help-13

[sinful zephyr]

Oh

A = 300/c

[frigid dust]

yes

so now we have a=100/b

and a=300/c

we can set them equal to each other

100/b=a=300/c
100/b=300/c

so now we have a "new" equation

\begin{equation}\frac{100}{b}=\frac{300}{c}\end{equation}
\begin{equation}bc=200\end{equation}

[wraith dagger]

Bonk

[frigid dust]

once again, rewrite (1) as b=??? and rewrite (2) as b=???

[sinful zephyr]

Hmmmm

B = 200/c

B = 100/a

[frigid dust]

where are you getting the a from?

[frigid dust]

here we only have b's and c's

[sinful zephyr]

Well

Then b = 300/c

[frigid dust]

300?

[sinful zephyr]

9h sorry

B = 200/c

[frigid dust]

yes, thats for equation (2)

now try equation (1)

[sinful zephyr]

ab = 100
B = 100/a

A = 100/b

100/100/b

[frigid dust]

Stop. Using. A.

[sinful zephyr]

Then what do i use

[frigid dust]

just use these 2 equations

you have already rewritten the second equation correctly

now you just need to rewrite the first equation

[sinful zephyr]

Hmmm

How can i rewrite ab

[frigid dust]

stop using ab

????

\begin{equation}\frac{100}{b}=\frac{300}{c}\end{equation}
\begin{equation}bc=200\end{equation}

[wraith dagger]

Bonk

[frigid dust]

only this

[sinful zephyr]

You told me rewrite ab

[frigid dust]

where?

[sinful zephyr]

Oh nvm

I misunderstood

200/b = 200/c

[sinful zephyr]

Right?

[frigid dust]

100/b=300/c

this one

[sinful zephyr]

What i do about thay

[frigid dust]

rewrite it

so that it is in the form of b=???

[sinful zephyr]

B = 300/c x 100

[frigid dust]

not quite

[sinful zephyr]

Why

Its in form of b =

[frigid dust]

yes, but its wrong

[sinful zephyr]

How

[frigid dust]

wdym how

it is not the same

[sinful zephyr]

Ehhhhh

btw one question
Ab + bc + ca can be written as
A^2 + B^2 + C^2

[frigid dust]

not in general, no

[sinful zephyr]

Dang

I cant really think

[sinful zephyr]

Can i write it as like

Ab x bc / ca
And cancel terms to get
B^2 then
B value is sqrt(b^2)?

Hmm

But i guess

[frigid dust]

yes, you could do that

[sinful zephyr]

Nvm

Oh

[frigid dust]

and then repeat for a and c

[sinful zephyr]

a^2, b^2 , c^2

100 x 200 / 300

200/3

B^2 = 200/3

A ^ 2 = 150

C^2 = 60

I mean 600

[frigid dust]

so then what are a,b,c?

[sinful zephyr]

Sqrt(a^2 + b^2 + c^2)?

[frigid dust]

no

a=sqrt(a^2)=sqrt(150)

[sinful zephyr]

Well there is no value of sqrt(150)

[frigid dust]

you can just leave it as sqrt(150)

,w sqrt(150)

[wraith dagger]

Results provided by WolframAlpha

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[sinful zephyr]

Yeah

[frigid dust]

it exist, its just not a whole number

[sinful zephyr]

Right

My answer sheet says
Sqrt(6050/3)

[frigid dust]

,w sqrt(6050/3)

[wraith dagger]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by .bonk__

[sinful zephyr]

So i guess i just have to plus all number

[frigid dust]

,w sqrt(150)+sqrt(300)+sqrt(600)

[wraith dagger]

Results provided by WolframAlpha

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

Requested by .bonk__

[sinful zephyr]

This bot is cool

Anyways i found the answer

Thanks

[frigid dust]

its not about the answer, its about the process

having the whole process correct but making a calculation mistake somewhere is much better than having the wrong process but correct answer

if you know how to solve it, then you can apply it to more problems

just having the solution doesnt teach you anything and might aswell have done nothing

[sinful zephyr]

I guess

I kinda hate math, because i found it hard to understand the concept

Like other subjects which is much easier to do

[frigid dust]

imo, most hate math because they havent been taught it properly

most of the people that "like" math are able to see it themselves and dont need as much help

[sinful zephyr]

I guess you are right, my math teacher just tells us solution but dont tell us the concept behind it

[frigid dust]

at least, thats what i believe

if you are able to understand how to solve it

and being able to go step by step

you will get much more enjoyment

because when you do another problem, you can see similar steps

and are able to apply what you did to other problems to the current problem

[sinful zephyr]

If you ask my class circle area they will know it, but they wont know why is it pi * r^2

Alright

I will close this channel now

.close

[cedar kiln]

Channel closed

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[cedar kiln]

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[fossil berry]

can someone explain me characteristics of the cumulative distribution function

[quartz coral]

okay so cdf is just a way to describe how probabilities accumulate as we move along the number line in the simplest way possible

imagine you have a random variable, like the roll of a dice, or the height of people in a group. The CDF tells you, for any number, what is the probability that the random variable is less than or equal to that number

[fossil berry]

right hand continuity

[quartz coral]

cdf starts at 0, and as we move along the number line, it adds up the probabilities.

[fossil berry]

can you explain this to me

can you explain right hand continuity and the last one

[quartz coral]

right

so

that entire limit just means as h gets closer to x from the right (so it values slightly larger than x), the value of F(h) approaches F(x)

in other words, there’s no sudden jump in the CDF when moving towards x from the right. The function behaves smoothly at this point, which is a key property of cumulative distribution functions.

