help-13

[past jungle]

i see, so can't use question assumptions during a proof

guess I'm worried in case a question appears where it's true, idk how i would approach that without using assumptions given in the question

[dreamy sleet]

no, you can’t use the conclusion to prove the conclusion

[past jungle]

oh shit right, yeah that.

mb

[dreamy sleet]

“P(A|B) =< P(A)” is not an assumption

[past jungle]

👍

[dreamy sleet]

The only things you can use are the assumptions given in the question (and prior knowledge)

So just be very clear about what’s an assumption and what’s the goal/conclusion

[past jungle]

got it

.close

[cedar kiln]

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[remote parcel]

I just need some pointers on how to get started

this is what I tried thus far but to no avail

[remote parcel]

I know that any vector that's the linear combination of v1... vk can be expressed as A(v1) +... A(vk) or A(v1+...vk) but I don't know if it helps

[next cloak]

You need to show the equivalency of sets

take any vector from R^m and prove that it can be written as that span

you know that any vector in R

[remote parcel]

take any vector from R^m and prove that I can get it from multiplying A by a linear combination of v1,...vk? (if I'm understanding right)

[next cloak]

Okay so: Any vector in R

sorry my keyboard is interrupting my texts

Any vector in R^m can be written as the linear combination of vectors in R^n and A. So to speak: You can get any vector in R^m by a multiplication A.v , where v is a vector in R^n.

And you can get any vector in R^n by the span of (v1, .... vk)

If you can get any vector in R^n with that basis, When you multiply it with A and take the linear combination you can get any vector in R^m

[next cloak]

This is because Rank of A is M

[remote parcel]

Wait, I'm trying to work it out

[next cloak]

I wrote a solution to the paper

if you want i can send you

[remote parcel]

sure

[next cloak]

can you open it?

[remote parcel]

yes am viewing it

I'm almost mentally there

[next cloak]

So A.w = v is because of Rank = m

[remote parcel]

So we have a vector v, in R^m space

[next cloak]

Yes and because the rank is full we can write it as an output of A and a random vector

[remote parcel]

Because of A being rank m, there exists some w such that Aw=v

[next cloak]

Exactly

[remote parcel]

And because the span of v1,...vk is R^n, w is a linear combination of those

[next cloak]

yess

[remote parcel]

And this works for any v, which means that the span of Av1,... Avk is the exact same as R^m

[next cloak]

yes

[remote parcel]

Because A is basically a translator between R^n and R^m

[next cloak]

Yeah

[remote parcel]

Alright, thank you

That was quite clear

.close

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[next cloak]

im glad you got it

[next cloak]

.close

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[drowsy obsidian]

why should i do this

[prisma pike]

I assume as preparation to use IBP

[drowsy obsidian]

by parts?

but how would i think of this approach

[cedar kiln]

@drowsy obsidian Has your question been resolved?

[drowsy obsidian]

<@&286206848099549185>

[regal talon]

you should use the parts formula for integration but its going to be lengthy

[cedar kiln]

@drowsy obsidian Has your question been resolved?

[drowsy obsidian]

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[crisp thorn]

Hey guys. I'm writing my IB math IA right now, and I wanted to compare three different attempts to model caffeine in the body using the Akaike information criterion. Namely, the michaelis menten equation (given by v = (Vmax*[s]) / (Km +[s]) where Vmax and Km remain constant), a first-order one-compartment model ( given by C(t) = (D*Ka) / (Vd(Ka-Ke)) * e^(−ke​(t))−e^(−ka(​t)). Where every parameter(?) is a constant ) and finally a polynomial regression model ( as y=β0​+β1​x+β2​x2+β3​x3+⋯+βn​xn+ε ). I'm learning stats and modelling basics for the very first time so please forgive me if im forgetting important information or giving unimportant ones. However, I read on the math stack exchange that if I'm comparing non-nested models like these, I shouldn't be using the AIC in Ripley's school of thought. Is he right in stating that, or for my purposes am I safe to use the AIC to determine the best fitting model?

https://mathoverflow.net/questions/249448/use-of-akaike-information-criterion-with-nonnested-models/249753#249753

[cedar kiln]

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[crude brook]

How do we call the red and blue lines in english? Sorry but my english terminology isn't very good

[dreamy void]

Isn't the blue a bisector and red a perpendicular?

[crude brook]

αα διχοτομος right

the blue one

the red line segment splits the bottom and the top line segment into 2 equal line segments

[crude brook]

midsegment?

[dreamy void]

never heard that before

apparently there is perpendicular and angle bisections

[crude brook]

wait

if they don't form right angles how are they called?

[dreamy void]

acute and obtuse angles?

[crude brook]

no i mean the red line segment would be a perpendiculer if it formed right angles with the other line segments, how would it be called if it just went through the middle of the other line segments?

[stiff totem]

just a bisector, or line segment bisector

blue is an angle bisector

[crude brook]

Ahh ok thank you

[crude brook]

Thank you too

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[dreamy void]

parakalo

[cedar kiln]

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[viral fox]

Hi, would you mind helping me understand the following sentence please? (I'm an undergrad student)

We also prove that if X and Y are complex normed spaces with dim(X) ≥ 2 and X admits a conjugation τ (i.e., a period-2 conjugate linear isometry), then every additive mapping A : X → Y preserving Birkhoff orthogonality is real-linear.
This is from the paper "Additive mappings preserving orthogonality between complex inner product spaces" (https://www.sciencedirect.com/science/article/pii/S0024379525000485?via%3Dihub). page 450

I don't understand the "X admits a conjugation τ" part, and its example as "a period-2 conjugate linear isometry"
Thanks.

