help-0

[whole shell]

you have the values at say f(5) = 6, and f(6) = 7, so you use interpolation to guess what f(5.5) would be

but if it was continous

[craggy nebula]

Yeah.

[whole shell]

you would already have the information for f(5.5)

there would be no jumps

you would know what f(5.000000000000000000000000000000000000005) would be

now that you know discrete vs continous

[craggy nebula]

Yeah.

[whole shell]

the integration symbol is used like the summation symbol

[craggy nebula]

I don't see why continuous data doesn't use linear interpolation though.

[whole shell]

you would have no reason to use interpolation

you only use interpolation to interpolate because you dont have the information

[craggy nebula]

Isn't this was interpolation is though?

[whole shell]

no, i probably shouldnt have used f(x) twice

but i meant, if f(x) was continous, you would have no reason to use any sort of interpolation

since you would always have the information for any number to use in its domain

[craggy nebula]

So which data type do integrals use?

[whole shell]

continous

[craggy nebula]

Both?

Oh ok.

[whole shell]

summation

sigma

.

[craggy nebula]

How is it used in a context though?

[whole shell]

uses discrete

[craggy nebula]

Yeah.

[whole shell]

the integral is the continuos version

of the summation symbol

integration litearlly just adds up a bunch of very small stuff, to find the overall thing

the very small stuff are infintessimally small

[craggy nebula]

You must use a variable like n with integrals, right?

[whole shell]

yh, integration is usually in the form of $\int f(x)dx$

[ocean seal]

doctor99268

[whole shell]

in this context

you would imagine

a rectangle,

where the base

is dx

as in very small

but the height is the value of f(x)

at the location the rectangle is in

@craggy nebula

like this

[craggy nebula]

Yeah. Thanks.

[whole shell]

then

for some reigion

[craggy nebula]

I don't see dx here though.

[whole shell]

we are getting there

if we pick some reigion, say between x = a and x = b

we can add up all those very smol rectangles

and those would add up to the area under the graph

[craggy nebula]

Mustn't we use the ⇒ symbol?

[whole shell]

i dont partically like using set symbols

[craggy nebula]

Is?

[whole shell]

remember here

imagine, that rectangle

was located at x = 3

[craggy nebula]

Yeah.

[whole shell]

and then we have another rectangle right next to it, at x = 3.000000000000000000000000000000000....000000000001

and so on

[craggy nebula]

Yeah, infinitesimals?

Isn't this infinitesimal calculus?

[whole shell]

pretty sure calculus is always infitessimal to begin with

[craggy nebula]

How can we determine the value of f(x)?

[whole shell]

we already know it

its like a given thing

if f(x) was x^2

now, if youre gonna wonder why, the process of integration itself

is the reverse of differentiation

that is something that confused me when i first learned about it

[craggy nebula]

So derivatives are the opposite?

[whole shell]

yh

integration is otherwise known as

the anti derivative

[quasi vector]

they are the opposite of indefinite integrals, to be more precise

[craggy nebula]

[whole shell]

infact, what integration actually does itself

is said

in its name

integrating

as in bringing things together

adding up small things to become one thing itself

those smol rectangles, coming together to combine to be the area under the graph

[craggy nebula]

Why is the slope so oddly formed, and what determines its curves?

[lament forge]

that's just what f is in this case

[whole shell]

thats not the oddest slope, plus i think thats just a random graph theyve drawn

@craggy nebula imagine, you were given the derivative of a function

so dy/dx

you were given the gradient at every single point

[craggy nebula]

Yeah.

[whole shell]

since the gradient is rise over run, if you times the gradient by run, you get the rise right?

in this case, our run is just an infitessimally smol change in x, called dx

so the rise would be $f'(x)dx$

[ocean seal]

doctor99268

[craggy nebula]

What is d?

[whole shell]

dx means very smol change in x

like how $\Delta x$ would mean a sizable change in x

[ocean seal]

doctor99268

[whole shell]

if we add up all the continous f'(x)dx for a certain reigion of a to b

so like, if you were walking down the street at t = 0 to t = 1 seconds, and i added up all the teeny bits of distance you have moved from then on

the total distance that i added up, would just be the distance you managed to walk from 0 to 1 seconds right? @craggy nebula

[craggy nebula]

Yeah.

[whole shell]

in this case

if we add up

all the smol f'(x)dx

which we know that f'(x)dx is the very smol change in height, at each x location

the total number we get, would just be the difference between height (which is f(x)) at the start (so f(a))

and the height at the end which is f(b)

$\int_{a}^{b} f'(x)dx = f(b) - f(a)$

[ocean seal]

doctor99268

[whole shell]

to go from f(x) to f'(x) you differentiate

so to go from f'(x) to f(x) which we have defined to be integration

the process would be the opposite

[craggy nebula]

f(x) and f'(x) are different? 😳

Is f'(x) the opposite?

[whole shell]

f'(x) is the derivative

of f(x)

so the gradient

youre thinking of

$f^{-1}(x)$

[ocean seal]

doctor99268

[whole shell]

that is the opposite of f(X0

[craggy nebula]

All right.

[whole shell]

@craggy nebula another thing about integration is that

its not only used for finding the area under the graph

or finding the total height change if we are given the derivative

if you have something youre interested in finding out that is from a continous function, you can make it infitessimally small, and then sum it up.

