help-49

[twilit field]

I don't really get this

[humble torrent]

Hint: ||think about picking a basis||

[twilit field]

I suppose I extend the basis

[midnight plank]

@twilit field Has your question been resolved?

[twilit field]

We assume $U$ to be a n-dimensional vector sub-space . We assume $V$ to be $(m+n)$ dimensional. We let $W$ be $r$ dimensional

Let the basis of $U$, be $\beta_1 = {e_1,e_2,\dots , e_n}$.
\
We can then extend the basis of $U$ to that of $V$ by taking $m$ linearly independent vectors in $V$, other than the ones already chosen.
\
\
Let the new basis be $\beta_2= {e_1,e_2,\dots, e_n,f_1,f_2,\dots, f_m}$
\
We now propose a linear transformation that results in an identical output.
\
As established earlier, an entire transformation, can be described by describing the transformation of the basis .
\
Let $v \in U$. Then $T(v) = \sum_{i=1}^{n} \alpha_i T(e_i)$ .
\
We now define $S(v)$ as follows , where $v = \sum_{i=1}^{n} \gamma e_i + \sum_{i=1}^{m} \beta_i f_i$
\
$S(v) = \begin{cases} \sum_{i=1}^{n} \alpha_i T(e_i) & \forall e_i \ 0_{w} & \forall f_i \end{cases}$

[grand pond]

A dense set

[twilit field]

How does this look?

[humble torrent]

Are u defining both S and T here?

Youve been given S arbitrarily

And your job is to define T

[twilit field]

I don't follow

[humble torrent]

Oh nvm youve just switched the letters

Yeah looks good

[twilit field]

I'm not too happy with how I defined s(v) tbh

[humble torrent]

There is some not needed cluttering imo but thats just taste, as long as you think the proof is convincing for anyone reading it

[formal blade]

Do you know that any lt is uniquely defined by where it sends basis vectors?

[twilit field]

Yes, I proved that a while ago

[formal blade]

If so you can make it simpler by using that.

[twilit field]

I've used that here

[formal blade]

Yes I know

[humble torrent]

You can simply state it, i.e "extend linearly"

[formal blade]

My point is if you don't like how you defined s(v) this is a bit more concise

[twilit field]

I'm listening

[formal blade]

So you map the basis vecs of ur subspace identically and map the other basis vecs to 0. It's the same map you have but you can state it this way more concisely.

"Extend linearly" is a good way to put it.

This way you don't have to explicitly talk about linear combos and sums like you do now.

[twilit field]

I see

Okie

[formal blade]

It's the same map just stated differently.

[twilit field]

Well, I think there is a slight mistake in how I defined S(v)

nvm

it's fine

Thanks!

LADR is so fun!

Re stating this

We suppose $V$ is a finite dimensional vector space, with non-zero dimension.
\
Let $W$ be infinite dimensional, then $\mathcal{L}(V,W)$ is infinite -dimensional.

[grand pond]

A dense set

[twilit field]

Right so far?

[runic hamlet]

We can then extend the basis of $U$ to that of $V$ by taking $m$ linearly independent vectors in $V$, other than the ones already chosen.
this sentence is bad. it makes it sound like you think just taking any m linearly independent vectors (except for e_1,...,e_n) is enough. thats not how that works.

you could just write "we extended the basis of U to a basis of V"

[twilit field]

We wish to prove the contrapositive: If $\mathcal{L}(V,W)$ is finite dimensional then $W$ is finite dimensional.
\
We assume $V$ is n-dimensional, and that $W$ has an arbitrary number of dimensions , say $r$
\
We know that all linear transformations $T : V \to W$ are of the form $\sum_{i=1}^{r} \sum_{i=j}^{n} \alpha_i \beta_j e_j$.
\
It follows then that there are $r$ basis vectors of $\mathcal{L}(V,W)$
\
It follows then that $W$ has a dimension of $r$.
\
As $r$ is finite, it follows that $W$ is a finite dim vector space

[twilit field]

Isn't that how it works

[runic hamlet]

no

[twilit field]

How does it work then

[runic hamlet]

look back at the proof of that theorem

[grand pond]

A dense set

[twilit field]

Ah, nvm

got my mistake

[twilit field]

How does this proof look?

[runic hamlet]

not good

you cant even write down that sum if r isnt finite to begin with

the dimension of L(V,W) certainly also isnt r

I am not sure how you are concluding that from the summation you wrote anyway. what are those variables

its very much easier to give a direct proof

[twilit field]

My idea is basically trying to exploit the general form a linear transformation from an n to m dim vector space

[runic hamlet]

but you dont know that W is finite dim

so how is that helpful

[twilit field]

oh right

So a direct proof it is

Let $W$ be infinite dimensional, then $\mathcal{L}(V,W)$ is infinite -dimensional.
\
As V is finite dimensional, and say of dimension $n$, a basis of it would look like ${f_1,f_2,\dots, f_n}
As $W$ is infinite dimensional , a basis would look like ${e_1,e_2,\dots,}$.
\
A transformation would thus look something like $T(e_i) = \sum_{i=1}^{\infty} \sum_{i=1}^{n} \alpha_{i,j} e_i$

[grand pond]

A dense set
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[twilit field]

nah, this is messy

and assumes the infinity is countably infinite

I think I'll push problems on infinite dimensions, until I formally study kinds of infinites

this feels really nasty

.close

[midnight plank]

Channel closed

Closed by @twilit field

Use .reopen if this was a mistake.

[twilit field]

.reopen

[midnight plank]

✅

[twilit field]

Yes?

