help-38

[viscid flower]

There is a fun exercise, I don't know if you say easy, but you can do it

Maybe if you want to push yourself

show that Q is open in Q, but Q is not open in R

or, just convince yourself

you already believe probably that Q is open in Q, we just reasoned through that a set will be open in itself

but is Q open in R?

[vast viper]

No

Each point will have some irrational points

[viscid flower]

yea

we wiggle even a little bit around a rational

we can find some irrational

so Q open in Q

but Q not open in R

anyways

Are you still up to continue? we are halfway done

we need to do X is closed in X

i'll be around for another few hours, if you want to continue later

[vast viper]

It is close because each element we wiggle we can include the boundary

[viscid flower]

you said you felt good about limit points right

we build a sequence out of points in the space, and make it so that it converges

a set is closed if it contains all its limit points

So, lets construct a sequence in R out of r in R, such that the sequence converges to some p in R

[vast viper]

Yes

[viscid flower]

this has the same thing as before, by design, by construction, all of these limit points will be in R

maybe you think the same thing, is Q closed in Q

lets make a sequence of rationals, out of the first Q, such that it converges to some q in the first Q

obviously q will be in the second Q

but is Q closed in R? @vast viper can you argue yes or no?

[vast viper]

Q is not closed in R@viscid flower

[viscid flower]

whys that?

let me make some sequence qn such that it converges

what issue might happen

[vast viper]

Every point we pick from rational it will contain some irrational and it fails

[viscid flower]

well, we can make a sequence of rationals that converge to something irrational

so, if we take Q in Q, any limit point of the first Q has to be in the second Q

if we take Q in R, any limit point of the Q has to be in R

[viscid flower]

htis is a big problem

[viscid flower]

here it doesnt matter

[viscid flower]

we end up with it being obvious that Q is closed

how do you feel about that

this should be everything, i think

if you feel okay

You agreed that X should be open in X

and you seem to agree that X contains all its own limit points

so it must be closed, too

[vast viper]

I want to compare open and closed with an another definition

A is open of it is equals to interior points

A is closed if it is set+limit points

[viscid flower]

there are two definitions you can use, if you want

but you need to be comfortable with pushing symbols around

[vast viper]

Let's take rational numbers...so

[viscid flower]

there are sequential definitions and set based definitions

like, the idea of a making a ball around some point is a set based one

and the idea of creating a sequence out of points is the sequential one

[vast viper]

How I show that Q is open in Q

[viscid flower]

do you want to use sequences or sets

[vast viper]

Sets

[viscid flower]

so, you will take some arbitrary q in Q

hmm lets say

[vast viper]

Yeah

[viscid flower]

do you think its okay if we write Q1 is open in Q2

do you see what i mean

I know that Q1 = Q2

but maybe it is helpful to be clear which Q we mean

when we are talking about things

[vast viper]

Two points??

Q1 and Q2

[viscid flower]

no

I just want to make it easier to talk about, instead of me saying first Q, second Q

when we say Q in R its very clear

but when we say Q in Q it is not clear

if i say q in Q

which Q

do you see what I mean?

[vast viper]

I see

[viscid flower]

so lets show Q1 is open in Q2

we mean Q is open in Q

so what we want to do is show this

we want to take some q in Q1

and build a neighborhood around q using points from Q2

if all of those points are in Q1, then we are done, and Q1 is open

[vast viper]

Yes yes

Ohh it is just rationals

So we can't have around irrationals

Each q will contain Q2 nbd and it will be Q1

[viscid flower]

yea, so maybe you see now

there is not much to do, other than be careful what we mean

[vast viper]

Hmm interesting

[viscid flower]

it will always work out

as long as Q1 = Q2

[vast viper]

Yeah

[viscid flower]

or in general X1 is open in X2, if X1 = X2

there is not much to do

[vast viper]

And for close

Q1 is close in Q2

[viscid flower]

you have choices

[viscid flower]

you can use sequences, if you want to work directly

[vast viper]

Sure

[viscid flower]

you can also use the fact (maybe you have proved??) that the complement of a closed set is open

[viscid flower]

okay

so lets show Q1 is closed in Q2

[vast viper]

Compliment of Q will be phi

[viscid flower]

[vast viper]

And phi is closed?

[viscid flower]

phi is closed and open

[vast viper]

Ohhh yes???

[viscid flower]

so again not much to do lol

[solid kiln]

jan Niku

jan Niku

[vast viper]

Buttttt

one second

Q1=Q2 and both are close

I mean Q is close because its compliment phi is open

😇

[viscid flower]

its nice it seems like its coming together for you

maybe you see once you start to understand a piece of it, then it spreads outwards

[vast viper]

I am really sorry actually i am taking so much time

😕😕

[viscid flower]

you can think of a lot of these things in terms of the other

[viscid flower]

the day we covered this in my class, i stayed in the library with another student until almost midnight

[vast viper]

My mind enjoying confusing

[viscid flower]

so he could help me understand

[vast viper]

😭

[viscid flower]

I am paying a debt

[vast viper]

That's nice

Actually there is no one around me who can help me more than such

[viscid flower]

this is not for a class?

[vast viper]

No professor, no students only books/internet

[viscid flower]

oh

well you will like this community lol

[vast viper]

Yes

[viscid flower]

these people really like to help with analysis

[vast viper]

The discord and community is amazing

I couldn't believe such applications exists

[viscid flower]

this community is really something

How do you feel about the idea now

X is open in X

X is closed in X

[vast viper]

Is this particular university server?

[viscid flower]

no, its general

there are people from all over the world

they usually say only english because its hard to make sure stuff follows the rules

if its in some other language

but people speak a lot of languages here

[vast viper]

Ah..my English is quite messy

[viscid flower]

i am not one of them 😭 i only speak english

[vast viper]

So i will show that X is open and its compliment phi is open/close

And phi is open then its compliment is close

---><----

[vast viper]

USA?

[viscid flower]

yea

[viscid flower]

yea, so its pretty easy

everything spreads outward, if you understand one part of the argument

you need make sure you believe phi is open and close

and be able to argue that complement of open set is closed

but then you are done

not much to do

I don't want to keep you too long

did you have other confusions about this?

