help-0

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[stuck sinew]

hi can someone help me with this: i asked for help before and i got the hint to use douvble angle formulas to rewrite th equation but i dont know how to. i'm stuck on which double andle identity to use and how to.

[wheat isle]

Weierstrass substitution maybe

[pallid scarab]

by double angle formulas, cos(x) = ?

and sin(x) = ?

[stuck sinew]

thats the thing i dont know what to use

[pallid scarab]

cos(x) = cos(2 x/2)

[stuck sinew]

ive heard of it but i dont know much about it

[pallid scarab]

does this make it clearer on what to use?

[wheat isle]

u = tan(x/2)

[stuck sinew]

i think so i need to try it

[pallid scarab]

I've said this previously to them in another channel, I think they don't know the tan(x/2) identities

so letting them go through it on their own

[wheat isle]

oh ok

[stuck sinew]

Is this correct?

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[viscid oriole]

call me crazy

[viscid oriole]

but if i have like (20-4)^2

-4 is squared

to be positive 4?

[ocean hawk]

No

[viscid oriole]

solve inside first

[ocean hawk]

I'd recommend that, yeah

[viscid oriole]

what do you call

the problems that are like csc(cos^-1x)

or sin(csc-sin)

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[meager cedar]

what does this ceiling funtion x mean in this funtion

[meager cedar]

like the funtion is repeated ceiling x times ?

like f(f(....(x))))) ?

[cinder compass]

interesting

yes it could mean that

or differentiation multiple times, depends on context

[meager cedar]

ohkk thanks

.close

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[pallid scarab]

it's definitely repeated ceiling x times

if it were differentiation, you would need parentheses

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[livid hazel]

I have Seen this in the other Channel and I’m interester what would the Solution be :
Find all Natural Numbers for Which : (x+2)^4 -x^4 = y^3 .
I have Found no solutions can someone confirm and maybe present a Solution themself

[round sable]

Well, you could try to prove that there are no such points.

That would also be an interesting result, if it's true

[limpid turret]

Recall that $a^4-b^4=(a-b)(a^3+a^2 b+ab^2+b^3)$

[ocean seal]

SWR

[lament forge]

it is at z = 0, but that corresponds to x = -1 which isn't a valid solution to the original problem

[modern sedge]

8x^3 + 24x^2 + 32x + 16 = y^3 shows that y would have to be divisible by 2

[round sable]

yeah there is a solution (-1, 0) to the original problem, but that's the only one I can find

[modern sedge]

if that happens, y = 2k and x^3 + 3x^2 + 4x + 2 = k^3

[lament forge]

z and z^2+1 are coprime, so they'd have to both be cubes, but then z^2 is also a cube so we have two consecutive cube numbers, which only happens at 0,1 (or -1,0 but that doesn't work because we can't have z^2 = -1)

[modern sedge]

Wait i misread one line

the proof is wrong, yeah i thought it was too easy

[round sable]

Yes, this is true

[livid hazel]

Anyone has a formal Proof to the Problem Statement

[modern sedge]

kepe and bee have one ig

[round sable]

What bee posted can be formalized

It basically already is

[thick lynx]

The approach is mostly taken from https://math.stackexchange.com/questions/1384931/diophantine-equation-without-using-fermats-last-theorem?noredirect=1, slightly adapted

[thick lynx]

I guess in its core, this is basically the same as this

[livid hazel]

This should be completalo formal now that I Read over it

[thick lynx]

Also include that k must be natural

[thick lynx]

One minor correction:

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[livid hazel]

What if z>0 then what happens with k

[livid hazel]

.reopen

[ocean seal]

Kepe

[thick lynx]

What do you mean what happens with k?

k is just some natural number

[ocean seal]

Kepe

[livid hazel]

How do we know there aren’t any other solutions I mean what if k=/z

[thick lynx]

Sry I wrote $\geq$ before, but we can actually say that it's $>$ with the justification I gave right now

Assuming in your definition of natural numbers, 0 isn't included.

[lone heart]

@livid hazel Has your question been resolved?

[thick lynx]

This is basically the same argument as https://math.stackexchange.com/a/1384937

Maybe that is clearer

[livid hazel]

It says that the solution is (0,0) but it should be (-1,0) right

[thick lynx]

Yeah, while we have the same equation as them, we got that by setting z = x + 1.

And we're really looking for x and y here

x = z - 1 = 0 - 1 = -1

[livid hazel]

Yh yh

Alright I think ur proof is formal enough for me to understand

Is there something to add to it

[thick lynx]

I don't think so except to specify that k is natural at the beginning

[livid hazel]

There are no special cases or anything

[thick lynx]

Nope. The most I would do is justify the last part a little more maybe

[livid hazel]

How would so do that

I*

[thick lynx]

The last paragraph of this

[thick lynx]

Uh

[thick lynx]

First of all, do your natural numbers include 0 or not?

[livid hazel]

Yh that should get added

But normally no

[thick lynx]

It mostly depends on the course

[thick lynx]

If not then you can change the >= sign to a > sign here

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[livid hazel]

.reopen

[lone heart]

✅

[thick lynx]

@modern sedge Since you just typed, did you want to say something regarding this?

[modern sedge]

Nope, i was just typing what I did to see if i can actually make it to a proof, i couldnt. And then you sent yours which is much better

[thick lynx]

Ah, alright. I thought you had some correction or something

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[shadow verge]

How to solve this?

[mossy laurel]

first write the equation defining of the y-intercept of function g

[shadow verge]

ax^2 +bx+c right?

[mossy laurel]

No, this is giving you the form of function g, not it's y intercept

[shadow verge]

Oh

Idk

[mossy laurel]

how do you define the y intercept?