[hard umbra]

it suffices to prove this for dim V = 1 and W with countably infinite dimension

since that'll be a subspace

[twilit field]

hmm, okay

thanks

[hard umbra]

[twilit field]

We suppose $V$ is a finite dimensional vector space, with non-zero dimension.
\
Let $W$ be infinite dimensional, then $\mathcal{L}(V,W)$ is infinite -dimensional.
\
Let $V$ have dimension $1$ and $W$ be a vector space of countably infinite dimension.
\
\
Let $e_1$ be a basis of $V$
It then follows that $\mathcal{L} (V,W)$ have elements of the form $T(v) = \eta \sum_{i=1}^{\infty} a_i e_1 ; \eta, a_i \in \R$

[grand pond]

A dense set

[twilit field]

Wait

no this is wrong

this doesn't increase the number of dimensions

[visual tiger]

what do you mean e1 is a basis of V

[twilit field]

I'm supposing $V$ is 1D

[grand pond]

A dense set

[visual tiger]

if V is one dim then L(V,W) has same size as W

it's when dim V > 1 that L(V,W) increases in size compared to both

[twilit field]

I'll havr to prove that, no?

[visual tiger]

since V is finite dim (suppose basis e1,...,en), to choose a linear map T of L(V,W) you need to choose T(e1),...,T(en)

and that's it

[twilit field]

Blake advised me to create a mapping from a 1 D space

[visual tiger]

Sure

if you start from a 1d space, let {e1} be your basis

give a name to T(e1)

[hard umbra]

there's no loss in that simplifying assumption i guess

[visual tiger]

find T(v)

[twilit field]

I'm trying to figure out how the LT would look for an arbitrary vector space

[hard umbra]

use the fact that linear maps are uniquely determined by how they act on bases

you dont need to write down the most general linear map V -> W

you just need to show that you can find infinitely many which are lin indep

[twilit field]

I know, but even for one basis it's messy

like just 1d to 2d

[hard umbra]

you don't need to write down all maps

if W has basis f1, f2

and V has a single basis vector e1

then i can define T(e1) = f1

alternatively i can define T(e1) = f2

so i've just given you two different linear maps V -> W

[visual tiger]

and two linearly independent ones at that

[twilit field]

I was told that a linear mapping had to be expressed as a function of the elements of the first set

unless I'm missing somethng

like (x) \to (2x,3x)

[visual tiger]

sure

if you want to give a name to the linear form that gives you the coordinate of x onto a certain basis vector

we usually call it $e^*$

[grand pond]

rafilou is not not born in 2003

[hard umbra]

so you want a function {e1} -> {f1, f2}

[visual tiger]

if $x = ae_1 + be_2$ for example

[grand pond]

rafilou is not not born in 2003

[hard umbra]

and you can boost it up to a linear map on the whole space

[visual tiger]

$e_2^*(x) = b$

[grand pond]

rafilou is not not born in 2003

[hard umbra]

eh you don't really need to bring duals into this

[visual tiger]

yes, and since T is linear

[hard umbra]

this is before axler discusses duals presumably

[visual tiger]

if he wanted the exact form of the linear map "in function of x"

[visual tiger]

But yeah didn't you already go through the fundamental theorem that says

T is uniquely determined by the images of a basis

?

[hard umbra]

you really just need that one theorem

[twilit field]

I do know that yes

[visual tiger]

ok

so you don't need the specifics of what that T is

[hard umbra]

define functions on the bases and boost it up to a linear map

[visual tiger]

if you know what the images of a basis are

[twilit field]

I can define $T(e_1) =\sum_{i=1}^{\infty} a_i f_i$

[grand pond]

A dense set

[visual tiger]

yes for example, with the a_i almost all zero

though you don't know if W is countably dimensional

so that sum is sketchy

oh wait you wrote it in your assumption ok

[hard umbra]

again, you really don't have to write down the most general map there is

just write down infinitely many that are linearly independent

[twilit field]

Suppose $dim(V)=1$ and $W$'s dimension is countably infinite. We then define $T(v)= \alpha T(e_1) =. \alpha \sum_{i=1}^{\infty} a_i f_i$.
\
a basis of $\mathcal{L}(V,W)$ would be ${f_1,f_2,\dots, }$, which is countably infinite. Thus $\dim(\mathcal{L}(V,W)$ is countably infinite too

[visual tiger]

f1,f2,... are not vectors of L(V,W)

[grand pond]

A dense set

[twilit field]

Hmm>

[twilit field]

Why not

[hard umbra]

they are presumably elements of W

[twilit field]

they are

[hard umbra]

f1, f2, ... is a countable linearly independent set in W

so they are not elements of L(V, W)

[twilit field]

Okay, so now I have to find elements of $\mathcal{L}(V,W)$ right

[grand pond]

A dense set

[hard umbra]

yes

i gave you an example earlier

\begin{itemize} \ii $\dim V = 1$ with basis $e_1$ \ii $\dim W = 2$ with basis $f_1, f_2$ \end{itemize}
you can define
[ T_1(e_1) = f_1, \quad T_2(e_1) = f_2 ]

[twilit field]

I don't see how that helps me

ah

I mean that gives me the same basis I gave earlier

[hard umbra]

,, T_1, T_2 \in \LL(V, W)

[twilit field]

oooh

okay

so $T_1, T_2,\dots$

[grand pond]

A dense set

[hard umbra]

you still should convince yourself that theyre linearly independent

i havent provided any proof of that

[twilit field]

I mean thease are

wait

oh

okay

yeah

makes sense

thanks

Can I close this now?

