help-28

[compact leaf]

Is this solution correct?

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@compact leaf Has your question been resolved?

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@compact leaf Has your question been resolved?

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@compact leaf Has your question been resolved?

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@compact leaf Has your question been resolved?

[thorny valley]

Hm you seem to be assuming the top limit exists in the first line

[compact leaf]

Can we do it? I mean in the first step we put cosx^2 as 1 and other things same then next steps we do l'hopital

[compact leaf]

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[compact leaf]

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✅

[torn jolt]

How did u make the (cosx)² = 1 by putting the limit but leaving the others in variable

[compact leaf]

Yes. That's why I was asking is it correct?

Someone did this

[torn jolt]

I don't think so hmm , either u put limits in all of them or none

[compact leaf]

Yes

Thank you

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[cloud violet]

.reopen

,w arctan27/48

[glossy valve]

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[tribal oxide]

#bots for this

[cloud violet]

kk

[wise wyvern]

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[last locust]

what am I missing? How do they get xv'+v in the first step when they plug in y/x? I just get v and v^2

[last locust]

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[serene pelican]

how do i find the 4th roots of negative 1?

[serene pelican]

as in solving z^4=-1

i can't seem to find an evident root of -1

cus -1¨^4 = 1 and i^4 = 1

and -i^4 =1

[wise wyvern]

-1 = i^2

= e^(ipi)

[serene pelican]

e^(i2pi) = 1 right?

[wise wyvern]

yeah.

[serene pelican]

so -e^(i2pi) = -1

[wise wyvern]

yeah.

[serene pelican]

oh i jsut need to solve, (-1)(z^4)=-1(1)

[wise wyvern]

No?

You need to solve z^4 = (-1)
Or z^4(-1) = 1

Not what you've written though.

[serene pelican]

oh, yeah so the roots are -e^(i2kpi/4) for k ([0;3])?

[wise wyvern]

Are they though?

They're not.

[serene pelican]

why not

[wise wyvern]

say k = 0

You have -1

(-1)^4 = 1

Not (-1)

You're doing it wrong...

[serene pelican]

huh

but the thing is i can't find an evident value that to the power of 4 = -1?

[wise wyvern]

Try this.

[serene pelican]

e^(ipi)^4 = e^i4pi

[wise wyvern]

NO!

[serene pelican]

how is that not the case

[wise wyvern]

z^4 = -1 = e^(ipi)

[serene pelican]

oh, i got it

[wise wyvern]

e^(ipi) is (-1), you don't have to raise -1 to the fourth exponent.
Rather 1/4

[serene pelican]

e^(ipi/4) ^4 = -1 right

[wise wyvern]

Yes.

[serene pelican]

the first root is e^(ipi/4)

so i do (z/e^(ipi/4)^4 = 1

then i just apply the formula

so we replace Z=z/e^(ipi/4)

and we get

Z = e^(i2kpi/4) for k ([0;3])

therefore the values are e^(ipi/4) x e^(i2kpi/4) for k ([0;3])?

and that makes e^(i2kpi/4+ipi/4)

[wise wyvern]

e^(i(2k+1)pi/4) yes

[serene pelican]

tyvm

but is there an easy way to identify the first root?

[wise wyvern]

What I did is the easiest way for me, atleast.

So that.

Though if you want easier, you're free to wait for what others have got to say.

[serene pelican]

so for z^4 = 1 + i for example, i need to find a simple root of 1+i

[wise wyvern]

1+i = sqrt(2) e^(ipi/4)

You can do this by first finding the argument of the given complex number on the right and then rewrite it.

Like I just did.

[serene pelican]

so that makes sqrt(2)^1/4 x e^(ipi/16)?

[wise wyvern]

Pretty much.

[serene pelican]

is there a way to simplify fourth root of sqrt(2)

[wise wyvern]

2^(1/8)

[serene pelican]

and then i just multiply the answers of z^4=1 by this

[wise wyvern]

Sure thing.

[serene pelican]

so in total, the formula is basically

to solve z1^n = z2 i need to do basically do (|z2|e^iarg(z))^1/n x e^i2kpi/n with k in ([0;n-1])

[wise wyvern]

Sounds about right.

Except k could obtain any n consecutive integers.

Not necessarily [0;n-1]

Just so you know.

It could be [1;n]

[serene pelican]

yeah,

but the value of 0 is the same as the value of n

[wise wyvern]

Yeah.

So any consecutive n.

[serene pelican]

ok

idk why my professor didn't explain it like this lol

tyvm

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[serene pelican]

how do i solve find the limits of these series

[serene pelican]

i managed to show that V(n) is strictly increasing and U(n) is strictly decreasing

how do i find the limits of V(n) and U(n) and that the limits are equal

i've also shown that if V(0) = U(0), lim(V(n))=lim(U(n))=V(0)=U(0)

i need to find the general limits for V(0)!=U(0)

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[dawn bluff]

how would you graph this?

