help-10

[final thunder]

Euler’s notation

[zenith epoch]

ok, thank u

[stable rain]

woo interesting

just learnt abt that rn

[final thunder]

Look up D operator as well if you’re interested.

[zenith epoch]

yes sir

[final thunder]

[zenith epoch]

how about this?

is it true?

[final thunder]

Think about the formal definition of derivative

[zenith epoch]

i can't connect them

ohh, is it "False"?

[kind hawk]

what is the limit in the definition of the derivative

and what is the limit in the definition of continuous

[zenith epoch]

Like this?

[kind hawk]

I'm gonna rewrite it slightly. $f$ is continuous at $c$ iff $$\lim_{x\to c} f(x)-f(c)=0$$

[warm shale]

Denascite

[kind hawk]

now what about the limit for the derivative

[zenith epoch]

Lim is a derivative of function

Like that?

[final thunder]

What is the formal definition of derivative at x=c?

[zenith epoch]

What?

[final thunder]

$lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$

This is the formal definition of the derivative at x

Ah wait

[warm shale]

Pure

[kind hawk]

which I will again slightly rewrite as $$\lim_{x\to c} \frac{f(x)-f(c)}{x-c}$$

[warm shale]

Denascite

[final thunder]

How do you get the limit thing under lim?

[kind hawk]

I don't think that is possible in inline math

or well it is but not as nice

[final thunder]

Ah ic

thanks!

[zenith epoch]

So what is the conclution?

[kind hawk]

I wrote two limits. can you see anything about how they are connected?

[zenith epoch]

The f(x)-f(c)?

[final thunder]

Yes

[zenith epoch]

Look at this

Ehh wait

[dire yacht]

what is the function?

[zenith epoch]

Wait, i didnt get it

[kind hawk]

you have to show that $\lim_{x\to c} f(x)-f(c)=0$

[warm shale]

Denascite

[kind hawk]

you know that $\lim_{x\to c} \frac{f(x)-f(c)}{x-c} = f'(c) <\infty$ exists and is finite

[warm shale]

Denascite

[kind hawk]

what can you do

[dire yacht]

the intuition to know what is the limit "0/0" would be to think which function is getting faster to the point? for example, for f(x)=sin/x
$\lim _{x\to :0}\left(\frac{sin\left(x\right)}{x}\right)=1$
this means that sin and x approach 0 "the same speed"

[warm shale]

Dav1d

[kind hawk]

we don't need to worry about any 0/0 stuff here

[dire yacht]

isn't the question about derivatives?

[final thunder]

The question’s about f being continuous at c if f’(c) exists

[dire yacht]

oh, I understand

[kind hawk]

we dont yet actually know that the limit has the form 0/0. while we do show that at the end, we don't really have to worry about that

[obtuse pebble]

@zenith epoch Has your question been resolved?

[zenith epoch]

So, like this?

[dire yacht]

we know that the following limit exists:
$\lim _{x\to :c}\left(\frac{f\left(x\right)-f\left(c\right)}{x-c}\right)$
we need to proof that this limit exists:
$\lim _{x\to :c}\left(f\left(x\right)-f\left(c\right)\right)$

note that:
$\lim _{x\to :c}\left(f\left(x\right)-f\left(c\right)\right)=\frac{f\left(x\right)-f\left(c\right)}{x-c}\cdot :\left(x-c\right)$

[warm shale]

Dav1d

[kind hawk]

forgot the limit at the end

[obtuse pebble]

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[timid silo]

How can I compute this ? I did not find the residue nor the Laurent development...

[dire yacht]

could you write this in latex?

[timid silo]

yes give me two minutes

@dire yacht I don't know how to write in latex directly in the chat btw

c_2 is a constant > 0

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

bro just ping the helpers

<@&286206848099549185>

the <@&286206848099549185> won't help

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[graceful crane]

can I get help doing this integral? I have did a few steps and ended up on the anti derivative of sec theta/tan theta and I do not know how to solve that

[obtuse pebble]

@graceful crane Has your question been resolved?

[graceful crane]

<@&286206848099549185>

[trail musk]

sec/tan = (1/cos)/(sin/cos) = cos/(cos*sin) = 1/sin = csc

@graceful crane

[graceful crane]

Can I integrate that?

