help-10

[subtle whale]

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[obtuse pebble]

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[silver plover]

need some help plz

done this so far

any ideas?

<@&286206848099549185>

[obtuse pebble]

@silver plover Has your question been resolved?

[obtuse pebble]

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[timid silo]

$f(x) = 200 - 20f^{\prime}(x)$

What is the maximum value of $f(x)$ ?

[warm shale]

trololol !

[timid silo]

(This is actually a physics problem which, when abstracted, reduces to this).

The problem, though, is that we haven't done differential equations yet (not even derivatives!), and not even how to calculate the global maximum of a function. How can I solve this without using derivatives, because I'm pretty sure our teacher won't allow that ?

[kind hawk]

without using derivatives
equation includes a derivative

[sharp pecan]

please include mentioned physics problem then

[timid silo]

Yeah, that was the only way I found to represent it using purely mathematical concepts.

Including the physics context, it actually is $v = 200 - 20a$, and we have to find the maximum value of $v$.

[warm shale]

trololol !

[kind hawk]

whole context

[robust sleet]

dimensions dont even match

is a, acceleration or a velocity?

[timid silo]

a is acceleration

Yeah, I've surely messed somewhere in solving this problem.

Anyways, the problem is:

An object with mass $m = 0.1kg$ is falling from height $h$. The air resistance is defined by $F_r = k * v$, where $k = 0.05 \cdot \frac{N \cdot s}{m}$. What is the maximum value of its velocity ?

[royal flint]

well just for knowledge, the relative maxima is obtained by putting f'(x)=0 (critical value), if f''(x) is less than 0 (holds true here)

[warm shale]

trololol !

[kind hawk]

if something is falling you can assume the acceleration is constant

[timid silo]

How can it be constant if the air resistance is not constant ?

[kind hawk]

ah whoops

nvm

[timid silo]

I also find it weird, because so far we've only done exercises with constant acceleration too.

[kind hawk]

I don't really wanna do the math but the acceleration might still be essentially constant

if the height isnt big enough

[fading quest]

the maximum velocity is given when $\sum F=ma=0$ as the air resistance is equal to the acceleration due to gravity. hence you have $$F_g-F_r=0\implies F_g=F_r\implies v=\frac{mg}{k}$$

[warm shale]

Duh Hello

[fading quest]

i think at least

[timid silo]

Hmmm, makes sense. I will try finding the value of the velocity when the air resistance is equal to gravity (thus acceleration is 0).

Thanks!

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[timid silo]

can someone show me how this goes to that

<@&286206848099549185>

[elfin burrow]

After 15 minutes, feel free to ping <@&286206848099549185>.

ah didn't mean to ping myself 💀

[rigid lintel]

divide top and bottom by 2n + 1

[shell wave]

2n/2n+1 is simply (1-1/2n+1)

[tardy epoch]

that's what !15m is for

[rigid lintel]

does anyone actually self assign the helper role though

seems annoying to get spammed all the time lmao

[elfin burrow]

$\left(\frac{2n+1}{2n}\right)^n = \left(\frac{2n+1-1}{2n+1}\right)^{-n}$

[warm shale]

tushar

[elfin burrow]

!15m

[obtuse pebble]

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.

[timid silo]

i still dont get it

[elfin burrow]

\begin{align*}\left(\frac{2n+1}{2n}\right)^n &= \left(\frac{2n}{2n+1}\right)^{-n} \&= \left(\frac{2n+1-1}{2n+1}\right)^{-n} \&= \left(1-\frac{1}{2n+1}\right)^{-n}\end{align*}

[warm shale]

tushar

[timid silo]

ty so much 😁

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[swift grove]

You know sohcahtoa

rules

[heady sphinx]

yh but for some reason i keep getting it wrong

[timber fox]

its not a right angled triangle

[heady sphinx]

it says i have to use sine rule

[grim mortar]

yoo

math lovers

does anyone know econnomatrics here?

[timber fox]

can u state the law of sines ?

[swift grove]

it was too see if he was using it

and was gonna tell him its wrong not to use it

[heady sphinx]

a\sin

b/sin b

c/ sin c

something like that

[nocturne minnow]

That's horribly typed out

[rigid lintel]

to what extent haha

[timber fox]

$\frac{sinA}{a}=\frac{sinB}{b}=\frac{sinC}{c}$

[warm shale]

Clarkie

[swift grove]

smh stealing my glory

[timber fox]

pair the angles and the sides and solve for the missing variable

[heady sphinx]

okay i’ll try and do it give me on second

[fleet abyss]

better to memorize it the other way

helps with extended los

[heady sphinx]

i got

7.516946389

for

[nocturne minnow]

Post your work

[heady sphinx]

i did it in my head and with a calculator 😭

[fleet abyss]

since the blue angle is less than 60, x has to be less than 6

so 7.52 is wrong

use los -> 6/sin(180-60-49)=x/sin(49)

[nocturne minnow]

That's not ideal, because one, no one can see what went wrong. And two, if you were taking a test, your teacher couldn't give you credit on what you tried

Don't do the work for people

[fleet abyss]

a lot of teachers don't give partial credit

mb

[heady sphinx]

this is just revision but yh i probably should be writing it down

my bad

[nocturne minnow]

Most teachers expect the work so they can see what the process was

[fleet abyss]

I'm not disagreeing with you, just saying

[heady sphinx]

what about this question how would i go about it

[nocturne minnow]

You can still use sine rule

You need to create the proper equation

[heady sphinx]

i don’t understand sorry, english isn’t my first language

[nocturne minnow]

This here is sine rule

[heady sphinx]

so since i have 38.5m and 65

i can use that to create the missing angle?