[boreal jetty]

i gues you first want to solve f(x)

[shadow verge]

When x= 0

[mossy laurel]

yes exactly, it's g(0)

now can you express this using f?

[shadow verge]

F(0+4)?

[mossy laurel]

yes

f(4)

Now you know what you want to compute

So your goal now is to compute f(x) in general (because you don't have f(4) in the table)

[shadow verge]

Don't we have to find g(x) first,

The function

[mossy laurel]

To find g(x), you just need to find f(x)

[shadow verge]

I got that the vertex is (6, z)

-6*

[boreal jetty]

and you can find f(x) by using substitution
y = ax^2 + bx + c

155 = a(-9)^2 + b(-9) + c
227 = a(-3)^2 + b(-3) + c
155 = a(3)^2 + b(3) + c

[shadow verge]

Oh

[shadow verge]

Is there a faster way or no

[boreal jetty]

are you aloud to use computer solvers

[shadow verge]

I see

[boreal jetty]

allowed*

?

[shadow verge]

I thought there was a faster way besides substituting

[boreal jetty]

not that I know of.

[boreal jetty]

three equations and three unknown variable.

do you need more help?

[lone heart]

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[vapid laurel]

Can someone simplify this please

[main junco]

?

Write it out as partial fractions

[vapid laurel]

It’s Indefinite integral

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[turbid minnow]

see for me as high school student i would tell you to divide by x^4

so 3-2-2x/x^4

1-1/2x^3

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[dim terrace]

how do i do number 4 and why is number 6 wrong

[pallid scarab]

to do number 4, factor

[cinder tundra]

What is 4i^2?

@dim terrace

[dim terrace]

41^2

and would number 4 be option 4?

@cinder tundra

[cinder tundra]

No you wrote

4 * i^2

What is i^2

[dim terrace]

huh

[cinder tundra]

Do you know what is i?

[dim terrace]

imaginary number

[cinder tundra]

i = sqrt(-1)

[dim terrace]

oh

LMAO

uh

[cinder tundra]

So what is i^2?

[dim terrace]

-1?

[cinder tundra]

And 4i^2?

[dim terrace]

-4

[cinder tundra]

You wrote 6 + 4i^2 is 10

But according to that

It should be 2

Do you see your mistake now?

[dim terrace]

wait where are you seeing this

that?

[rare crane]

[ocean seal]

paperpanda

[dim terrace]

how would it be 2

[cinder compass]

6 - 4 = 2?

[rare crane]

$6 -(3x - 2i)^2$
$6 -(9x^2 - 12xi + 4i^2)$

[ocean seal]

paperpanda

[dim terrace]

im talking about the answer

[rare crane]

So you get
$6 - 9x^2 - 12xi - (-4)$

[ocean seal]

paperpanda

[rare crane]

which would be
$ -9x^2 + 12xi + 10$

i think you did corrent

[dim terrace]

thats what im saying

[rare crane]

you should ask the teacher agian.

[dim terrace]

for number 4, would the answer be 4?

[rare crane]

you can factor out w,
w (x^2 +1) = 0
Since w is non zero you can divide both sides by w.
x^2 = -1
so x = sqrt of (-1) which is +-i

[dim terrace]

okayokay thanks

[lone heart]

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[lone heart]

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[alpine sable]

If this isn't the right place for this, I'm new and I read the how to so maybe I just don't understand. **Calculus 2, center of mass of a planar lamina region. **This is kind of a silly question, but I am pretty sure my professor's answer is backward here. In the notes the equations for x bar and y bar were swapped. So I think it's supposed to be (1/2, 8/5) not the other way around.

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[alpine sable]

.reopen

[lone heart]

✅

[alpine sable]

<@&286206848099549185>

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@alpine sable Has your question been resolved?

[lone heart]

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[uncut matrix]

i need help with functions

[uncut matrix]

One winter night, it snowed heavily in Lithuania. During the night, the thickness of the snow cover reached as much as 40 cm. The next morning the weather warmed up and the snow began to melt. The snow cover thinned by 4 cm every hour. Which linear function expresses the dependence of snow cover thickness y = f(x) (cm) on time x (in hours)?
A f(x) = 4-40x
B f(x) = -4x-45
C f(x) = -4x + 40
D f(x) = 40x-4

[cinder compass]

any ideas?

[uncut matrix]

i know for sure it isn't A

[cinder compass]

why

[uncut matrix]

because b ain't got an x in the back lmfao

but

i have a feeling it might be C but i'm not sure

[cinder compass]

what should f(0) be

[uncut matrix]

not given

[cinder compass]

the initial thickness of snow

[uncut matrix]

the initial thicness is 40

[cinder compass]

it's given

[uncut matrix]

which i assume is b

[cinder compass]

yea

[uncut matrix]

because that's where it starts thinning

[cinder compass]

yes

which options satisfy that

[uncut matrix]

only c

[cinder compass]

true

the answer is C

every hour it gets reduced by 4 so there's a -4x

[uncut matrix]

yea

and the function is decreasing?

[cinder compass]

yes

[uncut matrix]

alright

i have like

a whole page i needa finish, as preperation for a big test tomorrow

[uncut matrix]

would you mind helping me?

[cinder compass]

maybe not

i'd say just post ur q and if someone wants to help they will

[uncut matrix]

alright

i mean i already found

something i'm not sure about

1 x + 3. Given a linear function f(x) = -3x-
(a) Sketch the graph of this function.
b) Apskaičiuokite f(27).
c) Determine for which values ​​of the argument f(x) = 10.
d) Calculate for which values ​​of x the values ​​of the function are negative.
e) Determine whether the point K(-27; 12) belongs to the graph of the function.
f) Determine the coordinates of the point where the graph of the given function crosses the x-axis.

the function got translated wrong 1 sec

(a) Sketch the graph of this function.
b) Calculate f(27).
c) Determine for which values ​​of the argument f(x) = 10.
d) Calculate for which values ​​of x the values ​​of the function are negative.
e) Determine whether the point K(-27; 12) belongs to the graph of the function.
f) Determine the coordinates of the point where the graph of the given function crosses the x-axis.

i know how to do A and B

about c i'm not fully 100% because

ok so what i'm understanding is that

---

you have to find where x is when y is at 10?