I tried I regrating it but it came out with a weird value when I plugged it in the calculator so I wasn't sure if that was the way around it

Natural log of cos+sin x or something like that

[trail musk]

yes but it's not exactly the most intuitive thing to integrate

[graceful crane]

Hm idk what to do then

Not sure if I did something wrong but I ended up with sec theta/ tan theta in 2 different problems and when I change it to csc theta and integrate it still feels like I can go anywhere

I'll try writing csc theta in terms of X then integrating maybe that fixes it

[trail musk]

that's not gonna get you anywhere haha

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

you have to see it

Wait I actually found an even better method

https://www.youtube.com/watch?v=WYHCjfxYGjQ

[graceful crane]

alright that did it, ty

.close

[obtuse pebble]

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[fierce lagoon]

What have you tried

[dense plover]

Nothing yet bc i cant really get it wrong 😅

The past 2 problems were different from this one

[fierce lagoon]

Why can't you get it wrong

Is this homework

[dense plover]

Yes

Indeed

[fierce lagoon]

g of h just means g(h(x))

So you replace every x with h(x)

[dense plover]

So 4x+5+6/4x+5+3

?

[fierce lagoon]

Yeah just simplify that

[dense plover]

The composition is supposed to be simplified?

[nocturne minnow]

It's ideal to simplify

Like 5 + 3

[dense plover]

Its bc the past two problems weren’t simplified

[nocturne minnow]

You're saying it didn't simplify 5 + 3?

[dense plover]

It wasnt similar to this one

And the problem doesnt ask me to simplify it

[nocturne minnow]

Well, it's ideal to simplify 5 + 3

[dense plover]

So would the domain look like this

.close

[obtuse pebble]

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[frozen viper]

hi! may i get an explanation on why theres a x^2 on top at the last step

why is it not just x?

[zenith raft]

it's because 2ln(x) = ln(x^2)

[frozen viper]

oh! thank u sm!

.close

[obtuse pebble]

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[zenith raft]

np 🙂

[obtuse pebble]

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[unique kettle]

Is my solution here correct, because I feel like I did the math right but it doesnt seem correct for some reason

<@&286206848099549185>

[obtuse pebble]

@unique kettle Has your question been resolved?

[obtuse pebble]

@unique kettle Has your question been resolved?

[unique kettle]

<@&286206848099549185>

[obtuse pebble]

@unique kettle Has your question been resolved?

[unique kettle]

<@&286206848099549185>

[mild harness]

@unique kettle

Where ru stuck

[unique kettle]

I just want to know if my solution makes sense

[mild harness]

Sort of those except the second to last line

Hold up

First off why r u subtracting by zero first line

[unique kettle]

Its equal to 0

The bottom line is just barely showing for some reason

want to check the answer because I feel like its not right for some reason

what?

why you respond if you cant help lol. But its easy to understand when you have the ground work understood. But seeing something where you dont understand even the notation means you wont understand it

[unique kettle]

Its equal sign, bottom line is just barely showing for some reason

[mild harness]

Oh lol

[unique kettle]

yea you know if it looks right?

[mild harness]

I think it’s right

I didn’t know that the solutions could be expressed as free variables

I’m also taking linear algebra

[short umbra]

There should be a constant vector in your solution that isn’t effected by t and s I believe

[mild harness]

Something felt off with that

I wasn’t sure @short umbra

[unique kettle]

Yeah but how would I get the constant vector'

tghere

Because the vectors are equal to 0 so theres no way I can get a constant on the right side with only variables on the left

[mild harness]

I’m actually on clips side

Abt that

[short umbra]

Yeah that’s definitely right actually

I think it’s good

Makes sense to me

Got -t-s+t+s=0 for the 1st row

[mild harness]

@timid silo stop spamming

[short umbra]

And then second row is just always 0

Looks good!

[unique kettle]

Yeah but if we were to give that t=1 and s=2 it wouldnt hold is the problem

@mild harness ^

[mild harness]

I think it’s right

Hmmmm

[unique kettle]

Yeah but my solution set doesnt for some reason

[short umbra]

If u plug it into the original dot products up top it works

[unique kettle]

It doesnt though?

[short umbra]

I might be missing something

But

(-1 + -2) + 1 + 2 =0

[unique kettle]

where did you get that from

[short umbra]

I’m using your example of t = 1 s=2

And then v1 = -t-s

V2 = t

V3 = 0

V4 = s

[short umbra]

And then this is plugging that all into the 1st dot product u have at the top of the page

Confirming it is perpendicular to that vector

[mild harness]

Lol

[unique kettle]

-1-2 = -3 and that times 1 is -3

I dont know how you dont have -3 in your calculation

[short umbra]

(-1 + -2)

Sorry I wrote it weird

[unique kettle]

But then it should be -3 + 1 + 1

which isnt 0

[short umbra]

But v4 is s

Which is 2

So -3 + 1 + 2

[unique kettle]

Oh wait I wrote my solution system of equations wrong

[short umbra]

Or v2 is s*

[unique kettle]

I have v2 and v4 set to t

[short umbra]

Ah shoot right

[unique kettle]

Ok this should work now

I made the right edits

[short umbra]

Ya that looks better

Sorry idk how I didn’t catch that earlier lol

[unique kettle]

Ok bet, small little mistake that I have been looking at for 2 hours now lol

[short umbra]

Relatable

[unique kettle]

thanks

.close

[obtuse pebble]

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[obtuse pebble]

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[remote creek]

<@&286206848099549185>

[small thicket]

!15minutes

[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.