[lilac rampart]

can some one help me with this proplem ?

[dire geode]

! occupied

[cedar kiln]

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).

[quartz coral]

and this

this formula just explains how to compute the probability that a random variable X takes on a specific value x, particularly for discrete random variables. Let’s break it down step by step in simple terms now

[fossil berry]

what does this small minus on top means

[quartz coral]

where?

in the second formula?

[fossil berry]

yeah

on first here is plus

[quartz coral]

left-hand limit of the CDF at x

[fossil berry]

what do they mean

[quartz coral]

normal Fx(x) in that formula without any + or - is regular value of cdf at x

x- is the left hand limit of cdf at x

x+ the right hand limit of cdf at x

[fossil berry]

i see

[quartz coral]

although theres not one here

[fossil berry]

thanks.

.solvd

.solved

[cedar kiln]

Channel closed

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[quartz coral]

and yes btw the maximum value cdf can reach is 1

as u said earlier

[cedar kiln]

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[vast ermine]

Help please

[vast ermine]

So are they asking for me to do 20/4 and give them 5?

Or is the answer 20/4

[quartz coral]

Just plug in 10 instead of x

[frigid dust]

,rotate

[fresh quarry]

20/4 = 5 , so theyre same

[vast ermine]

Oh okay and yeah I was just asking how to write the answer

Also how do I do the next part?

It's an inverse function but I'm not sure how or what to substitute-

[pallid parrot]

express x as y and the value to be x

and find y

[vast ermine]

uhh so

[pallid parrot]

ex.. f(x)=23x+2/2

[vast ermine]

x is 5 or 10?

10*

[pallid parrot]

then x = 23y+2/2

[pallid parrot]

you don't find x as a number

you find out the inverse function

[vast ermine]

I mean like

[pallid parrot]

f(x)=2x, then f-1(x) =
x = 2y
so y = x/2 so f-1(x)=x/2

[pallid parrot]

oh yeah

but you need to find inverse first

[fresh quarry]

Its the best to write the most simplified answer
In this case its 5

[vast ermine]

Alright so like this?

[pallid parrot]

replace 5 with x

[vast ermine]

Ok

[pallid parrot]

now find y

[vast ermine]

umm how (I forgot)

[pallid parrot]

multiply both sides with y-6 to eliminate the denominator

[vast ermine]

Ok

This?

[pallid parrot]

yes put parenthesis

x(y-6)=2y

so multiply each side in the parenthesis by x

it's gonna be xy - 6x = 2y

[vast ermine]

Got it

[pallid parrot]

now rearrange

to find y

[vast ermine]

Ok

Do I now replace x with 5?

[pallid parrot]

no

[vast ermine]

What fo I do

Do

[pallid parrot]

from xy-6x = 2y
rearranging
xy-2y=6x

so it's gonna he y(x-2)=6x

@vast ermine

now divide by x-2 to find y

[vast ermine]

Qait how did u get y(x-2)=6x

[pallid parrot]

to get xy-2y

you have to multiply y by (x-2)

[vast ermine]

OHH OKAY

But why did u choose x-2

Sorry idk the method 😭

[pallid parrot]

look at xy-6x=2y

what's the common factor in the left side

or what do they share?

[vast ermine]

y

And 2 goes into 2 and 6

[pallid parrot]

no?

6x doesn't have y

in the left side

[vast ermine]

oh wait so on just left side

X

[pallid parrot]

if you remove the x

what is it gonna be

[vast ermine]

Y-6

[pallid parrot]

so x(y-6)=2y!

[vast ermine]

okayy

[pallid parrot]

but this is gonna give us nothing

[vast ermine]

yes

[pallid parrot]

i want y not x

so i made the y a common factor

do you know what a common factor is now

[vast ermine]

Yes

[pallid parrot]

xy-2y=6x
just in the left side they share y

[vast ermine]

yes

[pallid parrot]

so if we remove the y

it's gonna be x-2

[vast ermine]

Yes

[pallid parrot]

so y(x-2)=6x

do i want y or y(x-2)

[vast ermine]

y

[pallid parrot]

then can you get rid of the x-2

[vast ermine]

so ÷3?

[pallid parrot]

no like

[vast ermine]

÷x-2

[pallid parrot]

if 5 * 3 = 15
if i wanna get rid of the 3

i divide by 3

[pallid parrot]

exactly

what's the final result?

[pallid parrot]

CORRECT

now replace y with f-1(x)

and here's your function

you can now do direct substitution for x

[pallid parrot]

you dont need to find x

[vast ermine]

Does f-1 mean x-1?

[vast ermine]

yeah i remember it

[pallid parrot]

f-1 means inverse

$$ f^{-1}(x) $$

[wraith dagger]

harue

[pallid parrot]

it's gonna be the y you made

[pallid parrot]

it's gonna be 6x/x-2

you just express the X

[vast ermine]

ohh so y = 10?

[pallid parrot]

no

i meant yes

but

it doesn't ask for $$f^{-1}(10)$$

[wraith dagger]

harue

[pallid parrot]

it asks just for the raw equation

[vast ermine]

So it just asks for the function okay

[pallid parrot]

Yeah

Express means like express the formula or function you made

[vast ermine]

Alright and checked the mark scheme they wanted 6x/x-2 too, thank you for the help I appreciate it

[pallid parrot]

so the answer is gonna be
$$f^{-1}:x -> \frac{6x}{x-2}$$

[wraith dagger]

harue

[vast ermine]

gotcha

question uhh what's like

the steps to do this kinda question?

like uhh

is it

[pallid parrot]

just like what you did there

i have an idea

[vast ermine]

1. Find x
2. X = equation

and so on?