[crimson delta]

a map f:X->X with f^2=id, f(cx)=conj(c)f(x) and f(x+y)=f(x)+f(y) and ||f(x)-f(y)|| = ||x-y||

[viral fox]

Thank you sir, by f^2=id, did you mean the [f]^2 = I ?
And by f(cx)=conj(c)f(x) and f(x+y)=f(x)+f(y), so f is an antilinear transformer?

[crimson delta]

f(f(x))=x

antilinear and conjugate linear maps are the same, yes

its just a generalization of what the normal conjugation as a map C->C does

if we view C as a normed R vector space

doing it twice does nothing, is linear, doesnt change distances

[viral fox]

Interesting, like a rotation by pi for instance?

[crimson delta]

aka multiplying by -1

yes

[viral fox]

Awesome, thanks. really really appreciate your time sir.
Math bless us all.🤘🍻🤓

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[stiff bloom]

What is the simplest method / formula to find the sine or cosine of an angle Theta with the access of a calculator? (Excluding the common function)

[crimson delta]

using the sin/cos button

[stiff bloom]

exclude that, I want a method

[crimson delta]

why

[inland ether]

then whats the use of it?

no use probably

[stiff bloom]

curious if it's possible. I'm committing to a MAP test tomorrow and I want any second-hand methtod

[crimson delta]

just dont

waste of time

[stiff bloom]

is there any methods tho?

[oblique flare]

probably taylor series if you really don't want to use sin and cos

[inland ether]

manually it is more feasable uk

[crimson delta]

well of course there are methods. you can do what the calculator does

but it will be slower

[inland ether]

long and more hectic

[crimson delta]

and just not worth it

[lofty sand]

You can use a series

To approximate it

[stiff bloom]

Ooh. How can I work it out?

[lofty sand]

Or you can draw the angle, measure the sides of the triangle and just do the relevant fraction

[oblique flare]

some calculator do offer a whole button for sums/series so just approximate to maybe 15 terms

[inland ether]

what if geometry boxes are prohibited?

[inland ether]

yes

[stiff bloom]

aha

[lofty sand]

I need to understand your use case better. Is it for an exam ? Do you have a specific sin/cos to find ? Or do you just imagine travelling back in time and having to compute trig to build a caveman’s house ?

[stiff bloom]

the more I iterate, the more precise the result is?

[stiff bloom]

it's for an international test. I just want a method to find it out naturally

[oblique flare]

if the angle is small enough (x in radians) just use
sinx=x and cosx=1-x^2/2

[lofty sand]

Yes

[lofty sand]

But in this case

You should learn your usual trig and then use relevant formulas to solve …

This is what they’re expecting here

[crimson delta]

this is just a waste of test time

no test will require you to do this

use the time to go over the other questions again

[stiff bloom]

map test does it rarely

I remember 1 question back in g7

it probably costed me 3 points

in the MAP

I was working with 251-261 territory scores

rn I'm working 261+

[native heath]

what question lol

[crimson delta]

ther was a question in g7 (whatever that means) that required you to compute a sin without a proper calculator worth 3 points?

[stiff bloom]

I don't remember it throughly but I reached a part of working out where finding cosine of a binomial was mandatory

[stiff bloom]

3 points is significant

[crimson delta]

this just feels like you messed up the question

and there was an easier way

[stiff bloom]

270 is 2 times better than 240

[timid basin]

guys whats the interval that this function is increasing?

[stiff bloom]

maybe lol

[native heath]

!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).

[timid basin]

sorry guys

[cedar kiln]

@stiff bloom Has your question been resolved?

[stiff bloom]

Yes

[cedar kiln]

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[supple falcon]

I need help solving this system of equations

[supple falcon]

$\begin{cases} x(x + 2)(2x + y) = 9 \ x^2 + 4x + y = 6 \end{cases}$ for $x, y \in \mathbb R$

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

how

[queen stirrup]

x^2 + 4x + y = x(x+2) + 2x+y

now do

[supple falcon]

wait how?

[wicked mantle]

That’s insightful, impressive that you find it fast

[queen stirrup]

well this question is too complicated to solve just in x and y

clearly some substitution has to be made

so to make the second equation similar to the first

you can write x^2 + 4x + y = x^2 + 2x + 2x + y

(to get 2x+y)

furthermore

[supple falcon]

i learned 5 system of equations that can be solved using methods

[queen stirrup]

x^2+2x = x(x+2)

[queen stirrup]

but did you understand what i did

[supple falcon]

ah i see

[supple falcon]

so you turn it into $ab = 9$ and $a + b = 6$

[wraith dagger]

1 divided by 0 equals Infinity

[queen stirrup]

yeah you can solve for a and b

[supple falcon]

where $a = x(x + 2)$ and $b = 2x + y$

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

great method

thanks

i'll need more help later

[queen stirrup]

:)

[old swan]

beautifully done

[supple falcon]

lol

how about $\begin{cases} (x + 6y)(3x + 2y) = 12 \ 2x^3 + 6y^3 + 15x^2y + 19y^2x + x + 6y = 12 \end{cases}$

[wraith dagger]

1 divided by 0 equals Infinity

[timber arrow]

Try and factor equation (2)

[supple falcon]

i saw i wanted $3x + 2y$ in (2) im thinking

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

and i had $x + 6y$

[wraith dagger]

1 divided by 0 equals Infinity

[timber arrow]

Im looking at a potential x+6y in the cubic and linear terms, can you go from there?

