help-42

[frosty mica]

.reopen

[calm coral]

✅

[frosty mica]

I don't know if my teacher is saying. I wasn't in class and he won't answer me.

His reponse was "You should already know what this is saying"

[slate field]

oh.

your teacher is clearly not a mathematician

[potent igloo]

Check your book

[frosty mica]

Clearly not helpful.

A set is said to be finite if it is either empty or has n elements for some n in N.

So basically, if there's a countable amount of elements?

[potent igloo]

Not exactly. A set is countable if it is finite or countably infinite

[frosty mica]

If we start with the forward, then we are trying to prove similar cardinalities.

So we are given that T1 is finite, and want to prove that there is a bijection from T1 onto another finite set?

[potent igloo]

Think of the most trivial bijection you can

[frosty mica]

A linear function?

[potent igloo]

More trivial

[frosty mica]

Points?

[potent igloo]

Identity

[frosty mica]

So just the set of all natural numbers?

[potent igloo]

No

[frosty mica]

What is the identity of a bijection?

[potent igloo]

No. The identity function

[frosty mica]

If I know what the identity function is, it was called something else.

Because identity function isn't ringing a bell.

[potent igloo]

Identify function : f(x)=x

[frosty mica]

I thought this was what the reference was.

[potent igloo]

?

[frosty mica]

So we have f(x)=x. And we are bijecting it.

[potent igloo]

No

Bijection is not a verb

You don't biject something

[frosty mica]

I bet could if I f*kd up hard enough. But that's besides the point.

[slate field]

i do though

[potent igloo]

The identity function is a bijective function

[frosty mica]

Right, by default.

[potent igloo]

Yes

[frosty mica]

What do I do with that?

[slate field]

well you found a bijection, what is the image of that bijection?

[frosty mica]

The image of f(x) is f(x)?

I put in an x, I get x back?

[slate field]

yeah, so what is the set that’s mapped to by that function? what’s the range?

[potent igloo]

f(x) is not an image. It's just a single value

[frosty mica]

Range is the domain?

Is the range the set of all numbers?

[slate field]

no, the range is the thing that is mapped to by f. think about our goal. one could rephrase the goal as “find a bijection where the image/range is finite”. if i have a map f: X-> Y, and f is a bijection, then the image (or range) of f is Y, and the domain of f is X

[frosty mica]

Oh.

I don't really understand images. Hopefully it's not too important for the midterm.

[slate field]

they might be important, it won’t take much time to internalize. when is your midterm?

[frosty mica]

Friday. It's a 5 week class with two tests and a final.

Ill probably bug my teacher during office hours.

[potent igloo]

Do you get an extension since you have covid right now?

[frosty mica]

That is a great question.

[slate field]

better talk to them soon

[frosty mica]

I have class on Monday. I'll call into virtual hours before it.

But it's a 5 week class, so there's not a ton of room he can push it back. If I can take it on Monday instead of Friday, I may still be ok.

So the image of f(x) is Y?

[potent igloo]

The image of f is Y

[frosty mica]

I need to figure out the nitty gritty of what everything is called.

[potent igloo]

$f(x)\in Y$

[potent lotus]

SWR

[slate field]

yes if f is a bijection. Y is actually just called the codomain if f is not onto/surjective

[potent igloo]

Yes. You do

[frosty mica]

So what do I do with this Y?

[slate field]

depends on the problem. sometimes nothing

[frosty mica]

In the context of this problem.

[potent igloo]

Function, domain, codomain, range, image, preimage, surjective, one to one, injective, into, bijection, composition

All important

[frosty mica]

If you asked me what each of those individually was, I might be able to answer.

[slate field]

in this case, we are being asked to find a bijection where the image is finite, we found a bijection, and now we should ask “what is the image”?

[frosty mica]

Although I've never heard of into.

The image is Y. And we have Y from the bijection of X?

[potent igloo]

Onto. My bad

Even though into is another word for injection I think. But uncommon

[slate field]

we should be more specific. suppose that we have the identity function defined on T, given by f(x) = x. can you tell me the image of this function?

[frosty mica]

If I say something that makes no sense, it's because none of this is making sense to me, so I'm trying to make things until I find something that does.

An image is an interval?

[slate field]

the image will be a set. a function is a map between sets

[frosty mica]

The image is a set. So we're mapping x onto f(x)?

[slate field]

this is the case for every function. but the identity is special, since we’re saying f(x) is x

[frosty mica]

So the image is x?

[potent igloo]

x is a single value

[slate field]

no, the image will be the collection of all of the f(x) in general. x js just one element of the set

[frosty mica]

Because logically in my head, if the image is a set, I can't just say the image of this is N?

Or I guess R?

[potent igloo]

Where are you getting N and R from?

[frosty mica]

Becaus that's a set

I don't really know how else to represent every number that f(x)=x can.

To me "what is the image" is asking "what is the set of numbers that f(x)=x spans on the y axis

[potent igloo]

A set of just any collection of elements

[frosty mica]

I only just learned about images today, so I have virtually no knowledge of them.

[slate field]

picture i drew

the two dots on the right are the image

[frosty mica]

But if x is mapped onto y, isn't the image every single number in existence?

[slate field]

the arrows are f

[potent igloo]

Do you mean for identity, or in general?

[frosty mica]

Idk if this example is working, I'm just more confused on what an image is now.

[slate field]

which example?

[frosty mica]

If we have f(x)=x, then we have the x's on the bottom, which go onto its corresponding y going up. But it goes infinitely?

[slate field]

oh ic

[slate field]

but you’re implicitly assuming the domain here! i’ll give you an example that you can sink your teeth into: if f: N -> Q is given by f(n) = n, then you would get that the image is all of the natural numbers

[frosty mica]

Oh well the map was never actually specified. I thought we were just talking about all numbers since it was the identity.

[slate field]

but the identity can mean many things actually. for example if i take the “identity on {a,b,c}” that would mean the function where f(a)=a, f(b)=b, and f(c)=c.

[frosty mica]

So if someone asks for the image of f(x)=x, they're not asking for all of the y's associated with the x's? How do I know which elements they want?

[slate field]

they’re asking for the collection of all the outputs of f. so in the case of the example just above your message, the image of f would be {a,b,c}

[frosty mica]

So they have to specify the domain first?