[dawn bluff]

no luck with desmos

just shows a disc

id imagine it being like a bowl though

[kindred grove]

doesn't desmos only do 2D stuff ?

[white geyser]

desmos has 3d

[kindred grove]

ah ok

[white geyser]

desmos.com/3d

[dawn bluff]

yup, i also trie geogebra but still nothing

[white geyser]

$$x^{2}+y^{2}<4\left{0<z<x^{2}+y^{2}\right}$$

[glossy valve]

b0ngl0rd

[white geyser]

i got it to graph with that

[dawn bluff]

i tried this

[white geyser]

its a cylinder with a cone-like cut-out

[dawn bluff]

juast a q, but if we have x^2 + y^2 < 4, then how come its not a cylinder too since at z = a (any real nr.) the inequality would still be satisfied

[white geyser]

extend it to 3D

it'll show you a cylinder

[dawn bluff]

OHHh

im such an idiot

nvm

thanks

[white geyser]

without extending it'll only use the variables you provided and disregard the unused dimension unless you extend it

[dawn bluff]

that makes sense

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[pine plaza]

is the following predicate logic formula a tautology? ∀ x y z : N, x ≠ y → (x ≠ z ∨ y ≠ z)

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[left creek]

is this right? I haven't tried this yet, so ignore the x

[left creek]

or is it

it's not the latter, just tried it

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[gritty rose]

This is correct, but you might have to simplify

[left creek]

ah, thank you

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[torn jolt]

!status

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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

[torn jolt]

5, im getting 0.47 + 0.075t

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[steel fiber]

Does anyone know how to create tis graph? Can only be one equation no piecewise

I know the answer includes absolute values but idk how

[full marsh]

Try flipped the middle section, that way you get something that looks more like a polynomial

[steel fiber]

How do you mean?

[full marsh]

Find that polynomial and use some transformations to get it "break" like the picture

Wait let me draw it

[steel fiber]

isnt the graph a transformation of y=|x|^2

[full marsh]

what

[steel fiber]

like this isnt far from it

[steel fiber]

but what about the sharp points

at the top

[full marsh]

You need a cubic for this

You have 3 points where the derivative is 0

3 min/max points

[steel fiber]

I dont want to say 'no' but this is for a pre-calc course and we havent learned derivatives yet

im not against like arriving at the answer by whatever means neceesary

i just dont really know derivates

[full marsh]

Well I'm just saying, that this graph is a transformation of a cubic, since there are 3 min/max points

You don't need derivatives for this

Anyways, you see where the graph has pointy turns?

Those happened because of the absolute value

So start by "reversing" that, as I did in the picture I sent, and find the original polynomial

THEN use transformations you know to make it "break".

No, don't use any absolute values yet...

Just find the polynomial before it got "broken"

[steel fiber]

this u mean?

[full marsh]

Oh actually, just realized you need a quartic, not a cubic

[steel fiber]

how come

and what does that imply

btw i just wnana say i truly appreciate the help this is due tomorrow and ive been stuck on it for a long time

[full marsh]

3 min/max points mean you need a quartic. I accidentally said cubic earlier

[full marsh]

No problem lol

Why 1, 3, and 5?

[steel fiber]

cause those are the zero's, no?

or wait

cause its shifted down by 1

[full marsh]

Again, you are looking at the original picture

[steel fiber]

so the zeros ar ethose +1

right

[steel fiber]

i need this

is what ur syaing

[full marsh]

Yes

[steel fiber]

hows thiss

its not centered but

[full marsh]

Not quite

I'll brb in 10 mins btw so just keep trying for now

[steel fiber]

okay

thanks

i believe im getting closer

thats the right shape

the scales are just off

[full marsh]

Alr I'm back

Btw one thing that's weird is that the values are not really adding up

Like, if you also assume the graph passes through the points:
(-4, -1) and (4, -1)

(Min points)

Then you can't get a quartic to look like that

[steel fiber]

wdym

right now im here and kind of stuck

[full marsh]

At least you found the right idea

[steel fiber]

they ahve to be equidistant no

the 0's ?

[full marsh]

Giving absolute value to the factor of |x^2 - 4| corrects the sign of the function at the interval (-2, 2)

But now you have to scale it and move it up/down

And you'll see that SOME of the points wont match like the picture

[steel fiber]

im sorry i dont understand, i need to square an x?

[full marsh]

No

Ok let me show you what I mean

Give me a sec

You see how I got the points where it is 0, and I got the max point?

Just like what you sent

but it makes the min points not be the values we want

and there's no way to change it without messing up some other part

[steel fiber]

right that what im thinking

[full marsh]

So... buggy question

[steel fiber]

heres what im thinking

might be dumb

[full marsh]

You can make it be "approximately" the shape

[steel fiber]

but in the same way that x = 1/10(10x)

is there something in that

like

idk maybe that dumb

[full marsh]

Wdym

[steel fiber]

like

idek

[full marsh]

Just say it, it's fine

[steel fiber]

like

some way that restricts the mins at 1

whilst changing the zeros

in the same way that

u can pull a 2 out of 3x

and get 2(3/2x)

ok that makes no sense

[full marsh]

Any change that will make it (+-4, -1) will mess up other things

[steel fiber]

could u send me ur equation ?

