help-23

[lean otter]

Gotta find domain and range

I think it’s
Domain: [-infiniti,3]
Range: [-infinite,3]

It says I’m wrong tho

You can't put close bracket on infinity

[glass ruin]

make sure u enclose the infinity with parenthesis

[lean otter]

Is it right then?

(-Infiniti, 3], (-Infiniti,3]

[glass ruin]

yes

[lean otter]

What the heck

Range is [-6,3] and Domain is (-Infiniti,Infiniti)

Seems correct (I don't know much about them)

Nope it’s not

[glass ruin]

range is going to be infinite too since the arrows also point up and down

[lean otter]

Domain : set of real mumbers

So Sams answer for both

Same

[glass ruin]

yes

[lean otter]

Agree

Ok thank you

I got it figured out now

Thanks

.close

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[low galleon]

can someone please help explain this:

[low galleon]

this is what im trying right now

nvm i was in degrees 💀

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[crude star]

do I use induction for this

[fleet condor]

... assume there are not two integers that are consecutive, such that the subset is 1, 3, ...

[fair hound]

pigeonhole principle

[crude star]

brah

we never even did that in this course why does the prof expect us to kno it

[crude star]

im confused so I prove via contradiction?

[crude star]

yo if possible could you explain pigeonhole principle. I read about it last year in a different course and never rlly understood it

[fair hound]

if you have 8 socks and 7 drawers, you will necessarily have a drawer with at least 2 socks in it, since there are more socks that drawers

[fleet condor]

you can

[fair hound]

the pigeonhole principle generalises this simple fact

[crude star]

ah

[fleet condor]

pidgeonhole is like slightly less direct I think but it also works well

[crude star]

i dont see how that would help here doe

[crude star]

alright brb imma think abt it

[fair hound]

because when you pick the n+1 elements

you will necessarily pick two consecutive

because of pigeonhole principle

you don't have enough "drawers"

[crude star]

right but why cant I pick for instance

1 and 3

instead of 1 and 2

im still fitting in 2 elements

[fair hound]

because you have to pick n+1 elements

not just 2

[crude star]

ye I mean like

I could still pick every other element right

[fair hound]

no, that's the point

you will run out of elements not consecutive to another

[fleet condor]

this is the contradiction argument

you show that picking every other element is the max you can get, then show that you dont get enough

[crude star]

alright

so I could say

[fair hound]

in fact, what Bunny uses is the pigeonhole principle
he takes 1 3 5 ... as a drawer of size n, then another element in the other drawer must be picked to have n+1 elements

it will be consecutive to one of the odd number, which concludes bunny proof

[crude star]

fuck I rlly dont understand that proof

Why cant I pick only odd numbers

or only even numbers

oh

shit

[fair hound]

because there are only n of them

[crude star]

fuck

if S only consists of odd or even numbers then its size would be n

which is a contradiction

so there will always be that one extra element

which will come right after

[fair hound]

but you can pair up in a lot of ways
like take 1 2
3 4
...
as drawers
then there are n drawers and you have n+1 elements to pick
so two of the same drawer, you have one of these pairs of consecutive

[crude star]

one of the elments

right so if I were to pick

1 and 3

then the next would be 4

right

or if 2 and 4 then next would be 5

for n + 1 size

[fair hound]

you could also pick 2 if you want too, the point is that you have to do so

because you don't have enough numbers to avoid that

[crude star]

wait can u proofread what I got so far

could i literally j not say this

and end it

[fair hound]

hmm not proven

you're missing steps

[crude star]

before that line?

[fair hound]

you didn't prove that line

like the size of the subset matters, but you didn't use it in your argument

[crude star]

because if that wasnt the case

then it would have atleast 2 consecutive numbers

[fair hound]

yeah but because the subset is of size n+1

it you could take a subset of size n, you could just pick all odd without problems

the size matters

the whole point is that since you're taking n+1 numbers, you will necessarily have 2 consecutive

[crude star]

wait so

do I even need that line in my proof

abt even and odd numbers

because I only used it to prove that the size would be n if thats rthe case

[fair hound]

if you want to do a proof by contradiction, you have to "build" what you're doing

like proving why you couldn't have both odd and even numbers

[crude star]

oh

[fair hound]

you can't just say it you have to prove it

[crude star]

shit

I mean

we could still have size n with both odd and even numbers could we not?

oh shit im stupid

i forgot abt my assumption

that we dont have 2 consec integers

[fair hound]

yeah

[crude star]

ok one min brb imma reconstruct

my proof

[fair hound]

what happens is that

[crude star]

is this good

but yo I was thinking

about one more thing

u see that for n + 1 we need one more element

then lets say

[fair hound]

again you're just saying it

not proving it

[crude star]

our set consisted of only evens could I not just add another even for n + 1

[fair hound]

no, since there are only n evens

[crude star]

uh how would I prove it then

oh

so I could use that to prove that line right

that since we only have n even/odd therefore there will always be an extra element which will result the set in having 2 consecutive integers

[fair hound]

no because the case where you take n-2 even and 3 odd is not clear for example

you need a proof that this case too would have consecutive integers

[crude star]

shit

[fair hound]

either you make a constructive proof that shows that's necessary

either you use pigeonhole principle directly which is easier tbh

[crude star]

I dont understand how pigeonhole principle would help proving it doe

one sec lemme reread what u said earlier

so basically

since the size is larger then amount of even or odds there will always be 2 consecutive numbers?

