help-10

[astral ivy]

Oh to have an inverse?

[slow skiff]

Yeah

[astral ivy]

I thought generally when someone says f is invertible, they mean it’s both left-invertible and right-invertible

Just like “associative”

[slow skiff]

Should probably look those terms up

We learnt about inverses in discrete math before calc

in dm it had to be a bijection, but in calc, only an injection was enough

[astral ivy]

I mean yeah

I’ve only taken calc 1 and 2

[slow skiff]

Right right

[astral ivy]

And we never ever dealt with invertibility

Or injections or surjections or any of that

But it makes sense that calc only cares about left-inverses

[slow skiff]

True

[astral ivy]

Wait I’m like

Forgetting what a left inverse means

Why is that sufficient? I don’t know why I just said that it makes sense

g(f(x)) = x, right? Not necessarily f(g(x))?

[slow skiff]

Hmm...

Oh but

You can restrict the range R to the codomain and its surjective then

Right?

Instead of having f: R->R, do f:R->R* where R* subset R

and for every f(x) in R* you have an x in R that maps to y=f(x)

=> f is surjective?

[astral ivy]

Yeah

Ohhh

I feel like this relates to partial functions

Which is also a calc-only thing

[slow skiff]

What are partial functions?

[astral ivy]

It’s like a function, but not every element of the domain has to map to something

It’s like how f(x) = 1/x is considered to be R -> R

[slow skiff]

It is?

huh

No wait

[astral ivy]

Well not literally

But like

A partial function is often used when its exact domain of definition is not known or difficult to specify. This is the case in calculus, where, for example, the quotient of two functions is a partial function whose domain of definition cannot contain the zeros of the denominator. For this reason, in calculus, and more generally in mathematical analysis, a partial function is generally called simply a function.

[slow skiff]

Aha

"a partial function is generally called simply a function"

Lets define a whole term that we wont even use

[astral ivy]

lmao

That confused me for a while

So

If you think of functions instead as partial functions

Then for a “function” to have an inverse, it only needs left-invertibility

That didn’t really make sense lol

Whatever

[slow skiff]

No no

It does make sense

[astral ivy]

I still don’t get why calc only cares about left-invertibility

Oh rly

[strong sinew]

What's even the question ;-;

[slow skiff]

Would you believe me if I told you there isn't one x)

[strong sinew]

Yes I would belive u

[astral ivy]

I edited it to what we’re talking about rn

Oh, bruh the pinned message is glitched

[astral ivy]

Reopen it

[strong sinew]

I think the bot glitched

[astral ivy]

Yeah

[slow skiff]

I guess surjectivity will fix the swap between domain and range?

[astral ivy]

Hmm

What even is an inverse

Lemme draw this

I have absolutely 0 intuition right now

[slow skiff]

if f:A->B and f^-1: B->A

f: [0,2]->[0,5] w range [0,3]
f^-1 : [0,5]->[0,2]
but f^-1(4)=??? because it's not a surjection

[slow skiff]

There's a two way street between each element from A to B, would be the proper description I guess? ops, no

oh god ive confused range and codomain

[slow skiff]

Can I see?

[astral ivy]

Oh... g isn't a function

But if functions are really partial functions, then f and g are perfectly good inverses!

[slow skiff]

No, I don't think it is ok to do these definitions under partial functions

Also, didnt it say partial functions are functions? Not the other way around o.O

[astral ivy]

Ok here's my question: f is injective. That means there exists a function g such that g(f(x)) = x for all x in X. I've drawn g here, so what does 2 map to?

[astral ivy]

No it said in calc partial functions are often simply called "functions"

[slow skiff]

Nothing, because it's not a surjection

[slow skiff]

[astral ivy]

Wait can't I just make 2 map to whatever and it still works

Yeah

Oh

[slow skiff]

But then it's not an injection

[astral ivy]

The definition doesn't care what g is

Only that g(f(x)) = x

[slow skiff]

Oh right

Well yeah that's the way to go

I think I'm having some discrete math class flashbacks

[astral ivy]

Ohhh

Yeah

[slow skiff]

Here's a question though...

[astral ivy]

f can't possibly satisfy f(g(x)) = x

[slow skiff]

If f is an inverse of g, isnt g also an inverse of f?

[astral ivy]

Depends what you mean by inverse

Left and right inverse?

[slow skiff]

left is injection right is bijection?

[astral ivy]

No

Left is injection right is surjection

[slow skiff]

ye ye i just googled it

[astral ivy]

See the diagram, g is surjective so it can only have a right inverse, f (which is shown to the left of g)

God the diagram is very frustrating sometimes

Considering only the first 3 circles

f is on the left of g, but f is a right inverse of g

Man...

[slow skiff]

Hhahhaha

Ok I see

Anyway

We take codomain=range in calc

[astral ivy]

Wdym

[slow skiff]

So they're surjective by default

[astral ivy]

Like set the codomain to be the range

?

[slow skiff]

Yes

[astral ivy]

But in every calculus text I've seen, it talks about arbitrary functions f : D -> R, and they're including f(x) = x^2 in this theorem/Lemma/whatever

Ok I think

For sine, people understand that it doesn't have an inverse on R, so they don't call it "sin inverse" they call it "arcsin" (emphasizing that it is a different function). But everyone generally understands that injective functions like e^x have an inverse and so there's no name change

[slow skiff]

Ahh!!

