help-19

[runic swift]

(-6, 1)

[lyric lintel]

yes but then that is invers

so what i wrote is answer

[runic swift]

yes

[lyric lintel]

oooo ok ok thx sm

.close

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[lyric lintel]

@runic swift

.reopen

[odd edge]

✅

[lyric lintel]

@runic swift

I have one more question if that is okay?

[runic swift]

sure

[lyric lintel]

alr thx

how do you do this one, it rlly confused me to the point where i kinda guessed @runic swift

[amber schooner]

holy pings bro

[runic swift]

there are certain transformation effects

but honestly i didnt memorise it lol

[amber schooner]

it’s correct

[lyric lintel]

@runic swift we never learned about x in exponent

[amber schooner]

bruh

the ping

[lyric lintel]

i just looked at answer key. How is it correct tho

[amber schooner]

it doesn’t matter where it is

[amber schooner]

well f(cx) is a horizontal stretch

[wooden pasture]

in general f(x) when transformed to f(nx) compresses by a factor n

theres an intuition behind it but i cant really explain that in short

[amber schooner]

a good way to visualize it is with the following

,w x^2

[clever fjord]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by c.knief

[amber schooner]

omg

,w graph x^2

[clever fjord]

Results provided by WolframAlpha

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Requested by c.knief

[amber schooner]

,w graph (10x)^2

[clever fjord]

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Requested by c.knief

[amber schooner]

ok great it won’t show it

hold on

[wooden pasture]

💀

[lyric lintel]

i think just the x in exponent confused me

[wooden pasture]

just think abt the inputs

[lyric lintel]

thas like the only thing

[runic swift]

you can just let f(x) = 6^x

then do the things with it

[amber schooner]

i just think of it like it will get to the y values faster thus it’s skinnier

[wooden pasture]

yeha

[amber schooner]

and of course if the factor is less than one it has the opposite effect

it will fatten out

the derivative is less

[lyric lintel]

wait so what is the function herer

y is f of x

[amber schooner]

f(x) is x^2 in what i did

[lyric lintel]

but what is happening to the x like if u put in function notation

[amber schooner]

but for your question it would be 6^x

[lyric lintel]

f(x)= what?

[amber schooner]

if f(x) is 6^x then f(4x) is 6^4x

[lyric lintel]

how

[amber schooner]

bruh

[lyric lintel]

i dont get the second part

[amber schooner]

because the input is no longer x

it’s 4x

if f(x) = 6^x then f(smiley face) = 6^(smiley face)

[lyric lintel]

what happened to the 4x

[amber schooner]

dawg

i’m done here

you’re trolling

[lyric lintel]

im not but okay

im generally confused

[amber schooner]

generally

you mean genuinely

[lyric lintel]

autocorrect

[amber schooner]

right

[lyric lintel]

...

ok

L helper

[amber schooner]

$f(x) = 6^x \implies f(4x) = 6^{4x}$

[clever fjord]

knief

[amber schooner]

you’re hilarious

[lyric lintel]

my point exactly

[amber schooner]

this concept is understood by my 8 year old cousin

[lyric lintel]

my point exactly

[amber schooner]

my point exactly

[lyric lintel]

good for them ig

[amber schooner]

bad for you ig

[amber schooner]

^

review khan academy

[lyric lintel]

ur explaining just dont work for me @runic swift was better

but thx

i understand it mostly now

[amber schooner]

cant make it any clearer than this

[lyric lintel]

idk was just smoother before all g tho

appreciate ur time

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[compact delta]

<@&286206848099549185> could I please get assistance on limits involving infinity, I am cooked at calc pls have mercy on me

[true musk]

Send the question

[amber schooner]

what’s the issue

[lime grove]

split the fraction

[tacit haven]

divide the top and bottom by x

[lime grove]

so you’d get sqrt(10)/6 or smt

[compact delta]

i understand how to get sqrt(10)/6, how does the limit approaching inf involve in this problem

[tacit haven]

?

[compact delta]

I previously tired sqrt(10)/6 and it didnot accept this answer

[amber schooner]

did you divide everything by x?

[tacit haven]

i think the answer key is wrong then

or maybe its in a different form

[compact delta]

would I mulptiy by 6?

[odd edge]

@compact delta Has your question been resolved?

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[remote cloud]

Can someone explain how this jump happened?

[remote cloud]

Uhh is there a name for it lol

[dark kraken]

Shifting the index? Idk

[remote cloud]

kk

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[whole hazel]

Correct me if I'm wrong, my understanding of this is: We find algebraically that both m and n are even, and if they are both even then it is impossible because that would mean a is not in it's lowest form at a = m/n

[latent scaffold]

Yes, since you assumed it was in lowest form to begin with that's a contradiction.

[willow beacon]

Can someone help me

[whole hazel]

How does both of them being even mean it is not in lowest form exactly?

Does that suggest they can be divided by 1 and 2, not just 1 ?

[latent scaffold]

If they are in lowest form, the numerator and denominator don't have common factors.
If they are both even, then they both have a factor of 2, and hence are not in lowest form.

[whole hazel]

Okay I see, thank you

.solved

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[gleaming mason]

Even with direct substitution I’m still confused, because I kept getting DNE

4/0

Even if it was squared

[onyx cloak]

limit exists for the options you chose

it said limit shouldnt exist for both f(x) and g(x) but it should exist for f(x)+g(x)

[gleaming mason]

I’m going to being honest I guessed because I couldn’t tell from first glance.

