help-13

[stark pulsar]

:))

well

what if its like

5 (x-2y) = 3

[edgy spade]

you cant solve for x - y then

[stark pulsar]

ooo

[worldly chasm]

Then you might not be able to find x-y

[edgy spade]

it will have infinite possible values

[stark pulsar]

woah

5 (x-y) = 3x?

[worldly chasm]

Same deal

[edgy spade]

same again

[stark pulsar]

mm

5x (x-y) = 3

[edgy spade]

its gonna be the same deal

[stark pulsar]

oof

5 (-x+y) = 3

[edgy spade]

instead of us telling you

i'd much rather you try to figure them out on your own now that we've helped you with one example

[edgy spade]

you tell me

[stark pulsar]

hmmm

[edgy spade]

can we find (x - y)?

[stark pulsar]

−
5
3
​

​

-3/5

[edgy spade]

good

very good

can you tell me how you got it

[stark pulsar]

:>

mhm

simplfiy

then devide both by 5

find x-y

then get -3/5

[edgy spade]

nice

[stark pulsar]

thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you thank you

what did u want to say rudy

[edgy spade]

[stark pulsar]

awwwww

[edgy spade]

nothing just finding an emoji

[stark pulsar]

kk

[edgy spade]

!done

[cedar kiln]

If you are done with this channel, please mark your problem as solved by typing .close

[stark pulsar]

.close

[cedar kiln]

Channel closed

Closed by @stark pulsar

Use .reopen if this was a mistake.

[edgy spade]

if you have any other doubts feel free to open another help channel

[stark pulsar]

np

if u need someone to play chess with dm me

.close

.close

[edgy spade]

its closed lol

[stark pulsar]

oop

[edgy spade]

[cedar kiln]

Available help channel!

Send your question here to claim the channel.

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[stark pulsar]

ello

[stark pulsar]

i need intensive care unit for this

idk man

i cant even look at it

its scary

[lone dune]

ok so

we need to get it to a less scary format

how would we do that

[stark pulsar]

can we delete the scary 4z^2

or nope

[lone dune]

well

you can cancel stuff out...

[dusk goblet]

i mean you can but you should just factor the numerator and denominator to see what cancels

[lone dune]

[stark pulsar]

wdym factor

[lone dune]

do you know what factoring is

[dusk goblet]

is english not your first language?

[stark pulsar]

nope

whats factoring

[lone dune]

taking out a term

[stark pulsar]

im mind is blank

hmm

[dusk goblet]

reversing the distributive property in this case

for example

[stark pulsar]

mhm

[lone dune]

$a^2 + a = a(a+1)$

[wraith dagger]

satvik

[lone dune]

that's factoring

where you "take out" the a

[dusk goblet]

69x^2 + 138x = 69x(x + 2)

[stark pulsar]

how to factor ^3

my brain doesnt have enough brain cells to understand that

[lone dune]

factor out the greatest common factor

[dusk goblet]

sal khan is your friend

[lone dune]

on the numerator and denominator

[stark pulsar]

i forgot how to

[lone dune]

are you doing this exercise on khan academy

it seems like it

[stark pulsar]

maybe

[lone dune]

go back and watch whatever video you just did

[stark pulsar]

just help

[lone dune]

im guessing it probably explains this

[stark pulsar]

i dont understand

[dusk goblet]

"just tell me the answer"

[lone dune]

what dont u understand

how to factor?

[stark pulsar]

😭

[dusk goblet]

did you watch the videos before the exercises?

[stark pulsar]

i just want help

[lone dune]

2x speed demon

[stark pulsar]

how to factor z^3

[dusk goblet]

ya don’t

[stark pulsar]

how to make z^3

[dusk goblet]

both terms have a z^2

🤷🏼‍♂️

[stark pulsar]

z(???)

[lone dune]

factor out the greatest common factor

on the top

[stark pulsar]

what does it mena

[lone dune]

whats the largest power of z

[stark pulsar]

z^3

[dusk goblet]

replay the videos

[lone dune]

wait

i phrased that wrong

what's the largest power of z present in both terms

[stark pulsar]

z3

[lone dune]

is there a z^3 in the second term

[stark pulsar]

nope

[lone dune]

so then

whats the largest power of z present between the two terms

in $-12z^3 + 4z^2$

[stark pulsar]

isnt dat da same question?

[wraith dagger]

satvik

[lone dune]

like

what is the largest power of z that you can find in both terms

[stark pulsar]

isnt it z^3

[lone dune]

sorry i shouldve phrased that better

theres no z^3 in the second term

[stark pulsar]

ohhh

sowwy

z^2

[lone dune]

yes

so you can factor out z^2

[stark pulsar]

how does z^3 become z^2

[lone dune]

it doesnt

i have a question

[stark pulsar]

then

[lone dune]

what did the video you just watched go over

[stark pulsar]

chatgpt.com

[dusk goblet]

he didn’t watch it

[dusk goblet]

if that’s the case then why are you here

[lone dune]

[stark pulsar]

couldnt understand from hikm

[lone dune]

if sal khan cant teach you no one can

what part of it did u not get

[stark pulsar]

the whole thing

[lone dune]

what whole thing

[dusk goblet]

lol

do you know how to factor numbers?