[fair hound]

you pair up all numbers like that:
1 with 2
3 with 4
5 with 6
...
2n-1 with 2n
that's n pairs
but since you have to pick n+1 elements
you will necessarily pick two of the same pair

so two consecutive

that's the easiest proof I can think of really

[crude star]

How is it so sure that it will pick two of the same pair

for n + 1 elements

[fair hound]

because there are only n pairs

[crude star]

right

so for + 1

couldnt I pick any number

[fair hound]

yes, but it will be in one of the pairs

if you have 5 socks but 4 drawers, at least 2 socks are in the same drawer
likewise, if you have n pairs but n+1 elements to pick, at least 2 elements will be in the same pair

[crude star]

oh

Alr one sec brb

imma write it out

@fair hound

[fair hound]

you're mistaking several things

|S| = n+1 doesn't imply that there are n pairs

[crude star]

but u said we have n pairs right

[fair hound]

there are n pairs in the base set because there are 2n elements

so when you pair them up, that makes n pairs

then, you need to take n+1 elements

but there are only n pairs

so you have to take two of the same pair

also, you should specify how you construct the pairs such that they contains indeed consecutive integers

ie 1 with 2, 3 with 4, etc

let's take the case n = 3 as an example
then 2n = 6
the set is {1, 2, 3, 4, 5, 6}
you want a subset of size 4
you pair 1 with 2
3 with 4
5 with 6
since you have to take 4 elements, you will necessarily pick both of one of these 3 pairs, and therefore have two consecutive integers

[crude star]

OMFG

IT MAKES SENSE

[fair hound]

opposite, that's the pairs that play role of the holes here

and the elements that are the "pigeons"

[crude star]

oh i thought hole is amount of needed elements

so we have n holes and n+1 pigeons

[fair hound]

yeah

[crude star]

other thn that is everything else good

[fair hound]

so it's not a hole that has two pairs

[crude star]

ye lol mb

just fixed it

[fair hound]

it's that there is a pair where you have to take both elements

since there is more element to pick than pairs

[fair hound]

we need to fill n holes* with n+1 pigeons

and the other lines after have the same problem

you're not fully feeling the point yet

there are n pairs and n+1 elements to pick

so there is at least one pair where you take both elements

but yeah the conclusion is good after that

the point being that when you take two elements of one of your constructed pairs

they are consecutive

but the analogy is
n+1 pigeons, n holes -> at least 2 pigeons in the same hole
n+1 elements, n pairs -> at least 2 elements of the same pair

[crude star]

omg my pdf didnt update properly mb

[crude star]

basically

u pick n pairs

u are out of pairs

so you would have to go back

and pick a pair

that is alrdy in the set

which will result in 2 consecutive integers

[fair hound]

you pick one element per pair, it's not enough, so you have to take the 2nd element in one of your pairs

and you have 2 consecutive

yeah

[crude star]

alr got it tysm

bless u

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[pine thicket]

It’s to prove the inequality below

The first characters mean if

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@pine thicket Has your question been resolved?

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[lean otter]

\dm
There is a result for context-free languages analogous to the pumping lemma for regular sets. Suppose that $\map L G$ is the language recognised by a context-free language $G$. This result states that there is a constant $N$ such that if $z$ is a word in $\map L G$ with $\map l z \ge N$, then $z$ can be written as $uvwxy$, where $\map l{vwx} \le N, \map l{vx} \ge 1$, and $uv^i w x^i y$ belongs to to $\map L G$ for $i \in \N$. Use this result to show that there is no context-free grammar $G$ with $\map L G = \set{0^n 1^n 2^n \where n \in \N_0}$

[lean otter]

I have been eyeballing this for quite a bit, but i really really don't see how im meant to prove this at all yet

Maybe a proof by contradiction?

[devout shale]

proof by contradiction is not possible here

[lean otter]

why not

[devout shale]

idk I just wanted to contradict you

[lean otter]

LMAO

[cold aurora]

When i did it, i remember doing sort of constructive proof by using some graph theory stuff and then showing that a recurrence like that exists.

[lean otter]

i will keep thinking

[safe radish]

@lean otter Has your question been resolved?

[cold aurora]

I'll give you a little hint - ||parts of string which you can pump up or down - they must be leading to the same non-terminal again after deriving v and x such that non-terminal is between those v and x - which then facilitates pumping.||

Second hint - ||Remember how in DFA, you prove pumping lemma by making use of pigeon hole principle to say that at least one state repeats, similarly - by choosing an appropriate N, you can guarantee that some non-terminal in derivation tree of z must have repeated.||

@lean otter

[lean otter]

wait oh

you can do it similarly to the one in DFA?

oh this makes a whole lot more sense

[cold aurora]

By some tweaking.