[slate field]

yes, a function implicitly contains information about the domain, and in all of your previous classes, they were taking that fact for granted, but we will no longer do that.

for every function you should ask “what’s the domain” and be prepared for a wacky answer

[frosty mica]

So like in my first proofs class, when my teacher said "always ask the universe of discourse"?

[slate field]

yes

a lot of misunderstandings come from not being clear about the domains of functions.

[slate field]

in this picture, the domain of f happened to be the three dots in X

[frosty mica]

But two of them went to the same one for some reason

[slate field]

and the image of f was the two dots that are at the heads of the arrows

[slate field]

yeah, that’s just how f happened to be.

f of the first dot was the second dot

so f isn’t a bijection between X and Y, which is fine

[frosty mica]

So back to the original question. I have a nonempty set T1 that we are proving to be finite if there is a bijection from this set onto another finite set T2. Which means they have the same cardinality.

[slate field]

yes, so we can assume T1 is finite first, and try to get a bijection from T1 to a finite set

[frosty mica]

So if we're doing the forward direction where we're given that T1 is finite.

[slate field]

yep

[frosty mica]

How do I get bijection if I don't have any domains or anything to work with?

[slate field]

you do! you have a domain of T1

[frosty mica]

If T1 is finite, that means that there is n amounts of elements in N.

[slate field]

you don’t have any image in mind yet, so we will have to build a bijection to some set that we know is finite.

[frosty mica]

So a set of positive integers

Are we supposed to use an example image, or just generalize?

[slate field]

you should just generalize, but actually i had a different set in mind.

[frosty mica]

Ok shoot. Which one?

[slate field]

instead of using some subset of N, we can just use T1 itself

this is what we were trying to say with the identity function earlier

[frosty mica]

That example was completely lost on me, ngl.

[slate field]

but this is the bijection that you want, and i think we have spent enough time thinking about it where me telling you why this works will be helpful

because think about any set ( a weird set, like the set of leaves on a particular tomato or something)

and take a function that takes in a leaf of that tomato and outputs the same leaf

[frosty mica]

Ok.

Are we saying we can have a bijection from T1 onto T1?

[slate field]

then you’ve created a bijection between {leaves on that tomato} and {leaves on that tomato}

[slate field]

yep

[frosty mica]

Is the T2 a red herring then?

[slate field]

no, we will have to go backwards

and the power of the theorem is in the backwards direction anyways

[frosty mica]

But for the forwards direction, we can say that T1 is finite because we can biject T1 onto T1?

[slate field]

but we were allowed to pick any set to be T2, so why not let T2 just be T1 again

[slate field]

yes

[frosty mica]

I know that person said biject isn't a word, but I'm gonna say it anyways.

[slate field]

well

no we were saying that we could create bijection from T1 to a finite set T2 by letting T2 be T1. T2 was not determined from the start

[frosty mica]

And we are allowed to do that?

Because it states T2 is a finite set?

[slate field]

the statement is saying (formally):
IF (P) THEN (there exists Q such that R)

where Q is analogous to T2 here

so we are supposing P

[frosty mica]

If T1 is a finite set, then there exists a finite set T2 such that T1 is bijected onto it?

[slate field]

yes, except the wording at the end can be cleaned up

but that is the right intuition

[frosty mica]

Well yeah. I could use words that actually exist lol

[slate field]

but yes this is what the statement is saying

at least the forwards direction

[frosty mica]

Wait

If we are assuming that T1 is finite, then we are trying to prove that the bijection exists. Which we did since T1->T1?

[slate field]

yeah, and what does the arrow that you just drew mean

the ->

[frosty mica]

biject

[slate field]

but with what function?

[frosty mica]

The identity function?

[slate field]

yeah

[frosty mica]

How do I denote the identity function?

[slate field]

id: T1 -> T1 would be recognized as the identity on T1

you could also just say “let f be the identity function on T1”

[frosty mica]

Then do the arrow thing?

Are maps with the double line arrow or the single?

[slate field]

single

[frosty mica]

Shoot I messed that up.

[slate field]

double arrow is like the “implies” arrow

[frosty mica]

Is this the correct conclusion/logic?

[slate field]

remember the goal

we assumed that T1 was finite

[frosty mica]

The goal was to prove there is a bijection. But doesn't the identity function do that?

I just don't really know how to put that into words.

[slate field]

and we needed to show 2 things about f. first. that it was a bijection (you are fine by just asserting that) second, that the image of f was a finite set

[frosty mica]

Uh.

[slate field]

but the image of f was T1

[frosty mica]

Isn't the image of f T1?

[slate field]

yep

[frosty mica]

How do I even state that?

[slate field]

which is finite by the assumption

[frosty mica]

I don't really know what my concluding statement is here.

[slate field]

“the image of f is T1”

[frosty mica]

Because bijection, the image of f is T1?

[slate field]

because f is the identity, not because it’s a bijection.

[frosty mica]

Is it not a bijection?

[slate field]

it is, but that’s not why the image of f is T1

[frosty mica]

I don't realy understand what I'm even trying to prove any more.

[slate field]

yeah, i think you lost sight of it. the goal was to assume that T1 is finite, and prove that there is a bijection (which you guessed what that bijection might be) from T1 whose image is a finite set.

[frosty mica]

I have no idea how to turn any of that into words.

This hurts my head.

[slate field]

we sort of just did put it into words though. you named a function from T1 (by saying let f be the identity on T1), and then all that’s left to show is that f is a bijection between T1 and T1 (which you are pretty much allowed to assert) and that the image is finite (for you: why is that true?)

[frosty mica]

I don't know how to actually write the proof with the info given though. I have all the pieces, I just have no idea how to glue them together.

[slate field]

okay, well we can break it down a bit

[frosty mica]

I'm just confused what my final sentence is.

But so insanely confused, that I don't know how to structure any of the rest of it.

[slate field]

let’s just tackle this first

final sentence be damned lol

for now

[frosty mica]

Ok my first sentence is that I have a function that maps T1 onto T1.

[slate field]

first sentence is “suppose T1 is finite.”

[frosty mica]

Oh

WWait

Since when

At this point, I'm so confused I've forgotten how to write a proof entirely.

[slate field]

we are proving that if T1 is finite then (blah). the way to prove this is to assume the antecedent, and prove the consequent. what that means is we assume the first thing and prove the second and that’s a valid proof.

[frosty mica]

Assume T1 finite. Let there be f, identity thingy onT1

[slate field]

thanks layla

[slate field]

yes, let f be the identity is the second sentence

and then the third and fourth sentences will be the things about f.