[ocean hawk]

yes

[uncut matrix]

alright

on d) i would do for example - infinite till f(x) > 0?

wait i mean

[ocean hawk]

you want to find the values of x where f(x) < 0

[uncut matrix]

yes

mb i typed it wrong

so would it be like infinite until f(x) < 0 ?

[ocean hawk]

did you try solving the inequality?

[uncut matrix]

wait let me draw the graph, i can't do anything rn without it

[ocean hawk]

ok. but you can do it without the graph as well

[uncut matrix]

my brain doesn't work without one lmao

[cinder compass]

good habit to graph, or atleast imagine it

[uncut matrix]

this is what i'm imagining

[ocean hawk]

looks about right

[uncut matrix]

would i have to do like this

( some number that i get ; minus infinite)

to answer d) ?

[ocean hawk]

you just need to solve f(x) < 0. so you just substitute the equation of f(x)

[uncut matrix]

wait what

[ocean hawk]

part d) is asking you to find the values of x where f(x) < 0

you are given f(x)

[uncut matrix]

yep

[ocean hawk]

you need to solve the inequality for the set of x values that satisfy it

[uncut matrix]

so how would that be done

[ocean hawk]

well substituting the equation into the inequality gives you -(1 / 3) x + 3 < 0

[uncut matrix]

which would be 3 < 1/3x?

[ocean hawk]

yes

and simplify

[uncut matrix]

what would i do with this

result

[ocean hawk]

isolate x, and you get what?

[uncut matrix]

1 < 1/9x

9 < x

[ocean hawk]

yes

so those are the values where f(x) is negative

[uncut matrix]

oh?

[ocean hawk]

yes

[uncut matrix]

so i js have to do algebra in that case scenario

[ocean hawk]

yeah

[uncut matrix]

alright

for D

how would i calculate that mathematically

[ocean hawk]

didn't you just do D?

[uncut matrix]

e i mean

sorry

[ocean hawk]

so for e), you're given a point and you want to check if it's on the line

[uncut matrix]

yep

[ocean hawk]

the line being y = (-1/3) x + 3

[uncut matrix]

yep

[ocean hawk]

so you just plug in the coordinates of the point

see what you get

[uncut matrix]

ohh

wait i get it

so

12 = (-1/3) (-27) + 3

is that right?

[ocean hawk]

yes

[uncut matrix]

it would become 9+3

which means the coordinates do belong to the function graph?

[ocean hawk]

yes, since you get 12 = 12

[uncut matrix]

yea alr

hm

i could probably do f) in my notebook, if i drew the function graph and manually found where it crosses the x-axis

but i have no clue how to calculate it, since my teacher didn't explain shit

[ocean hawk]

well what are the coordinates for any point on the x-axis?

the y coordinate specifically

[uncut matrix]

0

[ocean hawk]

yes

[cinder compass]

guaranteed lifetime help if i become admin

[uncut matrix]

broo 😭

nah sorry

[cinder compass]

lmao okay nw

[uncut matrix]

okay so i know that f(x) = -(1/3)x + 3

wait so

is it literally just

0 = -(1/3)x + 3

[ocean hawk]

yup

[uncut matrix]

oh

so (1/3) x = 3

so x=9?

thats when it crosses the x-axis?

[ocean hawk]

yup

[uncut matrix]

oh alr

ima quickly do everything else and i'll ping u if i need help

From the sketched graph of the linear function f(x) = 3x - 3, determine:
a) range of function values;
b) the coordinates of the points where the graph of the function crosses the Ox axis and the Oy axis;
c) the interval in which the values ​​of the function are positive;
d) values ​​of x with which f(x) ≤ 0;
e) the value of the function f(2).

nvm i already need help wtf is the range of function values

@ocean hawk

[ocean hawk]

it's all the y values it can be

[uncut matrix]

ahh

[ocean hawk]

so for linear functions, it's going to be all real numbers

[uncut matrix]

(-3;0) ?

[ocean hawk]

that's the y-intercept

[uncut matrix]

waitt

uhh

(0 ; -3) and (1 ; 0)

this?

[ocean hawk]

lol yeah sorry, (0, -3) is the y intercept

[uncut matrix]

alr

[ocean hawk]

and (1, 0) is the x intercept

[uncut matrix]

wtf is b

[ocean hawk]

you just did b

[uncut matrix]

what's a then

😭

[ocean hawk]

the range is all real numbers

[uncut matrix]

oh yeah when i read it in my native language i js did b

[ocean hawk]

from -infinity to +infinity

[uncut matrix]

yeah that's what i wanted to say but

is that really the answer?

[ocean hawk]

for a? yeah, all real numbers

[uncut matrix]

oh alr

c would be

(0 ; +infinite)?

[ocean hawk]

No, it's asking when is f(x) > 0

[uncut matrix]

so

when y > 0

when x ∈ (1 ; +infinite)?

[ocean hawk]

Yup

[uncut matrix]

ah

ok so d would be

y =< 0

and i have to find which x'es

are for it

[ocean hawk]

Yeah

[uncut matrix]

so same thing

when x ∈ (1 ; +infinite)

[ocean hawk]

No, the inequality sign is the opposite of the one before

[uncut matrix]

no it's not an innequality sign it's like

y is equalled to or is less than 0

[ocean hawk]

That's an inequality

[uncut matrix]

oh wait it is

mb

wait so how would i solve this one

[ocean hawk]

Same as any other inequality. It's asking to solve f(x) <= 0

[uncut matrix]

3x - 3 <= 0?