[lean otter]

Okay i think i got the main idea now @cold aurora

i can proceed with the writing

thank you

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[lilac cape]

Question about infinity

[solid shell]

Sample space can be infinite, both countably and uncountably

So in that situation I presume it would be, yeah

[quiet juniper]

Did he delete a message or something

[solid shell]

Yeah lol sorry

[quiet juniper]

lol

[solid shell]

Nah I read their mind

[quiet juniper]

[quasi bison]

do you still have a question?

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@lilac cape Has your question been resolved?

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[north thunder]

im wondering why x and 3 are only multiplied once

[north thunder]

wouldnt you have to multiply x by 2x and -1

as well as multiply 3 to 2x and -1

[quasi bison]

do you think (a * b) * c = ac * bc?

[north thunder]

is it not

oops wrong photo

i actually dont understand this

excuse me my friends?

[ionic blaze]

it's not

(3 * 2) * 4 = 24, not equal to 12 * 8 or 96\

[north thunder]

oh right

thanks

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[wraith prism]

5.cos theta

What is the connection from sin to cos?

[dense prism]

Do you mean to ask sin²x+cos²x=1?

[wraith prism]

My approach here is
25 sin theta= 5

[dense prism]

Yes

[wraith prism]

Ohh trigo

Sin theta =1/5

[dense prism]

And then you just need to get cos theta

[wraith prism]

Cis theta =√24/5

[dense prism]

For the dot product

[dense prism]

Yes

[wraith prism]

√24/5 ×25

5√24

10√6

Yoo

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[grim minnow]

Find the value of 𝑝 for which the equation (𝑝−2)𝑥2+2(2𝑝−3)𝑥+5𝑝−6=0 has equal roots?

[flat frigate]

Dyssrupt

[calm bridge]

do you know when a quadratic has equal roots?

[grim minnow]

b^2-4ac = 0

right?

[calm bridge]

yeah

just apply it

[grim minnow]

im not able to

theres an extra 5p there

which my brain cant process

[fair hound]

the 5p doesn't have any x, it's just part of the constant term

like the -6

[grim minnow]

is the 5p - 6 supposed to be taken in c?

[fair hound]

yes

[calm bridge]

yes

[grim minnow]

ah i see

thx guys

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[mint veldt]

this proof doesn't seem right

I see what they did, it's clear.

but it just looks so trivial

I don't think any steps were skipped either right?

It looks like all he did was rewrite the lhs and rhs and that was it

[lean otter]

How do i prove that 2^n+3^n < 4^n is true for all n equal or greater than 2 with induction

[mint veldt]

.close

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[mystic vale]

https://youtu.be/LpW6zanbSf8?t=228

[mystic vale]

when he says "for mathematical induction you have that it follows for all values of n "

can anyone explain it in simple terms? first time i encounter this

[quasi bison]

timestamp?

[mystic vale]

3:48 if you click on the video i starts at right time

[quasi bison]

mathematical induction is a method of proving statements of the form "for all natural numbers n, ..."

it works by proving the base case -- verifying directly that your statements holds for, say, 0 or 1

and the inductive step -- showing that if your statement holds for n then it also holds for n+1

i like to imagine it as a stairway to heaven

to reach heaven you prove that you can climb the staircase as high as you want

base case = proving you can climb the first step

inductive step = proving you can get from any step to the next

[mystic vale]

ok

i dont see where he's proving the first step here...

THANKS for explaning

[quasi bison]

probably some earlier point

[mystic vale]

ok ill rewatch it

thank you

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[slate siren]

Hey math people

[slate siren]

I know how to compute the cross product - but I'm at a bit of a loss regarding "vector form" vs "component form"

[plucky elk]

Do you have a problem

[slate siren]

Yes

What is the difference between a cross product's vector form and its component form

[plucky elk]

<@&268886789983436800>

[slate siren]

.close

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[plucky elk]

The calculation is the same

[slate siren]

Hmmmm

Hehe, alrighty

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[tiny frost]

Hey im just looking for a nudge in the right direction on how to solve this

[tiny frost]

$$2x-24sq{x}+54=0$$

[flat frigate]

Totalani

[cold aurora]

$$2x-24\sqrt{x}+54=0$$

[quasi bison]

sq{x}?

[flat frigate]

Enemagneto

[quasi bison]

did you mean sqrt(x)?

[calm bridge]

is sq sqroot or square?

[tiny frost]

that sorry I missed the \

[quasi bison]

ok here's your nudge:

substitute t := sqrt(x)

[tiny frost]

I assume you start with divinding by 2?

[quasi bison]

can do that before or after my suggestion

doesnt matter when

[tiny frost]

I dont usually do math in english so forgive me if I ask dumb question

you mean move the 24sqx to the other side?

[calm bridge]

treat the sqroot(x) as another variable 't'

[tiny frost]

huh

[quasi bison]

what's your native language

[tiny frost]

so you mean same way you would treat when solving a x^4 with a x^2 equation

[quasi bison]

also please write square roots as sqrt, otherwise people may not understand you

[tiny frost]

so in that case you do t^2 and t

[quasi bison]

yes essentially

[tiny frost]

my bad, swedish

ok let me try that

alright so I get $$x-12t+27$$

[flat frigate]

Totalani

[tiny frost]

do I turn this into a equation system?

