

tres bon, Desmos was indeed tryna trick me

thank ya+

[split depot]

ln(0) just goes to infinity

[river arrow]

yuh, and e^infty's undefined

(well infty, to be precise)

[split depot]

$$|x|>2$$ is the domain

[hardy coral]

The limit is defined

[rocky lotus]

Rbit

[hardy coral]

f can be continued between -2 and 2, but yes, as a real function, it isn't defined in [-2, 2]

[river arrow]

is that through analytic continuation? (with the complex number stuff)

[hardy coral]

Ye

[river arrow]

i seeee, i should learn about that sometime umu

thanks again y'all++

.close

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[hardy coral]

(I said this because lim_{x -> 2} f(x) is 0, not +inf)

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[north scarab]

how do I solve e^x/x^2=0

[tawny moon]

$\frac{e^x}{x^2} = 0$?

[rocky lotus]

fox(x, y); ∂(fox)/∂x (Flower)

[north scarab]

mhm

[split depot]

theres no solution

[hazy pivot]

!noans smh

[vale dock]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[tawny moon]

I'm gonna need to see the original problem, because there might have been a misunderstanding

[split depot]

$$e^x>0$$

[rocky lotus]

Rbit

[north scarab]

B

[tawny moon]

you are asked for stationary points

not to solve for f(x) = 0

[north scarab]

I know but divided by x^2

[tawny moon]

ok, have you done a)?

[split depot]

stationary points just when derivative = 0

[north scarab]

yes

[tawny moon]

then I don't get your problem

because the question says stationary points, which is when deriv = 0

[tawny moon]

but the whole question you started with was neither the original function nor the derivative

[north scarab]

I found the stationary point, x=1

[north scarab]

I broke it up

[split depot]

that's not a point

(1,f(1))

[tawny moon]

you broke it up....?

[north scarab]

in parts

[tawny moon]

and how?

and why?

[hardy coral]

Probably to find other zeroes (split the factors)

[north scarab]

u saying this is wrong?

[hardy coral]

It's not wrong but you really should've started your question with this

[tawny moon]

ok that's fair, but

the two equations marked with a red arrow are the same thing

in short, you just did a 360

[north scarab]

yes but I cant find the x intercept

in the first form

[stark wedge]

i think last time he did this i called it samsara

[tawny moon]

here's another cycle then

[tawny moon]

wdym

[tawny moon]

so you have a fraction = 0

[hardy coral]

$\frac{ab}{c} = a \cdot b \cdot \frac{1}{c}$

[rocky lotus]

Nel

[tawny moon]

under what condition(s) will a fraction = 0?

[hardy coral]

This should be basic for you

[north scarab]

I cant set this equal to 0?

Im trying to find st. point with this first derivative

[tawny moon]

I didn't say you can't

I said the whole 360 was unnecessary

[tawny moon]

because it should be enough to think of it like this

anyway, you found x = 1 as a st. point

or rather, the x-coordinate of a st. point

so what is the y-coordinate? and what is the nature of the st. point you found?

[north scarab]

well I would need to

find the second derivative

which I use quotient rule

[split depot]

looking at the sign of f'(x) is enough

0<x< 1 and x>1

[north scarab]

wdym

[split depot]

the behavior of the function around the stationary point

it's negative to the left, positive to the right

rate of change

[stark wedge]

of... f'(x), you mean?

that's a bit of a high level insight for here tbh

[north scarab]

why cant I just find the second derivative 😔

[stark wedge]

you can and maybe you should in order to clarify the st. point as max vs. min

tho you will need a different procedure to find the y coordinate of the st point

[tawny moon]

btw, there is sometimes more than one way to do stuff. just because someone suggests another method doesn't mean your method won't work (unless they say so explicitly)

[stark wedge]

that also

locking yourself into one and only one procedure per "problem type" is no good

[north scarab]

This is what I did to find the second derivative

Evaluated it at 1

[tawny moon]

I'm 99% sure your 2nd deriv is wrong

because of this

in your u' you did not differentiate (e^x)x

[north scarab]

I did derivative of e^x is e^x

I believe

[tawny moon]

and xe^x?