[hard umbra]

sure but make sure you can explain why this is true

its a short calculation but its not trivial

[twilit field]

T_1(e_1), T_2(e_1)\dots are LI

[hard umbra]

e_1 not e_2

[twilit field]

and T(v) = \alpha T(e_1)

which is again LI

[hard umbra]

but why are the functions LI

we are talking about linear independence in the vector space L(V, W)

the elements are functions

[twilit field]

f_1,f_2,\dots are LI, so \alpha f_1,\alpha f_2 \dots are LI

[hard umbra]

these are the outputs of the functions, not the functions themselves

[twilit field]

yes

[hard umbra]

so that doesnt let you conclude that the functions are LI

you must make an argument here about the functions themselves

[twilit field]

Okay, so I have $\sum_{i=1}^{\infty} T(v)=0$

[hard umbra]

no infinite sums, only finite ones

[grand pond]

A dense set

[twilit field]

okay

[hard umbra]

and you are not looking at T(v)

the sum in question would be something like
[ a_1 T_1 + a_2 T_2 + \dots + a_n T_n = 0 ]
where the $0$ on the RHS is the $0$ linear map

[hard umbra]

this is an equation about linear maps

you must prove that all a_1, ..., a_n = 0

[twilit field]

$a_iT_i= T_{i,i}$ we then have $\sum_{i=1}^{n} T_{i,i} =0$

[grand pond]

A dense set

[twilit field]

I'm not sure

I'll do this after dinner

orry

*sorry

[hard umbra]

the argument essentially falls back onto what you've been saying

but theres one step that makes the connection that you should state

[midnight plank]

@twilit field Has your question been resolved?

[twilit field]

okay

so

if $\sum_{i=1}^{n} T_{i,i} =0$ , it follows that the transformation is the zero transformation

[grand pond]

A dense set

[twilit field]

What do I do now

[hard umbra]

hiding the a_i probably doesnt make this simpler

all you want to do is plug in e_1

,, a_1 T_1 + a_2 T_2 + \dots + a_n T_n = 0 \
a_1 T_1(e_1) + a_2 T_2(e_1) + \dots + a_n T_n(e_1) = 0(e_1) = 0

[twilit field]

$a_1 T_1(e_1) + a_2 T_2(e_1) + \dots + a_n T_n(e_1) =0$. We know that this is LI as the basis of one vector space, map to form a basis of another vector spac

[hard umbra]

typo, theyre all e_1

dim(V) = 1 by assumption

[twilit field]

yeah

[grand pond]

A dense set

[hard umbra]

no the justification is wrong

[twilit field]

hmm

[hard umbra]

remember what T_i(e_1) is

[hard umbra]

defined here

[twilit field]

Yes , it's a linearly independent set

[hard umbra]

what is?

[twilit field]

T_i

Sorry f-i

d_i

f_i

[hard umbra]

and how does this help us

[twilit field]

That's what you akse

[hard umbra]

yes but what is the conclusion from this

[hard umbra]

we have this equation right

[twilit field]

T_i is linearly indepndent

[hard umbra]

and you know this because?

[twilit field]

f_i is LI

[hard umbra]

i mean okay

to fully spell it out

,, a_1 T_1 + a_2 T_2 + \dots + a_n T_n = 0 \
a_1 T_1(e_1) + a_2 T_2(e_1) + \dots + a_n T_n(e_1) = 0(e_1) = 0

[hard umbra]

you have this, which rewrites into
[ a_1 f_1 + a_2 f_2 + \dots + a_n f_n = 0 ]

[twilit field]

yes, and this is the same as $a_1f+a_2f_2+ \dots _anf_n=0$

[grand pond]

A dense set

[hard umbra]

and from there, you conclude that a_1 = ... = a_n = 0

and hence the T_i are LI

[twilit field]

yes

[hard umbra]

okay

great

thats the proof

[twilit field]

Why can we just sub $e_1$ in though

[grand pond]

A dense set

[hard umbra]

because its a function

,, \overbracket {a_1 T_1 + a_2 T_2 + \dots + a_n T_n}^{\text{a linear map, element of $\LL(V, W)$}} = 0 \gets \text{the zero map}

[twilit field]

yes

I don't understand though

[hard umbra]

well this is an equality of functions

so
[ (\underbrace {a_1 T_1 + \dots + a_n T_n}_{\text{a function}})(e_1) = \overbrace {0}^{\mathclap{\text{a function}}}(e_1) ]

[twilit field]

ah, got it

thanks

[twilit field]

.close

[midnight plank]

Channel closed

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[midnight plank]

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[shut ingot]

3x +1 need help

[midnight plank]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[last slate]

?

What do you need help with 3x+1

[visual tiger]

they're talking about collatz conjecture aren't they

[fallow scarab]

indeed

[noble zinc]

Could someone send me the solution of problem 1,2, 3, 4 and 5

[visual tiger]

!occupied

[midnight plank]

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).

[visual tiger]

and btw we don't send solutions, we help you guide yourself to the answer

[midnight plank]

@shut ingot Has your question been resolved?

[midnight plank]

@shut ingot Has your question been resolved?

[midnight plank]

Channel closed

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[midnight plank]

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[steady trail]

What is the difference between $f$ and $f(x)$?

[grand pond]

crazytime

[merry pewter]

f refers to a function and f(x) refers to the value of the function at a point x

[steady trail]

So f(x) is a number?

[wide pebble]

can i get help

[wet pollen]

#❓how-to-get-help

[prime hornet]

not necessarily, but it could be

[steady trail]

For example, if a is a constant, then $f(a)$ is a number

[grand pond]

crazytime

[prime hornet]

f is a function, f(x) is the function f evaluated at the point x

[steady trail]

Why?

[prime hornet]

because f does not necessarily have to be a function whose inputs and outputs are numbers

define $f: {\text{cat}, \text{dog}} \to {\text{love}, \text{hate}}$ by $f(\text{cat}) = \text{love}$, and $f(\text{dog}) = \text{hate}$

[merry pewter]

needlessly complicating it for them

[grand pond]

higher!