[astral ivy]

What is it

[slow skiff]

I agree it's just that im having trouble typing sth

[astral ivy]

oh lol

[slow skiff]

What are you saying with this though?

I thought we were talking about functions not needing to be surjections to have an inverse

That is, the left right w.e one

Oh right

So then f: [0, +inf) - > R, f(x) =x^2 doesn't have a two sided inverse

I see what you meant with that now

Although what I meant was that we take it as a two sided because we make it surjective by restricting the codomain to the range

Those are my final thoughts

[astral ivy]

Ah yeah

[warm shale]

abs_0

[astral ivy]

God I hate typing on mobile

[slow skiff]

Right

But thats because the domain is R^+-{0}

so x cant be negative on the right side

if it were

e^ln(|x|) then itd be |x|

with domain R

which is also why
f(x)=e^ln(x)
is not the same function as
g(x)=x

[astral ivy]

Ah

Wait am I tripping

Isn't

Uhhh

[slow skiff]

Yes?

[astral ivy]

Oh I suppose it makes sense

e^ln|x| = |x|

Cause (red then green) : R+ -> R+ is really the identity on R+

So it doesn't matter what the blue function is, as long as its range is R+

[slow skiff]

Well...

[astral ivy]

So then we could say for any function b: D -> R+, we have e^ln(b(x)) = b(x) for all x in D

[slow skiff]

Yep

[astral ivy]

Hmm

Pretty pictures

I need to go play some stupid mindless game now

[slow skiff]

It's good cause I can barely read so I like looking at pictures

[astral ivy]

Literally

[slow skiff]

Although your pictures are confusing a lil

[astral ivy]

How so

Oh cause domain of ln x isn't R

But the circle includes all of R

[slow skiff]

Yeah

But it's fine

It gets the point across

x)

[astral ivy]

Okay so in short: Calculus doesn't care about right-invertibility because you can make any injective function surjective by restricting the codomain to the range at any time, and the function itself wouldn't practically change. However, if you've got a surjective function and restrict the domain so that it's injective, it's not really the same function anymore, because then you can't input as many numbers as before. Thus inverses must have a left inverse, but don't have to have a right inverse.

[slow skiff]

Yep

That's the conclusion

[astral ivy]

Awesome

That makes sense now

[slow skiff]

Take care now, I'm gonna debate whether I should sleep or just go 30h without

[astral ivy]

Haha ok

Thanks for the discussion

See ya

[slow skiff]

Bye

[astral ivy]

.close

[obtuse pebble]

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[karmic vale]

how do i prove that |a|=0 iff a = 0?

[karmic vale]

i know that it's really obvious but i don't feel good without a proof

and i know there's a property of the absolute value that says
|a|=0 <=> a = 0 (Positive-definiteness)
but how can i prove that?

[twilit loom]

you can individually prove the forward and backward implications of $\abs{a}=0\iff a=0$

[warm shale]

dino

[karmic vale]

sorry but i don't really get that

[twilit loom]

hmm, does it make sense if I say if we prove $\abs{a}=0\implies a=0$ and prove $a=0\implies \abs{a}=0$ then that proves your iff

[warm shale]

dino

[karmic vale]

yeah that'll make sense, but i don't know how to prove |a|=0 => a = 0 and the other way around

would appreciate if you can prove that

[twilit loom]

well let's try a = 0 => |a| = 0 first

[karmic vale]

alright

[twilit loom]

so we have a = 0, then it's clear that the absolute value of a is...?

[karmic vale]

by the absolute value definition if a>=0, then |a| = a. so if a = 0, then |a| = 0

[twilit loom]

if a>=0, then |a|=0 or |a|=a?

[karmic vale]

so if a>=b, then (what i think) a > b or a = b

if this is true, then what you said is also true i believe

[twilit loom]

yeah that's true but I'm just letting you double check your first sentence before

by the absolute value definition if a>=0, then |a| = 0

[karmic vale]

oh my bad

i did a typo

i meant if a>=0, then |a| = a

[twilit loom]

cool cool, this is good then

now |a|=0 => a=0

[karmic vale]

since 0>=0, then yes i believe

|0| = 0

[twilit loom]

that doesn't quite work, you seemed to have used the to-be-proved statement in your proof

[karmic vale]

wait did I?

[twilit loom]

you said 0>=0

but we don't know that a=0 yet

[karmic vale]

ohh right

[twilit loom]

unless you meant the 0 in |a| = 0? but that doesn't tell us much since in |b| = 2, 2 is greater than 0 but b could be -2

[karmic vale]

ahh yeah

good point

so we still don't know if a is only 0 if given |a| = 0

[twilit loom]

you could consider that |a|=b => a=b or a=-b (edit, we only need to consider the forward imp)

[karmic vale]

so |a| = b implies that a = b or a = -b?

[twilit loom]

yup, does that make sense?

[karmic vale]

aren't we kinda losing information? like the a>=0, then |a| = a or if a<0, then |a| = -a?

or is that not needed

[twilit loom]

the information of a>=0 or a<0 is only available if we know a, which we don't in this case, right?

[karmic vale]

i see

yeap we don't

[twilit loom]

so could you apply this to |a| = 0?