[north scarab]

U just multiply it by x

[tawny moon]

??

product rule

[north scarab]

Ohh

Bruh

but product rule

in quotient rule?

[tawny moon]

your turning point is correct and you got lucky your signs are correct
doesn't mean your second derivative itself is correct

[north scarab]

wait nvm

I got it wrong

ok so I need to

use product rule

inside the quotient rule?

[tawny moon]

for the xe^x term, yes

[north scarab]

cant i just epand it

[tawny moon]

?

[stark wedge]

there are a great many ways to work out f''(x), but if you're going the quot rule rule, yes you will need prod rule afterward for num'

[north scarab]

dont worry my mind is overthinking

ill come back to this

.close

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[vestal grove]

I just checked the solution, it says repitition isnt allowed

[stark wedge]

how tf are there this many answer options

[vestal grove]

How come? Its literally just 1s and 0s there has to be reptition, no?

[vestal grove]

The exam is like 50 pages long theyre all like this lmao

Its graded by a machine so theres not many options for writing

[stark wedge]

anyway

repetition is disallowed in picking which 3 of 8 positions the 1's go in

think about it: "I want 1's in positions 4, 7 and 7" makes no sense -- why are you specifying twice that the 7th slot has to be 1?

[vestal grove]

I understand this analogy but I dont get how it relates to the question sry

[stark wedge]

an 8 digit binary string with exactly 3 ones is specified completely by the positions that the 1's occupy

for example, "I want 1's in positions 2, 3 and 5" gives 01101000

[vestal grove]

1 moment I'm just rereading everything a few times

Alright I think i get it

Thanks!

❤️

.close

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[hybrid marten]

Hii could you help me with this

[hybrid marten]

Question 27

[hardy coral]

What's the question?

[karmic condor]

use a^2 - b^2 formula

replace (2x-1)^4 to [(2x-1)^2]2

[karmic condor]

What did the instruction say above the question

[hybrid marten]

Factories

[karmic condor]

then just minus or add the stuff inside, then you're done

[vale dock]

@hybrid marten Has your question been resolved?

[vale dock]

@hybrid marten Has your question been resolved?

[hybrid marten]

I don’t understand it

[ocean bloom]

lemme help
you know the formula right
a^2 - b^2 = (a +b)(a-b)
right
so
the first term (2x-1)^4 can be written as ((2x-1)^2)^2 ( here you can take a as (2x-1)^2
and the second term is (x-1)^2 and it will act as b
so now what we'll do is use the formula

so it becomes
the second step of the image solution
and then just the expansion was done

[hybrid marten]

Im sorry but can i do it tomorrow

Im feeling really done rn

[ocean bloom]

👍

[ocean bloom]

close the thread then, reopen it tomorrow, so that other person can use the channel 🙂

[hybrid marten]

Okay

Thanks

.close

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[azure raven]

help pleasee

[azure raven]

is this too blurry?

[abstract raptor]

what are you struggling with and what have you tried so far

[azure raven]

i believe im struggling to understand that concept with the graphs

I understand them on its own but

not using the graph

[abstract raptor]

so you understand how a limit works

but dont know how to use it with this graph?

[azure raven]

yep yepp

oh wait i lie there's some other stuff that i dont understand but ill figure it out

but the graph for me is hectic

[abstract raptor]

well yea its a discontinued function

so how would you take the limit of that function at say x approaching -3

[azure raven]

plug it into the equation-

but i dont see one?

[abstract raptor]

you look at the graph to see where x is and look at the corresponding y value

[azure raven]

is it 1?

[abstract raptor]

OH MY GOD

YOURE RIGHT

s9orry caps

[azure raven]

ohhh i was like did i do something wrong😭

ohhh so thats how?

[abstract raptor]

yes

try the first example

try a

[azure raven]

is it..

uhm 1

x approaching -1

is 1?

[abstract raptor]

YESSSSSSSSSSS

YOURE RIGHTT

OMGGG

do you understand it now?

[azure raven]

yussssssssssssssssssssssssssssssss

eek

eek ekk

[abstract raptor]

eek

[azure raven]

wait a moment

is the x approaching is strictly whole numbers?

[abstract raptor]

what?

