help-49

[last slate]

Im so fucking stuid

I dont think this makes sense

This

[polar star]

you factored it wrongly

m^2 - 16m = m(m-16)

[midnight plank]

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[stoic jolt]

how does it become 2^2

[stoic jolt]

or at least like

where does the exponent come from

[final laurel]

why arent there any calls

[rustic shale]

32=2⁵

[stoic jolt]

thanks!

.close

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[vestal moth]

Find x∈ℝ such that x⁴-2x²-[x]=0, where [x] is the integer part of x.

[midnight plank]

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@vestal moth Has your question been resolved?

[white gate]

graphing is probably the easiest way here, but there's also another way

substitute n = [x], solve for x² with respect to n.

[midnight plank]

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[vestal moth]

.reopen

[midnight plank]

✅

[vestal moth]

x⁴-2x²-[x]=0
⇔s²-2s-n=0, s=x², n=[x]
△=(-2)²-4⋅1⋅(-n)=4+4n=4(1+n)
s₁₂=(2±√△)/2
=1±√[4(1+n)]/2
=1±2√(1+n)/2
=1±√(1+n)

⇒1+n≥0
⇔n≥-1

n=-1⇒ s₁₂=1⇒x=-1 is a solution and x=1 is not because x≠1 due to n=[x]
n=0⇒ s₁₂=1±1=0 or 2⇒x=0 or x=±√2 which won't work so x=0
n=1⇒s₁₂=1±√2 but s>0 so s=1+√2 ⇒ x=±√(1+√2) are two potential solutions, only √(1+√2) satisfies n=[x]
n=2⇒s=1+√3⇒x=±√(1+√3) they don't satisfy n=[x]

the hard part is to prove that for n ≥ 1, only n = 1 satisfies

[white gate]

from here you can observe that "s" only exists when 1+n >= 0

giving n >= -1

n = -1 and n = 0 are all solutions which gives x = -1 and x = 0

the hard part is to prove that for n ≥ 1, only n = 1 satisfies

[vestal moth]

ok, let me think about that part. And thank you

for n=-1 we get that x=-1 and x=1, right?

[white gate]

x cant be 1

[x] = -1 but x =/= 1

[vestal moth]

right

[white gate]

even the 2nd case as well, u gotta stick to [x] = n

[vestal moth]

n=2⇒s=1+√3⇒x=±√(1+√3) the negative one won't satisfy n=[x] but the positive one does. x=√(1+√3) looks like a solution

[white gate]

,w floor(sqrt(1+sqrt(3)))

[grand pond]

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[white gate]

nope

[vestal moth]

√(1+√2) satisfies n=[x]

[white gate]

well, that one does

[dry merlin]

For x from 0 to 1, we have x^4-2x^2 which is x^2 (x^2-2). It won't cross except at 0.
LHL as x-> 1 is -1

For x from 1 to 2, we have x^4 - 2x^2 -1 which is satisfied at sqrt(1+sqrt(2))

RHL as x-> 1 is -2, LHL as x-> 2 is 16-8-1 >0

[white gate]

from n=2 onwards, pretty much the solution has no roots

prove it :)

[dry merlin]

4x^3-4x has roots at x= +- 1
it is >0 for all x > 1
it is the deriv. for given polynomia

[white gate]

graphing

it can be proven without using calculus

if calculus is involved then u can just graph from the beginning lul
(graph [x] is not even that hard)

[vestal moth]

x²(x²-2)≥2x²
⇔x²-2≥2
⇔x²≥4
⇔x≤-2 or x≥2

n≥2⇒x≥2
⇒[x]=x²(x²-2)≥2x²>x impossible

I think desmos did not recognize [x] as integer part

[white gate]

u gotta write floor(x) instead

i think it treats [x] like a set

there's also another way which might be quicker

The equation becomes: $\left( x-1 \right)^{2}\left( x+1 \right)^{2}=\left\lfloor x \right\rfloor+1=\left\lfloor x+1 \right\rfloor$

[grand pond]

TargetVN

[white gate]

easily see [x] >= -1

$\text{For }x\ge 2:\left( x-1 \right)^{2}\left( x+1 \right)^{2}\ge \left( x+1 \right)^{2}> \left\lfloor x+1 \right\rfloor$

[grand pond]

TargetVN

[white gate]

from here u consider cases as normal

[dry merlin]

analytical solution on top

[midnight plank]

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[shell summit]

Hello! Does anyone see where my mistake is for this Calculus 2 problem?

[shell summit]

This is Google's answer, but im not sure why we dont have the same coefficients for the natural logs

[wind oxide]

why did you partial fraction (2x-4)/(x+2)(x-3) ?

when u had 3x-4 in the numerator

[shell summit]

oh bruh that's where the mistake is

idk

[wind oxide]

i think a writing mistake

should fix your coefficients

[shell summit]

yeah ur right

thanks

.close

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[twilit field]

I suspect that the limit doesn't exist

approaching along y=0 gives us a limit of 0

approaching along x=0 gives us a limit of 1/3

[buoyant yoke]

looks good

[grim vector]

Ye

[twilit field]

Yay

I'm now going to approch it along x=y

[wind oxide]

HMMMM

[twilit field]

oops, wrong choice

[wind oxide]

if u prove its different along say x=y and x=1/2y, ur done right

[twilit field]

yeah

y=mx may work

Wait

[placid spoke]

it's convergent

[twilit field]

what about squeeze theorm

[placid spoke]

use polar coordinates

[buoyant yoke]

comparison

x<=sqrt(x^2+y^2)

[twilit field]

$0 \leq \frac{x}{\sqrt{x^2+y^2}} \leq 1$

[grand pond]

What a wonderful world it is !