[karmic vale]

|a| = b if and only if a = 0 or a = -0 = 0

is that true?

[twilit loom]

|a| = b? I think you meant |a| = 0

[karmic vale]

my bad

|a| = 0 if and only if a = 0 or a = -0 = 0

[twilit loom]

perfect

so |a| = 0 if and only if a = 0 or a = -0 = 0 and by the absolute value definition if a>=0, then |a| = a. so if a = 0, then |a| = 0

thus, |a| = 0 iff a = 0

[karmic vale]

ahh

thanks a lot

[twilit loom]

[karmic vale]

.close

[obtuse pebble]

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[glass palm]

what are all the values of x for which x^5-3x^3+2x^2>3

[obtuse pebble]

@glass palm Has your question been resolved?

[frosty spoke]

I assume you could find the roots and do derivative tests?

maybe finding the roots is going to be hard

[glass palm]

i still didnt take derivatives

i can get the answer choices if that makes it better

.close

[obtuse pebble]

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[gloomy carbon]

I don't know what to do next

[minor pilot]

I can help out

[lethal sand]

is this after or before you differentiated?

[gloomy carbon]

The x^2 + y^2 is before I diff

[lethal sand]

ok then, try again from the begining

remember, once you have started differentiating both sides, you differentiate both sides at once

also remember that y is a function of x

meaning the derivative of y wrt x is not simply 1

[gloomy carbon]

Okay lemme try again

I got this but what should I do after that? Should I transpose?

[lethal sand]

again, the derivative of y with respect to x is not 1

the derivative of y wrt x is dy/dx

if you are still having trouble, i'll do half of this for you as an example. In this problem, the derivative of x+y is not 1+1, but it's 1+dy/dx. Or for example, the derivative of y^2 is 2y*dy/dx

[gloomy carbon]

Oh okay you multiplied y' to the y variable as d(u)

[lethal sand]

yes, aka the chain rule

[gloomy carbon]

Ohhh okay, now I get it. Thanks!

.close

[obtuse pebble]

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[gentle turret]

What are some examples of theorems that were derived from the idea that there are infinitely many real numbers?

[onyx spoke]

what exactly do you mean

do you mean there being infinite real numbers in any line segment on the real number line

or that the real numbers go on forever

[gentle turret]

this

[onyx spoke]

well

limits to infinity wouldn't be as useful

because you could only do them on sequences

[gentle turret]

Why could you only perform them on sequences?

[onyx spoke]

but the real numbers not going on forever implies the natural numbers don't go on forever

so actually you couldn't

so you literally couldn't do limits to infinity

this would mean you couldn't have taylor series

which are incredibly important

[obtuse pebble]

@gentle turret Has your question been resolved?

[obtuse pebble]

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[pine sail]

Sure, what do you understand by this though?

What have you tried?

[zenith spade]

What do you know about the function?

[lunar mortar]

Please do my question I can't find direction ratios of first line

[pine sail]

#❓how-to-get-help

[zenith spade]

mate you’ve been spamming this in already-occupied channels for a while now
just find a free one

do you recognize the function?

How is log defined in the first place?

[lunar mortar]

Cz this Indian channel

[pine sail]

Exuse me?

@lunar mortar I want you to check #❓how-to-get-help

Nothing wrong with reading the rules as for how server works right?

[lunar mortar]

Not interested

[leaden ibex]

We're not really interested either

[pine sail]

Then no one is gonna help you.

[leaden ibex]

👋

@empty cobalt Have you tried defining log10 yet?

[obtuse pebble]

Channel closed

Closed due to the original message being deleted

[lunar mortar]

Can't u just help me rather than just arguing it's my exam on Monday

[pine sail]

It's not my problem. Besides, you don't know how this server works. I can't help you in a wrong Channel.

[leaden ibex]

No one is under any obligation to provide help

the least you can do is at least follow the rules here

If you want someone obligated to help you, you could pay a tutor

[lunar mortar]

U guys r not kind I think it's okay

[obtuse pebble]

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[timid silo]

when u rewrite numbers as polar form can’t we get infinite answers for everything

[timid silo]

might be a stupid question

like 2 = 2e^i2pi

if that’s even right

[leaden ibex]

What do you mean by infinite answers?

$e^{i\phi} = \cos\phi + i\sin\phi$

[warm shale]

Remavas

[timid silo]

well 2kpi

turn 360° we get same point right

[leaden ibex]

yes, cosine and sine are periodic

so is $e^{i\phi}$ consequently

[warm shale]

Remavas

[timid silo]

why does re^i*t work so easily

or better, how does it work

why is it cost + isint

[leaden ibex]

https://en.wikipedia.org/wiki/Euler's_formula#Using_power_series

[timid silo]

a bit too complicated for me i think

[leaden ibex]

It's just the good old power series for e^x

[timid silo]

wikipedia makes everything complicated

[leaden ibex]

and you plug ix in

[timid silo]

what’s the e^x power series??

[leaden ibex]

You didn't have calculus yet?

