help-27

[ocean haven]

nah

[shy osprey]

-# what are you trying

[ocean haven]

,w 1/30 (-60 x - 5 cos(3 x) - 15 cos(4 x) - 3 cos(5 x) - 120 sin(x) + 35 sin(3 x) + 15 sin(4 x) + 3 sin(5 x)), x = pi

[woven radish]

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[ocean haven]

idk

[shy osprey]

-# whats the Qsn

[ocean haven]

,w 16/3 + pi

[woven radish]

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Requested by hereisa_ghost

[ocean haven]

@shy osprey

i kill

[shy osprey]

.pin

[ocean haven]

.close

[devout snow]

Channel closed

Closed by @ocean haven

Use .reopen if this was a mistake.

[shy osprey]

-# ??

-# solved?

[ocean haven]

dont try its wrong

[shy osprey]

oh okay

[shy osprey]

-# that hurts btw when the Qsn is wrong

[ocean haven]

ikr

ty anyways

[devout snow]

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[sand mesa]

I would like to learn more maths stuff but I don't know what to do.

[upper urchin]

how to solve part b, pls help

[brave plaza]

!occupied

[devout snow]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[devout snow]

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[trail eagle]

This is very very very vague.

[shy osprey]

lol try another one @upper urchin

[sand mesa]

sorry

[upper urchin]

which one

[sand mesa]

im going to specify

[upper urchin]

im new to the server

[shy osprey]

#help-21｜아리스킨충1

[shy osprey]

.pin

[flint talon]

[shy osprey]

Shit

Sorry about that

[sand mesa]

i want to learn more, i'm in year 8 (uk) and i want to know more things

just in general

because i'm finding maths lessons quite boring because i already know

[trail eagle]

Have you looked into the curriculum for years above you? It might give you an idea of things to look into,

[sand mesa]

we don't have access to upper years

like

looking at the curriculum

well

on the website it gives vague things such as 'trigonometry', 'calculus' etc.

[trail eagle]

The specific curriculum may not be available like published from your school, but there's lots of information online about what a typical year 9+ learns

[sand mesa]

i'll try looking into that

but

what would you advise?

specifically

[trail eagle]

This document seems to have a related section

https://assets.publishing.service.gov.uk/media/621629ac8fa8f5490d52ee78/KS3_NonStatutory_Guidance_Sept_2021_FINAL_NCETM.pdf

And this for key stages 3 and 4

https://www.gov.uk/government/publications/national-curriculum-in-england-mathematics-programmes-of-study/national-curriculum-in-england-mathematics-programmes-of-study#key-stage-3

This can at least give you an idea of next subjects you could look into

[sand mesa]

i have a khan academy account (i learnt functions for fun)

would that be a good thing

to user

[sand mesa]

use*

[trail eagle]

Yes, that's a good ressource for learning

[sand mesa]

ok

so i'll read through it and learn it on khan academy

[trail eagle]

Yes, they probably have a nice progression on there

[sand mesa]

It's quite a fun and engaging format, so yes

ok then

.close

[devout snow]

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[trail eagle]

You can also ask more study recommendations in #study-discussion

[sand mesa]

ok

thanks

[devout snow]

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[devout snow]

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[trail eagle]

<@&268886789983436800>

[devout snow]

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[orchid oriole]

Hey guys i don't need much help i just need clearance on Why did you distribute the cosine across the brackets when I could provide the derivative inside a bracket? I'm watching a video and studying it, and it explains how to distribute the cosine instead of providing the derivative. So im wondering is its possible to just provide the derivative since its already there just add (-)?
Also, is there another method besides the difference between two squares? Just wondering because i don't wanna be stuck in solving with 1 way its like having to memorize the questions more than understanding it

[small geyser]

,rccw

[orchid oriole]

Sorry if my handwriting is cursed but thats how it usually is when i write fast

[trail eagle]

I don't know if that's what you mean, but could write it as $$\int (1+\sin(x))\cos(x) \dd{x} = \int (1+\sin(x)) \cdot (1+\sin(x))' \dd{x} = \frac{(1+\sin(x))^2}{2} + C.$$

[woven radish]

Azyrashacorki

[orchid oriole]

I see, now thats easier than the way he provided it to me

I thought of it but didnt know if i could use it

And about the 3rd step where i Solve by the difference of two squares is there another way i could use?

Instead of (1-sinx)(1+sinx). Cosx dx/1-sinx?

[narrow plank]

i'd prolly do what you did honestly

[orchid oriole]

So any other way will be more complicated you're saying😅

[narrow plank]

if there is a simpler way to do it, i don't know about it lol

i wish i knew everything - that'd be cool.

But I think cancelling is probably the simplest way.

my general approach is get f(sinx) * cos(x) dx as an integrand to prep a u-sub

[orchid oriole]

I mean hey, you're the boss here you know more than me so i guess imma leave it as its

[narrow plank]

i wouldn't give myself the title of "boss", but i appreciate it!

[orchid oriole]

Alright i guess thats it for me i just needed some clearance of other ways to solve that, thanks yall much appreciated you can close this because idk how

[narrow plank]

you can type .close in the channel

[trail eagle]

I think that's pretty much the clearest way forwards. The only other way I can think of would be to start off with u = 1-sin(x) at the very first line, but the process is not very clean.

[orchid oriole]

Well, guess sometimes the most unexpected ways is the easiest

Thanks though i will just use Difference between two squares whenever i had chance to

.Close

[narrow plank]

has to be lowercase

[orchid oriole]

(':

.close

[devout snow]

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[devout snow]

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[unborn depot]

hello! ive done this question and ive gotten a 25 degree. can someone double check if the ans is 25?