[azure raven]

for example

if x is approaching to 2

[abstract raptor]

yah.

go on

[azure raven]

are there instances where it can be x is approaching to the square root of lets say 4

[abstract raptor]

yes x can approach any real number

as well as infinity and negative infinity

[azure raven]

okkki

ermm

another oneee

what about of a value is indefinite

lets say 0/0

kks

but what do you plus in stuff like for exampe when x is approaching 2

and you use the tables

like 0.9

sorry

i mean

1.9

1.99

1.999 etc

mi no understand thattt

[abstract raptor]

im not sure what you're asking exactly

do you understand like what a limit is what it means

what it means for x to approach some value

[azure raven]

yups

but i saw an example

where people use a table

like it had

0.01

or 0.001

0.0001

etc

[abstract raptor]

okay so limit as x approaches some value just means that we're trying to find the answer as x gets really close to that value

so we can approximate the limit as we make a table getting closer and closer

so for example if we have the limit as x approaches 2, we can approximate it by plugging in x = 1.9999

[azure raven]

oki

but why not the whole thing tho?

infact

when should i apply it?

do they usually ask for it or it depends on the equation

[abstract raptor]

they usually ask for it

however you apply a limit when substituting the x value itself would give something undefined like 1/0 for example

lemme give you an example

the limit of y as x goes to 1 is that white dot

however we cant plug in 1 itself because we would get 0/0

so we use the limit to see what the value of that dot is

either via l'hopitals rule or approximating it with a table

you feel me ?

https://cdn.discordapp.com/attachments/1119741787280507121/1347102628089299017/image0.gif

[azure raven]

yus yuss

okiii im not worried for tomorrowwwwwwwwwww

[abstract raptor]

good good

[azure raven]

okiii tunx youuuu

have a nice dayayayay

.close

[vale dock]

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[abstract raptor]

you too good luck

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[gilded summit]

Im having some dificulty understanding this proof

[gilded summit]

To start, doesnt the use of the term apply to any interval close to s as there are inifnate elements?

like most of the terms are also 0.0000001 within s?

secondly, should the inequality be flipped?

[distant galleon]

Forgive me for being thick, but where does "apply to any interval" appear?

[distant galleon]

idts

Since $s_n$ converges to $s$, we have that for any $\epsilon>0$, there is an $N \in \mathbb{N}$ for which
$$n>N \implies |s_n-s|<\epsilon$$
picking $\epsilon=\frac{1}{2}|s|$ seems to match up with using the reverse triangle inequality $|a|-|b| \leq |a-b|$.

[rocky lotus]

Civil Service Pigeon

[vale dock]

@gilded summit Has your question been resolved?

[gilded summit]

Like, would this not be true for any element?

[distant galleon]

I believe it's b/c the initial terms of $s_n$ can be quite far from the limit - we only care about the "infinite tail" of the sequence that exists after a certain point

[rocky lotus]

Civil Service Pigeon

[distant galleon]

with the terms in that "infinite tail" being referred to by "most"

ex. $a_n=\begin{cases} n+1, & 1 \leq n \leq 100 \ 1+\frac{1}{n}, & n \geq 101 \end{cases}$

[rocky lotus]

Civil Service Pigeon

[distant galleon]

the $a_{101}, a_{102}, \dots$ are "most of" the terms in the "infinite tail" of the sequence

[rocky lotus]

Civil Service Pigeon

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[distant galleon]

.test

test

[distant galleon]

.close

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[honest stone]

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[rapid jewel]

Help

[rapid jewel]

Answer key says this is wrong

Where did I go wrong

?

Answer key says it’s 2x^2 cube root x

[safe fulcrum]

ah perhaps the answer key is the same as yours

[rapid jewel]

Can’t you combine since like radicals

?

[safe fulcrum]

I see the issue now wait

$2x \sqrt[3] {x^4} = 2x^2 \sqrt[3] {x}$

[rocky lotus]

south

[safe fulcrum]

you misread your work as having 2x^3

[rapid jewel]

Wait which line

Sorry I can’t see

[safe fulcrum]

third line

[rapid jewel]

Middle term?