[full marsh]

There's only a single quartic that passes through,
(-5, 0), (-3, 0), (3, 0), (5, 0) and (0, 7)

And that quartic happens to NOT pass through (4, -1) or (-4, -1)

I got:

$\frac{7}{225}(x^2-25)(x^2-9)$

[glossy valve]

RedstonePlayz09

[full marsh]

Really nothing special, just scaled it to pass through (0, 7) after getting the zeros.

[steel fiber]

why 0,7?

thank u tho

[full marsh]

Just looking at this

After reversing the effect of the absolute value, it should pass through (0, 7)

[steel fiber]

im sorry im confused

what do u mean

[full marsh]

This

should pass through (0, 7)

[steel fiber]

ok so

i found the function

[full marsh]

Holy shit

I see where I was wrong

You apply the transformation before finding the polynomial

Oh welp that's my bad

that's clever

[steel fiber]

yeah

thank u very much tho

i appreciate the help, genuinely!

[full marsh]

Man I feel bad now, wasted your time

[steel fiber]

no not at all

[full marsh]

So only terms of degree 2 or less?

[steel fiber]

no

[full marsh]

Must be a polynomial?

Or no?

[steel fiber]

just has to have started with x^2

[full marsh]

Alright lemme think

I'm assuming you're "creating" questions?

[steel fiber]

uh

no

[full marsh]

Alr

Maybe something involving 1/

So like..

$\frac{1}{\abs{x^2-1}}$

[glossy valve]

RedstonePlayz09

[full marsh]

Going with the spirit of the previous question and using an absolute value

Do you allow adding terms that are just x or |x| throughout the transformation?

Like, not ONLY having the original x^2?

[grim skiff]

Nesting a bunch of absolute values in there looks cool

[steel fiber]

uh

actually

sin(x^2)=sin(y^2)

is even cooler

[grim skiff]

So you can mess with the y too?

[steel fiber]

can do whatever u want

onto the next qeustion tho

it just says

[steel fiber]

might just do this

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[grim oak]

i dont understand this question

this is the first part which i already understand

[grim oak]

i dont get y they make it equal 4,5

4.5

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[odd dune]

I've given up and re-tried this problem so many times now

I first tried to calculate E(X) directly

that's not the image but I did try it, didn't rlly go anywhere

then I tried finding the MGF

That’s how I’d set up the integration but I kinda go into a dead end with my integration techniques

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@odd dune Has your question been resolved?

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@odd dune Has your question been resolved?

[gritty rose]

Your substitution of u = 1+x/theta almost works. Just do u = x/theta first then factor out the constants and you just need to integrate u/(1+u)^c. Then do v=1+u and use power rule to integrate

[odd dune]

Ah gotcha

Ye I see how that’ll work

Ty

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[nova island]

is this as easy as it seems?

[nova island]

is it just phi(gh)=(gh)^n=g^nh^n=phi(g)phi(h) where the second equality follows from G being abelian, or am i confused?

[fast peak]

yes

[nova island]

yes i am confused or yes that is correct?

[fast peak]

that is correct

[nova island]

okay awesome, thank you!

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[remote gulch]

Got this question for my discrete maths homework, and I have no clue how to go about this question. The topic we're currently studying is nunber theory and congruences. Any help?

[remote gulch]

Prove that if $x$ and $y$ are integers such that $x \cong 2$ (mod 8) and $y \cong 7$ (mod 8), then 8 divides $2(x + y)^{13} + y - 1$.

[glossy valve]

vin

[dense vector]

there are some properties of congruence

[remote gulch]

the addition, subtraction and multiplication one?

[fast peak]

in other words, you have to show 2(x+y)^13+y-1=0 mod 8

[remote gulch]

because besides that, we didnt learn any other properties

[remote gulch]

i think so

[fast peak]

what happens if you plug in what you know about x and y

[dense vector]

https://www.math.washington.edu/~greenber/Congruences.pdf

have a go before trying to solve ur question..will be much easier

[remote gulch]

👍

[dense vector]

most of the properties are quite similar to linear equations..but still try to understand them and then solve this question

[remote gulch]

im still lost 😭

my professor had to cut the lecture short because we were short on time, so he didnt explain much about congruences

so i have no clue what im supposed to be looking for

[stiff summit]

you can let x = 2 + 8m and y = 7 + 8n

[remote gulch]

ooo didnt think of it that way

alright lemme see if i can figure anything out

[dense vector]

@remote gulch were you able to understand the formulas mentioned in the pdf?

