help-19

[wooden python]

i.e. magic

[wide pollen]

ig

should i just put the equation of p(x) and p(1/x) wherever they are

it will be a ugly mess tho but i aint sure what i can do

[wooden python]

that's the masochist route

[wide pollen]

ik

@wooden python lol is it cool

[odd edge]

@wide pollen Has your question been resolved?

[odd edge]

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[frosty sinew]

Can anyone help me with how to graph slopes

[woeful briar]

Do you have a specific graph/problem in mind?

You want to know the slope from a graph? Or graph a line with some slope

[frosty sinew]

Graph the Cartesian plane with slope

[woeful briar]

Thats very vague

[frosty sinew]

It’s like uh

Lemme find a way to explain it

[woeful briar]

Screenshot?

[frosty sinew]

Imma show you my notebook

[odd edge]

@frosty sinew Has your question been resolved?

[frosty sinew]

Basically this but on the graph

On the Cartesian plane

Identifying the points

[tidal matrix]

so, graphing the line $-y = -8 + 8x$, as an example?

[clever fjord]

haseeb

[frosty sinew]

Yes

[woeful briar]

you choose two values for x, usually x = 0 is one of them

and calculate the y value. in the case of x = 0 its simply -8

[frosty sinew]

Yea but on the Cartesian plane I would start with 0 and then go down to -8?

[woeful briar]

yeah

so your first point is (0, -8)

your other is (1, 0)

draw a line between them using a ruler and extend it how much you want

[frosty sinew]

What about the other 8

[woeful briar]

I used it to get the point (1, 0)

0 = -8 + 1 * 8

[frosty sinew]

Ahh

Thank you I get it now

[odd edge]

@frosty sinew Has your question been resolved?

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[lost crest]

What does it mean by x cant be -1/3

But it’s just a straight line when x is -1/3

[woeful briar]

when x=1/3 the denominator is 0

[lost crest]

but on the graph when x = 1/3 , y still exists

[woeful briar]

does it?

its not defined for that value

[lost crest]

But y is just a straight line passing through 3/4

[woeful briar]

y is a function of x

it can't be defined that way if you want it to be defined for x=1/3

[lost crest]

sorry I just don’t really get it

[woeful briar]

y is defined as some function of x

the problem is that that function is undefined for x=1/3

so y=3/4 for every other value of x

[lost crest]

If it’s undefined what will happen?

[lost crest]

Does that make the straight line straight

😭

[woeful briar]

there is a difference between (3x+9)/(4x+12) and 3/4

y=3/4 is a horizontal line

[lost crest]

yes

[woeful briar]

y=(3x+9)/(4x+12) this is a horizontal line with a missing point

at x=-1/3

[lost crest]

Is it acc missing

[honest turtle]

x=-1/3 is the only point that affects the function, the rest is fine. So when x=-1/3, it's undefined, visually it's continuous, but there's a hole.

[lost crest]

OHHHHHHH

where would the hole go

Is that where complex nunber plays in

Like if I look at above

It’s gonna be curved in

[honest turtle]

Complex number can't help either, when denominator is 0, it's undefined.

[woeful briar]

no. thats unrelated

[lost crest]

Oh

[woeful briar]

put it on desmos

I think it shows the hole

[woeful briar]

try to click on the curve at x=-1/3

[lost crest]

Maybe dosent show up

[woeful briar]

yeah just ignore this then

[honest turtle]

Enter y(-1/3)

[lost crest]

How

[woeful briar]

call it f(x) instead of y

and then do f(-1/3)

[honest turtle]

The way you wrote y in desmos, write y(-1/3)

[lost crest]

Ohhh

Where do the hole go

Is it like a second dimension 🤡

[woeful briar]

no

[lost crest]

3D dimension lol

[woeful briar]

it is gonna be at (-1/3, 3/4)

but not really

there is nothing there

nothing from the curve on this line x=-1/3

[lost crest]

Do u think if we look from above

It’s gonna look like this

And Thats the hole

[woeful briar]

yeah its something like this

but what you showed is a differnet case

where the behaviour of the function is to blow up around the hole

[lost crest]

Oh

Okay

Thank you for teaching me @woeful briar and @honest turtle

Have a good day or night!!

.close

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[nocturne mica]

not again.

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[static locust]

@vapid leaf Did you try anything

[viscid flint]

tried getting banned

[static locust]

No like did he start the problem yet

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[last hearth]

im lost😔

[green elm]

there seem to be missing terms?

(a+b)^4 is not the same as a^4 + b^4

[last hearth]

Huh? i solved the other ones not on paper im on the final step of the equasion now

[green elm]

i am referring to this

the second line is not equal to the first

[last hearth]

wait how so

[green elm]

explain why you think it is true

you seem to be using (a + b)^4 = a^4 + b^4, but that is not true

try it with a = b = 1 if you don't believe me

[last hearth]

x is to the power of 4 so squaring it gets rid of the square root and theres ^2 left over so that makes it x^2 then i just put -7^4 and added the 8

[green elm]

yea but that's now how taking powers of expressions of the form (a+b) works

like (a+b)^2 is not just a^2 + b^2 right?

it's a^2 + 2ab + b^2

[last hearth]

oh do i need to foil?