[craggy nebula]

Yeah.

This is very confusing to me though, sorry, I seem to be missing the basics and I am taking a large step if you know what I mean.

I don't really know how derivatives work. 😅

[whole shell]

if it makes you feel better

derivatives are eaasier to understand

than integration

[craggy nebula]

Isn't it the opposite though?

[whole shell]

yh, but the concept of derivatives is pretty easy

[craggy nebula]

I haven't even learned about limits yet.

[whole shell]

derivatives are easier to do aswell

there are some things that you flat out cannot integrate, but all things can be differentiated

the video that was linked earlier, from 3blue1brown

is a great one

[craggy nebula]

Yeah.

[whole shell]

he has pretty good visuals and everything

my only issue was that he never showed me as to why the process of integrating is the opposite of differentiating

i had to figure it out myself

but he does clearly show what integration actually is

[craggy nebula]

Thanks a lot for your time and explanation.

I'll have a look.

.close

[lone heart]

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[quasi vector]

he did

in the later videos

[lone heart]

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[jaunty sedge]

hi can someone help me get the answer to this? watching a yt video and i'm getting a different answer to his

[whole shell]

start from the top

[vale wigeon]

show your answer and the video's answer

[jaunty sedge]

my thought process was that since it was sold for 5% less in 2019, i add 5% then take away 20% to get the original price of the house

[whole shell]

no

[jaunty sedge]

but that method did not align with what he was saying

[whole shell]

as it shouldnt

[jaunty sedge]

thought you were about to explain why

[whole shell]

oh right

well its best if i explain how to do it, then youll see

[jaunty sedge]

yes please do

[whole shell]

lets say the house alice bought

had a price of X

if she were to sell it for a 20% gain

then in 2014 the price of the house would be (1 + 0.2)X

you agree?

[jaunty sedge]

yes

[whole shell]

now lets call that price Y

so Y = (1 + 0.2)X

if the house was sold at a 5% loss

then the price of the house would be (1 - 0.05)Y right?

[jaunty sedge]

not sure

[whole shell]

you agreed it for this? @jaunty sedge

we are following the same logic

[jaunty sedge]

oh yes

i get it

[whole shell]

now that we know what the final price of the house is

we can find Y

since (1 - 0.05)Y = 182400

[jaunty sedge]

we do the opposite?

[whole shell]

yh

so what is Y

[jaunty sedge]

191520?

?

is that wrong

i'm adding 5%

[whole shell]

no

what you do is

as in, yes thats wrong

because

youre not supposed to do 182400 * 1.05

but rather

182400 / 0.95

you dont reverse 5% reduction with 5% gain

[jaunty sedge]

i see

[whole shell]

because

thats just not how it works

for example

if you had 10 things

and you reduced it by 50%

so now you have 5 things

[jaunty sedge]

so to reversing reduction, i don't use a multiplier

[whole shell]

to go from 5 things to 10 things, you need a 100% gain

[jaunty sedge]

ohh i see

yes continue also thanks for the help

[whole shell]

you dont really need to think about it that hard

we just established that

0.95 * Y = 182400

then Y = 182400 / 0.95

so what is Y

[jaunty sedge]

160000

?

192000/1.2?

[whole shell]

is that your Y or your X

[jaunty sedge]

x

[whole shell]

yh

that is what you would do

[jaunty sedge]

the original amount

[whole shell]

y = 192000

1.2X = Y

1.2X = 192000

[jaunty sedge]

i understand

[whole shell]

X = 192000/1.2

[jaunty sedge]

thanks!

have a nice day sir

.close

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[tired drum]

I found the equation of the tangent

And found the equation for the area of between the functions

But h is not 1/tan(h)

[lone heart]

@tired drum Has your question been resolved?

[tired drum]

<@&286206848099549185>

[gray isle]

!show

[lone heart]

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

[alpine sable]

okay uh

show your work

[tired drum]

Is there anything wrong in my working

<@&286206848099549185>

[gray isle]

you're missing 1/2 in the area of a triangle
where are you getting the numerical values for the max

since the area under the cos curve is constant you don't really have to worry about that
after getting the expression for area,
differentiate that wrt h
to determine when local extrema occur

[tired drum]

Is that the only mistake

[gray isle]

where are you getting the numerical values for the max
i don't see any attempt anywhere to determine when the min area occurs

[tired drum]

I found the expression for area by finding the area of the triangle

And minusing the area under the curve

I found the upper and lower bounds by finding the x and y intercepts of the tangent

You can see, where I let x=o and y=o that was me finding the values for max

But like there’s no point in finding those bounds

Because im finding the area between the two functions by finding the area of 2 seperate shapes

[gray isle]

why are you typing the letter o for 0...

anyway where you let x=0, y=0

that doesn't find the max

[tired drum]

Lol mb

[gray isle]

that determines the x and y intercepts of the tangent

[tired drum]

By max do u mean upper bound?

[gray isle]

you're the one using max

[tired drum]

Wdym by max?