[cold aurora]

Write x in terms of t as well.

I mean, in the expression.

[tiny frost]

$$t-12t+27$$

[flat frigate]

Totalani

[tiny frost]

then solve normally?

[cold aurora]

x is not t.

$\sqrt{x} = t$

[tiny frost]

im so lost

[flat frigate]

Enemagneto

[cold aurora]

that makes x = ?

[tiny frost]

no

[cold aurora]

I dunno what you mean.

[tiny frost]

x is not t

sqrtx is t

[cold aurora]

Exactly

So x is what?

[tiny frost]

t^2

[lapis pagoda]

how do u get rid of the sqrt

square both sides

[cold aurora]

How about you square this equation on the both sides?

[cold aurora]

Yes

[tiny frost]

you can quadratic or whats its called the sqrt x to get rid of it

[cold aurora]

Put x = t^2

[tiny frost]

right ofc

ooh

ok it just clicked

couldnt I have done that without turning sqrtx into t?

if I move everything to the right side except x

[lapis pagoda]

where is the original problem

[tiny frost]

and quadric everythiung

[tiny frost]

I mean this way $$x^2=(-27+12\sqrt(x))^2$$

[flat frigate]

Totalani

[tiny frost]

altho the t way seem simplier

[cold aurora]

You want to say "square" instead of quadratic.

[tiny frost]

got it

[cold aurora]

But yes, that would have worked.

[tiny frost]

not used to the english terms my bad

just gonna solve it the t way

How do you turn it back from t again?

this is what I got $$t=6\pm3$$

[flat frigate]

Totalani

[tiny frost]

well

since its sqrt it cant be minus right

so its 9, and since its t^2 its also 81

correct?

[lapis pagoda]

hiya how do u turn 54 into the t way

[tiny frost]

I didnt turn the 54 into t way

@cold aurora sry for the ping can you just double this what I wrote above ?

[cold aurora]

yes.

[tiny frost]

Thanks for the help !

[cold aurora]

These are your values?

Yes. t=9 and t=3 are correct.

[tiny frost]

thats what I got from $$t^2-12t+27$$

[flat frigate]

Totalani

[tiny frost]

Not 81?

[cold aurora]

Yes. That's one solution.

There's another as well.

[tiny frost]

So there is 3?

9, 81 and 3?

[cold aurora]

No.

Only 9 and 81.

[tiny frost]

got it!

Thanks for taking the time !

.close

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[safe radish]

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[little marlin]

I'm not sure what I'm doing tbh

[icy lance]

you could check each value of x and see which one makes 2x+1 the middle value of the 3

[little marlin]

How would I solve this inequalities tho?

[icy lance]

you dont need to do any inequalities

or solve

[little marlin]

This is what the solution says 😭

[icy lance]

there can only be two cases if 2x+1 is the median

either x-1<2x+1<3x-1 or 3x-1<2x+1<x-1

starting with x-1<2x+1<3x-1 you can split this into two inequalities to solve: x-1<2x+1 and 2x+1<3x-1

youll eventually end up with a set of boundaries for x that only one of those options falls into

[little marlin]

That would give me x> -2 and x> 2

[icy lance]

x>2 contains x>-2 so its just x>2

since 2>-2

then you do the second case for a second inequality

[little marlin]

Rightt

[icy lance]

(though only one of those options is >2 anyway)

how did you get those?

[little marlin]

Mb wrong inequalities 😭

[icy lance]

aha okay

[little marlin]

For case 2 I get
X<2
And x <-2

[icy lance]

yup which is just x<-2

so its either x<-2 or x>2

and only one of those options fits that

[little marlin]

It must be positive?

So 3?

[icy lance]

what do you mean by that?

[icy lance]

yeah

[little marlin]

How do we know which option is the correct fit?

[icy lance]

if you want 2x+1 to be the median there are two ways for it to be so:
x<-2 or x>2 , you have no options <-2 but there is one >2

if -5 was an option that would also be correct for example

[little marlin]

Oh right 😭😭😭

Sorry I keep making stupid mistakes

Thank u very much

[icy lance]

no problemo, mistakes are how we grow

[little marlin]

Thank uu

.close

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[vivid mortar]

Find how many different odd 4 digit numbers less than 4000 can be made from the digits 1, 2, 3, 4, 5, 6, 7 if no digit may be repeated.

[rustic needle]

360

[vivid mortar]

I have no clue on how to proceed for this one

[vivid mortar]

How?

It has to be above 400 I think

[misty coral]

try not to just give the solution

[vivid mortar]

Yeah

[rustic needle]

Count the number of choices for the thousands place:
Since the number must be less than 4000, the thousands place can only be 1, 2, or 3. So, there are 3 choices for the thousands place.

Count the number of choices for the hundreds place:
Since no digit can be repeated, there are 6 digits left (4, 5, 6, and 7) to choose from for the hundreds place.

Count the number of choices for the tens place:
There are still 5 digits left (1, 2, 3, 5, and 7) to choose from for the tens place.