[green elm]

and (a+b)^4 has even more terms

yep use foil

[last hearth]

okay thx

[last hearth]

i need help

[green elm]

tbh i would suggest not multiplying it out

[green elm]

if the software will accept it

it's just gonna be a mess

[last hearth]

this is where im at did I mess up

[green elm]

compare with the correct answer

,w (sqrt(x) - 7)^4

[clever fjord]

Results provided by WolframAlpha

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

Requested by bungo.the.original

[green elm]

oh cmon wolfram

[nocturne mica]

can't you leave it as is though

[last hearth]

what the flip

[north root]

Rootx*Rootx is just x

[nocturne mica]

this looks like a pain in the ass to expand

[green elm]

yea it's just gnarly

won't the software take it in the unexpanded form?

[last hearth]

let me try that

Oh, yeah that worked

thx

[nocturne mica]

thank goodness

can't imagine having a bajillion terms

[last hearth]

ik i wouldve been heartbroken

😭😭

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[vapid violet]

how to solve

[vapid violet]

find degree and roots etc

[nocturne mica]

degree is the highest exponent of x in the polynomial

[vapid violet]

ye i mean get in factired form

[nocturne mica]

so you want to factor this

have you learnt about the rational root theorem? if not, i'm not sure how you would find one root to start

[vapid violet]

dont think so

[nocturne mica]

this will be tricky

[vapid violet]

this the answer ig

[nocturne mica]

sure, but without access to the RRT, i don't know how you would quickly find one root to start

finding one root at least allows us to turn the cubic into a quadratic and makes it a lot easier

the other method is trial and error with polynomial division

[random pulsar]

May I suggest factoring by grouping

[nocturne mica]

oh yeah, that could work here

[vapid violet]

how

[random pulsar]

You can factor out a x^2 from the first two terms and then a -4 from the last two terms

[vapid violet]

k lemme try

got this

[random pulsar]

Looks good, now you can factor out the common term

[vapid violet]

wdym

[random pulsar]

well you have $x^2(x-3)-4(x-3)$ so what do the $x^2$ term and the $-4$ term have in common?

[clever fjord]

MrTuxedoSteve13

[vapid violet]

like (x^2 - 4) (x - 3)

[random pulsar]

yep

[vapid violet]

yea but those arent the number for the roots

she gave (x -3)(x - 2) (x + 2)

[random pulsar]

well the $x^2-4$ is a difference of squares so you can factor that and you should get the answer

[clever fjord]

MrTuxedoSteve13

[vapid violet]

dont they both need ^2

[random pulsar]

$2^2 = 4$

[clever fjord]

MrTuxedoSteve13

[merry finch]

You can always just force it by square rooting then squaring

For example, $x^2- 2 = x^2-\sqrt 2^2 = (x-\sqrt 2)(x+ \sqrt 2)$

[clever fjord]

frosst

[vapid violet]

think i got it

.close

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[warm pike]

I am trying to find the horizontal asymptote and I already checked the right bound so now I am checking the left. I simplified it as much as I could but now I don't know what to do

[odd edge]

Please don't occupy multiple help channels.

[marble laurel]

.close

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[nocturne mica]

you seem to have a channel already ongoing

[warm pike]

how does this work

its not helping me

I am trying to find the horizontal asymptote and I already checked the right bound so now I am checking the left. I simplified it as much as I could but now I don't know what to do

[twin echo]

I'd suggest using l'hopital. Take the derivative of the numerator and the denominator separately, then try to calculate that limit. If it is still infinity over infinity or the like, take another derivative. Let me know if you have more questions

[nocturne mica]

this channel is closed. in the future, please stick to one channel only

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[acoustic gazelle]

Can I get help on how to solve this? I don't understand the answer key

[forest sky]

which part are you having trouble with from the answer key?

[acoustic gazelle]

I don't understand why we're using the left limit as x approaches two

[twin echo]

Look at what side y is approaching two as x approaches zero

[forest sky]

because as f(x) approaches zero it approaches the value of 2 from values which are below 2

so the corresponding limit as x approaches 2 should also only consider values below 2

[twin echo]

What cloud said

[acoustic gazelle]

oohhhhh

alright, I see it now

I appreciate it

.close

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[stray cedar]

A museum is constructing a new wing with several rooms, whose layout is represented in the following map. The museum director needs to decide how to assign guards to these room, so that each room either directly contains a guard or is adjacent to a room where a guard is stationed. The goal is to minimize the number of guards.

Is it safe to assume adjacent refers to a connected room in this case?

[viscid flint]

yes

F is adjacent to D but not to J

[stray cedar]

Perfect, thanks!

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[violet sable]

For all composite m, let f_m(n) be the function that maps n to (n^n % m). Let T_m > 0 be the period of f_m, ie for all values n, let f_m(n) = f_m(n + T_m). My question concerns the period's value. Define \lambda(m) as the Carmichael function, which can be found online (such as at Wikipedia. It is related to Euler's Phi function, and can be thought of as the minimum exponent that sends all values to 1 modulo a modulus m).

For all m, does f_m have T_m = lcm(m, λ(m))? Can someone confirm or deny this result?

[violet sable]

$f_m \text{ has } T_m = \text{lcm}(m, \lambda(m))$

[clever fjord]

Yumdub

[violet sable]

I think I have a proof, so I can show some if there are questions

[woeful briar]

Well Tm <= lcm(m, lambad(m)) is true

[violet sable]

can you provide a counterexample? Maybe there's a smaller value?