[lone dune]

😭🙏

[stark pulsar]

nop

[dusk goblet]

so if i said find the prime factorization of say 42 you couldn’t do that

[stark pulsar]

uhh

i cant

how do u

[dusk goblet]

well

you should revisit rules of arithmetic before starting algebra

[stark pulsar]

i understand a lot in algebra but not dis

im more of a geometry and trigonometry person than algebra tbh

[lone dune]

go try the organic chemistry tutor

[dusk goblet]

do you know the distributive property?

[lone dune]

maybe he will help

[stark pulsar]

nope

[dusk goblet]

🤔

[stark pulsar]

well just teach me here

[dusk goblet]

a(b + c) = ab + ac

doesn’t ring a bell?

[stark pulsar]

mhm

axb and axc

ab + ac

[dusk goblet]

what

[stark pulsar]

a times b

ab

a times c

ac

ab+ac

[dusk goblet]

ok well factoring is just undoing the distributive property in this case

[stark pulsar]

like

ab + ac to a (b + c)?

[dusk goblet]

we have a sum : -12z^3 + 4z^2

[stark pulsar]

$-12z^3 + 4z^2$

[wraith dagger]

someone

[stark pulsar]

mhm

[dusk goblet]

do you agree that -4z^2(3z - 1) = -12z^3 + 4z^2

[stark pulsar]

$-4z^2(3z - 1) = -12z^3 + 4z^2$

[wraith dagger]

someone

[stark pulsar]

mhm

i agree

its too hard :((

[dusk goblet]

ok so you should see the idea is that we take the "greatest common factor" from the two terms in the sum and rewrite the sum using the distributive property

[stark pulsar]

whats the gcf

the largest number that both numbers have?

[dusk goblet]

as a factor yes

[stark pulsar]

aha

[slender ginkgo]

greatest common factor

[stark pulsar]

like 12 and 4 is 4?

[dusk goblet]

yep

[stark pulsar]

ohhh

oki

[dusk goblet]

z^2 and z is ?

[stark pulsar]

now back to da question

z

:>

$\qquad \dfrac{-12z^3 + 4z^2}{4z^2+16z}$

[wraith dagger]

someone

[dusk goblet]

now try doing what i did to the numerator but to the denominator

[stark pulsar]

hmm

abcd

then hmm

[dusk goblet]

first look at the constants

so 4 and 16

[stark pulsar]

z2 and z

so we do z

[dusk goblet]

what’s the gcf

[stark pulsar]

z(???)

[dusk goblet]

yes and what about the constants

[stark pulsar]

z(z+1)

then hmm

4z(z+4)

?

[dusk goblet]

uhh sure not sure how you get there but yes

[stark pulsar]

yayyy

first i

make the

image

then put the numbers

[dusk goblet]

like where did this come from

[stark pulsar]

we have z^2

so that means we have z inside and outside

z(z)

and we have another number with 1 z

which can use the outside

so

z(z+1)

z^2+1z

then i just put the numbers

4z

bc 4z times z is 4z^2

and 4 times 4 is 16

4z(z+4)

$4z(z+4)$

[wraith dagger]

someone

[dusk goblet]

youd get marked wrong if you wrote this on an assignment though

[stark pulsar]

yes ik

this is mental math only

[dusk goblet]

just look at the gcf of the constant multiples and of the variables them take them common and divide each of the terms in the sum by the gcf

[stark pulsar]

mhm

ok the numerator is hmm

12 and 4

is 3

z^3 hmmm

z^2(z+1)?

$4z^2(-4z+3)$

[wraith dagger]

someone

[stark pulsar]

wait mb

$4z^2(-3z+1)$

[wraith dagger]

someone

[stark pulsar]

$-12z^3+4z^2 = 4z^2(-3z+1)$

[wraith dagger]

someone

[stark pulsar]

am i correct?

<@&286206848099549185>

[tacit sand]

you can maybe try to plug in z as a number

I use that to check my answers

like let's say z=2

plug it in

[stark pulsar]

$\qquad \dfrac{-12z^3 + 4z^2}{4z^2+16z}$

[wraith dagger]

someone

[stark pulsar]

Which expression is equivalent to the given expression, for all
[z<-10]?

[wraith dagger]

someone

[stark pulsar]

so i can use maybe smt more to this?

[tacit sand]

wdym by that?

[stark pulsar]

$(Choice A)

[\dfrac{-11z}{4}]
A

[\dfrac{-11z}{4}]
(Choice B)

[-3z + \dfrac{z}{4} ]
B

[-3z + \dfrac{z}{4} ]
(Choice C)

[\dfrac{-3z^2 + z}{z + 4}]
C

[\dfrac{-3z^2 + z}{z + 4}]
(Choice D)

[\dfrac{3z^2 - z}{z + 4}]
D

[\dfrac{3z^2 - z}{z + 4}]$

[tacit sand]

ah

[wraith dagger]

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

[slender ginkgo]

oh dear

[stark pulsar]

i have a picture

much better

[tacit sand]

plug z as -11 ig

[slender ginkgo]

hmm

[tacit sand]

wait that's the right way right

[slender ginkgo]

have u simplified

[stark pulsar]

like

open them?

numerator:

$4z^2(-3z+1)$

[wraith dagger]

someone

[stark pulsar]

denomenator:

[slender ginkgo]

[ \frac{4z(-3z^2 + z)}{4z(z + 4)}]

[stark pulsar]

$4z(z+4)$

[wraith dagger]

someone

[slender ginkgo]

so like

[stark pulsar]

denomenator is wrong for u

[wraith dagger]

k

[stark pulsar]

perfect

[slender ginkgo]

[ \frac{-3z^2 + z}{z+4}]

[wraith dagger]

k

[stark pulsar]

whats this

[slender ginkgo]

cancel the 4z

[stark pulsar]

oh we delete same

mhm

can we delete z or nope

[slender ginkgo]

its choice c?

