help-26

[thorny flame]

jan Niku

[neon iron]

okay thanks

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[vernal vale]

👀 hope that helped 🙏

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[lost charm]

i need help with congruent or supplementary math

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[woeful jungle]

The day before two days after the ay before tomorrow is monday. What day is it today?

[woeful jungle]

i also need help how to justify the answer with a formula. My answer is saturday but others say its sunday

[charred rampart]

should be sunday

check again

[woeful jungle]

how is it so?

[charred rampart]

the day before tomorrow is js today

[woeful jungle]

oohh i see, oke tyy

is this right if i change the T into today, rather than tomorrow?

[dreamy drum]

,rotate

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[mossy agate]

I think my solution isn’t correct so what do u guys think of this ? Am I missing something?

[crystal goblet]

$du=\frac{sin(t/2)}{2}dt$, $du=sin(t/2)dt$ is incorrect

[thorny flame]

smidgin

[crystal goblet]

(chain rule)

[mossy agate]

oh i see

then multply by 2 right ?

[crystal goblet]

Yep

[mossy agate]

so i just need to add to my solotion 2 up there right ?

[crystal goblet]

Not add

Multiply

[crystal goblet]

.

[mossy agate]

sry i meant like this hold on

Is it fully correct now ?

[crystal goblet]

Yep

[mossy agate]

thx a lot

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[elfin seal]

Show that the number overline(352) in base b cannot be a perfect square whatever the number base b, b > 5, b is natural

[elfin seal]

I dont even know how to begin...

[gusty bane]

what is overline 352

[bitter hemlock]

well if 352 were in base 10 you can write it as 3 * 10^2 + 5 * 10 + 2 right?

[elfin seal]

like

[bitter hemlock]

what if it's in base b, how would you write that sum?

[elfin seal]

the number 352

[elfin seal]

generally 3* b^2 + 5*b + 2

[bitter hemlock]

yep, now that's a quadratic we can factor

[elfin seal]

is it uh (3b+2)(b+1)

[bitter hemlock]

yeah, now presumably we can hopefully draw some conclusion from that 😛

[elfin seal]

hopefully 😭

[bitter hemlock]

i guess one thing i see is that either b + 1 factor itself is a perfect square OR it has to be a factor of 3b+2

but 3b + 3 is divisible by (b+1) and 3b + 2 is only 1 apart from 3b + 3 so b + 1 cannot be a factor of 3b + 2?

so then if you can show b+1 being a perfect square implies 3b+2 isn't a square that would be enough

i'm not very good at this kinda stuff though so maybe this is completely the wrong way to think of it and theres something easier

[elfin seal]

Hm.

I was thinking we can prove that uh

(3b+2)(b+1) is a multiple of 3+2 or M4+2 or M4+3 or M5+2 or M6+2

therefore they aren't perfect squares

because we didnt use the information that b > 5

idk

[bitter hemlock]

i think that is just necessary because for b <= 5 we wouldn't even be able to use the digit 5 in the given number

[elfin seal]

Oh, right...

[bitter hemlock]

so if b + 1 is a pefect square
b + 1 = c^2
3b + 3 = 3c^2
3b + 2 = 3c^2 - 1

i dunno does that help 😅

<@&286206848099549185> anyone with an idea of how to prove that
b + 1 being a perfect square implies 3b + 2 isn't a perfect square?

[neon iron]

Hint: it doesnt matter that b+1 is a square

[elfin seal]

if nothing really matters

how can i finally prove

[neon iron]

||modulo 3||

[bitter hemlock]

can you elaborate?

[neon iron]

But b+1 doesnt necessarily have to divide 3b+2 or be a square itself

[bitter hemlock]

^ just interested in this question, we saw already that if b + 1 isn't a perfect square then the (b+1)(3b+2) number can't be a perfect square

[neon iron]

So this is wrong

[neon iron]

No that is wrong

You are saying that if xy is a perfect square then x is a perfect square or x divides y

But for example 144=8*18 contradicts your statement

[bitter hemlock]

yeah you're right, i guess the two numbers can share factors

so what i was saying is bogus

[neon iron]

Im gonna try to prove now that (3b+2)(b+1) can never be a square

[elfin seal]

alright

[bitter hemlock]

i'm also having a hunch that b+ 1 and 3b+2 might always be relatively prime

[neon iron]

Oh yeah ofcourse thats the solution

Nice!

[bitter hemlock]

eh?

but how to show that? : )

[elfin seal]

uh like

may the common divisor of them be d

and prove that d is 1

i forgot how to solve those

[neon iron]

Suppose d divides b+1 and 3b+2, then d divides 3(b+1)=3b+3 . So d divides 3b+3-(3b+2)=1

[crystal goblet]

||Euclid division algorithm||

[neon iron]

So d=1

[elfin seal]

yes, exactly

[bitter hemlock]

ok then we're back to my previous argument! 😄

the only way this could happen now is if both b+1 and 3b+2 are both perfect squares

[neon iron]

Yes

But obviously 3b+2 is never a square

[bitter hemlock]

no?

[neon iron]

And we are done

[bitter hemlock]

is it more obvious that i think or something?

[neon iron]

Do you know modular arithmetic?