AH ok I found a counter example

I wrote a program to confirm that T_21 = 42

$\text{but} \lcm(21, \lambda(21)) = \lcm(21, 6) = 42$

[clever fjord]

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

[violet sable]

wait nvm...

ok ok try m = 15

then T_15 = 60 experimentally

lcm(15, \carmichael(15)) = 60, so that also matches

[woeful briar]

You gave an example xD

Not a counterexample

[violet sable]

yeah I figured that out late

that's why the question is still open

[woeful briar]

if you let n=0
Then T^T = 0 mod m

[violet sable]

ok, that shows that T^T should need to be 0 modulo m I think

which shows T == 0 mod m

[woeful briar]

No

m=9, T=3

[violet sable]

what question are you answering, the original or are you disputing what I just said?

I acknowledge your edit

[woeful briar]

This implication

Anyway I need to go

Gl

[violet sable]

goodbye

[violet sable]

@woeful briar Just an FYI, I acknowledge that you originally answered a similar question in #elementary-number-theory, so I know you were at least somewhat interested. I don't mean to imply that I'm taking your question.

[odd edge]

@violet sable Has your question been resolved?

[violet sable]

ok ok, I was able to prove the result simply by myself. Thanks for your help, as flippant as it was. I will close the channel

.close

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[slender cave]

bro i need help im 100% lost rn

[vernal yacht]

What do you need help with?

[slender cave]

uh

piecewise functions

[shell haven]

What about them

[slender cave]

the top one

example 5

wait first is this a piecewise function

@shell haven

[shell haven]

Well yea you can express it as one

[slender cave]

ok so basically

i have a test tomorrow

[shell haven]

Rip

[slender cave]

ok inverses and domain and range of them and pieces functions

piecwise

idk what peice wise functions are

can you give me an example of one and maybe teach it to me

[shell haven]

Well like

A function defined by multiple functions

Where each function is like the value the function takes for a particular range

[slender cave]

?????

[shell haven]

Ok ok

Example

I've shamelessly stolen this photo

So when you look to the left

You'll see the piecewise function definition

[slender cave]

uh

[shell haven]

There are two functions in this definition

[slender cave]

definition?

oh yes

[shell haven]

2x and 3

[slender cave]

yes

[shell haven]

So you see that x<=0 to the right of 2x

[slender cave]

yeah what does it mean

[shell haven]

That means that if the x input into the function f(x) is less than or equal to 0

Then the function will output the value of 2x

Does that make sense so far

[slender cave]

no

why does 2x<0 how is 0 bigger

[shell haven]

Ok nvm

So let's say we had two functions

g(x)=2x

[slender cave]

yes

[shell haven]

And h(x)=3

If x<=0

Then f(x)=g(x)

So if the value of x inputted into the function is less than or equal to 0

[slender cave]

why f(x) i thought its h(x)

i don’t get it bro

[shell haven]

<@&286206848099549185>

Mb vro

Ion how to describe this fr

[slender cave]

it’s like 50% of my test

[shell haven]

Rip gng

[slender cave]

And i don’t get it

<@&286206848099549185>

[umbral jolt]

!15m

[odd edge]

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.

[slender cave]

help

@shell haven i guess ur my only option

[shell haven]

He kinda started more than 15m ago man

[slender cave]

ok wait

this example

it said write the piecewise function rule for the absolute value functions

and we have f(x)= |x+5|

so then it leaves us with like x+5 and -x+5

right?

[shell haven]

x+5 and -x-5

[slender cave]

how did you get -x-5

[shell haven]

Because if x+5>=0, then f(x)=x+5

[slender cave]

so how do you get -x-5

[shell haven]

But if x+5<0 then f(x)=-(x+5)

[slender cave]

thays if tho

it’s not like that i thought

[shell haven]

Ion what to say

[round lynx]

that's how absolute values work my guy

|x| can be -x or x

but i dont think it really matters for your piecewise stuff rn, what are you struggling with?

[shell haven]

He's struggling with interpreting their definitions

[shell haven]

As seen here

[shell haven]

And what they are in general

[odd edge]

@slender cave Has your question been resolved?

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[edgy holly]

I posted this previously too. No answer. Please someone help prove this

[fleet tapir]

The procedure should be fairly straightforward

Parametrize the lines passing through (x4,y4) and enforce the ratio condition

This will give you the parameter and hence the line

[edgy holly]

Could you please explain what is parametrizeing

[fleet tapir]

Which part of this are you having trouble with

[edgy holly]

I don't know where to begin

[fleet tapir]

If a line passes through (x4,y4), it must have some slope t right?

[edgy holly]

Yeah

[fleet tapir]

So the equation such a line will be (y-y4)=t(x-x4)

This line will intersect the two lines L3 and L2 at one point each

[edgy holly]

Ok

[fleet tapir]

Those are your B(t) and C(t)

[edgy holly]

Do I Assume those points?

[fleet tapir]

Now enforce the condition that D divides the line segment joining B(t) and C(t) the ratio m1:m2

This will give you t

[fleet tapir]

Put t back in this equation

And it will give you the line for BC

[fleet tapir]

They depend on t

[fleet tapir]

Anyways, did you understand till here?

[edgy holly]

Yeah

[fleet tapir]

Okay then it's just a matter of crunching through the calculations

[edgy holly]

Okay trying it.

Thanks

.close

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[edgy holly]

Gonna open again, if I mess up again .

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[low locust]

compact is equiv to closed and bounded

there are lots of other sets that are closed and bounded

not just intervals

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[mellow star]

How is (x+a)(x+b)=x²+x(a+b)+ab.
Like I can not understand. Is it basic multiplication or something?