[supple falcon]

oh okay

wait a moment

[timber arrow]

You can also set the equations equal to each other since they both equal 12

[supple falcon]

i don't know if that will work

[timber arrow]

why not

A=12 and B=12 implies A=B

it may not be helpful but you can certainly try

[queen stirrup]

it will

try to do long division after you equate them

[supple falcon]

polynomial division

it's been a long time since i've visited that

[queen stirrup]

left side is divisible

[supple falcon]

like a year or smth

[queen stirrup]

its not that difficult

you can google it up

[supple falcon]

yeah

wait up

[queen stirrup]

hm i dont know but it might help

[supple falcon]

can i draft here or smth?

i'll try setting both sides equal first

[supple falcon]

$(x + 6y)(3x + 2y) = 2x^3 + 6y^3 + 15x^2y + 19y^2x + x + 6y = 12$

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

distributing the LHS, we get

$3x^2 + 20xy + 12y^2$

[wraith dagger]

1 divided by 0 equals Infinity

[idle tusk]

,w (2x^3 + 6y^3 + 15x^2 y + 19y^2 x + x + 6y)/(x+6y)

[wraith dagger]

Results provided by WolframAlpha

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Requested by arte_m_etra

[idle tusk]

i'll spare you the misery

it simplifies

[supple falcon]

derivative 😭

[idle tusk]

2 x^2 + 3 x y + y^2 + 1

[queen stirrup]

did everything but the division

[idle tusk]

this

[supple falcon]

...

[idle tusk]

lmao ikr

[supple falcon]

check my display name lol

[supple falcon]

probably i'll try factoring at (2)

$2x^3 + 6y^3 + 15x^2y + 19y^2x + x + 6y = 12$

[wraith dagger]

1 divided by 0 equals Infinity

[opal hinge]

,w simplify (2x^3 + 6y^3 + 15x^2 y + 19y^2 x + x + 6y)/(x+6y)

[wraith dagger]

Results provided by WolframAlpha

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Requested by yobi_fx.

[idle tusk]

yup

[supple falcon]

thanks lol

you spared me the factoring

[idle tusk]

1 long divided by 0 equals infinity

[supple falcon]

anyways it's time to do it

[opal hinge]

This's why i hate algebra

[supple falcon]

now ik what to factor

[supple falcon]

im making a latex document

defining axioms derived from programming concepts

types

[supple falcon]

$2x^2 + 3xy + y^2 + 1$

[wraith dagger]

1 divided by 0 equals Infinity

[queen stirrup]

another hint

once you get 2x^2 + 3xy + y^2 + 1 = 3x+2y you can transpose and factorise further

[opal hinge]

Probably a bunch of square

Yeah or factor

[supple falcon]

lel

[queen stirrup]

damn this question is painful

[supple falcon]

😭

it's in my homework

dw

[opal hinge]

I had to deal with these things in the past

That's why i hate algebra

[supple falcon]

you a teacher?

[opal hinge]

No, I'm a student

[supple falcon]

lol

[opal hinge]

Just like you

[supple falcon]

you from vietnam or smth?

i remembered who you are if im not mistaken

[opal hinge]

Yeah you probably right

[supple falcon]

how the hell are you seeing this factor in your exam

💀

am i dividing polynomials just to get a factor 😭

[queen stirrup]

not exactly this

[queen stirrup]

im talking about this equation

like 0 on one side

and factorisation on the other

its the natural continuation

[supple falcon]

first case: $x + 6y = 0$ or you can't divide by $x + 6y$

[wraith dagger]

1 divided by 0 equals Infinity

[queen stirrup]

o yeah

[supple falcon]

leaving $x = 6y$ easy substitution

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

oh wait no

$x + 6y = 0$ does not satisfy equation (1)

💀

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

you are telling me that i can factorize this

,w factorize 3x^2 + 3xy + y^2 + 1 - 3x - 2y

[wraith dagger]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by iaminfinityiq

[supple falcon]

💀

[queen stirrup]

i meant 2x^2 + 3xy + 1 - 3x - 2y

[supple falcon]

what's that lmao

[queen stirrup]

,w simplify 2x^2 + 3xy + y^2 + 1 - 3x-2y

[wraith dagger]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by cherrymans

[supple falcon]

oh im dead

[queen stirrup]

bruh its so bad

,w factorise ( 2x^2 + 3xy + y^2 + 1 - 3x-2y)

[wraith dagger]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by cherrymans

[queen stirrup]

yeah

this is the factorisation

[supple falcon]

WHY CAN YOU DO IT WITH AN S

AND NOT WITH A Z

[supple falcon]

american-british

[queen stirrup]

spelling

so when you see 2x^2 + 3xy + y^2 + 1 - 3x-2y = 0

you know its gotta be factored into something like (2x + .... )(x + ,,, )

[supple falcon]

ah yeah

more like

$(2x + ... + y)(x + ... + y)$

[queen stirrup]

then you fill as (2x + y + .. )(x + y + ..) to get the 3xy

[wraith dagger]

1 divided by 0 equals Infinity

[opal hinge]

-# Imagin factorizing these things in the test when you only have 15 minutes

[queen stirrup]

then finally (2x+y-1)(x+y-1)

[queen stirrup]

i would skip this

so now we have (2x+y-1)(x+y-1)=0

so 2x+y=1 or x+y = 1

i think you can do it

from here

[supple falcon]

when you snuck a phone to the test

and use wolfram alpha

💀

yeah i have like 45 minutes on my clock

first case i got $5x^2 + 4x = 0$ 💀

[wraith dagger]

1 divided by 0 equals Infinity

[queen stirrup]

x=0,y=1 works

then

[supple falcon]

yeah i got that

and $x = \frac{-4}{5}, y = \frac{9}{5}$

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

that's for $x + y - 1 = 0$

[wraith dagger]

1 divided by 0 equals Infinity

[queen stirrup]

nice

any more questions?