[last parrot]

Can you show your working?

[devout snow]

@unborn depot Has your question been resolved?

[unborn depot]

one moment

let me get it rq

sorry this is so messy

[devout snow]

@unborn depot Has your question been resolved?

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[unborn depot]

<@&286206848099549185>

[fresh meteor]

yow

[unborn depot]

^

can someone double check my work?

[fresh meteor]

My answer is the same as yours, $$\angle A \approx \arctan(0.466) \approx 24.98^\circ \approx 25^\circ$$

[woven radish]

ShiroSharata

[unborn depot]

thank u so much

.close

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[devout snow]

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[devout snow]

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[sullen karma]

Sorry forgot how to rotate

But could I please get help started on 3a could I get a tutorial on how to do it please

[rotund verge]

Oooo thank god

Ok so for the value of k

We need to find the sum of all probabilities

And they will obviously equal 1

Cause probabillities

[sullen karma]

Yes alright

[rotund verge]

We expresses this as

[sullen karma]

So would it’s be k+k+…

[rotund verge]

Um the mathmaticall

God i cant find the sign

Ok your aware of stigma right?

Or just sum

[sullen karma]

Yes

[rotund verge]

Ok so stigma p( X = X ) =1

So basically at

X =0

Height is k

X = 1

Height is k2

X = 2
Height is k3

But x=3
Height is k4

[sullen karma]

Hold on what

[rotund verge]

Wait no

I mentk4

[sullen karma]

Yeah alright

[rotund verge]

So we add all of these

To get k10

[sullen karma]

Yes

[rotund verge]

Which can be represented as 1/10

Which is basically

[sullen karma]

Yes

[rotund verge]

0.1

[sullen karma]

So the answer is 0.1?

[rotund verge]

No

In a probability histogram the area of each bar represent the probabilith

[sullen karma]

Ohhh alright

[rotund verge]

The width kf each bar is 1unit

[sullen karma]

Yes

[rotund verge]

(0 to 1 , 1 to 2 etc )

[sullen karma]

Wait hold on

[rotund verge]

Ya

[sullen karma]

How is k10 = 0.1

??

Did u do k10 = 1

K =1/10

??

[rotund verge]

Ya

[sullen karma]

Ah alright

[rotund verge]

Ya

1/10

Probability

Can be 1

Maximum so we use that

Got it?

[sullen karma]

Yes

[rotund verge]

And the height is used for probability

[sullen karma]

Yes

[rotund verge]

For x 0 to 1
Height is k

From x 1 to 2
Height is k2

And so on

[sullen karma]

Yes

[rotund verge]

We can conclude

K+ 2k +3k + 2k + 1k = 1

8k = 1 as i mentioned

Before

Oh wait

[sullen karma]

Hm

[rotund verge]

You only want help for ka

I mean

3a

Not all

😭😭😭😭

[sullen karma]

I’m confused abit oh ye I just want a but it’s fine

[rotund verge]

I was explaining all and I got confused why so long for question

[sullen karma]

So what’s the answer for a abit confused

[rotund verge]

This one is for c

[sullen karma]

Alr alr

[rotund verge]

Wait let me look at the graph again

Oh wait no Iwas solving a now i did it in a weird order cause I looked at the sideways view

Ok

So

[rotund verge]

Here

8k = 1

[sullen karma]

So for a the answer is 1/8??

[rotund verge]

So k = 1/8

[rotund verge]

Yup

[sullen karma]

That’s what I thought but my textbook says it’s 3/8

[rotund verge]

Wait waaa

Wait for a?

[sullen karma]

Yes

[rotund verge]

1k 3k 3k 1k

8k

8k=1

K=1/8

Maybe my math is wrong or smth?

[sullen karma]

I think the textbook just has a mistake

Cause ur working makes sense

[rotund verge]

Hm

[sullen karma]

Ty for ur help though I appreciate it sm 🙂

😄

[rotund verge]

Ya my textbook does the same thing it said do until 2decimal point but the textbook does until 3 and rounds it off i got 1number off every time and I was so angry

😭😭😭😭😭

But ya sir gave me full makrs

[sullen karma]

😭😭

[rotund verge]

Later when I talked with him

[rotund verge]

Welcome also kinda sorry for extending this small easy answer into such a long thing i kinda got confused which question was which and went ahead solving them all

😅

[sullen karma]

All good atleast I now understand bc of u

Enjoy the rest of ur day / night 🙂

[rotund verge]

Hm thanks you too

[devout snow]

@sullen karma Has your question been resolved?

[devout snow]

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[tidal relic]

pls tell a shot trick to solve this

[rapid pewter]

The formula for compound interest is (\operatorname{CI} = P \cdot \left(1 + \frac{r}{n} \right)^{tn}) where (P) is the principal, here (P=10000), (r) is the annual interest rate, here (r=0.06), (n) is the compounding frequency, in your case (n=1) and (t) is the amount of time that you are compounding for, in this case (t=3)

[woven radish]

Ryan [She/Her] (TCC)

[rapid pewter]

Do you understand how I put the values into the formula from the board or are there any steps you would wish for me to explain?