[tacit elm]

Just expand the formula on the left ?

[atomic hornet]

yes

[mellow star]

Thats it

[atomic hornet]

have you learnt of how to express such form?

[mellow star]

I understood

[low locust]

do you know how to expand (x+a)*d

[mellow star]

No

But I understood

Expand the formula on the left

ANd it is done

[soft robin]

you just FOIL it

you're basically multiplying each term in the two quantities, in a set order

first, outer, inner, last

and be mindful of the ±, because (x + 1) or (x - 1) is the same as (x, 1) or ( x, -1)

so it would be like this:

(x²) + (xb) + (ax) + (ab)

and then find like terms and collapse to make it cleaner

produces:

(x² + (a+b)x + ab) = 0

[low locust]

\* for *

[soft robin]

(xd + ad). distribute d across both terms in the quantity. d(x) + d(a). when you multiply two variables you just put them next to each other.

[low locust]

like I was asking you...

[soft robin]

My apologies, I mistook you for someone with a similar question who used this forum to ask instead of making a new one.

The explanation should be useful for the original asker anyways since he said he didn't know how

[odd edge]

@mellow star Has your question been resolved?

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[mellow star]

.close

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[wooden oar]

Hi! Can someone check what i did wrong for letter (c)? Because i got 284.794 as the answer, but the solution says 36.77 days, so 37 days is correct.

[wooden oar]

Thanks in advance!

This is what i did

[quasi sparrow]

the question states 97.7% or 0.977 remains, but you put 0.997 into your calculator

[wooden oar]

Ohh, silly meee

You are right, thanks

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[misty fjord]

i could write in this sever now

[misty fjord]

freedommmmmmmmm

https://tenor.com/view/a23-gif-5165236331565184779

[quasi sparrow]

go to #chill

[wooden python]

do you have a question to ask

[misty fjord]

okay

[atomic hornet]

ye, I'll close this channel since you don't seem to have a problem

[vernal yacht]

.close

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[atomic hornet]

Have a good one

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[torn basalt]

help with a) i've gotten to x(one) = 3+2x(three) and x(two) = 3+2x(three) and x(1 and2) are basic and x(three) is free

[torn basalt]

not sure how to express my answer

and I'll prob need help with b also

[atomic hornet]

Have you tried row-echlon form?

row-echelon form*

[odd edge]

@torn basalt Has your question been resolved?

[torn basalt]

yeah

[lone elbow]

can you show your work

please

[atomic hornet]

could you write it down on a paper and take a pic of it?

[odd edge]

@torn basalt Has your question been resolved?

[fresh vector]

x−3(3+2z)+4z=−6⇒x−9−6z+4z=−6⇒x=3+2z.

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[severe abyss]

i got a hint to use rolle's theorem and induction. i tried induction but i ended up with the second ss which doesn't simplify using mvt

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[forest sky]

different only in the infinite case:
https://math.stackexchange.com/questions/39895/the-direct-sum-oplus-versus-the-cartesian-product-times

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[fleet oracle]

Pre-Calc honors

[odd edge]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[polar echo]

8th or 9th?

[fleet oracle]

Just these 2.

They are the last 2 questions so it’s not a lot.

[polar echo]

for the 8th, you're thinking correctly , use the distance formula

D= undr [ ( x-4)^2 + (x^2-0)^2 ]

let the function inside the root be called f(x) and then we'll differentiate f(x) because we need to minimise it

f(x) = (x-4)^2 + x^4
f'(x) = 4x^3+2x-8

huh, it doesnt have any ration root, thats a hassle

[polar echo]

uh, no, underoot doesnt work like that

what you did is you took underoot of (x-4)^2 and y^2 and added them

its underoot of their sum and not sum of their individual underoot

a way to proceed is you square both side, giving you D^2 = (x-4)^2 + (x^2)^2

[fleet oracle]

I’ll try that

[polar echo]

sure

[fleet oracle]

D^2 = x^2 -8x +16 + y^2

[polar echo]

right, and y=x^2 as its given, so you get x^4, cool?

[fleet oracle]

No I don’t understand.

[polar echo]

also, a bit of nuance, the ques has a language error, it should be 'curve' instead of 'line'

[fleet oracle]

How does it get to x^4?

[polar echo]

because the curve obeys the rule y=x^2 , and you got a term of y^2, try substituting y=x^2

[fleet oracle]

Ok

So it should be x^6?

[polar echo]

why?

[fleet oracle]

Photo loading.

But it doesn’t make any sense

[polar echo]

you cannot add x^2 and x^4 = x^6

they have different exponents, you cannot add exponents, had 2 and 4 be in coeffecient instead of power, you could've write 2x+4x=6x

[fleet oracle]

So what should I do?

Sq root the entire thing?

Actually, I’ll try

[polar echo]

no

[fleet oracle]

Nope…

Didn’t work

[polar echo]

.

you get D^2 as this

[fleet oracle]

Ok…

Wait then root the whole thing?

Nope…

Oh wait I wrote it wrong

It’s closer than y = root(x)

But it doesn’t match what it should be…X should be positive not negative.

X^2 +x -4, what’s wrong with it

[polar echo]

nah, you understood how D^2 equals x^4+(x-4)^2 right?