[onyx cloak]

then check if limit exists or not

for example the fourth one

limit doesnt exist for both f(x) and g(x)

but it does for f(x)+g(x)

[gleaming mason]

.

I must have the wrong idea of what f+g Is? @onyx cloak

[onyx cloak]

its correct

but

limit exists for both of them

so u just leave that option

because it doesnt satisfy $\lim_{x \to 0} f(x) = 0$ and $\lim_{x \to 0} g(x) = 0$

[clever fjord]

Astar777

[gleaming mason]

Oops wrote down the 5th one instead

[onyx cloak]

also

simplify that

instead of just writing f(x) + g(x) down

[brittle beacon]

(Indeed, do this before you try and work anything out)

[odd edge]

@gleaming mason Has your question been resolved?

[gleaming mason]

My friend came over and helped me , thank you guys @onyx cloak

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[quasi sparrow]

Your notation is pretty bad

[true bolt]

Could you elaborate more please?

[quasi sparrow]

y=-4

Sum = terms of the sum

[odd edge]

@true bolt Has your question been resolved?

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[wet sparrow]

why do they do this, what rule is it based on?

[onyx cloak]

its just the notation

lagrange's notation (y')

and leibniz's notation (dy/dx)

[wet sparrow]

this is what I was looking for! Didn't know what to google

thanks

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[onyx cloak]

why

if u want to close channel then use .close

[wet sparrow]

alright, thanks

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[placid oar]

my post in prealg-and-algebra channel

[mystic saffron]

u can multiply like how u multiply number

[onyx cloak]

if you're confused, do it one bracket at a time

then combine the terms in the end

[mystic saffron]

or like this

[forest sky]

any order will work as long as each step is done correctly

[placid oar]

ok thank you

[steady tide]

!occupied

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[placid oar]

i am done thank you

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[spark dome]

<@&286206848099549185>

[rancid gulch]

ok

so what is 8^2 @spark dome

[spark dome]

64

[rancid gulch]

ok

so what ur trying to do

is find the cube root of 64 then

do uk what cube root is?

[spark dome]

No

[rancid gulch]

a number multiplied by itself 3 times

is equal to that number

so lets say

if instead of 64

it was 8 under that square root sign

then the cube root would be 2

cuz 2 * 2 * 2 is 8

so what number multiplied by itself 3 times is equal to 64

[spark dome]

Ok

[rancid gulch]

?

[spark dome]

A let me think

[rancid gulch]

alr

[spark dome]

4

[rancid gulch]

yeah cuz 4 * 4 * 4 is 64

so ur answers 4

[spark dome]

Ok

Thank you

. close

[rancid gulch]

yw

[spark dome]

.close

[rancid gulch]

its already closed

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[frozen yarrow]

does anyone know how to do this???

[amber schooner]

yes

[fresh sphinx]

oh hey its knief

[amber schooner]

Jub

hi

[frozen yarrow]

hi

[amber schooner]

hello

[frozen yarrow]

what is it asking me to do???

i understand nothing

[amber schooner]

at your level you’ll use the vertical line test to determine if a graph represents a function

the vertical line test is merely drawing a vertical line | through any point on the graph, if the vertical line only crosses the graph once then it’s a function

and if it crosses twice or more as in question 3

then it’s not a function

a function has one and only one output for any given input

meaning no input can result in more than one output

hence why the vertical line test determines if it’s a function or not

because a vertical line is of the form x = k where k is some number

and x is our input

does that make any sense

[frozen yarrow]

yea kind of but what is a input or a output

i understand that for it to be a function it only can cross that line once

[amber schooner]

if it helps you can think of a function as a machine

i put something in

and it spits something out

input is what i put in

output is what comes out

it’s the result

[frozen yarrow]

what would be a input would it be the numbers on the graph??

what do they mean by all possible x or y value in a domain or a range

[amber schooner]

the input would be the x values

the output would be the y values

[amber schooner]

the domain is the set of input values

it’s just the collection of the x values the function/relation is defined for

and the range is the set of y values the function takes on/equals for some value x in its domain

[frozen yarrow]

for number one it would be function and it would be range -infinity and domain would be infinity???

[amber schooner]

so yes it’s a function

but the domain and range you can state in a few ways

you can say "all real numbers"

or you can use interval notation

(-inf,inf)

it’s not just infinity

it’s an interval

infinity seems more so like a singular value although it’s not a number

[frozen yarrow]

ok how do i know if it a negative or a positve??

[amber schooner]

if what is

[frozen yarrow]

the graph number thing

[amber schooner]

$(-\infty,\infty)$

[clever fjord]

knief

[amber schooner]

you mean this?

[frozen yarrow]

yea

[amber schooner]

well it’s just an interval meaning that for the domain we go all the way to the left towards negative infinity and all the way towards the right towards positive infinity and say x, our input takes on all values in that interval

and for the range it’s similar just -infinity is down and +infinity is up

[frozen yarrow]

ok so for exaple number 2 it would be function -inf ,2 and 6,-4

[amber schooner]

so it’s a function

where did you get the numbers though

for domain and range

[frozen yarrow]

isn't just where the it is???

in the graph??

[amber schooner]

what’s it?

so i agree that the domain goes to -infinity but it doesn’t end at x=2

[frozen yarrow]

the points and the arrow thing

so it would just be - infinity and the range would be 6,-4???

are the range numbers alawys in the right??? or is it the point closer to the x line???

[amber schooner]

no

let me draw for you

the point stops at x=6

so the domain is (-inf,6]

[frozen yarrow]

ok

[amber schooner]

closed bracket for inclusion

[frozen yarrow]

wait what is a domian???