[timid silo]

we do other stuff in school

[leaden ibex]

Well, it can be shown that $e^x = \sum_{n=0}^\infty \frac{x^n}{n!}$

[warm shale]

Remavas

[leaden ibex]

using calculus

Without knowing it, I can't further elucidate this

[timid silo]

how

[leaden ibex]

^

[timid silo]

OKAY i’ll look it up

[leaden ibex]

Unless you know calculus, you won't understand where it comes from

but you are free to try

[onyx spoke]

yes

for example
1 = e^(2n i pi)

[timid silo]

yea

[onyx spoke]

for n any integer

[timid silo]

yea

[onyx spoke]

you're correct

[timid silo]

u cool

both

[leaden ibex]

And once you know taylor series

You can come back and understand where Euler's formula comes from

[fierce lagoon]

You can use Euler's identity

I think that would be easier then trying to use the Maclaurin series for exp(x)

[timid silo]

help

what’s that

[leaden ibex]

$e^{i\pi} + 1 = 0$

[warm shale]

Remavas

[timid silo]

yea

but i mean the maclaurin thingee

[fierce lagoon]

I mean, i would just sneak around and then just use Euler's identity

[leaden ibex]

Without calculus you probably wouldn't understand it

(maclaurin series)

[timid silo]

what do u mean with calculus

[leaden ibex]

I mean calculus

[timid silo]

oh

[fierce lagoon]

you can rewrite 2e^(2iπ) as 2e^(iπ)^2

And then that's just 2(-1)^2

[timid silo]

u mean (2e^(ipi))²?

[fierce lagoon]

No

Because I'm rewriting e^(2iπ)

[timid silo]

yes

multiply powers

[fierce lagoon]

= (e^(iπ))^2

= (-1)^2

[leaden ibex]

But I mean that won't give him any insight into why it works. Might as well just use Euler's formula instead then

$e^{2i\pi} = \cos2\pi + i\sin2\pi = 1 + 0i = 1$

[warm shale]

Remavas

[timid silo]

ya

i watched a series about calculus 1 but he didn’t talk about power series

[leaden ibex]

I have no idea when the taylor expansion is taught under the calc 1 calc 2 calc 3 classification

Since I never took it in that format

[timid silo]

oh okay

[leaden ibex]

calc 2 it seems

[fierce lagoon]

Calc 2

Calc 2/BC is when it's taught

That's what I'm taking rn lol

[timid silo]

ooo

[fierce lagoon]

But, if you're like me and try to find shortcuts and shit, just know that e^iπ = -1

[leaden ibex]

yup, same

the equivalent of it anyways

[timid silo]

cool

[fierce lagoon]

Use this to your advantage because like

It'll become extremely useful

[timid silo]

yea it is

.close

[obtuse pebble]

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[drowsy cosmos]

Hello can someone proofread this:
Let $\omega,v$ be even differential-forms. Calculate exterior derivative of $dw\wedge v - w\wedge dv$

[warm shale]

Enoo58

Enoo58

[obtuse pebble]

@drowsy cosmos Has your question been resolved?

[drowsy cosmos]

<@&286206848099549185>

[drowsy cosmos]

<@&286206848099549185>

[obtuse pebble]

@drowsy cosmos Has your question been resolved?

[obtuse pebble]

@drowsy cosmos Has your question been resolved?

[vestal niche]

Trying to understand the rules of functions and when jt is continuous and not continuous

If anyone can direct me

Anyone?

<@&286206848099549185>

[tardy epoch]

read #❓how-to-get-help

[inland girder]

Definition of function continuity/discontinuity (page 3):
https://faculty.math.illinois.edu/~aydin/math220/lecturenotes/m220_Sec1_4.pdf

[compact shadow]

No d(wΛv)=dwΛv+(-1)^|w| wΛdv

[drowsy cosmos]

yeah and |w| is even

[compact shadow]

So you are correct when w is a n-differential form where n is even

Otherwise it’s zero

Okay then you are correct

[drowsy cosmos]

alright can i show you another one?

[compact shadow]

Sure

[drowsy cosmos]

thanks w and v are again even and $\sigma$ isnt specified

sorry wrong one

[warm shale]

Enoo58

[compact shadow]

It’s true when w and v are even

[drowsy cosmos]

alright thank you very much

.close

[obtuse pebble]

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[obtuse pebble]

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[tropic frigate]

Basic question but:
If there are 3 items and 20 people that could win it, how would i calculate the winning chance the person has to win one of the items, doesnt matter which

[violet compass]

1/20

right?

[timid silo]

5%?

or 33%

[violet compass]

if the item doesn't matter then 1/20

oh no

[tropic frigate]

woudnt that be the case if there was only 1 item to be won?

[timid silo]

its still 3/20

[violet compass]

it's (1/20)*3

so yh

[violet compass]

3/20

👍

[timid silo]

15%?

[tropic frigate]

yep, thanks

idk how i didnt figure that out 💀

[obtuse pebble]

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[versed zinc]

am I wrong? 🤨

[kindred rain]

ye 1/4 is the correct answer

[versed zinc]

ok sorry, my english level is not good

.close

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[rain coral]

i dont really get how i am meant to differenciate e^x

do i differenciate 4x first?

and then i differenciate e^x

[violet compass]

here's an example

e^2x

2e^2x

again would be

4e^2x

[rain coral]

ohhhh

so i just sorta ignore the x

[violet compass]

sorta

[rain coral]

so for me it would be 8e^4x and for d2y/dx2 32e^4x

[violet compass]

so your answer is B

[rain coral]

yep

thanks

[violet compass]

lemme show you a little tool

[rain coral]

but for some like if e appears in a product rule, do i differenciate e first?