[nocturne minnow]

Yes

[heady sphinx]

pero

but*

[nocturne minnow]

Wait

No

You're not finding an angle

You're finding a missing side

[heady sphinx]

is that harder?

[nocturne minnow]

No

The process is still the same

[heady sphinx]

so in the equation it would be

38.5x sin(65)?

[nocturne minnow]

That's not an equation

That's an expression

Equations have an equal sign

[heady sphinx]

so on bottom of the fraction

[obtuse pebble]

Please read #❓how-to-get-help

[heady sphinx]

what would i put

[nocturne minnow]

Follow the structure of the equation

You only need two fractions for that

Given the triangle you have, what is a?

Recall that in the formula a is a side length

[heady sphinx]

there are no side lengths

only 38.5

[swift grove]

yes

[nocturne minnow]

That is a side length

[swift grove]

but you have the legnth

and the angle

and yk the sine rules just need to rearrange it

[nocturne minnow]

a, b, and c, in sine law are side lengths while A, B, and C are the angles

You are given a side and you are looking for a side

You can fill in what you know based on that

[heady sphinx]

38.5/sin(65)?

[swift grove]

yes

[heady sphinx]

will that give me the answer

[nocturne minnow]

No

It will not give you the answer

[swift grove]

no but thats part of the equation

[heady sphinx]

72*sin(38.5)

[nocturne minnow]

It should look something like this

[heady sphinx]

over

[nocturne minnow]

No

[swift grove]

this is correct

for sinA over a

which is needed to find b

[nocturne minnow]

Don't confuse the person, B and b, depending what what variables you use they are not the same

B denotes the angle

Lower case b denotes the side

[swift grove]

my bad didnt realise i capitilised it

[heady sphinx]

so what exactly do i do now i put this in

and i got a number

[nocturne minnow]

Stop rushing and slow down

That's not an equation

You need an equation to solve

[heady sphinx]

i’ve been doing this for 4 hours straight and i still don’t get it

[nocturne minnow]

Do you know how to apply sine rule?

[heady sphinx]

no i don’t

that’s why im unable to do this

i don’t understand it because i do not know how to use it

[nocturne minnow]

https://www.youtube.com/watch?v=bDPRWJdVzfs&ab_channel=FurtherMaths

[nocturne minnow]

That's the first step, knowing how to use it

[nocturne minnow]

Here is a video on how to apply sine rule

[heady sphinx]

i literally don’t understand csn you please just tell me the equation i need to use

[nocturne minnow]

Literally this equation

You just need to plug in the proper info

[echo moss]

a is the length of side and A is angle opposite of it

[swift grove]

so look

youre trying to find b

and yk that sinA/a = sinB/b

and yk what sinA is, sinA = 65, you know what a is a = 38.5 and you know what sinB is sinB = 43

you have to rearrange

sinA/a = sinB/b to make b the subject

[obtuse pebble]

@heady sphinx Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[whole gull]

why is ECD isosceles

[nocturne minnow]

It's not?

[whole gull]

it is

[nocturne minnow]

Where does it say that?

[whole gull]

in the answers to the book

it says the answer is 8

[knotty crow]

now simiarity

but

it's drawn not to scale

[whole gull]

oh cool

thank you

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[obtuse pebble]

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[obtuse pebble]

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[whole plaza]

[knotty crow]

u know a^2 - b^2 formula?

[whole plaza]

err no

[warm shale]

Modus

[knotty crow]

apply this for your question

where a = 268 and b = 232

[whole plaza]

ok

36*500

correct?

[knotty crow]

yes

don't you think now it's easier to calculate it? even in your head

[whole plaza]

I guess so

thanks :D

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[obtuse pebble]

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[whole plaza]

.reopen

[obtuse pebble]

✅

[whole plaza]

what about b and c

[warm shale]

Result:

486

[tardy epoch]

,calc 548-469

[warm shale]

Result:

79

[whole plaza]

what? :o

[knotty crow]

it's not binomial expansion but on b) it's enough to take out common term which is 469

[whole plaza]

i dont get it :/

divide the whole expression by 469?

[knotty crow]

basically yes but 469 * (...)

[whole plaza]

1 * 548+ 469 *17

right?

[knotty crow]

469 * (1 * 548 + 469 - 17)

[whole plaza]

is that it or is there another step?

nvm hehe

[knotty crow]

this is what you should get after taking 469 out

now simplify expression within the bracket and you'll see smth nice

[whole plaza]

hold up why is that a -17

[knotty crow]

-469 * 17 divided by 469 is?

or in other words what multiplied by 469 gives -469 * 17

[whole plaza]

-17

oh

[whole plaza]

i got 469000

nice

[knotty crow]

,calc 469 * 548 + 469^2 - 469 * 17

[warm shale]

Result:

4.69e+5

[knotty crow]

it's ok

[whole plaza]

how do i start with this one?

theres more than one common term-

I actually have no idea lol

[solemn osprey]

Do this action: ab + ac = a(b+c)

[whole plaza]

what are a b and c though

[solemn osprey]

You have to do it multiple times in this fraction. The first thing you should do is the two terms that both have 65.1 in them

Cuz u can say 65x29 + 65x35 = 65(29+35)

[whole plaza]

ok

and take out 18 from the denominator?