[supple falcon]

second case i got $11x^2 - 28x = 0$

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

last one

$\begin{cases} x^3 - xy^2 - 6y = 0 \ (x + y)(x + 2y) = 3(xy + 2) \end{cases}$

[wraith dagger]

1 divided by 0 equals Infinity

[opal hinge]

People have no chill for doing this type of problem...

The first eq

Is quardratic in term y

[supple falcon]

my hw got 3 sections

all these questions are from section 3

i got 30 minutes on my clock

[honest field]

try to break down the 2nd equation

3xy gets eliminated

[supple falcon]

🤯

we have a genius in #help-13

[honest field]

what do you get?

[supple falcon]

$x^2 + 2y^2 = 6$

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

first equation annoys me

[honest field]

now replace the 6 into the 1st equation

[queen stirrup]

do you have some latex shortcuts or something

[supple falcon]

wait, sub that into the first equation

[supple falcon]

i use latex a lot

[supple falcon]

$x^3 - xy^2 - (x^2 + 2y^2)y = 0$

[wraith dagger]

1 divided by 0 equals Infinity

[honest field]

expand that

[supple falcon]

$x^3 - xy^2 - x^2y - 2y^3 = 0$

in vietnamese

[honest field]

wrong sign

[supple falcon]

phương trình đẳng cấp

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

probably @opal hinge is familiar with that

[honest field]

to make this easy you can divide y^3 on both sides

[supple falcon]

lol

i think i can do it from hể

thanks yall

[honest field]

dễ

[supple falcon]

💀

decarbonized

ik you lol

[honest field]

thì?

[supple falcon]

.close

[cedar kiln]

Channel closed

Closed by @supple falcon

Use .reopen if this was a mistake.

[opal hinge]

This guy has absolutely no chill for saying that

[supple falcon]

lol

.reopen

[cedar kiln]

✅

[supple falcon]

i'll keep this channel open until im done lol

[queen stirrup]

so how would you factorise a^3 - a - a^2 -2 = 0

[supple falcon]

you do $a^3 - a - a^2 - 2 = 0$

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

and that's your latex

[queen stirrup]

theres no style points in maths

[supple falcon]

shush

not like you would write

[opal hinge]

Use wolfram alpha

[supple falcon]

a^3 - a - a^2 - 2 = 0 in maths

💀

[queen stirrup]

i use latex for all my projects

[supple falcon]

,w factorise a^3 - a - a^2 - 2

[wraith dagger]

Results provided by WolframAlpha

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

Requested by iaminfinityiq

[queen stirrup]

BRO THATS NOT THE WAY TO FACTORISE

[supple falcon]

lol

anyways

i have a problem for yall

[indigo lagoon]

Yeah sure

[supple falcon]

given a polynomial $f(x) = \Sigma_{i=0}^{n}mx^i$ with degree $n$ ($2 \nmid n$) and $m \in \mathbb R, \abs{m} > 0$. Solve for $x \in \mathbb R$

[wraith dagger]

1 divided by 0 equals Infinity

[supple falcon]

@indigo lagoon

[queen stirrup]

wdym solve for x

so when f(x) = 0

[supple falcon]

yes

[queen stirrup]

or the roots of f(x)?

[supple falcon]

when $f(x) = 0$

[wraith dagger]

1 divided by 0 equals Infinity

[honest field]

isnt this just a geometric progression?

[queen stirrup]

ok m(...)= 0 or (...) = 0. as x is not 1 f(x) = 0 => (x^n+1 - 1)/(x-1) = 0 => x^(n+1) -1 = 0

[supple falcon]

(only factorisation is allowed, not anything else)

[queen stirrup]

wait a minute

x=1

wait

sorry

x=-1?

[supple falcon]

show your working

[queen stirrup]

so from here x^n+1 = 1

n+1 is even

so among all the n+1 roots of unity

only x = -1,1 are real

and 1 doesnt satisfy

[honest field]

also please do not give challenges in help channels

[supple falcon]

.

[queen stirrup]

right you should close this

we can do it in #math-discussion

[supple falcon]

yea

[honest field]

.close

[cedar kiln]

Channel closed

Closed by @honest field

Use .reopen if this was a mistake.

[supple falcon]

continue in there lol

[cedar kiln]

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[calm pine]

Can anyone reccomend a good channel that helps you understand general topics in math a?

[native heath]

#chill

[indigo lagoon]

Probably asking at #math-discussion would be a better idea since help channels are generally for specific math questions

[fossil dawn]

since you are asking for resources, #book-recommendations helps too

[native heath]

book recs for videos??

[fossil dawn]

as far as i know it's long been a spot to ask for general resources

[native heath]

ah

[cedar kiln]

@calm pine Has your question been resolved?

[cedar kiln]

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[fallen coyote]

for this, would i just plug in numbers close to 3 like 2.999?

[fallen coyote]

i tried this, and got to positive infinity, which is a VA?

[native heath]

yes

[fallen coyote]

.close

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[civic sage]

i have no idea what to do

[tough finch]

have you tried anything yet?