[violet compass]

https://www.derivative-calculator.net/

[rain coral]

oh lol

[violet compass]

https://www.integral-calculator.com/

and for antiderivatives

[rain coral]

ok

thx

.close

[obtuse pebble]

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[wintry stream]

how do i solve this?

[obtuse pebble]

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[harsh valley]

Hi, so if 2²=(-2)²
Than √2²=√-2²
The root cancel the square so
2=-2? Is there's something I'm missing here

[slow skiff]

Step 2 is wrong

And also

$\sqrt{x^2} = |x|$

[warm shale]

.nai

[harsh valley]

Yes i know the answer is always positive, so this means that i have mistake with the brackets (order of operations)

Like i should have solve (-2)² first which is 4 or something else

[slow skiff]

Yes

[harsh valley]

Oh thx

[slow skiff]

Or

But also, here:

Than √2²=√-2²
√2²=√(-2**)**²

[harsh valley]

K

. Close

.Close

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[obtuse pebble]

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[winter heron]

Hey guys I’m doing this assignment where u have to leap and calculate a launch angle. Is it the angle ur about to leap of the ground or when ur first in the air?

[spiral maple]

the launch angle is always measured from the horizontal

regardless of how high you launch it from

[winter heron]

So it’s from the ground?

[spiral maple]

from the horizontal..

oh you mean the angle they launch from

they launch (within approximation) straight up

[winter heron]

Like cos I have to leap right, I recorded myself in slow motion but idk which angle is the initial launch angle

so I’m not sure

@spiral maple

[spiral maple]

Yeah I'd think the launch angle is 90, cause it's straight up.

[winter heron]

the launch angle should be less than 90 under any circumstances right?

@spiral maple

[obtuse pebble]

@winter heron Has your question been resolved?

[royal solar]

most likely

but draw a curve and tangent to your curve at 0

easiest thing you can do is find the length you traveled and your max height

[winter heron]

But like is it the angle at which I’m about to leap at?

cos idk

[royal solar]

and model your jump

[obtuse pebble]

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[long phoenix]

can someone help me get the function

from the graph

i got (x+6)(x-3)(x+3) but its not right

[languid shard]

u need to have f(0)=-9

just multiply a reasonable constant

[long phoenix]

how would I use f(0)?

or the y int.

[somber vapor]

Try putting in x=0, y=-9 in the function you get from multiplying (x+6)(x-3)(x+3)

[long phoenix]

if x=0, then using my function you get -54

which I need to divide by 6

so do i just put ((x+6)(x-3)(x+3))/6

[somber vapor]

I think so :)

[long phoenix]

ok got it!!

thanks!!

.close

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[inland thicket]

guys

the first way is wrong

but i dont understand why I have to first do factorisation

in order to solve limits

[brave bramble]

0/0 isn't 0

[inland thicket]

hmm

so first way is also correct but its impossible to do it that way ?

because error 404

[brave bramble]

The first method just didn't get 0, but then pretended it did

So I wouldn't say that's "correct", no

[inland thicket]

so 0/0 is just impossible to calculate

therefore it cant be done that way

[brave bramble]

If you get 0/0, that's a hint that the limit may exist, but you'll need a better method

[inland thicket]

hmm

[brave bramble]

Often, cancelling something is the way to go

[inland thicket]

Ok

when x--> infinity

I just divide everything by the biggest x

is there explanation for that

[brave bramble]

Sometimes, haha. Depends on the limit in question

[inland thicket]

like i had this let me find

[brave bramble]

For example, limits with sqrts often don't use this trick

[inland thicket]

this is my test

i solved it right

but i dont get it

[brave bramble]

So multiplying by x^5/x^5 is like multiplying by 1. You're always allowed to do that.

Then, you have the lower x^5 cancel out the numerator

[inland thicket]

but why is it allowed when x-->infinity

and when x-->1 i need the different method

because if i gotten a task with x-->1 i woulda failed

i just thought its dividing by biggest x

[brave bramble]

You did skip a step. You do have this missing:

$\frac{2 - \frac{4}{x^2} + \frac{3}{x^4} - \frac{3}{x^5}}{1 - \frac{11}{x^4}}$

[warm shale]

Kaynex

[brave bramble]

So when you divide top and bottom by x^5, this is what happens

When you allow x to get very large, every term that's divided by an x goes away, leaving 2/1

[inland thicket]

howcome 11/x^4

and others

oh ye

its like

dividing numbers exponents go in subtraction right

[brave bramble]

Because $\frac{-11x}{x^5} = -\frac{11}{x^4}$

[warm shale]

Kaynex

[inland thicket]

ok ye

[brave bramble]

Now, this "disappearing act" only works because x is getting very large

If x is just approaching 1, this method doesn't work anymore

[inland thicket]

hmm

im trying to imagine this

how can it dissapear if its very large

[brave bramble]

$\frac{2 - \frac{4}{x^2} + \frac{3}{x^4} - \frac{3}{x^5}}{1 - \frac{11}{x^4}}$

[warm shale]

Kaynex

[brave bramble]

So you agree, you have this after dividing top and bottom by x^5

[inland thicket]

yes

[brave bramble]