[prime hornet]

f is the function, f(cat) is an output

[prime hornet]

perhaps

[merry pewter]

a function is just a mapping from one set to another, but most of the time you have functions from numbers to numbers

[steady trail]

Yes

[merry pewter]

so yes, depending on what f is, f(x) is a number

[prime hornet]

point is that f(x) is the output that f generates when the input is x

[prime hornet]

while f is the function itself

[steady trail]

But like

$f(x)=x^2$ Wouldn't this be the function itself?

[grand pond]

crazytime

[steady trail]

Oh no

[prime hornet]

technically no

but this is common to write

[steady trail]

Anyway

It's clearer for now

Thanks very much

[prime hornet]

[steady trail]

.close

[midnight plank]

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[gleaming spear]

I don't know where to start with this one.

[gleaming spear]

<@&286206848099549185>

[midnight plank]

@gleaming spear Has your question been resolved?

[gleaming spear]

<@&286206848099549185>

.close

[midnight plank]

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[terse mural]

how do i find the oblique asymptote of this function

[terse mural]

ive done up to here i dont rly know what to do with this

is the oblique asymptote just f(x) = 2x-1

[wintry eagle]

@terse mural

[terse mural]

yo

[wintry eagle]

figured it out?

[terse mural]

no

[wintry eagle]

so to find asymptotes

You need to explore different cases

There are 3 types of asymptotes you can in general find

Horizontal, vertical, slanted

[terse mural]

horizontal vertical and slanted right

yea

[wintry eagle]

Vertical is always braindead

and easy to find

[terse mural]

u just set denominator to = zero and solve x right

[wintry eagle]

Good job

for horizontal

You just take the limit of x to infinity or minus infinity

[terse mural]

i havent done limits yet

[wintry eagle]

💀

[terse mural]

the way i learned horizontal is if the degrees in the numerator and denominator r the same i gotta take the numerator leading coefficient / denom. leading coefficient

and if the bottom degree is larger its = 0

[wintry eagle]

I mean I just googled

apparently the rule is that if the degree in the nuemrator

is larger than degree in denominator

you have no horizontal asymptotes

[terse mural]

yea

[wintry eagle]

Okay, lastly slanted

[terse mural]

i learned that if the numerator degree is +1 the denominator degree then theres a slanted asymptote

[wintry eagle]

Correct

How you find it, is by doing polynomial division

[terse mural]

i did that

i got 2x-1 and the remainder was -6

i just dk what to do with those

[wintry eagle]

You just ignore the remainder

[terse mural]

is the asymptote just y=2x-1?

[wintry eagle]

This is the answer I got when I used an online calculator

But yes, if you did your long division correctly, the asymptote is just the answer you got, and you can ignore the remainder

[wintry eagle]

Why did you multiply by 2?

x^2 is what you are trying to eliminate

[terse mural]

yo i did it wrong

💀

hold on

[wintry eagle]

all good

But just important to know, that remainder term, you just take it and divide it by whatever was in the denominator

[terse mural]

alr i did it again and got x-1 instead of 2x-1

i think i just wrote 2 instead of 1 for some reason

[wintry eagle]

Ya, x-1 is your slanted asymptote

case closed, the question is solved

It's pretty much just a repeat of this

Now you just plot with the information you have

[fickle cradle]

Help me

Help

Help

[wintry eagle]

This channel is occupied

go into an empty help channel and ask for help there

[terse mural]

yo real quick

[wintry eagle]

Ya?

[terse mural]

how do i do the end behaviours for the slant asymptote

[fickle cradle]

☹️☹️

[wintry eagle]

wdym end behaviour?

[terse mural]

as x approaches infinity, y approaches [?]

[wintry eagle]

I mean it's a slaned asymptote

x-1

it's just a line

[terse mural]

so would i just put y approaches x-1

[wintry eagle]

y = x - 1

that's your slanted asymptote

[terse mural]

and then for when x approaches -infinity

[wintry eagle]

what happens if x goes to infinity?

[terse mural]

would i just put y = -(x-1) or something

[wintry eagle]

uhm no

Think about it again

y = x - 1

[fickle cradle]

HELPPP

[wintry eagle]

What happens when x goes to infinity

[fickle cradle]

It goes forever

[terse mural]

gets closer to but doesnt reach the asymptote

[wintry eagle]

what happens if you have 1 million minus 1?

[terse mural]

999999

[wintry eagle]

and if you have infinity minus 1?

[terse mural]

uhhh

huge number

[wintry eagle]

You still call it infinity

Because 1 is so trivially small

compared to infinity

y = x - 1

when x goes to inifnity, y also approaches infinity

because - 1 is so small, that it won't change the y value that much

[fickle cradle]

R u a bot??

[wintry eagle]

this is the line y = x - 1

[terse mural]

ye

[wintry eagle]

You can see it is strictly just growing

so for larger x, you get larger y

and for infinit x, you will get infinitely big y

[fickle cradle]

Help

[terse mural]

i dont get how i would write it tho

[wintry eagle]

<@&268886789983436800>

[fickle cradle]

Bruh

[terse mural]

cuz for postiive infinity u can just write y approaches (x-1)

[lethal owl]

!occupied

[midnight plank]

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).