[stark pulsar]

hmm

d

wait

c

soryr

sorry

they gave up on me

tysm k

.close .rate k infinite

[cedar kiln]

Channel closed

Closed by @stark pulsar

Use .reopen if this was a mistake.

[cedar kiln]

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!

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[hushed sonnet]

Does that sum of the inverse of reciprocals diverge or converge (to what...)?
$$\sum_{n=1}^{\infty} \frac{1}{t_n}$$
With $$t_n= \frac {n(n+1)}{2}

[slate lintel]

do a little algebra, write out the first few terms

also did you mean to start at n=1?

[hushed sonnet]

I have absolutely no idea how to deal with this, not to used to this type of problem. Not really looking for the answer, just wondering how you'd start with this problem/how to establish a plan

[hushed sonnet]

yeah my bad

[wraith dagger]

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

[lyric narwhal]

The sum can be written as $$2\sum_{n=1}^\infty \frac1{n(n+1)}$$

[wraith dagger]

kheer257

[slender ginkgo]

🔭 ?

[lyric narwhal]

I would think about partial fraction decomposition here

[crimson delta]

(if you just wanted to know about divergence/convergence, comparison would also work)

[slate lintel]

personally i'd start by investigating the partial sum up to n = 2 or n = 3

and see if you can spota pattern

[hushed sonnet]

Well I guess it's pretty safe to assume it converges, though I guess it needs proof yeah

[lyric narwhal]

Why is it safe to assume?

[slender ginkgo]

it is a monotonic and bounded

and u can do direct comparison

if u want to prove convergence

[slate lintel]

showing it's bounded seems nontrivial

[slender ginkgo]

its always greater 0

[nocturne ravine]

You would need to calculate zeta(2)

[nocturne ravine]

You also need to calculate upper bound

[nocturne ravine]

Oh no that’s bullshit

Partial decomposition really is enough

[hushed sonnet]

Hmm maybe not tha trivial, but just by looking at it, it seemed pretty insticyive that it does, seeing how it grows

[slate lintel]

yeah and comparing with 1/n^2 can easily get you convergence but finding the value is harder

try writing out the first few terms

[hushed sonnet]

Hmm I'd need to find some good transformation then

[slender ginkgo]

.

[hushed sonnet]

Hm

[slender ginkgo]

ive just realized ive confused sequence and series

brainrot has taken its effect

[nocturne ravine]

Which means decomposing your fraction as the sum of two fractions which denominators are in this case n and n+1

[hushed sonnet]

Is it 1/n - 1/n+1

[blazing zephyr]

yeah

[hushed sonnet]

Yeah that's interesting

Ah and then telescoping seems easier

[nocturne ravine]

Do you know the value of the sum of 1/n?

[slender ginkgo]

[nocturne ravine]

Bullshit again

[slender ginkgo]

-1/12

[nocturne ravine]

Nope that’s sum of n

[hushed sonnet]

Doesn't it diverge? Harmonic series right

[nocturne ravine]

But here it’s far easier to finish the proof

[hushed sonnet]

Though not sure how to do that with infinite terms? Like usually we're left with the first and last term and it's easy 😅

[nocturne ravine]

Partial sums and then taking the limit usually works

[slender ginkgo]

oh shi mb mb

another instance of brainrot

[hushed sonnet]

What do we call partial sums?

[nocturne ravine]

You only sum to an integer k

[slender ginkgo]

,w zeta(1)

[wraith dagger]

Results provided by WolframAlpha

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

Requested by nuclearinteraction

[nocturne ravine]

And make k go to infinity

[nocturne ravine]

There’s an equivalent of ln(n)

Plus euler’s constant

[slender ginkgo]

[ \lim_{n \to \infty} S_n = S_\infty 🤓 ]

[wraith dagger]

k

[nocturne ravine]

Works @hushed sonnet ?

[slender ginkgo]

oh i didnt know that. havent touched real anal yet.. seems interesting

[nocturne ravine]

Real anal

[hushed sonnet]

Partial sums is probably something I'm unacquainted to. What I understood from your message is that instead of n we have k that goes to infinity 😆

[slender ginkgo]

S_n represents the n-th partial sum

[nocturne ravine]

No $\sum_{n=1}^k \frac{1}{n} -\frac{1}{n+1}$

Is the k-th partial sum

[slender ginkgo]

?

[hushed sonnet]

Ah and then we take the limit of this *2 right?

1/1 no?

