help-43

[drifting drum]

-b <= t-a <= b

[azure vault]

so now ? <= t <= ?

[drifting drum]

a - b <= t <= a + b

[azure vault]

but we know 5 <= t <= 13

so...

[drifting drum]

9 - 4 and 9 + 4

so a = 9, b = 4

[azure vault]

Yes

[compact pewter]

@drifting drum Has your question been resolved?

[compact pewter]

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[compact pewter]

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[wary mulch]

whyd I get this wrong

[trail cave]

7/2...?

[karmic stirrup]

is (6+8)/2 = 7/2

[wary mulch]

below the line

[trail cave]

is 6 + 8 equal to 7

[wary mulch]

idk how I got that

[trail cave]

how many hours did you sleep last night

[fair cradle]

6-7

[wary mulch]

like

3h

[trail cave]

i got 4 😎

[fair cradle]

[trail cave]

rehearsing for ur movie 🙄

[karmic stirrup]

so do you see what went wrong now?

any more questions?

[modest egret]

-# vroomvroom

[wary mulch]

yes

how do we do this

[karmic stirrup]

are we supposed to assume that distance AB : BX = 3 : 1

or... what?

[karmic stirrup]

is this the full thing?

[wary mulch]

thats one subset

this is only for the first q

[karmic stirrup]

there seems to be a ratio between the distances that we dont know.

yeah I'm not sure what to do

good luck!

[wary mulch]

thanks

[wary mulch]

@modest egret do u know what to do?

[umbral niche]

maybe we can use section formulae?

did you try that?

[wary mulch]

no i didnt

[umbral niche]

External

[modest egret]

i saw this a few moments ago

[wary mulch]

whats the bst approach to this

[umbral niche]

Yeah this is internal section formulae

[wary mulch]

but internal section formula woudnt let us calc this right

[umbral niche]

use external

it's the same formulae

Just change the sign to minus

[modest egret]

what are the four diagrams

[umbral niche]

He needs to match it

[modest egret]

oh you have match an option which looks like the diagram

[umbral niche]

I think it's an incomplete question? Else he needs to substitute each and everything to get the answer maybe

[wary mulch]

its like the midpoint except

its scaled

I think

affine combinations is when the scalars add up to 1

[modest egret]

oh

[umbral niche]

wait I got it

It's just simple option verification

[umbral niche]

so in the 3rd Diagram X is between A and B , so both the coefficients should be positive

[wary mulch]

I dont get it

[umbral niche]

Can u share the whole picture

[wary mulch]

there is some hint

[modest egret]

the first one is incorrect , you can see B is closer to X so
AX/XB <-1

the coefficients are 4/3 , -1/3 so you can deduce m and n from here

m-n = 3

[modest egret]

view hint

[wary mulch]

ohh

ik how to do this

but

were not given the coordinates

for the position vectors

[modest egret]

we are not but from options you can match ratios of AX/AB and BX/AB with options

[wary mulch]

like assume it?

[modest egret]

yeah if Ato X is big then assume its the bigger ratio

else we cant do this they just want to match the diagrams which likely represent the option

[wary mulch]

so no calculation here?

[modest egret]

no idts

[wary mulch]

I also need help with this

[upbeat gorge]

what does it mean for two vectors to be colinear?

[compact pewter]

@wary mulch Has your question been resolved?

[compact pewter]

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[wary mulch]

I need help with this

[wary mulch]

it means the points lie in the same line

[magic wren]

you could cheese the problem by computing 6 2x2 determinants

it's not terribly mathematically sound but it is a pretty funny idea

There are 6 ways to choose two of those vectors

If you choose two of them and put them as the columns of a determinant and set the third column to be 1 0 0 then you can use that determinant to figure out whether the two vectors lie on the same line

that works because just by visual inspection 1 0 0 is "almost guaranteed" to not affect linear independence

but of course in general that may not be true which is why this is a cheesy method

you can maybe even get away with computing less than 6 determinants

because for each determinant that you get 0 you know that the two vectors are collinear

so if you find two zeroes then you might already be able to figure out which vector is not collinear just because being collinear is a transitive, symmetric and reflexive property

And after you suspect which vector is not collinear with the others, you can try to find the proper justification

this method, however, is kind of stupid. So I propose a second method

The second method is that if Azure and Ivory are collinear, then Ivory gets transformed to Azure after you scale it by 240/255

[wary mulch]

why cant we use uhm the

[magic wren]

And then you can just check whether or not that is true

[wary mulch]

scalar multiple method

thats the one I was taught

[magic wren]

The scalar multiple method probably corresponds to what I called my second method

it is highly advisable to use that one instead

[wary mulch]

yes but I coudnt think of a way to apply it here

normally we are given two vectors

[wary mulch]

do we do for each combination of pairs?

its ok Ill come back to it

.close

[compact pewter]

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[compact pewter]

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[clever fog]

int of x^2/x-1 dx

[dense mason]

what have u tried

[clever fog]

i did long division

[dense mason]

integration by parts

[clever fog]

and got x + x/x-1

23

[dense mason]

dont think thats right

[clever fog]

i messed up the long division

i got it now

ty

.close

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[dense mason]

np

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[clever fog]

35

[winged lion]

What have you tried

[clever fog]

i thought it was arctan

but it isnt

so idk

[thick comet]

Try multiplying the numerator and denominator by e^x

[compact pewter]

@clever fog Has your question been resolved?

[compact pewter]

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[compact pewter]

Please don't occupy multiple help channels.

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as this action is irreversible, and this channel may abruptly lock.

[compact pewter]

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[tulip elk]

triangle ACB and DEB are not similar triangles since <EDB is 90 degrees while CAB is 120 degrees so I can't just use a scale factor...

DE = 16 cm ( as mentioned )
and DB = AB/2 = 24/2 = 12
Since DEB is a right triangle.. I used the Pythagoras Theorem
to get EB which is the root of 12^2 + 16^2..
so EB = 20.. idk what to do from then on

[safe cairn]

You know the cosine rule?

[tulip elk]

yes, but I don't think I have to use a trigonometric ratio since this is a 7th grade problem.. + it's a no-calculator exam

[toxic cipher]

cos 120 can be evaluated without a calculator

[tulip elk]

but it's a 7th grade problem...