[vast viper]

No confusion

You gave me lots of things/learning

I am so much grateful you made it easy for me

Now i am ready to arguments and help my friends if they ask me about it😍

Do we have paid channels too?

Or just free

[viscid flower]

no it is all volunteers

no paid anything

[vast viper]

Amazing

[viscid flower]

there is a secret chat for people who help a lot

[vast viper]

Hats off

To the server creator

[viscid flower]

sometimes it is easier to get help there

but you cannot pay to get in, you just have to help a lot

[vast viper]

Are you in that chat...?

[viscid flower]

of course

[vast viper]

Wow☺️

So in that chat maybe high scholars help quickly?

[viscid flower]

You have maybe seen how smart the advanced members in this chat are

I am always so impressed

I am not very smart lol

you can almost always find someone who knows more about the thing than you

if you are patient

[vast viper]

To me You are...

Thank you very much

Can I ask lots of questions?

[viscid flower]

here? yea, there's almost always open channels

theres no limit how much you can ask

[vast viper]

I see

Nice to meet you

.close

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[untold phoenix]

Hi

[untold phoenix]

I need help with this question

[dull steppe]

proof by induction

[untold phoenix]

Im in calc 4 right now, and it's not supposed to be a "proofs" class,

But here we go,

I don't have any idea how to start

[untold phoenix]

I've never done proofs before I don't know what that means LOL

In class we've literally just gone over tensor notation and space curves

[dull steppe]

if it is true for n then it is true for n+1

[untold phoenix]

I did these ones before this one,

And I basically multiplied it all out, and then divided every answer by 3, and if my answer was divisible by 3 then N^3+2n was divisible by 3 was my proof

I don't know how that could help me here

[dull steppe]

so sub n=n+1 in this case and show that that is divisible by 6

[untold phoenix]

That's what I did on the ones above right?

[untold phoenix]

But how do I do that? Should I pick a random n?

[empty orchid]

this is a more modular arithmetic-esque approach.

[dull steppe]

no, instead of n, take n+1

[untold phoenix]

what does that mean 😭

[empty orchid]

like, its not a standard proof by induction.

since you have cases here that correspond to the divisibility of k.

so, you can either plug in 6k, 6k+1, 6k+2...6k+5, which would be tedious work, or manipulate the expression such that it becomes obvious that it is divisible by 6.

[dull steppe]

wait , or we can prove the divisibility of n^7-n to 2 and 3 and finish the proof

[empty orchid]

or, just prove that n^7-n is even, and that it is divisible by 3, and use the methods mentioned.

[untold phoenix]

Should I expand if all the way and then see if it’s divisible?

[untold phoenix]

Oh I see

[empty orchid]

what?

[untold phoenix]

Im so confused

Like this is my first time in my life encountering proofs like these

[empty orchid]

okay, just do this.

[dull steppe]

if n is odd
odd-odd=even
if n is even
even-even=even
n^7-n is divisible by 2

idk how to prove div by 3

[untold phoenix]

Yeah that's what I was thinking,

[sharp heart]

Well, even and odd are remainders after division by 2.

There are also remainders after division by 3.

Three of them, in fact: 0, 1, and 2.

[empty orchid]

you can also show that n(n^6-1)=n^7-n is even

why am i barred from reacting

do you see this

[untold phoenix]

but how do I use the fact that an even number is divisible by 2 to prove that it is divisble by 6

Also it says it has to be any positive integer n, so it could be odd?

[dull steppe]

but even then n^7-n is even

[empty orchid]

because to be divisible by 6, a number must be divisible by 2 and 3

[untold phoenix]

true

[sharp heart]

Well, your problem says: "You can use that fact that an integer k is divisible by m iff it is divisible by all the factors of m." m is 6 here and the factors are 2 and 3.

[untold phoenix]

So if its odd its divisible by 3 then right?

[empty orchid]

wait, do you guys see this

[untold phoenix]

And if its even its divisible by 2

[empty orchid]

like im reacting to chai's messages and its not letting me

[untold phoenix]

so N^7-n is divisible by 6?

[dull steppe]

not necessarily

[sharp heart]

No, 5 is odd but not divisible by 3.

[empty orchid]

no.

[untold phoenix]

true ok

uh

then what does showing its even help me do?

[sharp heart]

It shows that it's divisible by 2.

[empty orchid]

^

[sharp heart]

Now you need to show it's also divisible by 3. Then you're done.

[untold phoenix]

then I need to figure out how ot show its divisible by 3

[empty orchid]

bro please tell me if you see what i am seeing

[untold phoenix]

ah

[empty orchid]

i am literally reacting but it is not letting me

[untold phoenix]

I can see this

So whats a way I can show its divisible by 3?

Put in a number that's divisible by 3?

[empty orchid]

not that man

[empty orchid]

casework, but its gonna be tedious

[empty orchid]

no, the standard way is induction but since you haven't learned it use casework

[dull steppe]

To prove that ( n^7 - n ) is divisible by 3, we will use modular arithmetic.

---

Step 1: Rewrite the statement

We need to show that:
[
n^7 - n \equiv 0 \pmod{3}.
]
This means ( n^7 - n ) leaves a remainder of 0 when divided by 3.

---

Step 2: Consider cases for ( n \mod 3 )

When any integer ( n ) is divided by 3, the remainder can be 0, 1, or 2. Thus, we have three cases to consider:

1. ( n \equiv 0 \pmod{3} ),
2. ( n \equiv 1 \pmod{3} ),
3. ( n \equiv 2 \pmod{3} ).

---

Step 3: Analyze each case

1. Case 1: ( n \equiv 0 \pmod{3} )
   If ( n \equiv 0 ), then:
   [
   n^7 - n \equiv 0^7 - 0 \equiv 0 \pmod{3}.
   ]
   Clearly, ( n^7 - n ) is divisible by 3 in this case.

2. Case 2: ( n \equiv 1 \pmod{3} )
   If ( n \equiv 1 ), then:
   [
   n^7 - n \equiv 1^7 - 1 \equiv 1 - 1 \equiv 0 \pmod{3}.
   ]
   Again, ( n^7 - n ) is divisible by 3.