[wind oxide]

whaaaaaaaa surprising

[twilit field]

$0 \leq \frac{xy}{\sqrt{x^2+y^2}} \leq y$

[placid spoke]

let x = r cos ϴ, y = r sin ϴ, and let r, ϴ-> 0. it pops out right away

[grand pond]

What a wonderful world it is !

[wind oxide]

uhh r sin theta cos theta how do u knows that convergent for r,theta -> 0?

[buoyant yoke]

or just use $\frac{xy}{\sqrt{x^2+y^2}}\leq \frac{\sqrt{x^2+y^2}\sqrt{x^2+y^2}}{\sqrt{x^2+y^2}}$

[grand pond]

Bonk

[wind oxide]

this makes sense damn

[heavy falcon]

rsinθcosθ=1/2 rsin(2θ) and -r/2=<1/2 rsin(2θ)=<r/2

the squeeze theorem

[wind oxide]

oh squeeze

[wind oxide]

okay okay

[buoyant yoke]

with this you dont have to do anything fancy

[placid spoke]

crazy how you can get the same answer in multiple ways /s

all good ideas :)

[heavy falcon]

you can also say that $\frac{xy}{\sqrt{x^2+y^2}}\leq\frac{xy}{|x|}$

[buoyant yoke]

fr

[heavy falcon]

and this is equal to y/sgn(x)

huh why the

[wind oxide]

hows this true???

shouldnt it be x+y

oh wait

[heavy falcon]

$|x|=\sqrt{x^2}\leq\sqrt{x^2+y^2}$

[heavy falcon]

since y^2>=0 for all y in R

[wind oxide]

yeah i multiplied it out makes sense

[wind oxide]

could you slide me this assignment looks fun might do it instead of solving impossible integrals

[heavy falcon]

the effect of doing tons of integrals is moving to limits

[wind oxide]

next stop: trignometry

[wind oxide]

thats actually something i need to improve lol

[heavy falcon]

you cant run away i just saw what you wrote before editting 😏

[wind oxide]

[twilit field]

I was thinking of approaching this along (z^2,z^2,z)

[wind oxide]

if u approach along x=y=z u get 1

and if u approach along x=z^2, y=z^2, z u get 3/2

[twilit field]

yea

[wind oxide]

so diverges

damn this was easier than i expected

[heavy falcon]

wait how are you getting this

[heavy falcon]

shouldnt this be 1 too

[wind oxide]

what do you mean the z^4 cancel out

[wind oxide]

could be i did it in my head i make mistakes when i do it like that

[twilit field]

oops

[wind oxide]

yeah its 1 lol

[twilit field]

yeah

[heavy falcon]

$\frac{z^4+z^4+z^4}{z^4+z^4+z^4}=1$

[twilit field]

now I approach it along x=y=0

[twilit field]

That gives me 0

[wind oxide]

yeah okay i was right anyhow

[midnight plank]

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[twilit field]

This function is continuous on $R^2 - {(x,y) \mid e^x=y^2 }$

[grand pond]

What a wonderful world it is !

[dawn dagger]

Is there a question?

[grim vector]

Domain of continuity i bet

[twilit field]

this

sorry

[grim vector]

I think its ok

[twilit field]

Thanks

Now this

[wind oxide]

u have to check all possible combinations right

[midnight plank]

@twilit field Has your question been resolved?

[heavy falcon]

so first of all where could you face problems with the continuity of this function

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[prime garden]

If two equal chords of a circle intersect within the circle, prove that the segments of one chord are equal to the corresponding segments of other chord.

[prime garden]

It is to prove that LH = LJ and that LI = LK.

<@&286206848099549185>

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[prime garden]

<@&286206848099549185>

[keen saddle]

find the relation between their central angles

then use length of arc = radius * central angle (in rad)

[lyric charm]

wait why do you even need that

[keen saddle]

oh mb i thought we had to prove arc length is same

i think by equal segments, they prob mean area

so you can use ar(segment) = ar(sector) - ar(triangle)

💀

[lyric charm]

no, segments like line segments

not circular segments

[keen saddle]

can you draw and send?

[prime garden]

...

[keen saddle]

oh

you can that via congruency

i recommend erasing IK and drawing GI, GJ, GH, GK

[midnight plank]

@prime garden Has your question been resolved?

[earnest cairn]

You can use the intersecting chords theorem

So LH×LJ = LI×LK
Also IJ=KH (equal chords)

And since ILJ and KLH angles are verticals, we can say that LHK and LJI triangles are equal

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[last slate]

Say i have n keys and only one of the keys unlocks a door.Let random variable X denote the number of tries required to open the door,without replacement of the keys after each try And now i want to find out the probability distribution function for this.

[last slate]

P(X=1) = 1/ n

P(X=2) = (n-1)C1 * 1 / nC2

I am having trouble with P(X=3)

[fallow scarab]

why nC2 ?