[amber schooner]

i explained earlier

[amber schooner]

here

[frozen yarrow]

what does that mean??? ( im sorry for asking too many questions)

[amber schooner]

did you read all of it afterwards

[frozen yarrow]

yes

[amber schooner]

ok maybe it’s easier to consider something that’s not in math terms

let me see if i can find a nice picture hold on

[frozen yarrow]

by all numbers do you mean neagtive and positve numbers???

ok

[amber schooner]

ok this is in math terms but

still

maybe this helps you see

[amber schooner]

yes

[frozen yarrow]

im sorry im really tying to understand

[amber schooner]

no worries

[frozen yarrow]

why are we bring the domain over to the other one

[amber schooner]

the arrows are just for "mapping"

for example

in the image do you see the arrow going from 1 to 2

[frozen yarrow]

yea

[amber schooner]

so that means when i input 1 into my function, 2 comes out

[frozen yarrow]

so it adds up???

[amber schooner]

similarly the arrow from 4 to 3 means when i input 4 in my function i get 3

[amber schooner]

nono

like let’s say my function was f(x) = x^2

x is my input

so if x was 3

then f(3) = (3)^2 = 9

with the illustration that would be an arrow from 3 pointing to 9

when i put in 3

i get 9

that’s all it means

and they don’t define a specific function in the image of course

but the point still stands

some value in the domain will give me some value in the range when put in the function

[frozen yarrow]

so it used in boths side of the eqations because it almost as it was an x??

in a normal eqation

[amber schooner]

sorry what

[frozen yarrow]

i don't know anymore

ok so you got 9 when you input 3 becuase of this (^) thing that made it multipy itself

[amber schooner]

yes it squared the input

similarly if i put in 5 i’d get f(5) = 5^2 = 25

[frozen yarrow]

how does that make the 4=3

[amber schooner]

what

wdym

[frozen yarrow]

this way do you mean by this the arrows point to two numbers by why what does that have to do with a number being sqared???

[amber schooner]

oh you mean the image

[frozen yarrow]

yea

[amber schooner]

it has nothing to do with it

[frozen yarrow]

oh

[amber schooner]

i was just giving an example of a function

the numbers in the image are random

they don’t tell us any particular function

i came up with an example

[frozen yarrow]

ok so domian a just a number sqared???

[amber schooner]

nope

maybe i confused you

[frozen yarrow]

yea i don't know what we are talking about anymore

[amber schooner]

again the function i gave was an example

it could have been any function

[amber schooner]

the domain is the set or collection/group of numbers that are inputs of the function

let’s look at number 2

see how there’s an arrow all towards negative infinity and a point stopping at x=6

so that means that the function has no input for values after x=6

so for instance x=7 isn’t in the domain

and all values greater than 6 won’t be in the domain

so the domain is (-inf,6)

[frozen yarrow]

so far all im understanding is how to define when it

it's function or not

[amber schooner]

hmm

i’m not sure how much more i can help

id recommend watching khan academy

or something similar on youtube

[frozen yarrow]

i know im sorry

[amber schooner]

look up domain and range on youtube

[frozen yarrow]

ok

[amber schooner]

don’t be

[frozen yarrow]

i really am thankfull for trying to help me

[amber schooner]

you’re welcome

[frozen yarrow]

ill look up some videos and see if i can get a bit more

thank you for you time have a goodnight

[amber schooner]

you as well

[frozen yarrow]

bye bye

[amber schooner]

bye

!done

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[frozen yarrow]

.close

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[light sigil]

having trouble phrasing my answer to b

[light sigil]

what i got so far

ik its a contradiction because the number of prime factors of pb^2 would be odd while the number of prime factors of a^2 must be even given what i proved in (a)

i just dont know how to phrase that

[rotund hawk]

I mean, what you want is a proof by contradiction. So you make an assumption in the start (that sqrt(p) = a/b for some a, b \in N) and want to end up with a contradiction (namely a contradiction of the FToA).

Have you tried noting down the major steps you would need to take and then filling in the gaps?

In my experience, it is easier to have a sketch of a proof to build on, rather than just explaining away

[odd edge]

@light sigil Has your question been resolved?

[light sigil]

yeah im just unsure how to sketch out the proof

i added a ton of fat to what i was saying to get my understanding of it down

[rotund hawk]

Do you struggle with formulating the sketch or with finding the steps required for the sketch?

[light sigil]

i think its specifically with phrasing how to use the idea of the fundamental theorem of arithmetic with variables

like ik from earlier that if we square a an integer greater than 1, its prime factors double.

[rotund hawk]

So you want to know how to formulate the last step?

Cause I don’t think you need the theorem before that, unless I missed something

[light sigil]

oh i dont?

oh wait yeah ur righty

i need to figure out how to formally count the number of factors i guess

like can i say smth along the lines of "let k be the number of prime factors of a and j be the number of prime factors of b" or smth?

[rotund hawk]

What you could do is so it the way it was done in (a)

They got around this “number of prime factors” by focusing on each prime factor individually

This also means that your result for (a) only refers to a single prime factor

[light sigil]

lemme screenshot my answer for a

idk if its overcomplicatred

[rotund hawk]

Looks quite right. I guess you could add that p can’t be more than that because it itself can’t consist of other factors of n but that might be overkill here

I think for the proof of (b) you only need to focus on the prime factor p

That way you can use your results from (a)

[light sigil]

whys that?