[quiet nymph]

You can find the length of CB using the cosine rule BE using the pythagorean theorem. Finally you get:
CE = CB − BE

[toxic cipher]

he wants a solution that avoids cosine rule though

[odd birch]

Yeah I can’t think of any other solutions that don’t use trig. The figure kinda sucks lol and I don’t think there are any other special angles to exploit

[tulip elk]

damn...

[odd birch]

What have you learned in your class?

[tulip elk]

uh... math?

[odd birch]

Be more specific. Of course you learned math, but what math

Like, probability won’t help here lol

[tulip elk]

this is not a school-related problem but rather some math olympiad my mom dragged me into.. but.. ig probability... histograms. line graphs.. prisms and other secondary 1 lv stuff.. def not trigonometric ratios

[odd birch]

Oh.

[odd birch]

If it’s a math olympiad, then trig is fair play. 🙂

I had a feeling that trig was the only solution. Yeah, it’s going to be the expected solution

Here’s the law of cosines: $CB^2=CA^2+AB^2-2\cdot CA\cdot AB\cdot\cos(\angle CAB)$

I’ll give it to you since you don’t know it. It’s a generalized version of the Pythagorean Theorem

[boreal girder]

CST (reply ping for help)

[tulip elk]

Well.... I guess I'll just have to pray something like that doesn't show up on the exam day

.close

[compact pewter]

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[rotund sphinx]

.reopen

[compact pewter]

✅ Original question: #help-43 message

[rotund sphinx]

Extend AB past A to a point H such that CH is perpendicular to BH

and then it's standard 30-60-90 and pythagorean theorem

no cosine stuff required

[odd birch]

It’s not 30-60-90

[rotund sphinx]

triangle ACH is

[odd birch]

Because BDE is a scaled 3-4-5 triangle

[rotund sphinx]

ik

that's relevant as well

since CE=BC-BE

you use BDE to find BE

and BC comes from the aforementioned construction (||BCH is a right triangle||)

[tulip elk]

me go draw 👍

[rotund sphinx]

ok

[tulip elk]

like dis?

[rotund sphinx]

mhm

[tulip elk]

what do I do next..

[rotund sphinx]

I’m walking rn, but read what I said above

Preferably not all at once though

Because that is the entire solution

[odd birch]

PS I tried to do this and the side ratios are inconsistent when they should be consistent in principle
Thinking about it more, I think the figure is actually not geometrically possible

[tulip elk]

Gemini said the same thing too

[tulip elk]

welp... ig thx for the help

[odd birch]

I modeled it to scale on Geogebra

The point names may be different

But basically BA is the perp. bisector of CD

BA is 16 and CE is 12, both at 90° angles

But in the problem the angle is 120°, so it’s even further back

So it opens outwards and it doesn’t close

[tulip elk]

oh

[odd birch]

Now I’m not saying CSP’s approach is wrong. If the figure were actually geometrically possible, it should work

Nothing wrong with his idea in theory

It’s just that the problem sucks so whether you use Law of Cosines or similar triangles gives you different answers

[tulip elk]

the exam constructors had one job 💔
no way they messed up on an official exam paper

[tulip elk]

ah...

if that happens to me on the actual exam, im cooked 😭

Ima headout now, since it's 5:16 AM

.close

[compact pewter]

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[compact pewter]

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[worn wyvern]

Please anyone help me to do this

[worn wyvern]

<@&286206848099549185>

[eternal junco]

mmh

p and c

[loud sequoia]

math

[worn wyvern]

how to apply pnc here

answer is 6 here

pls anyone help to solve this SAMO Question

[eternal junco]

so <5000 and sum of the number is divisible by 9
and tens digit can only be from 1-9 and hundered's digit can be from 0-8

[fading flicker]

note all the properties first, a+b+c+d=9, 18, 27, but not 36 because 9999 has b=c, a__>__5, b=c-1, with that you can do diff setups for 9, 18, and 27,

[eternal junco]

mmh more accurate

[fading flicker]

oh ye that too

lowk wasn't thinking abt that

[worn wyvern]

I have tried this

but my answer is > 6

[fading flicker]

what does it mean by unique anyways

[worn wyvern]

can you pls check if the answer given is correct or not?

[fading flicker]

just by looking I'm pretty sure the answer is way more than 6

oh even

even number wait

[eternal junco]

oh forgot about even

lmao

[fading flicker]

d is either 0 2 4 6 8, I'm pretty sure you can just count this manually

[eternal junco]

that means last digit can only be 0 2 4 6 8

so

u can do like

[worn wyvern]

but my answer is >6

[eternal junco]

5 - - 0
try to 0 1 ; 1 2 ; 2 3 ; like this and see if which is divisible by 9

and then

i think its not hard

[fading flicker]

5 6 7 8 9, and 0 2 4 6 8
for which of these when subtracted from 9, 18, or 27 are odd(and less than 20) should get what you can get

[worn wyvern]

can you guys pls check if the answer is >6 or not??

[eternal junco]

mmh

[worn wyvern]

my answer is >6

[fading flicker]

5670
5562
5454
5346
5238
6120
7560(repeat)
9450
I don't think it's 6 unless it's asking for combinations w/ unique digits

I didn't enumerate all of them

[worn wyvern]

so answer is>6 right?

[eternal junco]

yeah its obviously more than 6 lmao

ans can't be 6

[eternal junco]

wait

i think i know why its 6

its asking for numbers when digits are different

then it will be 6

[fading flicker]

unique digits, yeah

[eternal junco]

mmh

[worn wyvern]

but unique locker combinations means different locker combinations right

[worn wyvern]

It didn't say different digits na

[compact pewter]

@worn wyvern Has your question been resolved?

[compact pewter]

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[worn wyvern]

Anyone pls solve this

[dense mason]

We won't solve it for you blud, we give hints 🤨

!status

[compact pewter]

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

[worn wyvern]

4

<@&286206848099549185>

[thorny urchin]

If you want your work checked, you'll need to post your work

also don't ping helpers before 15minutes, esp if there's already one here ready to assist you

[compact pewter]

@worn wyvern Has your question been resolved?

[paper rune]

@worn wyvern you need to post what you have tried so far if you want someone to help you.

[compact pewter]

@worn wyvern Has your question been resolved?

[compact pewter]

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[boreal girder]

Renato

[boreal girder]

Renato

[strange pendant]

can I get some help understanding example 3.1.5 and 3.1.6

[winged lion]

What would you like to understand?