[wintry eagle]

I already told this person, but they don't care

[solid iris]

@fickle cradle this channel is occupied, see #❓how-to-get-help

[wintry eagle]

That isn't how it works

[paper inlet]

bro died of dead

[terse mural]

why not

as x gets bigger, y gets closer and closer to the x-1 line

[wintry eagle]

You are correct, for the function as a whole

When the whole function you had from beginning

approaches inifinity

it approaches the line x - 1

aka the slanted asymptote

[terse mural]

yea

[wintry eagle]

That is correct

if that was your original question

I misunderstood it, since I thought you were asking what value the asympyote approaches

[terse mural]

idk what i need to write for negative infinity

[wintry eagle]

same thing, approaches the slanted asymptote

[terse mural]

so my final thing would be
slanted asymptote: fx = x-1
as x approaches infinity, y approaches x-1
as x approaches negative infinity, y approaches x-1

[wintry eagle]

Yes

[terse mural]

aight ty

🙏

[wintry eagle]

You also need to see what values it approaches

for the vertical asymptote

[terse mural]

i know how to do those

[wintry eagle]

when x approaches 2 from positive and negative direction

Then after that you have all the info you need to plot the graph

[terse mural]

i just plug in 2 numbers super close to the vertical asymptotes and then see if its negative or positive

[wintry eagle]

Pretty much, that's one way to look at it

[terse mural]

been told there was a way to do it with calculus stuff but i kinda just havent learned it cuz its not a calc class lol

[wintry eagle]

Ya, you'll understand it better when you do limits

But if you look in the image I sent

Blue line is our slanted asymptote, red line is our function, green line is vertical asymptote

[terse mural]

yea

[wintry eagle]

You can literally see

that for infinit x, and minus infinit x

it really does approach our slanted asymptote

just gets closer and closer, but never really touches the blue line

[terse mural]

yea

[wintry eagle]

And if you look on the green line, if you approach it from right side (the positive side)

you see that it just turns to minus infinity

and from left side it goes to infinity

so that's how you would plot it, you draw out your asymptotes

and then you just graph it, from the information you got

[terse mural]

right

[terse mural]

ty for the help

[wintry eagle]

Because your question did ask to plot it, I remember I had to plot those in the exam

so really practice plotting, it will be important

[terse mural]

yea

[wintry eagle]

Good luck my friend!

[terse mural]

thank u

.close

[midnight plank]

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Use .reopen if this was a mistake.

[midnight plank]

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• Show your work, and if possible, explain where you are stuck.
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[shrewd python]

helpppppp

[wheat steppe]

what is the definition of a prime number

[shrewd python]

a number w no factors

[wheat steppe]

not exactly

every number has 1 as a factor

[shrewd python]

ye 1 and itself

a number w only 2 factors is a prime number

[wheat steppe]

good good

over here

[shrewd python]

ye

[wheat steppe]

how can we apply that definition

[shrewd python]

make the right bracket on the RHS equal 1

[wheat steppe]

in particular if (n^2+\sqrt{2}n+1) is a integer

greater than 1

what does that say about n^4+1

[shrewd python]

I dont know

[lavish venture]

well if n^4 + 1 has those two factors and one of them is one what is the other?

assuming it’s prime of course

[shrewd python]

the other one is n^4+1

[lavish venture]

right

so set it equal to n^4 + 1

[shrewd python]

n^4-n^2-sqrt(2)n=0

n^3-n-sqrt(2)=0

n=sqrt(2)

[lavish venture]

oh wait the question says n^4 + 4

i’m blind

🤦🏼‍♂️🤦🏼‍♂️

[shrewd python]

im cooked

[lavish venture]

nah you’re not

hmm i don’t see how the first part is useful

unless they’re trying to tell you to factor n^4 + 4

oh maybe?

[shrewd python]

maybe n^4+4 is (n^2+sqrt(2)n+2)(n^2-sqrt(2)n+2)

n^2-sqrt(2)n+1=0

[lavish venture]

,w (n^2 + sqrt(2)n + 2)(n^2 - sqrt(2) + 2)

[grand pond]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by c.knief

[lavish venture]

ugh

guess not though

because that isn’t the anti derivative of n^4 + 4

maybe the middle term is wrong

[shrewd python]

ye

[lavish venture]

hmm

[shrewd python]

ye but im pretty sure if u expanded the top one it wouldnt be n^4+1 either

[lavish venture]

the original?

[shrewd python]

ye

[lavish venture]

,w (n^2 + sqrt(2)n + 1)(n^2 - sqrt(2)n + 1)

[grand pond]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by c.knief

[lavish venture]

why

[shrewd python]

bruh

[lavish venture]

wolfram so shit

[shrewd python]

ye ikk

[lavish venture]

,w expand (n^2 + sqrt(2)n + 1)(n^2 - sqrt(2)n + 1)

[grand pond]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by c.knief

[lavish venture]

no it is

[shrewd python]

oh right

[lavish venture]

,w expand (n^2 + sqrt(2)n + 2)(n^2 - sqrt(2)n + 2)

maybe the square root 2 is wrong then

oh i forgot the n

[shrewd python]

ye it is

[grand pond]

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[lavish venture]

oh it’s close

so i guess we need the 2n^2 terms to cancel with the sqrt(2)(sqrt(2)n^2

because we get

4n^2 - 2n^2

so the coefficient needs to be 2!!