[gray isle]

when i say it doesn't find the max, i mean it doesn't find anything close to what max could mean

whatever you meant when you said that

You can see, where I let x=o and y=o that was me finding the values for max

[tired drum]

Oh

So what do I do

If that doesnt find the max

[gray isle]

anyway where you let x=0, y=0
that determines the x and y intercepts of the tangent
which leads you to an expression for the area of triangle in terms of h

[tired drum]

Ye

[gray isle]

since the area under the cos curve is constant you don't really have to worry about that
(when finding the max area)
so the main focus would be to minimise the area of that triangle in with the restriction that it is in Q1

after getting the expression for area (in terms of h),
differentiate that wrt h
to determine when local extrema occur

[tired drum]

Oh

So i need to minimise the area of the triangle

And not the expression for the entire area

[gray isle]

you only need to the specified condition of the location where it occurs

[tired drum]

Mhm

Then what do I do?

[gray isle]

after getting the expression for area (in terms of h),
differentiate that wrt h
to determine when local extrema occur

don't make me copy paste again

And not the expression for the entire area
you could incorporate the area under the cos curve but its pointless as that area is constant and has no impact on optimisation

[tired drum]

Ok

[tired drum]

Is there anything else i need to do after this

[gray isle]

consider the condition that the triangle is in quadrant 1,

and you'll get what's being asked

to make your work less clunky and save ink you could do stuff like
let s= sin(h)
let c = cos(h)

[tired drum]

Ok thx

.close

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[nimble fern]

hello there!

! status

[lone heart]

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

[nimble fern]

have you learnt how to find volumes?

i see

so, you can try using the method that you've learnt to start with (c)

i see

I'll check

oh, you didn't change the subject correctly

y=x²/2+1
2(y-1)=x²

otherwise it's good

you're welcome!

[alpine sable]

Hey

i need help with this function question

after you find the gradient of BC

and convert it to find the gradient of L

[nimble fern]

!help

[lone heart]

Please read #❓how-to-get-help

[alpine sable]

!help

[lone heart]

Please read #❓how-to-get-help

[alpine sable]

sorry Ryoka

my question is incomplete tho

that looks right

mark scheme here

but i dont undersatnd

waht did u do after u found gradient

yes

yes

yes

to get M of L

yes

then what

yes

yes

bbut why do u put it into like that

y not put points A into Y = mx + c form

like what did u input the point a into0

ohh u did use y= mx+c

wha did the markscheme use tho

y-20=3(x+2)

where did that come from

yh me too

akrgiht fine thank u very much

u appreciate it

thxxxxxxx

[lone heart]

@alpine sable Has your question been resolved?

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[alpine sable]

hi

[alpine sable]

How do u do part b ii.

I figured out the first two parts

[woeful pulsar]

the regression line passes through (mean child age, mean adult age)

[alpine sable]

So to get mean child age I add 2+6+7+10+18/5 right

But what about mean adult age

Note that this is a non calculator exam

[lone heart]

@alpine sable Has your question been resolved?

[lone heart]

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[graceful quail]

When I'm using calculator in degree mode I'm still not able to get this answer

[graceful quail]

Is the book wrong with this

I can calculate A and B with law of cosines but when it comes to C

It becomes a problem

[last ether]

c = 7.2

It gives you

[hard inlet]

^

[last ether]

What c is

[graceful quail]

I know the formula

When I key it in

I'm getting different answers

[gray isle]

can you show what you're entering

[graceful quail]

Oh I see it now

I had the same numbers just different sign

So is it only law of cosines that we are able to take inverse negative of

Or is it possible with law of sines too

[gray isle]

you can take the inverse sin of negative values,
it just won't ever pop up when applying the sine law

[graceful quail]

Oh

So only with law of cosines

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@graceful quail Has your question been resolved?

[lone heart]

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[safe sail]

can someone determine the validity of this statement?

∀n ∈ N. (n ≥ 6 → ∃x ∈ N. ∃y ∈ N. 3x + 4y = n)

[safe sail]

<@&286206848099549185>

[timber needle]

yea you can do it by induction

start by saying it's true for n=6

[safe sail]

should i then look for x and y that will make 3x+4y=n true for n=6?

[timber needle]

yea start by doing that

have ytou ever done induction?

[safe sail]

not really

[timber needle]

you basically show that a statement si true for your first rank

then you show that for any k (you suppose that it's true) above or equal to that first rank

k+1 is true aswell

thus the property is hereditary

and since it's true for the first rank it's true for all of them

so when n=6 you can find x and y that will give you 3x+4y=6

with x= 2 and y = 0

[safe sail]

yeah i tried that a while ago. the problem is that x and y must be natural numbers, so y=0 is nope...

[timber needle]

0 is a natural number

it's N here not N*

so you can take 0

[safe sail]

oh, okay i was thinking about like the counting numbers 1,2,3

okayyy, i think i get it now

thank you so much

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[last ether]

I got another notation issue 💀

If I have a condition where z > -3 then something happens, and z is an element of [-2, 0], whose elements are all greater than -3, do I say:

z ∈ [-2, 0] > -3

[last ether]

Something about that seems janky

[rose sigil]

no i wouldn’t write that loll

[last ether]

How would I write it

Just say that the elements of [-2,0] are all greater rhan -3?