[frosty mica]

f is bijection of T1 onto T1. Therefore bijection exists?

[slate field]

no need for “therefore..” sentence, we can keep a better flow, and we still have to prove more stuff about f

[frosty mica]

What do we need to prove?

I thought the bijection was the conclusion

oh

bijection onto a finite image?

[slate field]

yep

and why is the image finite?

[frosty mica]

Because the image is T1?

[slate field]

great

and T1 was finite by assumption

[frosty mica]

?

I really really really really really have no idea how to wrap this up.

[slate field]

“then the bijection of..” sentence should read “f is a bijection between T1 and T1”

[frosty mica]

Can I say "Since T1 is a finite set, the proof is done?"

[slate field]

since T1 is a finite set, f is a bijection between T1 and a finite set.

[frosty mica]

Oh.

[slate field]

and then we’d be done

because that’s all we wanted

[frosty mica]

Man that took a full hour of brain pain to get to that.

[slate field]

that’s good

[frosty mica]

I'm starting to think I'm not going to pass this class if I'm struggling this hard.

[slate field]

we have gotten through a lot in this

[potent igloo]

You still have to prove the reverse too

[slate field]

that is true

[frosty mica]

I am aware.

This is only the first problem on this homework set 😦

Did I at least start the reverse correctly?

onto T2

[potent igloo]

No

[frosty mica]

No????

How no??

[slate field]

one thing that you should do is really review the definitions. do simpler problems to test your intuition. a lot of the reason people struggle on homework is because they don’t do easier problems

[frosty mica]

This is the simplest problem my teacher had 😢

[slate field]

onto a finite set T2

[slate field]

look in books i mean

[frosty mica]

I put the T2 tho

I'm not gonna lie, math textbooks confuse the everloving crap out of me. I have never been able to read them because the language is so scary.

[slate field]

you find them scary because you don’t read them

[frosty mica]

I also am extremely bad at reading in general. I can't even read a normal book without losing all interest within 10 seconds.

Even if it's the most interesting book in the world.

[slate field]

you absolutely have to read if you want to do well

in math

[frosty mica]

I am aware, I just have no idea how to develop the skillset of being able to read.

Or rather, to keep reading.

[slate field]

read simple things

[frosty mica]

Like what?

[slate field]

do you mean read regular books or math? are you willing to be bored for a bit?

[frosty mica]

All books.

[slate field]

to pass your classes?

[frosty mica]

I feel like this specific issue surpasses math, and is just a personal issue.

I've been trying extremely hard to get back into being able to read something for longer than 10 seconds, but it just hasn't worked out.

[slate field]

you don’t have to be interested, you just have to read. we all will have to do things we don’t enjoy in life, and you are in the fortunate position of choosing whether that’s reading books or whatever other hard thing

[frosty mica]

Massive walls of paragraphs have always been impossible for me, even for English classes though.

[slate field]

is it too hard, or just that you don’t want to?

[frosty mica]

It is legitimately too hard. I force myself to sit down and stare at the page, and then the words start floating off the page and I lose complete focus.

[slate field]

okay, maybe you should see a doctor about it then (not a joke, this might help you a lot)

[frosty mica]

It could be the most interesting story in the world and my brain knows it's interesting and actually wants to read it, but my brain cannot.

[slate field]

because you might be able to get the problem identified with that further

[frosty mica]

Doctors are very expensive though 😦 and I have no money.

A lot of my problems could be solved with a doctor.

I paid $55 for a covid test today just to be told I still have it 😢

[slate field]

you will not pass your classes if you can’t fix this problem, there is no way around it. does your university have a heath center? you absolutely have to have this problem sorted out before you take harder classes

[frosty mica]

I am not sure if my university has a health center.

Let me look.

It looks like they do. Not during the summer though.

[slate field]

okay. in any case, you do really have to get this sorted. i promise you that this is make-or-break for your academic success

[frosty mica]

I would love to get it sorted. It will be interesting to see what a doctor says.

[slate field]

i won’t say any more about it, i just want to make sure that you know that this is incredibly important for your success in the future, i would feel bad if i knew and didn’t say anything

because you’ll run into proofs that are literally paragraphs and pages long

[frosty mica]

I believe it is incredibly important for me as a person in general.

[slate field]

yeah, i think so too. books are really great (as you probably know). they’re sort of the cornerstone of a lot of conversations between my friends

[frosty mica]

I love books. I was heartbroken when I tried to read my favorites and I couldn't get past the first paragraphs.

[slate field]

i might try to sleep. i’m sorry that i can’t stay longer. good luck getting everything sorted out.

[potent igloo]

Zoomers have attention span issues thanks to smart phones, social media, and YouTube

[slate field]

that isn’t the issue here i don’t think

[frosty mica]

I'm technically a millenial by birthdate. I also didn't get my first smart phone until I was 20.

So if we assume there is a bijection from a nonempty set T1 onto a finite set T2.

That means that T1 and T2 must have the same cardinality?

What does that tell me though?

[potent igloo]

T2 is finite right? What does that say about its cardinality?

[frosty mica]

That it's countable?

That the elements of T2 are a subset of N?

[potent igloo]

That is cardinality is some n in N

[frosty mica]

Right.

Which means that the cardinality of T1 must also be some n in N?

[potent igloo]

Exactly

[frosty mica]

Which means it must also be finite?

[potent igloo]

Yes

[frosty mica]

So something like this?

[potent igloo]

I would explicitly mention that the cardinality of T1 must be the same n in N, which is why it is finite

[frosty mica]

Gotcha ok. Thank you.

.close

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[flint wasp]

Question 3, I did this...

[flint wasp]

,rcw

[tiny eagle]

the inductive step isn't to check the k=2 case, it is to assume the k=n case and then use that to show that k=n+1 also works

[flint wasp]

Okay, then how should I do it?

This is for my self, so your helping me specifically:p

Because I feel like this is painfully obvious, but idk how to prove it

[tiny eagle]

For this problem, if we assume the k=n case holds, then $\operatorname{span}(v_1,\ldots,v_n)=\operatorname{span}(w_1,\ldots,w_n)$. Using this, we need to show that $\operatorname{span}(v_1,\ldots,v_n,v_{n+1})=\operatorname{span}(w_1,\ldots,w_n,w_{n+1})$

[potent lotus]

evelyn

[tiny eagle]

Since these are sets, to show equality it is probably easiest to show that each is a subset of the other, which would then imply equality

does $\operatorname{span}(v_1,\ldots,v_n,v_{n+1})\subseteq\operatorname{span}(w_1,\ldots,w_n,w_{n+1})$ or $\operatorname{span}(v_1,\ldots,v_n,v_{n+1})\supseteq\operatorname{span}(w_1,\ldots,w_n,w_{n+1})$ seem easier to do first?