[fleet oracle]

Yeah

[polar echo]

now, we have to find the 'minimum' distance between the curve and point (4,0) which means we have to find such a x for which x^4 + (x-4)^2 minimises

the way we can do this is let f(x) = x^4 + (x-4)^2

and then differentiate this to find the value of x at which function minimises

[fleet oracle]

I’m asking ChatGPT and it’s saying the same thing, how am I supposed to “differentiate” to an entirely different equation?

How does x^4 become 4x^3?

Just how?

How am I supposed to do this 🙏

[shell haven]

Power rule

d/dx(x^n)=nx^(n-1)

Where n is a constant

[fleet oracle]

We never learned that.

[shell haven]

Oh then uh

Did you guys learn about how to identity the turning point of a parabola

[fleet oracle]

It’s been forever…

[shell haven]

Is this a calc class?

[fleet oracle]

Pre-Calc honors

My teacher got 14 emails in one hour over this homework 😬

Including me

So…yeah. Please help me 😓

[fleet oracle]

If I can’t use derivative, then what?

I’ve been stuck on this for the past hour…literally.

[shell haven]

Ehh

Tbh

Ion

[fleet oracle]

I n e e d h e l p

[shell haven]

Rip

What's the topic of this worksheet

[fleet oracle]

Pre-calculus

Honors

This is 12th grade math, right?

[shell haven]

This would require calc

I mean I don't see an alternative way

[fleet oracle]

I’m just gonna submit as is, 1/3

Also, thanks for helping @shell haven and @polar echo

.close

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[shy scarab]

Yo

[shy scarab]

How can i solve this

[unkempt lichen]

,rccw

[shy scarab]

How can i find this out

[lofty spade]

have you tried using contradiction?

[shy scarab]

not sure what is that

[lofty spade]

assume that exactly two of the numbers are odd, then prove False

[shy scarab]

i havent done that before

[lofty spade]

well, try it

assume this and the property in the question, then try and find a contradiction

[shy scarab]

can you just do it for me pls

[lofty spade]

no

[shy scarab]

y not

[atomic hornet]

Let's say you have
a, b, and c
With the given information , you'll get
ab = c Q1 + 2
bc = a Q2 + 2
ac = b Q3 + 2

[shy scarab]

yes

!help

[odd edge]

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

[lofty spade]

can you proceed from here?

[shy scarab]

not not really

[lofty spade]

if, say, a and b are odd, can you find a contradiction?

[shy scarab]

Whats a condradiction

[lofty spade]

if you haven't been taught it, do you have any teacher you can ask for help?

[shy scarab]

Nah sorry

This is due tomorrow

[lofty spade]

do you have a teacher who taught you this?

[shy scarab]

no

[lofty spade]

then who is it due to?

[tacit wasp]

What topics have you covered so far? At least we know what approach we can use for this exercise...

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@shy scarab Has your question been resolved?

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[blazing jolt]

1. Give 2 possible ex. of a function f that isnt cont at x = -4, but if f is redefined at -4 so that f(-4) = 10 then f becomes cont at x = -4

2. Determine if the following func is cont at a = 5 use the continuity checklist to justify your answer, if it isnt cont, identify points of discontinuity for g(x) = { x^2 - 25 / x - 5 where x =/ 5, or 10 where x = 5

[blazing jolt]

Am I allowed to treat it like the way I am with just doing x =/ -4 and x = 4, for 1 & Am I doing it correctly? (I know its a little messy)

And, for #2, Is this being done correctly?

[odd edge]

@blazing jolt Has your question been resolved?

[odd edge]

@blazing jolt Has your question been resolved?

[amber veldt]

its correct if you forgot to add parentheses when you typed up #2

you wrote $x^2-\frac{25}x -5$ but you surely meant it with parentheses

[clever fjord]

gfauxpas

[blazing jolt]

Oh, yeah. (x^2 - 25) / (x - 5)

sorry

[amber veldt]

i figured

all correct then!

[blazing jolt]

Same for #1 too?

[amber veldt]

[blazing jolt]

I'm allowed to do that? *Just making sure

[amber veldt]

your work is correct, but, define "that", whats your q exactly

[blazing jolt]

Give 2 possible ex. of a function f that isnt cont at x = -4, but if f is redefined at -4 so that f(-4) = 10 then f becomes cont at x = -4

Oh, sorry

I am allowed to treat it the way I am doing, just x doesnt equal -4, then x = -4, to solve the problem?*

[amber veldt]

you are allowed to define a function by cases like that such that it has one and only one discontinuity which is removable

[blazing jolt]

Okay, just checking!

[amber veldt]

in fact

any answer you give will just be that in disguise even if you try another notation

think about why

[blazing jolt]

Because they both lead up to 10

[amber veldt]

no, i mean in general, a function like tharlt

lemme give you an example

let P be a property , like , "x is rational" or "x = 7"

then Iverson brackets are defined by:

[P(x)]=1, if P(x) is true, 0 otherwise

let me define a function h

h(x)=x²[x != 6]+cos x[x=6]

this is just what youre doing, but written with a different notation or perspective

every answer to this type of question, continuous except for exactly one discontinuity that can be redefined

will be essentially the type of function you defined, blah for x!=constant, blah blah for x=constant

[blazing jolt]

Since its continuous, it just needs redefined if it's this sort of way, with < or > being the triggers, basically then

[amber veldt]

continuous everywhere else*

[blazing jolt]

yeahh

[amber veldt]

im saying that whether you use cases, >, <, !=, iverson brackets, etc

it will.be all.the same idea just with different notation

want a fancy example?