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[quaint nest]

help

[wild swallow]

[quaint nest]

so I need to use law of syllogism

to get like an extra line for the slogan

you call find them for these 2

[remote creek]

<@&286206848099549185>

[stable rain]

!15min

[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.

[stable rain]

its not even ur help channel

#❓how-to-get-help

[wild swallow]

lol

fake helper on the ball

[stable rain]

ballin

[quaint nest]

so can you guys help me out

<@&286206848099549185>

[obtuse pebble]

@quaint nest Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

suppose f was unbounded, then use the mean value theorem to show that f' is also unbounded

do you know what the mean value theorem is?

yup, so b-a is bounded, but f(b)-f(a) can be large

so what does that tell us about f'(c)?

differentiable on an open set though, so its fine here, just dont take the endpoints for a or b

f'(c) can get arbitrarily large, so unbounded

yup, thats how the proof would go

to make it formal, you could let N be arbitrary and show that we can construct a c such that |f'(c)|>N

since f is unbounded, we can find an a such that |f(b)|>N(x_2-x_1)|f((x_1+x_2)/2)|

its just a random number between x_1 and x_2

I just chose the midpoint

you can rearrange that to get the mean value theorem

oh mb

missing an absolute value at the end

$|f(b)|>N(x_2-x_1)\left|f(\frac{x_1+x_2}{2})\right|$

[warm shale]

Toby

[timid silo]

thats the contrapositive

(f' bounded => f bounded) <=> (f unbounded => f' unbounded)

note that (x_2-x_1) is certainly bigger than b-(x_1+x+2)/2

oh wait typo again ;-;

there should be a plus sign somewhere

$|f(b)|>N(x_2-x_1)+\left|f(\frac{x_1+x_2}{2})\right|$

[warm shale]

Toby

[timid silo]

sorry

[timid silo]

move the |f(...)| on the left, use triangle inequality, then divide by this ^

[warm shale]

HeyHey

[timid silo]

why did you split up f((x_1+x_2)/2)?

we cant do that here since the things is inside f

I meant something like this (let $a=(x_1+x_2)/2$): $|f(b)-f(a)|\geq \left||f(b)|-|f(a)|\right|$

[warm shale]

Toby

[timid silo]

then divide by x_2-x_1 and compare that to b-a

yeah

and the left is less than f'(c) (up to a sign) for some c

and the right is just N

so we've constructed c such that f'(c)>N

yup you can

yup

because N was arbitrary, then f' is arbitrarily large, so unbounded

yw :)

[obtuse pebble]

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[hot gull]

Hi, I'm working on an exercise which asks to compute the 4th degree MacLaurin expansion of this function:

[hot gull]

$log(1-x^2)-2cosx$

[warm shale]

Umma.Gumma

[hot gull]

which I did, and it seems to be:

$-2 -\frac{7}{12}x^4+o(x^4)$

[warm shale]

Umma.Gumma

[hot gull]

now, I'm asked - without doing further computations - to determine whether f(x) has extrema points, for x=0

how can this be understood from an expansion?

[kind hawk]

how do you compute extrema points

[hot gull]

starting from the roots of f'(x)=0 etc etc

[kind hawk]

good. what would be the derivative of that expansion

[hot gull]

$-\frac{7}{3}x^3$

[warm shale]

Umma.Gumma

[kind hawk]

plus some higher degree terms.

and is x=0 a zero of that?

[hot gull]

you mean o(x^4) ?

[hot gull]

yes it is

[kind hawk]

good. now what about a more general expansion $a_0 + a_1 x+ a_2x^2+a_3x^3+\ldots$

[warm shale]

Denascite

[kind hawk]

when is x=0 a zero of the derivative of that

[hot gull]

I'd say always

[kind hawk]

no

what is the derivative of that expansion

[hot gull]

you're right, we end up with a_1

all other terms are canceled

for x=0

[kind hawk]

good. so x=0 is a zero if a1=0

but for this we dont actually have to compute anything, we just have to look at the expansion

and this is what I assume they mean with "without doing further computations"

[hot gull]

to be precise, it says "without doing further computations on the function"

[hot gull]

but what if there's no a_1 in the expansion? like the first one I wrote

I'd really like to grasp this concept

[kind hawk]

well then a_1=0

[hot gull]

so we can conclude that f'(x) is 0, therefore there an extrema point at 0?