[remote gulch]

there were around 3 i didnt really understand, but the rest i could

the prime ones and the polynomial one

[dense vector]

u only need 4 and 6

[stiff summit]

Then when you expand it out, you’ll get a ton of terms, but most of them are divisible by 8, so you can ignore them

[remote gulch]

this is a bit off topic, but are there any websites/papers i could do as exercise for discrete maths, as its pretty apparent im really bad at this 💀

[remote gulch]

alright, will try

[remote gulch]

👍

[stiff summit]

you could ask in #book-recommendations for a book on discrete maths, it will have a bunch of problems you can try

[stiff summit]

or you could do the first few chapters of Abstract Algebra: Theory and Applications by Judson where he goes over the preliminaries of modulo stuff

http://abstract.ups.edu/aata/aata.html

[remote gulch]

cool, thanks a lot for the recommendations

back to scratching my head over the question

[remote gulch]

expand as in expand the polynomial? or just after substituting in x and y without opening the polynomial

[stiff summit]

Expand the polynomial

Just write out the first few terms for now

[remote gulch]

ah alright

[stiff summit]

and think about whether or not the rest of them are divisible by 8

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[remote gulch]

i feel so stupid 😭

still cant get it

[stiff summit]

Could you take a picture of your work?

[remote gulch]

i erase every time i hit a road block, but i could screenshot what i last tried

im probably doing something extremely wrong here

☠️

[dense vector]

can you use 4th property to get relationship of (x+y) with 8

[remote gulch]

as in the x + y = 9 (mod 8) or do i substitute in the x and y with 2 + 8m and 7 + 8n

now wait a minute

so something like this?

[remote gulch]

nah this aint right

if x = 10 and y = 15

hmmm

[remote gulch]

its a bit messy, but is this right?

read from red to blue*

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[digital chasm]

Help

[next sail]

!status

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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

[honest wing]

help

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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).

[digital chasm]

.close

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[torn jolt]

I don’t understand this

I don’t see any of this in my notes

I feel like it should be easy but idk

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[drifting cloud]

How i can proof that this serie is Cauchy sequence

[maiden marlin]

Start by using the trigonometric identity: sin(k) = (1/2i) * (e^(ik) - e^(-ik)), where i is the imaginary unit.

Rewrite the series as follows:

Un = Σ (1/2k) * (1/2i) * (e^(ik) - e^(-ik))

Split the series into two separate parts:
Un = (1/2i) * Σ [(e^(ik) / 2k) - (e^(-ik) / 2k)]

Focus on each part separately:
a) Consider the series Σ (e^(ik) / 2k). This is the sum of complex exponentials and can be challenging to analyze directly. To show it's Cauchy, you might need to rely on the Dirichlet test or Abel's theorem, which are techniques used to establish convergence of series involving complex exponentials.

b) Consider the series Σ (e^(-ik) / 2k). This series is essentially the complex conjugate of the previous series, and the real part of the corresponding series is the same as (1/2) times the series Σ (cos(k) / k), which is known as the Dirichlet series for the function cos(x). You can show that this series converges by using a well-known result.

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[drifting cloud]

Let me check

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[little gulch]

Could someone verify for me if this is correct

[little gulch]

I’m trying to solve this,

But the answer from the expert doesn’t look solid

solved it

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[vast solstice]

are vertical traces of a function examining how a function changes based off one variable since one of them is being held constant (Assuming we are in at least 3d)

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@vast solstice Has your question been resolved?

[azure grove]

yeah,they are essentially a way of examining how the function changes based on one variable while holding the other variables constant. These traces are often used for visualizing and understanding the behavior of multivariable functions.

If we are in a three dimensional space you are gonna have two independent variables for an example x and y and a dependent variable for an example z

When you take a vertical trace by fixing one of the INDEPENDENT variables (x,y) at a constant value and then you observe how the function behaves along that slice you are essentially exploring the function's behaviour

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[obtuse orbit]

hello

[obtuse orbit]

question 28

that’s what i’ve tried so far, but ive been unsuccessful

maybe like that? i still don’t think it works

.close

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[mental yoke]

Hello. I want to ask a question about the Fermat Numbers. Why can I define a Fermat Number into 6*k-1. I cannot comprehend why

[edgy path]

i need help with 1789 times 9

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[full forum]

@torn jolt Has your question been resolved?

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[grim skiff]

Nothing this is correct now

And?

That's a valid answer

You can have a no solution matrix

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[restive roost]

my working seems wrong for this question

3 cases

case 1: no 4 anywhere
case 2: a 4 anywhere but the end
case 3: a 4 at the end

4 total square numbers from 0-10, 3 not including 4

so case 1, 9^8 * 3

case 2, 8 * 9^8 * 3

case 3, 9^8 * 1

and then summate

whichb of those are wrong?

I feel like its case 2

[sharp lagoon]

Why do you think it's case 2?

[restive roost]

because case 1 and case 3 seem robust

wait

i'm not picking the 4 in case two

it should be 8 * 1 * 9^7 * 3

[fathom saddle]

I agree that looks better

[restive roost]

this is the other one I was having trouble with

BBC can be posititioned in 7 different places

and the subword BBC, the first letter has 3 options, the second letter has 2 options, and the third letter has 2 options

9 - 3 is 6

so i thought it would be

7 * (3 * 2 * 2) * 6!