@tidal relic

[devout snow]

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[tidal relic]

Thank you i got it

[rapid pewter]

Happy to help

[devout snow]

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[devout snow]

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[devout snow]

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[lost laurel]

Part(c)
Define $A_n$ to be that sequence consisting of $f_{n_{k}}(x_1),f_{n_k}(x_2) \dots f_{n_{k}}(x_{m})$ where the seqeunce are intereweaved in that order

[woven radish]

Wai

[sand quarry]

I think thats like kinda too complicated

[lost laurel]

Mhm

[sand quarry]

Well not for all x thats for sure

Cant you use cantor's

[lost laurel]

That's what I'm trying to do here

But first I need to come up with a sequence

Define $A_n(x)$ to be that sequence consisting of $f{n_{k}}(x),f{n_k}(x) \dots f{n{k}}(x)$ where the seqeunce are intereweaved in that order

[woven radish]

Wai

[sand quarry]

I fail yo see how it relates

Cuz like

[lost laurel]

hmm

Lemme think a bit more then I suppose, this is a problm worth thinking about

.close

[devout snow]

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[devout snow]

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[shy osprey]

Whats the maximum radius of a semicircle which can be contained in a square of unit length

[shy osprey]

-# ping me if any idea

[lapis pebble]

what did you try

[shy osprey]

Nothing js confused

<@&286206848099549185>

oh shit

15mins

sorry

[green crypt]

its alright

[thick lotus]

the side length/2?

[shy osprey]

idts

it could be inclined asw?

[finite fable]

umm you can place it along the diagonal

[shy osprey]

you cant

[finite fable]

umm. why not?

[shy osprey]

it would go out of the sqaure

[finite fable]

hmm

[thick lotus]

you do have an upper bound

[finite fable]

R^2 = 2a^2 + 2R^2 -4aR?

[finite fable]

ignore the top most diagram

[torpid sun]

yeah this looks promising

[finite fable]

lmao the expansion is wrong

i mean algebra

R^2 + R^2 = 2R^2 and it should 2ar instead of 2r

@shy osprey what do you think>

[shy osprey]

i didnt really understand how you got this😭

am bad at geometry sry

[finite fable]

i am good at geo only 😭 i am heavenly cooked (fr) in everything else

[finite fable]

you said about inclination

[shy osprey]

yeah

[finite fable]

and you said we cannot have diagonal

[shy osprey]

wouldnt the diameter be 45 degree with the edges

[finite fable]

yeah

congruency

[shy osprey]

oh i just went with symmetry😭

okay go on

[torpid sun]

(it's intuitively easy to see how that would be the figure with max radius but how do u prove that )

[finite fable]

yeah that's a good ques

i mean it should be somewhere from here only

[shy osprey]

also how do u apply this here to get that 45 degree statement

[finite fable]

lemme think with my small mind

[finite fable]

i mean you see two triangles in a square with a diagonal constructed. apply sss

angles then equal

[torpid sun]

(max radius is root(2)-1 after calc)

[shy osprey]

okay and how will that explain 45 degrre statement

[shy osprey]

isnt it supposed to 2 - sqrt2

[shy osprey]

this is smaller than the radius when you take edge as diameter

[shy osprey]

not sure tho i maybe wrong

[finite fable]

also i am not getting the relevance of 45 degree here?

we constructed a semi-circle with a radius greater than 0.5

[shy osprey]

arey i got this, i am saying how do you prove thhis

[finite fable]

so half a side of a square is certainly not the maximum radius

[torpid sun]

yeah that's true i made a mistake somewhere 😭

[thick lotus]

you can take a line that has end points on any two sides , the perpendicular distance from the midpoint to any side should be >= the length of line/2

[finite fable]

twin. i also made mistakes in expanding

but lemme think with my small mind again

[shy osprey]

wait could you explain again in more detail

-# geo is not my strong point sry

[finite fable]

i kinda remember this somewhere

i saw its application once

i cant remember it

[shy osprey]

oh wait i got it!

Tysm!!

.close

[devout snow]

Channel closed

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[finite fable]

you got what?

explain bro

[thick lotus]

the green showld be >= l/2

[shy osprey]

what prof paradox meant to say

[shy osprey]

yeah tysm gng🙏

[thick lotus]

that was crazy

[torpid sun]

I got root2/(root2+1)

which ig on rationalising gives 2-sqrt2

[shy osprey]

yeup

[devout snow]

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[elder bobcat]

Hii guys!! I'm just confused why this is true

[elder bobcat]

Is that even eligible 😭

[lunar harbor]

zoomed in version:

[elder bobcat]

Tqq

[sand quarry]

For continuous functions you can do [
\lim_{x\to a} f(g(x)) = f(\lim_{x\to a}g(x))
]

[elder bobcat]

Mhm

[sand quarry]

Yeah. So does that help clear your confusion

[elder bobcat]

I dont know. I dont think so?

[sand quarry]

Try to apply it to your problem

[elder bobcat]

How

[unique monolith]

plug in your equation given to that formula

[elder bobcat]

Ive never seen this before 😭

I don't know what I'm finding

[trail eagle]

Compute the limit at infinity of what's inside the logarithm.

[unique monolith]

this is worded very confusingly

[deft tapir]

If we have something like lim (x+4)² we can find the lim of x+4 then square that

[elder bobcat]

I see

[trail eagle]

Ok

[deft tapir]

So if you have a composite function, you can calculate the limit of the inner function, and then plug that into your function

[elder bobcat]

As t tends to invite, what would Ln (t+1) tend to. I can't picture it

[deft tapir]

You get that (t+1) goes to +inf

Then you calculate ln(+inf)

[elder bobcat]

I see

[deft tapir]

(for the sake of accuracy, it becomes lim ln(x) as x approaches +inf)

[elder bobcat]

I assumed it was 0 😭

Bur no

Hmmm

Okay so

[deft tapir]

The idea always remains that you calculate the inside part, then plug that into the outer function

Try to apply that to your problem

[elder bobcat]

Okok

Would the numerator tend to infinity and also the denominator?

Omg I feel like I don't know enough

😭

[deft tapir]

I'll help you out, there's a property of polynomials

When you have a fraction of two polynomials

The limit as x approaches ±inf follows the behavior of the largest exponent's term

[elder bobcat]

I see, 2?