[safe fulcrum]

this part is wrong

[rapid jewel]

Wait why

Oh

Nvm

Thx

[safe fulcrum]

no worries!

[rapid jewel]

Do I have to foil this

Or can I just apply the exponent

??

[safe fulcrum]

you need to FOIL it first

then one of the multiplications is sqrt(3) * sqrt(3) = 3

[rapid jewel]

I vaguely remember being able to apply to exponents

To sometjongmoone this

Like this

Is there like a rule

Or anything

When you have to foil

[safe fulcrum]

the rule is just FOIL

on $(2 - \sqrt3)(2 - \sqrt3)$

[rocky lotus]

south

[rapid jewel]

I

K

Thank you

[safe fulcrum]

no worries again!

[rapid jewel]

Wait when I multiply the two radicals

Do I I write it as +( crt 3)^2

?

[safe fulcrum]

this is the last term

so yes, -sqrt(3) * -sqrt(3) = (sqrt(3))^2, same thing

[vale dock]

@rapid jewel Has your question been resolved?

[rapid jewel]

How to do this?

Thx

Only got to this

[wraith summit]

i suggest using Heron‘s formula for area instead

[rapid jewel]

Yo thx

[safe fulcrum]

overkill

[safe fulcrum]

the easiest would be A = 1/2 ab sin C actually

[wraith summit]

"overkill" it‘s 3 multiplications lol. Pick whatever way you prefer yes

[safe fulcrum]

but actually, for an equilateral triangle, $b^2 + h^2 = (2b)^2$, so $h = b \sqrt3$

[safe fulcrum]

they will have learnt this but not Heron's formula in the IGCSE curriculum

[rapid jewel]

Bru

[safe fulcrum]

<@&268886789983436800>

oh the scammer... what

[wraith summit]

Was that automod or incredibly fast mods

[safe fulcrum]

that was quick!

[rapid jewel]

I don’t get it

[wraith summit]

Wtf hayley

[rocky lotus]

south

[sly fern]

multiple snipers pointing at chat

[rapid jewel]

Can’t u use outhrsogrean

Pythagorean

[safe fulcrum]

that's exactly what was used ^

[safe fulcrum]

let me fix it though

[rapid jewel]

This is what’s barley

Says

Answer key

16

But I don’t get it

[safe fulcrum]

like, b is half of the base is what I mean

[rapid jewel]

Says u can use Pythagorean

[safe fulcrum]

yes

[rapid jewel]

But how

[safe fulcrum]

can you apply Pythagoras here?

don't try to simplify anything

[rapid jewel]

What

I’m so confused

How does the area relate to any of the triangles sides

It’s bh/2

[safe fulcrum]

yes, and you want to find h in terms of the base, a

[rapid jewel]

Ohh

I forgot all sides are equal

[rapid jewel]

I’m stuck again

Where did I go wrong

[safe fulcrum]

how are you getting a^3 h/2?

[rapid jewel]

The base / sides

Oh

Rip

How do I do this

Please

So I need to find h

[safe fulcrum]

yes, and you need to apply Pythagoras

[rapid jewel]

A^2+A^2=c^2

?

Right

2(a^2) = c^2

[safe fulcrum]

no

that's for when you have a 45-45-90 triangle

[rapid jewel]

How then

[safe fulcrum]

what are the 3 sides of the right triangle?

[rapid jewel]

How do I find height 😳

[rapid jewel]

a?

base

And sides

[safe fulcrum]

a is one of the sides

is it the shorter leg, the longer leg, or the hypotenuse?

[rapid jewel]

it’s equilateral triangle tho

Or if you split it

It’s h in centre

The longer

Idk

Nvm I got it

Thank you

👍

[vale dock]

@rapid jewel Has your question been resolved?