[kind hawk]

yeah

hmm although technically we would need to exclude that this is a saddle point

which we can for example do by noticing that f is an even function

[hot gull]

ok so it's symmetric to the y axis

[kind hawk]

yes

[hot gull]

ok, will go through this again

thanks a lot, I really appreciated

.close

[obtuse pebble]

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[ivory cargo]

Hello 🙂 i would love some help going tru some tasks i made, to check if they are correct. Firstly

[ivory cargo]

follows into this

but should i say that the limit is 0 for r=1/2 or how do i finish the argument?

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

<@&286206848099549185>

.close

[obtuse pebble]

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[leaden shoal]

why does it end at ar^n-1? and not ar*n?

[stable rain]

hm

well

consider n=1

[leaden shoal]

for this one

[stable rain]

yes

consider n=1

[leaden shoal]

ok

it would be a

[stable rain]

yes

and so

whats the r term

[leaden shoal]

an integer

[stable rain]

no

like

its

r^(n-1)

which in this case is

r^(1-1) = r^0=1

so we get a * 1 = 1

aka the first term

[leaden shoal]

what does r term mean?

the common ratio?

[stable rain]

in geo series

its the

um i guesss

[leaden shoal]

the number u mutliply to get to the next term

[stable rain]

ye

that

[leaden shoal]

but whats this?

why to the power of n-1?

[stable rain]

the last term?

well

just consider how itd be

[leaden shoal]

isnt the last term ar^n?

[stable rain]

for

n=1

whats

sn

[leaden shoal]

s1

[stable rain]

if the last term was

[stable rain]

this

[leaden shoal]

ar^1

[stable rain]

ok

so

whats s1

[stable rain]

is this ur ans?

[leaden shoal]

it would be a

[stable rain]

ok

so

what happened to

ar^n

namely

ar^1

its not thr right

[leaden shoal]

whats thr?

[stable rain]

there

[leaden shoal]

ye thats strange

[stable rain]

no its

not

whats the formula

for the nth term

[leaden shoal]

im having trouble understanding what n actually means

[stable rain]

its just a number

saying how many terms there r

so like

n=3 means

we look at 3 terms

so

a3 would be

[leaden shoal]

n is the number of terms starting from the first term to the nth number of term

[stable rain]

3rd term

s3 would be

sum from 1st to 3rd term

[leaden shoal]

ok

hmm so why is the last term not ar^n?

[stable rain]

whats

the formula for the nth term

[wild swallow]

because its just a completely arbitrary choice

[stable rain]

of a_n

[wild swallow]

and it doesnt matter

[stable rain]

ye

u can just

shift the defn of what a is

anyway

[wild swallow]

they want it to be ar^n because the formula for the finite sum looks nicer

[stable rain]

i like doing that tbh

[wild swallow]

whereas if you end with ar^n

then you get a slightly different formula

it really doesnt matter

[stable rain]

oh

but

i wonder

how did u get that rhs exp

:O

[wild swallow]

🪄

[leaden shoal]

ye like this

hmm so how was the formula derived?

[stable rain]

my

google skilll

lv 2

[wild swallow]

sure

[stable rain]

ITS BLUR

NOOO

[wild swallow]

its mouldy

[stable rain]

ok thi is less blur

imagine

it was a svg instead

not as blur lol

[leaden shoal]

hmm this is quite hard to understand

[stable rain]

ok

which bit

[leaden shoal]

ive forgotten what ar^k mean

is it the rule?

[stable rain]

well

its summation

its just the kth term

[leaden shoal]

0<=k<=n

is this true?

[stable rain]

um

do uk whats summation

[leaden shoal]

ye i learnt it months ago

[stable rain]

just

treat it as

this thing

[leaden shoal]

if k=3 then in this case it would be ar^2

or ar^3?

[stable rain]

well

in this eg

[stable rain]

and here

the last term wld be

[stable rain]

this

[leaden shoal]

the last term will only be ar^3 if n=3 not k=3

k is the starting and n is the ending if im not mistaken

[stable rain]

well

in this case

we end with

ar^n

k is a dummy var

[leaden shoal]

and we start at ar^0

[stable rain]

yes

so

in this case

for some n

weve n+1 terms

[leaden shoal]

im confused in this part

why n+1?