[bitter hemlock]

its been like 15+ years since i took an abstract algebra class

most of my math has been collecting dust in my brain

[elfin seal]

3b+2 is M3+2

which cannot be a perfect square

as i said above

[bitter hemlock]

M?

[elfin seal]

multiple of 3

+2

[bitter hemlock]

those can't be perfect squares for some reason? 🤔

[elfin seal]

Yes.

[elfin seal]

as i said here

[neon iron]

Proof:
Every number is of the form 3k, 3k+1 or 3k+2.
(3k)^2=9k^2=3(3k^2)
(3k+1)^2=9k^2+6k+1=3(3k^2+2k)+1
(3k+2)^2=9k*2+12k+4=3(3k^2+4k+1)+1

So squares are of the form 3k and 3k+1

But never of the form 3k+2

[bitter hemlock]

sorry just went over my head i guess!

[neon iron]

Therefore 3b+2 cant be a square

[elfin seal]

Indeed!

[bitter hemlock]

cool, well we got there at the end with the proof : )

good one

[elfin seal]

thank you so much guys

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[solemn mountain]

I honestly have no idea where to start with this. This is a worksheet to prepare myself for a quiz tmrw. And would like guidance on how to figure out what to do! Thank you.

[woeful stone]

So the first thing would be figuring out what you don't understand

[solemn mountain]

So when I read it, it says ivt on the interval [3,5]

I look at the graph and I see theres lines going thru [3,5] but not points?

I don't know if I can still use that because I see a open circle at that point (-2,2)

but i'm just super lost tbh

[woeful stone]

a line consist of points

The [3, 5] is a set

[solemn mountain]

ik for it to be ivt it has to be continuous right

[woeful stone]

It refers to alll the values of x such that 3<=x<=5

[solemn mountain]

so the first one would be continuous because on the graph theres points that are between there correct

or i mean it would be an ivt

[woeful stone]

yes it would satisfy the ivt property

[solemn mountain]

that would mean the second one wouldn't

because there isn't a line connecting them

and the third one would

so then im stumped on question 4

[solemn mountain]

how would i go about the 4th one

[-2,3] ivt k = 1

[woeful stone]

You just find the x-values such that f(x) = 1

[solemn mountain]

also wanted to say thank you for the help with 1-3 because i had the answers written I just had no idea if I was right or not about them

[solemn mountain]

im still a bit confused by that

[woeful stone]

So x=-1 satisfy the equation. There is also another value of x between 1 and 2 that satisfies the equation but exact value is not clear

[solemn mountain]

i'm not sure how you got that answer thought thats what im trying to figure out

[woeful stone]

You just look at the graph

[solemn mountain]

we have [-2,3] -2<=1<=3?

[woeful stone]

[-2, 3], -2 <= x <=3

[-2, 3] is the domain

[solemn mountain]

im looking at the graph I see where -2 is i don't see where 3 is

sorry if I seem like so lost rn because i am

i'm very confused

[woeful stone]

when x= 3 y =0. It is the point (3, 0)

[solemn mountain]

mhm

okay so where that line is that we said [3,5] is right

(3,0)

[woeful stone]

yes

[solemn mountain]

mhm

so now where does x=-1 come from

I see the point (-2,0)

[woeful stone]

So when x=-1 what does f(x) equal?

[solemn mountain]

okay can i explain my thought process why im lost rn

the first 3 problems

where either on a line

or not

and i said yes or no

I see for #4

(-2,0) and (3,0)

not on a line together

K= 1

[woeful stone]

So the question is just asking for what values of c does, f(c) =1 for -2 <= c<=3

It is not about lines. It is about the graph

[solemn mountain]

yeah

okay so im looking at the graph

-2

is at (-2,0)

[woeful stone]

At what x values does the line y = 1 intersect or touch the graph f(x)?

[solemn mountain]

at -1 and 1

[woeful stone]

-1 yes but not 1

You can see the hollow circle. That indicates the function is not defined at the point

[solemn mountain]

so k = 1

is the same as y = 1

[woeful stone]

yep

[solemn mountain]

OH

that was my problem D:

jesus christ im dumb

[solemn mountain]

ty so much

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[solemn mountain]

.reopen

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✅

[solemn mountain]

oh oops

.close

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[alpine jetty]

Can someone solve this with full working out (no steps skipped please)

[alpine jetty]

heres the answer but i dont quite understand it. if anyone can supply a much more simpler working out I would greatly appreciate that ❤️

[sweet shard]

!noans

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[alpine jetty]

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[rotund hawk]

how do I do this?

[neon iron]

substitute x = y^6

so you get lim_{y-> 1} (y^2 -1)/(y^3-1)

y^2 -1 = (y-1)(y+1) and (y^3 -1) = (y-1) (y^2 + 1 +y)

simplify

[rotund hawk]

how do you go back to x tho

[neon iron]

you just need the limit don't you

so it doesnt really matter going back to x

because if x-> 1, y^6-> 1 so y-> 1

[fallow igloo]

[rotund hawk]

what if it was going towards something else

so x -> 4

[neon iron]

then y^6 -> 4 so y-> 4^{1/6}

[rotund hawk]

ok

thx broski

[neon iron]

yw

good luck

[fathom pendant]

Real

Wait actually

Direct substitution seems really difficult for this one

Have you heard of L’H? @rotund hawk

[rotund hawk]

no?