[potent lotus]

evelyn

[flint wasp]

Second once since it's just linear combinations

Of vi I'd assume

[tiny eagle]

yeah, there's really nothing to do there

so for the other direction, do you have any ideas what we'd do?

[flint wasp]

Il thinking. This is why I'm doing this again I'm not good at this stuff

Could I say that vi build the wi

So because it builds the wi, it's a subset?

[tiny eagle]

that's okay! it's often the case that the most intuitive things can be hard to prove rigorously.

[tiny eagle]

hmmm i don't think that suffices. we need to show that if we have some $v=a_1v_1+\ldots+a_nv_n+a_{n+1}v_{n+1}$, there exist coefficients $b_i$ such that $v=b_1w_1+\ldots+b_nw_n+b_{n+1}w_{n+1}$.

[potent lotus]

evelyn

[tiny eagle]

well, i guess that since the v_i's build the w_i's, you could make something work where you write the w_i's in terms of the v_i's (as they are defined), and then group coefficients. In other words, define each b_i in terms of the a_i's

[flint wasp]

Is this a linearly independent vs dependent argument?

[tiny eagle]

no, we just need to be able to write $v=a1v_1+\ldots+a_nv_n+a_{n+1}v_{n+1}$ as a linear combination of the $w_i$'s

and defining each $b_i$ in terms of the $a_i$'s will do that

[potent lotus]

evelyn

[tiny eagle]

i guess this isn't even really induction at this point

just replace all the n's with m's and you've got a direct proof

wait no

we do need induction (there is actually probably a way to do it directly, but I don't immediately see it)

[flint wasp]

Okay I'm trying to take this in.

I have to get my family ready for church.

I appreciate your help 🙂

[tiny eagle]

of course! If you work from this and keep the inductive hypothesis in mind, you should be able to work towards a solution

[flint wasp]

Thank you! Yeah I'm gonna fiddle with it. It seems right there 🙂

.closw

.close

[calm coral]

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[calm coral]

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[queen pendant]

anyone can help me out on the
Theorem (Corollary of the Intermediate Value Theorem)

[calm coral]

Please don't occupy multiple help channels.

[queen pendant]

idk if it translates this way

[stone flax]

Would help if you could send the exact snapshot where you are having the problem

[queen pendant]

i have all this

i can give the functions if needed

but basically was confused why we chose f(2) as our other interval

like i would have said the alpha we re serchine is betwwen -1 and + infitnity

[stone flax]

Am I missing some context?

[queen pendant]

i have to find f( alpha ) = 0

[stone flax]

Are you trying to find the roots

Okay, good and?

[queen pendant]

one solution

[stone flax]

Okay got it

[queen pendant]

i said it s fully upcreasing ı also said it s continued

[stone flax]

So can you see that the minimum value of f'(x) is at x = 1, f'(1) = 3?

[queen pendant]

yes

[stone flax]

So f'(x) has to be positive always

[queen pendant]

yes

[stone flax]

So f(x) will be increasing always

[queen pendant]

okay

[stone flax]

Now, you know that f(1) is negative, and f(2) is positive

So if you go from negative to positive

[queen pendant]

but why those specific values

[stone flax]

And f(x) is continuous (intuitively, you don't lift the pen while drawing the graph)

[queen pendant]

that s what ı don t get

[stone flax]

Basically you just want to check that f(x) goes from negative to positive

[queen pendant]

like why the 2

[stone flax]

You could have picked any values

You could have even seen that x goes to -infinity, then f(x) goes to negative infinity

And as x goes to +infinity, then f(x) goes to positive infinity

and since f(x) is continuous

Then by corollary of IVT, f(x) must have exactly one root

[queen pendant]

but can ı not just say this instead of using values

[stone flax]

Whats the question?

[queen pendant]

like lim f(x) on 0 = - infinity > alpha > + infitnity

[stone flax]

Does the question ask you to give a range for where the root could lie?

[queen pendant]

it s just prove that there is one unique solution

such as f( x ) = 0

[stone flax]

And what have you been told about f(x)

That figure you sent?

Is this the information given, or the solution?

[queen pendant]

solution but the apart from the 2 ur supposed to have everything

after the few questions

[stone flax]

Can you show the original question?

You could have chosen any 2 values

You just need that f(x) at one of the values if -ve and the other is +ve

So ideally you would choose 2 such values which are easiest to calculate f(x) at

[queen pendant]

but since the limit is + infinity and we have f(1) = -1 like isn t that enough

[stone flax]

Yup, intuitively it is, but the theorem asks you to pick real numbers

And infinity is not a real number

So to apply the theorem, you are better of using 2 easy to compute values

[queen pendant]

Really ? I learned that i can also just use limits

Is it wrong

[stone flax]

If you could show me the exact statement you are talking about, I can say

But the standard form of IVT uses real numbers

Not limits or "infinity"

[queen pendant]

this is the specific question

translated wait

it means show that the equation f(x) = 0 has one unique solution alpha on the interval

[stone flax]

And whats f(x)

[queen pendant]

?

[stone flax]

Yup

So the domain is from 0 to +infinity only

[queen pendant]

yes

[stone flax]

Where you want to check

So how about you just check 2 easy values?

The sign at them

f(1) and f(2) are pretty easy to compute right?

[queen pendant]

yeah but on the questions before they also asked it s limits

that s why ı thought it s easier to just concluse from the limits

[stone flax]

It is, but since you know the limit

You know that for a sufficiently large number

This functino would be positive

So why not just find that number?

IVT doesn't work by limits

It works by real number values

So to apply IVT particularly, you should pick 2 real values only

I don't see how is this such a big issue for you 😂

[queen pendant]

Okay what about the normal one

[stone flax]

Normal one what

[queen pendant]

Like for at least one solution

Do i also need to use real jumbers when i can?