[safe fulcrum]

.close

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[opaque tiger]

Help with 3 and 4 please I've been at it for like 45 minutes. I tried to solve the 3rd one in two ways first i expanded the trigonometric functions and then tried integrating but it just kept becoming longer and longer and the other method was to integrate it first and then solve it but that also didn't net any progress

[opaque tiger]

In the second one I tried it with partial fractions but maybe I got something wrong cuz I still didn't the answer

[honest stone]

the denominator in 3. becomes sin²(𝑥) - ¹⁄₄ I believe

$\int\frac{\sin2x}{\sin^2(x)-1/4}dx$

[rocky lotus]

𝙸𝚗𝚏𝚒𝚗𝚒𝚞𝚖³

[pure mirage]

$\sin (A)\sin (B)=\frac{1}{2}\l[\cos (A-B)-\cos (A+B)\r]$ and so \begin{align*}\sin \left(x-\frac{\pi }{6}\right)\sin \left(x+\frac{\pi }{6}\right)&=\frac{1}{2}\left[\cos \left(\left(x-\frac{\pi }{6}\right)-\left(x+\frac{\pi }{6}\right)\right)-\cos \left(\left(x-\frac{\pi }{6}\right)+\left(x+\frac{\pi }{6}\right)\right)\right] \ &=\frac{1}{2}\left[\cos \left(-\frac{2\pi }{6}\right)-\cos (2x)\right] \ &=\frac{1}{2}\left[\cos \left(-\frac{\pi }{3}\right)-\cos (2x)\right] \end{align*}

[rocky lotus]

Kepe

[honest stone]

welp.

[pure mirage]

Well ok I think that works too

[opaque tiger]

Did he just type that...

[honest stone]

there’s your answer straw

[pure mirage]

No it's not done yet

It's just the first step that I think would lead to the solution lol

[honest stone]

well I doubt im of any use anymore lmfao

[opaque tiger]

Oh let me see

[pure mirage]

Your idea might be simpler

Idk

[opaque tiger]

Id like to try it before i get the answer

[pure mirage]

The good thing is just that the pi/6 cancels like this and we get another 2x

[honest stone]

my idea was expand the numerator into 2 sin(𝑥) cos(𝑥) and that’s the derivative of sin²(𝑥) so let 𝑢 = sin²(𝑥)

[pure mirage]

Which might become sin(2x) after u-sub

[opaque tiger]

The provided solution used the indentity int(f(x)'/f(x)) = ln(f(x)) +c but I don't understand how they solved it cuz there's no steps

[honest stone]

$\int\frac{\sin2x}{\sin^2(x)-1/4}dx\Longrightarrow\int\frac{d\left(\sin^2(x)\right)}{\sin^2(x)-1/4};u=\sin^2(x);\Longrightarrow\int\frac{du}{u-1/4}$

[rocky lotus]

𝙸𝚗𝚏𝚒𝚗𝚒𝚞𝚖³

[honest stone]

This was my idea

[opaque tiger]

That's correct yeah

[honest stone]

maybe kepes is better

[opaque tiger]

Howd you get this

Mb if I'm asking really simple stuff I'm really struggling here

[honest stone]

product-to-sum identity

Kepe actually showed it

$\sin (A)\sin (B)=\frac{1}{2}\left[\cos (A-B)-\cos (A+B)\right]$

[rocky lotus]

𝙸𝚗𝚏𝚒𝚗𝚒𝚞𝚖³

[opaque tiger]

Oh alright

[pure mirage]

Yeah I think if we continue my idea we might get there too. So by the above we have [\int \frac{\sin 2x}{\frac{1}{2}\left(\frac{1}{2}-\cos (2x)\right)}\dd x=\int \frac{2\sin 2x}{\frac{1}{2}-\cos (2x)}\dd x = \int \frac{4\sin 2x}{1-2\cos (2x)}\dd x] and now I think $u$-sub would work: Let $u = 1 - 2 \cos(2x)$, then $\dd u = 4 \sin(2x) \dd x$ which is really nice because then [\int \frac{4\sin 2x}{1-2\cos (2x)}\dd x = \int \frac 1u \dd u]

[rocky lotus]

Kepe

[honest stone]

Oh wow

wait was mine wrong?

[opaque tiger]

I don't think so

[pure mirage]

Probably not, maybe it's even more elegant

[opaque tiger]

The answer is ln(4sin²x - 1)

Okay im gonna try it

Should I close this now?