[fathom pendant]

Ok

Do you know derivatives?

[rotund hawk]

yeah

we went over first principles tho

not the formula

[fathom pendant]

What are “the first principles”?

[rotund hawk]

using the f

[fathom pendant]

Ohh the limit?

[rotund hawk]

f'(x)= f(x+h)-f(x)/h

[fathom pendant]

Oh dear

Do you know the power rule then?

[rotund hawk]

he doesnt want us to use it yet

i know how it goes

actually i knew

[fathom pendant]

Ohh

[rotund hawk]

i forgo

[fathom pendant]

Hmm

[rotund hawk]

what were you going to tell me tho

@fathom pendant

[fathom pendant]

Its ok I could show you L’H

Dont use it though

Think of it as a tool to help you verify your answers

L’H states then when dealing with limits and using direct substitution gives you an answer that is either 0/0 or inf+/inf+ (this is called indeterminate form btw)

Then finding the derivatives of the numerator and denominator and using direct substitution on that gives the same result

For example, the problem you gave us is good

The numerator is: cuberoot(x) - 1

cuberoot(x) can be written as x^1/3

And using the power rule, we get the derivative of it, which is 1/3*x^-2/3

Which after some calculation is 1/(3*x^2/3)

Now the denominator is a lot easier: squareroot(x) is x^1/2

The derivative is 1/(2*squareroot(x))

And we divide these 2, and use direct substitution

To find the the limit is 2/3

I’ll send you some photos to make it easy to visualize

[rotund hawk]

yes pls

I understood getting the derivative

but i dont understad what you do next

[fathom pendant]

Anything tough to understand?

[rotund hawk]

oh so you just place the derivatives back into the equation

thank you so much!

[fathom pendant]

No problem

[rotund hawk]

why does this work tho

[fathom pendant]

Basically the trick to this is to find the rate at which the numerator and the denominator approach 0

Or infinity

Im not the best explainer for this

And the proof is absolutely insane

Convoluted too

Yeah thats about it

I should remind you that since you havent learned L’H or even the power rule, you cant use this

Its just a tool for now

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@rotund hawk Has your question been resolved?

[rotund hawk]

my teachers marking system is weird, so i might use it if im stuck

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[hardy meteor]

[m\dot{r}-\frac{l^2}{mr^3}=f(r)]\

[hardy meteor]

[m\dot{r}-\frac{l^2}{mr^3}=f(r)]\

[thorny flame]

yehuihe

[hardy meteor]

[m\ddot{r}-\frac{l^2}{mr^3}=f(r)]\

$m\ddot{r}-\frac{l^2}{mr^3}=f(r)]$

[thorny flame]

yehuihe

[hardy meteor]

$\dv{t}=\frac{l^2}{mr^2}\dv{\theta}$

[thorny flame]

yehuihe

yehuihe

[hardy meteor]

$\frac{1}{r^2}\dv{\theta}(\frac{1}{mr^2}\dv{r}{\theta})-\frac{l^2}{mr^3}=f(r)$

[thorny flame]

yehuihe

[hardy meteor]

hi all, how to substitute second equation to first one to get last equation

[hardy meteor]

this is second

[hardy meteor]

this is first

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[hardy meteor]

should be chain rule but i still can't get this one

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[uncut gale]

answers 2046

all i got was theres 2^10 max combinations but idk where to go from there

[gusty bane]

that's strings of length 10

what about .

or -

or ..

or .-

etc.

express it as a summation

[uncut gale]

and how would i go about doing that

im not too familiar with summations

[gusty bane]

how many ways are there to make a string of length 1

using a dot or a dash

[uncut gale]

2

[gusty bane]

or 2^1, yes

how about length 2

[uncut gale]

2^2

[gusty bane]

3?

[uncut gale]

2^3

ah

[gusty bane]

do you see the pattern

[uncut gale]

i see the pattern

increases by 1 each time

[gusty bane]

yes

[uncut gale]

idk how to use summation with the bot but i think i got it

[gusty bane]

this is essentially [\sum_{n=1}^{10} 2^n]

[thorny flame]

maximo

[gusty bane]

which has a well defined formula

[uncut gale]

yes okay

got it

[gusty bane]

but you could also just compute it by hand

[uncut gale]

wdym formula

[gusty bane]

[\sum_{n=0}^{k-1} r^n = \frac{1-r^k}{1-r}]

[thorny flame]

maximo

[gusty bane]

in our case, r = 2 and k = 10

so we'd get
[\sum_{n=0}^{10}2^n = \frac{1-2^{10}}{1-2} = 2^{10} - 1]

[thorny flame]

maximo

[gusty bane]

,calc 2^(10) - 1

[thorny flame]

Result:

1023

[gusty bane]

oh shoot what did i miss

[uncut gale]

yea this gives me 2046

[gusty bane]

ah i had an index wrong for the summation

but yes glad you got the right answer

[uncut gale]

thank you so much

[uncut gale]

and for this

is this a common formula?