[rotund hawk]

And the language used in the task

[light sigil]

oh since i gotta use some language from a

[rotund hawk]

Well you know that p occurs an even amount of times in the prime factorization of b^2, and since the prime factorization of b^2 is unique and p* the prime factorization of b^2 is a prime factorization itself that gives you the unique prime factorization of a^2, with an odd number of occurrences of p (contradiction)

[light sigil]

why can i jjust drop the other parts tho

[rotund hawk]

You don’t have to, but if you don’t know how to formulate it using the given formulation can help

[rotund hawk]

Because you don’t need them for a contradiction

[light sigil]

oh wait for a contradiction i can just assume specific things?

[rotund hawk]

See that (a) specifically focuses on the number of occurrences for a single prime number, not the total amount of prime factors

[rotund hawk]

No, you can assume one arbitrary thing in the start and if that gives you a contradiction through otherwise logical steps you know the assumption was wrong

[light sigil]

hm so why wouldnt i need them for a contradiction?

[rotund hawk]

Because you have a contradiction from the fact that p would occur and odd number of times in a^2

That’s not possible, because each prime factor must occur an even number of times in the prime factorization of a^2 (as per (a))

[light sigil]

ohhhhh

so saying that p appears an odd number of times is enough. i just need to say that if a number is squared, every single one of its prime factors has to appear an even number of times

[rotund hawk]

Correct

That’s the contradiction you need to prove (b)

[light sigil]

lemme try writing this down formally

wait shouldnt i immediately mention the fundamental theorem of arithmetic then?

eh ill write this down first

[rotund hawk]

The theorem is used to derive the “unique” I mentioned

[light sigil]

oops i meant odd

vscode autocorrect

should add a "therefore a^2 has an even number of prime factors" after the the pb^2 odd number part

[rotund hawk]

I think you could highlight that this “one unique prime factorization” means that the prime factorization of a^2 must be that of pb^2

This highlights why it’s a contradiction

Just to add clarity to the text

[light sigil]

at the end?

[rotund hawk]

Before the semi colon yes

There’s just some little tweaks I would do to the text, but if you’re first semester I don’t know that they’re strictly necessary

[light sigil]

so like this?

im assuming theres some stuff i can remove

probably could make it less wordy

[rotund hawk]

I gotta head out to school now though, just ping helpers if you want someone to have a look

[light sigil]

but if it gets the job done it gets the job done. i dont want to be writing your proof cuz thats against the point

ye thanks for the help dude

[rotund hawk]

Np, good luck

[light sigil]

ew

🫡

.close

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[wanton bison]

Is this proof ok?

[weary pelican]

how do you know this

[wanton bison]

shit

[weary pelican]

you're not too far off

you need to tweak delta a bit

for this to be valid

[wanton bison]

So instead I do delta/|2x|

and then I can do delta/|x|

I guess I can say

1/|x| < 1/|-delta+2|

[weary pelican]

what you're looking for is |x| > 1

which value of delta allows you to do that?

[wanton bison]

1

[weary pelican]

delta < 1 allows that

[wanton bison]

not delta=1?

[weary pelican]

<= 1 maybe

but

you took delta = epsilon

so if epsilon is too big...

[wanton bison]

my thoughts

[weary pelican]

how do we change delta?

[wanton bison]

hmm

[wanton bison]

,, \abs{x} > 1 \geq \delta \Leftrightarrow \frac{1}{\abs{x}} < \frac{1}{\delta}

[clever fjord]

bacc

[wanton bison]

hmm that seems to get rid of delta completely

which is not fine

[full nimbus]

Try making delta permanently close to 2 for large values of epsilon

Don’t need to even consider what happens on the other side of the axis

Remember, a smaller delta is always better, so take advantage of the local properties around x=2, by fixing delta to be close to x=2, unless a finer delta is necessary

[weary pelican]

yes

if delta = epsilon would let delta > 1, then don't let it

so pick a delta such that it's still smaller or equal to epsilon

[full nimbus]

Just remember, if delta = 0.5 works for epsilon = 10, then it also works for epsilon = 100 or 2000 etc

[weary pelican]

but doesn't go beyond 1 to let 1/|x| <= 1 or any other constant you need

[odd edge]

@wanton bison Has your question been resolved?

[wanton bison]

,, \abs{x-2} < \delta \Leftrightarrow \frac{1}{\delta} > \frac{1}{\abs{x-2}} > \frac{1}{\abs{x}}

[clever fjord]

bacc

[wanton bison]

\begin{align*}
\abs{\frac{1}{x}-\frac{1}{2}} &= \abs{\frac{2-x}{2x}} \
&= \frac{\abs{x-2}}{\abs{2x}} \
&< \frac{\delta}{\abs{x}} \
&< \frac{\delta}{\delta} \
&= 1
\end{align*}
But $\varepsilon > 0$...

[clever fjord]

bacc

[weary pelican]

it's not what you're supposed to do

because why is that

[wanton bison]

subtracting 2 should make the denominator less

and then makes the whole thing greater

[weary pelican]

and if x < 2 ?

[wanton bison]

then i am wrong

shit2

[weary pelican]

again

you need to take delta

such that you can make sure 1/|x| < 1 for example

[wanton bison]

I wanna bound it

[weary pelican]

yes

so don't take delta = epsilon when epsilon is too big

take, delta = min(...,..)

this way you could make sure that |x| > 2-delta always gives you |x| > 1

and still having delta < epsilon

[wanton bison]

I can choose 0 < delta < 1 and by that I would get 1/|x| < 1/delta < inf(1/delta) = 1?