3. Case 3: ( n \equiv 2 \pmod{3} )
   If ( n \equiv 2 ), note that:
   [
   2^2 \equiv 1 \pmod{3}.
   ]
   Thus, ( 2^7 = (2^2)^3 \cdot 2 \equiv 1^3 \cdot 2 \equiv 2 \pmod{3} ). Therefore:
   [
   n^7 - n \equiv 2^7 - 2 \equiv 2 - 2 \equiv 0 \pmod{3}.
   ]
   Once again, ( n^7 - n ) is divisible by 3.

---

Step 4: Conclusion

In all cases (( n \equiv 0 ), ( n \equiv 1 ), ( n \equiv 2 ) modulo 3), we have shown that:
[
n^7 - n \equiv 0 \pmod{3}.
]
Thus, ( n^7 - n ) is divisible by 3 for all integers ( n ). (\boxed{\text{Proven.}})

[solid kiln]

Lone Wolf

LaTeX source sent via direct message.
```Compilation error:```! You can't use `macro parameter character #' in internal vertical mode.
l.53 #
      ## Step 1: Rewrite the statement
Sorry, but I'm not programmed to handle this case;
I'll just pretend that you didn't ask for it.
If you're in the wrong mode, you might be able to
return to the right one by typing `I}' or `I$' or `I\par'.```

[empty orchid]

huh?

[silver glade]

isn't that chat gpt

[empty orchid]

[empty orchid]

oh yeah that confirms it

[past widget]

!nogpt

[trim jolt]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[empty orchid]

!nogpt

[trim jolt]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[sharp heart]

Well, you can say that n is 3k + 0, or n is 3k + 1, or n is 3k + 2. That covers all integers based on the ks you choose.

[empty orchid]

bro i swear

[dull steppe]

ye

[empty orchid]

why is it not letting me put reactions on messages

[sharp heart]

Like if k = 0, you get 0, 1, and 2. If k = 1, you get 3, 4, and 5. And so forth.

[empty orchid]

can any one of yall pls tell me if u see me tryna put reactions

[dull steppe]

ya, we can

[empty orchid]

why is it doing that wtf

[marsh forum]

you've been blocked

[empty orchid]

bro what

[untold phoenix]

Okok, so I can say n = 3k + 0, and that would be 3k^7 - 3k if k = 1 then it is divisible by 3 and 2 and thus divisible by 6 so n^7 - n is divisible by 6?

[empty orchid]

why

[empty orchid]

no

[untold phoenix]

😦

[empty orchid]

you have to "pull a 3 out" if you want to prove it divisible by 3

[past widget]

Well, for divisibility by 2, you can write that n odd => n^7 - n is odd - odd = even Also, n even is trivial

[empty orchid]

i.e., factor it out

[past widget]

the problem was to show 3 | n^7 - n

[untold phoenix]

If 3k^7 - 3k is divisible by 3

but not 2

[past widget]

.

[untold phoenix]

unless k is 2

[empty orchid]

but this is the first case

[untold phoenix]

so then I could do 2k^7 - 2k is divisible by 2 and with a second case? And prove that it is true because those two are factors of 6? if k = 1?

oh oh

no

Id have to do

3k+1

right

[empty orchid]

no, the point is to prove divisibility by 2 and 3 seperately, then join them to say that it is divisible by 6, since those are the requirements to do so

[untold phoenix]

So case 1 can be 3k and case 2 can be 3k+1

The only thing is I have to expand 3k+1 to the power of 7

[sharp heart]

n^7 - n

3k + 0: (3k)^7 - (3k) = 3^7k^7 - 3k = 3 * 3^6k^7 - 3k = 3(3^6k^7 - k) = 3u + 0
3k + 1: (3k + 1)^7 - (3k + 1) = ....

[empty orchid]

(and a case 3 too)

[sharp heart]

Yeah, you'd have to expand 3k + 1.

Note the trick I used with 3k + 0.

[untold phoenix]

thats painful

oh so you did

3k, 3k+0 and 3k+1

[sharp heart]

I could factor out a 3 from the whole expression, so it became 3u + 0, where u was the part left after factoring out 3.

Starting with 3k + 1 should also lead to 3u + 0 as should starting with 3k + 2.

You can use the binomial theorem if you know it to simplify expansion.

Once you show they all lead to 3u + 0, they're all divisible by 3.

[sharp heart]

Well, 3k + 0, 3k + 1, and 3k + 2.

With this kind of thing, you have 0 through the number multiplied by k minus 1.

So, 0 to 2 here.

Modular arithmetic makes this much easier.

You can essentially work with remainders after division.

Like let's say you're working modulo 3. That means the divisor is 3.

So, 28 has a remainder of 1 when it's reduced modulo 3.

That's because 28 divided by 3 is 7 r 1 from elementary school division problems.

What modular arithmetic does is it allows you to do arithmetic using these remainders. You can add, subtract, and multiply with them. You can't necessarily divide.

You can also do exponents, but the exponents don't get reduced to a remainder.

So, like let's say you have 35 * 5 + 71 modulo 3.

Well, 35 reduces to 2. 5 reduces to 2. 71 reduces to 2.

So, you have 2 * 2 + 2, which is 6, which reduces to 0.

So, 35 * 5 + 71 has a remainder of 0 when divided by 3.

,calc (35 * 5 + 71) / 3

[solid kiln]

Result:

82

[sharp heart]

As you can see, it has no remainder.

With your problem, you have (3k + 0)^7 - (3k + 0). Well, 3 reduces to 0 modulo 3, so you have (0k + 0)^7 - (0k + 0) = 0, which reduces to 0.

So, you know that it's divisible by 3.

Similarly, you have (3k + 1)^7 - (3k + 1) = (0k + 1)^7 - (0k + 1) = 1 - 1 = 0.

And you have (3k + 2)^7 - (3k + 2) = (0k + 2)^7 - (0k + 2) = 2^7 - 2 = 128 - 2 = 2 - 2 = 0.

[untold phoenix]

Chai I dont know what a modulu is 😦

[sharp heart]

A modulus is the divisor.

[untold phoenix]

oh

[sharp heart]

So, let's say you have 58 modulo 3.

[untold phoenix]

thats 58/3?

[sharp heart]

You divide 58 by 3 and look at the remainder.

19 r 1

So, you can just look at the remainder after division by 3.

Which is 1.

So, instead of dealing with 58, you can deal with 1.

[past widget]

Uhh cristian +_+

what's the status?

!status

[trim jolt]

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

[untold phoenix]

okay interesting, so because I have 3k+1^7 - 3k+1 and Im doing modulus 3, then the 3's reduce and 1-1 is equal to 0 so case 2 is proven

and now I have 3k+2^7 - 3k+2 modulus 3 turns my 3s into 0s and I get 128-2, why does that turn into 2-2?

[untold phoenix]

uhh, almost there?