Count the number of choices for the units place:
For an odd number, the units place must be either 1, 3, 5, or 7. Since no digit can be repeated, there are 4 choices for the units place.

Now, multiply the number of choices from each step to find the total number of different odd 4-digit numbers:

Total = (Choices for Thousands place) x (Choices for Hundreds place) x (Choices for Tens place) x (Choices for Units place)
Total = 3 x 6 x 5 x 4
Total = 360

[vivid mortar]

The second paragraph is wrong I think

[rustic needle]

true

the 6 right

[vivid mortar]

There can't be 6 choices, it has to be 5, since we need an odd in the unit value

[misty coral]

ai moment

[vivid mortar]

😭

[rustic needle]

hey im a hooman

[vivid mortar]

This is veryyy wrong actually

[rustic needle]

why

[vivid mortar]

Does it even make sense that 7-2 is 5 and yet you have 6 choices for the middle 2 box 😭

Man I'm fed up with this chapter of maths, it's overkillin

Does anyone else have any idea? 😭😭

[safe radish]

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@vivid mortar Has your question been resolved?

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@vivid mortar Has your question been resolved?

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[normal raven]

what do they mean by show

[thin bridge]

do the work that leads to/confirms that result

[normal raven]

yeah ik

i am taking a pic of my work

Just look at part ii

I am not sure what they mean by show

[thin bridge]

1 sec

[normal raven]

here's the answer

i don' understand the answer like well i don't get it by what they mean by show

[thin bridge]

two issues with your work

[normal raven]

what do u mean by tissye

tissue

[thin bridge]

problems

mistakes

[normal raven]

how

i still get a=11

[thin bridge]

i'm getting to that

firstly you are not being asked to solve / determine when
R_1 = 2R_2
you are being asked to do something along the lines of find
result 1: e.g. 14
result 2: e.g. 7
result 1 is clearly double result 2, and has been shown

secondly 2R_2 isn't 2a + b - 16

using the condition from part i),
the remainders were expressed in terms of a single unknown
allowing you to easily identify their relation

[normal raven]

no 2a+b-16 is correct

[thin bridge]

read what i said

[normal raven]

then how should i show the working

oh right

yeah

[thin bridge]

2R_2 isn't 2a + b - 16

[normal raven]

2()

ur right

my bad

[thin bridge]

using the condition from part i),
the remainders were expressed in terms of a single unknown

[normal raven]

ok so how should i show my answer than?

[thin bridge]

(using substitution as described in the solution outline)

[normal raven]

i don't get it

is it ok for u to show in a paper or in a word document

could be helpful for me to understand the working

[thin bridge]

you have
$$b = 5-a$$
$$R_1 = 3a + b - 27$$
$$R_2 = 2a + b - 16$$

[flat frigate]

ℝam()n()v

[normal raven]

ok so i did f(3)=R and f(2)=R2 right?

[thin bridge]

substitute $b = 5-a$ into those remainder equations

[flat frigate]

ℝam()n()v

[normal raven]

ok yeah right

[thin bridge]

that results in

the remainders were expressed in terms of a single unknown
allowing you to easily identify their relation
(whether one is double the other)

[normal raven]

and i will get value of R1 and R2 and i just show R1 is double of R2 right?

[thin bridge]

yes

[normal raven]

thanks man it really helped me especially im having add maths exam next week 🙂

.close

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[normal raven]

.reopen

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✅

[normal raven]

@thin bridge

Hey ahm

i get the value like in the mark scheme

so like should i solve it??

i shouldn't solve it right?

[thin bridge]

no,

all you need to confirm is that the remainder is double the other

that is all

nothing more

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[lean otter]

So ive been trying to convert cylindrical coordinates but i cant seem to succeed in the way cylindrical => cartesian.

Im currently stuck on the conversion of unitarian vectors : the problem is transform r er + theta etheta in cartesian.

I know r = sqrt(x^2 + y^2) and theta = arctan(y/x).

But when using the transform matrix wich lead to er = cos(theta) ex + sin(theta) ey and etheta = -sin(theta) ex + cos(theta) ey.

I cant find the right answer

[lean otter]

The answer is supposed to be the thing i circled in red

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[lean otter]

.reopen

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✅

[lean otter]

<@&286206848099549185> ?

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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.

@lean otter Has your question been resolved?

[lean otter]

.close

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[mystic vale]

hello!
for limit comparision test, i understand graphically why it works, but i don't understand the theorem itself

[mystic vale]

why the sum - integral < f(1) ?

how would i get to that conclusion?

[peak estuary]

interpret the sum as a riemann sum for the integral

[mystic vale]

lower or upper?

i understand i would get those "triangles" that are in excesss or defect with respect to the integral

that woudl be the left side

correct?

why would that be < f(1) ?

[peak estuary]

what do their heights add up to

[mystic vale]

the height of those shapes? i dont know

it would be the function at n-1 - the function at n

no?

[peak estuary]

I just mean graphically

[mystic vale]

i don't know i can't see any pattern or connection between those shapes and the rest

i only know that if i use left reiman sum, the integral would be larger than the sum, but i don't know how much

i just know that if the area of the integral is finite it means also the sum converges

but the area of the sum(left reinman) is actually f(1) + something else if i start at 1

i know that something else will be some value

so the area of the reinman will be between f(1) and f(1) + the integral from 1 to infinity

[safe radish]

@mystic vale Has your question been resolved?