[weary pelican]

1/delta is not < inf(1/delta)...

again you're rather banking on |x| > 2-delta

rather than |x| > delta

[wanton bison]

apologies that you are repeating yourself, i am trying really hard

1/|x| < 1 < delta if I choose delta > 1

?

[full nimbus]

That is incorrect

If delta > 1, it’s like saying let the maximum x values be atleast further than a distance of 1 away from 2, which includes rlly large and rlly small (negative) x

While the left inequality says that 1/|x| < 1 imply a the function is bounded between -1 and 1

This ofc doesn’t follow if x can be rlly rlly big or small, since it lets the functions value vary a lot

[full nimbus]

This will work, with some additional restrictions on delta (which is restricting the x values around 2)

[odd edge]

@wanton bison Has your question been resolved?

[wanton bison]

I could do 0 < |x-2| < 1 then I get 1 < x < 3 and because of that 1/|x| < 1

[full nimbus]

Mb, I thought u mean delta > 1

[wanton bison]

huh

[full nimbus]

Yh, delta < 1 works pretty well

[full nimbus]

.

[wanton bison]

that's my question now

[full nimbus]

I think u mean delta < 1?

[wanton bison]

no i didnt lol

you assumed i knew what i was talking about 😭

[full nimbus]

Um

[full nimbus]

Well, this much is right tbh

[wanton bison]

ok nice

i am still clueless

,, \frac{\delta}{\abs{x}}

[clever fjord]

bacc

[wanton bison]

This thing

I despise it

[full nimbus]

Um lol

[wanton bison]

Why do I have to approach from x < 1

why not bount it between 1 and 3 nicely

or even 1.9 and 2.1

[full nimbus]

You can

[full nimbus]

This won’t work

[full nimbus]

This will

[wanton bison]

ok

[full nimbus]

Remember, ur x values are around 2

[wanton bison]

now my problem is idk how to formulate it properly with delta

[full nimbus]

So delta is just some region around 2

Think of them as a collection of potential inputs

Now you’ve got some epsilon, a distance from which you can deviate from 1/2

[wanton bison]

I thought delta was a function of epsilon

[full nimbus]

It is, but let’s fix epsilon for a second

[wanton bison]

ok

[full nimbus]

You’ve got a delta, which you know, for the current epsilon

Then, a mega useful fact is that, by fixing this delta to be the biggest value delta can take, you only need to consider smaller epsilon

[wanton bison]

so now i fix delta too?

[full nimbus]

This is important because of the inequality |2-x||2x| in ur original solution

[full nimbus]

Yes, to be small enough, such that this inequality always holds

So lemme ask, what x values is it for which the inequality |2-x/2x| < |2-x| holds?

In ur original sol

[wanton bison]

x > 1/2

[full nimbus]

This is great, since a local region around x = 2 can be a subset of x > 1/2

Thus allowing complete use of the inequality

So

Say delta is at most 1

[wanton bison]

1.5 < x < 2.5

[full nimbus]

Or 0.5 even

This works too

We’ll say delta is at most 0.5

[wanton bison]

how do you say delta is at most 1

[full nimbus]

2-1 > 1/2

So the lowest input value still satisfies the inequality

[wanton bison]

that kinda makes sense

So I am choosing a cool interval and then gotta figure what delta would be?

[full nimbus]

This is exactly it

Choose a local region around ur limit point

Which plays nice

Then search for delta

[wanton bison]

1.5 = 2-0.5 < x < 2+0.5 = 2.5
then for this to hold delta < 0.5

[full nimbus]

Yep

[wanton bison]

cool

[full nimbus]

So, we set delta to be = min{0.5, epsilon}

Can u justify why?

[wanton bison]

oh

i am honest i watched a video previously where someone did something similar

uhm

i guess if epsilon is greater than 0.5 we pick 0.5 so that somehow our proof is valid 😂

[full nimbus]

Ig lol

[wanton bison]

otherwise we would leave the interval?

say delta = 1

[full nimbus]

Yes, but we know epsilon = delta works where the inequality holds

So if epsilon = 0.5

Delta = 0.5 works

Based of ur inequality sol

[wanton bison]

if epsilon 0.51?

we take 0.5?

[full nimbus]

Yep

[wanton bison]

but what'S the point of saying for all epsilon > 0 then just for me to say idc i take a lower value regardless

[full nimbus]

What’s key is the if delta works for epsilon = k then it also works for any epsilon > k

[full nimbus]

It’s more of a defining property rlly

Ofc

It doesn’t rlly matter what happens over the entire domain

But just a local region

[wanton bison]

i thought when we say for all epsilon, you can choose any epsilon regardless what and it will work

[full nimbus]

Like, what does the asymptote at x=0 say abt the limit at x=2

[full nimbus]

It will, but extending the delta that works in a local region works over the entire domain

Cus epsilon is only getting bigger

Actually

It’s easier to say that we cld literally restrict the domain

To be anything, as long as it is a region around our limiting point

Btw, have u seen graphical interpretations of limits?

It makes it a lot easier to visualise the distances epsilon and delta represent

[wanton bison]

hmm

If delta gets arbitrarily small, so should epsilon

[full nimbus]

No it need not

You can make delta smaller in the image

And maintain epsilon

[wanton bison]

why would you do that

[full nimbus]

All inputs in delta will still be mapped to a region within the epsilon

What I mean to say, a smaller delta wld work for the same epsilon

[wanton bison]

ok

[full nimbus]

And similarly, fix the delta, and let epsilon grow bigger

[wanton bison]

if epsilon grows bigger would I not go away from the limit?