[ocean hawk]

Yes

[uncut matrix]

oh ok

3x <= 3

x <= 1

[ocean hawk]

x <= 1, not equal to

[uncut matrix]

oh yeah ytou right

and the last one is easy

[ocean hawk]

Yup

[uncut matrix]

f(2) = 3(2) - 3 = 6 - 3 = 3

ok

you have to find k

what i would do here

is divide

1/3

and that's k

lmao

f(x) = (1/3)x + 1

[ocean hawk]

What k? Are you given an equation?

[uncut matrix]

this graph

that i sent

[uncut matrix]

k is the angle cofficient

like

how many y units the line moves per 1 x movement

for us its f(x) = kx + b

[ocean hawk]

Oh, so k is the slope of the line

[uncut matrix]

ye

so 1/3 right

[ocean hawk]

No

[uncut matrix]

wha

😭

WAIT

-(1/3)

mb

[ocean hawk]

no

[uncut matrix]

why not

i'm gonna cry

lmao nah jk

but why tf

[ocean hawk]

slope is "rise" over "run". i.e., the change in y divided by change in x

so if you change x by 1 (e.g., going from x=0 to x=1), how much does y change

[uncut matrix]

oh.

it changes by 3

so it's just -3

[rose sigil]

dun cri

[uncut matrix]

ok 9.

Write a linear function in the formula if:
a) f(0) = -2, or f(2) = -1;
b) the graph of the function includes the points A(2; 0) and B(-0.5; -5).

wouldn't you have to do a system

[ocean hawk]

Yup

I gotta walk my dog but if you get stuck someone should help

[uncut matrix]

{ k (0) + (-2)
k(2) + (-1)

[uncut matrix]

lright

[uncut matrix]

yo bro help 😭

[rose sigil]

sorry i'm too dumb to help with this

[uncut matrix]

oh

alr

[lone heart]

@uncut matrix Has your question been resolved?

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[uncut matrix]

(mostly resolved)

[lone heart]

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[stiff ore]

I am completely stuck on this one. It’s from Classic Set Theory by Derek Goldrei.

[stiff ore]

He says to use recursion, but I still need a way to map a natural number to the choice function. I’m not seeing how that makes the problem any easier.

[lament forge]

i don't know what you mean by "map a natural number to the choice function" but it probably isn't what the hint meant

the type of recursion it's referring to is that to define a function $g : \mathbb{N} \to X$, all you need is to specify, for each $n$, what the value of $g(n)$ is, given the values of $g(k)$ for all $k < n$

[ocean seal]

bee [it/its]

[lament forge]

specifically in this case we want g to be injective, so for each n we need g(n) to have some value that's different from all of the values before it

[lone heart]

@stiff ore Has your question been resolved?

[stiff ore]

For example, this is a well defined function, but I’ve hit a wall since I don’t know anything about the choice function. Typically I feel like I’m able to exploit some fact in the definition but I know so little about $f$.

[ocean seal]

covert_snorlax

[lone heart]

@stiff ore Has your question been resolved?

[lone heart]

@stiff ore Has your question been resolved?

[autumn epoch]

yo

[lone heart]

@stiff ore Has your question been resolved?

[lone heart]

@stiff ore Has your question been resolved?

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[safe vine]

3d

[safe vine]

i got stuck here

i skipped the base case

shouldnt matter though

[fallow marten]

3 (d)

[safe vine]

yep

question 3d

[fallow marten]

(I have a sneaking suspicion that since both sides are positive, there may be some raising stuff to an appropriate power to get an equivalent expression)

[safe vine]

hm ok

wait but how do you raise the power

because from my understanding of induction, you can only touch one side of the equation at a time

[fallow marten]

I don't have any paper 😭

so this was your attempt

[safe vine]

yea

[fallow marten]

and you want that the LHS is less than or equal to ((k+1)+1)/2

[safe vine]

yep

[fallow marten]

have you tried using logarithms?

[safe vine]

im not too familiar with them

ill try though

which step should i start using logarithms

[fallow marten]

making sure both sides are >= 1

(so that the inequality doesn't flip)

I'll get some paper too

[safe vine]

how do i do logarithm if i dont know what the argument is

becuase its all in k

[fallow marten]

ohhh I think I realised something

forget about logs for now

do you agree that if a < b

and a < 0 bad

then a^n < b^n?

[safe vine]

yea

[fallow marten]

oops I mean 1 < a

[safe vine]

yep

i get it

[fallow marten]

hmmm I'm not sure

in my head I was thinking

$LHS^{\dfrac{n+1}{n}}$ would get rid of the $1/(n+1)$ power

[ocean seal]

ekafeman

[fallow marten]

so like

[safe vine]

whoa

thats cool

whats the $ / { stuff lol

[fallow marten]

oh it's LaTeX

typesetting language for math

[safe vine]

cool

[fallow marten]

$\begin{aligned} \mathrm{LHS}^{(n+1)/n}
& = \left( (n+1)!^{1/(n+1)} \right)^{(n+1)/n}
\ & = (n+1)!^{1/n}
\ & = (n+1)^{1/n} n!^{1/n}
\ & \leqslant (n+1)^{1/n} \frac{n+1}{2} \quad \text{by assumption}
\ & = \dfrac{1}{2} (n+1)^{(1/n) + 1} \end{aligned}$

and then combine the exponents

so it'd be half of (n+1)^{(1/n) + 1}

actually maybe not

you want an n+2 in the end

[ocean seal]

ekafeman

[fallow marten]

that final exponent is (n+1)/n

[safe vine]

wowza

[fallow marten]

so it matches up with the LHS's exponent

[fallow marten]

but then yea I wonder how you get this

[safe vine]

dang why is this question so hard

[lone heart]

@safe vine Has your question been resolved?