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@mystic vale Has your question been resolved?

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@mystic vale Has your question been resolved?

[plucky elk]

Draw the rectangles

[mystic vale]

idid it

[safe radish]

@mystic vale Has your question been resolved?

[mystic vale]

ill try again tomorrow

[safe radish]

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[cedar lily]

Can you use calculus to find the length of a curve given by some function?

[dapper venture]

yes

[cedar lily]

How?

Is there a simple process or is it heavily dependant on the function?

[dapper venture]

you do an integration

so if the function is hard to integrate then it's not simple

[cedar lily]

What do you integrate?

[dapper venture]

for the 2d case

[cedar lily]

Yeah?

[dapper venture]

the length of f(x) is $\int_a^b\sqrt{1+(\frac{df}{dx})^2}dx$

[cedar lily]

When you say (df^2)/dx is that the second derivative of f with respect to x or is that the derivative of f(x)^2 with respect to x?

[dapper venture]

squared

[flat frigate]

WhereWolf(ping if needed)

[cedar lily]

Oh the whole thing squared ok

Ok, thanks!

.close

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[narrow oak]

why does it have the x does not equal 5 at the side?

[crystal palm]

consider what happens if x=5

by looking at the left side we get something divided by 0

which is not defined

on the right side we get something that is defiend

that means that for x=5, the two sides are not equal

but only because the left side isnt defined there

[narrow oak]

what do you mean by left side isnt defined?

[crystal palm]

on the left, if we choose x=5, we get soemthing divided by 0

which is not defined

[narrow oak]

ohh

ok thankx :D))

.close

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[bleak sandal]

I need help with qn 5

[plucky elk]

,rotate

[safe radish]

@bleak sandal Has your question been resolved?

[cold aurora]

!status

[safe radish]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

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[lean otter]

This makes no SENSE

[lean otter]

$$\frac{e^{k+1}}{(k+1)!} \frac{k!}{e^k}$$

[flat frigate]

What should I do

[lean otter]

$\frac{e}{k+1}$$

[flat frigate]

What should I do
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[lean otter]

$$<\frac{e}{k^1}$$

[flat frigate]

What should I do

[lean otter]

aaa

[marsh walrus]

whats the question

[split ether]

What makes no sense?

[lean otter]

i don't even know

[split ether]

I can't explain what needs to be explained if I don't know what causes the confusion

[icy lance]

hmm

[lean otter]

Oh

I see

It's because

I was thinking about

[split ether]

It's just an application of the ratio test

[lean otter]

$$\sum_{n=1}^\infty \frac{e}{k+1}$$

but

[flat frigate]

What should I do

[lean otter]

that's not what is supposed to be dun

[split ether]

Right

[lean otter]

it's $\lim_{k \to \infty}$

a

aaaaaaaaaaaaaaaa

anyways

[flat frigate]

What should I do

[lean otter]

Okay

I understand now

ty

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[formal zodiac]

how do i do this question 45d^2 - 5/4 im really stuck

[chilly valve]

Are you solving for d?

[stoic dune]

No equal sign anywhere?

[chilly valve]

I’m assuming it’s set to 0

[formal zodiac]

factorising the binomial

[chilly valve]

Factorise?

[formal zodiac]

yeah

[chilly valve]

Firstly, do you know what perfect squares and difference of two squares are?

[formal zodiac]

yeah

[chilly valve]

Which one do you think will apply here?

[formal zodiac]

difference

[chilly valve]

That’s right since there’s 2 terms and you are taking their difference

Do you know the general formula?

[formal zodiac]

yeah (a+b)(a-b)

[chilly valve]

good

Now, if you were to look at the question, the numbers are pretty ugly

You can’t nicely take the square root of 45

[formal zodiac]

yeah youd have to do 5(9d

but i dont know how to do the 5/4 part of the equation

also tysm

[chilly valve]

Okay so you take out 5 as a common factor right?

[formal zodiac]

yeah

[chilly valve]

What does that leave inside the bracket

[formal zodiac]

9d?

[chilly valve]

and

[formal zodiac]

1

[chilly valve]

1/4

Not just 1

[formal zodiac]

ah

[chilly valve]

So now you’ve got 5(9d^2-1/4)

[formal zodiac]

yeah

[chilly valve]

You could take out another factor

[formal zodiac]

3 and 2?

[chilly valve]

Nope

You can take out 1/4 as well

[formal zodiac]

oh

oh by multiplying?

[chilly valve]

You can pull the 1/4 from the 9d^2 which makes it 36d^2

Because when you divide 9 by 1/4, you are essentially multiplying 9 by 4

[formal zodiac]

ah

yeah

[chilly valve]

What happens to the 1/4 when you pull out the 1/4 from it

[formal zodiac]

its just minus 1 right

[chilly valve]

Good

So now, you got 5/4 outside the bracket

[formal zodiac]

yeah

[chilly valve]

What’s inside now?