[stable rain]

..

whats confusing abt it

ok

now

thats a btr q

[leaden shoal]

where did n+1 come from

[stable rain]

ye just

its written diff from b4

if in doubt

just

look at the base case

always

consider

n=1

well actually here the base case is

n=0

but well

n=1 can also work to show u how many terms there are

so

whats the sum look like

when n=1

[leaden shoal]

so it starts at a_0

[stable rain]

yes

kinda?

depends on how u define it

o well

just

define it how u like

as long as it works

[leaden shoal]

ok

[stable rain]

ok imma

dip

this

isnt fun

lol

[leaden shoal]

ok

[leaden shoal]

i am still a little confused by this, because it ends at the second last term

[leaden shoal]

ah thats very true!

i tried deriving it and i got it

.close

[obtuse pebble]

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[finite matrix]

help me

[finite matrix]

LCM(4,7,9) is 252
least 4 digit multiple of this LCM is 1008

i cant get any further

neither the remainders are same, nor the difference between divisors and remainders are same

[royal basin]

this sounds like a chinese remainder theorem problem

[finite matrix]

OH WWAIT

ohh congruence modulo

ahh why i didnt go for that procedure

[royal basin]

you need to solve the system of congruences $$\begin{cases} x \equiv 2 \pmod{7} \ x \equiv 7 \pmod{9} \ x \equiv 3 \pmod{4} \end{cases}$$ and then take its smallest solution that is between 1000 and 9999

[finite matrix]

i was trying to ur common sense

[warm shale]

Ann

[finite matrix]

hmm

@royal basin thanks
i completely forgot about CRT

i got it, answer is 1087 => 1+0+8+7 = 16

@royal basin how do i close

?close

/close

!close

.close

[obtuse pebble]

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[sage bluff]

hey, can someone help me with c) please?

[obtuse pebble]

@sage bluff Has your question been resolved?

[kind hawk]

so you need to find a bijection between all the equivalence classes and the integers

in particular you need to find a bijection which sends R_(a,b) to a-b

or rather, use that as your function and show that it is indeed a bijection

and that it is well defined

[sage bluff]

yeah got it

how do you do this?

[kind hawk]

we want that $h\circ f = g$. that means $h(f(a,b))=g(a,b)$. Using the definition of $f$ and $g$, we want $h(R_{(a,b)}) = a-b$

yes?

[warm shale]

Denascite

[sage bluff]

yes

how do we find such h and prove its bijectivity?

[kind hawk]

well we just define h that way

h(R_(a,b))=a-b

[sage bluff]

but why is that bijective?

[kind hawk]

well thats what you have to show

using b, how can we write each equivalence class

[sage bluff]

idk I did b one week ago and I don't remember...

[obtuse pebble]

@sage bluff Has your question been resolved?

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[ocean snow]

Hey, I have a question regarding mathematics in computing and trigonometry. I will be starting a course in computer science in university and I would rly appreciate some resources and books to read and study from. There’s lot to study and I’m quite overwhelmed on where to start on so some help please i would appreciate

[limber quartz]

Do you have less experience in trig or in programming?

[ocean snow]

Some experience in programming i Finished a course and starting in uni next year

But I’m still beginner

Trig not a lot of experience i needed help w friends

[limber quartz]

Are you comfortable with for and while loops? And recursion?

[ocean snow]

I recognise the terminology but not 100%

I wanna study up

[limber quartz]

Ok, fill in that gap

[ocean snow]

Would you have a book that contains everything I need

[limber quartz]

Hackerank is a good resource

[ocean snow]

Ohh ok ok

[limber quartz]

Do you know what language you will use

[ocean snow]

Python or either c#

[limber quartz]

Not MATLAB?

[ocean snow]

I heard Java will be taught as well

[ocean snow]

Not sure

[limber quartz]

Be sure about the language

[ocean snow]

I’m going off research and what my friends say

[limber quartz]

You make it sound like the course is going to be a mathematical programming course, is that not the case?

[ocean snow]

Isnr coding Lang quite easy once you understand one ? You can adapt to a different

[ocean snow]

They pile on the maths to weed out students

I wanna prepare for it

[limber quartz]

Please be clear

Is this like an Optimization Theory or Numerical Methods course?

Or not

Those are graduate level courses

[ocean snow]

Like if u fully understand C sharp starting in a different coding language is easier

[ocean snow]

It’s just computer science course 😭😭

[limber quartz]

Ok, you are just starting then

Don't worry about the math popping up seriously in your computer course

It will not.