[strange pendant]

the example entirely for 3.1.5 and 3.1.6

I want to understand the definition of convergence of sequences and both examples are using it

[winged lion]

Maybe start with the definition itself

[winter fern]

Are you familiar with limits by any chance

[compact pewter]

@strange pendant Has your question been resolved?

[winged lion]

People can only write the definition in other words, which may or may not cause you to understand something. But you need to ask a specific question fr

[strange pendant]

I am trying to understand the examples

I never asked for help with the definition

I mean I did, but like, I am more focused on the examples that use the definition

because, if I can understand the examples then I can understand the definition at least intuitively

[winged lion]

I don't think it works like that

The examples are built on the definition

[strange pendant]

I do understand the definition tho

at least somewhat

how is it being used in the examples can u help with that?

[winged lion]

You are given an epsilon from outside, and your job is to find a suitable n0(eps)

[strange pendant]

you mean n0

[strange pendant]

do you mind we go through the examples

[winged lion]

yes

[winged lion]

Ask your questions

[strange pendant]

I dont understand the examples

[winged lion]

You have (1/2)^n

[strange pendant]

that converges to what?

[winged lion]

What do you think it converges to?

[strange pendant]

n -> infty that converges to zero

[winged lion]

If you don't know that, forget eps/delta proof

[winged lion]

Good

So you are trying to find an n0(eps), so that when you assume n>=n0, you want to conclude |(1/2)^n-0|<eps

If you have an implication A => B, then you assume A and try to conclude B from that, but the issue here is we need to first prepare A, by finding n0

[strange pendant]

(1/2)^n = 1/(2^n)

|(1/2)^n-0| = |1/(2^n)| = 1/(2^n)

so what I want is that 1/(2^n) < e right

so 1/e < 2^n and so log(1/e) < n.log(2)

[winged lion]

Ok seems you are a bit skippy, but yes, you wanna do some scratch work

We are working backwards

[strange pendant]

then it follows -log(e)/log(2) < n

[strange pendant]

help me please

[winged lion]

What am I doing right now??

[strange pendant]

we might need to use arquimidean property, that forall x in R there exists some n in N such that n > x

n >= n0 <==> 2^n >= 2^(n0) <==> 1/(2^n) <= 1/(2^(n0))

[strange pendant]

|(1/2)^n-0| = |1/(2^n)| = 1/(2^n) <= 1/(2^(n0))

[winged lion]

I mean ya, if you are rigorous about it

[winged lion]

Yeah

I still don't see a question

Like you come

[winged lion]

say entirely but you already do know stuff

[strange pendant]

the hardest part is always finding n0 for the epsilon they give us

[winged lion]

Yeah

That's why we do scratch work

[strange pendant]

help

[winged lion]

Are you ragebaiting

[rich flower]

use archimedean principle to show that for all positive epsilon there exists natural n s.t 1/n < epsilon

[strange pendant]

what I want is that 1/(2^(n0)) < epsilon

so I can use transitivity

[winged lion]

Do you understand why tho

Intuitively

[rich flower]

is 2^(n0) not an natural number

[strange pendant]

right, my bad

[strange pendant]

I am trying to understand

this is why I am asking for help

I am trying to understand how to use the definition for convergence of a sequence on the example sequences

[winged lion]

You are literally using it rn

[fair cradle]

just sit and suffer

through suffering

we gain knowledge

[nimble needle]

If and only if we don't give up while suffering

[rich flower]

i can show you in vc @strange pendant

[fair cradle]

[strange pendant]

which vc?

[rich flower]

its trivial

[fair cradle]

https://discord.com/channels/268882317391429632/520310614988685312

[winged lion]

Cyrenux 🟨 when

[strange pendant]

wait, my pc is freezing

[rapid crescent]

At what point you are?

[strange pendant]

I am doing the first example

[rapid crescent]

have you already written something?

[strange pendant]

,rccw

[strange pendant]

@rapid crescent

[rapid crescent]

you have to show that 1/2^n <= epsilon

[strange pendant]

,rccw

[rapid crescent]

log(2^n0) is n0

[strange pendant]

what?

how?

[rapid crescent]

didn't you take the base 2 logarithm of both sides?

[strange pendant]

I think its just log_10

[winged lion]

log base 2 is lwk cooler

[rapid crescent]

so for n>n0 you can set n0 to approximately that quantity, in fact you have demonstrated that for every epsilon there exists an index n0 such that the definition is respected

[strange pendant]

dude

what

[rapid crescent]

you proved that the limit is 0

[strange pendant]

no i did not

[rapid crescent]

so what did you do?

[strange pendant]

I was trying to find an n0 such that 1/(2^(n0)) < epsilon

[rapid crescent]

yes

so for n>n0

for every epsilon >0

have you found an n0 such that for n>=n0 1/2^n <epsilon?

so that when all the epsilons vary, does this n0 exist such that the definition is respected?

[pulsar rose]

Gng what is the mistake here

I need help

Im bad at math

[compact pewter]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[compact pewter]

@strange pendant Has your question been resolved?

[rich flower]

@strange pendant vc?

[strange pendant]

why

[winged lion]

He wants to play mc with you

[rich flower]

you must prove

[strange pendant]

I cant I am doing anal homework

[winged lion]

Ok bro 💀

Might close your channel then and enjoy business

[strange pendant]

help please

[rich flower]

yes join

[strange pendant]

this is my progress

@rich flower

[winged lion]

log(1) 🥀

@rich flower

[strange pendant]

but dont play mc dude for the love of god

I cant play mc rn, I am doing anal

I should be done with this exercises this week, and I still have like 15 more exercises to go

and I am also taking like 3 other classes simultaneously so I cant play mc bro

[rich flower]

i was helping you while playing wdym, also im not playing rn

[winged lion]

bro you are not doing anal

[strange pendant]

calc*

[winged lion]

you are doing analysis

[mystic field]

What are you trying to do here?

[strange pendant]

im trying to proof this goddaym sequence converges to 0 as n -> infty

[mystic field]

What sequence?

[strange pendant]

1/(2^n)

[strange pendant]

i shee

[mystic field]

Yo you know the definition of convergence?

[strange pendant]

yes dude

I sent it a little while ago

basically forall n in N there exists some n0 in N such that

n >= n0 => |an - L| < epsilon

@mystic field @winged lion @rich flower

[mystic field]

Yeah, we know.