[pallid parrot]

did you understand the question how i did it

[vast ermine]

yes

[pallid parrot]

well

now

get an empty paper

and do it again

yourself

[vast ermine]

okay

[pallid parrot]

the whole idea

is to replace x in your function with y

and replace f-(x) with x

and find y

the idea not steps

[vast ermine]

got it and started myself

Is this fine so far?

[pallid parrot]

yes!!

i got a good method

just apply this triangle

if you want c then b/x
if you want x then b/c
if you want b then x * c

[pallid parrot]

yes!

you can multiply

[vast ermine]

Dobe

Done

[vast ermine]

Thank you I'll add it ti my notebook

[vast ermine]

i did it :D

[vast ermine]

wait so lemme try that with this question

what are b and c here?

[pallid parrot]

anything

b is numerator

and c is denominator

[vast ermine]

alright noted

ill try more questions with this then!! thank you i really appreciate the help (catching up a year's missed knowledge for an expensive exam in 4 months)

[pallid parrot]

good luck!

you could remember it with an example

$$ 5 = \frac{10}{2} $$

[wraith dagger]

harue

[pallid parrot]

10 is 5 * 2 and 2 is 10/5

[vast ermine]

alright then thank you <333

.close

[cedar kiln]

Channel closed

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[cedar kiln]

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[sand grotto]

This is what i did

it just needs checking ty

<@&286206848099549185>

[cedar kiln]

@sand grotto Has your question been resolved?

[cedar kiln]

@sand grotto Has your question been resolved?

[cedar kiln]

@sand grotto Has your question been resolved?

[pallid trout]

For bottom,

Equation will be

Horizonal motion >> F - f = Ma,

frictional force will Max as both crate move together.
Ie. f_max = (u) *mg

F- umg =Ma

a= F/(M+m)

So, we get : F - umg =M(F/(M+m)

FM +Fm - umgM - umgm =MF

Fm = umg(M+m)

F= ug(M+m)

Then putting value you will get the answer

[sand grotto]

so

my answer is correct?

[pallid trout]

Yeah.

[sand grotto]

ok

can i ask another question or do i have to do the process again?

needs checking check sig figs and units

[unborn imp]

I hate sig figs

[cedar kiln]

@sand grotto Has your question been resolved?

[cedar kiln]

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[cedar kiln]

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[molten ocean]

Yoooo what 1+1?

[flint plinth]

pls don't troll in the help channels

.close

[cedar kiln]

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[coral jewel]

any idea with this? i'm stucked

[worldly chasm]

@coral jewel with 10 dimensions, the lowest dim that V1 intersect V2 can have is 4. And then the lowest the intersection between this space and V3 can have is 1. Do you see why?

[coral jewel]

let me think a bit

[mighty shuttle]

Suppose that $\dim(V_1) \cap \dim (V_2) \cap dim(V_3) = {0}$. What would that mean about $\dim(V_1+V_2+V_3)$

[wraith dagger]

math rocks(wai)

[coral jewel]

i'll be right back, something cropped up

i'll reopen later

.close

[cedar kiln]

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[arctic vector]

Derivative of f(x) = cot^2 (x+5)

[arctic vector]

$f(x) = cot^2 (x+5)$

[wraith dagger]

Ski

[arctic vector]

I'm kinda confused with the chain rule here

[normal cipher]

We are to express this as a combination of functions

[arctic vector]

What do you mean?

[normal cipher]

Like f(g(x))

The innermost function is x+5

And then two cot

[arctic vector]

oh like composite function

[normal cipher]

Which is just 1/x^2 and tan x

[pseudo oasis]

yeah something like that

[normal cipher]

And then peel the thing like onion using chain rule

[pseudo oasis]

and treat the (x+5) as a function although it didn't really matter much

[normal cipher]

But dont cry while doing it

[pseudo oasis]

if you cry on derivitive then i don't know how you can surive integration

[arctic vector]

Well so based on that

2cot(x+5) right

is the first step

just bringing down using the power rule

[normal cipher]

Its a metaphor since i said peel onion

[normal cipher]

Uhh

Kinda

[pseudo oasis]

oh lol

[normal cipher]

But i like to approach from inside

[arctic vector]

inside?

[normal cipher]

Please interpret correctly btw

[pseudo oasis]

then solve cot(x+5)

[full urchin]

lol

[full urchin]

ayo pause

[normal cipher]

Yours is probably faster

[arctic vector]

derivative of cot(x+5)?

[normal cipher]

Transform it into 1/(tan(x+5))

And then use tan definition

[pseudo oasis]

or just recite the derivitive of cot(x)

[normal cipher]

Well ok

Can never memorize those

[arctic vector]

2cot(x+5) * -csc^2(x+5) * 1?

[pseudo oasis]

memory is key

[normal cipher]

Probably

[normal cipher]

...

[pseudo oasis]

nah in my school is basic requirement to recite the derivitive of inverse trigo 💀

[normal cipher]

Tf man

[arctic vector]

kinda same here

even the identities also

ion remember shit tho since my teacher just cramps all the topic

[pseudo oasis]

yes I am talking about arcsin something

[normal cipher]

Sad

Use integral-calculator.net to check ans im too lazy fr

[arctic vector]

(cot(x+5))^2

2cot(x+5) * d/dx cot (x+5) * d/dx (x+5)
2cot(x+5) * -csc^2(x+5) * 1
-2cot(x+5)csc^2(x+5)

[normal cipher]

K i gtg bye

[arctic vector]

oh

[cedar kiln]

@arctic vector Has your question been resolved?