[sharp heart]

Because 128 / 3 = 42 r 2.

Any time you get a large number, you can reduce it.

Except for exponents.

[untold phoenix]

42 with remainder 2? And I look at the remainder 2 -2 = 0?

[sharp heart]

Yes, you replace 128 with the remainder 2.

[untold phoenix]

ohhh

Okay

so using modulus

I can prove this thing is divisible by 3

[sharp heart]

Yes.

[untold phoenix]

and using the even case I can prove its divisible by 2?

And then because its divisible by both voila its divisible by 6?

[sharp heart]

You use both even and odd cases to prove divisibility by 2.

2k + 0 and 2k + 1.

If you prove it's divisible by 2 with 2k + 0 and 2k + 1 and you prove it's divisible by 3 with 3k + 0, 3k + 1, and 3k + 2, then it's divisible by 6.

[past widget]

Alternatively,

So, n^7 - n = n(n^6 - 1)
Now, either n is of the form 3k and 3 | n => 3 | n(n^6 - 1)
Or, n² is of form 3k + 1
=> n^7 - n = n[(3k + 1)^3 - 1] = 3n(9k^3 + 9k^2 + 3k) which is divisible by 3 again

Therefore, 3 divides n^7 - n```

To show that 2 divides n^7 - n is trivial

[untold phoenix]

So I can do modulus 2 and modulus 3?

[sharp heart]

Yes, but make sure you do all 2 cases with modulo 2 and all 3 cases with modulo 3.

[untold phoenix]

ooo okok

[sharp heart]

Here's a better introduction: https://brilliant.org/wiki/modular-arithmetic/.

[untold phoenix]

Man Its so uncool this prof is making us do proofs when its not supposed to be a proofs class and we've never done proofs before LMAO

[empty orchid]

proofs are essential to math.

[sharp heart]

Well, calculus 4 is not too far from when you do courses that have lots of proofs to do.

It's good to have some practice going in.

[untold phoenix]

This is my last math class LOL

Im a physics major

[sharp heart]

Oh, OK.

[untold phoenix]

with 2k+1 I got 2^7 -2

Which is 128 - 2

and then I divided by 128 by 2 and got remainder 0

so now I have 0-2

Unless I can use modulu to get rid of that -2 as well?>

and then its 0 - 0?

[untold phoenix]

is that valid?

[sharp heart]

No, you just do normal arithmetic and then reduce big numbers by the modulus.

[untold phoenix]

ah ok

[sharp heart]

You would have 2 - 2.

Since 128 reduces to 2.

[untold phoenix]

so 128/2 64/2 32/2 16/2 8/2 4/2 -2?

[sharp heart]

Oh, I see, you switched moduluses.

[untold phoenix]

well dont I need to do modulus 2 and modulus 3?

[sharp heart]

Yes, I just didn't notice.

[untold phoenix]

LOL

[sharp heart]

So, 2k + 1 has you doing (2k + 1)^7 - (2k + 1), modulo 2.

Then 2 reduces to 0, so you can replace all the 2s with 0s.

(0k + 1)^7 - (0k + 1).

[untold phoenix]

yeah then my 2ks go to 0k and I get 1+1 on both sides because its n+1 on each one

and n = 2k + 1

so I end up with 2^7 and 2

[sharp heart]

No, you shouldn't.

You don't get 1 + 1.

[untold phoenix]

oh

Idk why I had n+1

[sharp heart]

You get 0k + 1, and 0 times k is 0.

[untold phoenix]

How's that?

[sharp heart]

There are two problems.

One is that you need parentheses on the second lines of each section and the 7 exponent.

(0 + 0)^7 - (0 + 0), for example.

Otherwise, you change the meaning of things.

Also, you need to divide by the modulus, and only once.

In the last one, you keep dividing by 2 instead of dividing by 3 once.

[untold phoenix]

Oh yeah my bad

I got carried away

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@untold phoenix Has your question been resolved?

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[lethal lodge]

i'm so stuck on b

[lethal lodge]

we have to do it geometrically not using the fact that modulus = squareroot(a^2 + b^2)

so what i did is plot it on the complex plane

and i made a rhombus

[safe loom]

never saw anyone use cis before

what is it

[lethal lodge]

cos(x) + isin(x)

or e^ix

[empty orchid]

shorthand notation, so people don't die writing it all out.

[lethal lodge]

this is what the markscheme has as the diagram

i managed to derive the same thing

and pulled out that triangle, turned it into a rhombus

but now my math is failing me or smth

[safe loom]

for the first one u can just forget the 2 and only focus on z^(-1)

[lethal lodge]

the first one isn't my issue

its the second

[empty orchid]

think about how 1-z would be defined in the cartesian world.

[lethal lodge]

its like

a vector

so first you go 1 unit to the right

then you go down to the left (z)

like diagonally

and the diagonal also has a length of 1

so you make a new point (1-z)

and connect the origin to that point

and that's the vector for 1-z

[trim jolt]

@lethal lodge Has your question been resolved?

[prisma bronze]

https://www.tiktok.com/@satoru_gojo42069123/video/7334386585930190123

[lethal lodge]

huh

<@&286206848099549185>

nvm got it

.close

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[vast viper]

how should I start integration?

[marsh forum]

Partial fractions

it will be messy

but it can work

[vast viper]

I see

[marsh forum]

,w partial fractions 1/(x^5+1)

nvm

[orchid wagon]

[marsh forum]

I was wrong

[marsh forum]

Don't use partial fractions

[vast viper]

Ohh why

[marsh forum]

have you seen the decomposition

,w partial fraction decomposition of 1/(x^5+1)

[solid kiln]

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[marsh forum]

if you want to integrate this be my guest

[safe loom]

Well you can still partial fraction it tho

[past widget]

why'd u want to use partial fraction

[safe loom]

Not that monstrous partial fraction lol

[past widget]

XD

[trim jolt]

@vast viper Has your question been resolved?

[safe loom]

Hint: $x^{5}+1=(x+1)(x^{2}+\frac{1}{\varphi}x+1)(x^{2}-{\varphi}x+1)$

[solid kiln]

TargetVN

[safe loom]

with phi is the golden ratio

[past widget]

https://www.youtube.com/watch?v=v4-ljCe8igY

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[analog pier]

Find the remainder.