How about ration test?

ratio?

[strange pendant]

I need to use this def isaac

[strange pendant]

I cant 🥀

[mystic field]

You HAVE to?

[strange pendant]

ye

[mystic field]

|1/2^n|<epsilon

remove bars as 2^n is alway positive.

[strange pendant]

wait

wait

[mystic field]

Take the log base 2 of both sides.

[strange pendant]

,calc log(1)

[boreal girder]

Result:

0

[strange pendant]

log(1) is 0

[mystic field]

n>log_2(1/epsilon)

N>n

[strange pendant]

I get that n0 > -log(epsilon)/log(2)

[winged lion]

You can't use calculus in real analysis vro

[rich flower]

no vc?

[strange pendant]

n0 > -log(epsilon)/log(2)
n0 > log(1/epsilon)/log(2)
n0 > log(1/ epsilon - 2)

[mystic field]

So, what exactly is your problem?

Just pick an N such that n>N and formalise the proof.

Preferably ceil(log_2(1/epsilon))

To get 1/2^n < epsilon

as n approches +infinity,

[strange pendant]

ceil?

[mystic field]

2^n>n does too.

[mystic field]

Least integer greater

[mystic field]

So, there exists an N in Z such that 2^N > 1/epsilon

So, for any n in R,

there exists N such that n>N a

and 2^n>2^N>1/epsilon.

@strange pendant

[mystic field]

Reason being that if it were log_2(1/epsilon) the condition would still hold but N has to be a natural number so we set it to be the least integer greater than that for the condition to still hold.

@strange pendant

@strange pendant

Hello?

@strange pendant

[strange pendant]

yes

[mystic field]

Did you see what I sent you?

[mystic field]

n>N

[strange pendant]

n0 = N?

[pliant scroll]

blud really pinged OP 4 times in 8 minutes

[mystic field]

HAHAHAHAHAHAHAHAHAHAHAHAHA

[mystic field]

What?

[winged lion]

Karma

[mystic field]

Indeed, anything else?

MAYBE A MATH COMP PROBLEM OR TWO??????????

[compact pewter]

@strange pendant Has your question been resolved?

[compact pewter]

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[glossy oyster]

i know its not exactly math but i have some kind of competition where i need to write proofs can you give me some advice to study

[hushed magnet]

its not exactly maths
I have to write proofs
what do you think it is then

[mystic field]

What math competition? I'm ok at those.

[glossy oyster]

i mean its not a math question like if i ask u how much is 1+1

[glossy oyster]

olympiad dsmc

[mystic field]

Have you heard of AoPs?

[glossy oyster]

no

oh like the ytb channel

[compact pewter]

@glossy oyster Has your question been resolved?

[compact pewter]

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[worn wyvern]

,calc log(100)

[boreal girder]

Result:

4.6051701859881

[worn wyvern]

.close

[compact pewter]

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[rigid perch]

If all you need isto use a bot, use #bots

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[jaunty sky]

for this question i simplified the square root into $\sqrt{x^2 (x^2 + 2x)}$

[boreal girder]

g

[jaunty sky]

and then i simplified it further by x$\sqrt{x^2(1 + \frac{2}{x})}$

[boreal girder]

g

[jaunty sky]

am i allowed to do that

[acoustic stump]

yes

[jaunty sky]

i'm kinda confused on this operation because can't i just repeat this process and get infinitely many x^2 out of the square root?

and that just wouldn't make sense

[thick comet]

You really only need to take $x^2$ out, since then you have $x^2 - x^2(\text{something})$ in the denominator

[boreal girder]

Azyrashacorki

[thick comet]

It would probably help to multiply and divide by the conjugate in this case.

[jaunty sky]

yes, but i just feel like this just wouldn't make sense if i can just take out another x^2 out of 1 + 2/x, and make it something like 1/x^2 + 2/x^3

and repeat to take out infinitely many x^2

[thick comet]

Yes of course but the expression you would get wouldn't help.

[jaunty sky]

I know for this question i don't need to do that, but this is kinda confusing me

[thick comet]

The usual goal here is that you want to force a factor of the highest degree present in the denominator and numerator.

[blissful tangle]

One can alternatively also use the binomial expansion for (1+x)^n for small x

[thick comet]

The fact that you can do it doesn't mean that it's helpful is my point. There's nothing wrong with it but now you end up with a form that's even more indeterminate than it was

[jaunty sky]

i see

thank u very much

.close

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[modest egret]

hm..

t = 1/x
x-> infinity , t -> 0

this might work as well

[compact pewter]

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[round pier]

since it says leaf unit 0.01, is the first observatiuon 19.01, or 1.901?

[round pier]

my prof says 19 | 5 = 1.95 but is it 19.5 instead?

[compact pewter]

@round pier Has your question been resolved?

[compact pewter]

@round pier Has your question been resolved?

[rotund sphinx]

"A leaf unit of 0.01" means that the leaf digit represents the hundredths place. So to find the value, you combine the stem and the leaf and multiply the leaf unit (which is precisely what was done to obtain $195 \cdot 0.01=1.95$). As a sanity check, you have a 2 litre container, so an answer of $19.5$ is ridiculously large.

[boreal girder]

Civil Service Pigeon

[compact pewter]

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[compact pewter]

Please don't occupy multiple help channels.

[solemn wasp]

<@&268886789983436800> look at this 💀

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[terse mauve]

Like this?

[vivid breach]

what angle do you need

just googled one: https://www.geogebra.org/m/t76ctzgb

put a paper over the screen on the left then sketch out the net

np

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[dry wraith]

Good morning/good afternoon everyone, could someone help me with statistics? I'm in the 4th year of secondary school.

I wanted to understand how to draw the graphs.

[upbeat gorge]

try one of the channels in the Available category, like #help-14
also it would be helpful if you had a specific problem to work off

[compact pewter]

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[modern totem]

If I just compare the height and say that by observing since the equation has a linear part and a func part, I will try u= ln (tan(u^bar/2)) and then check for it(which turns right), would that be a solid proof, or should I do the more methodical method of equaling for r and clear the equation

[compact pewter]

@modern totem Has your question been resolved?