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[languid trail]

What is the quotient for the following question (x²+16)÷(x+4)

Solve with long division

Solve with synthetic division

Solve by factoring

I can do the long division, and the synthetic, but when it comes to factoring I get complex numbers which I don't think is correct, as we havnt learned about them yet. Could anybody offer some advice?

[languid trail]

This is an intro to algebra in trig, or ig trig identities work sheet btw

Then next question is a similar format of (sin²x+16)/(sinx+4)

when i solve by long div or synthetic i get x-4 r32

would i just leave it as (x²+16)÷(x+4) and consider it fully factored?

[crimson sedge]

Try completing the square

For x^2+16

@languid trail

Would that come under factoring though 🤔

[languid trail]

not sure tbh, havnt used that method in a minute but lemme try

hold on how would i complete the square with no b term?

[pastel vault]

are you sure it's not (x^2 - 16) / (x + 4) ?

[languid trail]

That was the previous question

Low key kinda lost

[pastel vault]

yeah then I don't know what they mean by solve by factoring for this question

[languid trail]

Ig ill just leave it in that form, lemme know if anybody else has any ideas

Nah that equals x²-16

[prisma pike]

Oh mb

[languid trail]

Cuz 4*-4=-16

[prisma pike]

So suggestive

Welp

[languid trail]

Ig ill ask on Monday lol

[prisma pike]

Oh

I think I know what it is

It's a typo

Wait nvm

That's the previous qn😭

[languid trail]

But the next question is the same form, and it asks for only "solve by factoring" I doubt it's a typo

Are u able to simplify (sin²x+16) into anything?

I don't think so right

[prisma pike]

Yea

That's kinda weird

Can we see the whole page

And the previous one

I just wanna see the context

[languid trail]

It's all been super simple.up till this point lol

It's just algebra trig connections as an intro for trig identities

[pastel vault]

yeah definitely seems like a typo

[prisma pike]

Nah but the x^2-16 was the previous qn

[languid trail]

Yea

So this is like a follow up

[pastel vault]

hhmmmm

[cedar kiln]

@languid trail Has your question been resolved?

[languid trail]

it hasnt but i dont think it can be resolved, so ill close lmao

.close

[cedar kiln]

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[cedar kiln]

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[vestal sandal]

can someone explain the second paragraph? How is the dimension of a matrix MxN same as the vector MN and while you are at it can you simply explain the idea of dimension in vector spaces thank you very much

[hollow trail]

the dimension is the length of any basis of a given vector space (noting that every basis of a given vector space has the same length)

the dimension of the vector space of m x n matrices, and the vector space of vectors of length m*n are the same (i.e., we need the same number of basis vectors to describe them)

for example the vector space of $2\times2$ matrices ($\R[2\times2]$) has this basis:
[ \ab{\begin{bmatrix} 1 & 0 \ 0 & 0 \end{bmatrix},\begin{bmatrix} 0 & 1 \ 0 & 0 \end{bmatrix},\begin{bmatrix} 0 & 0 \ 1 & 0 \end{bmatrix},\begin{bmatrix} 0 & 0 \ 0 & 1 \end{bmatrix}} ]
and the vector space of vectors of length $2\cdot2=4$ ($\R[4]$) has this basis:
[ \ab{\begin{bmatrix} 1 \ 0 \ 0 \ 0 \end{bmatrix},\begin{bmatrix} 0 \ 1 \ 0 \ 0 \end{bmatrix},\begin{bmatrix} 0 \ 0 \ 1 \ 0 \end{bmatrix},\begin{bmatrix} 0 \ 0 \ 0 \ 1 \end{bmatrix}} ]

[wraith dagger]

cloud

[hollow trail]

so they are both 4-dimensional vector spaces

[cedar kiln]

@vestal sandal Has your question been resolved?

[vestal sandal]

Not really though let's say the basis vector is (0,1)

It's dimension is 1

And length is 2

Isn't it the number of independent columns

Vectors

[hollow trail]

by 'length of the basis' i mean the number of basis vectors

[vestal sandal]

Ohhhh aight okay gotchu

[hollow trail]

since the basis is really the set of basis vectors

[vestal sandal]

What is the dimension of the set of matrices

4

Ohhh

Hmhmhmhnhmhmynehsiwhsjs

Okay I get it

Thank you so much

.close

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[vital torrent]

help

[vital torrent]

hi i have a cuestion, im doing this exercise and i tried to obtain the basis resolving f1(x) = 0 and f2(x)=0 and i have obtain three vectors it is the correct way or there is another way to do it

<@&286206848099549185>

[pastel vault]

yes that seems like you're describing the correct way

use Gaussian elimination / row reduction to obtain the RREF form

then you will have three free variables, say s, t, u

express x1, x2, x3, x4, x5 in terms of s, t, u

like if you have (2 + t, 2 + s, 5), this is (2, 2, 5) + t(1, 0, 0) + s(0, 1, 0)

a basis is (2, 2, 5), (1, 0, 0), (0, 1, 0)

[vital torrent]

okay i did that well but i wasnt sure if i did it right because after i need to calculate the algebraic complement and it should be U right ? and i need to calculate also a basis of the subspace annihilated by this complement. And all this things seems the same

thats why my question hahah

[pastel vault]

ah you have more questions

hmmm

[vital torrent]

yes

haha

[pastel vault]

I'm not the best at explaining those sorry

[vital torrent]

dont worry

[cedar kiln]

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@vital torrent Has your question been resolved?