Like 2²

The exponent of t or

[deft tapir]

So something like $$ \frac{5x^2 + 6x -2}{x^7 + x^6 + 8x^3 - x} $$ can simplify to just $$ \frac {5x^2}{x^7}$$

[elder bobcat]

Oh I see why 2 thou

[woven radish]

ScoobySnacks

[elder bobcat]

Ahh okok

[deft tapir]

I suggest simplifying your numerator and denominator first

[elder bobcat]

So here would it be 2t²/t

Or would I need to simply first

[deft tapir]

Although remember, the highest exponent term by itself is sufficient

[elder bobcat]

Okok

[deft tapir]

@elder bobcat I insist on this because simplifying (t+1)⁴ can be tedious

[elder bobcat]

Hm would it be 4t⁴/t⁴

Sorry for the late reply, had to take care of something rq

Hmm then u cancel out to get 4

Okok I get it tqqqqqqq

Sm

I didn't know the polynomial rule but thinking about it seems intuitive, I didn't catch on 😭

[trail eagle]

In general you can always do this to get a simpler limit to analyse. It stems from the fact that you can divide the numerator and denominator by the highest power of your variable and make terms like 1/x^n vanish in the limit.

[elder bobcat]

Hm I see I see

That makes sense

[trail eagle]

And as for the original thing about bringing the limit inside the logarithm, this is a property of continuous functions in particular.

[elder bobcat]

This?

[trail eagle]

Yes

[deft tapir]

Remember that you only apply that as x approaches ±∞

[elder bobcat]

Hm oh like finding out the limit for t+1 first

[elder bobcat]

Okok I see

If it approaches 0, I would only see the constants?

[deft tapir]

You just sub the t values

And check what you get

[elder bobcat]

Hm okok

Thanks

Everyone

.close

[devout snow]

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[deft tapir]

Was a pleasure! Keep mathing

[elder bobcat]

Tqqq!! I'm one step closer to getting that A for my finalss

Lowky kinda fun

[lilac moat]

.reopen

[devout snow]

✅ Original question: #help-27 message

[lilac moat]

also

though this is not necessary in this particular example

a technique that is useful sometimes for limits at infinity is

to make a substitution t=1/u, then t->+infinity is interchangeable with u->0+. if that confuses you, nevermind it, but it can be helpful

.close

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[elder bobcat]

Hmm I think I get it cause 1/invite gets smaller and smaller

And becomes insignificant

No

I'm so stupid

I read it wrong

give me a sec

[lilac moat]

simple example:

[elder bobcat]

Nope o do not get this

[lilac moat]

lim t->+infinity ln(t)= lim u->0+ ln(1/u)

[deft tapir]

I'm not sure that quite helps there tho

[lilac moat]

i use it mainly for inequalities

squeeze theorem

[elder bobcat]

Hm I might need a minute 😭

[lilac moat]

okay nm if it confuses you, its not necessary!

maybe youll get it later :)

[deft tapir]

Can we discuss this further in #calculus ?

[elder bobcat]

Hopefully!! I really shoukd learn how the functions behave

[devout snow]

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[inland carbon]

P and Q are polynomials such that deg P<= deg Q - 2, and Q has no real roots\
In the first part, I had to show that $\int_{-\infty}^{+\infty} \frac{dt}{(t-z)^k}$ equals 0 if $k\geq 2$ and $i\pi \epsilon(z)$ where $\epsilon$ is the sign of the imaginary part of z when k=1

Then I had to show $$\int_{\mathbb{R}} P/Q = i \pi \sum a_k \epsilon(z_k)$$ where the sum is over the roots of Q, and $a_k$ is the coefficient in front of $(x-z_k)^{-1}$ in the partial fraction decomposition. \

Then I need to show that in the case where P and Q are real this is in fact equal to $$2i\pi \sum_{Im(z_k) >0} a_k$$ but I don't see why the $a_k$ are necessarily purely imaginary

[woven radish]

bloubbloub

[inland carbon]

Also I don't know any complex analysis

[devout snow]

@inland carbon Has your question been resolved?

[sullen island]

well if Q is real, then the roots appear in conjugate pairs, so that kinda explains accounting for only half the roots by only looking at the ones w/ positive imaginary part

by the looks of it, if P is also real, there should be a nice relation between the coeff of (x-zk)^-1 and (x-conj(zk))^-1 in the PFD, which gives you the answer

[devout snow]

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[sullen island]

.reopen

[devout snow]

✅ Original question: #help-27 message

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[cyan scarab]

oh wtf

[cyan scarab]

=))

[lean crater]

do you have a question? otherwise, type .close

[last parrot]

có câu hỏi gì không? ;-;

[sand quarry]

.close

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[cyan scarab]

người việt hả

[cyan scarab]

=))

[small geyser]

!occupied. and you have a forum post ongoing, so please wait for help there.

[devout snow]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[last parrot]

Ờ

[last parrot]

.pin

[last parrot]

Biết xài mod ko

Môđun

[devout snow]

@cyan scarab Has your question been resolved?

[cyan scarab]

=))

[last parrot]

Vậy thử với 2 trường hợp n lẻ và chẵn đi

[cyan scarab]

ok

[devout snow]

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[orchid oriole]

Hi, im back again my exam is in 2-3h and i finished most of integration but there's only 1 thing that i struggle with no matter how i revise or how i solve equation because except its hard to remember i don't know where and when and how to use these laws

[orchid oriole]

Lets say i know a bit about cos2x=1-2sin²x and cos2x=2cos²x-1 since its all about if theres cos or sin next to it we use them To provide a derivative within an arc but the rest? Idk ig

[devout snow]

@orchid oriole Has your question been resolved?