[full nimbus]

No, the limit is the centre of the epsilon region

The boundaries tend away

But we are simply giving a bigger net to catch the outputs from delta

But they already fit into the smaller epsilon

So trivially they fit into the bigger epsilon

[wanton bison]

So I wanna find a suitable interval of x.
For that I must choose delta wisely, like we did.
And then I have to make sure, no matter for the choice of epsilon, delta doesn't exceed its boundaries, so that then my interval for x holds, so the proof holds

[full nimbus]

Yep

Yh

[wanton bison]

hmm i actually expected a no

well that minimum is like a cheatcode

[full nimbus]

Yes, it’s probably the most important ‘trick’ to these problems

[wanton bison]

min{epsilon/(who cares), 0.5}

[wanton bison]

my problem is then to understand, why that is then a proof in the first place lol

[full nimbus]

Why would it not be a proof?

[wanton bison]

because we are making epsilon look useless with that minimum trick

or rather redundant

[full nimbus]

Well, it’s just how it was defined tbh

I suppose a local region

Is up to us to decide and implement

But tbh

[wanton bison]

I mean okay, "for all epsilon > 0"
You can choose any epsilon, and if it don't suits I take an appropriate delta anyway.
I broke no rules, and you chose still epsilon, so nobody violated anything

[full nimbus]

A definition like for all 10 > epsilon > 0

Also works for example

[wanton bison]

ah

[full nimbus]

In fact

If I asked u the same problem

But said, the domain is {x: x > 1/2}

Then ur range is limited to {y: y <2}

Then the maximum epsilon that should be considered is epsilon = 1/2

Any larger, and your out of the range of the function

So 0 < epsilon < 1/2 would be what ur proving on

[wanton bison]

it's kinda weird that epsilon has to do with y but is treated like x

oh wait

is it

[full nimbus]

Ye, the picture comes to mind for me to distinguish

[wanton bison]

Say I chose an epsilon and basically bounded the limit

[full nimbus]

Yep

[wanton bison]

then it must follow that the distance of |a-delta| gets smaller

[full nimbus]

No

Wait

|a-delta| getting smaller implies delta gets bigger

[wanton bison]

what

ah

wait how

if delta gets bigger i get farer away from a

[full nimbus]

a-delta is the lower point of the boundary

|a-delta| is it’s distance from 0

If it gets closer to 0, away from the limit point, the delta is getting bigger

Ahh

It seems you’ve interpreted |a-delta| as the distance of the region

[wanton bison]

yes

[full nimbus]

The distance is delta itself

No modulus, no a

Just delta

[wanton bison]

ah ok

[full nimbus]

Then a - delta, is like travelling from the limit point a a distance delta

To the boundary of the region

[wanton bison]

so what i mean to say is if epsilon decreases then so does delta

[full nimbus]

This much is usually true

Ofc, when we set delta = min(0.5, …)

[wanton bison]

i am just trying to gain intuition on whats happening during that proof

and what i am showing

[full nimbus]

Then a change from epsilon = 1000 to epsilon = 100 doesn’t mean much

[wanton bison]

I am showing 0 < |x-a| < delta implies |f(x)-L| < epsilon

[full nimbus]

Yes

[wanton bison]

so if i bound the region i approach then it follows that my limit is also bounded?

[full nimbus]

You mean to say delta is bounded?

[wanton bison]

no, how, if delta is a bound itself and varies

[full nimbus]

A bound on regions around, say x=2, in this instance?

Such as delta being 0.5 is the bound

[wanton bison]

ok

[full nimbus]

Ok yh I see what u mean

[wanton bison]

so we wanna show that if we restrict x, that the f(x) is within some arbitrary interval

[full nimbus]

Yh that’s exactly right

Just that x=2 ofc is in that interval

Or x=a even

[wanton bison]

and if we can do this arbitrarily then f(x) will always be bounded and thus be approached

[full nimbus]

For any arbitrary epsilon yes

[wanton bison]

weird idea

[full nimbus]

Why analysis is touted to be super hard tbh

[wanton bison]

i always thought as a->0 so does b->0

[full nimbus]

No

It’s x -> a, then f(x) -> b

Where a and b are independent of each other

[wanton bison]

oh yea

then i meant actually

delta -> 0 then epsilon -> 0

[full nimbus]

So, I sps to be clear, we do think of delta as some function of epsilon

But that’s more for working out a value of delta that works for given epsilon, so we know it exists

[wanton bison]

aH

[full nimbus]

But, there are lots of deltas that work for any epsilon

If u no delta = d works, then delta < d works as well

For any 0 < delta < d

[wanton bison]

we choose epsilon now we gotta always find a delta so that we find an interval that is for some reason sufficient

[full nimbus]

Yes, it’s abt finding a sufficient interval that we know for sure works

[wanton bison]

integral haha

[full nimbus]

And this is usually done by finding delta = f(epsilon)

Where f is some random function

[wanton bison]

damn how do i find this delta then

[full nimbus]

Ofc, we use inequalities to prove some specific function of delta in terms of epsilon works

As ur original solution, delta = epsilon was one such function

[wanton bison]

but something was missing in the proof

|x| can be anything

[full nimbus]

Ofc, that’s why delta = epsilon can’t be your final answer

It’s just something that works for small epsilon

[wanton bison]

well if epsilon is not small we make it small

that's prob the attitude

[full nimbus]

Yh

I’d say if epsilon is not small then it doesn’t matter

Cus we can make it work for small epsilon

[wanton bison]

so i simply say min of epsilon and 1/2

[full nimbus]