[fallow marten]

Maybe you could multiply both sides of the inequality by (n+1)^something

My internet died

[safe vine]

f

[tight marten]

WTS n! <= [(n+1)/2]^n

if n = 2k + 1
WTS (2k+1)! <= [(2k+2)/2]^(2k+1) = (k+1)^(2k+1)
(2k+1)! = 1*2*...*k*(k+1)*(k+2)*...*(2k)*(2k+1)
so its the same as showing (2k+1)! / (k+1) = 1*2*...*k*(k+2)*...*(2k)*(2k+1) <= (k+1)^(2k)

consider multiplying pairs (k+1 - r)(k+1+r) = (k+1)^2 - r^2 <= (k+1)^2 for r = 1, ..., k
and in these pairs we get all of the terms of (2k+1)! / (k+1)
as they are all less than (k+1)^2 and there are k such pairs we get our result

perhaps you can try to case n = 2k now and see how it goes

[fallow marten]

By induction though...

[tight marten]

oh it has to be by induction ?

mb

whoops haha

[safe vine]

lol its fine

[fallow marten]

np it's a nice proof too

[safe vine]

thanks for trying

i asked ai and it gave me this

your somehow allowed to raise power of the inductive hypothesis?

or is the ai wrong

lol

[fallow marten]

not the LLM 😭

[fallow marten]

yea you should read what we tried already

you've basically parroted us haha

🦜

[fallow marten]

maybe we can just finish it off with basic stuff

[safe vine]

oh dang

mb

i thought by what you meant u raised the power of the assumption case when p(n) = k+1

[fallow marten]

omg I mixed you guys up 😭

we end up with $LHS^{\frac{n+1}{n}} \leqslant \dfrac{1}{2} (n+1)^{\frac{n+1}{n}}$

[ocean seal]

ekafeman

[fallow marten]

so just undo that original exponentiation

$\mathrm{LHS} \leqslant \dfrac{n+1}{2^{n/(n+1)}}$

[ocean seal]

ekafeman

[fallow marten]

that's less than n+2 over all that stuff

[tight marten]

WTS n! <= [(n+1)/2]^n

result is immediate for n = 0 as both sides are 1
assume we have k! <= [(k+1)/2]^k for some k
now WTS (k+1)! <= [(k+2)/2]^(k+1)

from our assumption we have (k+1)! <= [(k+1)/2]^k (k+1) = 2 [(k+1)/2]^(k+1)
so if we can show that 2 [(k+1)/2]^(k+1) <= [(k+2)/2]^(k+1) then we are done

raising both sides ^1/(k+1) gives 2^(1/(k+1)) (k+1)/2 <= (k+2)/2
doing some rearranging we arrive at 2^(1/(k+1)) <= 1 + 1/(k+1)
then 2 <= (1 + 1/(k+1))^(k+1) which we know is true by bernoullis inequality
and so we are done

[fallow marten]

and maybe the 2 falls out somehow 🤣

ah nice you did the 2 stuff

[safe vine]

what what happened with the square root

[tight marten]

where?

nth root of x = x^(1/n) if thats what you mean?

[safe vine]

on the LHS

[tight marten]

i just started with both sides raised ^n

[safe vine]

ohhh

[tight marten]

(n!)^(1/n) <= (n+1)/2 is the same as n! <= [(n+1)/2]^n

although i suppose consider the way the original question is posed i shouldve started at n=1, but this is also quick

since (n!)^(1/n) at n = 0 is undefined

[safe vine]

i see now

thank u so much

[tight marten]

Nws

[safe vine]

this question took exactly an dhour lol

look at the pinned comment 😂

.close

[lone heart]

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[tight marten]

This is a fairly common question so worth remembering how to do

[safe vine]

definitely

[lone heart]

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[vocal tapir]

what is gamma here?

[vocal tapir]

and/or where'd it come from?

[lethal belfry]

,w gamma constant

[ocean seal]

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[lethal belfry]

,w gamma

[ocean seal]

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[lethal belfry]

a constant

[vocal tapir]

Ah, alright

thank you

.close

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[fallen shuttle]

for the partial derivative of:
f_x = 3x^2+2y
the answer is 6x, but why does the 2y become zero?

[mossy laurel]

Because you derive with respect to variable x

So you consider y like a constant

[fallen shuttle]

so even if its was 3x^2+2y^4

it would still be 6x because y^4 is treated as a constant?

[mossy laurel]

Yes

[fallen shuttle]

I saw another example where its x^2y^2+y^2+5 and the answer is 2xy^2

why did the first y^2 stay?

[mossy laurel]

Because it is a constant

When you multiply by a constant, the derivative keeps the constant

[fallen shuttle]

Oh ok, I think I get it! thank you, sorry to bother with such a simple question

thank you

.close

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[misty tendon]

I dont kbow if my solution is correct

Can someone help?

[lone heart]

@misty tendon Has your question been resolved?

[misty tendon]

No

[misty tendon]

<@&286206848099549185>

[lone heart]

@misty tendon Has your question been resolved?

[lone heart]

@misty tendon Has your question been resolved?

[lone heart]

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[bold silo]

did i use small angle approximation incorrectly?

[pseudo ice]

You're fine, just bear in mind that because theta is small, you can e.g. ignore theta^4 etc as those are even smaller

[limpid turret]

I can't read the start of your second line

[pseudo ice]

2theta + 3(1 - theta^2/2)^2 - the start is crossed out

[bold silo]

oh didnt know that

[limpid turret]

Also, your $\theta$ looks more like $\emptyset$

[ocean seal]

SWR

[limpid turret]

Oh you crossed it out. OK

[bold silo]

ngl idek what the second one is

[limpid turret]

A set that is empty

[bold silo]

ngl skipped the set notation stuff so hopefully i wount need it

[limpid turret]

One day you will

[bold silo]

could u also sub in 1- sin^2 and do 1- theta^2?