[solemn osprey]

You got it!!!

You can do the same thing with the two terms that are top right

[whole plaza]

Ok :)

does the bottom term give 18(18-5)?

[solemn osprey]

Yup!

[whole plaza]

but when i do the whole thing i get 4/9 and the mark scheme says it's 15

[solemn osprey]

Wdym “Mark scheme”?

[nocturne minnow]

Answer key

[whole plaza]

When i get a homework i get the answers with it

[nocturne minnow]

That shows the step by step solution and how much each line is worth

[whole plaza]

but it doesnt have any of the working out

[solemn osprey]

Hm idk what went wrong

[whole plaza]

perhaps the mark scheme is wrong

thanks for your help

<3

.close

[obtuse pebble]

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[fading quest]

the answer is 15, can be easily checked with a calculator

must have done a mistake somewhere

[obtuse pebble]

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[royal karma]

does this proof looks correct?

[warm canopy]

Bunch of rubbish

[royal karma]

😦

[warm canopy]

Made zero effort to prove anything, lazy chatgpt

[royal karma]

yep, its bad imo

[tardy epoch]

chatgpt is a waste of time.

[royal karma]

ohhh new color

nice!

[warm canopy]

He requested to be godly golden

[royal karma]

x^4 is one to one

great

[timid silo]

@tardy epoch how to become golden

[tardy epoch]

shit on chatGPT enough

[upbeat plinth]

if u wanna dunk on chatgpt go to chill

[timid silo]

@upbeat plinth how to become golden

[upbeat plinth]

if u actually want math help then post the problem

[upbeat plinth]

we dont have a set criteria

[rigid lintel]

can i be gold

[timid silo]

how do you become helpful

[robust sleet]

Have 1000k+ messages

[timid silo]

i want to become helpful

[royal karma]

can i get another color?

lol

[upbeat plinth]

helpful is auto assigned

[timid silo]

to whom

[robust sleet]

helpful ppl

[timid silo]

wouldn't have guessed this in a million years

[upbeat plinth]

the bot has a metric for determining who gets helpful. we intentionally dont reveal how its calculated

[rigid lintel]

can i get unhelpfulled

[timid silo]

can you reveal it to be if I bribed you

[upbeat plinth]

no

[rigid lintel]

is that a thing

[robust sleet]

Yes

[timid silo]

what if i bribed you with cute pictures of my dog

[rigid lintel]

so i can give shitty advice and still be helpful

[timid silo]

i have one of him eating a toy piggy

[rigid lintel]

you may bribe me with that

what would you like to have

[robust sleet]

#chill

[upbeat plinth]

.close

[obtuse pebble]

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[royal karma]

😦

why

[obtuse pebble]

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[solemn swallow]

hey! i'm trying to find this real integral using complex integration. usually when doing this the orange integral goes to 0 as a->infty. however it's not the case here, so I'm not sure how to approach this problem. should I take another contour? which one?

[obtuse pebble]

@solemn swallow Has your question been resolved?

[obtuse pebble]

@solemn swallow Has your question been resolved?

[obtuse pebble]

@solemn swallow Has your question been resolved?

[pastel meadow]

hey sorry just seeing this - I'll take a look at what you've got

[solemn swallow]

thank you there's no rush ❤️ i've looked a bit into it, seems like it's a ML inequality problem. still don't know how to deal with it, but at least i can put a name on it

[solemn swallow]

btw this is how I got my upper bound

[pastel meadow]

oh sorry for seeing this only now - yeah your computations look right and just double-checking rn

can I ask where you came across this one? I'm not convinced it converges absolutely

[tardy epoch]

Yea the integral grows like 1/x which diverges

,w int -inf to inf (x^3-1)/(x^4+4)

[warm shale]

Results provided by WolframAlpha

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

Requested by riemann#2529

[pastel meadow]

Yeah that's what I thought

There needs to be some principal value thing here

But I haven't totally worked through the details

[tardy epoch]

Have you learned principal value yet

[solemn swallow]

nope I don't think so

the previous question was the integral of (z²-1)/(z⁴+4), I had no trouble with this one

the question says "give an expression of J in terms of residues (you don't need to calculate them explicitely). you can use previous results in your answer"

6.1 is the previous question
6.2 is the one I have trouble with

[torn badge]

I don’t understand what the “solution” means and how to get it

<@&286206848099549185>

[tardy epoch]

!help

[obtuse pebble]

Please read #❓how-to-get-help

[hot galleon]

emmm

change it into an equation?

maybe,D?

[obtuse pebble]

@solemn swallow Has your question been resolved?

[meager swift]

Rewrite it as $\int_{-a}^a \frac{x^3}{x^4+4} dx - \int_{-a}^{a} \frac{1}{x^4+4}dx$

[warm shale]

edwardborn

[meager swift]

Is there anything "odd" with these integrals?

[neon eagle]

i see what you did there

[meager swift]

ty . Nice name btw

[neon eagle]

they might have trouble with the second integral tho?

[meager swift]

nah they'll be fine. The second one is similar to the first one they solved

[neon eagle]

they did the first one in the picture?

[meager swift]

They solved one where the polynomial in the enumerator was of 2 or more orders lower than the one in the denominator

This is a similar case

[solemn swallow]

ohhhhhhhh

I feel stupid now lol

thank you so much!!!