[fresh meteor]

!show

[devout snow]

Show your work, and if possible, explain where you are stuck.

[orchid oriole]

Should i provide examples or?

I mean i will provide with what i struggle with as in equations if thats what u mean

Q.no7 is not that hard but its the laws that i mentioned is what making it confusing because idk why he used that instead of other ways same for Q.No4 when i couldve used something else i think?

[fresh meteor]

Actually, I meant, show how far you've tried to do it, but btw I've tried and it turns out I don't really understand it either ahahaha
Let someone else help you

<@&286206848099549185>

[orchid oriole]

Oh bruh i don't know where to start thats the problem, is i knew the start the rest is like a sip of water

Its the start that needs these laws that makes me confused like, when how where do i use them? Isnt there other ways to solve them? If so why isnt it provided such as power reduction formula

[devout snow]

@orchid oriole Has your question been resolved?

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[orchid oriole]

I gotta go now cya yall

.close

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[rain coral]

what is up witb this q

[dire thistle]

Hi, can anyone provide the solution manual for Applied Numerical Methods with MATLAB for Engineers and Scientists (5th Edition) by Steven C. Chapra?
I really need it for my studies. Any help would be greatly appreciated!

[trail eagle]

!occupied

[devout snow]

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@rain coral Has your question been resolved?

[supple knot]

What part don't you understand

[rain coral]

everything

like

i solve it

attempt i mean

and it jst

idk

[supple knot]

Uh huh

[rain coral]

it jst gets too long and it doesnt work out

like oaky

first get z in terms of w

put it into u+vi

real=x

imaginary=y

plug into the circle eq

and jst

its too long

[supple knot]

Mistake here probably. Easier to just leave z in terms of complex. Do you know how to write equation of a circle in modulus form?

Like |z| = 1 is a circle of radius 1 is the same as x^2 + y^2 = 1

[rain coral]

wdym

[supple knot]

Don't substitute x and y for z

[devout snow]

@rain coral Has your question been resolved?

[rain coral]

wgat

what

wld it be in this case

[supple knot]

|z - z0| = r is a circle of radius r centered at z0

Find z0 and r

[rain coral]

thats it

man thats so much easier

man

ok i see

[devout snow]

@rain coral Has your question been resolved?

[devout snow]

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[cyan barn]

hello i want help in

[devout snow]

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

[cyan barn]

1

should i do the sign of a then start?

[cyan barn]

1

could anyone help me?

please in finding the integral

[lament frigate]

does the form of these look familar

perhaps in the form of certain derivatives?

[cyan barn]

no

[lament frigate]

(hint, inverse trig functions)

[cyan barn]

arctan?

[lament frigate]

,w derivative of arctan(x)

[woven radish]

Results provided by WolframAlpha

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Requested by yeetusdeletus5

[lament frigate]

compare with b)

[cyan barn]

so if i took a =1 its her

[lament frigate]

and what if a ISN'T one?

[cyan barn]

but like should i do that if its zero its false and all the sign or i what

[lament frigate]

(If a is 0 then the result is just reverse power rule)

[cyan barn]

sorry ik i am dump

[lament frigate]

(hint: u-sub)

?

[cyan barn]

sorry didnt catch it well so i need to take another value variable changement first) right?

[lament frigate]

if a isn't one then yes, you should use a u-sub

You know which sub to preform?

[cyan barn]

yes if its not 1 or 0

no

tbh

but

[lament frigate]

you know u-sub, yes?

[cyan barn]

yes

variable changement

sould i take if its not 1 or 0 1 on the racine of a

??

the x

[lament frigate]

Well wouldn't it be cool, even if a wasn't one, we could factor out 1/a^2 from the integral and make it 1?

[cyan barn]

can i send u a small solution and tell me what do you think abt it?

[lament frigate]

ok

[cyan barn]

should i delet the cases ?? is it right?

[lament frigate]

we were talking about part b,

[cyan barn]

ahh i taught its a

sorry

[lament frigate]

It's ok

do you get what I was saying about it tho?

[cyan barn]

yeah but what should i take as u..sub

[lament frigate]

Well, we want to factor out the constant right?

[cyan barn]

yeah

[lament frigate]

so it's just 1 and it's arctan right

[cyan barn]

yeah sir

if its 0 its normal

[lament frigate]

we would need u to have a factor of a

so we could factor a

right?

[cyan barn]

yes

[lament frigate]

So what do you think of the sub u=ax

[cyan barn]

wait let me think abt it for a sec

ok thank u soo much

.close

[devout snow]

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[devout snow]

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[compact kiln]

find the equation of the tangent circle that is normal to the line

[compact kiln]

I need help with this task guys

<@&286206848099549185>

[dull parrot]

rearrange 3x - 4y = 0 to make y the subject

then substiute it into the other equation

and solve for x

[torpid sun]

I'm not too sure I understood the problem here

is that supposed to be x^2+y^2+2x+4y+7?

because you've written x^2-y^2

which would not give a circle

is the question to find the equation of the tangent to the circle [as given in question] such that it is perpendicular to 3x-4y=0

[dull parrot]

I didn't realise it was x^2 - y^2

[torpid sun]

Mhm, I think it's most likely just a typo

[devout snow]

@compact kiln Has your question been resolved?

[devout snow]

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[compact kiln]

my teacher wrote it on the table

[compact kiln]

she could do a typo i guess

I think this is a good solution, no?

[devout snow]

@compact kiln Has your question been resolved?

[unique monolith]

^ og question

[devout snow]

@compact kiln Has your question been resolved?