[last slate]

would it be (n-1)C1 * (n-2)C1 * 1 / nC3 or (n-2)C2 * 1 / nC3 ?

[last slate]

total number of possibilities ?

[fallow scarab]

does P(X=2) simplify to 1/n

[last slate]

it simplifies to 2/n 👀

huh

[fallow scarab]

oh yes 2/n

might be missing something then

sum P(X=i) = 1

[last slate]

original question

[fallow scarab]

yea 13 should be samples with replacement since they're independent and at random

[last slate]

with replacement means (n-1)^k-1 * 1 / n^k

for without

it might be (n-1)C1 (n-2)C1 or (n-1)C2

one is exactly half of the other

so order

order does matter

ahh

so its

P(X=3) = (n-1)C1 (n-2)C1 * 1 / (nC3 * 3!)

e

[last slate]

this should be

P(X=2) = (n-1)C1 * 1 / (nC2*2!)

@fallow scarab

correct?

[fallow scarab]

yea that's 1/n

the way i think of it is the experiment resets and there are n-1 keys left so P(X=2 | X != 1) = P(X != 1) * P(X_reset = 1) = (n-1)/n * 1/(n-1)

X_reset = same experiment with remaining n-1 keys

[last slate]

what is P(X!) ?

[fallow scarab]

X != 1 is read as X is not equal to 1

And that probability equals (n-1)/n

my justification is connecting consecutive events makes finding the pattern easier

maybe not as robust as doing the proper combinatorics like you're doing above

[last slate]

thanks!

.close

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[fallow scarab]

||https://math.stackexchange.com/questions/246855/probability-problem-with-n-keys||

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[solemn flint]

Given that F(t) is a filtration for the Wiener process W(t) and X(t) is adapted to F(t), how do we show that X(0) is degenerate?

[solemn flint]

It is probably by showing that F(0) is necessarily trivial but i dont really know how to do this (since that would lead to the sigma algebra generated by X(0) also being trivial and ...)

[lyric charm]

what's the definition of a random process being adapted to a filtration

[solemn flint]

X(t) is F(t) measurable for every t

This would be trivial if we were talking about F(t) being the natural filtration for W(t) but that wasn't specified so if I have understood the definitions correctly, then F(0) might not be degenerate

but that can't be the case

The only confusing part (as in I am having trouble seeing what this changes and kinda some intuitive meaning) is that a filtration for a wiener process F(t) also has to statisfy: W(u)-W(t) is independent of F(t) for every u>=t

I know that this is due to the definiton of wiener processes

but idk

omfg

this is so obvious

nvm

[solemn flint]

nvm nvm this is trivial

wtf

how did i not notice it

or

nvm

nvm nvm nvm

huh

this doesnt even have to be true

because F(0) might be {empty, sample, x, sample\x} such that this is independent of every W(t) but X(0) might then not be degenerate

[midnight plank]

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[solemn flint]

.close

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[prime garden]

I wanted a geometric proof.

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[vivid thicket]

how to do this

as per basic trigno identities

[lethal path]

try making a common denominator on the LHS

[dusty portal]

Get everything in terms of sines and cosines, work from there.

[lethal path]

the easiest way to do that actually would be $\frac{\sin}{1 - \cot} \frac{\tan}{\tan}$

[vivid thicket]

tried that but i end up getting cosA + sinA - 2sin3a - 2cos3a/2sinacosa - 1

[grand pond]

south

[dusty portal]

$\frac{\sin}{1-\frac{\cos}{\sin}}+\frac{\cos}{1-\frac{\sin}{\cos}}$

[grand pond]

;(

[vivid thicket]

yea i did that then i got at a stuck mentioned bfr

[lethal path]

try from here

you should get $\frac{-\sin \tan + \cos}{1 - \tan}$

[grand pond]

south

[vivid thicket]

okey

[lethal path]

(ask yourself how tan x - 1 and 1 - tan x are related to see why)

[dusty portal]

I will work this out in #latex-testing because why not

[subtle gulch]

this works for any two fractions including that one

[dusty portal]

That works

I had a different approach

$\frac{\sin^2}{\sin-\cos}+\frac{\cos^2}{\cos-\sin}$

[grand pond]

;(

[dusty portal]

I just simplified each fraction; nothing fancy

@vivid thicket What do you notice about the denominators?

[lethal path]

yeah I get something that's similar

[lethal path]

.

[lilac estuary]

can anyone help me with mymaths personally please i am idiot at maths but i am willing to learn.

[vale zephyr]

!help

[midnight plank]

To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[lilac estuary]

for wit

for wut

[vale zephyr]

see #❓how-to-get-help

[lilac estuary]

i cant open thissssssssssssssssssssss

lol

am i dumbbb

but it cant be

[lethal path]

ask in #help-11

[dusty portal]

@lilac estuary https://discord.com/channels/268882317391429632/1018700472988749904

[lethal path]

or that

[dusty portal]

Or that

[lilac estuary]

thankyou so muchh

[midnight plank]

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[dusty portal]

Let $g:\mathbb{C}\mapsto\mathbb{C}$, $f:\mathbb{C}\mapsto\mathbb{C}$, $\omega^3=1$, $\omega\neq 1$, and $z\in\mathbb{C}$. Show that there is one and only one function such that:
$$f(z)+f(\omega z+a)=g(z)$$

[grand pond]

;(

[dusty portal]

So I already got $f$ in terms of $g$

[grand pond]

;(

[dusty portal]

Only problem is proving the uniqueness

“One and only one”

$f(z)=\frac12(g(z)-g(\omega z+a)+g(\omega ^2 z+\omega a +a))$ if I recall correctly

[grand pond]

;(

[lyric charm]

did you mean $g: \bC \to \bC$? not $\mapsto$?