[civic sage]

no

i could write tan as sin/cos but that leads nowhere

[terse kettle]

Its very simple...Firstly think how can I convert this sin to tan....Exactly by dividing with cos....So the answer is just divide everything with cos in denominator and numerator....You'll get the answer

[native heath]

have you learnt the ancient art of 'divide by cosine'

[civic sage]

no sir

[terse kettle]

🤣

[upper laurel]

suppose you try it

and find out?

for example if you had (5 - 3)/(5 + 3),

you can divide by 3/3

[last apex]

this is exactly what you should do!

[upper laurel]

and get (5/3 - 1)/(5/3 + 1)

[civic sage]

the others seem to be recommending dividing the LHS by cos

[terse kettle]

Exactly

[last apex]

its the same thing but in the other directon

but what you did is correct

[terse kettle]

Yes....Both ways you will get RHS or LHS respectively

[civic sage]

ok here id prefer working with the RHS

.close

[cedar kiln]

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[cedar kiln]

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[bronze dome]

i’m back 😞

[bronze dome]

idk what i’m doing 💔

[modern compass]

you're 5 line starts at x=0 instead of x=1

and there's a closed dot on your -x+8 line at x=6

[bronze dome]

do i open the dot for the 5 line too

[modern compass]

yes, since it's given -5 < x < 1, neither endpoint is included in your line

[bronze dome]

ohh okok

does this look correct ? 😭

[wooden wagon]

yeah

[bronze dome]

okay thank you

does this look correct ? 😓

[final river]

what's the first part of that piecewise

[bronze dome]

my bad

[final river]

You got it!

[bronze dome]

oh jolly i’m starting to understand this :D

.close

[cedar kiln]

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[harsh hazel]

$P(m,n)=\sum{i=1}^m $

[indigo lagoon]

remove the space

[tropic oxide]

did you cut yourself off by accident?

[indigo lagoon]

$P(m,n)=\sum{i=1}^m$

[wraith dagger]

This is sad 😢

[harsh hazel]

I need a double summation

Check #latex-testing

[tropic oxide]

$P(m,n)=\sum_{i=1}^m \sum_{j=1}^n (i+j)^7$

[wraith dagger]

Ann

[tropic oxide]

this thing? (from there)

[harsh hazel]

Yes! Thanks

[harsh hazel]

So if |P(3,3)|=M find M

[neon falcon]

Oh man I was here for basic math help and one look and I’m cooked

[tropic oxide]

well that's like, what, 9 terms?

wait

P(3,-3)?

[harsh hazel]

Should I do binomial theorem?

[tropic oxide]

how are we allowing n to be negative here??

[harsh hazel]

It should be 3

[tropic oxide]

so P(3,3)?

[harsh hazel]

Yes

[tropic oxide]

i think it is easiest to just write the sum out explicitly ngl

[harsh hazel]

What would be a general approach? (If m,n aren't small)

[tropic oxide]

honestly probably some kind of algebraic magic to reduce this to summations of the form $\sum_{k=1}^n k^p$

[wraith dagger]

Ann

[tropic oxide]

this feels like it's going to end up with like, m^9 type shit

[harsh hazel]

Where p is?

[tropic oxide]

a natural number

[harsh hazel]

Okay, how can we reach it?

[tropic oxide]

im kinda suspicious tbh of how youve presented the problem

can you share a pic of the original

like why'd they ask you for the absolute value of sth that is already obviously positive, for example...

[harsh hazel]

Okay, I'll try. (The thing is that I found this in my desk while cleaning..appears to be a asked 3-4 years ago)

@tropic oxide Leave this, this problem is faulty.

(3,-3) is simply not defined.

Domain of P is Z+

[tropic oxide]

ok no hold on

ok so the summation only defines P for m, n in N that's correct

[harsh hazel]

So, we can't use this form for P(3,-3)?

[tropic oxide]

but it is true that it'll be a polynomial in m and n, and that makes sense for any values of the variables

so yes we cannot use the summation form directly for P(3, -3)

nor do we need to, i think

[harsh hazel]

Yup

[tropic oxide]

i think the idea is rather that $P(m,n) - P(m,n-1)$ has a nice expression (which can be used for $n \leq 1$ as well)

[wraith dagger]

Ann

[tropic oxide]

maybe write put P(3, n) and then P(3, n) - P(3, n-1).

[harsh hazel]

Oh alright and then we can plug in n=-3

[tropic oxide]

not quite, you'll need to take it in several steps.

you'll need to "roll it back" starting from e.g. P(3, 1)

[harsh hazel]

That's painfully calculative..

[tropic oxide]

is it now

can you write me an expression (with one sigma symbol) for P(3, n)?

[harsh hazel]

Yeah, $P(3, n) = \sum_{j=1}^{n} [ (j + 1)^7 + (j + 2)^7 + (j + 3)^7]$

[wraith dagger]

L'empereur

[tropic oxide]

good, now the backward difference?

[harsh hazel]

Starting from n=1?

$P(3,n)-P(3,n-1)= (n+1)^7 + (n+2)^7 + (n+3)^7$

[wraith dagger]

L'empereur

[harsh hazel]

So, we have a recurrence relation?

[cedar kiln]

@harsh hazel Has your question been resolved?

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Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[crimson sedge]

Is this right

[last apex]

you created the graph for this function?

[gray roost]

what

is negative 1

[crimson sedge]

yea

[tropic oxide]

nope

[crimson sedge]

how

[last apex]

there are 3 mistakes

[crimson sedge]

I used desmos

[tropic oxide]

fails the vertical line test for one

[gray roost]

you can put -1 into a function???