[wraith dagger]

Matthias

[nocturne ravine]

Yes exactly

[hushed sonnet]

Ah I see, so it's basically 2* lim of 1/1 - 1/k+1

Which means the whole thing is 2

[slender ginkgo]

👍

[nocturne ravine]

Yup perfect

[hushed sonnet]

Ah great! I'll need to look more into partial sums and fraction decomposition

Thanks! ❤️

[nocturne ravine]

Partial sums is really just a way to calculate limits of series most of the time

[nocturne ravine]

Ur welcome

[hushed sonnet]

One last question actually.. sorry 🙂
Can we simply conclude that the finite sum (till n) is <2 ?

[nocturne ravine]

Yes since the sequence of partial sums is strictly increasing

[hushed sonnet]

Right, makes sense

Thanks again!

.close

[cedar kiln]

Channel closed

Closed by @hushed sonnet

Use .reopen if this was a mistake.

[cedar kiln]

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• 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!

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[shadow shore]

does the binary operation need be closed on the finite set for it to be an algebraic structure? are all agebraic strucutres closed? then why do terms like "groupoids" exist, isnt that for closure ?

[shadow shore]

every algebraic strucutre is groupoid because every algebraic strucutre must be closed

then why do terms like groupoids exist

[brave copper]

yes this is primarily a historical thing

[shadow shore]

isnt that a bit redundant

[brave copper]

some people who really really love to be general take algebraic structure to be any set with any operation.
most people however say closure is needed.
These people call "algebraic structure" what the first people call "groupoid" or "magma"

[shadow shore]

every algebraic structure must be closed on some operation right

[crimson delta]

a binary operation on a set S by definition is a map SxS -> S

proper groupoids do not have binary operations

they just have partial functions

[brave copper]

dont worry this wont be the last time you learn that the same thing means something different to different people

[shadow shore]

which terminology is preferred by authors

i mean what shall i write in the notes

[shadow shore]

idk what this means

[brave copper]

writing a few extra words doesnt kill anybody, i think in most books i know of the author says somethging like "an algebraic structure is a set equipped with at least one binary operation that is closed (i.e a+b in S for any a,b in S)"

[shadow shore]

ok

thnaks

.close

[cedar kiln]

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[cedar kiln]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[wanton marsh]

Can anyone help me with this question by chance?

[wanton marsh]

You gotta find the value of x

[tropic oxide]

as in, EF?

also, obligatory:

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[wanton marsh]

I don't know where to begin and I need to know like how to do it and the answer (the answer in the book is 2.87 however ai said this not possible 💀)

[tropic oxide]

!nogpt

[cedar kiln]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[tropic oxide]

ai can go fuck itself

anyway you should begin by finding the length of DB

then you will be able to cook something in triangle EDB, which is a right triangle (as explicitly marked)

think about the role of line DF in triangle EDB, too

[idle crystal]

if you wanna know if any geometry question is possible at all you should ask yourself: "is what the question asking me fixed?"

[wanton marsh]

Wait wym by that

[tropic oxide]

sometimes questions ask you to find something thats not determinable from the given data

e.g. you can imagine like... your diagram kinda warping in some way while all of the data in it stay fixed (ie known angles and lengths are rigid)

not the case in this problem though.

i can assure you this problem is doable and has a definitive answer

[wanton marsh]

O okay

[tropic oxide]

have you found DB?

[wanton marsh]

4√2?

[tropic oxide]

indeed

you can also find EB, as you will need it

[wanton marsh]

EB=2√6 right

[gritty galleon]

isnt this just repeated sin and cosine law

[tropic oxide]

p sure that's wrong, how did you get that?

[tropic oxide]

you can do it without

[wanton marsh]

Pythagoras?

[gritty galleon]

[tropic oxide]

yeah show work for Pythagoras

[past wind]

angle bisector theorem i think also works

[tropic oxide]

gonna answer in #「helpers-lounge」

[wanton marsh]

√(4^2+4√2^2)

[gritty galleon]

DANG

[tropic oxide]

missing brackets.

(4sqrt(2))^2

[wanton marsh]

Oh ok

So 4√3

[tropic oxide]

indeed

[wanton marsh]

What do I do next

[tropic oxide]

think about the role of line DF in triangle EDB

maybe look for some angles that you could pin down

[wanton marsh]

How? 😭

[tropic oxide]

BD is a diagonal in a square

what is angle BDC?

[wanton marsh]

45?

[tropic oxide]

indeed

can you find and mark some other angles at vertex D

[wanton marsh]

Can you give me some idea how?

I can't find the angles for triangle AFB and the angles in the triangle fbd

This is the solution my friend gave me

[tropic oxide]

i had a different idea in mind

[wanton marsh]

How?

[tropic oxide]

do you know this theorem?

[wanton marsh]

Yeah

[tropic oxide]

yeah DF is a bisector bc angles FDB and FDE are both 45°

thus it splits side EB in a ratio equal to ED:DB

the length of EB is known and the ratio ED:DB is also known to be 1:sqrt(2)

[wanton marsh]

Wait lemme try

[wanton marsh]

How did you get sqrt2?

Wait nvm

So it's x:4√3-x=4:4√2?

[tropic oxide]

eh i guess but that's somewhat overcomplicating it

[wanton marsh]

Wait I got the ans

Tyyyy

Tysm

.close

[cedar kiln]

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[cedar kiln]

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[edgy spire]

Does anyone know if "use your calculator to estimate" means "use it to find the exact value of your answer" or "use it to evaluate the integral itself"?