,w expand (n^2 + 2n + 2)(n^2 - 2n + 2)

[grand pond]

Results provided by WolframAlpha

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Requested by c.knief

[lavish venture]

😤😤😤😤

[shrewd python]

oh nice

and then make that right bracket equal 1

so then u get (n-1)^2=0

[lavish venture]

so it’s prime for n = 1 for sure

[shrewd python]

n=1 makes n^4+4 prime

so the answer is b prime for exactly 1 postive integer

[lavish venture]

$n^2 + 2n + 2 = n^4 + 4$

[grand pond]

knief

[lavish venture]

$n^4 - n^2 - 2n + 2 = 0$

[grand pond]

knief

[shrewd python]

n^2(n^2-1)-2(n-1)

[lavish venture]

yea

$(n-1)(n^3 + n^2 - 2) = 0$

[grand pond]

knief

[lavish venture]

n = 1

which we already had

and

n^3 + n^2 - 2 = 0

,w n^3 + n^2 - 2 = 0

[grand pond]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by c.knief

[lavish venture]

ahh no other solutions

[shrewd python]

ye so it should be b'

nice

[lavish venture]

fun problem

[shrewd python]

ye it from MAT which I taking tmr

[lavish venture]

is that oxford?

the math admission test

[shrewd python]

yeppp

[lavish venture]

good luck

[shrewd python]

thx

[lavish venture]

tom rocks math

[lavish venture]

nice

could’ve completed the square too ig

[shrewd python]

ye

this year they changing the format of the MAT there are more multiple choice now and less longer questions

which sucks cause I like the longer questions more

[lavish venture]

multiple choice can sometimes be more tricky

[shrewd python]

ye so real

[lavish venture]

because you look at the answers instead of trying to solve it yourself

going in reverse

[shrewd python]

ye

at there aint loads of logic questions like TMUA

I get so confused with that only if and if and if and only if stuff

but thanks for the helppp

.close

[midnight plank]

Channel closed

Closed by @shrewd python

Use .reopen if this was a mistake.

[lavish venture]

you’re welcome

[midnight plank]

Available help channel!

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[hasty rampart]

I dont understand

[hasty rampart]

Nvm

.close

[midnight plank]

Channel closed

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[gaunt gust]

#5

[gaunt gust]

to start
how am I supposed to get the y coordinate

[oak slate]

to find the equation for the line AB you need two points, you can use geometry to find the height at L(0) and it's given that L(1) = 0

[terse mural]

"sketch a graph of each function. describe each special case."
what is a special case

[gaunt gust]

what are you referring to with L

point A

?

wait no

just the leg lengths

basically right

[oak slate]

L is the line function for AB

[gaunt gust]

oh

[oak slate]

just see at 1 the height touches the x axis, which means the height is 0

[gaunt gust]

yeah

oh wait

point B is 0,1

I thought the right isosceles triangle it was talking about was a smaller triangle bruh i thought line AB was the hypotenuse 🤦‍♂️

[gaunt gust]

sorry I had to help my group on another project
slope is -1 and so the equation for line AB is y=-x+1

wait so the y coordinate of P in terms of x would just be plugging in x for x which would just give you -x+1 right

area = 2x*(-x+1)

a=-2x^2+2x

got the answer

.close

[midnight plank]

Channel closed

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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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[austere pawn]

yo

[austere pawn]

i have a test tommorow and its on all the toipics

its factoring

[twilit pelican]

what is your question?

[austere pawn]

i need to review all of this

ive done a couple questions but i need help on the 3 term equations

because i struggle when there's 3 terms

[midnight plank]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[austere pawn]

but i didnt

[twilit pelican]

you still haven't asked a question. I'd recommend you to send a problem that youre struggling with and explain what you've tried and where you get stuck/confused so that people can help you

[austere pawn]

I JUST WANT TIPS TO STUDY FOR THE EXAM I HAVE TOMMOROW

[austere pawn]

AND IM STRUGGLING WITH THE 3 TERM EQUATIONS IN THIS IMAGE

sorry for the caps

if you cant help me and dont have the helpers role

then

idk what to say

honestly

and its taking a minute to type

just let it out

[oak slate]

find a question related to the problem, try solving it and if you can't post it here

[austere pawn]

k

[oak slate]

and be more respectfull

[austere pawn]

i needed a explanation like that

[austere pawn]

forsure but when i repeat what i said 2x times

[wintry eagle]

It would help if you asked more clear questions

Rather than an image and ''lecture me on these topics''

[midnight plank]

@austere pawn Has your question been resolved?

[austere pawn]

totally what i asked forsure

and once again someone without the helper role

[wintry eagle]

I don't need the helper role

to teach you these topics

Secondary school level math

But, you maybe should work with your attitude

Good luck on your exam tomorrow

[twilit pelican]

how do you think people get the helper role

[wintry eagle]

He thinks they do an exam for it

[austere pawn]

never said that or thought of it

if you guys are gonna make wrong claims

[wintry eagle]

yes yes mister politically correct

Let us hope your questions are as correct in the exam tomorrow

[austere pawn]

i mean the helper gave me a better response then all of you did so i think he has more experience in helping students?

[twilit pelican]

the helper role is self assignable

it's in order to get pings

[midnight plank]

@austere pawn Has your question been resolved?

[brave stag]

The best tip is to study now

and stop going on discord

[austere pawn]

man can we just talk in dms atp

[midnight plank]

@austere pawn Has your question been resolved?

[midnight plank]

Channel closed

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• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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

[twin anvil]

yo

[twin anvil]

THis would be A right

[lethal path]

nope

[twin anvil]

how

[lethal path]

wait let me read again

[twin anvil]

this is the graph of the deriative of f

so we look where its going from negative x values to positive, or vice versa not the slope

[lethal path]

oh wait

[twin anvil]

and 3 points of inflection cause 3 points where slope is 0

[lethal path]

I agree that there are 3 points of inflection

ah right A is correct then

[twin anvil]

Oh ok

@lethal path

i have another question

[lethal path]

ok

[twin anvil]

how would we do part A

i understand B and C

[lethal path]

oh concave up is just when the second derivative > 0

[twin anvil]

Yea

x^2 is always positive

[lethal path]

yep

[twin anvil]

so we can just focus on the cos(blahblah) part

how to find intervals

[lethal path]

cool so if cos(x^2 + pi) = 0, we could have x^2 + pi = 5pi/2
or that x = plusminus sqrt(3pi/2)