[stone flax]

Yup you do

Its easier

Than finding the limits

To just evaluate f(.) at 1 and 2

[queen pendant]

Okii thanks

.close

[calm coral]

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[calm coral]

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[feral prawn]

Im not sure on where to begin

I think I need to find vector AD and compare its norm with that of vector AC and if |AC| is bigger than |AD| then C is to the right

<@&286206848099549185>

Im getting C is to the left left but I want someone to check

[calm coral]

@feral prawn Has your question been resolved?

[calm coral]

@feral prawn Has your question been resolved?

[glad parrot]

In fact for the french national maths exam(end highschool, where this question comes from) they assume that the IVT works with the limit and in this point they just want us to say that f monotonely increased, continuous, and the limit at zero is -inf and at +inf it's +inf therefore, there is exacly one solution to f(x) = 0 (since 0 is in -inf; +inf), in this question she's not asked to find the value of alpha. They don't require necesserly to use real values.

[glad parrot]

tu ne peux que utiliser le corollaire pour cette question, pas le simple théorème, car il demande une unique solution et pas au moins une, la différence se trouve dans les strictes variations (strictement décroissante ou strictement croissante)

[stone flax]

You can show that if what you are saying for limits is true, then IVT follows

It's pretty easy from the definition of limit

But the point is that definition of limit is generally not taught in high school

[glad parrot]

it's done one time for the example (with epsilon) and never again, they just learn the operations of the limits

[stone flax]

Yup exactly my point

Epsilon delta wasnt taught to us in high school at least

[calm coral]

@feral prawn Has your question been resolved?

[feral prawn]

.close

[calm coral]

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[novel token]

!help

[calm coral]

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

[novel token]

is this right

[vagrant oak]

oh its simpel

yes, its right

[novel token]

how do u get better at math

[vagrant oak]

in case of this question and similar, practise ig

[calm coral]

@novel token Has your question been resolved?

[calm coral]

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[manic grove]

Does the solid bounded really exist or does it just have an infinite volume?

I tried to draw it using graphing app

And got this, it doesn't have any boundaries for y axis

[calm coral]

@manic grove Has your question been resolved?

[calm coral]

@manic grove Has your question been resolved?

[calm coral]

@manic grove Has your question been resolved?

[worldly lodge]

bro

is it a question of triple integral or double integral?

[calm coral]

@manic grove Has your question been resolved?

[manic grove]

They asked for the volume, so triple i guess?

[calm coral]

@manic grove Has your question been resolved?

[calm coral]

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

Hi

[calm coral]

Please don't occupy multiple help channels.

[nocturne osprey]

I got left on my other channel so I picked another channel

I need help with this

[modern hemlock]

pic?

[tiny monolith]

hello

[modern hemlock]

which question

!status

[calm coral]

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

[nocturne osprey]

No idea what to do

[modern hemlock]

what 7

oh

ok

[nocturne osprey]

It shows a list of the things

[modern hemlock]

but which question

or all

[nocturne osprey]

All

If your ok with it

[modern hemlock]

ok

so for the first one

you know the law of exponents

[nocturne osprey]

A little

[modern hemlock]

$a^na^m=a^{n+m}$

[potent lotus]

Amby

[modern hemlock]

so you have 4 numbers

[nocturne osprey]

That’s a multiplication rule

[modern hemlock]

you can do it two by two

[nocturne osprey]

From what I know

[modern hemlock]

starting from the left

$2^62^3$ should equal what

[potent lotus]

Amby

[modern hemlock]

can you tell what a is

[nocturne osprey]

Umm

[modern hemlock]

in this rule

[nocturne osprey]

You want like what it equals?

[modern hemlock]

yep

[nocturne osprey]

Can I use a call

Calc

[modern hemlock]

best not

it can be done by hand

[nocturne osprey]

Ok

Would take a minute

[modern hemlock]

its fine

take your time

[nocturne osprey]

Ty

[modern hemlock]

just make sure you dont expand it

you can use the rule to combine them into one

[nocturne osprey]

2 to the power of 9?

[modern hemlock]

yes

perfect

$a^na^m=a^{n+m}\2^62^3\a=2, n=6, m=3$

[potent lotus]

Amby

[modern hemlock]

6+3 would be 9

perfect

so you have another 2 and 2^2

$a^na^m=a^{n+m}\2^92^22\a=?, n=?, m=?$

[potent lotus]

Amby

[nocturne osprey]

2 to the power of 12

[modern hemlock]

yep

perfect

thats your answer

[nocturne osprey]

I expanded it to 4096

[modern hemlock]

you dont need to expand it

[nocturne osprey]

Because it says to simplify and then evaluate

[modern hemlock]

i think you can jsut leave it as 2^12

cuz normally you woldnt be able to evaluate it by hand

[nocturne osprey]

Oh

So I erase the 4096?

[modern hemlock]

if its a noncalculator test

youwould leave it as 2^12

if it was calculator allowed

you probably could just ignore the rule and simplifcation and plug it in

[nocturne osprey]

it’s a calculator cpt

[modern hemlock]

to get the answer

oh

then put everything in the calculator

🤣

[nocturne osprey]

So evaluated would be 4096?

Yes I’m pretty sure I’m correct

[modern hemlock]

yep

👍

[nocturne osprey]

Wait on the top of the sheet it says to “explain the following rule in your own words”

[modern hemlock]

oh

try doing that

and ill help u if u need it

[nocturne osprey]

But idk if I should say “ if two exponents are multiplied it becomes a addition”

Ik it’s a multiplication rule tho

[modern hemlock]

so

when two numbers with the same base but different exponents are multiplied

what happens

[nocturne osprey]

They don’t get multiplied

The base stays the same

But the exponents are added

[modern hemlock]

yep

perfect

say that

[nocturne osprey]

Ok

1 sec

Ok

I wrote it down

I’m now on 11) b

[modern hemlock]

ok

b is very similar to a

can you try doing it yourself first?

[nocturne osprey]

Yes

I have it written down but correct me if I’m wrong

[modern hemlock]

yep

good

[nocturne osprey]

Ok

[modern hemlock]

but you shouldnt expand the 2^4

you can just leave it as 2^5

its more "simplified"

[nocturne osprey]

Ok

Wait my friend did this

Tell me if it’s correct

[modern hemlock]

looks all g

[nocturne osprey]

Ok I’ll copy it down

For this

It’s similar to the top one right

[modern hemlock]

basically

example is

$3^{3^{2}}\=3^{3*2}\=3^6\=729$

[potent lotus]

Amby

[nocturne osprey]

I don’t get

[calm coral]

@nocturne osprey Has your question been resolved?