[grand pond]

ann.in.a.teacup

[lyric charm]

also wait your wording of the problem is confusing af

@dusty portal show original?

[midnight plank]

@dusty portal Has your question been resolved?

[dusty portal]

Oh ok

@lyric charm

[midnight plank]

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[lyric charm]

uh

.reopen

[midnight plank]

✅

[lyric charm]

that image is pretty hard to read bc it's so low res

ok so f is not given

f is under an existence and uniqueness quantifier

[lyric charm]

so how did you find this

[dusty portal]

I will give you pdf then

[lyric charm]

i was able to read it fine

[dusty portal]

Ok

[dusty portal]

I did $z\to \omega z+a$ twice

[grand pond]

;(

[dusty portal]

So I got a system of equations

Like so

[lyric charm]

ok so you got a system of linear eqs in f(z), f(ωz+a) and f(something else)

3 equations 3 unknowns

and you solved it

correct?

[dusty portal]

$f(z)+f(\omega z+a)=g(z)\ f(\omega z+a)+f(\omega ^2 z+\omega z+a)=g(\omega z+a)\ f(\omega ^2 z+\omega a +a)+f(z)=g(\omega ^2 +\omega a +a)$

fucking

Why did it not??

Gah

[grand pond]

;(

[dusty portal]

There

[dusty portal]

Yes

I did (1)-(2)+(3)

(1), (2), (3) represent first, second, and third equations respectively

So then $2f(z)=g(z)-g(\omega z+a)+g(\omega ^2 z+\omega a +a)$ and it's kinda obvious to see now

[grand pond]

;(

[dusty portal]

How it is this

[dusty portal]

This

I'm having problem

[lyric charm]

you have already done so

[dusty portal]

Let me see if I can find the message

[lyric charm]

if you can prove that the system $\begin{cases} x_1+x_2 = p \ x_2+x_3 = q \ x_3+x_1 = r \end{cases}$ has a unique solution no matter what $p$, $q$ and $r$ are, then you win

[grand pond]

ann.in.a.teacup

[dusty portal]

I think I was incorrect in what I said earlier sorry

[dusty portal]

How to do this?

[midnight plank]

@dusty portal Has your question been resolved?

[dusty portal]

<@&286206848099549185>

[midnight plank]

@dusty portal Has your question been resolved?

[midnight plank]

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[midnight plank]

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[lyric charm]

!msgdel

[midnight plank]

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[amber zenith]

So hear me out

I was thinking n! = n(n-1)(n-2)....

which is just 1.2.3.4.5.6 etc

so It would mean that this is just sum of 2.4.6......2n?

wait divided by 2n+1?

i am not so sure anymore 💀

[sullen grove]

trying to find convergence?

[amber zenith]

or divergence

[sullen grove]

yeah

just use factorial

rewrite it and use regular tools

[fallow scarab]

Find the limit of the terms

[amber zenith]

test for divergence?

[sullen grove]

do u know stilings formulae?

Stirling

[amber zenith]

no

[sullen grove]

okay uh

[amber zenith]

but i do think

[amber zenith]

that 1.3.5.7...(2n+1) is just (2n+1)! ?

[sullen grove]

thank you

do this

[amber zenith]

hmmmm

[sullen grove]

use stirlings formula to find an equivalent

[snow steeple]

It's not.

[amber zenith]

I wasnt so sure then i asked AI and they said yes ;-;

AI griefing

[sullen grove]

Chat gpt is horrible at math i recommend otherwise

[snow steeple]

7! is not 7.5.3.1.

[buoyant yoke]

!nogpt

[midnight plank]

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

[buoyant yoke]

you know better than this

[amber zenith]

But I used it for myself

:-:

[amber zenith]

but what can I do with this ;-;

[amber zenith]

:/

[buoyant yoke]

dont read only the second sentence

[dawn dagger]

n!! does exactly what your denominator does, it skips every second number, depending whether n is odd or even, the formula was a suggestion

[amber zenith]

oh so

n!! is just 1 * 3 * 5 * 7 etc

so n!! is smaller than n! thats crazy

[dawn dagger]

yea if n is odd

[amber zenith]

aaah okay

and if it was even

2.4.6 etc

got it!

[dawn dagger]

yes

[amber zenith]

the more u know

[dawn dagger]

7!! = 1×3×5×7 for example

[amber zenith]

Alright thanks :)

[fallow scarab]

Use Wikipedia, not AI
https://en.m.wikipedia.org/wiki/Double_factorial

[amber zenith]

Aaah okay thanks

i am kind of stuck now tho

[amber zenith]

hmmm got something maybe

dont say the answer yet please

[dawn dagger]

i think my suggestion was trash, maybe stirling is better to use

[amber zenith]

oh i get inf / 0

;-;

[buoyant yoke]

this is wrong

should be (2n)!

[amber zenith]

aaah true

Still get the same tho :/

[buoyant yoke]

!show

[midnight plank]

Show your work, and if possible, explain where you are stuck.