[analog pier]

I suppose I have to use crt for 8^7^7 but im not sure how to find modulo for 3 and 5

[marble wharf]

you started with the problem (huge number) mod (something) and reduced it to (still huge number) mod (something else). why do you now want to switch strategy

[analog pier]

I want to write 8^7^7 as 30* something so I can find the remainder

since 8^30 modulo 31 is 1

and I thought about using crt on 8^7^7

but Im not sure how to go from here

[marble wharf]

so you have to compute 8^7^7 mod 30

[analog pier]

yes

[marble wharf]

previously you had to compute 8^8^7^7 mod 31

which is the exact same kind of problem

which you managed to reduce

so do the same thing to 8^7^7 mod 30

[analog pier]

I cant use crt on 7^7

since it's modulo 29

[marble wharf]

which theorem did you use to go from mod 31 to mod 30

[analog pier]

fermat

[marble wharf]

good

[analog pier]

little theorem

[marble wharf]

which works because 31 is prime

is there a non prime version of fermat

[analog pier]

i dont think so

yeah so going to 29 wouldnt have worked anyway

[marble wharf]

there is

eulers theorem

a^phi(n) = 1 mod n

if a and n are coprime

[analog pier]

what about phi

how do i find it

[marble wharf]

for example $\varphi(n) = n\prod_{p\mid n} \left(1-\frac1p\right)$

[solid kiln]

Denascite

[marble wharf]

where the product is over all prime factors of n

otherwise you could also just count the number of coprime numbers less than 30

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[analog pier]

thanks

.close

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[uneven oyster]

Please I need help with these questions. They are practice questions. I have a really hard time with differentiation and integration. Could you please work me through how to solve these?

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@uneven oyster Has your question been resolved?

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@uneven oyster Has your question been resolved?

[void berry]

Hi are you there ? I think I can help with some of the number 2 , 3 and 4

[uneven oyster]

Oh thanks

I am

[void berry]

Soo let's start with some of the easier one 2c

[uneven oyster]

K

[void berry]

I believe it's y =e^(x^2+1 ) right ?

The problem

[uneven oyster]

yh

[void berry]

So you could just use the chain rule to find it's derivative

[uneven oyster]

Alr

This is the chain rule right?

It's been a while since I've visited the topic

[void berry]

Yes

Or you could see it as

[uneven oyster]

I get but I'm more familiar with the first format we didn't use this much

[void berry]

Just use what you're familiar with it's basically the same just a different notation

[uneven oyster]

Also when you differentiate an exponential it stays the same right?

[void berry]

only for base e

[uneven oyster]

Ok

Am I on the right path?

[void berry]

Yes that is correct

[uneven oyster]

Final answer?

[void berry]

That's correct

[uneven oyster]

Alright

[void berry]

Next let's go to the number 3a

[uneven oyster]

OK I have no idea where to start there

Integration isn't my strong suit

[void berry]

I believe the function is x^2 e^(2x) ?

[uneven oyster]

Yes

3a right?

[void berry]

Yes

[uneven oyster]

It is

[void berry]

To do it just use integration by parts 2 times

[uneven oyster]

Not sure how to integrate by part

[void berry]

It's hard to explain it by words I think it's better for you to watch some yt video I'm gonna go search About it right now

[uneven oyster]

ok thanks I'm here

Also if you know any youtube channels that can explain this area of maths I would really appreciate

The other ones I find are either too basic or too complex

[void berry]

https://youtu.be/xFk9cZYhFrw?si=p-CY3jNMt9I_815o

Here I think this is pretty good

[uneven oyster]

Alright I'll check it out

[void berry]

And DI method is basically just a way to make the integration by parts more easy to think of but it's basically the same

[uneven oyster]

OK I think I understand

3b is the same right?

[void berry]

Wait really 😅 ?

[uneven oyster]

Like the method I'll use to solve it

[void berry]

Yes

[uneven oyster]

OK

[void berry]

4a too

[uneven oyster]

Noted

Should I reopen the channel while I watch the vids?

[void berry]

It's up to you

[uneven oyster]

OK I understand

[void berry]

And maybe some of the problems are included in the video as well so when you understand the main formula please try doing it first 😅 (I forgot to check)

[uneven oyster]

I will, thnx

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[uneven oyster]

.reopen

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✅

[wraith hinge]

@north forge

?

[north forge]

please ping the Moderator role, not individual mods

[wraith hinge]

👍

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[ember ivy]

How do derive lnx^2 ?

[gaunt wave]

Are you familiar with the chain rule?

[ember ivy]

somewhat

so first I derive x^2?

And then multiply it by lnx derived?

[gaunt wave]

Not quite, x^2 is the "inner" function. You first derive the outer function then multiply it by thederivative of the inner function

[ember ivy]

1/x is the derivation of the outer?

[gaunt wave]

almost

ln(f(x)) derived is 1/f(x) * ??

[ember ivy]

1/x^2

[gaunt wave]

yep, and the derivative of the inner function?

[ember ivy]

2x

[gaunt wave]

multiply them together?

[ember ivy]

2x/x^2

[gaunt wave]

simplify

[hybrid vine]

||Also you can use properties of logarithms first||

[ember ivy]

ln x^2 ----> 2 / x

is this correct?

[gaunt wave]

yes

[ember ivy]

Can you give me another test of chain rule? I struggle with this the most?

[gaunt wave]

https://tutorial.math.lamar.edu/problems/calci/chainrule.aspx

[ember ivy]

thank you

.solved

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[wraith hinge]