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[twilit kraken]

help

[twilit kraken]

Hello i want help in this question is Logarithmic function and he wants me to find its derivative

[prisma pike]

You can use a combination of chain rule and quotient rule

[cedar kiln]

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[cerulean garnet]

You have a gun that holds 6 bullets maximum. It is fully loaded.
You shoot at a target with a 20% chance of hitting it
Every time you hit the target you reload the gun with 2 more bullets.
What is the expected value for the number of shots taken before the gun is empty?

[cerulean garnet]

Not looking for a solution but just what topic in probability this would fall under

[steel crest]

expectation

maybe

reccurence

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[stray otter]

why

[hushed folio]

They are just writing f(x) in the form you want to demonstrate

Where the denominator has all the variables multiplied

[stone kettle]

A ladder, leaning against a wall, makes an angle of 60° to the ground. If the foot of the ladder is 9.5m away from the wall. The length of the ladder is (a) 16 m (b) 10m (c) 19m (d) 18m

Can anyone help

[hushed folio]

First, draw a picture of what they are describing

Id draw it so that you're looking at the ladder from the side

[noble flame]

There's already a question in this channel already, ask in an available "help" channel

[stray otter]

ah I see

alr

.close

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[stone kettle]

Okay sry

[cedar kiln]

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[final spruce]

hello i dont know how to solve the following

[final spruce]

i have already f' and f''

[hushed folio]

Set f' and f'' to 0

And find all the values of x that satisfy

[final spruce]

my bad both the f' are wrong :c

can i write f(x) as x*2ln(x)

[hushed folio]

Sadly not, as the exponent isn't inside the logarithm

[final spruce]

how should i do it than?

[hushed folio]

Don't forget the negative

Use product rule and chain rule

[final spruce]

and i wrote it wrong its x*(ln(x))^2

[hushed folio]

Mhm

[final spruce]

sorry for the missunderstanding

[hushed folio]

No problem

[final spruce]

so f' is (ln(x))^2+2ln(x)

thats what i got rn

[hushed folio]

Yep but don't forget the negative

[final spruce]

there is nothing negative

i think

[humble arch]

there is

[final spruce]

where?

[humble arch]

this -

[final spruce]

thats my bad the wrong excersice

[hushed folio]

Oh

So just the same but without the negative

In that case then ur derivative is right

[final spruce]

the one in the picture top left

[humble arch]

alr

[final spruce]

but f'' is looking a little weird

can i make it more simple?

[humble arch]

wanna factor out 2/x?

its very simple alreayd

[final spruce]

yea but how?

[mortal ledge]

Heyy

[humble arch]

do you see 2/x

[final spruce]

yea

[humble arch]

heres another

i wouldnt factor it but it can

[final spruce]

ohhh

[final spruce]

i think i will let it be

so now i put f' = 0

[humble arch]

yes cuz the point with f' = 0 is a peak with a flat slope

[final spruce]

Okay

so now i got x1 and x2

so now i put x1 and x2 into f''?

what do i do with this now?

<@&286206848099549185>

[cedar kiln]

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[crimson sedge]

After the last step, Can I divide by e^2x on both sides to simplify further?

[idle gulch]

This seems fine !

[idle gulch]

Yes you can surely do but that will make the constant part c/e²x

[crimson sedge]

Why?

[idle gulch]

So it's absolutely unnecessary

[crimson sedge]

Isn’t it arbitrary constant?

[idle gulch]

e²x isn't a constant right ?

[crimson sedge]

Oh

Wait my teacher did it here

See last step

[idle gulch]

I think he has defined A = e^x then

[crimson sedge]

I don’t think so

Defined how?

This is the entire question:

Find the equation of a curve passing through the origin, given that the slope of the tangent to the curve at any point (x, y) is equal to the sum of the given point.

[crimson sedge]

Do you think she made a mistake?

[idle gulch]

Idk honestly, but that's not the way it should be ! It's fine to keep the e^(-x)

[crimson sedge]

Please tag me

[crimson sedge]

Okay^^

Thank you

You think I should wait and see if anyone has another opinion on what she might have meant to do?

[idle gulch]

Yes you can surely keep it open, but if the bot closes it

[mighty shuttle]

which part

that seems to be a different question

[crimson sedge]

I’m just trying to figure out what the motive behind the last step could have been?

[mighty shuttle]

Pretty sure A would be a variable if you divide across by $e^{-x}$

[wraith dagger]

math rocks(wai)

[crimson sedge]

No same question, I had trouble in the simplification part, so I hadn’t supplied the question before because I didn’t deem the context important then

[mighty shuttle]

so we have $e^{2x}= \frac{e^{5x}}{5}+C$

[wraith dagger]

math rocks(wai)

[mighty shuttle]

Pretty sure A has to be a variable here

[crimson sedge]

This as well?

Is it incorrect?

[mighty shuttle]

Depends on what A is

[cedar kiln]

@crimson sedge Has your question been resolved?

[crimson sedge]

Wdym?

[cedar kiln]

@crimson sedge Has your question been resolved?

[mighty shuttle]

I feel A should be a variable for this to make any sense

[crimson sedge]

Bro that A?
That’s just her habit to mark out the final answer

I’m so sorry it took me so long to answer what A you were talking about

[mighty shuttle]

Oh

Then it makes no sense to me

[mighty shuttle]

I could be wrong

[crimson sedge]

Okay

Thank you so much

.close

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[inner solar]

how to solve $\int\frac{,dx}{x^4+1}$?