[nocturne osprey]

Yes

Thanks @modern hemlock

[calm coral]

Channel closed

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[nocturne osprey]

.close

[calm coral]

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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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[brazen elbow]

Given the tetrahedron O.ABC with OA, OB, OC all perpendicular to each other. OA = OC = $a\sqrt3$, OB = $a$. Let M be the midpoint of BC, calculate the distance between AB and OM

[brazen elbow]

not sure how to approach this

[clear delta]

wait are those lines even parallel?

anyway idk just put the origin at O and add a coordinate system

bash out the algebra

annoying but should be doable

[brazen elbow]

which lines parallel?

[clear delta]

AB and OM

i guess they want minimum distance

[brazen elbow]

they arent parallel, but they are uh, skewed?

[clear delta]

ye

[brazen elbow]

not sure what to call it

[clear delta]

skew yes

[glacial nacelle]

this is the most coord bash heavy question ever

just assign O to be (0,0,0)

find M's coords

find the equ of AB

[brazen elbow]

hm, didnt think of making it a coord, good idea

[glacial nacelle]

and then just bash out distance using the formula

[brazen elbow]

thanks yall, imma do the rest

[glacial nacelle]

making it a coord is very easy because you can just make OA, OB, and OC to be the axes

[calm coral]

@brazen elbow Has your question been resolved?

[brazen elbow]

got it, asqrt15/5

thanks!

.close

[calm coral]

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[calm coral]

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[hasty creek]

To start this off, should I plug in -x everywhere since it's going to -infinity?

[winter elbow]

That works

Because
x——> -inf is equivalent to -x——> inf

[hasty creek]

oh i see

and then from there i can just divide everything by the highest power in the denominator right

thank you

.close

[calm coral]

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

$\phi_1 \lor \phi_2 \lor ... \lor \phi_n$

[potent lotus]

kytsu

[snow spire]

what is the unknown?

a strategy

what are the data?

i don't know

statements connected with OR relationship, disjunction

what is the condition?

What is the concept of condition in mathematical problems?

condition: for a disjunction to be true, one of its statements must be true

$\exists \phi[ \phi \in true ]$

[potent lotus]

kytsu

[slate field]

well, you can just go through each $\phi_i$ and see if any of them contain contradictions (like x and not x)

[potent lotus]

smay

[slate field]

and then if any of then don't have one, then you're done

the problem is called DNFSAT

it's in P, which you can verify by analyzing this algorithm,

[slate field]

you should ask questions that make sense instead of randomly guessing at a question. Something like "What does this statement mean" is way better than "what are the data" because the first question makes sense and we can help you, and the second is just you grasping at stuff, but actually it makes it harder to help, becuase it obscures what you don't know

[calm coral]

@snow spire Has your question been resolved?

[snow spire]

what is P?

[snow spire]

thank you! is this the only strategy we can adopt for disjunctions?

[slate field]

don't worry about it if you don't know what P is

[slate field]

well, what else would you want to do? I think it's just fine to do it this way, it's pretty quick and you can stop early if you find something that works for one of the phi_i

[snow spire]

i'm curious about different ways to approach a problem

[slate field]

okay sure, well you could just naively try all the possible combinations of the variables and seeing if any of them work

by that i mean if there are like x_1, x_2, .... , x_n, you can try the combination "x_1 = T, ... , x_n = T" and see if the formula works

and then say, set x_n to F and see if it works

and so on trying all the combinations

[snow spire]

we could use induction also

[slate field]

oh, are you seeing this in a computer science class or a math class

[snow spire]

but for disjunction we only need one statement to be true

[slate field]

I was assuming that you were seeing this in a computer science class lol sorry, you can ignore everything that I just said. Is this an intro to proofs class?

[snow spire]

this is from "introduction to mathematical thinking" https://www.coursera.org/learn/mathematical-thinking/home/week/1

[slate field]

oh i see, okay

[snow spire]

but i'm learning computer science also

which is an interesting perspective to this

[slate field]

yes, this is right. You can probably look for the easiest statement to prove and try proving that one

[snow spire]

learning LEAN, which might be relevant!

the point of this course is to develop an inventive way of thinking with mathematics, instead of procedures, for example, we should think in terms of relationships

[slate field]

this is indeed how you should think about mathematics

[snow spire]

how does this way of thinking apply to this problem, do you see?

how do you apply it to approaching this problem?

i'm still trying to grasp what that "mathematical thinking" is about

[slate field]

I think this is already a meta enough question. We can try something simpler. Suppose that I have two statements, S_1 and S_2, and I want to prove S_1 or S_2

[snow spire]

ok

[slate field]

then what do i have to do, well I'm trying to show that either S_1 is true, or else (if S_1 is false) that S_2 is true.

so all I'm trying to do is show that if i DON'T have S_1, then I can prove that S_2 must be true. this might be a weird way to think about it at first but a perspective like this might be useful.

we can make it more concrete

suppose that we want to show that "for a natural number n, then either n is even or n is odd"

then what we can do is suppose that n isn't even, and show that n must be odd (or the other way around, depending on if you thought that was easier)

[snow spire]

a = ["S1", "S2"]
for i in range(2):
  if a[i] = True:
    return True

[slate field]

well, yes, but sometimes you'll have statements whose truth values depend on other things, such as a natural number

so maybe sometimes S_1 will be false for some values, or maybe S_2 will be false for other values

but you could prove that S_1 or S_2 is always true

by supposing that our input is such that S_1 is false and showing that S_2 must be true

but that kind of thing needs human verification or reasoning

[snow spire]

because we suppose that the disjunction statement is true, so we can reason that one is true

[slate field]

we don't suppose that the disjunction is true

be careful what you're supposing and what you're trying to conclude

but if we suspect that the disjunction of two statements is true, then we don't suppose that it's true

that's sort of a fallacy that we don't want to commit

i might have worded things a bit weird, but that is the nature of talking about math without a concrete problem

it really is good to solve a problem, and then get general lessons from doing that problem later when you solve it

[snow spire]

thank you for your help @smay
my brain definitely contorts in different shapes at the moment 😌

[slate field]

but seriously i think this approach is gonna be better, and you'll develop the same sort of intuition that you're trying to get from that

so like the sort of meta question "what do you do when there is a conjunction" is not as helpful for you as "prove that for every integer n, either n is even or n is odd"

[snow spire]

do you mean making the problem more concrete, as we have seen in the natural number example, solve the concrete problem, and gain insight into the abstract ideas?