Yh

[wanton bison]

could i also say delta = epsilon^0/2

😂

[full nimbus]

I don’t think that works lol

[wanton bison]

you dont or do what

😭

[full nimbus]

Well

[wanton bison]

this is as genius as min

[full nimbus]

Yh, the min function works very well, and now it certainly works

[wanton bison]

So again if delta is set to 1/2 or smaller than i am within that interval of x

for which that estimation works

2-delta = 1.5 < x < 2.5 = 2+delta

[full nimbus]

Ofc

If for small epsilon, you use delta = epsilon

[wanton bison]

did i do a mistake

delta = 0.01

2-0.01=1.99

not 1.5

[full nimbus]

It’s not a mistake

What you’ll find is for small epsilon, th 2-delta < x < 2+ delta gets closer and closer to 2

But for large epsilon, the interval is immediately set to 1.5 < x < 2.5

When I say small epsilon, I talk abt epsilon < 0.5, where where change the choice of argument in our min function

[wanton bison]

ahhh

ok

So is this basically the play ground I have?

[full nimbus]

Yes, just alpha needs to be bigger than 1

[wanton bison]

i realized it should been delta * alpha too

[wanton bison]

hmm why

[full nimbus]

Not sure why u introduced it tbh, but making it rlly small (close to 0), makes epsilon/alpha rlly big

So ur min function cld end up choosing 1/2, even when epsilon is smaller than 1/2

And ur proof doesn’t facilitate this

[wanton bison]

yea i know, i was trying to experiment

i wanted to see if it matters

cause of the min function

[full nimbus]

The usual way of represent the variety of deltas is to say delta < min(..), instead of delta = min(…)

Is saying we know the delta = min(…) works

But any delta smaller than it shld also work

[wanton bison]

ok

damn ok idk what to say

it's not even complicated

it's just weird

well after all it's a definion

and you just trying to use that definition and that's your proof

[full nimbus]

Yh, that’s abt the size of it

[wanton bison]

hmm ok

i guess acceptance is the best way to deal with this rn

thank you for your time, i appreciate it

[full nimbus]

Np m8

Gl

[wanton bison]

I will try to practice more of this

.solved

[odd edge]

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[heady plover]

Cool, start analysis

[wanton bison]

too bad i had analysis 1,2 and 3

guess this then real analysis

[heady plover]

Just realized that you've been awake for more than 24 hours💀

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[charred mauve]

I don't get how you can make a line between A and B, because A and B is on the same y axis how can there be a Perpendicular Bisector

[late dust]

!original

[odd edge]

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.

[late dust]

Is your problem that the bisector of AB is vertical and thus not a function?

[charred mauve]

Yea, Im wondering how to get the edge even if its vertical

[late dust]

Just x=2

It's a line equation, it doesn't have to be a function

[charred mauve]

Oh ok, but the thing is geogebra shows me something diffrent so i wasent sure

[late dust]

Idk how Geogebra works but this is definitely wrong

[charred mauve]

alr lol, thanks tho

.close

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[pastel dew]

hey how are you? do you remember me ?

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[heady plover]

this circle is the D area, I dont understand this step

[wanton bison]

x = rcos(theta)+1
y = rsin(theta)+1

r(sin(theta)+cos(theta))

if you integrate over a circle meaning 0 to 2pi then that will be 0

cause cosine and sine are periodic

so you remain with 2

[heady plover]

You're the best

.close

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[fathom steppe]

Hello, I’m a high schooler. Can someone help me solve this?

[mystic saffron]

cool

i think square both sides?

[lyric dust]

note some restrictions first

[mystic saffron]

wait

$x^2 - 1 \geq 0$

[clever fjord]

icannotdoanymorecauchy

[mystic saffron]

dont do this

[fathom steppe]

This is a restriction right?

[mystic saffron]

yes

[lyric dust]

or |x| >= 1

i would suggest doing this by cases

ie, when x is positive or negative

since |x| >=1

analyse when x is <= -1

and when x is >=1

[fathom steppe]

I think I did something wrong (the book says that the answer should be x<=-1 V 1 < x < 5/3

[lyric dust]

,rccw

[lyric dust]

your cases aren't quite right i think

should be x>=1 and x<= -1

[fathom steppe]

Yea sorry I was distracted

[lyric dust]

all good

[fathom steppe]

What mistakes did I make @lyric dust?

[odd edge]

@fathom steppe Has your question been resolved?

[odd edge]

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[fathom steppe]

Hello, I’m a high schooler. Can anyone help me with this?

[clear current]

well where have you gotten so far

[fathom steppe]

I keep calculating the wrong result

the book says the answer should be x<=-1 V 1 < x < 5/3

[clear current]

the thing is you cant just square both sides

cause the equality might flip

for example think about x=-10

[elfin zodiac]

I have i difficulties reading your writing. Can you tell me the domain of sqrt(x^2-1)?

@clear current

[clear current]

he wrote that part correctly x <= -1 or x >= 1

[elfin zodiac]

Ok and the quadratic equation gave u which results?

[elfin zodiac]

Did u get 1 and 5/3? I can’t read

[clear current]

he got -1, and -5/3

[elfin zodiac]

That looks like a mistake

[fathom steppe]

Why?

[elfin zodiac]

What is your quadratic?

[fathom steppe]

My quadratic equation you mean?