[wraith dagger]

Slowaq

[crimson delta]

partial fractions

[inner solar]

but what should be in the denominators?

should i somehow factor x^4+1?

[crimson sedge]

https://youtu.be/to0bkvIl2cc?si=6Kw91gsQFzupwRkh

This video should help

Literally watched it last week

If u wanna do factoring

https://youtu.be/X-36E6vdcP8?si=0RSV78phoNIUpTOA

You can use this as well

Personally I find the second solution to be a bit more intuitive

[cedar kiln]

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[graceful viper]

.close

.close

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[limber yoke]

Im losing my mind, somethings wrong with the derivative for x<0 I think but Ive checked it 7 times over

[native heath]

...yeah

what is this

i think you just ate a lot of steps to save writing space and ended up missing negatives

[south tundra]

Looks like they replaced x with -x (for whatever reason)

[native heath]

pretty much

did you do that because x<0

cuz that aint it fella

[limber yoke]

ah

yes thats y i did it

how should I proceed

[native heath]

well no just change the sign of the mod term

[limber yoke]

why just that?

[native heath]

because $f(x) \neq f(-x)$

[wraith dagger]

Percy

[native heath]

which i know doesn't sound like it's saying a lot but yeah no literally just that

you dont change the entire function for negative values

but the modulus opens with a negative yes

[limber yoke]

right so if I had |x|-x it would be -x-x for x<0 and x-x for x>0

cuz I only change the value inside the module since its that which Im trying to determine

[humble arch]

(x+1)^2 is NOT (-x+1)^2

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[dark girder]

this may be a stupid question but on part b why do we not care abt adding c? its not a definite integral but the markscheme just ignores it and I dont think you can solve it if you put c in?

[dark girder]

*the value for t for part b is 3.5 and it integrates to this

[cedar kiln]

@dark girder Has your question been resolved?

[cedar kiln]

@dark girder Has your question been resolved?

[humble arch]

it is a definite integral

youre integrating from 0 to 3.5

[dark girder]

oh right

thanks

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[signal sigil]

Could anyone please tell me what to do here? i'm blanking

[cedar kiln]

@signal sigil Has your question been resolved?

[signal sigil]

ah well, no worries.

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[keen quest]

alpha is 0, beta is pi.

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[crystal merlin]

help pls

[cyan eagle]

is it asking you to prove it combinatorially or something? (if so, think about picking l objects from a set of p objects that are part of a set of k objects)

[cedar kiln]

@crystal merlin Has your question been resolved?

[crystal merlin]

yes I understand the interpretation of the left part but I can t understand how to prove the right part is equal if i only know that k >= p >= l

[cyan eagle]

so, it is a combinatorial proof? if so, then try to find a way to describe the same interpretation of the left with the expression on the right (or, i mean, find a way to count the objects in a different way)

[crystal merlin]

is it that In both cases, we select a subset of p elements from k by breaking down the choice into two steps: the selection of the first l elements, then the p - l elements among the remaining k−l.

[cyan eagle]

yeah, that's what i thought of (and should be correct)

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[crystal merlin]

thank you

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[sand grotto]

needs checking all answers should be to 3sig figs

[sand grotto]

<@&286206848099549185>

[echo dagger]

what are you having difficulty with here?

[fringe folio]

I can give you an answer but i don't think that would be particularly helpful

I agree with you about the tension. I think you lost precision along the way as my decimal places were different to yours (be careful of that).

The reaction force on the wall will be the horizontal component of the tension

Hopefully that will give you enought to find the answer @sand grotto

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[strange vessel]

I need help with question 8a

[dusk goblet]

the set of all points such that y = 3

(0, 3), (0.1, 3), (-69, 3), (420, 3) etc

notice anything

[strange vessel]

Yeah

Y is always 3 however I can set x as whatever I want?

[dusk goblet]

mhm

what would that look like

graphically

you can always plot a few if you’re still confused

[strange vessel]

Wdym

[dusk goblet]

plot a few points

[dusk goblet]

^

if you can’t tell what it will look like yet

[strange vessel]

I’m not sure

[dusk goblet]

^

[strange vessel]

Alright gimme a sec

[dusk goblet]

in fact

plot as many as you can until you realize

(something, 3)

[strange vessel]

Well with 0, 3 it’s just a straight line going through the y axis

[dusk goblet]

what

i said points not lines

[strange vessel]

Ah frick

💀

I’m a bit special sorry

[dusk goblet]

sorry brother i don’t have much patience for this but good luck i’ll give the answer and cover it if you want to check your answer

or just use desmos

[strange vessel]

Bruh

I just want the formula that’s it

[dusk goblet]

||y = 3 looks like a horizontal line with all points having a y coordinate of 3||

[dusk goblet]

formula?!