[crimson sedge]

bro theres no way

wait is desmos

[tropic oxide]

ok so show what you put into desmos.

[crimson sedge]

yea im cooked

I put a :

[tropic oxide]

and what did desmos give you

[crimson sedge]

v

ah

I see

[last apex]

desmos is correct

[tropic oxide]

oh wait hold on

you typoed

-4x < x < 1?

[crimson sedge]

yes

oh

[tropic oxide]

that and you then didn't copy the graph correctly anyway. mind your axes and coordinates

[crimson sedge]

how about open

and closed circles

[last apex]

for this you should look at the function it self, and ask yourself what is f(1)

[tropic oxide]

let's focus on the shape first. recreate the graph now, and fill the circles in however you think is right, but don't worry about them

we will get the shape right and only then talk about the end circles

[crimson sedge]

now this is

better

right

[tropic oxide]

yes now it is

now look at the horizontal piece, the -1

[crimson sedge]

thats open

[tropic oxide]

its "subdomain" is -4 < x < 1

[crimson sedge]

ah

[tropic oxide]

on which end?

[crimson sedge]

so if its greater or less than equal to

its closed

right

[tropic oxide]

strict inequality (> or <) means open, weak inequality (>= or <=) means closed, yes

[crimson sedge]

ye

[crimson sedge]

so open circle

is at (1,3)

[tropic oxide]

that's one of several open circles that will need to be put down.

how about you put down all the open circles that you think should be open, and show what you got

[crimson sedge]

1,-1

[tropic oxide]

rather than saying stuff piecemeal like you're doing rn

[tropic oxide]

far left

[crimson sedge]

so theres

3

[tropic oxide]

-4**<**x<1
that end also wants an open circle!

[crimson sedge]

open

circle

[tropic oxide]

can

you

type

more

than

one

word

per

message

[crimson sedge]

so theres 3 open circles

[tropic oxide]

really annoying to read this typing style

[tropic oxide]

yes there are

[crimson sedge]

ok

thanks

[crimson sedge]

can you help with level 2

because level one for this is easier

and desmos isnt very helpful for the second level

[tropic oxide]

so you have another question like this

send it here ig

[crimson sedge]

yes

[tropic oxide]

ok so you have three pieces this time

make an attempt

then we will see how you did

[crimson sedge]

ok

[crimson sedge]

would desmos work or no in this scenario

[tropic oxide]

i mean, desmos CAN handle functions composed of 3 pieces. (in fact you can put cases in one pair of braces, separated by commas)

[crimson sedge]

I did last time

but the graph was wrong

[tropic oxide]

cause you made a typo lmao

[crimson sedge]

ill show you

[tropic oxide]

or like alternatively you could graph these straight line bits manually

if you know how to graph y=mx+c it is not terribly difficult in comparison

[crimson sedge]

ok

[tropic oxide]

i would say relying so strongly upon desmos here is a bit silly. it's like learning basic arithmetic in class, and then shoving every single q into a calculator

[crimson sedge]

its a pass or fail class

nothing serious

I plugged in the third

graph

and nothing showw

s

is that a error

oh x>0

[tropic oxide]

are you in like college algebra

[crimson sedge]

no

highschool

precalc

im a junior

[tropic oxide]

which is... 11th grade?

[crimson sedge]

yes

[tropic oxide]

are you gonna do anything STEM in uni

[crimson sedge]

never seen this question before

[crimson sedge]

im majoring in biological sciences

im into science

[tropic oxide]

right

[crimson sedge]

medicine primairly

[tropic oxide]

you'll need math then. suggest taking a less lackadaisical approach to this stuff.

[crimson sedge]

the issue is my teacher isnt good

he is 77 years old

[tropic oxide]

-3 for x=5 means the graph has a single point jutting out

[crimson sedge]

?

wym

[tropic oxide]

select "closed circle" and click at (5, -3) and then don't click anywhere else

[crimson sedge]

can you demonstrate visually

[crimson sedge]

okay

Like this?

[tropic oxide]

yes good

the other bit of the graph is just the entire line y=2x-7 except for the point at x=5 (on which you should put an open circle)

[crimson sedge]

okay

@tropic oxide

Like this?

[tropic oxide]

• make the line go ALL the way across the graph. don't stop it anywhere
• open circle at x=5 on the line

[crimson sedge]

k

[crimson sedge]

so its infinite

right

[tropic oxide]

yes, it goes on forever

[crimson sedge]

Like this?

@tropic oxide

[tropic oxide]

no

that circle should be on the line y=2x-7

[cedar kiln]

@crimson sedge Has your question been resolved?

[cedar kiln]

@crimson sedge Has your question been resolved?

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[cedar kiln]

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[sonic fossil]

Question is find the velocity of ball and triangular wedge after impact

Question is given by teacher in this form so that anyone cannot internet it

But I am confused to identify line on impact

of*

[cedar kiln]

@sonic fossil Has your question been resolved?

[cedar kiln]

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[strong dawn]

can anyone help me with question 12 abc

[past wave]

What have you tried?

@strong dawn

[strong dawn]

I expressed the equation like this and dont how to go

[past wave]

Nice

Do you know how to draw the graph for x²-2x-8?