[formal zodiac]

36d^2 - 1

[chilly valve]

Great

Does that look more familiar

[formal zodiac]

yeah

[chilly valve]

With the difference of two squares formula

[formal zodiac]

yeah

[chilly valve]

Factorise it

[formal zodiac]

tysm

[chilly valve]

All good

And put the 5/4 at the front

[formal zodiac]

you taught me more then my teacher

[chilly valve]

Lol

[formal zodiac]

ty took an hour trying to figure that out then you helped me work it out in a minute

is there a way to vote or thank you?

[chilly valve]

For these types of questions, you try to get the expression to fit those general formulas like perfect squares

Lol no

But I’m happy to help

[formal zodiac]

tysm

[chilly valve]

👍

[formal zodiac]

.close

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[next inlet]

Youll need to give more detail so they can help ya

[gusty ridge]

like?

i know basic trigonometric functions

<@&286206848099549185>

[tranquil steppe]

Oh wait so they are asking for the co tangent of 180 degrees

But cotangent 180 degrees is undefined

[gusty ridge]

it's a jee advance level problem

we have a set of 12 questions and we have a month to solve it, everything is allowed except putting the values 💀

a month 💀

[tranquil steppe]

Rip lol this is so weird do you have any lectures or anything to approach this question with? I’m trying to find some kind of way you can use for this

Like a trig property that will give you this in the format you need

[dense spoke]

Looks like an allied angle question

[gusty ridge]

i have a set of hints

which has first step of every question

we were asked to see that only after thinking about the problem for a day..

[dense spoke]

It's of the form Cota + cot(90-a) + cotb + cot(90-b)

[tranquil steppe]

There are 4 values in this problem would you just use 9 and 27 or make a and b to plug in from the 4 variables?

[dense spoke]

a=9 b=27

[tranquil steppe]

Oh okay cool and we don’t need 63 or 81 to solve? That’s cool

[safe radish]

@gusty ridge Has your question been resolved?

[safe radish]

@gusty ridge Has your question been resolved?

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[marsh relic]

I have more of an undergraduate question rather than a math question, but here it is: I was having a really bad week and flunked by calc 2 test, the course is really cutthroat and is an online course. I would have to get a 100 on everything I do from here on out in order to get an A (I have had straight A's for 2 semesters now and have a 4.0 GPA); she has no grace or anything like that, no retakes no nothing, and the tests are 10 questions so if I get one wrong I get a 90 and a B in the course. Is tanking a W in the course and saving my GPA better than getting a B in this scenario? My courseload is also extremely tough and I have a few leadership positions that require a lot of attention too, so idk if I can afford to keep pouring resources into an almost guaranteed B. What do y'all think?

[fading mango]

there's no curve?

[marsh relic]

Nope

literally 0 grace all around

[stoic dune]

"Getting a W?"

Dropping without affecting GPA?

[marsh relic]

I'm going to meet with her in a few days just to clarify but her syllabus is very clear on no grace

You can drop a course mid semester and get a "W" which means withdraw, it doesnt effect your gpa but is like a stamp that everyone will see saying you dropped

From what I see dropping one course isnt too bad, especially if it affects a max GPA, but I figured I would ask around to see what others think

[viscid salmon]

guys someone know what is the area of the circle?

[marsh relic]

please open your own help channel

[cold aurora]

Ever heard of Google? Also, try opening your own help channel. Refer #❓how-to-get-help

[quasi bison]

!help

[safe radish]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

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).

[marsh relic]

damn oh well

.close

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[rain gulch]

Please help

[surreal flare]

yes

just extend them

[rain gulch]

Thank youu

.close

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[wraith prism]

So here I'm putting random numbers and checking

It's decreasing

So surely 0

Am I right

[tired forge]

I think so. infinity - infinity would be 0

[wraith prism]

@devout shale see

😅

[wraith prism]

True

But how?

[tired forge]

how what?

[devout shale]

Don’t let others misguide you through your battle Arjunn. This is not always the case when dealing with limits

[modern mauve]

infinity - infinity is indeterminate

[wraith prism]

Yes

[devout shale]

This limit could equal 0, but it is not necessarily 0 just because it evaluates to infinity - infinity

Must you evaluate it rigorously?

[wraith prism]

How was my method?

[tired forge]

Then you get 1/infinity, which is 0

That's not super rigorous, but should hold true for powers < 1

[wraith prism]

I was looking for a genuine method

[dire fjord]

wdym by genuine method

[west hedge]

What is that symbol

[dire fjord]

for

[west hedge]

But it follows by the “or”

It’s fragmented sentence

[dire fjord]

it is not a symbol it is literally the word "For"

[cold aurora]

You can write $(n+1)^{\alpha}$ as $n^{\alpha}\left(1 + \frac{1}{n}\right)^{\alpha}$ and then use binomial theorem for fractional indices.

[flat frigate]

Enemagneto

[west hedge]

Ohh

[cold aurora]

@wraith prism

[dire fjord]

he doesnt need binomial theorem he can just take n^alpha as a common factor then write it as a fraction and perform l'hospital'a rule

[dire fjord]

i didnt evaluate this maybe it wont lead somewhere from the first l'hosp

[cold aurora]

Doesn't mean it can't be done using binomial.
Also, I think binomial is easier to understand.