[edgy spire]

Because I don't think you can evaluate $\int_1^2 2 \pi \left(t - \frac{1}{t^2} \right) \sqrt{1+ 9t^4 + 6t^2 + 1 + \frac{4}{t^6} + \frac{4}{t^3}} \dd t$

[wraith dagger]

Sarin

[slender ginkgo]

erm

what the sigma

,w simplify sqrt{(1 + 3t^2)^2 + (1 + 2/t^3)^2}

[wraith dagger]

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[slender ginkgo]

,w integrate from 1 to 2 2pi(t - 1/t^2)sqrt{(1 + 3t^2)^2 + (1 + 2/t^3)^2} dt

[edgy spire]

What

[slender ginkgo]

might be me

but i think they u want to plug the thing into a calculator

a normal scientific calculator has an integral symbol

[wraith dagger]

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[edgy spire]

Yeah that's the correct answer

Thanks lol

.close

[cedar kiln]

Channel closed

Closed by @edgy spire

Use .reopen if this was a mistake.

[slender ginkgo]

https://tenor.com/view/realistic-cat-smurf-meme-funny-gif-5895795080802392404

[cedar kiln]

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[crimson sedge]

Hi

[crimson sedge]

[wintry monolith]

Hello

[crimson sedge]

How to integrate -1^x e^x ?

[wintry monolith]

hmm

[idle tusk]

(-1)^x e^x?

[crimson sedge]

Yes

It’s nightmare

[dire geode]

oof

[wintry monolith]

U mean

[dire geode]

,tex .exp rules

[wraith dagger]

riemann

[wintry monolith]

The entire thing to the power

Or

(-1)^x (e^x)

[gentle echo]

$(-1)^x e^x$ is what you mean?

[wraith dagger]

parabolicinsanity

[grand idol]

$\int (-1)^x e^x$

[wraith dagger]

Bleach_Enjoyer

[wintry monolith]

I hope so 💀

[grand idol]

UHH

that whole thing raised is crazy..

[crimson sedge]

Yes , but the original e^n^c+ln cos(yln n)

[idle tusk]

treat $(-1)^x$ as $e^{i \pi x}$

[wraith dagger]

artemetra

[grand idol]

I see euler formula

most beautiful formula

[idle tusk]

yessir

[wintry monolith]

Yes

[crimson sedge]

Complex integral ?

[idle tusk]

no integration at all: -1 = e^(i pi)

[wintry monolith]

i is just a constant

[grand idol]

me: Oh just put it as $(-1)^x+1/ [-ln 1]$

[crimson sedge]

X is not constant

[gentle echo]

so you have the integral of $e^{(1+i\pi)x}$ which should be trivial now?

[wraith dagger]

parabolicinsanity

[gentle echo]

i guess

[wintry monolith]

U can

Write the whole thing

As e^x(1+ipi)

[grand idol]

Uh i found it as undefined

[crimson sedge]

So I get e^c x + i pi x + ln cos(y ln x) we’re c is constant

[wintry monolith]

Hmm

Why that term

1+ipi is constant

[gentle echo]

you integrate it as $e^{cx}$ which is just $ce^{cx}+C$

[wraith dagger]

parabolicinsanity

[grand idol]

My Photomath put it as this:
$\int (-1)^x e^x$
=$\int (-e)^x$
=$\frac{(-e)^x}{ln (-e)}$

[gentle echo]

well that's just wrong

ln(-e) isn't a thing

😔

[wraith dagger]

Bleach_Enjoyer

[wintry monolith]

😔

[crimson sedge]

The expontional power will be ( i pi x + c ln x + ln cos(y ln x) )

[wintry monolith]

No

[grand idol]

.tex logarithms rule

[crimson sedge]

C and y is constant

[echo arrow]

It’s a ,

[grand idol]

,tex logarithms rule

[wraith dagger]

Bleach_Enjoyer

logarithms rule

[gentle echo]

cinema

✋ 😐 🤚

[torn island]

bro did not have to use tex

[gentle echo]

you have to make it beforehand

[grand idol]

me who doesnt know how to use the commands..

[gentle echo]

anyways

[gentle echo]

this approach is incorrect

[echo arrow]

,tex .logarithms rule

[wraith dagger]

Laïko

[crimson sedge]

The expontional power will be ( i pi x + c ln x + ln cos(y ln x) ) after this I have to integrate e to complex power !!

[echo arrow]

No 🥲

[dire geode]

!original

[cedar kiln]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[wintry monolith]

Where did u get this

[crimson sedge]

No

[gentle echo]

bruh

[dire geode]

[gentle echo]

you can't just say "No"

[crimson sedge]

It’s not the original function I want to integrate

[dire geode]

everyone just got xy'd

[crimson sedge]

I told it’s nightmare

[wintry monolith]

What

how

[echo arrow]

Give the entire context

[wintry monolith]

😭

U cant just half of an integration question

And expect an answer

[echo arrow]

Is it the context ?

[crimson sedge]

-1^x *( e to power ( c ln x + ln cos(y ln x ))

This is the entire context

[wraith dagger]

Laïko

[dire geode]

Doubt

[crimson sedge]

Exactly

[dire geode]

Screen shot or pic

[crimson sedge]

It’s this one

[wintry monolith]

Gamma function

I dont think

U can manually integrate that

[echo arrow]

e^{cln(x} is easy to integrate but e^{ln(cos(yln(x))} is just ugly

[crimson sedge]

If we treat -1^x is e^i pi x , only one e will still but to complex power , so e it’s self * power derivate

[wintry monolith]

I think

Ur integration is wrt y

😔

[crimson sedge]

So just complex number derivate ?