[pure mirage]

Well I'd get ln(|1 - 2cos(2x)|) but that's probably equivalent

Yeah WA says it is

[opaque tiger]

Nice

Thanks a lot

[pure mirage]

Ok but I do get absolute values in there

It's weird that your solution doesn't have them

[honest stone]

I got log(|4 sin²(𝑥) - 1|)

[opaque tiger]

What

[pure mirage]

Yep that's it, nice

[honest stone]

are they equal?

[pure mirage]

Yes

See the WA query I posted

[honest stone]

interesting

who doesn’t like more identities

[pure mirage]

This is most likely just by cos(2x) = cos^2(x) - sin^2(x) = 1 - 2sin^2(x), probably not too spectacular

[opaque tiger]

I don't

[pure mirage]

which follows from cos(a + b) = cos(a)cos(b) - sin(a)sin(b) (and the last from sin^2 + cos^2 = 1)

[pure mirage]

I'm saying that both Infimimum and me got ln(|4sin^2(x) - 1|), with absolute values

Since integral of 1/u du = ln(|u|) + C

[opaque tiger]

Ohh alright

[pure mirage]

Also of course the + C shouldn't be missing

[opaque tiger]

Yeah I didn't type that in

Mb

I GOT IT

How to do 4?

[honest stone]

OH SHIT I FORGOT THERES ANOTHER PROBLEM

uh

[ocean bloom]

try adding and subtracting x in the numerator

[opaque tiger]

Alright

[honest stone]

rewrite the integrand as $1+\frac{1-x}{x+x^2\log(x)}$

[rocky lotus]

𝙸𝚗𝚏𝚒𝚗𝚒𝚞𝚖³

[honest stone]

then rewrite again as $1+\frac{1/x-1}{1+x\log(x)}$

[rocky lotus]

𝙸𝚗𝚏𝚒𝚗𝚒𝚞𝚖³

[honest stone]

$\int1+\frac{1/x-1}{1+x\log(x)}\d x$

[rocky lotus]

𝙸𝚗𝚏𝚒𝚗𝚒𝚞𝚖³

[honest stone]

seperate the integrand

$\int1\d x+\int\frac{1/x-1}{1+x\log(x)}\d x$

[rocky lotus]

𝙸𝚗𝚏𝚒𝚗𝚒𝚞𝚖³

[honest stone]

$x+\int\frac{1/x-1}{1+x\log(x)}\d x$

[rocky lotus]

𝙸𝚗𝚏𝚒𝚗𝚒𝚞𝚖³

[honest stone]

$u=\log(x)$

[rocky lotus]

𝙸𝚗𝚏𝚒𝚗𝚒𝚞𝚖³

[vale dock]

@opaque tiger Has your question been resolved?

[opaque tiger]

Hey man I got this what am I supposed to do now

Oh also there's a +1

[ocean bloom]

now you see infinium has done some steps over here
you can again part the second integral in two parts
in the first integral take lnx = u
and then you will get
$\int\frac{1/x}{1 + x\log(x)}\dx$ as $\int\frac{1}{1 + u.e^u}du\ $

[rocky lotus]

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

[ocean bloom]

nooo
it will keep getting lengthier

leave it

lemme think of something else

[opaque tiger]

Okay okay

[pure mirage]

\begin{align*}
\int \frac{1+x^2\log(x)}{x+x^2\log(x)}\dd x &=\int \frac{1-x+x+x^2\log(x)}{x+x^2\log(x)} \dd x\
&=\int 1 \dd x + \int \frac{1-x}{x+x^2\log(x)}\dd x\
&=x + C +\int \left(\frac{1-x}{x+x^2\log(x)}-\frac1x\right)\dd x+\int \frac1x \dd x\
&=x + C' + \ln(|x|) + \int \frac{1+\log(x)}{1+x\log(x)}\dd x\
&=x + \ln(|x|) + \log(1+x\log(x)) + C''
\end{align*}

[rocky lotus]

Kepe

[pure mirage]

Where the last = comes from identifying that integral as integral of u'/u

@opaque tiger

[opaque tiger]

Yippee

Let me see

[ocean bloom]

hey how did you reach the 4th step ? ( the last term of the 4th line )

[pure mirage]

You mean the step from the third = to the fourth?

[ocean bloom]

yeah