[rotund sphinx]

Seems fine to me

your "methodical method" is a way to find the substitution, but the crux of the proof is showing that the subsittution actually works

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[remote bane]

hi

[remote bane]

for this question, how can we use the curl test to dictate that the field is conservative because the region is not simply closed right?

so the second choice is iffy right

or is it a simply closed region

[upbeat gorge]

well the curl is only defined for three dimensions, not two, so you'll want a different test to see if it's conservative. (hint: directly related to the potential function)
moreover these tests dont depend on the region of integration being closed, all that's required is that the vector field is defined over an open, simply connected subset of R^2

[upbeat gorge]

you can sketch the region or desmos it to see if it's closed

[remote bane]

that's true actually with the 3d

but we can just assume this form <x, y, 0>

[tranquil prawn]

Remember that theres scalar curl

[remote bane]

i see so it's

i guess the region of the field

region of integration is for greens thorem

[remote bane]

what's that

[upbeat gorge]

i mean yeah, which gives you $\pdv{F_2}{x} = \pdv{F_1}{y}$ iff the field is conservative

[boreal girder]

haseeb ♥

[remote bane]

yes right

that's in the k component

[tranquil prawn]

For $F(x,y) = \langle f, g\rangle$, the scalar curl (equivalent to curl for flat 2d regions) is $\frac{\partial g}{\partial x} - \frac{\partial f}{\partial y}$

[remote bane]

i,j,k

[tranquil prawn]

Its just the divergence of the 90º degree rotated field.

So instead of outness you measure "rotativeness"

[upbeat gorge]

regardless of how you curl your fields, is F conservative?

[remote bane]

yeah it is then

if the field is defined on R^2

R^2 is simply connected

[upbeat gorge]

we'll consider it to be defined and smooth on R^2 since why wouldn't it be

then can you find a potential function for F?

[remote bane]

yeah

integrate with respect to x for the x component for the field

then with y for the y component

the constant you get out is a function of y and x respectively

[upbeat gorge]

and then are there any stipulations that forbid us from using the fundamental theorem of vector calculus?

[remote bane]

no

so you just integrate with the f(B) - f(A)

i mean

you just do f(B) - f(A)

with B being the point and A being the other point

thank you I just wanted to know about the curl test really

.solved

[compact pewter]

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[upbeat gorge]

yeah exactly :)

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[placid oak]

How would I do this?

[olive hazel]

the limits exist for all monomials

which is x^k

By linearity, they exist as polynomials.

[short ferry]

wait are you claiming that the lim 1/n sum x^k always exists?

[olive hazel]

No, The problem assumes that for this particular sequence

that the limit 1/n sum x^k exists

using that assumption, we prove the limit exists for every continuous f

so we can say monomial is there but not proved

I hope it helps :)

[short ferry]

Yeah, i get your point now

[olive hazel]

@placid oak you want anything more?

[placid oak]

ah hang on

let me read this

[placid oak]

ah we say it exists for all polynomials

[placid oak]

and then argue that because every continuous function can be approximated by a polynomial, the image continuous functions at those x values exist

[olive hazel]

yeah

[placid oak]

cool I will write a proof and get back if I have any troubles.

thanks

.close

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[strange pendant]

ok please can I get some help with this exercise
I kind of understand but at the same time I dont
the idea is for example, if a) is a definition of primality, then
a) holds for p <=> p is prime
so what I have been doing so far is that, I pick random values for p (both prime numbers and composite numbers) and try to see if a) holds when p is prime and if it doesnt hold if p is composite
after this, I get a general idea of whether a) is a possible definition or not (if for example I try many p primes that make a) be true aswell as if I try many composite p that make a) be false)
after this, I get a general idea of whether a) is a definition of primality, suppose every p i try if it is prime it makes a) be true and if p is composite then a) be false, after all this p I try then I would like to prove this a) is a definition of prime number, so I would like to prove
a) <=> p is prime
the left direction <= is simple, suppose p is prime then d | p means d is 1 or d is p, but since d < p then d = 1, so if p is prime a) holds
what I am struggling with is the right direction =>
that is,

a) holds for p => p is prime

same for the other letters from a to e

can I get some help with this? the exercise is asking that after seeing this predicates I see if its a definition of prime number > 1

[thick comet]

For the direction "a) holds for p => p is prime", it's easier to work with the contrapositive and prove "p is not prime => a) does not hold for p".

[strange pendant]

why

[thick comet]

I mean you can argue the other direction as well if you want, but it's just easier with the contrapositive.

[strange pendant]

why it is easier I ask

[strange pendant]

can you elaborate

[thick comet]

I just feel like it is. In any case if you were doing "a) holds for p => p is prime" directly, the easiest argument would be contradiction.

Because you want to show that p has no divisors but 1 and itself. Unless you want to test every number to see if it's a divisor less than p, it's easier to show that such a divisor can't exist.

So it would all boil down to the contrapositive anyway

[strange pendant]

but in the case of a) it is a def, so no counterexample exists correct?

@thick comet

how to prove it tho

[thick comet]

The contrapositive is an equivalent statement. It's not a counterexample.

"p is not prime => a) does not hold for p"

• What does it mean that p is not prime?
• Once you've defined it explicitly, then why can't it satisfy a)?

[strange pendant]

p => q and the contrapositive is not q => p

[thick comet]

The contrapositive is not q => not p.

[strange pendant]

i see, this is very useful

[strange pendant]

but I dont understand suppose P is p is prime and Q is a) holds then by countrapositibe logic if a) does not hold then p is not prime

@thick comet

[thick comet]

This should make sense. If you want to show that a) describes primes, and only primes, it suffices to show that :

• primes satisfy a) (i.e. p is prime => p satisfies a)) -> this means that a) describes primes
• composites don't satisfy a) (i.e. p is not prime => p does not satisfy a)). -> this means that a) describes only primes, no other numbers.

[thick comet]

This would be the contrapositive for the statement "p is prime => a) holds for p". You've already proven this statement directly.

[strange pendant]

ok help me with the left direction then

a) => p is prime

[thick comet]

Answer those

[strange pendant]

p is not prime => not a)

[strange pendant]

there exists some d | p such that d =/= 1 and d =/= p

[strange pendant]

the cardinality of the positive divisors of p is greater than 2

[strange pendant]

unsure

d < p and d | p forall d in N => d = 1

[thick comet]

Well use the first one. If there is some divisor of p which is neither 1 nor p, does a) hold?