[gusty bane]

yes but i have it off by 1

this should be the actual one
[\sum_{n=0}^{k-1} r^n = \frac{1-r^k}{1-r}]

[thorny flame]

maximo

[uncut gale]

ahhhhh

so r = 2 and k = 10

but if the top is 10

and its k-1

am i saying its 11?

why am i getting 2047

(1 - 2^11)/(1-2)

sorry

[gusty bane]

k should be 11 yes

[gusty bane]

because the sum starts at 0

but our original one started at 1

so we have to remove the n = 0 term

which is 2^0 = 1

[uncut gale]

ahhh

so thats where its 2046

okok got it

tysm

[gusty bane]

👍

[uncut gale]

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[woven otter]

For linear sequence this is the general formula for nth term

If we have a example like

20,17,14,11

And use the formula Tn=an+b

Can someone tell me what a,n and b are

this is the example in the book

I dont understand what is a,n and b

[dusk blade]

Look at the note on the side. It's similar to a line, y = mx + c

Let's take your data, 20, 17, 14, 11

Think about it as a line. What would it look like?

It's going down 3 points every next value, meaning the slope is ?

And it's starting at 20, meaning the c is ?

[woven otter]

🥲i dont understand

[dusk blade]

You know what a line is, yes?

[woven otter]

Yes

[woven otter]

What do u mean what the c is

[dusk blade]

It's the equation on the right side

y = mx +c

what does the c stand for

[outer carbon]

what sort of tn is this

[frosty prairie]

yo I never imagined an arithmetic progression as a line this is cool damn. It's obvious in hindsight

[woven otter]

Ah im not sure

[outer carbon]

The one I know is tn = a +(n-1)d

[frosty prairie]

Yeah but that's the nth term for AP in general, I think this topic is trying to relate it with a linear function which is better for inuitively understanding it

[dusk blade]

Same

It works tho

[woven otter]

Yes i like using that one but my teacher uses this one and i dont think i should use that for the test

[dusk blade]

This has a constant change between each value

[frosty prairie]

Yes

[dusk blade]

It's like a slope between them

[woven otter]

Yea

[dusk blade]

Think about it like a line

Look at the points visualized. You see something going on between them?

What if we put a line in the middle of them

That's similar to what you're finding. It's basically telling to you find the "line"

[woven otter]

Ok i understand

[dusk blade]

If you look closely, each value is _ points less than previous number @woven otter

20 becomes 17, 17 becomes 14

Their difference is 3 points right?

[woven otter]

Yes the difference is 3

[dusk blade]

We can infer that the slope is -3

Because it's going down

[outer carbon]

But why are they relating it to a line in the first place?

it is simple as it is

[dusk blade]

Visualization, I assume. Regardless, I think it's genius

[woven otter]

Mhm

[dusk blade]

Now, think about it. It's going down by 3 points, meaning a should be -3 right?

[woven otter]

Yep

[dusk blade]

-3x + b

Because each next value they're less 3 points

And what value do we start at? isn't it 20?

So what if the sequence is -3x + 20

(20 - 3x is better way to write it imo)

Does that make sense or

[woven otter]

Wait wait wait im confooosed

Brain not braining

[frosty prairie]

No it's actually pretty smart to relate it to a line. Any arithmetic progression can be represented by a line because the slope always remains constant which signifies that the difference between each subsequent term in an AP also remains constant. If you move on to more difficult stuff like 2, 7, 14, 23, where the difference BETWEEN the terms is in AP i.e. 7-2=4, 14=7=7, 23-14=9, etc. i.e. 4,7,9,... then you can represent the nth term for that sequence by increasing a degree and denoting it as tn=an^2+bn+c and so on and so forth

[dusk blade]

Why is brain not braining

You know they're reducing by 3 points

What's the first value that you start with

[woven otter]

The formule is Tn=an+b

T1=20 becUse its the first twrm to the sequence

[dusk blade]

Oh, then b should be 23 actually

-3 + 23 = 20

Because 20 is the first term

[woven otter]

Wait my twacher did something

Let me send another one

This is a diff sequence

[frosty prairie]

Yeah this is what I was talking about when I said difference of each term is in AP (quadratic), difference of difference of each term is in AP (cubic) and so on. But first try to understand why the basic AP is represented by ax+b

[dusk blade]

Ok, with the quadratic and cubic sequences... this doesn't seem cool anymore

[woven otter]

😂i want the linear sequence only

[dusk blade]

It would work as a good visualization method, but this is certainly quite a different approach on teaching sequences

[frosty prairie]

Let's try it with an easier linear sequence. Like 0,1,2,3,4,5,6...

The 0th term is a*0 + b

b=0 this means the intercept is 0

so at the x=0 y=0 as well so the line starts at the origin

[woven otter]

Hoooooold up why is there x and y.y is b 0 and why is a*0 sjdksndbsksknsbdbdsjks

[woven otter]

Can we skip the line thing

[frosty prairie]

Nope

What do you understand by t_n?

(t_n)

[thorny flame]

Usopper

[woven otter]

Its the nth term

[frosty prairie]

Yes

nth term for an AP, which happens to be a line

forget about my previous example for a sec

so each n denotes a point in the sequence right?