[meager swift]

np

[solemn swallow]

.close

[obtuse pebble]

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[timid silo]

i need help on a math question

[mental solstice]

!help

[obtuse pebble]

Please read #❓how-to-get-help

[chilly nebula]

What do you want?

[timid silo]

idk how to do this one

[chilly nebula]

let me translate it

[timid silo]

ok

[dim bobcat]

Use trigonometry

[frank hamlet]

this isint your channel

#❓how-to-get-help

[timid silo]

ok

[tardy marsh]

please help me with this question

[chilly nebula]

I ll write now 21rad3

[timid silo]

@chilly nebula help 12

[tardy marsh]

.close

[obtuse pebble]

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[clear girder]

Why is it possible to multiply on both sides dx? I thought df/dx is just notation? I don’t find a good definition of d, I only find the definition of df/dx. And I thought that the d in the integral has nothing to do with the d in the differential quotient.

[chilly nebula]

Armin

Where r u from?

[timid silo]

ah yes the classic argument between mathematicians and physicists

[clear girder]

Why is this important?

[swift grove]

im gonna start it

[chilly nebula]

Im irainiam mybe i can tell you better if u r too

[bleak maple]

yeah mathematicians just want answers but physicists want your address

[clear girder]

Well, I’m from Germany my parents come from Persia, but I am very poor in Persian. So I would prefer to get the answer in English.

[timid silo]

it is just less rigourous

but you can avoid it

you can integrate both sides and then take the derivative

it would give you the same form of "multiplying both sides by dx"

[clear girder]

Lol

Wtf why are the profs are doing that? It just confuses me…

Thank you very much

.close

[obtuse pebble]

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[obtuse pebble]

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[exotic ledge]

what is the principal part of this function as x goes to 0

i factorised x^2 inside and i get it out of sqrt as abs(x)

then i have $x+abs(x)((1+x^2)^1/2)$

[warm shale]

lirdi

[exotic ledge]

from here i use the limit to take sqrt 1+x^2 as 1+ 1/2x^2

then idk

[obtuse pebble]

@exotic ledge Has your question been resolved?

[obtuse pebble]

@exotic ledge Has your question been resolved?

[elfin burrow]

what do you mean by principal part?

[obtuse pebble]

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[analog sky]

help

pls

[high lily]

show the work that led to your asnwer

[obtuse pebble]

@analog sky Has your question been resolved?

[analog sky]

4 - 1 = 3 5/8?

1 10/8

3 10/8

[obtuse pebble]

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[frosty valve]

I need explanation for 3 and 4

[opal marlin]

For 3:

We are given that X is bigger or equal than 0

[frosty valve]

so how do I solve it?

[opal marlin]

Range means what values can g(x) have

Or rather

[ruby path]

The range is all possible values that the function can output

[opal marlin]

What values can it not have?

[ruby path]

So find the minimum and maximum

It's intuitive for this one

[frosty valve]

Can you send me a simple working of it?

I still don't quite get it

Is the answer for num 3, two

[opal marlin]

What is the lowest value the function can have?

[frosty valve]

0?

[opal marlin]

How did u get to 0?

[frosty valve]

which one is the function?

[opal marlin]

g(x) = x^2 + 2

[frosty valve]

So the lowest is 2?

[opal marlin]

yes, this happens when x is zero

[frosty valve]

So what number do I put for x?

[opal marlin]

Well, now u know that the lowest value the function can have is 2. This happens when X is equal to zero, since: g(0) = 0^2 + 2 = 2

[frosty valve]

oh, so how do I get the final answer?

[opal marlin]

Range means find the interval of values the function can have, now that we know its minimum we need to find the maximumm

what is the biggest value it can have?

[frosty valve]

alot?

[opal marlin]

If u tell ur teacher the function has the biggest value alot

u will get a 0 on ur test

[frosty valve]

sorry

[swift grove]

okay so lets first

before doing that

know function rules

so

if x = 0 in this function

what does y =

@frosty valve

[ruby path]

You'd just put infinity with an open bracket

[frosty valve]

2?

[swift grove]

yes

good

[frosty valve]

how bout q4?

[swift grove]

we didnt finish that question

[frosty valve]

oh

my parents are calling, do u mind if we continue later?

can I ping u later?

[swift grove]

sure thats fine

[obtuse pebble]

@frosty valve Has your question been resolved?

[frosty valve]

@swift grove

[frosty valve]

<@&286206848099549185>

[chilly nebula]

Hi whats matter?

[frosty valve]

I'm done with all but I'm stuck at 6 and 7

[chilly nebula]

Opps sorry

Its incorrect

First find x

Then put it in f(x) to give y

After that draw the chart

Find Range function from chart

And this the last one

Did you get it ._. ?

[obtuse pebble]

@frosty valve Has your question been resolved?

[obtuse pebble]

@frosty valve Has your question been resolved?

[obtuse pebble]

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[timid silo]

Trying to compute what $\lim_{x \to 0} \frac{\sin(x) - x}{x^3}$ is without taylor nor l'hopital

[warm shale]

♡LexQa♡

[timid silo]

obvious chance of there being some application of the sinx/x limit, but that would leave us with a 1/x^2 that explodes to infinity, so not plausible

thoughts?

I think there can be some sort of plausible substitution

oh i may have figured it out

wait maybe not

hmm

[buoyant silo]

yeah idk of any other method for this

[timid silo]

this is definitely solve-able without either of them though

it requires a smart substitution

i believe

[mental solstice]

except, belief is not proof

[buoyant silo]

sinx=x 😎

[timid silo]

Well, i would say not knowing isn't proof of the opposite either

[timber fox]

??