If x is really big, then something like 4/x^2 becomes really small

[inland thicket]

oh yes

[brave bramble]

Every term with an x below it becomes 0

Leaving 2/1

[inland thicket]

ok like

basically what im trying to do with this

is determine the limit on y axis

[brave bramble]

,w graph \frac{2 - \frac{4}{x^2} + \frac{3}{x^4} - \frac{3}{x^5}}{1 - \frac{11}{x^4}} between 0 and 10

[warm shale]

Results provided by WolframAlpha

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

Requested by Kaynex#7520

[brave bramble]

So you can see on the graph how it trails off to the right

[inland thicket]

yeah

[brave bramble]

It is slowly getting near y = 2

Which is the answer to your limit

[inland thicket]

wait is the y horizontal

y axis

[obtuse pebble]

@inland thicket Has your question been resolved?

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[wanton dagger]

I need help with finding the domain and range for the first question. I found the inverse though.

[lone jetty]

Talking about part a?

[wanton dagger]

The inverse function:

[wanton dagger]

Yes

[lone jetty]

Sure

So you know what the domain is?

[wanton dagger]

It’s all the x-values that a function can have?

[royal solar]

and range ofc

[lone jetty]

Exactly

So can you think of a number that can't be cubed?

[wanton dagger]

0?

[lone jetty]

0^3 = 0

0 * 0 * 0

[wanton dagger]

Ye

[lone jetty]

So it can be cubed 🤷‍♂️

So there just .. isn't a number you can't cube

[wanton dagger]

Yes

[lone jetty]

So then if the function is cubing

And you can cube all numbers

We know the domain is ...

[wanton dagger]

-infinity to +infinity

[lone jetty]

Yes, but you would write it as $$x \epsilon \mathbb{R}$$

[warm shale]

Alfie

[lone jetty]

x is in the set of real numbers

(I can't latex well)

[wanton dagger]

Oh so how about in interval notation

[storm bronze]

i think in a lot of highschools they don't require that notation

so interval notation is fine

[lone jetty]

I can only talk about what we do in the UK 🤷‍♂️

[storm bronze]

yeah makes sense lol

[lone jetty]

If they accept -infinity < x < infinity

[wanton dagger]

Ok

[lone jetty]

And then the range, can you think of a number you can't take the cube root of?

[wanton dagger]

So would that be right in interval notation?

[wanton dagger]

Mmm..

0?

[lone jetty]

So you're saying that no number, when multiplied by itself 3 times, equals 0?

[wanton dagger]

No I’m saying the number 0

[lone jetty]

So the cube root of 0 is 0?

[wanton dagger]

Ye

[lone jetty]

So it is valid?

[wanton dagger]

Yes?

[lone jetty]

Yes. The cube root of 0 is 0

[wanton dagger]

Ok

[lone jetty]

0 cubed is 0, therefore the cuberoot of 0 is 0

So... all numbers can be cuberooted

[wanton dagger]

Yep

[lone jetty]

Therefore..

[wanton dagger]

Yes

Y x R

[wanton dagger]

I mean this

But y….

Is all real numbers

[lone jetty]

Yep

[wanton dagger]

Ok I get it

Thanks

.close

[obtuse pebble]

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[neat rivet]

map g is such g(x, y, z, w) = (2x - y, 3z + w, x + z - 2w)

what i did was to represent the map as x(2, 0, 1) + y(-1, 0, 0) + z(0,3,1) +w(0, 1, -2)

and found whether they are linearly independent or not

nvm

.close

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[wild forge]

3. The weekly allowance of all the Senior High School students in the schools in a certain city has an average of Php 350 with a standard deviation of Php 20. A random sample of 60 students was asked of their weekly allowance.
   A. Determine the mean of the sampling distribution of the mean weekly allowance of Senior High School students in this city. State the rule/concept that you have applied (1 point)
   B. Determine the standard error of the sampling distribution of the mean weekly allowance of Senior High School students in this city. Assume an infinite population. (1 point)

[wild forge]

A.)
u=X̂
u=350
X̂=350

B.)
20/√60
standard error = 2.58

is this correct?

[obtuse pebble]

@wild forge Has your question been resolved?

[wild forge]

<@&286206848099549185>

<@&286206848099549185>

[obsidian isle]

What is a php

[obtuse pebble]

@wild forge Has your question been resolved?

[wild forge]

Currency

[obtuse pebble]

@wild forge Has your question been resolved?

[wild forge]

.close

[obtuse pebble]

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[shell hull]

In what case is this triangle equilateral?

[mighty geyser]

which triangle? triangle DCG? what are the cases you have in mind?

[shell hull]

Yes DCG
GD=DC and angle ABC=120

Is it enough?