[paper light]

not rlly its the same thing right?

since all of the Bs are the same

and all the Cs are the same

[restive roost]

hmm

[paper light]

So if actually theres only one way to form BBC

[restive roost]

i guess

[paper light]

and if i consider BBC as one letter

then im left with AAABCD

[restive roost]

ahhhh that makes sense

and so it would be

7 * 6! / 3!?

[paper light]

wait

but there might be duplicate cases

hmmm

or is there

[restive roost]

i think the only duplicate case would be the 3 A's right, thats why the 3!

[paper light]

yes

ok yeah theres no duplicates except for the A's

[paper light]

I think you need to multipy this by the number of arrangement of BBC

[restive roost]

thats why i had the 3 * 2 * 2

[paper light]

no what i meant is the number of places u can put BBC

like (BBC)AAABCD and AAABCD(BBC)

[restive roost]

yee

i think it would be 7 right

[paper light]

yes i think so

[restive roost]

i think i get it ty 🙂

theres just one more of the same style

i think the difference with this one is that

with the previous one, one occurence of BBC means there cannot be another occurence of BBC

but for this one

there are 3x A, 3x B, 2x C

so CBA could appear twice

[paper light]

i dont think it would make a difference

im not too sure

[restive roost]

its like a practice test so i can see the answers

7 * 6! / (2! * 2!) was incorrect

so something is finicky

[paper light]

is the correct answer more or less than ur answer

[restive roost]

sorry correct answer isnt given

it just tells you whether you;'re right or wrong

[paper light]

ohhhh

ok

Im not sure perhaps some other helpers can come and help u

sry

[restive roost]

no worries 🙂

ill repost cheers for other help

.close

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[shrewd hamlet]

Show what you’ve tried

[swift sand]

oh I got it sorry

.close

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[crimson lance]

dude

[echo crest]

sry

[crimson lance]

np

[crimson lance]

how does the absolute value affect 1/n - 1/n^2

i thought it would make the - into a + but i think thats just wrong math lmao

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@crimson lance Has your question been resolved?

[slate violet]

you just have $|1/n - 1/n^2|^{n \cdot 1/n} = |1/n - 1/n^2|$

[glossy valve]

south

[slate violet]

the reason why you can ignore the absolute value sign here is

$\frac{1}{n} - \frac{1}{n^2} \ge 0$ for all $n \ge 1$

[glossy valve]

south

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[burnt socket]

i'm not sure what this means nor how to begin

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[misty lake]

I'm currently studying for a physics quiz involving the maxwell equation and induced electric fields, I'm a little confused about how the equations are to be used. The related equations are in the picture sent If someone can go through them with me to help me understand them that would be amazing!

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@misty lake Has your question been resolved?

[misty lake]

<@&286206848099549185>

<@&286206848099549185>

[misty lake]

<@&286206848099549185>

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[astral fractal]

How do we do these <@&286206848099549185>

[astral fractal]

We were also given this

[glass crystal]

first find the inf of the set

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@astral fractal Has your question been resolved?

[astral fractal]

So 1/11?

[glass crystal]

whats (1/11)^2?

[astral fractal]

The next term?

[glass crystal]

whats the value i meant

[astral fractal]

1/121?

[glass crystal]

yeah

so can 1/11 be the inf of the set?

if 1/121 is inside

[topaz geode]

This MATA31?

[astral fractal]

I figured it out. Thx

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[full forum]

@torn jolt Has your question been resolved?

[rough plaza]

because k-9 are the number of terms

you can plug in n=10 to see

it considers the case when r=10 as the first term

it is just because it's simpler to calculate

you can use a as the 10th term for the second if you want

it depends on your habit dealing with such questions

you have a formula when geometric series add up all the way to infinity

yep

for simplicity let's consider 2 cases

for Q8

Method 1:
as written in the answer,
we start from r=10 to k,
so there are (k-10+1) terms.

and
a=first term in the sequence,
l=last term in yhe sequence.
the formula is
(number of terms)*(first term + last term)/2

you can try to count some with numerical values

e.g.

1,3,5,7,9

term 1 to 5

so we have 5-1+1 terms

11,13,15,17,19

term 6 to 10

so we have 10-6+1 terms

so, we have
ending term number - starting term number + 1 = number of terms

k is not sum of the terms

yes

yes

[rough plaza]

read this again.

yep

that's correct

so, k-9 is the number of terms for Q8

oh this

why is it 20

and this is 10

i bet it's a typo on the Q

otherwise it's correct

if you have any other questions regarding similar topics, feel free to close this channel and open a new one, since it's another question:)

and bot will pin your first message, so it'll be easier to get help

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[worn matrix]

it would be nice if if you tell us what the actual question is rather than just posting a quesetion

@fervent kelp

[fervent kelp]

Is there a point on the graph where the tangent slope is equal to -1? If so, find it

is the question it just took me time to translate it

[worn matrix]

oh ok

[torn jolt]

remember what f'(x) gives us?