WE MOVE

[lost tree]

but the limit 1/x^2 as x goes to 0 does not exist right

[timber fox]

yeah idk how to deal with it

[lost tree]

so u cant just write it as a product of teh limits

[timber fox]

[tawny fog]

$\lim_{x \to 0} {\frac{\sin(x)}{x³}}-\lim_{x\to 0}{\frac{x}{x³}}$

[warm shale]

Arnab

[tawny fog]

Dunno actually

🥲

[ruby path]

$\infty$

[warm shale]

NEONPerseus

[ruby path]

Why aren't you hospitaling it

[tawny fog]

Call an ambulance

[timber fox]

hospitalization is banned

[ruby path]

💀

[timber fox]

along with taylorization

[slim linden]

US moment

Random thought, multiply by the complex conjugate?

[timid silo]

oh thats interesting

[slim linden]

That’s like one of the only other tricks I learned for doing limits

[tawny fog]

Maybe this works maybe

[timber fox]

1/x^2 leads to dead end

I dont think this is possible

[timid silo]

it does not

[timber fox]

im betting 5 quid

[timid silo]

you cannot "split" limits when one of them is exploding to infinity

[tawny fog]

The limit is actually 0

[alpine raven]

why not factor by 1/x ?

[slim linden]

Lemme go work it out for a sec

[ruby path]

Usually is

[timber fox]

maybe

[timid silo]

What?

of 1/x^2?

[tawny fog]

The whole thing

[timid silo]

No

[lost tree]

might introducing a sin3x help

[timid silo]

the limit evaluates to -1/6

[lost tree]

coz it has a sin^3

but idk

[tawny fog]

Oh wait

I am a ducker

[timid silo]

you put infinity..

[tawny fog]

Yeah

[slim linden]

It’s to zero

[timber fox]

summoning stuff out of nowhere might be a move

[tawny fog]

Yk I put limit tends to 0 and I got all the steps @timid silo

[timid silo]

probably l'hopital or taylor

[tawny fog]

But they used L'hôpital

[timid silo]

no calculator shows other methods lmao

[tawny fog]

Yech

[timid silo]

I will definitely say; this has been solved without l'hopital or taylor, i am just looking to figure it out on my own

[tawny fog]

They applied L'hôpital 3 times

[timid silo]

How about tediously sketch the graph

[tawny fog]

💀

[ruby path]

Y'all L'Hospital didn't die a virgin for you to not use his math

[timid silo]

lmao just plug in numbers, smartest method i see

[tawny fog]

,w plot (sin (x)-x)/x³

[warm shale]

Results provided by WolframAlpha

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

Requested by 𝓐𝓡𝓝𝓐𝓑 𝓟𝓐𝓛#6985

[slim linden]

Good enough for an engineer

[timid silo]

Also like if integral bros can look at integrals for an hour trying to find a suitable substitution, then they should shine their abilities here

😵‍💫

[timber fox]

wait

why dont you just manually plug in numbers tending towards 0

also banned?

[tawny fog]

You find me a girl I solve this problem trade ☠️@timid silo

[timber fox]

works perfectly here

[timid silo]

yeah i mean sure, but i would expect that would result in using a calculator, which isn't really uh allowed

[slim linden]

That does, yeah

[mental solstice]

that be the same as tediously sketching graph lol

[timber fox]

[timid silo]

i think thats a losing battle either way for me

[tawny fog]

Yeah that's why this problem and my problems didn't have a solution if we don't use L'hôpitals and money trick

[timid silo]

wait

wait

wait

wait

guys

guys

i think i have an idea

[tawny fog]

Ok

[timid silo]

what if we multiply the numerator by cosx?

[lost tree]

i was thinking that too

[timid silo]

that is equal to 1 as x approaches 0, so it is not diffeerent

lets do it

[tawny fog]

Yeah do it

[lost tree]

but idk if it helps

[mental solstice]

i shall watch and be entertained

[lost tree]

and then i considered multiplying numerator and denominator by sin^2x

but idk i cudnt make any progress, maybe u can

[timid silo]

\begin{align*}
\lim_{x \to 0} \frac{\sin(x) -x}{x^3} & = \lim_{x \to 0} \frac{\sin(x)\cos(x) -x \cos(x)}{x^3} \
& \lim_{x \to 0} \frac{\frac{1}{2}\sin(2x) - x\cos(x)}{x^3}
\end{align*}

[warm shale]

♡LexQa♡

[timid silo]

we can definitely go from here

we need to split it up somehow

i think we need to make it so it is something something - 1-cosx/x

maybe

ah pain wait

[slim linden]

Yeah that’s kiiiinda what I’m thinking too

[timber fox]

bro im just trying every identity in the book atm

[timid silo]

lmaoo sure is a brain teaser huh

imagine if we didn't have l'hopitals or taylors

[slim linden]

What if you just cross out the part where you’re not allowed to use lhopital’s rule and gaslight your teacher

[timid silo]

oh this is not a homework

i am just masochistic

[slim linden]

Ah

[timid silo]

and stubborn too tbf

but anyways

[slim linden]

I’ve done that before with an integral I generated from wolfram alpha, eventually just gave up

[timid silo]

although this is an actual limit problem

it is from advanced JEE tho

[timber fox]

WHERE IS THE NO LHOP

[slim linden]

Iirc lhop is never strictly necessary anyway

[timid silo]

oh i got it from a site that says to not do it with either

i think it can be solved

keep going guys!