[wintry monolith]

🙏🙏

[crimson sedge]

natural log (ln)

[remote nacelle]

https://math.stackexchange.com/questions/2114774/what-is-1x

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

Yes

[cedar kiln]

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[gray lynx]

R these equivalent?

[gray lynx]

,rotate

[grand idol]

is it negative tan(x/2)

[gray lynx]

Yes

[tropic oxide]

no they aren't

[gray lynx]

Why not

[tropic oxide]

tan(t + pi/4) is not in general equal to -tan(t)

if you want i can give an example

or show you graphs

[gray lynx]

In the second fourth of a circle tan is negative

Isnt it like moving tan 1 foirth of a circle

Into the second quadrant

[tropic oxide]

fourth of a circle
in english we call this a quadrant

[grand idol]

i see bunch of straight lines...

[crimson sedge]

guys is this for edexcel pure math asw? - this server?

[tropic oxide]

also pi/4 is not 1/4 of a circle but 1/8

[gray lynx]

Oh fak

[tropic oxide]

you are allowed to post edexcel questions here if you want. but do it in an unoccupied channel. read #rules, #info and #❓how-to-get-help

[gray lynx]

If it was π/2

Would i be right?

[tropic oxide]

@gray lynx even so, tan(x + pi/2) = -1/tan(x).

not -tan(x)

just because the sign flips does not mean the magnitude stays the same.

[gray lynx]

Damn it

[crimson sedge]

okay thank you

[gray lynx]

So tan -pi/4 isnt equal to tan pi/4

Mm

I'm asking for this problem

,rotate

[gray lynx]

The derivative of ln(tan(0.5x+0.25pi)

Is is equal to 1/cosx

Gotta prove that

[tropic oxide]

your prime symbol is pointing the wrong way

[tropic oxide]

-1 and 1 are not equal to each other indeed

[gray lynx]

Its just a difference in signs

[gray lynx]

Doesn't rly matter rn

[tropic oxide]

anyway

what's your progress so far

[iron wren]

in the domain from 0 to 2pi i get 4 answers btw

[tropic oxide]

OP was asking if his original thing was true for all x. it isn't.

[iron wren]

i must have used the wrong trig identity bruh

[gray lynx]

,rotate

[gray lynx]

I feel like a reddit user

When u say OP

😃

[grand idol]

Original poster thats what it means so...

[gray lynx]

I never rly thought about what it stood for

Good to know

[tropic oxide]

your tan derivative is wrong

d/dt tan(t) = sec^2(t) not sec(t)tan(t)

[gray lynx]

Its sec²

[tropic oxide]

sec(t) tan(t) is the derivative of sec(t)

[gray lynx]

Omg im dumb

[tropic oxide]

the abbreviation OP did originate from reddit but this server is a fitting context for it so there's nothing wrong with using it

[gray lynx]

So ingot sec² / tan

[tropic oxide]

ingot

[gray lynx]

Ingot bad typing skills

[tropic oxide]

you should probably write sec^2 as 1/cos^2 here

[gray lynx]

I was gonna do that

And change tan to sin/cos

I got 1/sin × cos

Sin

I meant sin

[tropic oxide]

ok write out what you get in full

cause the details are about to matter

[gray lynx]

1/sin(π/4 + x/2) × cos(π/4 + x/2)

How do I change sin and cos to where the angle is just x

There has to be some formula

[edgy spire]

yes

actually wait

[gray lynx]

And that is?

[gray lynx]

I'm waiting

[edgy spire]

nvm

There is one for pi/2

not one for pi/4

[tropic oxide]

there's a constant 2 factor you missed somewhere...

[gray lynx]

U mean half?

[tropic oxide]

should be $\frac{1}{2\sin(x/2+\pi/4)\cos(x/2+\pi/4)}$

[wraith dagger]

Ann

[tropic oxide]

2 in the denominator

so yes sure half

anyway $2 \sin(\theta) \cos(\theta)$ should look familiar

[wraith dagger]

Ann

[gray lynx]

Yup it does

Its cos

[tropic oxide]

act on it

[edgy spire]

No

[tropic oxide]

ok gonna ask you more fully then

[gray lynx]

Sin2x

[tropic oxide]

sin(2θ) the way i wrote it just now but yes

so apply that here

see what you get

[gray lynx]

1/sinx+ 0.5p

Pi

[tropic oxide]

brackets 💔

[gray lynx]

Which is 1/sin

[gray lynx]

Yk what i mean

[edgy spire]

No

[gray lynx]

Ur smart

[tropic oxide]

you should never drop brackets

no that's not a valid excuse for it

1/sin(x + pi/2)

[gray lynx]

1/(sin(x + 0.5pi)

Which is the same as sin(x)

Cause second quadrant

[edgy spire]

No it is not

[tropic oxide]

sin(x + pi/2) ≠ sin(x).

[gray lynx]

Its positive for sin(x)

[tropic oxide]

sin(x + pi/2) ≠ sin(x).