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[elder bobcat]

I'm stuckk

[elder bobcat]

This is what I've done so far

Amd im not sure if I've made a mistaje

*mistake

[uncut crow]

[elder bobcat]

Slay

[willow helm]

It seems you have made a mistake

[elder bobcat]

I see where

[willow helm]

oh I was making a joke on mistaje 🙂

[elder bobcat]

Omgg sorry

I dont have any humour in me rn the work is draining me

[uncut crow]

[elder bobcat]

@helper ty, any help would be appreciated

Opf

<@&286206848099549185>

[last parrot]

Isn't the derivative of coshy is sech^2x

What is sec^2hy here

[harsh stream]

Nope

[last parrot]

Similarly to coshy = sqrt(1+sin^2h) it must be sqrt(1+sinh^2)

[harsh stream]

Derivative of cosh is sinh

Oh ok

U confused me with the sech^2y

😭

[elder bobcat]

Ohh

Let me see

Oof yea soryy

Let me change it

[last parrot]

Ye

[simple grotto]

By the way, sech is one function not secant if h?

[uncut crow]

hyperbolic secant

[elder bobcat]

Ye

[simple grotto]

Just making sure you are doing the right thing. X E.

Carry on. X E.

[elder bobcat]

Kkkk

I don't know here to go from here 😭

[last parrot]

What is X.E.

[last parrot]

Sub sinhy=tanhx

[elder bobcat]

Kkkkk thank youuu

[last parrot]

https://tenor.com/view/meow-dancing-cat-cat-dancing-meow-meow-meow-gif-14608759981751543562

[elder bobcat]

Oh

I see

😭

Sighhh

Omg

Thanks guys

I was js being really stupid

Kk

Thanks again

.close

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[granite island]

can i pls have a hint for the problem i legit don't know where to start

[granite island]

maybe we can represent the star as a function instead

not really usefull tho

[supple knot]

Try replacing a with b * a

[granite island]

ok

[supple knot]

And maybe another substitution if that doesn't complete the proof. The only thing you have is closure so you have to use it creatively

[granite island]

ok thx

[granite island]

wait u can just do this

does that mean a=b*a?

[supple knot]

No?

[granite island]

or is it let a=b*a for some a and some b

[supple knot]

Also no

[granite island]

oh

[supple knot]

The identity holds for all a and b in S and since ab is in S, you can pick ab to be one of a or b

[granite island]

oh

[granite island]

can you explain it a little more im kinda dumb i don't rly understand it yet

[granite island]

so i can let ab=a or ab=b?

[uncut crow]

bad way to express that

[granite island]

ok

so how should i express it?

[uncut crow]

well you shouldn’t write “let ab = a” anywhere

[granite island]

yea i know

just kinda confused abotu what he said

[uncut crow]

one way to think of it would be to imagine that the given condition was (c*d)*c = d for all c,d in S. then you are given arbitrary a and b and you want to prove a*(b*a) = b

[granite island]

yes

[uncut crow]

then ‘letting c = something’ makes more sense

[granite island]

ok

[uncut crow]

so you can pick some c and d (in terms of a and b) and get a true statement. hopefully one that helps you prove the equality that is requested if you pick them well

[granite island]

ok

so i got (ab)(cd)=a(b*(c*d))

wait the stars are gone oops

like i want to show taht

[uncut crow]

\*

[supple knot]

I don't understand how you have 4 variables now

[granite island]

sub a= c * d

[uncut crow]

there should only be a’s and b’s in the equality ‘derived’ from putting things into (c*d)*c = d

[granite island]

what

yea i don't understnad pls elaborate more

[uncut crow]

you want to set c and d equal to something (in terms of a and b) so that a*(b*a) = b pops out of (c*d)*c = d

put c = something that only has a’s and b’s

same with d

replacing c and d in the equation will give you something with only a’s and b’s

[granite island]

ok

so let c= a * b

d=b * a

wait what

[uncut crow]

that is a legal thing you can do

[granite island]

ok

[uncut crow]

it may or may not be helpful but that is the idea

[granite island]

yea uh how is this usefull

[uncut crow]

maybe your choice was not helpful. you may need something else

[devout snow]

@granite island Has your question been resolved?

[devout snow]

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[drowsy parcel]

can anyone teach me every trigonometry basics that can be used in linear algebra

[devout snow]

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[drowsy parcel]

.close

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[fickle moon]

is all the work in this done correctly??

[glossy dew]

more or less yeah

if $v'(t) = \vec{a}(t)$ is a reasonable enough expression to the teacher

[fickle moon]

why wouldnt it be

?/

[glossy dew]

hmm that's okay then i guess

[devout snow]

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[warm aspen]

can someone explain me path integrals,

[placid minnow]

in context to a problem or just in general

like are you talking about the feynman path integral or something

[warm aspen]

just in general, I don't understand why a particle moving to a certain place over time means the sum of all possible paths it takes

[sand quarry]

Probably line integrals

In context of multivariatr calculus

Oh ok nvm

I feel like this is better asked in a physics server

[placid minnow]

well as you know things dont have a definite position in the quantum world really

[warm aspen]

yes

[sand quarry]

I mean usually the stepping stone idea is the double slit experiment

[placid minnow]

yeah

actually thats probably the experiment to read up on for this

im not that qualified for this

[warm aspen]

I've asked it and not got a reply yet, but since it may be considered as maths I decided to ask it here

[warm aspen]

ah alright

[sand quarry]

If you can formulate it into a mathematical problem sure

As it stands your question is not obviously of that sort

But like

Conceptually speaking

If you want to calculate the probability of an electron or particle or whatever going from point A to B, you have to sum up the probabilities of it taking every single conceivable path in the universe

[warm aspen]

but why

and also

[sand quarry]

I mean, it's like total probability right

,, \vb P(A) = \sum_n \vb P(A\cap B_n)

[sand quarry]

That's how I see it

[warm aspen]

when you think about it like that, while doing the summing up, in a specific case abcdef, if the particle moves zig zag for like 500km and THEN reaches point b,

do you not have a problem with THAT abcdef case being a total summation of all OTHER possible paths being made into that abcdef?
why do you only have a problem with the real path which the electron takes and not have a problem with the other infinitely imaginary paths it would take according to you

[sand quarry]

The 500 km path is not a sum of other paths tho

It is its own path

[warm aspen]

then why isn't the real path which the electron takes not its own path?