[rose sigil]

z \in [-2,0] so z > -3?

idk depends what you’re trying to say

[heady pollen]

$[-2,0]\ni z > -3$

[ocean seal]

~Martin

[heady pollen]

🙂

[last ether]

That looks cursed

[heady pollen]

it is haha

[mortal trellis]

$[-2, 0] \subseteq {z: z>-3}$

[ocean seal]

Denascite

[last ether]

I have to use subset?

[mortal trellis]

$z\in [-2, 0] \implies z>-3$

[ocean seal]

Denascite

[mortal trellis]

give proper context about what you actually want to say

[last ether]

Well basically I have a 3D vector field whose divergence is 12 + 4z. And then I have a sphere whose radius is 1 and the center is (0,0,-1)

And so z is an element of [-2, 0]

And I have to explain why I predict the net flux of the field across the sphere is going to be from inside to outside

My teacher wants me to explain fully so

I'm trying to explain as much as I can 💀

Because his definition of "fully" is vague

[rose sigil]

just add one or two english phrases like so, since, we have, if then, implies, etc to “z \in [-2,0]” and “z > -3” that make it sound good

[last ether]

Alright then

Hopefully that's enough

Imma keep this open in case something real stupid happens

[rose sigil]

maybe you can even add

z >= -2 > -3

[last ether]

Something more like this

Although maybe I could word it out

z in [-2,0] whose elements are all less than -3

It just sounds nasty

[rose sigil]

i feel like i’ve told you before you shouldn’t be afraid to write words

[echo socket]

"If it's greater than -2, then it's greater than -3"?

[last ether]

I am using words, it just sounds very runny as a sentence

I also wanna practice my understanding of symbols since things like this pop up

[mortal trellis]

as z is in [-2, 0], it is greater than -3 and therefore ...

[last ether]

Yeah that's what I did

I resorted to words when referencing sinks and sources in the field

[mortal trellis]

unbelievable, using words in math. imagine that

[echo socket]

Using only words or only symbols for a proof is hard

[last ether]

No I used both

[rose sigil]

not using a lot of words for a proof is hard

[last ether]

But then sentences :///

[rose sigil]

yes, writing sentences is part of math

[echo socket]

Just noticed this is help-0

The legendary help channel

[last ether]

Yeah I snagged it for funnies

Anyways ty

.close

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[last ether]

I hate writing sentences

[rose sigil]

too bad

[last ether]

I almost microwaved a fork 💀

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[vivid quiver]

.reopen

[ember sigil]

How do you call this rule? Or how do you do to get this ?

[heady pollen]

$(33)^{3/2}=(33)^{1+1/2}=(33)^1+(33)^{1/2}=33\cdot\sqrt{33}$

[ocean seal]

~Martin

[heady pollen]

i would call it power properties maybe

[ember sigil]

Oh wow, thank you, that's better to understand

[ember sigil]

So here

It shows this answer

We used lcf, but why 33 sqrt/ 33 didn't change?

Like just the other 1/9 changed to 1 because of the factorization

[heady pollen]

lcf?

also, are you sure the answer is true?
it looks like they messed up a minus

we can use the distributive rule here
if that looks weird, we can also write a+ab as a⋅1+a⋅b=a(1+b)

[ember sigil]

yeah, a minus was messed lol

[heady pollen]

nice integral btw, classic case for substitution

[ember sigil]

it's a fun one, I just always have trouble at the end

Thanks bro

.close

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[alpine sable]

What is l'Hopital's rule for limits?

[lone heart]

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[ember sigil]

This is when your answer is 0/0 or infinity/ infinity, which is not valid, so you derivate the problem and plug the limit again

[ember sigil]

https://tutorial.math.lamar.edu/Classes/CalcI/LHospitalsRule.aspx

[prisma elbow]

*indeterminate form

[ember sigil]

Check this, this page is really helpful

[prisma elbow]

there are most than just 0/0 or inf/inf

[ember sigil]

oh oh okay

[prisma elbow]

there's like 7 in total

[ember sigil]

Like how, I've just learned this lol

[prisma elbow]

if i remember correctly

0/0 and inf/inf are just most common

[ember sigil]

Now I'm curious, I'll search

[alpine sable]

Sorry to bother but are 0/inf and inf/0 determinate?

[alpine sable]

Thanks for this useful ressource!

[ember sigil]

I think just inf/0

0/inf should be 0? Not 100% sure

[ember sigil]

No worries, there's much more apart from the L'hospital rule ( Integrals, derivatives, etc.)

[alpine sable]

I guess it depends on whether the denominator approaches 0 from the left or the right...

And if it oscillates around 0, then it would be indeterminate

So, all of this put together, inf/0 is indeterminate but gives you more information than 0/0 or inf/inf

[ember sigil]

Hmm, I understand, I should check all the indefinite forms, sounds interesting :)

.close

[lone heart]

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[hard inlet]

!status

[lone heart]

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

[whole shell]

@river canopy

find the angles in the order

that i listed

acc

[hard inlet]

1->2->3->x

[whole shell]

the order doesnt even matter

you can go 1 3 2 x

or 1 2 3 x

2 and 3 are independant of each other

but 3 needs 1

[lone heart]

@river canopy Has your question been resolved?