[dire fjord]

i didnt say it doesnt work

i said its not necessary to do it that way

but yea he could do any of the 2

[wraith prism]

For

[wraith prism]

I'm on the way thanks

[safe radish]

@wraith prism Has your question been resolved?

[wraith prism]

@cold aurora @dire fjord

[cold aurora]

You used the wrong binomial expansion.

Alpha is not a natural number.

[wraith prism]

Which binomial then?

I only know this one

[cold aurora]

Binomial theorem for fractional indices. Look it up.

Also works for negative indices.

[wraith prism]

It is same as I did

@cold aurora sorry for a ping

[cold aurora]

It's not really. n choose k is not defined for a fractional n. Also, it has infinite terms.

[cold aurora]

That's alright.

[safe radish]

@wraith prism Has your question been resolved?

[flat frigate]

Enemagneto

Enemagneto

[grim plover]

@wraith prism close this if ur done

[wraith prism]

You did the same as i did previously

[wraith prism]

Here

I want to know what is the reason behind 0 behind all coefficient aren't constant

Got it now denominator n is bigger than numerator and a(0to1)

[wraith prism]

.close

[safe radish]

Channel closed

Closed by @wraith prism

Use .reopen if this was a mistake.

[safe radish]

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[grand veldt]

Can someone explain to me how the formula works??? If two solutions of the equation ax² + bx + c = 0 is by s¹ and s² then

[surreal flare]

-coefficient of x/coefficient of x^2 is the sum of roots

c/coefficient of x^2 is the product of roots

[grand veldt]

Sorry, can i get an explanation step by step

[surreal flare]

do yk abt -b/a and c/a proofs?

[grand veldt]

Yess

But i need to do explanation on this part

Forr school hahha

[feral relic]

I have it on my textbook

but Chinese

[grand veldt]

Ohhh, can i see, atleast a picture can help, thanksss!!

[grand veldt]

Sorry this isn't the one im looking for

[surreal flare]

this should be it

[grand veldt]

I'm very sorry, it's not very much related to this

[surreal flare]

bruv its literally the proof

[grand veldt]

I already know them, but the thing is, how do i do this formula or solve

Sorry my english isn't good

[surreal flare]

nah nah u good

the formula is just ax^2 - (sum of the roots)x + (product of roots) = 0

for example u have x^2 -5x +6 = 0

the roots of the equation are 3 and 2

the sum of 3 and 2 is 5 hence the coefficient of x is -5

and the product of 3 and 2 is 6

hence c = 6

[safe radish]

@grand veldt Has your question been resolved?

[grand veldt]

I need to explain the formula itself sorry

[safe radish]

@grand veldt Has your question been resolved?

[safe radish]

Channel closed

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[safe radish]

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[lean otter]

Is this correct?

[quasi bison]

$x \land {...}$ is sus

[flat frigate]

Ann

[plucky python]

does (a) mean for any odd natural number?

in that case, you could do $\forall y \in {1,3,5,7,9,\ldots}, \exists x \in \mathbb{N} \text{ s.t. } x>y$

[flat frigate]

lpieleanu

[lean otter]

I have to use the predicate odd(y)

What about this

[crude bear]

No

Some odd natural number

not all odd natural number

Your first one was almost correct

[crude bear]

just this

[lean otter]

What can i change with the C

X*

??

[vapid dirge]

what does x(N) stand for? cant really figure out what it is there for from the given syntax

[lean otter]

Natural numbers

[vapid dirge]

but thats given from the $\exists x,y \in\mathbb{N}$

[flat frigate]

Alvirus

[vapid dirge]

if you leave that out, it should make sense

[lean otter]

This was my first answer

[vapid dirge]

yes that seems fine

you could also write that y is odd via $\neg (\text{y mod }2 = 0)$

[flat frigate]

Alvirus

[lean otter]

My TA is picky, your only allowed to use the predicate odd(y) :/

[vapid dirge]

ah fair

but ye your last sent pic should be a valid answer

[lean otter]

Someone said the x was sus

[vapid dirge]

there exists an x, such that at least one y exists, where x is bigger than y, and y is an odd number

would be the given answer in natural language

[lean otter]

Yes

Now i have Q2b

[vapid dirge]

alright, every integer = for all which is element of N , for some integer m = exists m which is element of N

[vapid dirge]

you need to change the order of the quantors

you need "for all x exists an m", but right now you have "there exists an m, so that for each x ..."

[vapid dirge]

id write it as $\forall x \exists m, x,m\in\mathbb{Z}(\text{odd(x)} \wedge x = 2\cdot m+1)$

[lean otter]

Ah I see

[flat frigate]

Alvirus

[vapid dirge]

sorry had to fix the TEX

[lean otter]

Np

Thanks for the help

[vapid dirge]

np, glad i could be of assistance

[lean otter]

.close

[safe radish]

Channel closed

Closed by @worldly minnow

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[safe radish]

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[lean otter]

Need help to write a logical formula for a and b

[safe radish]

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'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above