[sand quarry]

It is

[warm aspen]

so why is it a summation of all other infinite paths?

it's just a unique path independent to all other paths the electron could take

[sand quarry]

You have to understand that this statement is talking about PROBABILITIES

[warm aspen]

where do u bring probability from/to? an electron reaches point b from a, tell me where probability is, in here

[sand quarry]

Again, you need to really look up the double slit experiment

If you make an electron go through double slits, where it appears on the other side is dictated by a probability function because it is not a deterministic process and cannot be known a priori

Feynman realised that empty space can be thought of as an infinite number of slits, which is why it ties back to your original statement

[devout snow]

@warm aspen Has your question been resolved?

[devout snow]

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[smoky gyro]

im having trouble with finding the bounds for theta

i arrived that the inequality r<=tan(theta)sec(theta) from letting x =rcos y=rsin

and then solving tansec =1 we get that sin(theta) = sqrt(5) - 1 / 2

so theta = arcsin ( sqrt5) -1 / 2 ) but this is only one of bounds for theta and im not sure how to find out the other one

[sand quarry]

You mean (sqrt5-1)/2 to be clear right

[smoky gyro]

yes golden ratio

[sand quarry]

If so I think thats correct

[smoky gyro]

but thats only one bound for theta

how can i find the other one

[faint gorge]

vro

[smoky gyro]

what

[sand quarry]

Anyway for the other angle, try to think of where the parabola in y >= x^2 intersects with your circle

[flint talon]

[faint gorge]

did you try to draw the region vro

[smoky gyro]

nah

i was never good at art

i js rawdog it w algebra basically

[faint gorge]

do both

you dont need to be an artist

it's just for reference so you understand the algebra

[smoky gyro]

polar stuff is messy here isnt it

like the other theta bound isnt just +- theta right that would be wrong do we subtract pi for the other bound

(brb dinner)

[faint gorge]

You can make use of symmetry

[devout snow]

@smoky gyro Has your question been resolved?

[faint gorge]

dinner too good apparently 🙏🏻

[devout snow]

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[smoky gyro]

soo good i got pizza and garlic bread yo

[smoky gyro]

like symmetry about y axis?

[faint gorge]

yes

So you can just subtract your angle from pi

[smoky gyro]

i see.. thank you

[faint gorge]

https://www.desmos.com/calculator/nk3louz3fo

[smoky gyro]

so y axis symmetry is all that matters like it doesnt need to be x symmmetric or

[faint gorge]

Well if you draw it, you will realize that the angle is mirrored/reflected at the y-axis

That's all, so yeah no symmetry with the x-axis

[smoky gyro]

hmm i see

my doubts is solved

.close

[devout snow]

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[faint gorge]

🔡

[sand quarry]

wait but you still need to construct the integral

[smoky gyro]

thats true

my doubts is unsolved

.reopen

[devout snow]

✅ Original question: #help-27 message

[sand quarry]

its not going to be one integral

in polar

[smoky gyro]

hmmmm

how do i construct

[sand quarry]

try to like trace the rays as you sweep the angle

like it will hit different boundaries

which will need different integrands

[smoky gyro]

hmmm

[sand quarry]

yes so there are three 'phases'

the green, red, and black

[smoky gyro]

righr

[faint gorge]

you can formulate black and green into one integral

[smoky gyro]

is because they are symmetric

so if we have small theta our ray will hit the parabola

if our ray is like in the middle we hit the circle

and if theta big close to pi then we hit the parabola again

[devout snow]

@smoky gyro Has your question been resolved?

[faint gorge]

yes

[devout snow]

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[devout snow]

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[devout snow]

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[sour rampart]

Is anybody there from class 10th cbse i need some tips..

[jade pecan]

!da2a

[devout snow]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[jade pecan]

and also #study-discussion

[sour rampart]

Im new to this server is dont quite understood what u said..

[jade pecan]

you should use #study-discussion for tips

[sour rampart]

Ooh

[jade pecan]

if u have specific questions, use the help channels

[sour rampart]

Ok thx

Thx

[daring basin]

21. If a ≡ b (mod n), show that a^n ≡ b^n (mod n^2). Is the converse true?
    Solve this anyone
    Without binomial theorem

[hasty cargo]

!occupied

[devout snow]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[daring basin]

Aahh ok

[flint talon]

does that mean there's no question here

[devout snow]

@sour rampart Has your question been resolved?

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[long trout]

This is a problem that I thought of, so there might be some inconsistencies. Let S be another sequence that is permutation of the sequence: 1,2,3,…,n. In each move, one can chooses a consecutive subsequence of the previous sequence (initial one being S), and either performs a reversal within the subsequence (flips the order of elements in the subsequence), or a transposition of the subsequence (moves the subsequence around within the sequence). Out of all possible S, what is the largest minimum number of moves to permute from S back to the sequence: 1,2,3…,n ? Ik this may sound confusing, so feel free to ask for clarifications and examples of how it works

[placid minnow]

for operation 2, is it "cut a block and paste it anywhere else" (one block moves), or "swap two adjacent blocks" (the stricter genome-rearrangement definition)? these will give different answers

[long trout]

The first one.