[lone heart]

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[primal galleon]

,, \begin{gathered}\int^{}{} \left( 2x+3\right)^{2} \ln \left( 2x+3\right) dx\\ \int^{}{} udv=uv-\int^{}{} vdu\ \left( 2x+3\right)^{2} \ln \left( 2x+3\right) -\int^{}{} \ln \left( 2x+3\right) \frac{\left( 2x+3\right)^{3} }{3} dx\ \end{gathered}

[ocean seal]

dgh

[primal galleon]

Can someone tell me if this is the correct?

[quasi vector]

well it is

but

you're just making things more complicated for yourself

you can just use substitution

infact

[primal galleon]

@quasi vector Initially I started with substituting 2x+3 with u

[quasi vector]

what you wrote is also wrong

I just noticed

[quasi vector]

yes

what did that get you

[primal galleon]

So I kinda tried the DI method that you told me yesterday

and LIATE

I took the integral of (2x+3)^2 <=> u^2 = u^3/3

and then the derivative of ln(u)

I stopped at the first row

And got what you see on latex

[quasi vector]

that's correct

[quasi vector]

no

[primal galleon]

You mean I should've gone for more rows?

[quasi vector]

no

you stop at one row

but your multiplication is wrong somewhere

can you show me how you did it?

on your paper or whatever

[primal galleon]

Ok one moment

We start by making the substitution $u = 2x+3$. This means that $du/dx = 2$ and $dx = du/2$, so we can rewrite the integral as:

$$\int (2x+3)^2 \ln(2x+3) dx = \int u^2 \ln(u) \frac{du}{2}$$

[ocean seal]

dgh

[quasi vector]

that's correct

[primal galleon]

Now, we focus on solving the simpler integral. We'll use integration by parts, with $u = \ln(u)$ and $dv = u^2/2 \cdot du$:

$$\int u^2 \ln(u) \frac{du}{2} = \frac{u^2 \ln(u)}{2} - \int \frac{u^2}{2} \frac{1}{u} du$$

From the formula $$\int^{}{} udv=uv-\int^{}{} vdu$$

[quasi vector]

well

[ocean seal]

dgh

[quasi vector]

since you're already using u, using u for the IBP is kinda wrong

but glossing over that, your first term is wrong

$$\int^{}{} udv=uv-\int^{}{} vdu$$
Here, $$dv = u^2/2 \cdot du$$ as you said
so $$v = \int \frac{u^2}{2} \cdot du = \frac{u^3}{6}$$

[ocean seal]

kheerii

[quasi vector]

so your first term has to be u^3ln(x)/6

and similarly for the second term

$dv \ne v$

[ocean seal]

kheerii

[quasi vector]

to get v, you have to integrate dv

also, you're already using u for your substitution, so u = ln(u) doesn't make sense

either use a different variable for your IBP or for your substitution

[primal galleon]

Okay I will need some time to process this, I will open another ticket if I'm still stuck.

[lone heart]

@primal galleon Has your question been resolved?

[lone heart]

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[vague coral]

you can take a counterexample i guess

[echo socket]

Pick a_k = (-1)^(k + 1)/k

[lone heart]

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[rotund shoal]

How many unqiue number plates can there be in the form XXYYYY wher X represents a number from 0 to 9 and y represents a letter from the alphabet?

[rotund shoal]

Would the answer be 10 x 10 + 26 x 26 x 26 x 26

Is there meant to be a + separating the digits and the letters since they don't overlap

as in, the digits are set to be on the left side of the numnber plate only?

I cannot decide if there should be a plus or multiplication separating the 10 and 26

[quasi vector]

since the two probabilities are different, it would be a x sign

it's the same as why you have done 26x26x26x26

since they are all independent events

[rotund shoal]

I see. In what case would it be a plus?

Could you give an example?

[quasi vector]

why would it be a +

[carmine reef]

If you could choose either a 2-digit license plate OR a 4-letter one

then it would be +

[rotund shoal]

Ohh right okok

since they are separate?

[carmine reef]

since it's or

or means +

[rotund shoal]

ohh right or/and

[carmine reef]

and means *

[rotund shoal]

gotcha

thank you

.close

[lone heart]

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[carmine reef]

no problem

[lone heart]

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[vagrant lantern]

what single equation could i use to calculate a value between 0 and 10(y) to smoothly convert to a value 0.3333 to 1.5(x)?

[vagrant lantern]

im programming in python and normally id just solve this with several expressions but i need to optimize as much as possible and express it as one value. the expression im modifying is dynamic_weight = 0.5 * ((running_avg / weight) - 1)

or
0.5 * ((A/B) - 1) = Y if that makes it any easier to read

and if one of you hugebrains happens to be experienced with ai stuff, google is not great when it comes to notation regarding the effects of modifying the l1 regularizer value after its been put through the lambda function on over/underfitting

[marsh rapids]

so how to find a line passing through 2 points ?

[alpine sable]

2 point form?

[marsh rapids]

you call it whatever you want

[alpine sable]

idk if thsts related to the question being asked

[lone heart]

@vagrant lantern Has your question been resolved?

[vagrant lantern]

sorry

got sidetracked there

@marsh rapids maybe? dunno, i dont have big mathbrains at all. im kinda dumb tbqh im just looking for a single equation that i can convert into a python expression to translate y into x

[marsh rapids]

so this ?