[limpid turret]

Imo you just can't go through life without a basic understanding of axiomatic set theory

[bold silo]

im doing cs in uni so ima prob come across it

ima delay the process of actually understanding it

[pseudo ice]

And you could

[limpid turret]

That is a good question. Yes you could, but you are putting an approximation into a new function rather than approximating the function itself, so your result will be less accurate

[bold silo]

im doing another q and didnt want to start a new channel so ima put it here

how would i do b ii? its not the discriminant right, icr how to find points of inflection

[pseudo ice]

What happens at points of inflection?

[bold silo]

gradient=0, and changes from concave to convex right?

[pseudo ice]

Gradient needn't be zero, but yea, concave to convex

[bold silo]

how come the gradient isnt 0?

[pseudo ice]

There are examples of inflection points which aren't stationary points (that I would have to think about for an example )

[keen plinth]

x^3 - x

no need to think

[elder forge]

1/x?

Y tho

x=0 is critical point but not stationary right

[keen plinth]

1/0 mmm my favourite point on the graph of 1/x

[elder forge]

Mhm

[bold silo]

anyways, how do i show its a point of inflection?

[elder forge]

Oh yea nvm inflection different from stationary

[elder forge]

f''(x) = 0

[pseudo ice]

Show that f'' changes sign there

[elder forge]

Wot thats the definition of inflection point right

[bold silo]

so i have f''= 60x^2 -144x +84

[pseudo ice]

Also, rather coincidentally-

[elder forge]

Yes f"(x) is 0

-_-

[keen plinth]

x^4 would like to have a word with you

[pseudo ice]

"and that it has different signs on either side of that point"

[elder forge]

Yea atleast what i said is one of the condition

Y say its completely wrong

[pseudo ice]

You implied that it's the only condition

[elder forge]

Oooof

U reacted bfr i could type the whole thing and i got confused

[keen plinth]

one of the conditions of a number being an even number is that it has to be an integer

that isnt exactly a correct definition

[elder forge]

Yea fine

[bold silo]

so 60(1)^2 -144(1)+84=0, how do i prive that either side has differnet signs

[keen plinth]

factorise f''

[bold silo]

(5x-7)(x-1)

[keen plinth]

should be clear that it changes sign at x=1 then

[bold silo]

how

[keen plinth]

have you ever graphed a parabola?

[bold silo]

i mean i put into desmos

i wouldnt be able to graph it myself

[elder forge]

This does not look like parabola wot

[pseudo ice]

[not the original, f'']

[keen plinth]

im talking about (5x-7)(x-1)

[keen plinth]

i think you can see quite clearly that it has an intercept at x=1

and that it changes sign because well

its a parabola

[elder forge]

Its pretty simple
U found the roots right
Here 7/5 and 1
So the curve touches x axis at these points
If coefficient of x^2 is greater than zero u draw upward parabola else downward parabola

[bold silo]

so since it has 2 roots, it has to reverse somewhere between 1 and 7/5

[pseudo ice]

[as a tiny side note, remember that for quadratic inequalities, you'll want to graph them, so make sure you're happy with them!]

[bold silo]

ight i think i understand now, thanks guys

[lone heart]

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[wanton mural]

Just a quick question, is it a bad idea to take diff eqs right after calc II or should I wait until I have finished calc III to do diff eqs?

[azure needle]

#math-discussion or #advanced-lounge

[lone heart]

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[lone heart]

Please don't occupy multiple help channels.

[azure needle]

Substitute q for the changing variable

🤷‍♀️

[alpine sable]

k

.close

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[olive oar]

If V is a vector space and $V= W_1 \oplus W_2$, what can I say about a matrix of a transformation from V to itself if the $ W_i$ are isomorphic ?

[ocean seal]

Zander

[lone heart]

@olive oar Has your question been resolved?

[mortal trellis]

what do you want to say

the W_i being isomorphic says basically nothing

there are lots of ways to write a n dim space as the direct sum of two n/2 dim spaces

[olive oar]

I wanna describe the blocks that I'll get, but I'm pretty sure I wont be able to

In a general setting at least

[mortal trellis]

unless you know that L(W_i) subseteq W_i, you can say nothing about the blocks

it could be literally any matrix

[olive oar]

Yea that's what I was expected

Ty

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[alpine sable]

Hi! If X and Y are independent exponential random variables with rates lambda and mu respectively, and M = min(X,Y), I'm really struggling to see how E(MX | M = X) = E(M^2) = E(MY | M = Y). I understand that the exponential has the memoryless property, but how do I apply it here? Thanks!

[median oar]

Seems obvious

[alpine sable]

Wait really? I'm not sure how to see it

[median oar]

If M = X then X < Y and E(MX| M=X ) is just E(X^2) = E(M^2)

[alpine sable]

Sure, but what about the conditional aspect of it

I can see E(MX | M = X) = E(M^2 | M = X)

[median oar]

And if M = Y then Y < X

Isn’t that obvious or am I missing something

[alpine sable]

Wait, how do we know that E(M^2 | M= X) = E(M^2) ?

[median oar]

Yeah I must be missing something I never used that it’s exp distributed

[alpine sable]

I think we must use the memoryless property, but I'm not sure where.

[alpine sable]

This isn't an obvious step to me.