[ocean snow]

Ya, just trying to study up on maths and I’m also using a guide to fully understand C sharp

[ocean snow]

They pile on trigonometry students are weeded out 😭

And it’s my weak point

[limber quartz]

I can't imagine how they plan to do that in C#

Python has modules for plotting, I don't know about C# (but it could)

[ocean snow]

I’m just preparing for every possibility

I can’t fail this course

[limber quartz]

I just can't imagine them giving you difficult math

[ocean snow]

Soo ya, any maths relating to computing and trigonometry would be immense help

[limber quartz]

Airquotes "Difficult"

[ocean snow]

Well I’m not sure either

Just preparing and building my maths knowledge

[limber quartz]

There is no math you will see in a first year computer science course that you can't read about in 5 minutes on Google

[ocean snow]

There’s so much I don’t know where to start, I wish there was a book I can read

Just a decent book would be everything i need 😭

[limber quartz]

Ok ok

[ocean snow]

I have hackerrank open so thanks for that btw

Will start on that

[limber quartz]

Rent any pre-calculus text and you will be miles ahead of where you need to be in terms of math

The focus of courses like that IS NOT MATH

[ocean snow]

👀

Ohh

Alr gonna try Google it

[limber quartz]

paulsonlinemathnotes
Is an excellent online review resource for calculus...which will NEVER come up

[ocean snow]

This looks good

[limber quartz]

Uh-huh.

[ocean snow]

Oh nice thank you sm

Getting that up rn

[limber quartz]

You really want to spend more of your time being familiar with using the language

[ocean snow]

Finally something to grind and read on

[ocean snow]

Mhm

I mainly use C sharp rn

Trying to master it

[limber quartz]

That's OK, it takes time

If you are also taking a math course, study that as well

[ocean snow]

It will def come up this book will help a lot

🙏🏼

[limber quartz]

Although you should already be comfortable with pre-calculus by the time you start uni

[ocean snow]

I have about 10 months

[limber quartz]

Pre-calc (college alg 2 iirc) is required

To get in

[ocean snow]

To really study and grind rn

[ocean snow]

Ohh ok

I screen shot this to remember study on it

Alrrr

[limber quartz]

What is the highest math you finished?

[ocean snow]

GCSE 😭

Maths is weak point so I came here for advice

Glad I asked

[limber quartz]

Ok, yeah

[ocean snow]

Is there anything else I should study on

[limber quartz]

Math is honestly your most important thing to study now tbh

Since you have a math class coming up, right?

And being solid in loops and recursion

hackerrank has you there

[ocean snow]

Yess

Alright

Now I just need to study

How do I close this ?

Also thnx 🙏🏼

[obtuse pebble]

@ocean snow Has your question been resolved?

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[eternal stream]

not asking for a problem to be solved but basically, here is the markscheme for a paper

[eternal stream]

my final answer came out to be y=1/4x+31/4

I just want to know if the examiner would give a tick for the correct answer

i checked it is indeed correct

Side note: I'm taking Cambridge AS level maths

[obtuse pebble]

@eternal stream Has your question been resolved?

[eternal stream]

Literally a yes or no question

<@&286206848099549185>

nvm

.close

[obtuse pebble]

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[somber apex]

hi! i need help with these questions, i have not taken trig before and im lost
i do not understand what they are asking and how to illustrate it like they asked

[somber apex]

^ this is technical math btw

[obtuse pebble]

@somber apex Has your question been resolved?

[lean trail]

Which question do you need help with?

[somber apex]

17-20, but i think if i figure out 17 i shouldbe set for the rest

[lean trail]

Do you know how to draw angles in the cartesian plane?

[somber apex]

nope </3

gonna be real w u i dont know what that is

[lean trail]

The 2D plane which is formed by the intersection of x and y axis

If you have ever drawn a graph, you already know this

[somber apex]

ohhh

i just searched it up, i know what it is i just never knew the name HAHA

[lean trail]

Yeah, I figured

Take the x axis as the base of one ray and draw another ray such that the angle between them is 150 degree

[somber apex]

ahh

wait hold on

would it look like this?

[lean trail]

not quite

Let the point of intersection of your ray be the origin of the plane

[somber apex]

okay

so i need to draw two rays from the origin poijt thats 150 degrees but not make it a truange?

but uhm also! for c in the instructions, how would i find the other angles (one pos and one neg) and whattt does coterminal mean

[obtuse pebble]

Channel closed

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[lean trail]

You don't really need the third line to make it a triangle

[somber apex]

so would it just look like this

.reopen

[obtuse pebble]

✅

[lean trail]

one ray needs to coincide with the x-axis

[somber apex]

wdym

likw

[lean trail]

coterminal means that both the rays of both angles coincide

[lean trail]

coincide means overlap

[somber apex]

the line must be on rhe x axis

ohh okok

[lean trail]

like the angles in this image are coterminal

[lean trail]

yup

[somber apex]

so on the graph id illustrate it likw this

[lean trail]

Yes!