[strange pendant]

F -> F

[thick comet]

Say this divisor is d.
Is d > 1? Does d | p? Is d=1?

[strange pendant]

i told you if p is composite then there exists some d in N such that 1 < d < p such that d | p

[thick comet]

Okay, so does a) hold or not?

[strange pendant]

it does hold yes

[thick comet]

Why?

[strange pendant]

F => F is true.

[thick comet]

Why is the premise false?

d > 1 and d | p

[strange pendant]

the d < p and d | p dude

[thick comet]

Okay sure. Is d < p? Does d | p?

[strange pendant]

right, mb so d < p and d | p holds so T => F is false as fk

[thick comet]

So then a) does not hold.

[strange pendant]

[strange pendant]

ye

what is your point though?

[thick comet]

...

You've shown that if p is not prime, then a) doesn't hold for p.

So you've shown that if a) holds for p, then p is prime.

[strange pendant]

p is prime <=> a)

i proved already
p is prime => a)

i want to prove a) => p is prime

[thick comet]

And you haave

[strange pendant]

that is
p is not prime => not a)

[strange pendant]

i see, so we just proved the left direction

we are done then

p = 2
d > 1 and d | 2 => d = 2 => 4 does not divide 2

p = 3

d > 1 and d | 3 => d = 3 => 9 does not divide 3

p = 4
d > 1 and d | 4 => d in {2,4}
d = 2 => 4 | 4
d = 4 => 16 does not divide 4

b) <=> p is prime

[thick comet]

You'll notice that for primes the argument is pretty much always the same.

You've only tried 1 composite though.

[strange pendant]

this shit is confusing though

[reef oasis]

try 6

[strange pendant]

right?

[thick comet]

We've already gone through this if you just try the first two composites you'll find it.

[strange pendant]

p = 5
d > 1 and d | 5 => d = 5 => 25 does not divide 5

[thick comet]

5 is not a composite.

You don't need to try every number. The fact that you've argued the same way for each prime so far should give you a hint that primes satisfy b).

The issue is composites.

[strange pendant]

p = 6
d > 1 and d | 6 => d in {2,3,6}
d = 2 => 4 does not divide 6
d = 3 => 9 does not divide 6
d = 6 => 36 does not divide 6

[strange pendant]

i dont understand anything

I want to prove that
p is prime <=> b) holds

[thick comet]

If 6 satisfies b), can b) be a definition of prime numbers?

[strange pendant]

for primes p, b) holds everytime

[strange pendant]

something related to the contrapositive

I cant figure it out

p is prime => b) holds is proved

[thick comet]

Well it's pretty clear that "p is prime => b) holds for p" is true yes.

[strange pendant]

suppose b) => p is prime
then p is not prime => b does not hold

but I found p = 6, such that b holds

T => F indicated contradiction

p = 2
d | 2 => d in {1,2}
gcd(1,2) = gcd(1,0) = 1
gcd(2,2) = gcd(2,0) = 2

p is prime <=> c) holds

p is prime then d in {1,p}
gcd(1,p) = gcd(1,0) = 1
gcd(p,p) = gcd(p,0) = p

@thick comet

suppose p is prime then c) does not hold

because T => F is contradiction

[thick comet]

Yes, in this case no prime satisfies c). You could have just stopped at your computation for p=2.

[strange pendant]

help with d)

@thick comet

[thick comet]

Have you tried some numbers? What have you decided on proving?

[strange pendant]

i dont know how to do it

d) is hard

@thick comet

[thick comet]

Well Euclid's lemma states the direction

If p is prime, then d) holds for p.

[strange pendant]

no

[thick comet]

Yes.

[strange pendant]

no

we didn't covered euclid lemma

we need to prove both directions from scratch

p is prime <=> d) holds

[thick comet]

That's fine I'm just stating that Euclid's lemma is that specific direction so you can prove it.

[strange pendant]

we CANT escape proving the rigjt direction

[thick comet]

I never said not to prove Euclid's lemma

[strange pendant]

ok

then yes.

[strange pendant]

can you help

[thick comet]

Can you write down in detail what you want to show?

[thick comet]

Not that.

You're showing that p is prime => d) holds.
What does this mean in words?

[strange pendant]

what

[thick comet]

.

[strange pendant]

p is prime then euclids lemma holds

[thick comet]

More detail. What does d) say about p?

[strange pendant]

p is prime then d holds

[thick comet]

What does "d) holds for p" mean?

[thick comet]

In words.

If p is prime, then if .... p must ....

[strange pendant]

if p is prime then forall a,b in N p divides the preduct of a and b and then p divides a or p divides b

@thick comet

p => q = not p or q

if p is prime then there exists some a,b in N such that p does not divide ab or p divides a or p divides b

[thick comet]

You don’t need the contrapositive here.

[strange pendant]

no

[thick comet]

The statement is that if p is prime and p divides a product ab, then p must divide a or b

[strange pendant]

is not the contrapositive...

[thick comet]

Yes nevermind. Although this formulation doesn’t clarify things.

[strange pendant]

so?

you forgot the forall a,b

no?

is not p divides a product ab, is that p divides every product ab with a,b in N

[thick comet]

It’s implicit

The point is that if you have a product ab and p divides it then p must divide a or b

So start with p being prime, assume you have some ab which p divides.

Try to argue that p must divide a or b.

[strange pendant]

how

[thick comet]

Do cases on whether p divides a or not.
If p divides a you’re done.
If p does not divide a, what can you say about gcd(a,p)?

[strange pendant]

who tf knows at this point

[thick comet]

This doesn’t help. Think about the case where p does not divide a and what it entails for gcd(a,p)

Problems in maths exercises aren’t made to be easy, they’re made to make you think.

You’re skipping the thinking part

[strange pendant]

gcd(a,p) divides both a and divides p

[strange pendant]

i am not skipping anything

is just that proving euclids lemma is hard

[thick comet]

In assuming that you’ll know what to do right away, you are skipping parts of the process

You have to try things.

What can you say about gcd(a,p) if p is prime and p does not divide a?

[strange pendant]

suppose euclid lemma is true

then the negated predicate should be false

maybe its easier to show that the negated is false

@thick comet

[strange pendant]

if p does not divide a then a does not have a factor p in his prime factorization, so p and a are coprime

.close

[compact pewter]

Channel closed

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[thick comet]

Use Bezout and think about it.