[woven otter]

Ye

[dusk blade]

Usopper is cooking 👨‍🍳

[frosty prairie]

Noice

wait a bit

ok I got confused for a moment there as well but now I understand

[woven otter]

Oki

[frosty prairie]

Note that the x co-ordinates starting from 1 denote n

[woven otter]

Yes

[frosty prairie]

for 1st term i.e. n=1 the first red dot will be the number you get for a particular line

each sequence can be denoted by a unique line

for n=2 you will get the second dot

[woven otter]

Mhm

[frosty prairie]

see that the difference between them is constant because the slope remains constant

The curvature always remains straight, there is no change

it's linear

[woven otter]

Ohhh ok

[frosty prairie]

now let's use your initial example

[woven otter]

20,17,14,11?

[frosty prairie]

This one

[frosty prairie]

Yes

[woven otter]

Oki

[frosty prairie]

I pasted it again so that it will be easier to reference

[woven otter]

Ok

What abt the a and b .in the equation its tn=an+b

What happened to the n near the a for t1 and t2

[frosty prairie]

yeah getting there

[woven otter]

Oki

[frosty prairie]

In this example, not that the sequence is increasing, it's probably something like 3, 5, 7, ...

In your example the sequence is decreasing

[woven otter]

Yea

[frosty prairie]

(t_n=an+b) is nothing but (y=ax+b) which is clear from the graph

[thorny flame]

Usopper

[woven otter]

Mhm

[frosty prairie]

So to find the first term of sequence you just put x i.e. n=1. y=a+b. It's already given to you that it is 20

(a+b=20)

[thorny flame]

Usopper

[frosty prairie]

This is first equation, you just need another to find b

just see the 2nd term

[woven otter]

Mhm

[frosty prairie]

put n=2

2a+b=17

(2a+b=17)

[thorny flame]

Usopper

[frosty prairie]

Solve these both equations

Tell me what do you get

[woven otter]

Simultaneous?

[frosty prairie]

yes

[woven otter]

Ah oki waiy

[frosty prairie]

Still waiting lol

[woven otter]

Ok wait

Done

Wait

Ill send

Sorry

[frosty prairie]

No worries lol

[frosty prairie]

((2a+b)-(a+b)=2a+b-a-b)

[thorny flame]

Usopper

[frosty prairie]

You wrote (2a+b-a+b)

[thorny flame]

Usopper

[frosty prairie]

Although your final answer is correct but it's better to be safe

in the steps

[frosty prairie]

Correct

[woven otter]

Ohh oki

[frosty prairie]

Now just put it in the eqation

equation

y=ax+b

OR here

t_n=an+b

[woven otter]

So tn=-3n+23?

[frosty prairie]

Yes

[woven otter]

Ohhhhh

[frosty prairie]

Note that the graph is not required, it's just to understand why you use t_n=an+b and then plug values to figure it out

and not just something else like t_n=an+bn or t_n=an^3+b

For any arithmetic sequence you just need to make an equation from the first and second term (or any two distinct terms really) and then solve

find a, b you're done

hope you get it now

[woven otter]

Oh yea

Thank u very very much

Hope u have a nice day

[frosty prairie]

Welcome

Thanks! Same to you as well ☺

[woven otter]

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[neon iron]

How exactly do you define unit vectors for $\theta$ and $\varphi$ in cylinderical and spherical coordinates?

[neon iron]

it seems kind of cursed but i dont see how it is done

[ionic oar]

lol

axial vectors?

[neon iron]

hmmmm

this is how my book is notating the orthogonal unit vectors for example

[neon iron]

idk if thats exactly what it is

maybe

[ionic oar]

doesnt look like

the azimuth would be along the same direction as z

[neon iron]

https://www.youtube.com/watch?v=xmMGz5iAlqw

oh okay wait this seems kind of related

probably easier to start there

[neon iron]

oh wait wdym

[ionic oar]

like

we'd assume $\hat z$ to function in the same way as $\hat k$ right

[thorny flame]

NEON

[neon iron]

yes

[ionic oar]

and if the azimuth was an axial vector

then by the right hand curl rule

or whatever

it would also point along $\hat k$

[thorny flame]

NEON

[neon iron]

so what we havw is not that then

[ionic oar]

yeah probably something else

let me know if you find out btw

[neon iron]

apparently like the theta is the unit vector tangential to the cylinder

[ionic oar]

interesting

[neon iron]

although like

idk how uh

[ionic oar]

me neither

[neon iron]

thats orthogonal to r and z?

i mean

[ionic oar]

I went with the axial assumption cuz angular displacement and it deirvatives are axial vectors

[neon iron]

okay if its tangential to r then obviously it is orthogonal to r

so thats that

how tf is it orthogonal to z tho

[ionic oar]

it would be right

z points straight out of the plane containing r and theta

[neon iron]

oh ic yeah fair enough

okay i mean i guess this makes sense

kinda weird

but makes sense

ok until another time

time to close this

.clsoe

/;cplse

,close

.close

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[ionic oar]

i dont see why you just cant use regular unit vectors

if the end result is the same

[neon iron]

i mean i guess its important for actually defining the coordinates?

[ionic oar]

Perchance.