[somber apex]

OKAY!! THANK YOU

so for the

[lean trail]

This image shows 45 degree in the standard form

It has the arrow of rotation

And it has shown one positive and one negative coterminal angle

for 45 degree

[somber apex]

c part it would look like this or did i get it wrong?

if c is the negative where would the positive coterminal be

oh wait now i see where the posiitve co teeminal would bw

like this?

and ro calculate the degrees for rhe negative it would be

150°-360°
and for rhe positive it would bw

150°+360°

righr?

[lean trail]

Yes

[somber apex]

YESSS

[lean trail]

Also correct

[somber apex]

and rhe ways to calculate rhe degrees r right too

[lean trail]

Good job

[somber apex]

YAY

thank u so so mich

[lean trail]

Happy to help ^^

[somber apex]

:D

.closw

.close

[obtuse pebble]

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[open shale]

It is given that p and q are supplementary angles such that p > q. If the difference between p and q is 52 degree, calculate the values of p and q.

[high lily]

what happened to your other channel

[open shale]

i .close

[high lily]

why

[open shale]

cause its not responding

[high lily]

someone responded...

and you should wait patiently even if someone does not

[open shale]

ok

[thick oracle]

hi

if p and q are supplementary

then p+q equals what?

[high lily]

if you had questions about the responses, clearly explain which part you had issue with

[open shale]

I have issue at the diffrence between p and q is 52 degree part

[high lily]

do you know what difference means

[open shale]

yes

[high lily]

what's the difference between 2 and 5?

[open shale]

2-5=-3

[high lily]

not quite

[open shale]

5-2=3

[high lily]

yes, you'd subtract the smaller from the larger number

similarly what's the expression for the difference between p and q (i don't want you to give the numerical value of 52 for this)

[open shale]

p-q

[high lily]

yes

and you are told this has the numerical value of 52 degrees

which gets you an equation

[open shale]

ok

[open shale]

it says the difference between p and q is 52 and idk how to get p and q

<@&286206848099549185>

[royal basin]

it also says p and q are supplementary angles, and someone else here asked you if you knew what that meant, but you conveniently ignored their question

[open shale]

its 180 degree

[royal basin]

too imprecise

can you translate the sentence "p and q are supplementary angles" into an equation?

[open shale]

p+q=180 degree

[royal basin]

ok

great

so you now have two equations:
p + q = 180
p - q = 52

are you able to find the values of p and q from here?

[open shale]

yes

[open shale]

but why will we get 2 equations

[royal basin]

we have two pieces of information

one is that p and q are supplementary angles, the other is that their difference is 52 degrees

each of these gets translated into an equation

does this answer your question?

[open shale]

yes

can i type .close now?

[royal basin]

if you have nothing else left to ask then yes you can

[open shale]

ok

.close

[obtuse pebble]

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[empty bison]

Hi guys, I don't know if this counts as maths but I'm gonna explain my problem anyway
What does it mean that an analogic signal is infinite and a digital signal is not?

[empty bison]

I mean the digital signal too is infinite, 'cose It can assume infinite values and can last infinite time

The analogic signal is infinite too, cuz it can assume infinite values and last infinite time

So I don't really understand why analogic is called infinite and digital is not

[alpine raven]

isnt digital signal a sequence of 0 and 1 ?

[obtuse pebble]

@empty bison Has your question been resolved?

[empty bison]

No

Thats binary

OH

WAIT

I understood

Analogic --> it can assume infinite values in infinite time
Digital --> it can assume a defined number of values in infinite time

.close

[obtuse pebble]

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[minor belfry]

not sure how to do this

<@&286206848099549185>

[chrome mesa]

<@&268886789983436800>

[chrome mesa]

look at the highest power in the numerator and denominator (when expanded)

[minor belfry]

bruh

dammit i have to eexpand

i was hoping not to do

[chrome mesa]

you don't have to

you can tell visually

for example whats the highest power if you had

$$(x+1)^{1000}$$

[warm shale]

Tronsi

[minor belfry]

uhh

it would be positive inf

but if its odd number its neg inf

[chrome mesa]

huh? Let's just go back to the original question here, what's the highest power in the numerator and the denominator?

remember, you only need to look at the term that grows the fastest

[obtuse pebble]

@minor belfry Has your question been resolved?

[minor belfry]

x^9

[chrome mesa]

yeah

[minor belfry]

so

thats neg inf

it approaches?

[chrome mesa]

yes

[minor belfry]

ok well then

thats not too hard

ty

.close

[obtuse pebble]

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[proven karma]

How does X here equal 3/5?

I tried doing it myself and got x = -3

i rewrote the base 8 as 2^3 and got

2^2x = 2^3-3x

equated 2x = 3-3x

substracted 3 from both sides

2x - 3 = -3x

subtracted 2x from both sides

-3 = -5x

oh

oh my god

.close

[obtuse pebble]

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[alpine raven]

what is the common factor ?