[placid minnow]

and for each move it is either a reversal or a block move and i can choose right

[long trout]

yes

[placid minnow]

and if i reverse entire sequence that still only costs 1?

meaning it costs 1 per move regardless of block length?

[long trout]

Yes

[placid minnow]

and you want max_s min(moves to sort S) as a function of n

which is just the worst case starting perm

[long trout]

yes

[long trout]

correct

[finite briar]

what the hell does largest minimum mean

supremum esque?

[placid minnow]

so a single reversal can eliminate at most 2 breakpoints, which happens if the elements at the ends of the flipped segment manage to plug the gaps perfectly in their new orientations

and for second way to get rid of breakpoints a block move involves cutting the sequence at 3 places and rejoining them, so a transposition can eliminate at most 3 breakpoints

so upper limit is going to be using transposition

cause if you think about it if a permutation
$S$ has
$b(S)$ breakpoints, and the best possible move removes 3 breakpoints, then the minimum number of moves
$d(S)$ must satisfy:

$d(S) \geq \frac{b(S)}{3}$

[woven radish]

thecrumbeler2

[placid minnow]

so this suggests the number of breakpoints is approximately
$n$ and max distance grows linearly

[woven radish]

thecrumbeler2

[placid minnow]

because we know for a permutation of length
$n$, the maximum number of moves
$D(n)$ required to sort it is:

$D(n) \approx \lceil \frac{n+1}{2} \rceil \text{ or slightly lower depending on the mix of moves.}$

[woven radish]

thecrumbeler2

[placid minnow]

but we are dealing with specifically the combination of reversals and transpositions

so if only transpositions were allowed, the maximum distance is
$\lfloor \frac{n+1}{2} \rfloor$

[woven radish]

thecrumbeler2

[placid minnow]

If only reversals were allowed, then the maximum distance is just
$n-1$.

[woven radish]

thecrumbeler2

[placid minnow]

with both the transposition is dominant, which is what i said before

and at this point idk the right constant

[long trout]

Oh ok, so it’s better to take transpositions rather than reversals

[placid minnow]

yeah

after some research the worst case after a lot of debate seems to be n/2

so ill just go with that

since even though a transposition can remove 3 breakpoints, it is impossible to find a move that removes 3 breakpoints in every single step for the most difficult permutations. Most moves will only remove 2 or 1

[long trout]

alright

[placid minnow]

and just thinking about this in small $n$:

[woven radish]

thecrumbeler2

[placid minnow]

$n=2$: Max moves = 1 (e.g.,
$(2, 1)$ needs 1 reversal).

[woven radish]

thecrumbeler2

[placid minnow]

$n=3$: Max moves = 1 (e.g.,
$(3, 2, 1)$ needs 1 reversal).

[woven radish]

thecrumbeler2

[placid minnow]

$n=4$: Max moves = 2.

[woven radish]

thecrumbeler2

[placid minnow]

and $n=10$: Max moves is usually around 5 or 6.

[woven radish]

thecrumbeler2

[placid minnow]

so we already established that for a sequence of length
$n$, the maximum number of moves (the diameter) is approximately: $f(n) \approx \frac{n}{2}$

[woven radish]

thecrumbeler2

[placid minnow]

But going more precisely, for large
$n$, it is bounded by: $\frac{n+1}{3} \leq \text{Max Moves} \leq \frac{n}{2}$

[woven radish]

thecrumbeler2

[placid minnow]

(if you remember what i said before)

which is my answer

and the most difficult permutations to sort are those that are interleaved, like
$(2, 4, 6, \dots, 1, 3, 5, \dots)$, because they maximize the number of breakpoints that cannot be fixed simultaneously by a single move

[woven radish]

thecrumbeler2

[placid minnow]

but yeah this is my general answer

[long trout]

thanks for your input

.close

[devout snow]

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[devout snow]

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[mystic arrow]

For this, I’m not too sure on whats happening on the line i’ve underline.

Is it taking n^3 -(n-1)^3 and then adding the (n-1)^3 to both sides? And is the section above displaying that for when you substitute (n-x) in as x approaches n the values follow a progression downwards?

[arctic yarrow]

It is a telescoping sum

$n^3 - 1 = \sum_{k=1}^n (k^3 - (k-1)^3)$

[woven radish]

Lin Xia

[mystic arrow]

How does it re-arrange to that line tho?

[arctic yarrow]

from Newton formula we have $(k^3 - (k-1)^3) = 3k^2 - 3k + 1$, for all $k>0$

[woven radish]

Lin Xia

[arctic yarrow]

I dont know if that is what you asked

[mystic arrow]

I mean for this part sorry, the main part im confused on is where n comes from and how the 1 is dropped

[arctic yarrow]

Sorry i did a mistake btw there is no - 1 here

[arctic yarrow]

after replacing terms in the sum $n^3 = \sum_{k=1}^n (k^3 - (k-1)^3)$ you just split it into 3 sums. The last one is $\sum_{k=1}^n 1$ that is exactly $n$.

[woven radish]

Lin Xia

[mystic arrow]

Ohh okay, thank you for your help

.close

[devout snow]

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[sinful sand]

i need help

[sinful sand]

so

about infos k is parallel to l

what is the degree of showed with ? symbol

[untold ravine]

maybe it helps to extend the line l till it intersects the line below

[lethal wraith]

try tracing this line which is also parallel to k (ignore the shitass drawing)

[sinful sand]

hm

[lethal wraith]

(notice how there's a triangle)

[sinful sand]

i see it