[timber fox]

https://tenor.com/view/breaking-bad-walter-white-devastated-crying-gif-4290107

[timid silo]

turn the cogs in your brains!

[timid silo]

No way 😂

[timber fox]

NOTHING IS HELPING

[slim linden]

LOL

[timid silo]

clarkie going insane

[mental solstice]

which site

[timid silo]

stock exchange 😵‍💫

[tardy epoch]

||y = 3x||

[slim linden]

didn’t know NASDAQ did limit problems

[timber fox]

oh its ok

riemann is here

no problem anymore

[timid silo]

[timid silo]

elaborate to us, divine yellow genius

[novel knoll]

You found the question on stackexchange? If you google this question 1st hit is stackexchange with answers.

[timid silo]

oh i know i just didnt look at the answers

[tardy epoch]

Let L be the limit. get a linear equation in L after the substitution

[timid silo]

i am trying to do it on my own

[timid silo]

oh hmm

[mental solstice]

oh, that's clever

[slim linden]

Wha

[timid silo]

"linear equation"?

[mental solstice]

||i'd do 2x instead of 3x though||

[tardy epoch]

lmao can't believe i didn't try that first

[slim linden]

Huh, I’ve never learned this method

[timid silo]

i actually see the t = 2x

i did try that at the beginning

but didnt go anywher

e

[timber fox]

neither

[mental solstice]

you have to kill the x term

[tardy epoch]

i think you know what linear equations are

[mental solstice]

the idea is to factorise into sin y / y forms

for 2x sub

[timber fox]

where do you read up on this

[timid silo]

oh right

https://math.stackexchange.com/questions/4605733/computing-lim-limits-x-to-0-frac-sin-x-xx3

here is the link

i didnt go through the solutions though

you can check them out tho

[timber fox]

what about the x=something, i've never been taught or read how to solve limits this way

wouldnt mind practicing a couple questions on it :_:

[timid silo]

It is what i meant by a substiution

it is a common thing

well

"common"

[mental solstice]

it's the same technique as finding integrals by letting I = weird integral

[timid silo]

just a common way to do advanced limits

[timber fox]

uhhh pain

[slim linden]

The stack exchange accepted answer isn’t right

[timber fox]

not gonna be taught it for a while I guess then :((

[timid silo]

also omfg

i just realised

did i really call it stock exchange earlier

kill me

why did i just make sense of it now

[slim linden]

YES THATS WHY I MADE THE JOKE

[timid silo]

BRUH i was so focused i was confused

anyways i guess this was fun

thanks everyone

[mental solstice]

i dunno about common, most places don't teach it in standard curriculum afaik

[timid silo]

yeah but common for like high-ended individuals i meant

[mental solstice]

fair

[timid silo]

snow is the first person to introduce me to it, i was so confused lmao

but i guess olympiad level it is kind of...required i suppose

[timber fox]

do u have a book that teaches it or a video series? the book that my course follows hasn't gone over it

[mental solstice]

eh, i never did it until today when riemann gave the sub

and then i was like

[slim linden]

Can confirm, never learned it

[mental solstice]

oh shit you can do this

[timid silo]

no there is not

[timber fox]

pain

[timid silo]

its like by the time u do it, you kind of exit the territory of

"hey i am too dumb teach me it"

because it is usually for very advanced limits

or stuff without L'H

[mental solstice]

honestly i'd love to see a limit that can't be done by the usual methods but instead is amenable to this method

[timid silo]

oh that would be sick

but i think taylor and l'hopital are in fact very very strong tools

would be hard to find something that doesnt work with them

[mental solstice]

exactly the point

[timid silo]

yep

[mental solstice]

you need some sort of pathological function probably

[slim linden]

I got it by searching “variable substitution in limits”

Here’s the top link from that search https://math.fel.cvut.cz/en/mt/txtb/5/txe4ba5h.htm

[timber fox]

thank you

[timid silo]

dont expect anything elementary probably lmao

[timid silo]

yeah

oh i did find another interesting one guys

[mental solstice]

i consider a day good if i learn one new tool by the end of it, and today is now a good day

[slim linden]

Evil

[timid silo]

[
\lim_{x \to 0} \frac{x-\sin(x)}{x\sin(x)}
]

[warm shale]

♡LexQa♡

[timid silo]

similar kind of but still without l'h or taylor

i think this one is easier looking

[timid silo]

exaaaaaactly

build up that intuition 😎 👍🏻

[timid silo]

first thoughts: factoring out x from the numerator

that would give us a sinx/x expression but

still dead

[mental solstice]

i should probably grab paper, i've been doing things mentally lol

[timid silo]

same

but i dont like writing down problems

[slim linden]

I’m gonna head out, but it was fun. Good luck!