[strong dawn]

I know how to draw that graph, I just don't how to deal with the u3(x)

[past wave]

Ohk

Do you see that for x>3, the output of u3 will be 1

So f(x) will just be that quad

[strong dawn]

so I draw a graph of x^2 -2x-8 and erase the part that is less that x = 3

[past wave]

Yup

[strong dawn]

but how about u3(x)=0

[past wave]

Yes for x < 3

[lavish linden]

i hate when people do this shit

[strong dawn]

I think I just get it, I just draw y=0 (x<3)

both graph

[past wave]

Yess

Share what you have drawn

[strong dawn]

.close

[cedar kiln]

Channel closed

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[past wave]

.reopen

[cedar kiln]

✅

[past wave]

@strong dawn

The minima is not at x=3

Oh shi mb

It's correct

.close

[cedar kiln]

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[cedar kiln]

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[orchid salmon]

hi

[opal hinge]

wassup

[orchid salmon]

are u australian

[opal hinge]

nahh

[orchid salmon]

ok no prob

i got like the big test for year 11 coming up

so like

early entry for uni test or wtv in otgher countrys

coutnies

im not bothered to spell sorry

i dont understand how to get something else to a subjectof the formula

its so confusing

this from my study queystions

my teacher gave

[woven ruin]

which one of these do you not understand

[orchid salmon]

making r the subjeect of the formula

[woven ruin]

the context of a rope is basically irrelevant in this case

[woven ruin]

make it so that R = sonetging

instead of T = something

[orchid salmon]

r = 1 ig

[woven ruin]

R is a variable

move R to the left side

and everything else to the right side

by squaring, dividing and multiplying

[orchid salmon]

so first ik u square both sides yes?

[woven ruin]

yes

[orchid salmon]

ok

now divide

divide what 😭

[woven ruin]

in no particular order

[orchid salmon]

oksy

[woven ruin]

what do you have after squaring

[orchid salmon]

T^2 = 8R over 2.5

[woven ruin]

okay what will you do next

[orchid salmon]

i have no idea

[woven ruin]

what would you have if you multiplied both sides by 2.5

[orchid salmon]

T^2 = 20 R

wait

[woven ruin]

think harder

[orchid salmon]

oh

the 2.5 will move to other side

and since it sa fraction

the other will disappeasr

[woven ruin]

indeed will

so what will you end up with

[orchid salmon]

so T^2 x 2.5 = R

[woven ruin]

you dropped a number

[orchid salmon]

8

20

SHJIT

8 R

not 20

T^2 x 2.5 = 8 R

[woven ruin]

yes

and what’s the last step left to do so that you only have R = …

[orchid salmon]

divide both sidr by 8?

[woven ruin]

yes

[orchid salmon]

actually

i just guiessed

[woven ruin]

well

[orchid salmon]

l;olk

[woven ruin]

so part a done

is part b clear

[orchid salmon]

lemme read it rq

b What is the length of rope if it takes 16 seconds for a single revolution?

i measn its just putting it in to the formula u just helped me get

OLH MY GOD

IT ACTUALLY WORKED

80

YES

c What is the time taken for a rope of length 18 metres to complete a revolution? Answer correct to
one decimal place.

ok

YES

i got them right now

thanks

i might have more questions in a bit when i do other quesitons

[woven ruin]

ye feel free to ping me

for now close this channel

[cedar kiln]

@orchid salmon Has your question been resolved?

[cedar kiln]

Channel closed

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[cedar kiln]

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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[nocturne yacht]

Hey guys can anyone explain me my doubt

[nocturne yacht]

It's more physics than maths but still

[indigo lagoon]

Yeah, just post

[nocturne yacht]

You know that equation f=ma

Which earlier is f (directly proportional to )ma

So we add that k and make it f=kma

And then we say that let f,m&a be 1 so k will be 1 too

[floral arrow]

It only depends on the units you use

[sonic fossil]

It's not like that

[nocturne yacht]

?

[sonic fossil]

What is logic of saying f is directly proportional to ma

[nocturne yacht]

I don't know if it's directly proportional but this is the sign

Sorry the image changed

[sonic fossil]

Yes it is sign of directly proportional

[livid hound]

$\propto$

[wraith dagger]

ραμOmeganato5

[nocturne yacht]

So then it's correct

[floral arrow]

Yes it is

[livid hound]

F = ma = kma with k=1

[nocturne yacht]

We take k as the constant of proportionality

See that's in my book

[sonic fossil]

if there is acceleration then this is due to force.
=> $$ F $\propto$ a
=> F = ma where F is net force on the object and m is mass of com and a is acceleration of com

[wraith dagger]

BlackidoZΣ
Compile Error! Click the reaction for more information.
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[sonic fossil]

Ahh my latex

[nocturne yacht]

Yeah but my question is not that

[floral arrow]

The force F is defined such that k = 1

[sonic fossil]

alr

[floral arrow]

That's what your book says

[nocturne yacht]

I wanna know when we take f,m&a 1 the k them to 1 so what if we all of that 2

[sonic fossil]

It's defined there with different way

[sonic fossil]

Can you rephrase your wording please

[nocturne yacht]

Ah sure
We know that after we got f=kma , we take / suppose value of f , m & a to be 1N , 1kg & 1m/s² respectively and obtain k = 1 and then we write f=ma .
But what if i take all those 2 (/ or any other constant) wouldn't that make the value of k different

Like in the case of 2 ,k will be 1/2

[sonic fossil]

It would make a difference if you change magnitude but here we take k = 1 to balance the proportionality sign

[floral arrow]

What you don't seem to understand is that 1 Newton is defined to be 1kg * 1m/s²

[nocturne yacht]

Nope i understand that

[nocturne yacht]

Exactly

Why 1

[floral arrow]

If you take 2kg and 2m/s², then F = ma because F = 4N

[nocturne yacht]