[buoyant yoke]

i hope you didnt do n!/(2n)!=1/2

[amber zenith]

its this but then with the constant 1/2

OH

imagine

if i did that

would be weird

wouldnt it

[buoyant yoke]

n!=1 * 2 * 3 * .... * n

[amber zenith]

I KNOW

[buoyant yoke]

(2n)!=1 * 2 * 3 * .... * (2n-1) * (2n)

(2n)! has twice as many terms as n!

[amber zenith]

i get 0

so i did all of this for nothing ;-;

[buoyant yoke]

wdym

[amber zenith]

i notice i forgot brackets for the n!/(2n)!

[buoyant yoke]

uhm

wtf did oyu do?

n!/(2n)! * n! * 2^n = 0?

[amber zenith]

well n!/(2n)! is 0 right?

[buoyant yoke]

no??

what

write it out

pls

[amber zenith]

okay

[dawn dagger]

hes begging u bonk

[amber zenith]

Bro sometimes

math makes me rage so ahrd

[buoyant yoke]

what

[amber zenith]

i now have n(n-1)(n-2)....
2n(2n-1)(2n-2)....

but i cant just factor out 2 then i will have -1/2 -1 -1.5 etc

[buoyant yoke]

n! and (2n)! dont have the same amount of terms

[amber zenith]

i am trippin

[buoyant yoke]

do you have to find the value or just determine whether it converges or not?

[amber zenith]

just determine

[buoyant yoke]

i would just do a term comparison

[amber zenith]

but what is n!/(2n)! ?

[buoyant yoke]

if you write out n! and write out (2n+1)!! you get
1 2 3 4 5 6 ...
1 3 5 7 9 11 ...

btw, im not sure if this is going in the correct direction

then $\frac11\cdot\frac23\cdot\frac35\cdot\frac47\cdot\frac59\cdot\cdots$

[grand pond]

Bonk

[buoyant yoke]

and this is smaller than $1\cdot\frac23\cdot\frac23\cdot\frac23\cdot\frac23\cdot\cdots$

[grand pond]

Bonk

[buoyant yoke]

which is equal to $\left(\frac23\right)^{n-1}$

[grand pond]

Bonk

[amber zenith]

;-;

Its time for a break

thank you very much bonk

[buoyant yoke]

this way i think is easier 🙂

[amber zenith]

yep i hate math rn

imma close it thank you

.close

[midnight plank]

Channel closed

Closed by @amber zenith

Use .reopen if this was a mistake.

[buoyant yoke]

🤣

we all have those moments

[amber zenith]

I hope so

[buoyant yoke]

i think this is a common technique for factorials

this works because the numerator and denominator have the same amount of terms

if the terms in the numerator grow more than the terms in the denominator, you start getting issues

unless you can combine two or smth

[amber zenith]

aaah alright

[midnight plank]

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[safe onyx]

(proof for above picture)

[dusty portal]

Uh

The term with cx+d should have only an x^2+1

It is not squared

[safe onyx]

I thought it goes out twice?

[dusty portal]

What the

Nevermind

[safe onyx]

,w Partial fraction[(5x^2+7x+5)/(x^2+1)^2]

[grand pond]

Results provided by WolframAlpha

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Requested by iamsmeagol

[dusty portal]

I have not seen that before

Anyways continue

[safe onyx]

it is strange

(cropped out dx sorry)

a in this case is 1

[dusty portal]

Very obvious

[dusty portal]

Good job!!

[safe onyx]

😭

[dusty portal]

I hope that |- looking this is a plus sign

[safe onyx]

i forgot to move the negative down

[safe onyx]

i only had this in my head because the video from 3 problems ago used it

[safe onyx]

it is a plus

[dusty portal]

You are fine then

[safe onyx]

this - needed to go down

sorry bad crop XD

thank you!

.close

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[midnight plank]

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[dusty portal]

?

[safe onyx]

dx = du / 2x

[flat veldt]

1/2x

[dusty portal]

Me sus of that

[dusty portal]

I see it now nevermind

x^3=x*x^2=1/2 du*(u-7)

[flat veldt]

,w int x³/⁷√[x²+7]

[grand pond]

Results provided by WolframAlpha

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Requested by ishigamisenku_

[dusty portal]

Why

[safe onyx]

I think my u sub is wrong

should I do u = (x^2 + 7) ^1/7

[dusty portal]

No

[dusty portal]

^

[safe onyx]

but these are different

[dusty portal]

aaaaaaaaaaaa

I give up

[dusty portal]

You did NOT change the bounds

Remember that whenever you u-sub

[safe onyx]

oh

do I put x^2 +7 instead of the x?

[dusty portal]

You think about it

[dusty portal]

What did we sub?

[safe onyx]

u = x^2 + 7

[dusty portal]

What will the lower bound be?

[safe onyx]

7

[dusty portal]

It is still probably wrong

😔

[safe onyx]

oooh

no it is

thank you!!!

[dusty portal]

Yay!