[slate field]

yes, and if you are ever stuck with how to do the first part of what you said, the key phrase is "this problem feels too abstract. Are there any concrete problems that exemplify this idea?"

becuase the question that you posted at the beginnning was VERY abstract

and sometimes you like to work in this sort of abstract setting, but for now you don't want that

since you're trying to learn about mathematical thinking

[snow spire]

okay, i see! but apparently some people learn, and enjoy / make sense of mathematics, more easily, approaching from the very abstract
perhaps it is practicing how to move levels of abstraction

[slate field]

the abstraction will come naturally, and the thing that you're trying to do is a different kind of abstraction than the one that mathematicians are doing

you're practicing metamathematical thinking in a way, or sort of strategy planning

but this won't be as helpful to you

as trying to solve problems or just prove simple statements

becuase you will get the planning part naturally

[snow spire]

nice

[slate field]

and you'll look back on this question in a year and say "wow this is easy now that I did all those proofs"

[snow spire]

it is similar to the idea of learning functional programming by doing Haskell, you gain insight by learning the language

[slate field]

yeah, that's right. do you know about functional programming?

[snow spire]

a little

[slate field]

okay, and you know this from doing haskell?

so the question that you posted was something like "how do i do this thing in functional programming" but it's way harder when you have no practice with any language or making anything with a functional language.

[snow spire]

i read the very first pages of Haskell Programming from First Principles Christopher Allen, Julie Moronuki

[slate field]

but you know haskell so what you would do is translate it into haskell in your mind, and then you'll understand what's going on in that framework

okay well suppose that you knew haskell

[snow spire]

that's where i read this idea

[slate field]

but yeah, find some concrete problems. What is your math education before trying this course out? any calculus, algebra, etc?

and since you're on coursera I am assuming you're trynig to teach this to yourself (good on you for wanting to learn more btw)

[snow spire]

thanks

[slate field]

specifically any linear algebra?

[snow spire]

i learnt math in high school, but abandoned it for many years, and then picked mathematics up because i started learning computer programming, and then i got addicted to it, and i am learning everything

[slate field]

okay, this is good. do you have any calculus or linear algebra under your belt? these are two, very concrete subjects that are extremely useful in computer science stuff

[snow spire]

and now i'm applying for a BSc course also

[slate field]

especially if you're gonna try to learn LEAN

[snow spire]

yes, some

[slate field]

linear algebra specifically?

[snow spire]

only the basics, i think

not yet 😄

i don't know

[slate field]

the reason that I mention is because linear algebra is GOOD for learning mathematical thinking

[snow spire]

ok

[slate field]

i am biased because I really like linear algebar

algebra

[snow spire]

i have the books and how to learn it

[slate field]

okay, that's good. Then you should try going through your book, and whenever you're given a fact in the book, you should ask yourself, 'do i know why this statement is true?"

[snow spire]

"the why is the what"

[slate field]

and if the answer is no, then usually it'll be the case that there is a proof somewhere online

but proofs are sort of central to mathematics, and by reading a lot of proofs you'll understand metamathematics stuff way more (like the question in your original picture)

okay i am gonna sleep now, wishing you well for your math journey.

[snow spire]

https://tenor.com/view/makasih-gif-64826954323601588

[calm coral]

@snow spire Has your question been resolved?

[calm coral]

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[calm coral]

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[novel token]

!help

[calm coral]

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

[novel token]

is mistake on step 1

isnt it 9x^2 + 12xy + 4y^2

i dont really trust myself

ok thx

WOOHOO

how can i trust myself more

in math?

[calm coral]

@novel token Has your question been resolved?

[calm coral]

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[timid canopy]

How is this possible?

[timid canopy]

It must be the second quadrant so how can tax also be positive?

[hasty fiber]

Tax is always positive

Nah that's indeed impossible

[karmic radish]

whoa what

[timid canopy]

lol so my textbook is just stupid

[karmic radish]

yeah something wrong with the question

[hasty fiber]

It happens

[timid canopy]

thanks

.close

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[calm coral]

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[sly needle]

Shouldn't subtract 50 from both sides and get -40 why is it +40

[tacit linden]

subtract 10 from both sides

+40

[vagrant oak]

they subtracted 10, 16x and added x^2

[tacit linden]

3x + 50 - 10 = 10 - 10 + 16x - x^2
3x + 40 = 16x - x^2

[sly needle]

oh right

I always get it wrong when the + - get flipped around

.close

[calm coral]

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[sly needle]

.reopen

[calm coral]

✅

[sly needle]

wait

why didn't we subtract 50 from both sides

wouldn't that be right also?

[tacit linden]

same thing

[sly needle]

that would be -40 instead of +40

[tacit linden]

3x + 50 - 50 = 10 - 50 + 16x - x^2
3x = -40 + 16x - x^2
3x + 40 = -40 + 40 + 16x - x^2
3x + 40 = 16x - x^2

you get to the same result

[sly needle]

got it

[calm coral]

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[cosmic vault]

I’ve been struggling with question 28 unsure what I could be missing, seems like the answer should always be 0

[clever ruin]

,rotate

[clever ruin]

for part a), you cross the x axis when y=0

So you need to solve $\sin(2t)=0$

[potent lotus]

Max

[clever ruin]

For part two just use[\int_{t_1}^{t_2} y\frac{dx}{dt}dt]

[potent lotus]

Max

[cosmic vault]

it seems like the answer to this should always be 0 though no for t?

[cosmic vault]

These are the answers in the booklet for 28

[cerulean cradle]

It shouldnt

sin2t is a periodic function and achieves 0 at multiple points

But you must see that At point A although y = 0 , x ≠ 0

Taking that in consideration

2t = pi [ sinx is 0 at every n(pi) where n is integer]
or t = pi/2

[calm coral]

@cosmic vault Has your question been resolved?

[cosmic vault]

ahhhh okay I see, that makes a lot of sense thank you

[calm coral]

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[magic olive]

hi

[magic olive]

need help to find all maximal solutioins

[clever ruin]

what have you tried?

[cerulean cradle]

This is a linear differential equation

The best approach would thus be to consider y/x = z and cont.

[magic olive]

rewriting it

into different forms

[cerulean cradle]

Try this

[magic olive]

will do

[calm coral]

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[carmine dawn]

Is it correct to apply set operations on n-tuples?