[vagrant lantern]

if what i described is that then yes

[marsh rapids]

passing through (x1, y1) and (x2, y2) is the line y(x) = y1 + (x-x1)(y2-y1)/(x2-x1)

[vagrant lantern]

not ignoring just trying to sort and test

x and y in that example = x and y in mine?

[marsh rapids]

applying it to these values gives x = 1/3 + 7/6 * y / 10

you might want to review basic algebra and more btw

[vagrant lantern]

thanks for the advice

so i end with
x = 1/3 + (7/120) * (A/B - 1)?
i feel like ive screwed it here

[marsh rapids]

I guess

[vagrant lantern]

you know, rather than being condescending id rather you just let someone else try to help

would have saved the both of us some time and effort

[marsh rapids]

I meant that as in correct

not as in "you messed up"

[vagrant lantern]

sorry, im fried out right now i misread the tone there

my bad, ive been wrestling with overly complicated programming stuff all day every day for like four days and im getting frustrated with myself because im getting worse at simple stuff the longer i keep going for

but the worse i get at it the longer its taking

[marsh rapids]

when you struggle to find the equation of a line, you know it's time to take a break

take some rest

[vagrant lantern]

yeah, i know but im so enamored with the project im just obsessing over it when im not working on it

[lone heart]

@vagrant lantern Has your question been resolved?

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[vagrant lantern]

.reopen

[lone heart]

✅

[vagrant lantern]

i thought i had it down but testing is showing that im not getting expected values

dynamic_weight = 0.5 * ((running_avg / weight) - 1) is the expression i started with

or y=0.5*((A/b)-1)

combined with this to find a line for variable y
x = 1/3 + 7/6 * y / 10

using this formula y(x) = y1 + (x-x1)(y2-y1)/(x2-x1

to convert y into a spread of numbers ranging from 0.3333-1.5

with a tolerance for y to be between 0-10

i simplified it into this string

dynamic_weight = 1/3 + 7/60 * (0.5 * ((10 / 5) - 1)

or

x = 1/3 + (7/120) * (A/B - 1)

with x being the new variable created from calculating the previous dynamic_weight(y) variable

the issue im having is my own fault because i asked for the wrong expression
im trying to figure out how to phrase the expression i need

assuming y calculates at most to 7 and at minimum to 0.1428 how can i apply this so that the resulting number lands between 1 and 0.3333 for values under or = 3.157 and between 1.0001 up to 1.5 for values over 3.157?

the formula for y being y=0.5*((A/b)-1)

the key here is that im trying to express the entire calculation, including the one for y in a single expression

[lone heart]

@vagrant lantern Has your question been resolved?

[lone heart]

@vagrant lantern Has your question been resolved?

[lone heart]

@vagrant lantern Has your question been resolved?

[violet flare]

I'm having a bit of a stroke and if someone could help me with this quesiton would be much appreciated. f(x) = x if x εQ, and |x| otherwise

the range would be -infinity, infinity right?

.open

@lone heart

[slate monolith]

this channel is occupied

#❓how-to-get-help

[violet flare]

ah, thank you

[lone heart]

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[keen trellis]

I need asap help for c), d), e)

[surreal meadow]

can you take a guess at the domain of f o g?

[violet flare]

lmao

[surreal meadow]

whats qld

[violet flare]

misspelt

[surreal meadow]

!nosols

[lone heart]

As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

[violet flare]

ah whoops

[surreal meadow]

that’s not quite it btw

[violet flare]

ok how would you solve it then?

[surreal meadow]

i’ll walk the asker through it if they come back

[violet flare]

now im curious

[surreal meadow]

can you del this

ty

[violet flare]

yes

why did you want me to delete it

[surreal meadow]

because it has a good amount of the answer in there

[violet flare]

ha

*ah

can you walk me through it

im curious now

[surreal meadow]

i’ll just dm you

[violet flare]

ok

[lone heart]

@keen trellis Has your question been resolved?

[keen trellis]

@surreal meadow hi

How did you get fog as well?

[surreal meadow]

do you have any guesses as to what the domain would be

[keen trellis]

All real?

[surreal meadow]

yes

and how about the range

[keen trellis]

The same?

[surreal meadow]

not quite

can you think of any numbers that give you -1 as an output?

[keen trellis]

I’m having trouble understanding how you find fog cuz it’s a pie wise function

[surreal meadow]

think about what happens when you plug in a number in either of the pieces

i.e. a rational or an irrational

[keen trellis]

Well if it’s rational it be psoyive or negative

Can only be irrational positives

[surreal meadow]

well let’s take it step by step

f o g (x) = f(g(x))

[keen trellis]

Yeh

[surreal meadow]

let’s say we plug in a rational number

what is g(x) then?

[surreal meadow]

i assume you mean |x|

[keen trellis]

Yeh

[surreal meadow]

then yes

and is |x| rational or irrational (given that x is rational)

[keen trellis]

Rational

[surreal meadow]

yes

so we get f(|x|)

where |x| is rational

so what is f(|x|)

[keen trellis]

Uhhh g(x)?

[surreal meadow]

sort of

f(x) = x when x is rational

[keen trellis]

Yep

[surreal meadow]

so f(|x|) = |x|

when x is rational

so f o g(x) = |x|, when x is rational

[keen trellis]

Oh okay

So when it’s irrational it would be x?