[mighty geyser]

are the lines AG and BC parallel

[shell hull]

No

[mighty geyser]

then i don't think you can show that the triangle is equilateral

[shell hull]

I have these options

[mighty geyser]

yeah seems like all the options don't give that DCG equilateral

[shell hull]

lol

.close

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[woeful monolith]

well hello, i'm kinda stuck on my math homework about solving a murder mystery. the lore is basically:

you have arrived at the murder scene at 4:35 am. theres the lead detective there and he says he's been there since 3:45 am looking for clues. you're trying to figure out the time of death. you take the temperature of the body and its 84.5 degrees. more detectives come, blah blah, you see that the room's temperature is 73 degrees. you take the temperature of the body for the second time and find that its 74.9 degrees, noting that the time is 6:00 am.

we have an equation given to us to help with the problem, which is:

https://cdn.discordapp.com/attachments/957541993431584838/957542858871697428/unknown.png

[woeful monolith]

what i know/have done so far is

rooms 73 degrees, which is Ta
the first time you take the temperature is 84.5 degrees, which i think is T0
the second time you take the temperature is 74.9 degrees, which i think is Tf
and the time is 85 minutes

74.9 = 73 + (84.5 - 73) e^k85 is what i get when i plug everything in (im really sleepy rn so something i plugged in might be in the wrong place) which if you add and subtract everything, and leave k on one side becomes -9.6 = e^k85

[glossy sparrow]

use log

[woeful monolith]

which is where everything goes wrong, i cant take the natural log of a negative number

and leave k85 by itself

[woeful monolith]

i can't, taking the log of a negative number is impossible

[glossy sparrow]

why did you get -9.6

[woeful monolith]

74.9 = 73 + (84.5 - 73) e^k85

[glossy sparrow]

when i tried it, i get 74.9=84.5 x e^85

[woeful monolith]

true

wait

[glossy sparrow]

you dont subtract 74.9 to 84.5

because 84.5 is multiplied to e^k85

[woeful monolith]

oh shit

i need to divide it not subtract

[glossy sparrow]

yes

[woeful monolith]

got it

my bad

thank u for help

[glossy sparrow]

ok

[woeful monolith]

.close

[obtuse pebble]

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[fluid meadow]

How do u prove that sqrt(2), sqrt(3) and sqrt(5) can never be terms in an arithmetic series

[buoyant coral]

help

somebody here?

[astral estuary]

?

[nocturne minnow]

#❓how-to-get-help

[buoyant coral]

this

[nocturne minnow]

#❓how-to-get-help

[buoyant coral]

question24

[nocturne minnow]

Read it

It means don't post in an occupied channel

[buoyant coral]

so where I should?

[nocturne minnow]

Read #❓how-to-get-help and you would know

[buoyant coral]

but I do bad in English

[nocturne minnow]

Literally the first sentence in #❓how-to-get-help

[warm shale]

.76

.76

.76

[fluid meadow]

Why are we doing n-1, o-1, p-1

[astral estuary]

nvm its arithmetic series

[obtuse pebble]

@fluid meadow Has your question been resolved?

[obtuse pebble]

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[glossy sparrow]

hello

[glossy sparrow]

$\lim _{x\to 0}\left(\frac{tan^33x}{x^2}\right)$

[warm shale]

robic

[glossy sparrow]

how i find the limit?

i get a 0/0 so i used a different approach

when direct substitution

[potent dagger]

you could use the L Hopital rule

if ur aware of it

[high lily]

consider finding a way to apply the limit identity for
lim as x→0 sin(x)/x

[glossy sparrow]

i did actually

[high lily]

,rotate

[glossy sparrow]

it leaves me with 9, but in the desmos website shows no graph

[high lily]

you seem to be missing a factor of sin(3x)/cos(3x)

dunno what you did that

[glossy sparrow]

i tried to make ^3 to ^2 for both the deno and numer

[high lily]

but you can't just multiply by cos(3x)/sin(3x) like that

[glossy sparrow]

oh

what shall i do then

[high lily]

as that isn't 1

[glossy sparrow]

make it ^4

[high lily]

don't overthink it

and apply basic exponent laws

just like how you expressed sin^2(3x) as sin(3x) * sin(3x)
sin^3(3x) = sin(3x) * sin(3x) * sin(3x)

[glossy sparrow]

ok

i will think about it

thank you i gtg

/close

.closee

.close

[obtuse pebble]

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[drowsy cosmos]

alright i just saw it nvm

.close

[obtuse pebble]

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[loud egret]

I need help

[spiral maple]

recall basic algebra

[loud egret]

hmm

do I just rearrange

?

[spiral maple]

yes

you solve for t

[loud egret]

but it does not work

=v

[spiral maple]

then you did it wrong

post your algebra.

[loud egret]

kk

[spiral maple]

???

[loud egret]

=v

[spiral maple]

the hell happened in the 2nd line?

[loud egret]

=v

uhh iam dumb

forget the conjurgate

[obtuse pebble]

@loud egret Has your question been resolved?

[loud egret]

@spiral maple

lol far out

=v

my brain is now full of NFTs and crypto

[spiral maple]

$z-zit=1+it$

[loud egret]

I grind too much

[warm shale]

Mosh

[spiral maple]

this is a linear equation in t...

[loud egret]

yeah

I am so behind in uni

I am f

Thanks anyways

@spiral maple

[spiral maple]

this shouldn't be uni material

this is like.. grade 9

it's solving a linear equation

[loud egret]

=v 9 grade in your country

my one is different

[spiral maple]

regardless, it's still well before post-secondary

[loud egret]

.close

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

how do i work this out

[obtuse pebble]

Channel closed

Closed due to the original message being deleted

[timid silo]

oh

shu

make the table

bigger

lesh

oh

.CLOSE

.close

[proper coyote]

.