[twin anvil]

wait what

wdym we could have

where did the cos go

[lethal path]

cos(x^2 + pi/2) = cos(5pi/2)

[twin anvil]

huhh

[lethal path]

so that implies x^2 + pi = 5pi/2

cos isn't an injective function I know

so we also could have x^2 + pi = pi/2

but then this new equation doesn't have any real solutions

actually I looked a solution too far

[twin anvil]

bro

im so lost

[lethal path]

only x^2 + pi = 3pi/2 has solutions in the domain

[lethal path]

it's like how if x^2 = 2^2, then x = 2

[twin anvil]

yes

[lethal path]

or in exponential function problems you must have done at some point, if e^(2x + 5) = e^3, then 2x + 5 = 3

that solution is unique cause exponential functions are 1-to-1

[lethal path]

for cos it's like this where this is only one solution

we could have x^2 = (-2)^2 ofc

[twin anvil]

im so lost i cant lie

idk what u j said

i just read everything

im so lost

[lethal path]

so everything I mentioned is an example of if f(a) = f(b), then a = b is one solution

I used f(x) = x^2, f(x) = e^x for example

and now f(x) = cos(x)

and with a = x^2 + pi/2, b = 3pi/2

f(a) = f(b) is the equation cos(x^2 + pi/2) = cos(3pi/2)

a = b is the equation x^2 + pi/2 = 3pi/2

[twin anvil]

wait

just go back to the original question

ur confusing me

[lethal path]

you want to find when cos(x^2 + pi) > 0 right

it would be easier to see where cos(x^2 + pi) = 0

then make a sign diagram

[twin anvil]

Yes

[lethal path]

okay so cos(3pi/2) = 0

cos(x^2 + pi) = cos(3pi/2)

x^2 + pi = 3pi/2

[twin anvil]

where are u getting x^2 bro

bro what

[lethal path]

it's in the original question lol

[twin anvil]

ohhh

[lethal path]

$f''(x) = x^2 \cos(x^2 + \pi)$

[twin anvil]

yea

[grand pond]

south's secret twin brother

[twin anvil]

cs both equal 0

[lethal path]

you said x^2 >= 0

[twin anvil]

ohhh

i get it

but shldnt it be an indeuqality

not a equation

to find where is it greater or equal to 0

[lethal path]

yes, so that's when we use the sign diagram

[twin anvil]

ok

[twin anvil]

what do we do

now we have this

[lethal path]

x = plusminus sqrt(pi/2)

,calc sqrt(pi/2)

[grand pond]

Result:

1.2533141373155

[lethal path]

remember our domain is [-1, 2] though

[twin anvil]

ok

[lethal path]

so we only have two regions actually to test

-1 < x < 1.2533 and 1.2533 < x < 2

(the other roots of the function are not in the domain)

[twin anvil]

so we cant include -1.2

[lethal path]

yeah that's outside the domain

[twin anvil]

so 1.2 is the only critical point?

[lethal path]

yep

[twin anvil]

ohhh

but i have a question

[twin anvil]

when we did this

why cant we do cos(7pi/2) or (pi/2)

[lethal path]

oh we absolutely can

they just happen not to be in the domain

so try x^2 + pi = 5pi/2 for example

[twin anvil]

x^2 = 3pi/2

x = squrt 3pi/2

[lethal path]

,calc sqrt(3pi/2)

[grand pond]

Result:

2.1708037636748

The following error occured while calculating:
Error: Unexpected type of argument in function multiplyScalar (expected: number or Complex or BigNumber or bigint or Fraction or Unit or string or boolean, actual: function, index: 0)

[twin anvil]

ohh

[lethal path]

yeah and also x^2 + pi = pi/2

[twin anvil]

but how did u know we had to use 3pi/2 in the first place

[lethal path]

well I tested on Desmos beforehand

[twin anvil]

oh

[lethal path]

you'd see that this is impossible firstly

then test 3pi/2

then test 5pi/2 and see it's outside the domain

so then the rest wouldn't work, cause x^2 would equal some larger number

[twin anvil]

so concave up on [-1,sqrt(pi/2)) U (sqrt(pi/2),2]

[lethal path]

also like x^2 + pi = -pi/2 is impossible, whenever RHS is negative

[lethal path]

it's just (sqrt(pi/2),2)

[twin anvil]

why not form the left?

[lethal path]

cause cos(x^2 + pi) is negative on the left

you can test x = 0 for example

[twin anvil]

oh

yea

srry i just assumed

but yea its negative

@lethal path

for part B

ik how to do it

[lethal path]

yeah

[twin anvil]

so no need to go in full depth

[twin anvil]

here

[lethal path]

oh right

[twin anvil]

we can use second deriative test

[lethal path]

you told me yes, you did b and c already

[twin anvil]

Yea

i just wanna go over it again so i understand concepts

we can use second deriative test

to plug in 1 in second deriative equation

if f''(1) is < 0 there is a relative max

where there is right

[lethal path]

oh right that's smart, so you just need to test f''(1) and find the sign

yep

indeed the sign is negative so concave down

[twin anvil]

btw we have to justify our answers, the question states that it is a twice-differentiable function do I need to state that anywhere for any parts of the question

[lethal path]

yes

[twin anvil]

Where and for what part

[lethal path]

justify basically means give a reason instead of just stating the answer

[lethal path]

for part b you need twice-differentiability

but for these problems you won't have any weird edge cases

you can't be expected to find the concavity of a function which isn't twice differentiable to begin with

[twin anvil]

Ohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh

function needs to be twice differentibale to use second deriative test

ohhh

[lethal path]

yeah

that's the assumption

in real analysis you look at these assumptions in more detail

for now it doesn't matter all that much

[twin anvil]

Oh yea lol

thx sm

[lethal path]

but it helps to be aware that MVT for example, the condition is that f is continuous on [a, b]

differentiable on (a, b)

[twin anvil]

oh yeea

bc if its differentibale that means its continous

[lethal path]

oh nice, you know that

so we don't want or need differentiability on the endpoints

otherwise we'd have to use one-sided limits

[twin anvil]

ye

can i dm u for more help if i need u?