[safe onyx]

I know this step is good now

[dusty portal]

😄

[safe onyx]

i found it

I didn't spread out the 1/2

I only did it on the first integral

the second one needs -49/12

it should be

Thank you for helping!
(thank you for the bounds adjust I forgot about that, will keep in mind

.close

[midnight plank]

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[vocal briar]

conceptually how do i think of this

i think in the unit circle so i dont rly get what it menas when it says its approaching from the right

[buoyant yoke]

mby easier to see with a u-sub

[vocal briar]

what does u rep

[buoyant yoke]

u is just another variable

[vocal briar]

yeah

[buoyant yoke]

for example, i take u=tan(t)

then u->?

[vocal briar]

tan of pi/2 from the right?

[buoyant yoke]

yes

whats $\lim_{t\to \frac{\pi}2^+} \tan(t)$

[grand pond]

Bonk

[vocal briar]

im so confused

how am i supposed ot know that from knowing u = tan(t)

[buoyant yoke]

i replace t with a new variable which is in terms of t

as choice i that u(t)=tan(t)

i need to transform my limit thusly

$\lim_{t\to \frac{\pi}{2}^+} e^{\tan(t)} \rightarrow \lim_{u(t)\to u\left(\frac{\pi}{2}^+\right)} e^{u(t)}$

[grand pond]

Bonk

[buoyant yoke]

does any of this make sense?

[vocal briar]

yeah i get this

so its js $e^{u\frac{\pi}{2}+\right}$?

[grand pond]

gamer75431
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[buoyant yoke]

yeah kinda

you know the chain rule?

[vocal briar]

yeah

[vocal briar]

so this x pi/2?

[buoyant yoke]

wdym?

[vocal briar]

cz e^upi/2 + or wtv is f'(x) and then the power differentiated with respect to u is pi/2

[buoyant yoke]

$e^{u\left(\frac{\pi}{2}^+\right)}$

[grand pond]

Bonk

[buoyant yoke]

this is what we are basically evaluating

[vocal briar]

yeah this times pi/2

[buoyant yoke]

why times pi/2?

[vocal briar]

wait

o

[buoyant yoke]

we dont actually use the chain rul ehere

[vocal briar]

oh

[buoyant yoke]

i wanted to ask if you knew the chain rule so that i could use it to compare

[vocal briar]

oh

lmao

yea

[buoyant yoke]

but you see how this is the case, right?

[vocal briar]

yeah i get that

[buoyant yoke]

okay great

does that help giving a better intuition about the original limit?

[vocal briar]

im still confused lol

like

if we go back

to without the u sub

how am i supposed to know what tan(pi/2 from the right) even means

[buoyant yoke]

$\lim_{t\to \frac{\pi}2^+} e^{\tan(t)}$

[grand pond]

Bonk

[vocal briar]

yes

[buoyant yoke]

,w graph tan(x)

[grand pond]

Results provided by WolframAlpha

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Requested by .bonk__

[buoyant yoke]

look at x=pi/2

[vocal briar]

infinity yeah cuz its an asymptote right

[buoyant yoke]

*-infinity

[vocal briar]

pi/2?

[buoyant yoke]

from the left is +infinity

[vocal briar]

oh

yeah

[buoyant yoke]

from the right is -infinity

[vocal briar]

true

so js e^-inf

[buoyant yoke]

yes

[vocal briar]

which is 1/e^inf

so 0

[buoyant yoke]

yup

[vocal briar]

yeah makes sense

thank you

.close

[buoyant yoke]

its not closing?

bot ded?

[vocal briar]

uh

idk

.close

[buoyant yoke]

.close

strange

just give it some time then

[vocal briar]

okok

cya

[midnight plank]

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[buoyant yoke]

ah, there itg oes

[midnight plank]

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[uneven sandal]

How do I find

Gal(F_27/F_3) ?

[uneven sandal]

or maybe first [F_27 : F_3]

.close

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[worn depot]

Grades are normally distributed with μ = 5 and σ = 1.5. It was decided that the lowest 20% would receive a failing grade. What minimum grade was required to pass the exam?

[worn depot]

So

we´re finding 80%

z score for that

0,84

so 80% of people fall between 0 - 0,84σ

im getting 6,3...

@cedar patio

[midnight plank]

Please do not ping individual helpers unprompted.

[cedar patio]

Its the same method as the previous question. Since we were asked to get the minimum grade which is 20%, then get the z score of the 20th percentile and use the formula again

[worn depot]

why 20thÐ

were finding for 80th

@cedar patio

<@&286206848099549185>

[midnight plank]

@worn depot Has your question been resolved?

[worn depot]

<@&286206848099549185>

[midnight plank]

@worn depot Has your question been resolved?

[midnight plank]

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[last slate]

yo

[last slate]

Q2:-"Check if +x-2x^2 , 2+5x-x^1 , x+x^2 span P^2 or not"

[last slate]

i can solve this question in the same way i solve it for Q2 right ?

because

[hard shard]

yes

[last slate]

okay

thanks

.close

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[midnight plank]

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[last slate]

Say i have this set

It doesnt span R^3 because we need 3 vectors but we only have 2 here

bacc

what about R^2 ?

i say no

[nova yoke]

are those elements of R^2?

[last slate]

because the vector should be of the form (a,b) only

[last slate]

no

[sharp coral]

well they certainly can't span a space of vectors they don't belong to

[nova yoke]

what you can say is that they span a 2-dimensional subspace of R^3

(i.e. a plane)

[last slate]

okay

thanks everyone

.close

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[last slate]

{(1,1)} ?

or should i also include(0,0) ?