[neon iron]

[neon iron]

altho its kinda weird in the sense that $\vc*\theta$ is going to be changing direction

[neon iron]

like everytime

[knotty ledge]

Yes it's weird it's like a local coordinate system, maybe people sometimes use the phrase "frame"

[neon iron]

yoo thanks for the spherical corods hook up

[knotty ledge]

https://en.m.wikipedia.org/wiki/Frenet–Serret_formulas

Generalisation here

[neon iron]

aight thanks

oh by the way sigma

nvm i answered my own question

LOL

[knotty ledge]

Also note θφ are flipped here, blame physicists

[neon iron]

i was about to say

which one should i stick with

[knotty ledge]

Whichever is in your text

[neon iron]

i remember snow saying if u switch them it becomes like a left handed coordinate system

or something

probably misinterperting

[knotty ledge]

Eh maybe idk

I just prefer the other one than what I posted

"Extends" polar coords

[neon iron]

yeah same

fuck spherical tho ong

i hate it sm

[knotty ledge]

It's not overly fun

[neon iron]

its more so i cant get it to my brain ill be honest

well i think i get it now

but its just more so

i cant remember the formulae for like the conversions

[knotty ledge]

Just have to look at a lot of pictures

[neon iron]

so i hate deriving them each tinme

[knotty ledge]

No one remembers those unless you use them lmao

[neon iron]

too true

oh actually sigma can i ask u like a few questions regarding defining surfaces

ok lemme reopen this if u say yes

[knotty ledge]

I have like 10 mins before teaching sure

[neon iron]

oh boy ok

speedrunning

.reopen

[topaz sinew]

✅

[neon iron]

so uh in their derivation of like the formula for the surface area of an implicitly defined surface

why is their parameterisation like this

its super weird

(we can continue this later whenever u r freer haha)

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[knotty ledge]

I'm sorry I got nerdsniped lol

I'll look later if you haven't a answer

[neon iron]

alrighty haha

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[wide bane]

Not sure how to begin

The one mentioned from the book

Actually, I would prefer to just be given or referred to a place where I can find the calculation rules needed for this proof, since it requires some things like linearity, etc

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[sweet shard]

use 25 = 5^2 and get a quadratic equation in 5^t

[neon iron]

,, a^{m+n} = a^m\2a^n

[icy sky]

5^(t-1) = 5^t * 5^(-1)

[icy sky]

oh lmao

[frosty prairie]

Garlic!

[icy sky]

hello

[frosty prairie]

Hello

Fancy seeing you here. Your Discord profile is really memorable. I know you from MODS

[icy sky]

ah

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[native flume]

How would u solve this other than using a system of equations?

[rigid ivy]

What can you say about the quadrilateral LMNP? What features does it have?

[topaz sinew]

@native flume Has your question been resolved?

[native flume]

It’s a parallelogram that has 360 degrees

The opposite sides are equal

Yeah that’s all I know

[rigid ivy]

It is a parallelogram. Yes.

[rigid ivy]

If it is a parellologram, what else does that say about the opposite sides?

[native flume]

Oh it would be that opposite angles are equal so x = y

[native flume]

Which would mean that 3a-7 = 4a+6

[rigid ivy]

yup

there you go

[native flume]

Okay cool thank you

I couldn’t figure it out when tutoring lol

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[cursive sparrow]

Hi, im having problems figuring out why the the numerators can be disregarded in the eqn when finding x.

-1/16 = -1/x^(4/3)

[sweet shard]

show how you got x

[cursive sparrow]

-1/16 = -1/x^(4/3)
16 = x^(4/3)
4root(16) = 4root [x^(4/3)]

2^3 = (x^1/3)^3

8 = x

I think it would have been fine to take a picture

[rocky gulch]

$-\frac{1}{16} = -\frac{1}{x^{\frac{4}{3}}} \
16 = x^{\frac{4}{3}} \
\sqrt[4]{16} = \sqrt[4]{x^{\frac{4}{3}}}$

how do i go on the next line

[cursive sparrow]

wdym

[rocky gulch]

with the bot

[cursive sparrow]

im not sure

o

[rocky gulch]

great

like this?

[cursive sparrow]

uhhh

4th root

my bad

and no coefficient before the root

[thorny flame]

dqvidutzul

[cursive sparrow]

yes

[rocky gulch]

do you need this answer checked?

[cursive sparrow]

it should be right but I was confused about the numerators

[rocky gulch]

seems fine to me

[cursive sparrow]

do you know why the numerators can just be taken out from the equation?

I think I might have missed it entirely

cross multiplying

🤦‍♂️

ok thanks for the help

[rocky gulch]

you flipped the fractions on the first lines

and multiplied by -1 to get rid of the minus

[cursive sparrow]

yep

I overthought it

thanks

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[bitter hemlock]

you can also think of it as...well if the numerators are equal then the fractions are only gonna be equal if the denominators are also equal

5/2 = 5/x is only gonna work for x = 2

[bitter hemlock]

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[faint trail]

Hey, I just wanna make sure I'm doing this right, it seems wrong because usually the graph isn't put in a certain quadrant if the answer is in a different quadrant 😭
This is rotations with a non-origin center
The points in the column on the left is for the distance from the point and the right set of points are after they're put in the counterclockwise form
We don't have to graph these, but I've just been graphing the point and the prime figure because it's easier

[faint trail]

Let me know if the image is blurry

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[whole geode]

@faint trail your technique for rotations by 90 degrees would be correct if you were rotating around the origin (0, 0). But the problem specifies that you are intended to rotate around (4, 1)

So what you need to do is map (4, 1) to the origin, do the rotation, and then map the origin to (4, 1).