[lethal path]

nah no thanks

you can always ask in a help channel though

[twin anvil]

kk

whats ur pronouns btw

[lethal path]

he/him or they/them is fine

[twin anvil]

Oh okk

im he/him or they/them is fine

Thans for help cya

[midnight plank]

@twin anvil Has your question been resolved?

[midnight plank]

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• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
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• Be polite and have a nice day!

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[severe scroll]

Might be a very dumb question, but how is the antiderivative of 1/t^4 equal to 1/t^3 and not my guess -1/3t^3 ?

[dense harness]

It is -1/3t^3 but you multiple that by some arbitrary constant

And -1/3 times that constant is just another arbitrary constant

So you call the whole thing just C2

[severe scroll]

Ah, so bascially C2 = -1/3 C1 ?

[sudden yacht]

Yep

[severe scroll]

But is it mandatory to give it a new constant? Or would have -1/3C1 have been fine in this case too?

[dense harness]

Its fine

[severe scroll]

And now that I think about it, why isn't it then -1/3t^3 C1 + C2?
Or is that the same aswell?

[dense harness]

What do you mean exactly

[sudden yacht]

That they forgot to add the constant of integration

[dense harness]

Yeah they did

[severe scroll]

Yeah some how I'm used to doing it is as follows, for example in a double antiderivative, you get a constant C1, for your first integration, and an another one C2 for your second one.

Like y'' = 1, then y' = x +c1, then y = 0.5x^2 +c1x +c2.

Like in this case you would have two constants.

[dense harness]

Yes you would

[severe scroll]

So then why isn't it 1/3t^3 C1 + C2 in this case?
Or is that still the same as replacing the constants?

Oh yeah I see now that's the same, as we found earlier that C2 = -1/3 C1.

Thank you both very much for your help.

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[midnight plank]

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[midnight plank]

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[last slate]

The general form of triangle number is

idk

I know the triangle numbers are like 1, 3, 6 and 10

<@&286206848099549185>

[junior flower]

hi rosy

[last slate]

omg hi layla

[junior flower]

hiii

[last slate]

[junior flower]

the nth triangular number is n(n+1)/2

[last slate]

is that like a fact everyone should know

or do you work it out

[junior flower]

probably

[last slate]

oh!

help my teacher never told me that

[junior flower]

loll

[junior flower]

but also you can do that. there is actually some folklore surrounding how to work it out haha

[last slate]

o rlly

[junior flower]

yea totally off topic and probably not even true but people say gauss' teacher when he was in school told him to compute 1 + 2 + ... + 500 (or something, the story changes every time i hear it) for some reason and he came up with that formula

[last slate]

oh lol

so like now we have the formula how do we put it in a ratio

[junior flower]

anyway yea this is usually helpful for working with triangular numbers

[last slate]

ok ima memorise it

[junior flower]

consider n(n+1)/2 and the next triangular number, (n+1)(n+2)/2

(just putting n+1 in for n in the formula)

[last slate]

oki

[junior flower]

the ratio between them is $$\frac{n(n+1)/2}{(n+1)(n+2)/2}$$

[grand pond]

generating function courtesan

[last slate]

Oh and then you just cancel down the fraction right

[junior flower]

yea

[last slate]

ohhh ohk i get it now

Tysmmm

🤍

[junior flower]

[last slate]

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[calm quest]

for this does anyone know why they used this instead of just pie/5

[midnight plank]

@calm quest Has your question been resolved?

[brazen moth]

Hello

[brittle shoal]

cz they've given that they need the answer in decimals(read the last sentence in the q)

[calm quest]

but in the formula wouldn't it be pie/5 without in sin(pie/5)

[brazen moth]

I don't think that's the case @brittle shoal

Here they are asking for area of segment

Not sector

So you must subtract area of triangle from sector

To get the area of segment

[brazen moth]

They need the purple guy

So the formula is basically

Area of sector minus area of triangle

[calm quest]

ohhhh

[brazen moth]

That's it

I think That's all

?

Anything else you wanna ask?

If not, pls .close the chat

@calm quest ?

[calm quest]

thank you makes sense!

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[left badge]

How to do this

I dont really understand it

[midnight plank]

@left badge Has your question been resolved?

[left badge]

Nope

[midnight plank]

@left badge Has your question been resolved?

[midnight plank]

@left badge Has your question been resolved?

[midnight plank]

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[viral dagger]

i need to find EF where its the divisor of CED (CED is 90 deg)

turns out this is nastier than i thought D:

im pretty sure this blue line is 12sqrt2

so j need to find the rest of it (spoiler:i have no fucking clue how)

[paper inlet]

je ne comprends pas?

[viral dagger]

?

[paper inlet]

Im drawing this out pause

[paper inlet]

is that a square and a triangle

[viral dagger]

uhh yea

E is outside the square

[paper inlet]

Ok so what are you trying to find

[modern sapphire]

Blue triangles are similar

[grave moth]

[modern sapphire]

dafuq thats not a good embarrassed emote mathcord

[paper inlet]

its pretty good

[viral dagger]

oh

tyy

.close

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[placid spoke]

!status