💀

[strange vessel]

Like how to do it or whatever

[dusk goblet]

brother

i told you

as i said

i don’t have the patience for this

have a nice day

[strange vessel]

K

.close

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[dire viper]

I am stuck with the following problem

[dire viper]

Is this serie uniform convergent $\sum \frac{nx}{1+(nx)^2}$ for $x \in \mathbb{R}$

[wraith dagger]

Erijoni

[dire viper]

I found that x=-1/n and x=1/n are points where max and min of the function

I’ve tried to bound it to use Weierstrass Test but I don’t think it works

As for Abel’s and Dirichlet’s Tests I haven’t tried

[soft holly]

$$\lim_{n \to \infty} \qty|\frac{(n + 1)x}{1 + ((n + 1)x)^2} \div \frac{nx}{1 + (nx)^2}| < 1$$
$$\lim_{n \to \infty} \qty|\frac{(n + 1)}{1 + ((n + 1)x)^2} \div \frac{n}{1 + (nx)^2}| < 1$$
$$\lim_{n \to \infty} \qty| \frac{1 + (nx)^2}{1 + ((n + 1)x)^2}|$$

[wraith dagger]

Idli Robot

[dire viper]

I don’t think this works

[soft holly]

Yeah me neither

I think it equals 1, so idk

[celest gyro]

maybe integral test?

[dire viper]

I’m not even sure if these test can be used for functional series

[soft holly]

They definitely can

In fact, the [ratio test](#help-13 message) is use very often to find the interval of convergence of a power series

[dire viper]

Yeah thats true

[soft holly]

(A power series is usually of the form $\sum^\infty_{n = \dots} f_n(x)$

[wraith dagger]

Idli Robot

[soft holly]

I think you can try integral test

[dire viper]

You are right but in the problem I have seen abel dirichlet and wiererstrass were

Used more often

[soft holly]

abel dirichlet and wiererstrass
i havent heard of this, so i cant help if you specifically need to use that

[celest gyro]

same abel dirichlet is unkown to me haha

[dire viper]

I think i solved it

[celest gyro]

tell us

[dire viper]

For the series to be uniform convergent

[celest gyro]

i'm waiting in awe

[dire viper]

We must have [ \lim_{x\to\infty}
sup(f_n(x))=0 ]

[wraith dagger]

Erijoni

[dire viper]

You can find that f_n(x) has a maximum of 1/2

[celest gyro]

i literally was thinking about "wasn't sup of something a thing?"

took this course a year ago haha, i'm a bit rusty

[dire viper]

But it is pointwise cont

I think

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[green crow]

Need help on this exercise, I'm new to proofs

[mighty shuttle]

Is this from abbott

Okay, can you find an upper bound for $B$

[wraith dagger]

What a wonderful world!

[mighty shuttle]

Recall that every non-empty set of real numbers that's bounded from above has a least upper bound (why?)

[green crow]

I don't know where to start but I wrote down , since A is bounded below then there is a number l≤x where x is in A, but I just wrote down a definition I believe

[mighty shuttle]

Have you done the axiom of completeness yet ?

[green crow]

No

[mighty shuttle]

I suggest you do that first

[green crow]

There's just a section that says Axiom of Completeness: Every non empty set of real numbers that is bounded above has a least upper bound. And that's all it says

[mighty shuttle]

Yes

Now define B( write down the definition rigorously)

[green crow]

From the question if b is a lower bound of A could that mean A is bounded below

[mighty shuttle]

Yes, and what does that mean?

Write it down using inequalities

[green crow]

There exists a number b where b≤a although I don't know what a is, I used a because it's part of definition

[mighty shuttle]

$B={x:x \leq a, \forall a \in A}$

[wraith dagger]

What a wonderful world!

[mighty shuttle]

Now what would the supremum of $B$ be?

[wraith dagger]

What a wonderful world!

[green crow]

Is B also a subset of R

[green crow]

I'm not sure because so far you've been describing a lower bound but the definition of supremum wants an upper bound

[mighty shuttle]

yeah, nut notice how the set of lower bounds is defined by saying it's all reals less than or equal to a given number

[hushed folio]

It might be helpful to read the definition of B out in words

B is the set of all real numbers that is less than or equal to all values in A

[green crow]

I'm having some trouble visualizing the set B is it the interval [x,a] ?

[hushed folio]

I wouldnt look at it as an interval

Just look at how B is restricted by A

Every element in B has to be less than every element in A

[green crow]

The way I'm thinking of it isina number line x comes before a

Well if I want to find supremum of B I have to first find an upper bound of B but that would mean B is bounded above possibly?

[hushed folio]

Yep

Since B is defined by being less than all the elements of A, it has to be bounded above by those elements

[green crow]

If I use the definition of bounded above will it need to include this number a or not. I'm confused in the sense of introducing these variables

[hushed folio]

I guess let's look at a particular case

Say we had a set S that was defined as being a lower bound for the set of all real numbers 3 or above

What would be the infimum of that 3 or above set?

[green crow]

Would it be 3

[hushed folio]

Yep

Now what would the supremum of S be?

[green crow]

Would there be no supremum

[hushed folio]

There is a supremum

[green crow]

Is x≥3 is that same as x is in the interval [3, infinity)

[mighty shuttle]

yes

[hushed folio]

Yep

So we have a set that is all the real numbers that are lower bounds for [3, infinity)

What is that set's greatest possible element

[green crow]

Well right of the number line it's going infinitely so you can't find a greatest possible element I would say

[hushed folio]

It's not going infinitely to the right

If B is the set of elements that are lower bounds to [3, infinity), then asking what that set's supremum is means you're asking the "greatest lower bound" of [3, infinity)

[green crow]

But I thought supremum is least upper bound

[hushed folio]

It is

But B's supremum is the infimum of the other set

Because B is all the lower bounds of the other set

So the greatest B is the greatest lower bound of the other set

Which is the infimum

[green crow]

By other set you mean A?