[gray lynx]

Then its cos(x)

[tropic oxide]

you should think before you speak

[gray lynx]

Idk my trig identities

At all

I'm going based of what I know

[tropic oxide]

ok then study trig identities

this is a known gap in your knowledge

[gray lynx]

Idk how

I type in trig identities in Google it gives me gibberish

[tropic oxide]

practice simplifying trig expressions i guess

maybe go to khanacademy

[gray lynx]

Is this right?

[tropic oxide]

there's some trig exercises there

yes sin(x + pi/2) = cos(x) is correct

[gray lynx]

I guessed cause then I would have solved the question

Not cause I remembered

[tropic oxide]

trig is not a guessing game

[gray lynx]

Cause its the only thing that fits

[gray lynx]

Trig is cool but tedious

[tropic oxide]

here is a list of trig identities that i compiled myself many years ago. you can ignore section 6 (which is mostly me being a showoff) but the others are handy.

obviously this alone is not a substitute for practice, practice and practice!

[gray lynx]

Ignore that

I thank you dear friend

Thats better

.close

[cedar kiln]

Channel closed

Closed by @gray lynx

Use .reopen if this was a mistake.

[cedar kiln]

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[gray lynx]

Lets say I have a function F(x) and I get the derivative f(x)

[gray lynx]

And the question tells me that to find the slope of the tangent line when x =1

Does F(x) touch f(x) at 1 and f(1)?

(1,f(1))

The tangent point

[tropic oxide]

!xy

[cedar kiln]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[tropic oxide]

gonna need some more specifics here

[gray lynx]

Its in arabic

[tropic oxide]

tangent line to what

[gray lynx]

The function F(x)

[tropic oxide]

big F?

[gray lynx]

Ye

[tropic oxide]

ok

then the point is (1, F(1))

[gray lynx]

Should I use 2 dif letters?

[tropic oxide]

and f(1) [small f] is the slope of the tangent line

i mean tbh it would be better if you said F and F'

or f and f'

[gray lynx]

Ok I'll do that

[tropic oxide]

less confusing this way

[gray lynx]

But not for any x value I plug it

Just 1

How do I know what the tangent point is

[tropic oxide]

you plug in the x-value at which you want the tangent line

in your case, 1

[gray lynx]

Wdym

[tropic oxide]

you want the tangent line at x=1

so you calculate the derivative at x=1

and not overthink

[gray lynx]

Ohh

So a tangent line can change

For some reason I thought it could only exist at 1 point

Only 1 tangent line for each function

So F'(1) is the slope of F'(x)

[tropic oxide]

overthinking

idk how to rescue you from this particular overthink

[gray lynx]

M

[tropic oxide]

you should leave this problem and return to it tomorrow

[gray lynx]

So what's equivalent to the slope of F'(x)

[gray lynx]

U said it here

The tangent line is F'(x)

Isn't it

Fak

What is then

[tropic oxide]

sorry i have to go to sleep now

[gray lynx]

Damn okay

.close

[cedar kiln]

Channel closed

Closed by @gray lynx

Use .reopen if this was a mistake.

[flint cape]

You did read the "If the original problem is not in English, then post I anyway!" bit, right?

[gray lynx]

I used to do that

But no one can read it

Why bother

[flint cape]

I'd still suggest doing so

There are 30k people online here and this is an international server; it's always worth a shot

[gray lynx]

The handwriting is horrible as well

And i can't send a voice message

[twilit escarp]

.reopen

[cedar kiln]

✅

[flint cape]

I'm also busy (with some assessment) but I could give at least translating this a shot...

If so defined is the function $f(x)$ equal to $\frac{3}{^5 \sqrt{x}} - \frac{1}{\sqrt{x}} + 2$ ...?

[wraith dagger]

Waes (Wires)

[flint cape]

Determine:
(1) the gradient of the tangent (to this curve) where x = 1
(2) the point of intersection (with the axes - [I'm guessing the y-axis?])
(3) the equation of the line (the tangent)

[NB for @helpers - the circled and the pink texts are power rules involving fractional powers and surds]

Rough explanation to start:
When you have a curve of the form y = f'(x), the expression f'(x), the derivative of f(x), is also called the gradient function - give me an x-value, and I can tell you the gradient (AKA "slope") of the tangent to the graph at this x-value, by plugging it into this gradient function

e.g. if my graph is y = x^3, then the gradient function 3x^2 tells me the gradient at an arbitrary value x on this graph

So at the point on this graph where my x-value is, say, 4, the gradient of the tangent at this point on the graph is 3(4)^2 = 42

[cedar kiln]

@gray lynx Has your question been resolved?

Channel closed

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[cedar kiln]

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[zealous palm]

can someone explain wtf this is

[hidden apex]

a piecewise function

[zealous palm]

hello ava welcome back lol

[hidden apex]

helo

a piecewise function is a function where there are conditions for some normal equation

like for f(x)={2x if x<3, 3 if x>=3 this means it is y=2x for the values of x<3 but when it gets to x=3 or greater it will be y=3

[zealous palm]

so for the first number in the formula is it just where the point is plotted?

[hidden apex]

if theres just a number and the condition thats the y value

if theres a equation and the condition you need to see the condition then find the equation for that condition

if it says “10 if x>2” that means for every value of x greater than 2 the y value is 10

[zealous palm]

if the dot is closed its greater/less than if its open its = to/greater than/less than?