[median oar]

No it’s not and I also don’t think so

Let’s see, if M = X then X < Y

[alpine sable]

My best guess was to use E( M^2 | M = X) = E(M^2) / P(X<Y)

and M is exp(lambda + mu) , and P(X<Y) = lambda / (lambda + mu)

but this is not the same as E(M^2)

[median oar]

$E[MX| X<Y] = \int_{0}^\infty \int_0^y x^2 f_{X,Y}(x, y),dx,dy$

[ocean seal]

Frosst

[median oar]

But X and Y are independent so you can split the joint pdf up into its marginals

$=\int_0^\infty \lambda e^{-\lambda y}\int_0^y x^2 \lambda e^{-\lambda x},dx,dy$

[ocean seal]

Frosst

[median oar]

Perhaps find the pdf of M as well using transformations

But this seems like the hard way of doing this question

[alpine sable]

I considered using that approach. But there must be an intuitive line of reason

[median oar]

I thought so too but I can’t think of it 😦

[alpine sable]

I tried some numerical experiments with R

the identity to prove seems correct

set.seed(1)

experiment <- function(){
  c(rexp(1,10), rexp(1,5))  
}

Nsamp <- 10000
matr <- matrix(nrow = Nsamp, ncol = 3)
for(i in 1:Nsamp){
  ex <- experiment(); matr[i,1] <- ex[1]; matr[i,2] <- ex[2]; matr[i,3] <- min(ex)
}

v1 <- {}
for(i in 1:Nsamp){
  if(matr[i,1] == matr[i,3]){
    v1 <- append(v1, matr[i,1]^2)
  }
}

v2 <- {}
for(i in 1:Nsamp){
  if(matr[i,2] == matr[i,3]){
    v2 <- append(v2, matr[i,2]^2)
  }
}

c(mean(matr[,3]^2), mean(v1), mean(v2))

# In theory E(M^2) = 2/(lambda + mu)^2 = 2/15^2 = 0.00888...

Here's my code.

Here's another idea, is it possible to show that the event M=X and the random variable M^2 are independent?

[alpine sable]

Are there other strategies ?

[median oar]

I don’t see one (doesn’t mean there isn’t one)

[alpine sable]

hmm

maybe there's some intuitive explanation about how M^2 is independent of M= X. I will try this one.

I might close out this channel. Thanks for your help!

.close

[lone heart]

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[median oar]

Perhaps E[M²] = E[M² | M=X]P(X < Y) + E[M² | M = Y]P(Y<X)

[alpine sable]

That could work. I’ll have a think about that as well

[lone heart]

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[thin rock]

Jeff’s cell phone plan has limited talking minutes and text messages. He knows that three times the sum of his minutes and texts is equal to his number of texts combined with eight times his number of minutes. He also knows that if he doubles his minutes and then divides by 25, he will get a result of 8. If he has sent 75 texts, how many does he have left?

[thin rock]

I am getting 250 but it says it is wrong

😭

!close

.close

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[reef karma]

Can anyone help me to parameterize the given surface

[lone heart]

@reef karma Has your question been resolved?

[reef karma]

Anyone?

<@&286206848099549185>

Help!

[reef karma]

<@&286206848099549185>

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@reef karma Has your question been resolved?

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[deep kiln]

Hello i have to calculate the magnetic field of the straight parts using biot-savart any ideas, cause the way the point is placed i think it doesnt induce any magnetic field

[remote hazel]

there is a magnetic field in the point

[deep kiln]

yes overall there is but they are asking if there is a magnetic field induced by the even straight parts of the circle

Cause the way P is placed relative to the straight lines its parallel or not?

[remote hazel]

the straight lines will have no effect on the field

[deep kiln]

Are you 100% sure cause we got a formula for a similar issue

where it isnt 0

[remote hazel]

i am like 99% sure

what formula?

[deep kiln]

one sec ill send it

[remote hazel]

thats amperes law, wich is not that well suited for tasks like these

[deep kiln]

yh but even this would explain why its zero

[remote hazel]

it can, just not practical at all

[deep kiln]

ok so using biot i can just say cause di x e_r(vektor) with di || e_r right?

[remote hazel]

$\vec{dl} \times \vec{r} = 0$

yes

[ocean seal]

caspar

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[wheat jackal]

One message removed from a suspended account.

[modern sedge]

hint: cos^2(a) + sin^2(a) = 1

[wheat jackal]

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[modern sedge]

How did you manage to get 2 answers?

How did you get 8/9?

[wheat jackal]

One message removed from a suspended account.

[modern sedge]

hmm

why do you think this should work?

(sin^2(a))^2 - 1 = cos^2(a)?

[wheat jackal]

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[modern sedge]

If cos^2(a) + sin^2(a) = 1

then cos^2(a) = 1 - sin^2(a)

and since sin^2(a) = 1/3

cos^2(a) = 1 - 1/3

so cos^2(a) = 2/3

no need to square anything

[wheat jackal]

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[modern sedge]

np

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[rose patio]

Hi guys, this isn't math but it's related. It's a challenge that my teacher proposed to students with top marks in geometry, the challenge is to create a pentagon from a midpoint of the sides and a point, without using non-"obvious" angles ig, I've been trying to figure it out for a few hours If anyone has any ideas I would appreciate it!!

[rose patio]

i tought i could find EA with a 45 angle but M isnt exactly in the midle so the angle its 50 and i cant use that so i cant get MC either

[manic agate]

Pentagon or pentagram?

[rose patio]

idk im not english

its that thing

[rose patio]

this

[manic agate]

what language u speak

[rose patio]

portuguese

[manic agate]

KKKKKKKKK

ent ta bom

[rose patio]

hahah

[manic agate]

é um pentagono ent

[rose patio]

isso

[manic agate]

Mas o q vc quer dizer por angulos ñ obvios?

tipo, sem usar 45, 90?