[median trench]

Wdym?

[alpine raven]

You know that to factor an expression, you find the common factor right ?

[median trench]

Yh

How can I find it

If that number is the power of 2x

[alpine raven]

Hint : 6x = 2x + 4x

[median trench]

I didn't get it

I mean

I understand

But idk what to answer

[alpine raven]

wdym you dont know what to answer ?

You have factored things like :
ab + ac = a(b+c)
right ?

the same concept here

just you need to know how to work with exponent

[alpine raven]

thats why I gave you this hint

[median trench]

Allright one sec

I'll think about it more

6x is the power right?

[alpine raven]

yes

[median trench]

5^3x + 5^3x?

Noo no

I mean times

[alpine raven]

$5^{3x} \times 5^{3x} = 5^{3x+3x} = 5^{6x}$

[warm shale]

Ｈｅｒｅｌｓ

[alpine raven]

yea but you have 5^{2x} on the left

[median trench]

I can do the same

[tidal geode]

hi

[median trench]

5^2 x 5

2x*

[median trench]

Hi

[alpine raven]

?

[tidal geode]

yep

[median trench]

Idk I just can't understand that

[median trench]

We did that first for the second one

And the first one I tried to do the same

[tidal geode]

im having trouble with this

[alpine raven]

#❓how-to-get-help

[alpine raven]

but did you just look at the hint I gave ?

[median trench]

You can go and ask on available channel

[tidal geode]

yes

my bad im new on this server

[median trench]

It's OK, u will learn

[median trench]

Yh I did, but didn't understand why 6x

[alpine raven]

because 5^{6x} has 5^{2x} as a factor

[median trench]

So instead dor 3x

And 3x

So I have to take 5^4x x 5^2x?

[alpine raven]

Yes

Like I said, try to look at both terms their common factor

the more obvious one

[median trench]

And it will be 5^2x_5^2x x 5^4x?

[alpine raven]

eh, you have never done a factoring exercise before or...

you was almost there and you did it bad at the end

[median trench]

Nvm

Sorry for bothering u

[alpine raven]

5^{2x} (1-5^{4x})
is what you should get

[median trench]

Thanks anyway

.close

[obtuse pebble]

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[timid silo]

yo can someone help explain this

[cedar lichen]

What about it is confusing

[timid silo]

the chain rule bit

the change of variable

[cedar lichen]

We let u = x²

And the derivative of f(u) is f'(u) * u'

[timid silo]

how com the variable doesnt change inside

[cedar lichen]

Wdym

[tardy epoch]

x is the variable being substituted, not t

[timid silo]

oh

thx

[timid silo]

riemann sum is confusing asf

i was cruising until integration

@cedar lichen @tardy epoch

can ya help with this too?

[cedar lichen]

Whats the problem

[timid silo]

i dont get the "let G_b also be an anti derivative" and the last line

:/

how do u interpret F(X) as 2 sums, isnt F(X) a function in x

[cedar lichen]

G(x) is being defined as ∫f(t) dt from a to x, right?

[timid silo]

yeah

[cedar lichen]

So G'(x) = f(x) by the FTC, right?

[timid silo]

yeah

[cedar lichen]

So it's an antiderivative of f by the definition of an antiderivative

[timid silo]

yeah

[cedar lichen]

So what's your next question

[timid silo]

btw persona 5 royal just came out on game pass right 👀

[cedar lichen]

Idk lol

[timid silo]

so I get that G(X) must be an antiderative and so is F(X)

meaning their derivatives are =

and also thatthat means they are parallel functions

ie F(X)=G(X)+C

so that's the 2nd last step

but i dont get the last step at all.

how they get F(B)-F(A) and then (G(B)+C)

is it related to the prior step or supposed to be equivalent?

[cedar lichen]

That's not actually true in general buttt it's true like 99% of the time

What are you trying to prove btw?

[timid silo]

I believe this

[cedar lichen]

So we're trying to prove F(b) - F(a) is actually equal to ∫f(x) dx from a to b

Well, what is F(b) - F(a), since we know F(x) = G(x) + C?

[timid silo]

Actually can we get back to this, why is it true?

[cedar lichen]

What was the theorem in the previous slides

[timid silo]

Actually nvm I think I get it, because the anti derivative is not unique right

[cedar lichen]

Yea

[timid silo]

Idk tbh

[cedar lichen]

(Although F(x) = G(x) + C iff F and G are continuous)

[timid silo]

Yeah

[cedar lichen]

Anyways, F(x) = G(x) + C

So what's F(b)

[timid silo]

G(B)+C 👀