So what will be k then

[sonic fossil]

To balance the proportionality sign otherwise it wouldn't be generalized

[floral arrow]

Still 1

You can define your own unit if you want, say a T, where 1T = 2N
Then you can have 2T = 2kg * 2m/s²

[nocturne yacht]

Can you explain more briefly

[nocturne yacht]

But wouldn't there be a k in there too

Sorry guys if i am wasting your time but i just can't get my head over it

[floral arrow]

Yeah, it would be 1T = 2 * 1kg * 1m/s²

k = 2

The point of a Newton is that k = 1

[nocturne yacht]

Yeah so then we should take k in the equation too

[nocturne yacht]

Why and how

[floral arrow]

Because we define it that way

[nocturne yacht]

And why do we do that

[floral arrow]

Convenience

[nocturne yacht]

Where tf did it come from i am asking

[nocturne yacht]

Sorry for the language

[sonic fossil]

For different situations in physics you deal with different forces and as far as putting different magnitudes for m and a you will get a force on that body with it's acceleration, here the value of k is equal 1,the reason is that if you take a mathematical constant to be any value( as you're saying 2,3,4etc) then 1)it won't give you the exact force acting on particle. 2) if you multiply k to m and a force then will be a multiple of that constant, in short force will be different from its original calculation.

[floral arrow]

... from the definition?? I don't understand what you don't understand about this

[sonic fossil]

Don't get confused, read this carefully

[nocturne yacht]

Sure sir

[floral arrow]

We already use kg for mass and m/s² for acceleration, these are our standard units

Since force is the product of these two, we use the unit kg*m/s², and we call it N

[nocturne yacht]

Ahh but i still don't understand this

[floral arrow]

Rather, since force is proportional to the product of mass and acceleration, we use the unit kg*m/s², instead of 2kg*m/s² or something else

Because we don't want to keep a useless constant around

[nocturne yacht]

I am saying that why

[sonic fossil]

I think you will understand it, if you can't comprehend right now then leave it for now and understand it next day maybe you will understand more better

[sonic fossil]

Oh that's the fundamental definition

[nocturne yacht]

But it will be inaccurate according to maths

[floral arrow]

Sorry, if you can't understand that we define things such that they are convenient to use, I can't help you further

[nocturne yacht]

Why

[nocturne yacht]

Sure i don't wanna waste your time
Have a good day

[sonic fossil]

It's good to ask why but not for the fundamental definition, physicists observed nature and then wrote it in mathematical terms and so definition is fundamental

[nocturne yacht]

Ok sorry but i just wanted to understand it

Sorry again for wasting your time @sonic fossil

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[cedar kiln]

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[glacial warren]

Hey i need help understanding logs, im in calc 2 but my teacher says they are review from calc 1(but my teacher in calc one barely touched them)

[indigo lagoon]

What sort of log

Like from the basics?

[glacial warren]

yes

natural logs and regular logs

[indigo lagoon]

Alright, do you have any prior knowledge about it?

[glacial warren]

very little

[indigo lagoon]

It's okay, just tell evrythung you know about log

everything

[glacial warren]

so

i know that log base 2 (8) = 3

so its like 2^3 is 8

but then they started to put xLn(x) / x and asked to take the deriv

[indigo lagoon]

$log_{2}^{8} = 3$

[wraith dagger]

This is sad 😢

[native heath]

lmao

[indigo lagoon]

oops

@native heath can you fix it 4 me?

[native heath]

this is hideous latex

$\log_2(8) = 3$

[wraith dagger]

Περσυ

[glacial warren]

yes that^

[indigo lagoon]

Alright, anything else aside from this?

[native heath]

see @indigo lagoon the helpee knows better than you

[glacial warren]

not really

my calc one teacher didnt touch more than that

[indigo lagoon]

Alright

Let's talk about how to add two 🪵

[glacial warren]

sorry if im kinda a lost cause

[indigo lagoon]

So basically $\log_{a} b + \log_{a} c = \log_{a}{bc}$

[wraith dagger]

This is sad 😢

[indigo lagoon]

nah, it's good

[indigo lagoon]

Is it acceptable?

[glacial warren]

okay so if it was log base 2 of 8 and log base 2 of 4 it would become log base 2 of 32

[indigo lagoon]

yup

exactly

[indigo lagoon]

Can you find the formula of deduction according to this?

[glacial warren]

so 2^5

so it would be log base 2 of 32 = 5

[indigo lagoon]

First formula? yes

[glacial warren]

okay cool!

[indigo lagoon]

✅

[glacial warren]

so i can kinda understand natural logs

its just another way of exponents

but what exactly are natural logs

[indigo lagoon]

it's just log with base e

nothing special

[glacial warren]

what is e exactly ...

[indigo lagoon]

,w e

[wraith dagger]

Results provided by WolframAlpha

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[glacial warren]

sorry if thats a dumb question

[indigo lagoon]

e is a constant....but for further explaination or application you should look up some youtube videos for it

[glacial warren]

anyone you recommend?

[indigo lagoon]

overall, e is a irrational constant

[indigo lagoon]

Not really, might consider asking at #math-discussion

[glacial warren]

okay will do! also i really appreciate the help

[native heath]

should i just repost the video from yday lmfao

https://www.youtube.com/watch?v=rBhJhD0ITuY&t=508s

[indigo lagoon]

Anything else? We've only undergone one property

[glacial warren]

i was gonna show you the math problems ive been attempting

[indigo lagoon]

Sure!!

More than welcome

[indigo lagoon]

@glacial warren yeah, you should check this out