[carmine dawn]

My professor applied set theory operations on arrays (which represent n-tuples in programming)

[vagrant oak]

Did he do that in programming?

[carmine dawn]

Yes

[vagrant oak]

the standard set operations are obviously defined on sets

but i dont see an issue with extending them to n-tuples, although i dont exactly see how and its not done often in set theory i believe

did he do like array intersection array?

or what operations exactly

[carmine dawn]

Yeah it doesn't make much sense, but i guess she just wanted to practice doing that in programming

And it was a standalone thing, not involved in a bigger problem

[carmine dawn]

Yes

[vagrant oak]

the problem with intersections and unions on arrays is that arrays can have more than 1 equal elements and that they have a defined order

but as long as she gives a proper explanation of what she means, i think its fine

[carmine dawn]

Yes

[vagrant oak]

e.g. union could be defined as joining those 2 arrays

[carmine dawn]

I think the intention was to simulate set operations, even though internally it's still n-tuples

[vagrant oak]

ah

[carmine dawn]

She could have made that a little bit clearer, but no issue

[vagrant oak]

Also, it's just programming, it doesnt need to be so rigorous about where the set operations are defined and where they are not

[carmine dawn]

Yes

Thanks for the response. I'll be closing this room

.close

[calm coral]

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[calm coral]

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[young granite]

I have this language here;
L4 = {w1w2 | w1 != w2} over the alphabet {a,b} and I need to find if L4 is regular or not. When I used pumping lemma, I could not find a case where we leave the language so that means L4 is regular. But as you know pumping lemma cannot be used to prove a language is regular but rather when it is not regular (got points deducted because of it in an assignment so i know i need to do it some other way). Do you have any ideas on how to approach this? Or maybe it is not regular but I made mistakes?

[calm coral]

@young granite Has your question been resolved?

[young granite]

<@&286206848099549185>

[young granite]

if L4 is regular or not

i need to know how to prove that :)

lol <@&268886789983436800> sorry for tagging you guys i just thought @remote mural is trolling

[whole hinge]

don't apologize; thanks!

[young granite]

thank you 😁

[clear delta]

i don't think it's regular though

[whole hinge]

Yes, you are correct; the pumping lemma can't be used to show that a language is regular. The only way to show that a language is regular is to exhibit a regular expression/regular grammar/finite state automaton for it

However, if you believe that it is not regular, the pumping lemma is a good way to go!

[young granite]

I picked a few different words (s in pumping lemma) to see if i can get w1 and w2 to be equal but i guess it is hard to pick a good starting word for that, I assume it should be something like (ab)^p ab^(2p) and when i pump y up i should get them to be equal?

[whole hinge]

I would try something simpler first, like a word that only contains "a"s in it.

The more complicated your word is, the more cases you'll have to deal with in your proof.

[young granite]

yeah that makes sense

i also considered that but since we cannot know how many a's will be in y, we won't be sure if w1 and w2 have the same number of a's
for example if we were to choose s = a^3p such that w1 = a^p and b = a^2p, then we can say that xy is in w1 for sure since |xy| <= p. we can take n and m for x = a^n and y = a^m so we have z as a^p-n-m+2p

we can pump k down to 0 so we'll have a^p-n+2p = a^p-n but we dont know if p-n is even or not

[calm coral]

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[limber plume]

How can I find the value of ln 5 if ln 2 and ln 3 is given.

[limber plume]

What is the value of $\ln5$

[potent lotus]

Avirup

[limber plume]

Please help

<@&286206848099549185>

[whole hinge]

Hmmm are you supposed to calculate an exact value? Or just approximate?

It doesn't seem possible to calculate an exact value here

[limber plume]

In the question the value of ln3 and ln2 is given

[whole hinge]

Can you send a picture of the entire question

[clear delta]

yeah I don't think it's possible to build ln5 out of ln2 and ln3

[limber plume]

Given
ln3=1.1
ln2=0.7

[limber plume]

Why

[clear delta]

because 5 is prime, ultimately

,tex .log rules

[potent lotus]

ad nihilum

[limber plume]

So can I write ln5 = ln(2+3)

[cerulean cradle]

Ln 5 or ln 1.5 ?

[clear delta]

you can yes

that doesn't really help you much though

[limber plume]

ln 5

[limber plume]

Cant I just multiply ln with 2 and three in the brackets

[clear delta]

what are you proposing?

[limber plume]

ln(2+3)=ln2+ln3

[clear delta]

ln(2*3) = ln2 + ln3

[limber plume]

Oooo

So how can I get the value of ln5

[clear delta]

from a calculator I guess

i don't think it's possible to build it from the things you were given

[limber plume]

.close

[calm coral]

Channel closed

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[limber plume]

.reopen

[calm coral]

✅

[limber plume]

.reopen

@clear delta can you help me in this

[limber plume]

Please give me an approach

[blazing coyote]

change of base me thinks

[clear delta]

use change of base formula to turn everything into ln

[limber plume]

How can I do that

[clear delta]

,tex .log rules

[potent lotus]

ad nihilum

[clear delta]

base change rule

[limber plume]

Lemme try

[limber plume]

Am I right

[clear delta]

yep

[calm coral]

@limber plume Has your question been resolved?

[remote mural]

@limber plume there is a shortcut for this

[limber plume]

Oh Hi. What's the shortcut

[remote mural]

See the numbers are in a wave, 3 at the bottom and 4 at the top then 4 in the bottom and 6 at the top, then 6 at the bottom and 9 at the top

It's like a chain ,did you notice?

[remote mural]

The. The answer would have base as the left most base ( 3 here) and power as the rightmost power( here 9)

And log base 3 power 9 is 2

@limber plume , did you get it?

[calm coral]

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[limber plume]

I didn't get the img

[remote mural]

use the base change rule

oh you got the answer too

great

[remote mural]

it is basically base change, just that he knew that it is going to get cancelled out

[remote mural]

here also you notic at the last step
it is log3 ^9

[calm coral]

@limber plume Has your question been resolved?

[calm coral]

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[muted shoal]

is this wrong? i cant understand where the + sign came from

on the expression before last

[reef nimbus]

subtracting (6 - a_{n+1}), so it's subtracting a negative

[calm coral]

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[thin fossil]

Calculate the value of the expression

[thin fossil]

I don't know where to begin with this

I thought about starting with this

[pure kayak]

thatll work, sure

try write each side as a^(u) where u is some value