[rose patio]

tipo nao posso medir os angulos

[rose patio]

acho q é isso

n posso medir tipo os anglos que fazzem dentro do pentagono

eu ja tentei

[manic agate]

se eu n estiver enganado, se o pentagono n for regular, n precisará ter angulos "certinhos"

[rose patio]

se eu usar 40 consigo mc

[rose patio]

tem o 72, 36 e 108

que é o que se usa em pontos normais

[manic agate]

hmmm

complicado kkkk

[rose patio]

pois é

tou á 2 horas a rabiscar

e nao acho nada

[manic agate]

Portuguesa?

[rose patio]

sim

[manic agate]

ataa

Se vc colocar pontos médios nos lados do pentagono orignal, e entçao traçar as retas não dá certo?

[rose patio]

isto?

n dá porque n tem ligaçao com os lados

[manic agate]

pera, tem q usar os lados do pentagono?

[rose patio]

nao

tem de se usar um pontpo

e um ponto medio

[manic agate]

ahh, isso pra todo lado do pentagono q é para criar?

[rose patio]

a partir de esta reta

MA

[manic agate]

entendi

acho

vou tentar mandar o que pensei

[rose patio]

ok

[manic agate]

Por pentagono, pode ser um poligono qualquer cinco lados ou tem que ser igual a figura?

[rose patio]

tem de ser regular

[manic agate]

ataaaaa

[manic agate]

só para confirmar, tem que ter um dos ângulos que você citou aqui?

[rose patio]

nao

[manic agate]

parece ser impossivel de fazer o pentagono sendo regular

So se forma poligonos de cinco lados estranhos

o mais proximo de um pentagono normal q connsegui chegar seria isto

[lone heart]

@rose patio Has your question been resolved?

[rose patio]

Pois eu também

[manic agate]

Pois acho que será isso, visto que do ponto A não é possível traçar uma reta semelhante a um lado de pentágono.

[lone heart]

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[distant surge]

how am I to do 7.1.2? i split it into sin/cos, but i have nothing more

[placid zinc]

That's the same as tan(β + β)

[distant surge]

i need it in terms of cosB

[placid zinc]

Your solution method is going to depend on whatever identities you're allowed to use

[distant surge]

yeah

[cinder tundra]

It says in terms of k

[distant surge]

i broke it up using our theorems and got 2sinBcosB/1-2sin^2B

[cinder tundra]

Ups

I didn’t read

Like, im dumb

[distant surge]

anyone got a thought?

[cinder compass]

use sin^2 + cos^2 = 1 here to replace sin with cos

[distant surge]

i shouldve startedd with the 2cos^2B-1 come to think of it

youre so right

thanks

[cinder compass]

wl

tbh technically you don't know the sign of sinB so there's not enough information

[distant surge]

a problem remains though

yeah thats what i was thinking

ill ask my tutor about it 2mrw morning

thanks anyway tho < 3

.close

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[upbeat pilot]

I don't know how to find the result

[wide tartan]

is there anything that you notice about the expression that you think can help you? What have you tried so far?

[upbeat pilot]

Yes

242424 = x and 121212 is equal to x/2

[wide tartan]

okay, that is a good start

what about the expression in the numerator?

[upbeat pilot]

one second

I am back, this is what I've done: ( x^2 / 1 - x^2 / 4 ) / ( x/1 * x/2 )

[wide tartan]

okay, that's good. so we should focus on the numerator for now. Do you know about differences of squares?

[near chasm]

I dont think we need that here

[upbeat pilot]

I have already heared about

But I don't know what it is

[wide tartan]

we don't need any technique here, since this can be done with a computer. idk what your point is, this is surely the easiest say to do it

doesn't really matter. if you haven't heard of difference of squares, then we can just break up the fraction. I'll rewrite your work in tex so that it's more readable

you got $$\frac{x^2/1 - x^2/4}{(x/1)(x/2)}$$

[ocean seal]

smay

[upbeat pilot]

Sure

[wide tartan]

okay thinking about it more yeah we for sure don't need it here. sorry for being an asshole to you, I'm just tired.

[near chasm]

nah its okay, things happen

[wide tartan]

okay so we can simplify the bottom part

[upbeat pilot]

sure

Sorry, but if possible, I would like just to understand how I have to make all the things because I really want to think a little bit to learn more about that. I think you already know, but it really would help me a lot

[wide tartan]

sure, what do you mean by "make all the things?"

like typing in latex?

or the thought process behind solving the question?

[upbeat pilot]

I would like to do the calculations

[wide tartan]

you're already doing the calculations!

[near chasm]

Opinion: I would call 121212=x and 242424=2x

This way, you don't have to deal with fractions so you get a simpler form.

[upbeat pilot]

Alright, thank you 😊

[upbeat pilot]

Yes, so we can call 121212=x. thx

[near chasm]

you can call anything you want tbh

[wide tartan]

all we are really doing is trying to guide you a little bit until you feel confident enough to finish out the problem by yourself. Once you get enough problems under your belt, you won't need such guidance.

[wide tartan]

so where are you at right now?

then we can work on simplifying it, and some magic will happen

[upbeat pilot]

I think i started to understand with x = 121212

Wait

instead of 4 * 2x + 2x * x^2

I think we should do 4x^2 - x^2 = 3/2

RIght?

[wide tartan]

okay and close it out

[upbeat pilot]

I tried with x = 242424 too, just for practice, but I am struggling with ( 3x^2/4 ) / ( 2x^2/2 )

[wide tartan]

we can cancel the x^2's

[upbeat pilot]

( 3 / 4 ) / ( 2 /2 )

[wide tartan]

wait why are you getting (2x^2/2) in the denominator

if you multiply x*(x/2) you'll get x^2 / 2

[upbeat pilot]

alright

so ( 3 / 4 ) / 2

[wide tartan]

not exactly