[placid spoke]

any vector aside from the 0 vector can be considered the basis since it's in 1 dimension

[last slate]

1 dimension

missed that!

[sharp coral]

if a set contains the 0 vector it's automatically linearly dependent

[last slate]

so no

[tidal turret]

(x,y)=(y,y) = y(1,1)
U = <(1,1)>, y is free, x is constrained

every vector space contains 0, notice if y is free, y = 0 , y(1,1) = (0,0), no?

[lyric charm]

as said earlier, the zero vector can never be part of any basis, because any set containing the zero vector is linearly dependent (prove this if you haven't already).

[last slate]

found my mistake

thanks everyone!

.close

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[prime garden]

I need help.

[prime garden]

angle ABD = angle DBC and angle BCD = angle DCA. If angle BAD = 40 deg, find angle DAC.

Note: It is not allowed to use the fact that the angle bisectors of the angles of a triangle are concurrent.

[prime garden]

Yes.

[lyric charm]

are you allowed to drop perpendiculars from D onto all three sides

[prime garden]

Got it, thanks.

.close

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[viral dagger]

find all integers $a,b,c,d$ such that $a+b+c+d=ab=cd$

[grand pond]

skissue.in.a.teacup

[viral dagger]

ok idk why but for some reason i want to turn this into a polynomial??

$\sqrt{abcd}=a+b+c+d\geq 4\sqrt[4]{abcd}\implies abcd\geq 256$

[grand pond]

skissue.in.a.teacup

[long dagger]

(a+b+c+d)^2 >= 256

|a+b+c+d| >= 16

|ab| >= 16

|cd| >= 16

plenty of relations

[viral dagger]

oh right am gm doesent work for negative ints

i think you can just slap on a bunch of absolute vals tho so $|abcd|\geq 256$

[grand pond]

skissue.in.a.teacup

[spring wave]

you are assuming ab is positive

[viral dagger]

ok i dont think this qorks

[spring wave]

first assume that one of them is 0

then you get an easy to find set of solutions

[long dagger]

if one of them is zero then all of them are zero

[spring wave]

no

[long dagger]

or well

[spring wave]

(0,x,0,-x) is a set of solutions

[long dagger]

at least two of them have to be zero

[spring wave]

yep

[long dagger]

and the other two have to be of the form k and -k

[spring wave]

yep

[long dagger]

yeah

[spring wave]

then assume all are non 0

so there are 3 cases
a+ , b+, c+ , d+
a-, b+, c-, d+
a+ , b+ , c-, d-

[spring wave]

so in cases

1. A+B+C+D = AB = CD
2. A-B+C-D = AB = CD
3. A+B-C-D = AB = CD

[viral dagger]

wlog a is 0, then either c and d is 0, wlog c is also the 0, them b=-d

[spring wave]

for the third case
AB < A + B
AB - A - B + 1 < 1
(A-1)(B-1) < 1
SO A is 1, or B is 1
but then
A+B = AB+1
so C+D = 1, ehich is impossible

[viral dagger]

oh crap sorry my battery is low

ill open another one later

.close

[midnight plank]

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[tidal turret]

If V and W are vector spaces of the same finite dimension and V is a subspace of W, V is isomorphic to W?

and a followup question, is R3[X] isomorphic to R^4?

[tidal turret]

hey, wassuup bro

[grim vector]

You don't need that V is a subspace of W iirc

Being same finite dim is enough

[lyric charm]

V is a subspace of W and dim(V) = dim(W) < ∞ ?

this makes them straight up EQUAL not just isomorphic.

[tidal turret]

True

[tidal turret]

thanks 💖

what about follow up?

[grim vector]

Whats the dim of the first and second one ?

[tidal turret]

dim(Rn[X])=n+1

dim R3[X] = 4

dim R4 =4

[tidal turret]

xd

[tidal turret]

I appreciate the help to all you guys and gals

for helping with my shit

I didn't knew people reads other's descriptions

[grim vector]

I've forgot it was in it to be honest

[tidal turret]

is based

[grim vector]

Anyway you good with this question ?

[tidal turret]

ye, tysm

[grim vector]

Gut

[tidal turret]

I appreciate the help that is provided everyday

.solved

[midnight plank]

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[sly steppe]

Hello could I get help on this, i'm stuck

[sudden yacht]

Why did you change help channel?

[sly steppe]

nobody was there lol

[sudden yacht]

The people are the same in each channel 😅

[sly steppe]

💀

oh

lol im desperate

[sudden yacht]

Sometimes you just have to wait

[sly steppe]

myb it's just late and i'm trying to sleep lin algebra got me stressing

[sudden yacht]

You have to find a path that arrives at the "point" b, starting at the origin

[sly steppe]

yeah so i'm unsure on how to do that

[sudden yacht]

Of course you have to move by multiples of u and v

[sly steppe]

how do I do that?

[sudden yacht]

By looking at the graph

Nothing more than that

[sly steppe]

is it -u + 2v?

[sudden yacht]

For example, where do you arrive if you move 1u - 1v?

[sudden yacht]

That's w

[sly steppe]

wow you got that fast

[sudden yacht]

It is pretty straightforward actually 😅

[sly steppe]

um u-v involves something like the parallelagram rule doesn't it?