Original point: (x, y)
Map (4, 1) to origin: (x-4, y-1)
Perform rotation: (1-y, x-4)
Map origin to (4, 1): (1-y+4, x-4+1)
Simplify: (5-y, x-3)

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[toxic aspen]

$[(X^T - B)A]^T = 3(X-B^T)$

[thorny flame]

Merineth

[toxic aspen]

How do i solve for X?

matrix equation

[outer ledge]

If you know the properties of transpose, then it's easy to solve.

[toxic aspen]

I think so, we are given a formula for the properties of transpose

Is that what you mean?

$[(X^T - B)A]^T = 3(X-B^T) \
[AX^T -AB]^T = 3X-B^T$

[thorny flame]

Merineth

$[(X^T - B)A]^T = 3(X-B^T) \\
[AX^T -AB]^T = 3X-B^T$
```Compilation error:```! Missing number, treated as zero.
<to be read again> 
                   A
l.50 [AX^T -AB]
               ^T = 3X-B^T$
A number should have been here; I inserted `0'.
(If you can't figure out why I needed to see a number,
look up `weird error' in the index to The TeXbook.)```

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[proud sluice]

hello i badly need help in a student organization there are 4 positions in the executive committee; president secretary vice president and treasurer. these officers will be chosen from among 12 finalists who were elected by the members at large. in how many ways can the positions be filled up if one finalist will only serve in the executive committee if he is chosen president?

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[vale oriole]

What it mean J/I ?

[opal vault]

J/I is the image of J by the quotient map

[topaz sinew]

@vale oriole Has your question been resolved?

[vale oriole]

It is extension of J ?

[opal vault]

yes

[vale oriole]

But how to prove exercise

I try set function R/I->R/J but i couldnt find kernel

[opal vault]

if you have $\pi_I:R\to R/I$ and $\overline{\pi}_J:R/I\to R/J$, prove that $Ker(\overline{\pi}_J) = \pi_I(J)$

[thorny flame]

rafilou2003

[opal vault]

Of course $\pi_I(J) = J/I$

[thorny flame]

rafilou2003

[opal vault]

and then you'll be done

[vale oriole]

It is equivalent to that preimage of Kernel is J ?

[opal vault]

if you prove that preimage of kernel by pi_I is J then you're done yes

because pi_I is surjective

[vale oriole]

For sujective maps image of preimage is identity?

[opal vault]

For $f:X\to Y$ surjective then $f(f^{-1}(A)) = A$

[thorny flame]

rafilou2003

[vale oriole]

Okay

[opal vault]

so indeed try to prove that $\pi_I^{-1}(Ker(\overline{\pi}_J)) = J$

[thorny flame]

rafilou2003

[vale oriole]

These two sets are the kernel of πJ ?

And thats why they are equal?

Because πJ = overline πJ(πI)

[opal vault]

oh yes

they are both Ker(pi_J) without the overline

$Ker(\pi_J)$

[thorny flame]

rafilou2003

[opal vault]

yep

[vale oriole]

Thats it?

[opal vault]

$\pi_I^{-1}(Ker(\overline{\pi}_J)) = Ker(\overline{\pi}_J\circ \pi_I) = Ker(\pi_J) = J$

[thorny flame]

rafilou2003

[opal vault]

that's it

[vale oriole]

Oooo there is a beautiful relation
With kernel preimage and composition

[vale oriole]

Thanks

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[ivory sorrel]

I know I have to perform a sub

[ivory sorrel]

but what would that be here?

Or do I graph it

[opal vault]

my hunch says 0

the sub can be tricky

[ivory sorrel]

Desmos gives 2

[dry otter]

Don't you get -1=1 then?

[opal vault]

positive real solutions

[reef fjord]

still 2?

[ivory sorrel]

3 soln, my bad

[opal vault]

i meant 0 solutions

[opal vault]

yeah 2 solutions

[dry otter]

Oh wait didn't read the question properly 😅

[reef fjord]

it looks like there's 2 positive solutions

[ivory sorrel]

oh, +ve

ok, so do I do this via a sub

or a rough sketch

I mean the solns are neat, so I could probably sub smal values of x, but that wouldn't be very rigourous IMO

[reef fjord]

might be an IVT sorta thing

[opal vault]

for starters

[ivory sorrel]

hmm

So I have to learn IVT to solve this?

[reef fjord]

weren't you doing taylor stuff earlier?

lmao

[ivory sorrel]

yeah, but I just know the expansions for standard functions

[reef fjord]

ah right

there's probably a convoluted way without IVT to do this

[gleaming thunder]

if you don't know IVT, do dammit

[ivory sorrel]

I will learn it now, sorry

.close

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[gleaming thunder]

undergrad math major + not knowing IVT are basically incompatibly

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[ivory sorrel]

Ok, how would I apply IVT here?

[haughty gale]

Can someone please help me with this

[reef fjord]

let f(x)=x^2-x-1 and g(x)=2^x-log_2(x^2+2x), then define g(x)-f(x) or smth. Plug in values for f(x)-g(x), until you reach the case where it changes sign from positive to negative or negative to positive, then you can basically use some derivative/always decreasing/always increasing argument

[ivory sorrel]

hmm, l think about this, thanks

do close this now, or do I leave it open for Azram?

because I claimed it first, but it's not obvious

[reef fjord]

azram already has a channel open

[ivory sorrel]

where?

[reef fjord]

#help-27

[ivory sorrel]