[strange pendant]

gcd(a,p) = 1 so

aX + pY = 1

has a solution and exists

.reopen

[compact pewter]

✅ Original question: #help-43 message

[strange pendant]

then what?

[thick comet]

Then you think. You want b in there on the RHS optimally.

[strange pendant]

wdym?

[thick comet]

Problems in maths exercises aren’t made to be easy, they’re made to make you think.

You’re skipping the thinking part

[strange pendant]

why

[strange pendant]

baX + bpY = b

[thick comet]

Now you think.

About what you want to show. About what you have assumed and/or have derived. About how those could help you show what you're meant to show.

[strange pendant]

(ba).X + (bp)Y = b

(ba).X + (bp)Y = 0 (mod p)

b = 0 (mod p) <=> p | b

[thick comet]

See? It was only 5 minutes and you found it.

[strange pendant]

I got help though

but can you believe this is my first uni class?

[thick comet]

Yes

[strange pendant]

shit is getting abstract real fast

[strange pendant]

i mean keeping track of many stuff is hard

[thick comet]

The point is that now you know you're able to come up with things on your own if you just think about them for more than 5 seconds.
This isn't calculus anymore. There isn't a nicely drawn path to solve any problem. Unless you learn how to try things and fail you're going to hit a wall very soon called exams. It takes time to solve exercises in mathematics, not because it's long to write a solution, but because it's long to come up with it in the first place.
Each time you ask for help without trying anything, you're missing out on an opportunity to learn.

[strange pendant]

you are making it seem like a hardship to just study maths

i try to not ask questions without trying stuff

but sometimes I dont know even how to start

[thick comet]

If you picked maths as a degree because you thought it would be easy and you'd solve everything in 5 minutes then you're in for a surprise.

[thick comet]

Maybe you're not realizing but you do that a lot!

[thick comet]

And that's why you try things. If it's been a while and you can't make progress, then you ask for a hint or a head-start and you try more things from there.

It doesn't mean that the hint will give you a clear way to the end, but it gives you some more spaghetti to throw on the wall until it sticks.

[strange pendant]

who would think dat?

[strange pendant]

only when I dont know what to try out

I have been days trying to figure out this simple exercise in definining primes but I think after hitting my head against the wall and the explanations u guys gave on the logic behins this I am finally starting to figure it out

like it seems easy but for me it isnt that simple, this is why I have been trying to do this over the week

[thick comet]

Just in this conversation, I've given you ways to go forwards like 7 times and for all of them your response was within 5 seconds and it wasn't "I'm going to think about what you just said for a few minutes and try things."

[strange pendant]

depends

sometimes I think bigger hints would get me less stuck so often

you said, use bezouts!

but also we needed to recall that p | ab and p | px

[thick comet]

I said "Use Bezout and think (...) About what you want to show. About what you have assumed and/or have derived. About how those could help you show what you're meant to show. "

This thinking process includes knowing that you assumed p|ab

[strange pendant]

yes but this is what I was saying out earlier

when you are deep in the casework that p does not divide a you also need to remember the assumptions from the premise, like p | ab and recall basic stuff like p | px forall x in Z

[thick comet]

But this is ubiquitous in maths

It's part of arguing something is true

[strange pendant]

I guess, but is hard, when I barely recalled the bezouts lemma by himself

I had to shuffle my memory for that def

[thick comet]

Again, that's part of the thinking process when you're given a hint.
You try to understand what it means and how it may help.

And case in point after a few minutes you made it through!

[strange pendant]

I am just saying that is not easy by any means and is ok to ask for help when you get stuck

[thick comet]

I didn't say that you shouldn't ask for help. I said that if you're given help you should take the time to think about what is being said before asking for more help.

Take the moment I first told you that for a) you may want to use the contrapositive to show the other direction of your proof for instance. That would have been a good time to think about what contrapositive means, about what it would look like in that case, about how you may argue then. If you've done those, have given it an honest attempt and can say where you're stuck, there's no issues asking for help.
Chances are just like with this last one, you would've ended up finding it yourself and learning more in the process.

And then the next time you would be proving something similar to this, you would think back to the time where you were stuck and used the contrapositive of the argument.

[strange pendant]

.close

[compact pewter]

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[thick comet]

Again this isn't a plead against you asking for help. It comes from a place of concern because you're missing out on so many learning opportunities by assuming stuff will come to you in seconds and giving up suuuuuuper fast. At the risk of repeating myself, it takes time to prove things in maths.

[strange pendant]

I have been at least trying for a week to figure out this exercise

who said I gave up on it super fast?

[thick comet]

You give up on trying super fast.

Not on the exercise

[strange pendant]

no it's not that, sometimes I can't just think of approaches for the exercise, then ask for hints close the channel and try it for myself

[thick comet]

Do of it what you will. I'm just telling you that's how it comes across when you ask for help.

[strange pendant]

this is my first year of university. I am still figuring things out, is not that I give up trying super fast, is just that I am not that creative to be frank

[thick comet]

And I totally understand that. First year uni is lots of new abstract concepts, and it involves getting familiar with types of problems that you've not seen before, building up intuition that you don't necessarily have.
My point is about how you approach getting help, which at the end of the day is hindering your progress.

[strange pendant]

There was some times when I asked for help and nobody helped, I still tried stuffs out and asked classmates and somehow stuff ends up clicking if you keep trying you will understand everything, is just that sometimes I dont even know how to start or am tired of trying things out and I ask for help here and without much context it might seem like I didn't do much, but is not like you guys can comprehend my thinking process just because I ask for questions here sometimes you know, I am also a real person in life

[compact pewter]

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[fallen needle]

can someone teach me how to multiply matrices? I keep getting confused on what to multiply and add

[paper rune]

do you have a question you were stuck on?

[dull moon]

khan academy

go to khan academy, learn the concept and then come back here.

with a doubt in a particular question

or libretext math also works

[fallen needle]

okay

[fallen needle]

not really just the concept

[paper rune]

oh then yes go look online and come back if you are stuck on a particular question and need help

[fallen needle]

alr

[slate sand]

You “dot” the ith row with the jth column for the i,j th entry

Where the dot product mean you multiply the corresponding entries and the add it all together

[compact pewter]

@fallen needle Has your question been resolved?