[elfin zodiac]

Yes write here

[fathom steppe]

-3x^2+8x-5>0

[elfin zodiac]

$-3(-1)^2+8(-1)-5=-3-8-5=-16$

[clever fjord]

Samuel

[elfin zodiac]

-1 is not a solution

-5/3 is neither a solution

U made a mistake in your manipulation

But I can’t ready your writing you you better do it again

[fathom steppe]

Got it, I’ll tell you when I finish

@elfin zodiac is this correct?

[clear current]

I told you the issue before

$\sqrt{x^2-1}>2(x-1)$

[clever fjord]

parm

[clear current]

you cannot just square both sides at this step

[elfin zodiac]

What is the issue now? I can’t see

[clear current]

when you want to remove the square root

$x^2-1>\abs{2(x-1)}^2$

go from here

[elfin zodiac]

?

[clever fjord]

parm

[elfin zodiac]

That doesn’t make sense

x^2=|x|^2 over real numbers @clear current

I think you are getting confused with the definition of absolute value

[clear current]

nevermind im stupid

[elfin zodiac]

Looks good unless i am not seeing the obvious

[fathom steppe]

Thank you so much for your help!!!

[elfin zodiac]

!done

[odd edge]

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[fathom steppe]

.close

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[brave torrent]

Hello friends,

I'm not sure to understand how to go from there to there (see pic).

I feel like I'm missing some steps or summation formulae. Someone can elaborate for me?

[brave torrent]

I understand how to go from the summation to n(n-1)/2. But not the step before.

[uneven osprey]

it is like a change of variable

like n-1-i = k

[brave torrent]

How can you assume such thing?

I kinda see it's like a change of variable, but I feel there is something more.

[uneven osprey]

we can assume that to be anything provided we change the bounds according to that

[brave torrent]

Not sure to understand.

[echo ginkgo]

what do you expect by "something more" ?

[brave torrent]

I think I would do stuffs like

summation of n - summation of 1 - summation of i

And hten use some properties of the summations.

[echo ginkgo]

well you can do that yes

they just went another way

[brave torrent]

Wait there is an another way?

[echo ginkgo]

I'm talking about what's on your screenshot, it's the other way

[brave torrent]

Yes.

But I mean, how can you just "skip" to the end.

I feel like we are totally changing the whole equation if we skip to the summation I'm having issue.

[echo ginkgo]

cause they're clever in their change of variables

it's just summing in reverse here really

[uneven lodge]

Yeah. In the first sum they count from n - 1 and down by a different i each term, in the next they count up.

The first sum is (n - 1 - 0) + (n - 1 - 1) + (n - 1 - 2) + … + 3 + 2 + 1 and the next sum is 1 + 2 + 3 + … + (n - 3) + (n - 2) + (n - 1). For finite sums the order doesn't matter, so they are the same.

[brave torrent]

Jus tto be sure, you would have to "calculate some values" to see if it does count down and seea pattern?

Or should this be that obvious? 😄

[echo ginkgo]

if you just wanna check that someone didn't fuck up, then yeah just compare what is being summed in the two cases

[uneven lodge]

It gets obvious with practice.

[brave torrent]

Ah I see!

[echo ginkgo]

as to why they did the change of variables, it's cause they didn't wanna deal with that n-1-i or something

[brave torrent]

Yeah I totally understand now

[uneven lodge]

In this case it should be obvious that the first sum counts down by 1 with each term and the second counts up by 1 each term. Then the thing you just need to check is if it starts and ends the correct place.

[brave torrent]

I did not see that it was a countdown.

[echo ginkgo]

but here the sum is quite simple, the "just distribute everything" way is completely viable

[brave torrent]

Yeah. So basically, either I use properties, or see if there is a simplier form of summation.

And to do the convertion, I need to calculate hand by hand the summation.

Just to be sure.

[echo ginkgo]

yeah if you're unsure just check manually

[brave torrent]

Perfecto, thank again aPlatypus and mikkel 🙂

[uneven lodge]

Whenever in doubt, just write out a few terms until a pattern catches on for you.

[brave torrent]

Yeah makes sense.

And I should see both ends jus tot be sure, right?

Like i = 0 and i = n-2

[uneven lodge]

[brave torrent]

Perfecto!

.close

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[meager sail]

Hi all. I'm working on this problem from a mathematical methods book (couldn't find the solutions anywhere online).

My first idea was to solve it similar to how you would a system of linear first orders (ie. differentiate the top equation and substitute in values from the second) but then expanding that out seems to give an unforgiving amount of algebra
Is there a neat way to solve this question?

[clever fjord]

Fiery

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@meager sail Has your question been resolved?

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[meager sail]

.reopen

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✅

[meager sail]

<@&286206848099549185> I would really appreciate any help with this

[quasi sparrow]

nope. unforgiving amount of algebra is the way to go

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[manic nymph]

not sure how to start this question

[proven harbor]

i dont know limits very well

but can we say that x = sinx?

then subsititue all the sinx as x

sin(sinx) = sinx = x

@manic nymph

[halcyon lava]

You can use LHopital

[halcyon lava]

This also works as sinx tends to x when x tends to 0

[halcyon lava]

Nevermind you arent allowed to

[manic nymph]

okay i will try that

[halcyon lava]

Or you could multiply and divide by x again and again

Inside the sin functions

[proven harbor]

ohh thnx

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[spiral crane]

how do i solve the tree diagram

[mystic saffron]

Balls

[spiral crane]

im kind of stuck if not white would be n-3/3

[mystic saffron]

Hello

[spiral crane]

Hiii

[mystic saffron]

Are you sure?

[spiral crane]

im not

[mystic saffron]

With the denominator

Why did you choose 3 ?