uhh, as in why will you be differentiating in the first place

[fervent kelp]

It gives 2x +9

[torn jolt]

.

it gives the gradient function, we went through this a while ago

[fervent kelp]

to find the slope that shows the angles of the tangents

[torn jolt]

do you understand i said

[fervent kelp]

I did f'(-1) = -2 -9

I understand what it gives just not how to phrase it

[torn jolt]

why x=-1?

your question
point on the graph where the tangent slope is equal to -1

[fervent kelp]

because I need to find a point on the graph that the tangent slope is equal to -1

[torn jolt]

yes

so what will we let f'(x) be equal to

[fervent kelp]

-1

[torn jolt]

okay go on

[fervent kelp]

and this is what I did

[torn jolt]

no you did not

f'(-1) is not the same as f'(x) = -1

[torn jolt]

here you're assuming you already know what the x value is

which is what we want to find

[fervent kelp]

I thought if I swap x then I should do it with all the x's

so its f'(x) = -11

[torn jolt]

what? wdym

[torn jolt]

where is -11 coming from now

[torn jolt]

here you just found the slope when x=-1

which is not what we're looking for

[fervent kelp]

-11 is the slope?

[torn jolt]

reread your question and explain to me what we're looking for

[fervent kelp]

I am looking for (x,y) = f'(-1)

[torn jolt]

no no

[fervent kelp]

or f'(x) when the tangent slope is -1

[torn jolt]

f'(x) gives us the slope/gradient

you input a point on the graph, lets say i want the gradient when x=2 i'd f'(2) and that would give me the slope/gradient

if this is still unclear you might want to review your calc fundamentals

[fervent kelp]

I dont understand

[torn jolt]

what part is unclear

[fervent kelp]

what to do with -1

[torn jolt]

what did i say f'(x) gives us

[fervent kelp]

I understand that f' gives slope

[torn jolt]

and what is -1

[fervent kelp]

the slope

[torn jolt]

and we want to find the point right

so f'(x) = -1

[torn jolt]

finding x

[fervent kelp]

do I use this? f'(x) = 2x +9

[torn jolt]

yes

[fervent kelp]

f'(x) = -2 +9

[torn jolt]

?

what did you do now

[fervent kelp]

2x -> 2* -1

[torn jolt]

[fervent kelp]

forget it..

[torn jolt]

why would x=-1

[fervent kelp]

so its -1 = -2 +9

[torn jolt]

f'(x) = 2x +9

finding x when slope is -1 so

-1 = 2x + 9
....

go on

[fervent kelp]

x= -5

[torn jolt]

okay now lets find y

[fervent kelp]

I put this in the original equation

[torn jolt]

you then plug that x into the original

[torn jolt]

yes

[fervent kelp]

I got y = - 70

[torn jolt]

hmm?

,w (-5)^2 + 9(-5)

[glossy valve]

Results provided by WolframAlpha

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

Requested by 0izne

[fervent kelp]

right

(-5, -20)

[torn jolt]

yes

anything else you dont get?

[fervent kelp]

I understand it thank you

I will do more similar questions now

.close

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[torn jolt]

all the best

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[novel dragon]

Hello

[novel dragon]

<@&286206848099549185>

[odd lantern]

bro second one

[odd lantern]

for this

[novel dragon]

Yea

[pseudo frigate]

You clearly did not read #❓how-to-get-help as you say you did

!15m

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Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[novel dragon]

He’s alrighty helping me

Why you so mad

[pseudo frigate]

@novel dragon is this a quiz

[odd lantern]

He's just a math

[novel dragon]

No it’s a warm up for the test Tuesday

[novel dragon]

What about this one

??

Help or what

[pseudo frigate]

You seem to misunderstand how this works.

Helpers are volunteers. They have no obligation to help you

Wait patiently and someone will probably help you, eventually

[novel dragon]

Oh

Ok

[pseudo frigate]

While you wait, say what you have attempted and where you got confused. Show your work

This will make it more likely that someone help you

[novel dragon]

There’s no work To be done

It’s graphing functions

[pseudo frigate]

Show what you have attempted to graph.

[novel dragon]

I haven’t been to school bc of covid so I haven’t been able to learn it

^

[glass crystal]

whats the definition of a function?

[novel dragon]

I got no clue

Stuck at home with covid very complicated

[glass crystal]

look it up in your course perhaps

or class material

[novel dragon]

It’s something relationship involving one or more variable

[glass crystal]

it should be written somewhere

you cant guess maths

use what your teacher gave you

[novel dragon]

How you do this

[novel dragon]

Can’t give me nun I got covid

[glass crystal]

you got material

like a book

or a pdf no?

[novel dragon]

Nope

All he told me was test code

[glass crystal]

you got until when to do this?