[hushed folio]

Yes

[green crow]

Do you have to use sets for this problem. It makes more sense to me saying b≤a because from the problem b is a lower bound of A

[hushed folio]

I mean $b \leq \text{min}(a)$

[wraith dagger]

hiidostuff

[hushed folio]

And min a is inf(A)

[green crow]

But that's not part of definition

This the definition I'm using

[hushed folio]

I guess let's take this step by step then

$B = {b : b , \text{is a lower bound of} , A}$

[green crow]

The only part I understand is A is bounded below and then using the definition so far

[wraith dagger]

hiidostuff

[hushed folio]

So this is the set B

Which means that $B = {b : b \leq a , \forall , a \in A }$

[wraith dagger]

hiidostuff

[hushed folio]

Now suppose we look at the infimum of A

We know that all b in B must be less than or equal to the infimum of A

Because of how B is defined

[green crow]

Is the infinum of A and element of A?

[hushed folio]

And now that there exists a value such that all b is less than or equal to, we know B is bounded above

[hushed folio]

Not necessarily

The infimum of a set is just the greatest value that bounds the set below

[green crow]

Could you instead use the definition of infimum and work from there

[hushed folio]

Sure

Actually instead I'll do a simpler two parter

This is what I would say is the simplest proof that sup B = inf A

Ok so we're gonna do two things

[green crow]

Wait, are A and B different sets or do they equal each other

[hushed folio]

Different sets

Anyways

Let's show that sup B is a lower bound of A

[green crow]

We know A is bounded below but I don't know if B is bounded above or below

[hushed folio]

And that b is less than or equal to sup B

[hushed folio]

If A consists of finite-valued elements then B is bounded above

Ok lets do part 1

Since B is the set of all lower bounds for A, then every b is less than or equal to every a

Where a and b are elements of A and B respectively

Because of this, sup(B) must also be less than or equal to every element in A, making it a lower bound for A

And of course, sup(B) is greater than every other b, making it the greatest lower bound of A

And the greatest lower bound of A is inf(A)

Thus, sup(B) = inf(A)

[green crow]

So are you saying b=supB

[hushed folio]

No

I'm saying that sup B is greater than all b

Remember b isn't one value

It represents all elements of B

[green crow]

How can I visualize this

From a picture

[hushed folio]

Not sure honestly

[hushed folio]

Actually there might be a way

[green crow]

The way I tried doing this problem at first is since A is bounded below there is a number l≤a , where a is in A. From set B, b≤a, So we have l≤a and b≤a but that would mean l≤b.

[hushed folio]

@green crow here's the real line, and suppose the shaded red i drew is the set A

[green crow]

But I don't know what l and b could be

[hushed folio]

Hold on

On the left, I drew some blue points to show instances of lower bounds to A

[green crow]

So is A half the real line

[hushed folio]

I mean "half the real line" isn't really a thing

But its some portion of the real line

Do the blue points make sense tho?

[green crow]

So something like [1,2] for example?

[hushed folio]

Sure

Or like [3, infinity)

But suppose it has an infimum

[green crow]

If the blue points are lower bounds to A then lower bounds can't be inside the set A?

[hushed folio]

Uhh that's kinda difficult

One lower bound can be in the set A

The infimum

But the infimum could also maybe not be in A

@green crow does my drawing make sense tho?

[green crow]

So if it was [1,2] 0 could be a lower bound of that set or -1 and so on

[green crow]

Kind of

[hushed folio]

I guess just a quick recap

The shaded red area is the set A

And the points I marked in blue are lower bounds to the set A

Since they are less than all the elements of A

[green crow]

I can understand that

[hushed folio]

Alright

Look at that right most blue marker i put tho

Do you agree that we can keep putting points in between the blue mark and the leftmost portion of A, and it'll still be a lower bound of A?

[green crow]

So the closest blue marker near set A?

[hushed folio]

Yep

We can keep putting markers between that marker and set A, and all of those markers will be lower bounds to set A

Does that make sense?

[green crow]

I can see that

[hushed folio]

As soon as we mark a point in blue that crosses A, however, that point isn't a lower bound of A anymore

And so its not in the set B

Basically meaning that every element of A bounds the set B above

Do u see that?

[green crow]

Well In the picture there is only set A I don't see set B

[hushed folio]

Well every point in blue that i marked is an element of B

Because it's a lower bound to A

But if I were to mark a point in A, then it's not a lower bound of A anymore

So it can't be in B

So if a point is in A, then it's not in B

[green crow]

So the set A comes after the set B?

[hushed folio]

Yes you could think of it like that

But more importantly, it comes directly after set B

[green crow]

So are these two sets connected in a way

[hushed folio]

I mean by definition yes

B is literally defined according to A

[green crow]

Maybe I can think of it like a border

[hushed folio]

Yes

And the largest value that B can take stops right as soon as it gets to A

And that largest value would be its supremum

(We do have to prove that sup B is in B tho)

[green crow]

So if b≤a then is b in set B and not A

[hushed folio]

Yes UNLESS b is equal to inf A

And if inf A is in A

Which it doesn't have to be

[green crow]

When you say as soon as it gets to A is it near A or in A

[hushed folio]

It can either be infinitely close to A or right at the border of A, depending on if inf A is in A

[green crow]

From the picture the supremum of B is near the same place as the infmum of A

[hushed folio]

Not just that but sup B is exactly equal to inf A

The largest that B can get is at the border of where A starts

Both of which are some visual definitions of the supremum and infimum

[green crow]

It makes sense visually but actually proving it although I mentioned I'm new to proofs is really difficult

[hushed folio]

It is

The challenge of doing math really well is u gotta be smarter than ur eyes

You gotta let go of the dependence on visualizations