[timid silo]

alright have a nice one! was fun with you :)

we continue struggling

@mental solstice

i think maybe

WAIT I JUST REALISED

LMAO

LMAO

WE CAN USE THE LIMIT FROM BEFORE

$\lim_{x \to 0} \frac{x- \sin(x)}{x\sin(x)} = \lim_{x \to 0} x \cdot \lim_{x \to 0} \frac{1-\frac{\sin(x)}{x}}{\sin(x)}$

[warm shale]

♡LexQa♡

[timid silo]

so like

$\lim_{x \to 0} x \cdot \lim_{x \to 0} \frac{x-\sin(x)}{x\sin(x)}$

[warm shale]

♡LexQa♡

[timid silo]

wait

[mental solstice]

good idea, wrong execution

$\frac{x - \sin x}{x\sin x} = \frac{x - \sin x}{x^3} \cdot x \cdot \frac{x}{\sin x}$

[warm shale]

Camilleone

[mental solstice]

tadaa

[timid silo]

ah there it is!

so ultimately 0?

wait

lets wolf it up

[mental solstice]

seems like, yeah

[timid silo]

,w limit of (x-sinx)/(xsinx) as it approaches 0

woo

oh wait

[mental solstice]

your sin x in the wrong place lmao

[warm shale]

Results provided by WolframAlpha

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

Requested by ♡LexQa♡#5050

[timid silo]

yeah

doesnt matter

wooooo

but ty i will end this insanity now

.close

[obtuse pebble]

Channel closed

Closed by @native knoll

Use .reopen if this was a mistake.

[obtuse pebble]

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[obtuse pebble]

Available help channel!

Send your question here to claim the channel.

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[silver plover]

i proved i

need help for ii

[spiral knot]

just multiply x and y

x = k* n + a
y = l * n + b
x * y = (k * n + a)(l * n + b)

[dim bobcat]

Do the same thing which you did in 1, just multiply this time

[silver plover]

wait thats it

bruh

ok

[silver plover]

what now

oh

wait

[spiral knot]

use mod n

anything multiplied by n will be 0

[royal basin]

xy = (some shit that's a multiple of n) + ab

[silver plover]

here

SUIIIIIIIIIIIIIIIIIIIIII

[silver plover]

ann do u have a masters in maths

just curious

u seem very cracked

[royal basin]

i am a semester away from a master's.

[silver plover]

damn

u gonna do phd?

[royal basin]

if circumstances allow it.

[covert forum]

PHD

LIKE THE THING FROM HULK

[silver plover]

dang cracked

[abstract beacon]

hi

[covert forum]

ew

[silver plover]

do u go oxford?

[abstract beacon]

na

[silver plover]

cuz ur very cracked as well

[covert forum]

whats an oxford

[abstract beacon]

its funny that u say that bc i dont even do a maths degree at all

Im just an interested woman

[silver plover]

imagine being cracked at maths and not even doing a maths degree

[spiral knot]

wasted talent 🥲

[silver plover]

@abstract beacon

this tbh

[abstract beacon]

that is correct

[silver plover]

hbu mehdi

u doing masters?

[abstract beacon]

when it came time to select though i was more interested in physics

so i picked a science degree

sad that i couldnt keep up the interest until the end of the degree...

[spiral knot]

nah , last year I got my bachelor degree
in programming

[silver plover]

dang

[abstract beacon]

so now im doing philsoophy of science

[silver plover]

wow

[burnt sphinx]

cambridge 🤓🤓🤓

[spiral knot]

nice

[red loom]

Egg

[abstract beacon]

it might have been a waste of talent but whatever,

got more important things to think about in my life

[silver plover]

@burnt sphinx is the most cracked

he did his a level further maths in year 11

[silver plover]

hey sunshine

[marble burrow]

average mathematics help channel

[silver plover]

perfect for u then

[marble burrow]

?

[silver plover]

ur average

[marble burrow]

ah nice

i think you missed the joke though

[obtuse pebble]

@silver plover Has your question been resolved?

Channel closed

Closed by @silver plover

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Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[acoustic pawn]

The figure below shows the graphs of the parameter curves x = sin t, y = 1/2 sin(2t) where 0 ≤ t ≤ 2π and the polar curve r^2 = cos(2θ), with 0 ≤ θ ≤ 2π. Which curve corresponds to which curve?

[acoustic pawn]

I tried verifying using the bounded areas of both curves, clearly the red curve has a larger area

although im not sure what bounds i should pick for my integrals

for the polarcurve r i chose the bounds 0 to pi/2 and then multiplied by 4

For the parameter curve i chose the bounds 0 to pi and then multiplied by 4

Are the bounds correct? I get the result that the polar curve has area 1 unit and the parameter curve has area 4 units

I assumed area symmetry bcs well look at the graph, and for the polarcurve its pretty clear that theta 0-->pi/2 is the first quadrant while for the parametric curve i wasnt as sure. I dont think the second bounds r correct since i thought well the first zero is pi but the next zero wouldve been 2pi which wouldve only given half the curve (first and fourth quadrant)

[visual plover]

seems so. but are they asking for areas or just knowledge of which curve is which?

[acoustic pawn]

knowledge of which curve is which

another approach couldve been calculating the max for both curves but i got no idea how to do that

[visual plover]

you mean a stationary point?

maxima?

you'd want to derive x and y then

get dy/dx and equate this to 0

[acoustic pawn]

ehh i like the area approach better

i feel like my parameter bounds r off tho

[visual plover]

it's asking to find which curve applies to which though

[acoustic pawn]

i know and finding the area is a way to figure that out

[visual plover]

surely you'd be better off finding the two distinct maxima?

[acoustic pawn]

Thats more work then the simple integrals and also not why i opened this channel

So there are 4 x intercepts which means y = 0 for 4 t's. T goes from 0 to 2pi and sin(2t) thereofr has t = 0, t = pi, t = 2pi as zeroes

im missing one

ahha t = 3pi/2

okey my bounds are correct i think, anyone that can confirm?