[hidden apex]

and if it says “2x if x<=2” it will have the function 2x until x=2

[hidden apex]

closed is >= or <=

open is > or <

[zealous palm]

okay cool

so for this one its A

where did the 2x come from

i understand the rest of the formula

oh is it the slope?

[remote nacelle]

2x is part of the definition of the function. For the first option you have, it means that if x > 5 then f(x) = 2x

[zealous palm]

but where did we get 2x from

like how do we know its 2x and not 4x

[remote nacelle]

you could compute the slope

like it goes from 20 to 30 between 10 and 15

so slope 2 by (30-20)/(15-10)

[remote nacelle]

we don't see all the answer on this btw

[zealous palm]

its A

skipped the graphs bc got too complicated

getting nauseous looking at the rest of these units no joke

going to replace f(x) with y and then replace that with x and then solve for y

honestly have no clue what im doing

[oblique flare]

provide your work

[maiden halo]

it sounds like you were onto something here

[zealous palm]

tbh im trying to distribute the fraction is there something special i have to do?

atp they are just making shit up the solution to this makes 0 logical sense

[maiden halo]

so you got to the point where you said

y = 1/3 (x-5)

now what did you do?

[zealous palm]

I gave up

this is the solution and it looks crazy simple but like how tf are we getting +5 and how are we moving x around

we moved 3y to x and then changed y to x

makes no sesnse fuck this class

[maiden halo]

so all that’s happening here

is that they are making x the subject

do you see how it ends in “ x = “

the 3y came about because we multiplied both sides by 3 to get rid of 1/3 on the other side

[zealous palm]

thats the only part i understand

[maiden halo]

@zealous palm do you see this!

[zealous palm]

yes

[maiden halo]

okay great thats one part

[zealous palm]

thats as far as i got myself

[maiden halo]

now its x-5 and we got rid of -5 by adding 5 on both sides

[zealous palm]

but it would just cancel out

[maiden halo]

on the x-5 side it cancels out and becomes just x

on the other side we just had 3y

now its 3y+5

[zealous palm]

so they flipped the equation

why would they not just write it backwards like a normal person

[maiden halo]

They made x the subject

[maiden halo]

i guess you could say that

it’s an inverse after all

[maiden halo]

write it backwards?

[zealous palm]

3y+5=x

instead of x=3y+5

[maiden halo]

oh its just a thing in math to write the subject first

writing it the other way is messy

normally write it this way when we know the value of the subject

like 2+2=4

[zealous palm]

how would i know to put x/1

does it even matter if i do?

just bc its a fraction and i use 1 to simplify?

the unit after this looks EZ

im considering ignoring this ^ entire unit and just learning the other 7 and taking the exam

dont have the capacity to learn whatever this bullshit is in one hour

if its not hard and im just not understanding let me know

[slate drift]

No

That step is completely pointless and unnecessary

You can just go straight to multiplying by (-8y+2)

[cedar kiln]

@zealous palm Has your question been resolved?

[cedar kiln]

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[tight veldt]

How should I start this?

It is a simultaneous linear differential equation

[cedar kiln]

@tight veldt Has your question been resolved?

[cedar kiln]

@tight veldt Has your question been resolved?

[flat mica]

no initial conditions?

you can solve this the same way you solve a systems of 2 equations with 2 unknowns

Using that $(D_t + c_1 I)$ commutes with $(D_t + c_2 I)$

[wraith dagger]

gfauxpas

[tight veldt]

Oh ok

Thank you

[flat mica]

[cedar kiln]

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[cedar kiln]

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[lunar oak]

Hi how do I solve this

[split ice]

,rccw

[lunar oak]

will there be a bot to help or people's will help?

[lone dune]

people will help

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[lunar oak]

I don't know where to begin.

[lone dune]

do you know what the angle of depression and elevation are?

[lunar oak]

I don't actually...

[lone dune]

in this diagram

x is the angle of elevation

y is (90 - the angle of depression)

basically the angle of elevation is the angle thats like opening upwards

and the angle of depression is the angle thats opening downwards

[lunar oak]

I'm not good with English at these maths stuff ;-;

[split ice]

wouldn't the angle of depression be made with the horizontal? i.e. this?

[lunar oak]

yo @split ice can you give me the answer

[split ice]

uhh yeah i guess satvik left

[lunar oak]

like I'm Chinese, if U explain in Chinese I ofc understand but English hard for me to understand

[split ice]

unfortunately i do not speak chinese :( but i will try to explain feel free to ask for clarification

so you want to draw a horizontal line from the start, and you draw the line segment (for example, CD)

then you draw the angle between them

[lunar oak]

or can just draw me the answer I can understand quicker

[split ice]

i think it's helpful if i draw the first one

[lunar oak]

okay

[split ice]

and depression = down

elevation = up

[lone dune]

oops yeah you're right

thats my bad

[split ice]

so

• horizontal from start
• segment
• angle between them

[split ice]

hehe happens to the best of us ^-^

[lunar oak]

@split ice what's segment?

[split ice]

line from C to D

[cedar kiln]

@lunar oak Has your question been resolved?

[lunar oak]

@split ice so number 2 is like this?

[split ice]

this is the angle of depression

you want angle of elevation

starting at B this time