Determine the y-intercept of the line representing the following affine function: x = 5 f(x)= -11 x=6 f(x)=-13****

[trim jolt]

Please don't occupy multiple help channels.

[past widget]

!original

[trim jolt]

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.

@wraith hinge Has your question been resolved?

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[ocean quail]

the fact that the expansion has 3 terms instead of two makes me not know how to solve this

[ocean quail]

i thought of grouping 1+x and then 1/x but then we would get that T(k+1)=C5^k * (1+x)^(5-k)*x^(-k)

[clear cloud]

Imagine (a+b) but b = x+1/x

[magic eagle]

multinomial theorem 🗿

[clear cloud]

You can group stuff

[ocean quail]

so you wanna group them that way

[clear cloud]

It seems the most natural to me

x with x and constant with constant

[ocean quail]

but, wouldnt that mean that k=0?

if we group them your way

[ocean quail]

how does this help us

[teal vale]

Just make it the numerator and then the denominator and move on

you just need to know what is the last term

[ocean quail]

hm?

what is that even supposed to mean

@past widget pls help

[zinc ginkgo]

Did you try this yet

[clear cloud]

I don't get the quotient thing

[zinc ginkgo]

Yea that doesn't help much

I was asking Gustavo sorry if that wasn't clear

[limpid siren]

!NOPING

[trim jolt]

Please do not ping individual helpers unprompted.

[ocean quail]

i alreaedy asked earlier if this wouldnt mean that k=0

[clear cloud]

||Also some Laurent series at x = 0 but i don't think this is the purpose here||

[ocean quail]

because the b would have to be 1

[clear cloud]

It was clear chief dw

[zinc ginkgo]

No

[ocean quail]

ma bad

[zinc ginkgo]

Whatever k is

[ocean quail]

so how would you find k

[zinc ginkgo]

No?

a=1

[ocean quail]

it asked for the term independent of x

[zinc ginkgo]

b=x+1/x

[ocean quail]

yea

b^k

[zinc ginkgo]

Expand (a+b)^5

[ocean quail]

wait one second

[zinc ginkgo]

Without expanding b

[ocean quail]

cant we just write the formula for a random term in an expansion?

[ocean quail]

like (a+b)^5 => (the thing im replying to)?

[magic eagle]

🗿

lmao

[ocean quail]

what do you think the answer is?

lets see if that formula got u the right answer

[magic eagle]

ok gimmi 5 mins

[zinc ginkgo]

No?

[ocean quail]

why not

[zinc ginkgo]

x can pair with 1/x in many different ways

[zinc ginkgo]

Try it and see if you get the right answer

[ocean quail]

i told u

k=0

[zinc ginkgo]

x * (1/x) = 1

[ocean quail]

but its wrong

[ocean quail]

ok?

[zinc ginkgo]

Yea because you're missing terms that multiply to a constant

[zinc ginkgo]

.

[ocean quail]

so how do i do it?

[zinc ginkgo]

Do this

[ocean quail]

just expand this?

and then what?

[zinc ginkgo]

Find the constant term in each summand

[zinc ginkgo]

This would be the expanding b term

[magic eagle]

sheesh i got 50

[zinc ginkgo]

b^k rather

[limpid dawn]

how

[magic eagle]

nvm

is question asking the coefficient of the term independent of x?

[zinc ginkgo]

It's also not helpful to just give numbers without any explanations

[ocean quail]

good question

[ocean quail]

ill just show you the expansion

[zinc ginkgo]

Do this too

[magic eagle]

ye i got 51

using multinomial theorem

lemme try using binomial theorem

[ocean quail]

are the constants: 1, 5, 10, 10, 5, 1?

[zinc ginkgo]

No

[ocean quail]

how

what

[zinc ginkgo]

Read this again for the third time

[ocean quail]

whats wrong if i wrote it like this?

instead of the way u said

shouldnt matter

[zinc ginkgo]

It's not wrong

[ocean quail]

then?

[zinc ginkgo]

This is wrong though

[ocean quail]

what are the constants then?

[zinc ginkgo]

Find out one term at a time

Use binomial theorem again

[ocean quail]

????????

[zinc ginkgo]

https://www.mathsisfun.com/algebra/binomial-theorem.html

[ocean quail]

ffs

[zinc ginkgo]

I mean clearly the last term has power -5 so why are you thinking it contributes 1

[ocean quail]

cuz i thought that the constant term is the coeficient buddy

thats why

[zinc ginkgo]

Coefficient of the constant yes

[ocean quail]

this one exercise alone is taking way too much time

[zinc ginkgo]

Not coefficient of random powers of x

[ocean quail]

right right

good for you bud

[zinc ginkgo]

Why are you still talking

You haven't explained single thing beyond "multinomial theorem" which op clearly doesn't need to learn

[naive otter]

If you have a better solution, you can surely type it ! But riemmans is good enough

[zinc ginkgo]

Please go anywhere else

Jesus christ please leave

[ocean quail]

so whats the constant

[zinc ginkgo]

This shouldn't take 20 minutes

[ocean quail]

i started working on other exercises since noone wants to explain to me how to figure this exercise out

if anyone's got any explanations, go ahead

[naive otter]

(1+(x+1/x))⁵=
1 + 5(x+1/x) + 10(x+1/x)² + 10(x+1/x)³ + 5(x+1/x)⁴ + (x+1/x)⁵

[ocean quail]

you can write it like that too

i guess

[naive otter]

Can you do it now ?

[ocean quail]

the thing i had problems on is finding out the constant term or whatever

[naive otter]

I actually got an exam tmrw !! So struggling with it !

[naive otter]

(x+1/x) yk both terms have to be of the same power to get constant

So yk you need to check for even powers only

Like (x+1/x)² middle term and (x+1/x)⁴ middle term

Then you add

[ocean quail]

and then what

[naive otter]

That's the answer!

[ocean quail]

nah

i just dont fucking get it

too damn bad

[magic eagle]

Can i expl 🗿

[naive otter]

1+ 10×2c1 + 5×4c2

[ocean quail]

if you feel like it

[magic eagle]

(1+(x+1/x))⁵

now the rth term of this expansion wud be given as:

[ocean quail]

are you asking me?

[magic eagle]

im drawing it, since i cant use texit and stuff, gimmi a min

[zinc ginkgo]

Constant term is when the power on x is zero