[blazing jolt]

Sure

[amber veldt]

let s(x)=sin(x)/x, where this expression describes a continuous function, and defined by limits otherwise

but this is just saying s(x)=sin x/x, if x!=0, and 1 if x = 0

there's no way around it

any answer will.be that type of function fundamentally

[blazing jolt]

I see.

[amber veldt]

the limit of sin x/x as x to 0 is 1 if you didnt know

[odd edge]

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[robust glade]

I've been trying to learn this for the longest time I don’t understand how we know what to put for the n’s I get the numbers hes just substituting but how about the n’s

[woeful briar]

He is just plugging n-2, n-1 and n into the sequence you sum

[robust glade]

But why specifically n-2 how do we know that we’re supposed to start with that

[woeful briar]

We do not start with n-2

We start with 1 which is the lower limit of the sum

But we can't write the sum up to n

So we skip some elements with ... and continue at end

So we know where we start, a few terms after that and then a few terms at the end

[paper pendant]

you're just trying to find a pattern

he could've written ... then started after that from n-3 or n-1

it doesn't matter really

how i learned method of differences, i wrote it all in columns which i do think is probably better (atleast imo)

Σ (r - 1/r + 1/r-1)

= 2 - 1/2 + 1

• 3 - 1/3 + 1/2
• 4 - 1/4 + 1/3
• 5 - 1/5 + 1/4
  ...
• (n-1) - 1/(n-1) + 1/(n-2)
• n - 1/n + 1/(n-1)

then we'd cross things out when recognising a pattern

and circle the terms that we keep

[robust glade]

We keep the terms that don’t get cancelled out right

[paper pendant]

in this case, it would be a case of
n 1
Σ r - ---
r = 1 n
i think

[paper pendant]

yea

do u do a level fm

[robust glade]

Yeah

Ice

Cie

[paper pendant]

oh, iAL?

[robust glade]

Yeah

[paper pendant]

i did british a levels edexcel

i just finished it

anyone whos done al maths would recognise tlmaths 😭

[robust glade]

😭 he carried me through p3 when I self studied that fm isn’t going as smoothly but it’s ok eventually I’ll hopefully learn everything

How about the 2 3 4 n-2 n-1 aren’t they left over?

[woeful briar]

They are in the sum

[robust glade]

Wdym

[woeful briar]

There is a sum on the last lines

It contains all of them

[robust glade]

So they get cancelled out in those?

[woeful briar]

They dont cancel

Look closely

[paper pendant]

the thing with method of differences is that we use it when we recognise that the summation will result in a pattern which consistently brings out a cancelling out pattern.

i think it would be useful to set out the working like this

i think it would be useful to set out the working like this

[odd edge]

@robust glade Has your question been resolved?

[odd edge]

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[gritty furnace]

for non singular matrix $A$ is it the case that $| A^{-1}| = |A|^{-1}$? I have one direction, $1 \leq |\mathbf{I}| = |AA^{-1} | \leq |A||A|^{-1}$ is the reverse true?

[clever fjord]

uwu

[tacit elm]

Is this for any norm ?

[gritty furnace]

yes

[tacit elm]

Ok consider the infinity norm, and consider the diagonal matrix (1,2,…,n)

Norm of A = n
Norm of A^-1 = 1

[gritty furnace]

ok tyvm!

.close

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[swift heron]

i dont really get the point of this, arent these just variations of vector space axioms

[blazing shuttle]

no theyre CONSEQUENCES of vector space axioms

[amber veldt]

for example, (c) follows by distributivity only once you prove -v = -1v for all vectors v, it's not an axiomn that for every v, -v = -1v

it's true but not axiomatic

I think you should prove that first, for every vector v, -1v = -v

it will give you a powerful tool for the parts of the problem @swift heron

[swift heron]

ic

[swift heron]

for this i did something like

(1 - 1)v = 0
v + (-1) * v = 0 (distribution and identity)
(-1) * v = 0 - v = -v (additive inverse and closure of addition?)

[amber veldt]

excellent, though I'd write an additional step 1v+(-1)v=0 after the first line

so now you've proven that you can replace "-v" with "-1v" and so problem a is to prove:

-1(-1v)=v

much easier, right?

[swift heron]

yep!

alright i think i know how to do these

this question also tricked me, but i dont really know how to read this notation

[amber veldt]

this means the set of all functions with domain {1,2,...,n} and codomain a scalar ring

[swift heron]

so functions

f: N -> F?

[amber veldt]

yes precisely

well no

[swift heron]

hmm

is the domain a single natural number

[amber veldt]

$$\bigcup_{i =1}^\infty \mathbb N^i \to \mathbb F$$

[clever fjord]

gfauxpas

[swift heron]

hmm

so like

[amber veldt]

easier to think about a space that's very isomorphic

[swift heron]

f(1), f(2), f(3) ?

[amber veldt]

which is the set of sequences with only finitely many non-zero

it's well-established that these two spaces are "almost the same", if I were a teacher I'd let you assume it without proving

[swift heron]

ah i see

[amber veldt]

because now it's clear how to add functions

[amber veldt]

this is an abuse of notation, reeally it's a direct product, but you probablyt didn't ge tto direct products yet

[swift heron]

yeah we havent

[amber veldt]

because not only are you unioning it you're also adding a vector space structure, so "union" isn't a good enough of a symbol

buit don't worry about it

[swift heron]

so at the end of the day these are functions

so if we wnat to find a basis

we want to make

[amber veldt]

think about the bases for R^n and C^n, you basically want something like that but for an infinite dimensional space

[blazing shuttle]

functions $\cbr{1,\dots,n}\to\mbb F$

[clever fjord]

ロケット・ジャンプ

[swift heron]

$c_{1}f_{1} + c_{2}f_{2} + ... + c_{n}f_{n} = 0$ where c1 = c2 =... cn = 0

[clever fjord]

toast

[amber veldt]

write out the functions as infinite tuples

(x1,x2,x3,x4,...)

and only finitely many are non-zero

[swift heron]

hm

[amber veldt]

so take a guess, intuitively

the standard bases for R^n and C^n

are what?