[obtuse pebble]

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[vapid thistle]

Hello, how do I go about simplifying this? I know I have to use the Fundamental Theorem of Calculus [F(t = inf) - F(t = 0)] but I can't seem to get the value of F(inf) because it results in an indeterminate of inf * 0. I know F(0) will simply be 0 when evaluated.

[vapid thistle]

This is what I tried so far, doesn't seem to get me anywhere?

[novel knoll]

is s>0?

[vapid thistle]

Yep

I forgot to mention that, sorry

[novel knoll]

exponential grows faster than polynomials

[vapid thistle]

So, the limit would be 0?

[novel knoll]

si

[vapid thistle]

Oh okay, I thought of that too earlier when I checked online graphing calculators. But I wasn't sure if just mentioning that is enough

[novel knoll]

you can prove limit of t^(a*n)/(e^(b * t)) is 0 for any b>0

[vapid thistle]

Hmm how do I do that though? I tried manipulating the fraction to be a different indeterminate form earlier, but that got me nowhere.

[novel knoll]

induction in n or write exp as its taylor series

are two ways

[vapid thistle]

oh alright, guess I'll try induction first

[novel knoll]

could also apply L'H n times to make numerator a constant

and denom still having e^t

[vapid thistle]

The problem didnt really state a specific n though, just that its from the set of n=1,2,3,...

[novel knoll]

its still a fixed unkown constant

[vapid thistle]

Oh, hmmkay hmmkay

I think I got it now, thanks!

.close

[obtuse pebble]

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[hallow bough]

6a help pls

[teal prawn]

that doesnt seem like a math question

[robust sleet]

doesn't matter

[hallow bough]

It is

[teal prawn]

ok

well we need to use an equation in this case

[robust sleet]

h=(v^2)/2g use this

[leaden ibex]

it's from a book titled

"A level mathematics"

[teal prawn]

ohk

[leaden ibex]

plus kinematics is mostly mathematics anyways

[teal prawn]

true

[obtuse pebble]

@hallow bough Has your question been resolved?

[obtuse pebble]

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[fading juniper]

.close

[maiden flint]

Is this right?

[high lily]

yes

[fading juniper]

yeah

i think

[maiden flint]

Yes but -2^2 = -4 though

[fading juniper]

no

its 4

[high lily]

-2^2 is indeed -4

[fading juniper]

-2x-2=4

[high lily]

-2^2 is not the same as (-2)^2

[maiden flint]

So why is my scientific calculator when I square negative 2 coming up as negative 4?

[high lily]

because you're not entering stuff in the calculator properly

[solar depot]

think gc interprets it as -(2)^2

[maiden flint]

Right I’ve just figured it out

[high lily]

yeh, you're not telling it to calculate what you want

[fading juniper]

it is saying 2 square times -

[high lily]

respectable calculators follow the order of operations

[maiden flint]

Sorry for wasting your time again 😂

[fading juniper]

you must put brackets

lol

[pine sail]

It's again very amusing to me how many people get stuck at this.

[maiden flint]

Thanks for being patient again with me…….

[pine sail]

It is alright.

You're not the only one.

[maiden flint]

I have an exam in may I will be using this channel a lot 😂

[fading juniper]

just remember not to forget those brackets

lol

[pine sail]

Right.

[fading juniper]

i can imagine how many times i got questions wrong because i typed it wrong in the calculator

*cant

.close

[maiden flint]

So can someone please remind me why putting the brackets around the number changes the answer

[fading juniper]

so like this

-2^2 is same as this

-(2^2)

which equals to

-(4)

how ever if you put brackets

(-2^2)

it is showing that you are squaring the whole -2

not just 2

[maiden flint]

Right so why don’t they just put the ^2 on the inside of the bracket in the first place?

[fading juniper]

i mean it means the same thing i think

[pine sail]

wdym that changed everything.

[fading juniper]

im confused now

[high lily]

how ever if you put brackets
(-2^2)
no

that changes nothing

[fading juniper]

i meant (-2)^2

sry

[pine sail]

Right that is 4.

[fading juniper]

yeah

[pine sail]

(-2^2) or -(2^2) is same thing.

Thank you, I started doubting my existence tbh.

[fading juniper]

lol

[maiden flint]

Me again

In this one I understand why we have negative 8 but why has the plus symbol gone?

Actually I’ve figured it out

[obtuse pebble]

@maiden flint Has your question been resolved?

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[lusty vigil]

Two functions are given by the regulations f (x) = 2x^3−150x and g (x) = x^2 + kx + 25−500, where k is a constant. The function f has a local maximum at the same point as g has a local minimum. Based on this information, determine the value of k.
@robust sleet sorry let's go here

[lusty vigil]

u questioned the +25-100

orrrr?

[obtuse pebble]

@lusty vigil Has your question been resolved?

[lusty vigil]

<@&286206848099549185>

[spiral maple]

Yeah.. why wouldn't the 25-100 be simplified in the question?

[obtuse pebble]

@lusty vigil Has your question been resolved?

[obtuse pebble]

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[paper socket]

Given the slope and a point, find the equation of the line in m = 3 ( -7,2 )

[paper socket]

how do i find the y=mx + b given the slope and point?

[spiral maple]

slope-point formula

[potent dagger]

u have the point and m tho

you can find b