[visual plover]

yeah they are as if you plug it in it'll equal to 0

[acoustic pawn]

damn, its pretty weird tho from the picture i wouldnt think the red curve encompasses 4times the area of the green one

especially since its to scale

but fk it i solved it hha idc

[visual plover]

what language is that questin in

[acoustic pawn]

dutch

i translated it above tho

[acoustic pawn]

this is an exact translation

[visual plover]

I know you're moroccan for sure my boy

[acoustic pawn]

no u dont

[visual plover]

got lots of maroc family dow there

[acoustic pawn]

cool, there's plenty of arabic speaking diaspora here tho

but you were right i am moroccan lol

[visual plover]

true ngl

love that country

[acoustic pawn]

hey me too

Anyway im closing the channel now

[visual plover]

nice one

[acoustic pawn]

.close

[obtuse pebble]

Channel closed

Closed by @acoustic pawn

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[timid silo]

i want to know how.

[haughty coyote]

Use algebra to be able to use (e^u - 1)/u -> 1 as u -> 0

Basically rewrite the exponential in a useful way

[timid silo]

i didnt understand

[haughty coyote]

you know that result right ?

[timid silo]

i tryed many things and failled

u isn't x?

[haughty coyote]

doesn't have to be

in that case it obviously won't be since x -> +inf

[timid silo]

hm

[haughty coyote]

e^((x+1)/x ln x) = x e^(ln x / x)

[spiral knot]

what is under ln(x) ?

[haughty coyote]

looks irrelevant to the question

especially since it looked like the expected difficulty level as is

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

well just true and false

[haughty coyote]

So I take it that wasn't enough of a hint ?

[timid silo]

yeah

and u need to show how

that to get point

[haughty coyote]

subtract x from that and you get x (e^(ln x / x) - 1)
That's when the limit I mentioned at the start comes into play

[timid silo]

no wait

e^((x+1/x)*lnx)-x

u didnt saw it complety

[haughty coyote]

it goes to 1 though

[timid silo]

i want to know how he got 2

i dont give a fuck about true and false

[haughty coyote]

He got 2 by making a mistake

[spiral knot]

,w limit x to infinity (e^((x+1)ln(x)/x) - x)/ln(x)

[timid silo]

ln

[warm shale]

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Requested by Mehdi_Moulati#1210

[timid silo]

not log

[spiral knot]

log is ln

for them (wolfram)

[timid silo]

but with numbers and tactics

i mean

ah nvm

[obtuse pebble]

@timid silo Has your question been resolved?

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[pale ice]

help pls ion understand

[warm mantle]

https://courses.lumenlearning.com/waymakercollegealgebra/chapter/summary-linear-functions/#:~:text=The initial value%2C or y,and has a positive slope.

[obtuse pebble]

@pale ice Has your question been resolved?

[obtuse pebble]

Channel closed

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[obtuse pebble]

Available help channel!

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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[timid silo]

is the answer (4, 11) or (-4,5)

[obtuse pebble]

Please don't occupy multiple help channels.

[rare wyvern]

,rotate

[nocturne minnow]

Don't open multiple channels

[timid silo]

i just closed prev one

[nocturne minnow]

You already have a channel open

[timid silo]

mb i just closed it

[rare wyvern]

its closed, the bot hasnt moved it yet dld

[nocturne minnow]

They opened a new channel before closing the old one

[timid silo]

been trying to figure this question out for like at least an hour

[rare wyvern]

1:2 ratio means that 1 of PQ = 2 of the new two segments

so if R was the midpoint, PQ=PR+RQ

[timid silo]

(-4, 11) or (-4,5)*

[nocturne minnow]

Do you have work that supports those answers?

[timid silo]

yes for the answer(4,11)

[rare wyvern]

to find the midpoint use the formula

(x,y)
where x = ( x_1 + x_2 ) / 2
and y = ( y_1 + y_2 ) / 2

[timid silo]

ye but not midpoint

in ratio of 1:2

[nocturne minnow]

.

[nocturne minnow]

,rotate

[nocturne minnow]

Where is 1/3 coming from?

[rare wyvern]

from the 1:2 ratio

[nocturne minnow]

Yeah, I see now

[nocturne minnow]

This looks right

[timid silo]

ok

wait nvm (-4,5) is correct

ty

.close

[obtuse pebble]

Channel closed

Closed by @dusk anchor

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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• Be polite and have a nice day!

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[worn yacht]

Hi, I am doing this calculus course https://www.coursera.org/learn/introduction-to-calculus/quiz/kVlbQ/the-theorem-of-pythagoras-and-properties-of-the-square-root-of-2/attempt but I am being very slow while doing it, I am a couple of days behind, I would really appreciate it if someone in this server who has a decent understanding of calculus would like to do this course with me, we could set a time every day to do some of the course, If you would like to help me do this please dm me the times that you are available, thanks!

[patent jetty]

,rotate

[obtuse pebble]

@worn yacht Has your question been resolved?

[obtuse pebble]

@worn yacht Has your question been resolved?

[worn yacht]

<@&286206848099549185>

[serene fossil]

You’re asking for free tutoring, I’m pretty sure no one is willing to do that sorry

You could bring individual questions from the course here and someone might help

But the odds of someone “doing the course with you” seems pretty slim imo

[worn yacht]

lol

I understand now

thanks