{e1,e2,e3} for C^3 for example is

[swift heron]

((1,0,0) (0,1,0) (0,0,1))

[amber veldt]

yes

now , take a guess

gut instinct

[swift heron]

hmm

[amber veldt]

what would it be for our space to make infinitely many?

i bet your instinctual guess will be correct

[amber veldt]

this notation means an ordered basis btw

which isn't bad, but just so you know

use curly brackets if you want an unordered set

[swift heron]

for R^n i'd imagine

{e1, ... ,en} where e_j represents (0, 0, .... 1, 0, 0) where 1 is in the jth position

[amber veldt]

very good

now, how many coordinates do we have in our space?

countably infinite, with only finitely many non-zero

so what's e_i ?

[swift heron]

n coords?

[amber veldt]

well

remember I'm considered an isomorphic space with infinitely many zeros because otherwise

it's not so obvious how to add vectors

oh my !

hold on

did I misread the question?
🛑

[blazing shuttle]

r u sure theyre isomorphic?

n is fixed in the problem

[amber veldt]

yes I just realized that

so sorry toaswt

[amber veldt]

this is correct

[swift heron]

ahhh

[amber veldt]

just need to prove it, sorry for the silly mistake

anyway you need to prove it

do you know kronecker delta notation? it makes the math easier to write

[swift heron]

ahahah never heard of it

[amber veldt]

it's a function of 2 variables

[swift heron]

like a piecewise?

oh

wait

[amber veldt]

$$\delta_{ij} = \begin{cases} 1 , \text{ if }i=j \ 0 \text{ otherwise}\end{cases}$$

[clever fjord]

gfauxpas

[swift heron]

hm i see

[amber veldt]

so you can write your basis vectors for R^n, C^n, for example, as

[swift heron]

what is i and j?

[amber veldt]

indices

$$\mathbf{e}i = (\delta{in})$$

[clever fjord]

gfauxpas

[blazing shuttle]

using this notation try to pick a basis of functions $\cbr{1,\dots,n}\to\mbb F$

[clever fjord]

ロケット・ジャンプ

[amber veldt]

the identity matrix is

$\mathbf I = [\delta_{ij}]$

[clever fjord]

gfauxpas

[swift heron]

this makes sense actually

bc the diagonals of I is 1

[amber veldt]

yup

and for the basis e_2 for example is

e_2 = (delta12, delta22, delta32) = (0,1,0)

[swift heron]

kk that ameksk sense too

[amber veldt]

so, anyway, teaching it to you because i think it will help you find the answer,

[amber veldt]

as roketto said

[swift heron]

i think, im a little confused by the nottation, what is an example vector from this vector space of functions

[amber veldt]

let's fix n=5 to make notation easier

$(a,b,c,d,e) \mapsto a+b+c+d+e$

[clever fjord]

gfauxpas

[amber veldt]

$(a,b,c,d,e) \mapsto abc^d\cos(2e\pi)$

[clever fjord]

gfauxpas

[amber veldt]

assuming the scalar ring is real numbers here, or complex

which it usually is in linear algebra

one of those two

[swift heron]

ohhh hm so we take a tuple of natural numbers as the domain and output some element of a field?

[amber veldt]

yes

and forget my nonsense about adding tuples of different sizes that was me beind dumb

n is fixed

[swift heron]

Ah bc it’s from {1 to n} -> F

[amber veldt]

wait

I think im still using my mistake

I'm so sorry

yes okay scratch that

I'm very tired rn

you wanted examples

the domain is simple

any number from 1 to n

not a tuple

[swift heron]

honestly 😂 it’s all good It’s the first time I’ve seen this kind of notation befor

each

Ahhh

That makes sense

[amber veldt]

so it might be

$n \mapsto e^{in}$

[swift heron]

so it’s a single number -> element of field

[clever fjord]

gfauxpas

[amber veldt]

$n \mapsto n$

[clever fjord]

gfauxpas

[amber veldt]

$n \mapsto \log|n^2+1|$

[clever fjord]

gfauxpas

[amber veldt]

assuming the codomain is real numbers here, again

here's the problem

wait no problem

just be careful which is the domain which is the codomain

[swift heron]

hm alr

[amber veldt]

alright let's think about what it means to be a vector space

let f, g be some functions in this space

f+g? alpha f?

with alpha in F

[swift heron]

f+g should also be in V and so does alpha f

[amber veldt]

$(n \mapsto e^{in}) + (n \mapsto n) = ?$

[clever fjord]

gfauxpas

[swift heron]

( n -> n + e^in)?

[amber veldt]

yup!

[swift heron]

Ok lemme think about the original problem

[amber veldt]

$5(n \mapsto 2^n)$?