1-1=0, 2-2 = 0

[zinc ginkgo]

Which generalizes this a bit

[past widget]

that's trivial tho

[ocean quail]

ah, right

my bad then

[past widget]

General term: $\frac{5!}{p!q!r!}x^{q-r}$ subject to condition: $p + q + r = 5$

[past widget]

constant term is q = r.

Constant term: $\frac{5!}{(5 - 2q)!q!q!}$ subject to condition: $p + 2q = 5$

[past widget]

sum over q = 0, 1, 2

Xor has a better sol?

@ocean quail . Do you know multinomial expansion?

[ocean quail]

no

[past widget]

or multinomial coefficient / theorem, anything with multi in it

Alr np

Do you know (x + 1/x) raised to odd never has a constant term?

[ocean quail]

nope

[ocean quail]

why not

[naive otter]

I didnt use multinomial

[naive otter]

.

[past widget]

because, for (x + 1/x)^{2n + 1}, for any general term, the sum of powers of x and 1/x should be odd

[naive otter]

Sry i was just using the space here to copy paste somewhere else lol

[past widget]

so both powers cannot be equal, and hence, (x + 1/x)^{2n+1} does not have a constant term

[ocean quail]

sure

ive never heard about this before

ever

in my life

[past widget]

so in the expansion of [1 + (x + 1/x)]^5, you're only interested in 5(x + 1/x)^4 + 10(x + 1/x)^2 + 1

pick out constant terms from both binomials and you have:

Constant term = 5 x 6 + 10 x 2 + 1 = 51

[ocean quail]

whered you get the 6 from

the thing yall miss is that ive never done this thing before

[naive otter]

4c2 = 6

[past widget]

pascal's triangle

[naive otter]

(x+1/x)⁴ the middle term. Will be 4c2 x²×(1/x)²

[past widget]

(x + y)^4 has coefficients in pattern: 1 4 6 4 1

(x + y)^5 has 1 5 10 10 5 1

(x + y)^2 has 1 2 1

[naive otter]

I thought you would cont with (x+y)⁶ 😕

[past widget]

1 6 15 20 15 6 1

so anyways, I used pascal's triangle to go `5 x [middle term coeff = 6] + ...

[zinc ginkgo]

It's just binomial theorem

[zinc ginkgo]

If you bothered to expand any term here like I said you would have seen the constant terms

[ocean quail]

.close

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[leaden hawk]

how do i show this is differentiable [f(x, y)=\ln (1+x+y)]

[solid kiln]

dghf

[red mountain]

for all x,y?

[leaden hawk]

yes

[dense breach]

it's not defined on all x,y

[leaden hawk]

like which x and y?

[clear cloud]

1+x+y > 0

[leaden hawk]

it's defined when 1+x+y > 0?

[clear cloud]

Y

[leaden hawk]

so how do i show it's differentiable?

[dense breach]

differentiate it

[leaden hawk]

∂f/∂x and ∂f/∂y?

[dense breach]

hmm

like

ln() is differentiable

[leaden hawk]

∂f/∂x = 1/(1+x+y) and same for ∂f/∂y = 1/(1+x+y)

[dense breach]

1+x+y is differentiable

it's a composition of differentiable functions

that automatically makes it differentiable

technically just showing ∂f/∂x and ∂f/∂y exist isn't enough, you need all directional derivatives to exist

[leaden hawk]

do i need to find the lim?

[dense breach]

you're not allowed to use that compositions of differentiable functions are differentiable?

[leaden hawk]

this is how they show differentiability

in point (a, b)

how did they do it

[dense breach]

i assume that's some definition of differentiability they have

[leaden hawk]

[\lim_{\left( h,k \right) \rightarrow \left( 0,0 \right)} \frac{f\left( 1+h,2+k \right) -f\left( 1,2 \right) -hf_{x}\left( 1,2 \right) -kf_{y}\left( 1,2 \right)}{\sqrt{h^{2}+k^{2}}}]

[solid kiln]

dghf

[leaden hawk]

this is how i would do it?

[
\lim_{\left( h, k \right) \to \left( 0, 0 \right)} \frac{\ln(4 + h + k) - \ln(4) - h \cdot \frac{1}{4} - k \cdot \frac{1}{4}}{\sqrt{h^2 + k^2}}.
]

[
\ln(4 + h + k) - \ln(4) \approx \frac{h + k}{4}.
\implies \frac{h + k}{4} - \frac{h}{4} - \frac{k}{4} = 0. ]

[solid kiln]

dghf

[keen panther]

how do u even type this shit🛐

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[barren crescent]

Ok I think I fixed it

[barren crescent]

disregard the min max stuff on the long sc, it should be correct in the separate one

.close

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[vocal mist]

A cone of height H with a base of radius r is cut by a plane parallel to and h units above the base where
h < H.
Find the volume of the solid (frustum of a cone) below the plane.

[vocal mist]

i tried using disk method and i've ended up with pih/3(r^2+r(r2)+(r2)^2)

where r2 is the upper radius

but i need help getting r2 in terms of r H and h

and also i tried doing big cone minus little cone and that didn't work

also i tried plugging (H-h)r/h in for r2 and that didn't work

<@&286206848099549185>

please im cooked 😭

[fleet bear]

$$ R(y) = \frac{r}{h} (H - y) $$
Low key should have started with the full equation

[solid kiln]

StrangeQuarkAL

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[wet nest]

Can someone explain to me what a field is? My professor covered but neither the textbook or my personal book covers it formally

[wet nest]

I will also accept any resources or videos

[zinc ginkgo]

Field means different things in different contexts

What book or subject are you learning

[wet nest]

Real Analysis, its the thing where you do operation on a set I think

like +: FxF -> F

[zinc ginkgo]

https://math.libretexts.org/Bookshelves/Mathematical_Logic_and_Proof/An_Introduction_to_Proof_via_Inquiry-Based_Learning_(Ernst)/05%3A_New_Page/5.01%3A_New_Page

9 axioms in 5.1

[ionic pendant]

a field is a set of numbers equipped with operations of addition and multiplication that behave "nicely" (with the precise definition supplied in the axioms posted above)

[wet nest]

Thats what I got out of the lecture, it just seems a bit weird

obviously I know what a set is its just the operation part that throws me off

[zinc ginkgo]

The addition or multiplication?

[ionic pendant]

if you're at all familiar with linear algebra, it's very much similar to the vector space axioms

[wet nest]