[dull moon]

https://math.libretexts.org/Bookshelves/Linear_Algebra/A_First_Course_in_Linear_Algebra_(Kuttler)/02%3A_Matrices/2.02%3A_Multiplication_of_Matrices

this link may be useful

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[compact pewter]

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[hard solar]

hi

[hard solar]

my teacher posted these notes and I’m not sure if the formula is right

and what is the right formula

and is the example she provided correct?

and if wrong how do I do it

[hard solar]

can you explain how I got this wrong then

[thorny urchin]

can you show your work

[compact pewter]

@hard solar Has your question been resolved?

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[hard solar]

I just used the teacher formula

[thorny urchin]

can you show your full work

ok as mentioned above teachers formula was wrong

[compact pewter]

@hard solar Has your question been resolved?

[compact pewter]

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[round mountain]

I tried workin on this with a previous tutor but they weren't able to make much progress.

[thick comet]

Are you familiar with geometric series?

I see you've used them on the previous problem

[round mountain]

Maybe? But I'm not sure if I'm familiar enough with them for it to matter. I think we were like, trying to find the derivative to get the pattern for the power series representation but I have zero clue how I'm going to find the second derivative or the pattern at all from something as complicated as this.

[thick comet]

The reason why I'm bringing it up is that you can rewrite the function so that it looks like a geometric series.

[round mountain]

then no, for these purposes

[thick comet]

Well since you've used it above I don't see why it wouldn't be used here. It's not particularly intricate.

And there won't be a nice pattern for the derivatives

[round mountain]

Well, I did work on that prior problem with another tutor

[thick comet]

The only thing you need to do really is to divide the numerator and denominator by 4.

Then you're left with $\frac{x^2}{4} \cdot \frac{1}{1+\frac{x^2}{4}}$. Can you see how you'd write this as a geometric series?

[boreal girder]

Azyrashacorki

[thick comet]

The second factor in particular, $\frac{1}{1+\frac{x^2}{4}}$

[boreal girder]

Azyrashacorki

[round mountain]

Yeah, that is part of the maclarian series x^n, which could work I guess.

hmm, but where would x^2/4 go?

[thick comet]

You essentially just need to write this part as $\frac{1}{1+\frac{x^2}{4}}=\frac{1}{1-r}$ for some $r$.

[boreal girder]

Azyrashacorki

[thick comet]

In particular, consider that $\frac{x^2}{4} = - (-\frac{x^2}{4})$

[boreal girder]

Azyrashacorki

[round mountain]

now you've lost me again.

[thick comet]

Are you aware that $\frac{1}{1-r} = \sum_{n=0}^\infty r^n$ for $|r|<1$?

[boreal girder]

Azyrashacorki

[round mountain]

yes! I do not know how to rewrite this part

[thick comet]

Well if you use the fact that $\frac{x^2}{4} = -\left(-\frac{x^2}{4}\right)$, then $\frac{1}{1+\frac{x^2}{4}} = \frac{1}{1-\left(-\frac{x^2}{4}\right)}$.

[boreal girder]

Azyrashacorki

[thick comet]

Can you see what r is now?

[round mountain]

-x^2/4?

[thick comet]

Yes

[thick comet]

So can you rewrite it as a series using this

[round mountain]

This it then for our power series representation?

(bottom left)

Doesn't feel right to me

but if it is, I won't argue

...I guess the silence is supposed to be my answer?

[thick comet]

Sorry it’s not loading nor letting me respond on my laptop

[round mountain]

ahh, sorry

I didn't know that was actually that big of a problem

now I feel like an asshole lol

[thick comet]

You just need to use this with r = -x^2/4

For the factor on the right

[thick comet]

No don’t worry about it hahah

[round mountain]

Yeah, I did. But wait, is it just sum (-x^2/4), or do I need x/n! and (x-0)^n?

We're going for a full maclarian, right?

[thick comet]

Yes exactly. No n! or factor on the right

[thick comet]

It gives the same thing as the MacLaurin series yes

[round mountain]

Alright, so our power series is sum (-x^2/4)^n?

[thick comet]

That's for the factor on the right, you can still multiply it by x^2/4

[round mountain]

oh

[thick comet]

To get the whole series for the function f

Which was this

[round mountain]

how do I multiply it then? Just, what, -x^4/8?

or just, since it's already beneath the power of n, just (-x^2/4)^n * x^2/4?

[thick comet]

The latter

[round mountain]

sounds good! I'm just never sure if I have the answer

[thick comet]

So you can like multiply it in and combine exponents and all to get your terms

Now to determine the radius of convergence, you had $r = -\frac{x^2}{4}$ so you can check what condition you need on $x$ so that $|r| < 1$.

[boreal girder]

Azyrashacorki

[round mountain]

I've actually found someone else in meatspace to work on this with. Will come back later when I need more help. thanks!

.close

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[fluid sparrow]

Hello, I was hoping to confirm my answers to the following two highlighted questions:

[fluid sparrow]

My answers are as follows:

7(c): 14x - 10.
7(d): the orthogonal basis is {1, x, x^2}.

[compact pewter]

@fluid sparrow Has your question been resolved?

[hushed magnet]

I can confirm d) but c) is hard to confirm without seeing the work

[oak juniper]

,w (p(0) + p(1) + p(2))/(3) * 1 + (p(0)(-1) + p(2))/(1 + 1) * (x-1) where p(x) = 3x^2 + x - 2

[boreal girder]

Results provided by WolframAlpha

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

Requested by 0lante

[oak juniper]

should be 7x - 3

[compact pewter]

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[near bluff]

<@&268886789983436800>

[compact pewter]

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[fluid sparrow]

Thank you again, it is indeed!

[compact pewter]

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[visual flint]

((y-1)/y)dy=dx

y-log(y)=logx+c

y-log(xc)=c

y(0)=1 so how do I apply it for C

[blissful tangle]

Integrating dx gives you only x

[visual flint]

Ohh right

y-log|y|=x+c

When x=0 y=1

so we get c=1

y-log|y|=x+1

[blissful tangle]

Yeah right

[visual flint]

How many solutions?

@blissful tangle

[blissful tangle]

How many solutions of what?

[blissful tangle]

Shouldn't it be a unique solution?

[compact pewter]

@visual flint Has your question been resolved?

[compact pewter]

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[visual flint]

Please tell me this concept

[blissful tangle]

It's from the existence and uniqueness theorems