[sudden yacht]

It does

But you don't need to construct any parallelogram, they're already drawn within that grid

[sly steppe]

usually it would be like u and v are vectors and u-v is the middle part that's dotted

u-v = b?

[sudden yacht]

Nope

Can you mark the line path that makes you arrive at b?

[sly steppe]

I begin at the origin?

[sudden yacht]

Of course

[sly steppe]

like that?

[sudden yacht]

Is that vertical line a multiple of u or v?

You are only allowed to move on the "railways", which are multiples of u and v

[sly steppe]

so i can only make the line through the blue lines?

[sudden yacht]

Yes, exactly

I thought this was quite obvious 😬

[sly steppe]

it goes through -2v

ohh wait

[lyric charm]

u and v

not u and c

[sudden yacht]

Oh my god bad typo

[sudden yacht]

Perfect

[sly steppe]

so it's the middle dashed line of the parallelagram

-2v + u

[sudden yacht]

-2v + 2u

[sly steppe]

wait where does the 2u come from?

[lyric charm]

you moved along two gridlines' worth in the u direction.

[sly steppe]

ok, so when I get to the u point that's 1u and then 2u would be at the next intersection?

now it's c

-3v + 2.5u?

.close

[midnight plank]

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[midnight plank]

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[fleet sonnet]

How do I do (a) where do I start

[shadow osprey]

Wow

seems "fun".

[fleet sonnet]

idk what going on

h(x) = -1 / x-2 is the asymptote

and the other equation for the half perabola

porabola

where u go

.close

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[charred fable]

Hi, I have to "Model the velocity of a skydiver throughout descent with variations in air density with altitude?" I know how I can model the skydiver with velocity function models, however, I'm not quite sure how I'm supposed to model the changes in velocity with changing air density. Could someone please help me with this?

[polar star]

I mean it gets super complicated but you could try

ma = mg - (1/2)ρ(h(t))C_dAv^2

ρ(h(t)) being the density function dependent on height

[charred fable]

oh wow ok, do you have some other idea suggestions for a topic related to this then: Modelling the velocity of a skydiver throughout the descent

[polar star]

what exactly is this for?

[charred fable]

my math aa hl ia

[polar star]

I would not recommend aerodynamics for math ia

its super common and people end up getting most of the theory wrong because they simplify very complicated stuff too much

but if you do want to pursue it

you could try finding how accurate the quadratic drag model is in modelling a skydiver's descent

which is of the form:
ma = mg - kv^2

you can use irl data to check the percentage errors maybe. you shouldve done error propagation in sciences

[charred fable]

ok ty

[midnight plank]

@charred fable Has your question been resolved?

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[last slate]

guys how to do this

2). Graph using SLOPE-INTERCEPT FORM
-2x+3y=10

[dreamy lichen]

You should probably start by converting it to the slope-intercept form

which means you want to have the equation written as:
y = something

so you will need to somehow rearrange it to that form, do you know how to do that or do you need help with that?

[last slate]

is it y=-2x+3?

[dreamy lichen]

nope

you can do one of the following things:

1. Add something to both sides of the equation
2. Subtract something from both sides of the equation
3. Multiply both sides of the equation by something
4. Divide both sides of the equation by something

[last slate]

i add 2x?

to both the first one and the 10

[dreamy lichen]

yep, thats a good start

what will you get?

[last slate]

3y=10

[dreamy lichen]

remember that you have to add the 2x to 10 as well

[last slate]

3y=10+2x

[dreamy lichen]

yep

and now?

[last slate]

i divide all by 3?

[dreamy lichen]

correct

[last slate]

so its y=10+2x
3

?

[dreamy lichen]

yes

$y=\frac{10}{3}+\frac{2x}{3}$

[grand pond]

MathIsAlwaysRight

[dreamy lichen]

we could also split it up like this

and btw when you write fractions in text, you can use /
e.g. what you wrote would be written as (10 + 2x) / 3

[last slate]

okk

[dreamy lichen]

anyway, from this we can now read the intercept and the slope

[last slate]

so thats the answer?

[dreamy lichen]

not yet

[last slate]

what do i do next

[dreamy lichen]

you are supposed to graph it

[dreamy lichen]

now you need to find the intercept and the slope

do you know what these terms mean?

[last slate]

yup

do i need to find both x intercept

?

[dreamy lichen]

Not necessariyl

the y-intercept should suffice

[last slate]

ok

[dreamy lichen]

the y-intercept gives you one point on the line

[last slate]

soo (0,10/3) and (2/3,0)?

[rustic garden]

the y intercept (0, 10/3) is good, the x intercept is wrong

to graph an equation means to plot all of the points such that if you plug in (x,y) the equation is true

if you plug (2/3, 0) into the equation

it doesnt work

[dreamy lichen]

Yeah, only the (0, 10/3) is correct

the slope isnt the same as the x-intercept

[last slate]

what do i do for the x

[dreamy lichen]

the slope tells you how fast the line is growing

[last slate]

ohh ok

[dreamy lichen]

Slope of 2 / 3 means that if you move 3 units to the right, the line will rise 2 units up

[last slate]

ohh ok

[dreamy lichen]

so you can either use that to get your second point (it's pretty easy to move right by 3 units and up by 2 units on a graph paper)

or alternatively, you can just plug in e.g. x = 1 or x = 2 to get your second point on the line

and then by connecting the 2 points you have, you'll get the line