both? I just dont get what the point of it is

$+ : F \times F \to F$

[solid kiln]

TheKingPin

[zinc ginkgo]

It's not very deep

[wet nest]

the notation is just a bit sudden for me. From my previous knowledge this is an ordered pair that goes into the set of F through addition?

[zinc ginkgo]

If you know things like 1+1=2 then you know addition

And things like 2*3=6 then you know multiplication

[ionic pendant]

so that notation means + is a function which takes two elements as input (the two numbers you're adding) and returns one number as output (their sum)

since F x F is the set of pairs of elements of your field (pairs of numbers)

[wet nest]

oh, that is very simple.

[zinc ginkgo]

The point of this notation is to set foundation for generalizing to more complicated binary operations

https://math.libretexts.org/Courses/Mount_Royal_University/Higher_Arithmetic/1%3A__Binary_operations/1.1%3A_Binary_operations

[wet nest]

thank you for the resources

so whats the point of it?

[zinc ginkgo]

Point of what

[wet nest]

obviously im using it in RA but why use fields?

[ionic pendant]

the utility of introducing fields is that if there is any theorem about the real numbers relying only on the field axioms, then the same property holds for any other field

[wet nest]

it just seems like a function

like for the one i wrote just say f(x) := 2x

oh wait no

cause FxF would be any combination from the set of F

{f(x,y) := (x+y) | x,y is in F}

[zinc ginkgo]

to do all the common operations you need in real analysis

addition, multiplication

[wet nest]

ya, what I mean is that it seems there simpler way to express this

[zinc ginkgo]

sure, try to do real analysis without a field

say (Z, +)

[ionic pendant]

you'd have to introduce formal definitions of addition and multiplication even if you only plan to use the real numbers. using the field axioms allows you to generalize certain results to other number systems

[wet nest]

makes some more sense

ill read through the resources and come back if I have more questions

thank you both

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[glad pumice]

How do I calculate DG/EF? I know the measurements of the other side lengths but idk how to calculate DG/EF

[glad pumice]

So far, I know

Area of triangle DAE= 24

DE=10

Area of AEFGD is in the form 24+10x, where x is the measurement of DG/EF

CD= 6 and EB = 8, which are both equal to the diagonal of the square when you reflect the two triangles

I applied the formula 1/2d^2, where d is the diagonal for the area of a square

I got area of triangle CGD = 9

I got area of triangle EFB = 16

[exotic pine]

AEFGD=ABC-EBF-DGC?

[glad pumice]

Yeah, that was what i tried and got it wrong

[exotic pine]

how did you get this?

[glad pumice]

1/2 (1/2*6^2)

you get a square when you reflect triangle CGD

and the hypotenuse (CD) is the diagonal of that square

we know CD is 6

[violet gust]

no you dont

[exotic pine]

you get a rectangle, not a square

[glad pumice]

how do you know that

[violet gust]

Youd get a square only if CGD were isosceles right triangle

But its similar to ABC, and ABC is not

[exotic pine]

if its a square, then it would imply CD=CG (if you work it out its not)

[glad pumice]

then idk how to solve the problem then

[exotic pine]

do you know simmilarity?

[glad pumice]

yes, triangle CGD is proportional to triangle CAB

[glad pumice]

it's like proportional triangles, right

[exotic pine]

mhm

[glad pumice]

how would we know if side length GD is an integer or not

[exotic pine]

its not

[glad pumice]

yeah cause no triangle has hypotenuse 6 if it's legs are integers

we know CB is 20 too

[exotic pine]

mhm

[glad pumice]

CG+FB=10

yea idk

[glad pumice]

how can i calculate GD

[exotic pine]

simmilarity on hypotenuse and base

[glad pumice]

base of triangle ABC , right

nvm

[glad pumice]

proportion is 16/6=8/3

[exotic pine]

no

[glad pumice]

AB=16, CD= 6

[exotic pine]

hypotenuse/hypotenuse=base/base

[glad pumice]

so area of the figure is 72?

yeah it is

thanks for helping

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[inland bear]

can someone help me? im just confused on how they figured out the values of b for these parts. eg, for part A how they found 1.2

[dull temple]

black is x^2
blue is the triangle that they found the area of

@inland bear

[inland bear]

yeah but how did they find the value of b? for example in part A, i see that they did 4-1.2 but where did they get 1.2 from

[dull temple]

oh wtf

ok that doesn't make sense to me hold on

[empty orchid]

me neither.

base should be 4 units.

[dull temple]

wait none of these make sense

are you sure this is the right problem lol

[tardy aurora]

mif 👀

[empty orchid]

[ionic pendant]

,tikz
\begin{axis}[axis lines = center, xlabel=$x$, ylabel=$y$]
\addplot[black]{x^2};
\draw[ultra thick, blue] (1.2,0) -- (4,0) -- (4,16) -- (1.2,0);
\end{axis}

[solid kiln]

cloud

[dull temple]

nice plot

[ionic pendant]

it seems like they estimated a point that would sort of fit

[empty orchid]

this is real coding 😭

[ionic pendant]

gotta add pgfplots to the preamble

[dull temple]

you do it pls i want to see what the √x one is about

[ionic pendant]

,tikz
\begin{axis}[axis lines = center, xlabel=$x$, ylabel=$y$, xmin=-1]
\addplot[black, samples at={0,0.001,0.005,0.01,...,0.2,0.3,0.4,...,5}]{sqrt(x)};
\addplot[ultra thick, blue] coordinates {(0,0) (3,0) ( 3,2.2) (0,0)};
\end{axis}

[dull temple]

ok i have no idea

[ionic pendant]

no that's just my plot being broken

[solid kiln]

cloud

[inland bear]

so i'd just have to draw the graph as accurate as possible then?

[ionic pendant]

i suppose so

i'm not entirely sure how they came up with these triangles myself

[inland bear]

alright, thanks

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[vast geyser]

Question