[surreal meadow]

let’s do it step by step again

f(g(x))

when x is irrational, what is g(x)?

[keen trellis]

x

[surreal meadow]

so what is f(x)

[keen trellis]

x

[surreal meadow]

no

look at the definition of f(x) again

[keen trellis]

Oh so |x| again?

[surreal meadow]

yes

[keen trellis]

When irrational

[surreal meadow]

so f o g(x) is always = |x|

so what’s the range

[keen trellis]

[0,infinity)

[surreal meadow]

yes

[keen trellis]

That was a huge help

So then with fog it is multiplying by x?

For the next question?

[surreal meadow]

for d or e?

[keen trellis]

D?

D*

Wait no

[surreal meadow]

d is asking if it’s one-to-one

[keen trellis]

Yeah

And then e is multiplication

[surreal meadow]

e is multiplying the two functions yeah

[keen trellis]

So how would you multiple a piece wise

It’s the piece wise part that confuses me

[surreal meadow]

multiply it piecewise

[keen trellis]

When doing all this

[surreal meadow]

multiply the rational parts and the irrational parts

[keen trellis]

So x^2 and

No wait

Bruh I’m so dumb

[surreal meadow]

you’re not

what would the multiplication be?

[keen trellis]

x|x|?

[surreal meadow]

yes

[keen trellis]

For both

[surreal meadow]

and for the other piece?

yup

do you know how to show if they are one-to-one?

[keen trellis]

I think so

You sub in numbers right?

To prove that x1 and x2 have the same values

Or not

[surreal meadow]

that’s to show they are not one to one yeah

[keen trellis]

Yeh I’ll give them one to ones a shot on my own

Can I dm you if I get stuck?

[surreal meadow]

ask here

[keen trellis]

Alright

[surreal meadow]

i’ll help if i’m available

[keen trellis]

You’ve been an amazing help

Thank you

[lone heart]

@keen trellis Has your question been resolved?

[candid obsidian]

Hello anyone can please solve this question?

[keen trellis]

@surreal meadow I’m having touble with the 1-1

[surreal meadow]

!help

[lone heart]

Please read #❓how-to-get-help

[surreal meadow]

for which one

[keen trellis]

The x|x|

And the |x|

@surreal meadow

[surreal meadow]

for the |x|, find 2 different values of x that give the same output

for |x|x, you’re not asked to prove it, so i’d graph it and check whether or not it is one to one

[keen trellis]

For |x| tho it’s 3 makes

How would we write it so it’s worth 3 marks

[surreal meadow]

if it’s one to one then f(x) = f(y) implies x = y

so find two different x and y such that f(x) = f(y) that’s all there is to it

[keen trellis]

So any negative and positive value have the same y values

[candid obsidian]

!help

[lone heart]

Please read #❓how-to-get-help

[surreal meadow]

i get what you mean yeah

specifically, x and -x have the same output

[candid obsidian]

Can anyone solve this question please

[surreal meadow]

!help

[lone heart]

Please read #❓how-to-get-help

[wraith bramble]

Hold on!

[surreal meadow]

please read #❓how-to-get-help @candid obsidian

[wraith bramble]

The solution is long but the answer is 1.

If you want the specific solution, I will be writing it

[keen trellis]

@surreal meadow also (e) actually does say to prove it, I cut it off in the photo

I’m v confused how to do that

[surreal meadow]

suppose |x|x = |y|y. then (|x|x)/(|y|y) = 1

squaring you get (x^2 * x^2)/(y^2 * y^2) = 1
so x^4 / y^4 = (x/y)^4 = 1, so x/y = +-1

then either x = y or x = -y, consider what happens when x=-y with the original function. it’ll be an extraneous solution (unless x = 0), so the only actual general solution is x = y, but you need to show that

[keen trellis]

I’m very confused on this question

How do you work out x1, x2

@surreal meadow

[surreal meadow]

i’m confused what are x1 and x2

[keen trellis]

I do one_one if like f(x1) = f(x2)

[surreal meadow]

your x1 and x2 are my x and y

[keen trellis]

Can you step me through how I’d do it

[surreal meadow]

i did here

[keen trellis]

Why Are we diving it to prove 1-1

[surreal meadow]

to show 1-1 you show f(x) = f(y) implies x = y

the steps we did let us arrive at the conclusion that x = y

[keen trellis]

Which prove it’s one-one

[surreal meadow]

yes

[keen trellis]

I don’t understand when you said it’s an extraneous solution, but you need to show that

[surreal meadow]

if x = y we’re done, it’s 1-1

then we need to test the other case

so we assume x = -y

plug that back into |x|x = |y|y

[keen trellis]

-x = -y

X = Y

[surreal meadow]

|-y|*-y = |y| * y

so |y| * -y = |y| * y

suppose y is not 0 so we can divide by |y|

then -y = y

what is the only solution to this

[keen trellis]

So your saying x1 = -x2

[surreal meadow]

yes

[keen trellis]

Why is it |y| * y

[surreal meadow]

so

when we did the calculations

we arrived at 2 possibilities

either x = y

or x = -y

[keen trellis]

Yeh

[surreal meadow]

if x=y then it’s 1-1

so we consider the other case

x=-y

or x1=-x2

so we go back to the original assumption

|x|x = |y|y

and plug in x = -y

[keen trellis]

Yep

[surreal meadow]

so we get this

[keen trellis]

Oh right

Okay

[surreal meadow]

read on from there

you get up to -y = y

what’s the only solution to that

[keen trellis]

Does that mean it’s not one to one

[surreal meadow]

answer this

[keen trellis]

0

[surreal meadow]

yes

so if y=0

then x = -0 = 0

so x = y

[keen trellis]

So one -one

[surreal meadow]

yes

[keen trellis]

Ahhh alright

Thank you

[candid obsidian]

Can you please send me the whole solved question

[keen trellis]

.close

[lone heart]

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[dry tartan]

Someone help me with this question pls

[lone heart]

@dry tartan Has your question been resolved?