[novel dragon]

2;50

[glass crystal]

i mean if you dont have material and no knowledge how are you expected to do this

[novel dragon]

I figured the problem out

[glass crystal]

how did you do it?

its still up

[pseudo frigate]

This user has been banned

[glass crystal]

oh ok

[pseudo frigate]

.close

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[glass crystal]

understandable

they are mping me

[pseudo frigate]

DMing?

Yeah we've banned them for that reason.

They shouldn't be able to anymore unless you share a different server with them

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[tired oak]

Suppose $\sigma=(a_1,a_2,…,a_n)$ and $\tau=(b_1,b_2,…,b_m)$ are cycles that are not disjoint, but $a_i\neq b_j$ for one or more $(i,j)$. Prove $\sigma(\tau(x))\neq\tau(\sigma(x))$.

[glossy valve]

jsidind810

[tired oak]

$\sigma$ and $\tau$ are in $S_X$

[glossy valve]

jsidind810

[trail barn]

you can reduce to a case where $a_1=b_1$ and $b_2\notin{a_1,\dots,a_n}$ (justify this), and then consider $x=a_1$

[glossy valve]

Edward II

[trail barn]

you could also do a proof by contradiction, so show e.g. that if the cycles are not disjoint and στ=τσ then you cannot have your second condition on the cycles

or equivalently assume second condition and στ=τσ, and then show the cycles must be disjoint

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[tribal tide]

Hi, I need help with a linear algebra question.
Questions were originally in french but I translated them into English. I already found the solutions for a) and b). a) = (1800, 1800, 950) and b) = (38400, 51200, 32000). I don't really understand how to solve c) and d)

[tribal tide]

here are the solutions for a) and b)

here is my attempt at doing c)

not sure if I used linear combination and if I gave the the result in a vector of R3 properly

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.close

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@wicked frost Has your question been resolved?

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[tired oak]

do u think u could help me a bit more?

[full forum]

@tired oak Has your question been resolved?

[tired oak]

<@&286206848099549185> or anyone else

[trail barn]

I was trying to go to bed but have fully woken up (unrelated to ping obviously)

Which route do you think you'll find easiest

[tired oak]

Im not sure man

id like to do it directly

without cases

[trail barn]

sure

btw, can you send the original version of what you sent as $a_i\neq b_j$, because that can't quite be the right condition

[glossy valve]

Edward II

[trail barn]

(e.g. (1 2)(2 1) obviously don't satisfy what you are trying to say, but they satisfy what you did say as 1 =/= 2)

[tired oak]

it just means {a1,a2,…,an} and {b1,b2,…,bm} are not disjoint

[trail barn]

cool

[trail barn]

what you originally sent just means I can pick an ai and a bj that are different, which isn't really at all the same as this

[tired oak]

oh ye sry

[trail barn]

ok I think assuming that $\sigma\tau = \tau\sigma$ and ${a1,...,an}, {b1,...,bm}$ not disjoint and then showing ${a1,...,an}={b1,...,bm}$ is the way to go

[tired oak]

ye

[trail barn]

we want to show that $a_i\in{b_1,\dots,b_m}$ for any $i$

[glossy valve]

Edward II

[tired oak]

mhm

[glossy valve]

Edward II

[trail barn]

(I missed a 'not')

what do we know about how ai behaves?

[tired oak]

call {a1,..,an}=A and {b1,…bm}=B

btw

uh

well its a cycle so

its mapped to by a_{i-1}?

$\sigma(a_{i-1})=a_i$

[glossy valve]

jsidind810

[trail barn]

and how does this interact with $\tau$?

[glossy valve]

Edward II

[tired oak]

well in the $\tau\sigma$ case

[glossy valve]

jsidind810

[tired oak]

if $a_i\in B$

[glossy valve]

jsidind810

[tired oak]

and if it is equal to say $b_j$ then $\tau(a_i)=b_{j+1}$

[glossy valve]

jsidind810

[trail barn]

if $a_i\in B$ we've done what we wanted, because we're showing $a_i\in B$

[glossy valve]

Edward II

[trail barn]

.

[tired oak]

so how do we show ai in b

[trail barn]

so the other case is the more interesting one

[tired oak]

oh

sry yes

[trail barn]

we show that's it's not not in B

[tired oak]

if $a_i\notin B$ then

[glossy valve]

jsidind810

[tired oak]

uh

what happens

is that even in the domain of $\tau$

[glossy valve]

jsidind810

[trail barn]

yes because $\tau$ has domain $X$, but it doesn't do anything to it

[glossy valve]

Edward II

[tired oak]

oh sry yes

but then $\tau(a_i)$ still in A

[glossy valve]

jsidind810

[trail barn]

you can be more precise if $\tau$ doesn't do anything to it, rather than 'in $A$'

[glossy valve]

Edward II

[tired oak]

$\tau(a_i)=a_i$

[glossy valve]

jsidind810

[trail barn]

so, we've established $\tau(\sigma(a_{i-1}))=a_i$

[glossy valve]

Edward II

[trail barn]

what else does the LHS equal (look at the assumptions)

[tired oak]

switch sigma and tau

[trail barn]

mhm

[muted junco]

Can I please get some help

[trail barn]

!occupied

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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).