ah,ok

.close

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[reef fjord]

well, actually, it might be easier to just say way the 2 zeros are(they're nice numbers), then basically show past those the function f(x)-g(x) is always increasing/always decreasing for positive reals

[ivory sorrel]

, IVT seems to be a way more intuitive way of doing it though. Thanks again!

[crystal goblet]

What moosey said here guarantees that there will be no other solutions other than the 2 you can find via guesswork for x>0

So it's more correct

You cannot guarantee just by the IVT that there are only 2 solutions for x>0

[ivory sorrel]

why not though?

let me think, don't answer that please

that was rhetorical

hmm, so after applying IVT

if I don't prove it's monotonic

I wouldn't have proved that they aren't more solns/

[crystal goblet]

Yep, you would've just proved that there is a solution between x=whatever and x=whatever else

[ivory sorrel]

got it, thanks!

.close

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[white grotto]

How they did this

[topaz sinew]

@white grotto Has your question been resolved?

[sweet shard]

they factored the stuff in the square root in the denominator

(1 - a^2) = (1-a) * (1+a)

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[neon iron]

im trying to find a parameterisation for the cap cut by $z=\3{x^2+y^2}$ for [
x^2+y^2+z^2=9
]
but i dont know how to go about doing this

[neon iron]

would appreciate some guidance

[sweet shard]

find the intersection

[neon iron]

so z^2 + z^2 = 9 -> 2z^2 = 9 -> z^2 = 9/2 -> z = \pm 3/sqrt(2)

well the cone is defined only for positive values of z so we can drop the negative of that

ok cool z = 3/sqrt(2)

so where do i go from here

[loud oasis]

find the intersection in terms of x & y

[neon iron]

so x^2 + y^2 = 9/2

i guess?

[loud oasis]

so then you can parameterize as a circle

[neon iron]

i need to include the interior of this circle right

ok wait this might be easier if i just convert all of this to spherical

[sweet shard]

the "cap" of the sphere r^2 = 9 is just the shell. don't know why you'd include the interior

or is cap different for you

[neon iron]

i imagined it to be like

[sweet shard]

then it'd be r^2 <= 9

[neon iron]

you know the hat

of the sphere

because the question does say cap as well

but nonetheless

we have $\rho^2 = 9$ bounded by $\phi = \4\pi4$

if i did the conversion correctly

[pastel salmon]

if:

then:

[neon iron]

so uh we just restrict phi between 0 and pi/4

and then thats the standard equation of a sphere

[pastel salmon]

that is yoru cap in spherical

[sweet shard]

psi

[pastel salmon]

yes psi )

[sweet shard]

rebel joanna

[neon iron]

i dont understand ur restrictions?

isnt it jsut

[sweet shard]

her psi is probably pi/2 - theta for you

[pastel salmon]

in my interpretation : phi is angle on OXY, and psi is in 3d angle

[neon iron]

yeah but like why is the radius varying even its a sphere

[pastel salmon]

ro is in 3d distance

[neon iron]

oh ok i see

[pastel salmon]

analyse this pciture:

[sweet shard]

is the parametrization you're looking for a hollow sphere or are you parameterizing the interior as well?

[neon iron]

oh you are taking the latitude instead of the colatitude

thats kinda based

[neon iron]

interior yeah, i just verified with my book's solution

phi is not fixed to pi/4

ok i guess thats it

ill do another exercise on my own tho just to see if i actually got the idea one sec (ill keep this open)

ok so

[pastel salmon]

my description is referring to such a region:

such an " ice -cream" 🙂

[neon iron]

yeah its what the question intended it to be

a wide ice cream

but anyhow

[pastel salmon]

ahah es

[neon iron]

i get the following:

[2\textwidth][
\vj\rho(\theta, \phi) = \33\6\sin\phi\6\cos\theta\vc*\imath + \33\6\sin\phi\6\sin\theta\vc*\jmath + \33\6\cos\phi \vc*k \q -\4\pi3 <\theta < \4\pi3, ; 0 < \phi <\pi
]

would that be correct

[pastel salmon]

a wedding ring :

[sweet shard]

Cheese wheel

[topaz sinew]

@neon iron Has your question been resolved?

[pastel salmon]

that is description of the "cheese", using my spherical cords definition:

[topaz sinew]

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[twin marsh]

hey

[twin marsh]

Need help with this

why is it not 11

[snow nimbus]

Why would it be?

[topaz sinew]

@twin marsh Has your question been resolved?

[twin marsh]

because count around the shape it 11 btw i dumb

[snow nimbus]

Let's count the horizontal length. What did you get?

[twin marsh]

4

[snow nimbus]

I think you mean the bottom one. And the top ones?

[twin marsh]

top on 2 and 1

[snow nimbus]

Left top is 3 units long

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@twin marsh Has your question been resolved?

[topaz sinew]

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[finite mortar]

Don’t get why these parts repeat like where does this formula come from ,trying to understand the reasoning behind it

[keen raptor]

the first column: all the x_i terms represent what's directly produced