[dry tartan]

<@&286206848099549185>

[exotic belfry]

!show

[lone heart]

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

[dry tartan]

completly stuck

why is A1 = 1.002

where did it come from

[exotic belfry]

first think of this: what is the amount at the end of the week, when starting with 100 and a weekly compounded interest from 10.4 percent?

[lone heart]

@dry tartan Has your question been resolved?

[lone heart]

@dry tartan Has your question been resolved?

[lone heart]

@dry tartan Has your question been resolved?

[exotic belfry]

whats about this?

@dry tartan

[lone heart]

@dry tartan Has your question been resolved?

[mystic kestrel]

help

[slate monolith]

#❓how-to-get-help

[lone heart]

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[covert cobalt]

claim

[covert cobalt]

i think in english name of cours is Straight in plan

but in french droite dans le plan

[marsh rapids]

lines in the plane

or lines in 2D

[covert cobalt]

yes

i have course next week but idont know what i do here

[marsh rapids]

AM = M - A

[covert cobalt]

it is that u vector = M ?

[marsh rapids]

no ?

u is a fixed vector, M is a point. There's no way they can be equal
Also M isn't fixed so since u is, there's no way u = AM could always hold

[covert cobalt]

i think AM = xA i + yA j

[marsh rapids]

AM = (x-1, y-1)

[covert cobalt]

aah yeah

AM =(xb - xa ; yb - ya)

AM = xi +yj

AM=(1 i + (-3) j )

[marsh rapids]

do you know how to find a vector from 2 points ?

[covert cobalt]

yes

[marsh rapids]

so why are you struggling

[covert cobalt]

doing the same projction

no just puting my ideas there 💀

like to try

im working just from pc no paper

[marsh rapids]

it doesn't take 10 minutes to apply that formula

[covert cobalt]

fr im in another exercise

[lone heart]

@covert cobalt Has your question been resolved?

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Remember:
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[languid barn]

help

guys

[plain flame]

$a^x = e^{\ln (a) \cdot x}$

[ocean seal]

heavy

[languid barn]

you sock

[plain flame]

excuse me

[languid barn]

not you brother

[languid barn]

what is that money sign for?

[plain flame]

its how you initiate a math environment in LaTeX

which is documentation software

[languid barn]

oh

[plain flame]

which is useful for math notation

[languid barn]

ok

[plain flame]

hence this server has a LaTeX bot

[languid barn]

I need to solve this before I eat my lunch

[plain flame]

$e^{\ln (x)} = x$

[ocean seal]

heavy

[plain flame]

and more generally

$e^{\ln (x) \cdot c} = (e^{\ln (x)}) ^c = x^c$

[ocean seal]

heavy

[plain flame]

which you know how to intergrat

[lone heart]

@languid barn Has your question been resolved?

[languid barn]

yes

[tacit arch]

.close

[lone heart]

Channel closed

Closed by @tacit arch

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Remember:
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• Show your work, and if possible, explain where you are stuck.
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[last ether]

D is the entire circle?

Also shouldn't they be double integrals

[vale cliff]

im not completely sure

[last ether]

Hm well I'm assuming they're supposed to be double integrals because like

Yeah

[vale cliff]

why

[last ether]

Or are they path integrals?

It wouldn't make sense to single integrate a region and have dA

Typically dA is for double integrals

$$\iint_D\dd A$$

[vale cliff]

hmm

[ocean seal]

Umbraleviathan

[last ether]

This just measures the area of D

$\int_D \dd A$ to me, makes no sense

[ocean seal]

Umbraleviathan

[tacit arch]

some books are like that, unfortunately

[last ether]

So it's supposed to be a double integral?

Well @vale cliff I think you can answer the 1st question. The second one, I would sketch out the region it's being integrated over, and then sketch the integrand if need be

But notice how the region is R (which is all positive x values), so 5x within R must all be positive

[vale cliff]

hmm okay

ty

.close

[lone heart]

Channel closed

Closed by @vale cliff

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[lone heart]

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Remember:
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[cursive cosmos]

can i have help

[lone heart]

Please don't occupy multiple help channels.

[alpine sable]

How to cite this paper?
https://journals.sagepub.com/doi/abs/10.1177/10943420221136848

in bibtex file I am sesing 0:
year = {0},
volume = {0},
number = {0},

[lone heart]

@cursive cosmos Has your question been resolved?

[serene junco]

@alpine sable Someone else opened this channel before you. You should ask your question in one of the channels in the available section (although I'm not really sure that's a math question but whatever)