[tired oak]

thinking? @trail barn

[trail barn]

oh I was thinking you were thinking

we now have $\sigma(\tau(a_{i-1}))=\tau(\sigma(a_{i-1}))=a_i$

[glossy valve]

Edward II

[trail barn]

can you draw a conclusion about $\tau(a_{i-1})$?

[tired oak]

well we have $\tau(\sigma(a_{i-1}))=\sigma(\tau(a_{i-1}))$

[glossy valve]

Edward II

[tired oak]

wait

no?

[trail barn]

remember $\sigma$ and $\tau$ are bijections by definition

[glossy valve]

Edward II

[trail barn]

so in particular, they're injective

or one to one

(same meaning, different words)

[tired oak]

ik

just tell me

lol

[trail barn]

$\sigma\tau(a_{i-1}))=a_i$

[tired oak]

if it wasnt late i would have the capacity to think

[glossy valve]

Edward II

[trail barn]

apply $\sigma^{-1}$ to both sides

[glossy valve]

Edward II

[tired oak]

oh

tau is identity

[trail barn]

well, $\tau(a_{i-1})=a_{i-1}$ it doesn't necessarily mean it's the identity

[glossy valve]

Edward II

[trail barn]

but it does imply something about $a_{i-1}$

[glossy valve]

Edward II

[tired oak]

it cant be in B?

[trail barn]

exactly

[tired oak]

so..?

[trail barn]

so by uhh I guess induction, what do we know about any element of A if we assume this for just one element

(treat subscripts as looping around)

[tired oak]

idk

[trail barn]

well if ai is not in B, then a_(i-1) isn't in B, and then a_(i-2) isn't in B, and then so on for all the elements of A

but! we assumed that some element of A is in B, as we assumed they're not disjoint

[trail barn]

contradiction

[tired oak]

are we proving contrapositive or by contradiction

i thought we were proving A=B

[trail barn]

we're proving that ai is in B for any i by contradiction

[tired oak]

what did we assume

i thought we only assumed $\sigma\tau=\tau\sigma$

[glossy valve]

jsidind810

[trail barn]

For the whole question, we assumed sigma and tau can be switched, and that A and B are not disjoint.
We want to show A = B, so in particular A is a subset of B
We took arbitrary ai in A, and assumed it's not in B, and then showed this would mean A and B are disjoint, a contradiction.
So ai is indeed in B, and A is a subset of B

[tired oak]

but not(assumption1 and assumption2) = not(assumption1) or not(assumption2)

[trail barn]

if assumption 1, then we've forced not(assumption 2), so that not(assumption 1 and assumption 2) is true

otherwise, not(assumption 1) directly

I guess you could also think of it as a contradiction, that if [assumption 1 and assumption 2 and (sigma and tau commute)] then not(assumption 2), and we just never use assumption 2 in the proof

well

swap the numbering in the order you introduced them

[tired oak]

can you tex this entire proof lol

i need to sleep and i cant think

[trail barn]

I'll compile the bits in the chat

[tired oak]

pls make it as cohesive as possible

thx man

[trail barn]

https://cdn.discordapp.com/attachments/717123612125167628/1167613249177522236/365017527484416001.png?ex=654ec393&is=653c4e93&hm=0fde40ef64bf58d889751850d65a86ca8509da476979af0259c30bd543cc7d42&

https://cdn.discordapp.com/attachments/717123612125167628/1167614326127988887/365017527484416001.png?ex=654ec494&is=653c4f94&hm=9f5fe5c567cdeb2a157d052231c2329bf967f2c9c11342ca1ff72e3dd14d976b&

[trail barn]

https://cdn.discordapp.com/attachments/717123612125167628/1167614772347416666/365017527484416001.png?ex=654ec4fe&is=653c4ffe&hm=5c3495682cd292372dac1dd5d66027912b5ce223f8d609e468c7bd75926f357f&

[trail barn]

correction: induction is used either way, the question is actually how to understand a_0 when we get there by going down if we start with arbitrary i

I might also explain why exactly a_{i-1] isn't in B in a proper write up

(As for the centre of D_8, it suffices to check if x commutes with the generators to check if x is in the centre. The reflection / generator of order 2 (usually s or sigma I think?) is the problem here..., you should get that the centre has either one or two elements depending on n)

[tired oak]

well i tried to see under what conditions

$r^nsr^ms=r^msr^ns$

[glossy valve]

jsidind810

[tired oak]

but i dont really know where to go from there

i guess take off the right s $r^nsr^m=r^msr^n$

?

[glossy valve]

jsidind810

[trail barn]

I wouldn't use $n$ for a power of $r$ because $n$ is involved in the dihedral group

[glossy valve]

Edward II

[trail barn]

btw are you using $D_8$ for symmetries of a square / 8 elements or of an octagon / 16 elements

[glossy valve]

Edward II

[tired oak]

yse 16 elements

yes

order 16