[clever fjord]

gfauxpas

[swift heron]

(n -> 5 * 2^n)

[amber veldt]

yeah

[swift heron]

So if we want a basis our basis should be a list of functions

now I’m not sure what those functions exactly are but f_n maps from some natural number and outputs some elements f(n)

[amber veldt]

this is confusing indeed because we dont normally think about vectors like this

or about this like vectors

the thing is

this function has n parts

what it does for i=1, for i=2, for i=3, ..., and for i=n

so any basis element

[swift heron]

My best understanding is that for example if i = 2, then we get a function like f: 2 -> f(2)?

[amber veldt]

also has to have a rule for all n inputs

b(1), b(2), ..., b(n)

[swift heron]

oh hm could we map each f(j) = 1?

wait

[amber veldt]

not quite

thinking in the right direction though

[swift heron]

Hmmm

[amber veldt]

you want f(j) = 1 or 0

when 1, when 0?

there are many choices, but what';s the simplest most straightforward choice

and why am I picking 1 or 0? because those are the only two things I know for sure are in the field

because the simplest field is {0,1}

[swift heron]

I guess we can use the delta function if I = j then f(j) = 1 otherwise f(j) = p

0

[amber veldt]

YES

[swift heron]

Ohhh

[amber veldt]

$${\delta_i(j) : 1 \le i, j \le n}$$

[clever fjord]

gfauxpas

[amber veldt]

writing it as delta_i(j) because of function notation, but you dont have to

you could write $\delta_{ij}$ like how it's usualyl written and that's fine too

[clever fjord]

gfauxpas

[swift heron]

ohh I see why it’s n parts now, like we could technically have f_3 : 3 -> f(4) but here f(4) = 0?

[amber veldt]

so

these are actually how tuples are defined

not necessarily how we think about tuples

we think about tuples as a list

but really, something like

$(4,5,i, 2-i)$

is

[clever fjord]

gfauxpas

[amber veldt]

$1 \mapsto 4, 2 \mapsto 5, 3 \mapsto i, 4 \mapsto 2-i$

[clever fjord]

gfauxpas

[swift heron]

Ohhh

[amber veldt]

so even though this is the definition of a tuple, it's not usually how mathematicians use and think about tuples

but sometimes we do!

[swift heron]

I see 😂

[amber veldt]

so when I say

$\delta_i(j)$

as a function

[swift heron]

Ye

[amber veldt]

it means the same thing as

[clever fjord]

gfauxpas

[amber veldt]

$(\delta_{ij})$

[clever fjord]

gfauxpas

[amber veldt]

as a tuple !

[swift heron]

Ahhh that makes sense now I think!

[amber veldt]

and you can just write $\delta_{ij}$ and mathematicians won't care which one you mean because they're equal

[clever fjord]

gfauxpas

[amber veldt]

it's just a matter of perspective

so, I personally would write delta_i(j) as a function because the problem presented the vector space as a space of functions

but they're the same creature

[swift heron]

oh kk I see :)

I need to write this down when I get back

💀

Pretty interesting tho!

when I try to think of Lin alg, I try to visualize what I see so what made this so hard was that I had 0 intuition as to what I’m looking at

[amber veldt]

yeah

here's a cool thing

let's say we consider the bijection

{1} ~ 1, {1,2} ~ 2, {1,2,3} ~ 3, ..., {1, 2, ..., n} ~ n

obvious bijection, right?

[swift heron]

Yep

[amber veldt]

the normal notation for

"all functions with domain X and codomain Y" is

$Y^X$ . this matches cardinality rules

[clever fjord]

gfauxpas

[amber veldt]

$|Y^X| = |Y|^{|X|}$

[clever fjord]

gfauxpas

[amber veldt]

but check this out

$\mathbb F^{{1,2,\cdots,n}} = \mathbb F^n$

[clever fjord]

gfauxpas

[amber veldt]

which is nice

[swift heron]

ye

hm I see

alright this is very cool :)

I’ll have to save this convo 😂

[amber veldt]

yeah linear algebra is the first time students get exposed to more abstract stuff so it's great

[swift heron]

Definitely 😂

[amber veldt]

also, just a warning

many texts do:

N = {0,1,2,3,...}

[swift heron]

like sitting in lecture feels like sitting in a foreign language class compared to real analysis

[swift heron]

Ah

[amber veldt]

which has many advantages

the disadvantage is that the "first" element of a sequence is the 0th one

which can be confusing

[swift heron]

Maybe a better notation is Z^+ for 1 - n

[swift heron]

True

[amber veldt]

oh there are so many notation attempts of people trying to figure out what the best notation is

[swift heron]

💀

[amber veldt]

same idea

[swift heron]

😂😂

That’s pretty funny ngl

[amber veldt]

some people want a subscript

$\mathbb N_0, \mathbb N_1$

[clever fjord]

gfauxpas

[amber veldt]

but that's bad because

these are used for something else

if you believe N should contain 0, then

N0={}, N1 = {0}

if you believe N should start at 1

N0={}, N1 = {1}

and in general

N_n = {0,1,2,3,..., n-1}
OR
N_n = {1, 2, 3, ..., n}

so those symbols are ambiguous

$\mathbb N_1$ has 4 meanings

[clever fjord]

gfauxpas

[amber veldt]

rather, 3

I think Munkres uses Z^+ = {1,2,3,...}

[swift heron]

Agree 😂

Honestly it’s pretty funny

Well at the end of the day

It’s notation 😂

[amber veldt]

okay need to do laundry now

[swift heron]

Kk Ty!!!

.solved

[odd edge]

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