help-17

[severe stone]

nee lm

dat wordt denk te ingewikkeld aangezien je da ni heb gehad

oke dus nu hebben we 4^x²

[burnt lily]

ja

[severe stone]

wat ik altijd doe bij exponentiele functies de afgeleide

eerst gewoon overschijnen

dus ik schrijf 4^x²

[burnt lily]

ja doe ik ook meestal alleen dit is digitaal dus dan is er iets minder vrijheid over wat ik kan opschrijven

anders zou ik ook [x]' gebruiken

om het op te delen

[severe stone]

aaah oke

[burnt lily]

maar het moet in een stap

[severe stone]

dus eerst schrijf je dit op

4^x²

dan doe ik de ln(4)

dus 4^x² * ln(4)

snap je die stap nog?

[burnt lily]

ja dat is voor g

zegmaar

[severe stone]

en daarna doe ik de afgeleide in de macht in dit geval x²

[burnt lily]

in f(g(x)

[severe stone]

f

g is x²

[burnt lily]

hoe

hoe

het is toch

f = ^2

en g = 4^x

want het is (4^x)^2

[severe stone]

dit is geen functie

[burnt lily]

jawel x^2

[severe stone]

ooho zo

[burnt lily]

en x is 4^x

[severe stone]

alleen

is

(4^x)² is niet gelijk aan 4^x²

[burnt lily]

niet?

dat dacht ik

ik dacht eigenlijk ook dat

4^x^2

hetzelfde is als

4^2x

[severe stone]

nee

helaas niet

[burnt lily]

het was toch altijd zo dat

[severe stone]

wil je dat ik hem voor je uitschrijf?

[burnt lily]

(a^b)^c

=

a^b*c

dat dacht ik altijd

is dat echt niet zo?

[severe stone]

ja klopt alleen nu zit de kwadraat alleen op b

ja beetje fucked up

[burnt lily]

die haakjes maken uit?

[severe stone]

ja :/

maar wiskunde heeft wel vaker kut notaties

[burnt lily]

tering

[severe stone]

wacht ik probeer een vb te verzinnen

[burnt lily]

ok

dit is wel complex man

[severe stone]

je went eraan

jij kan dit

[burnt lily]

het is ook zo kut om dat in discord zo te schrijven

met ^enzo

[severe stone]

ja eens je hebt zo'n bot die het netjes kan maar die nsap ik ni

[burnt lily]

ja die snap ik ook niet haha

das latex vgm

[severe stone]

maar ik zl ff het antwoord voor je uitschrijven

ja klopt

[burnt lily]

huh

dus

[severe stone]

je kan ln(4) nog omschrijven naar ln(2²)=2ln(2)

[burnt lily]

waarom wordt die u^x

[severe stone]

das een 4

sorry man

[burnt lily]

oh shit

ja

ok

[severe stone]

hahaha

[burnt lily]

maar maakt niet uit voor dit

waarom wordt die 4^x^2

ln(4)*4^x^2

en niet

ln(4) * 4^x

het hoort toch alleen dat binnenste te zijn

[severe stone]

aah oke

dus h(x)=f(g(x)) toch?

[burnt lily]

of is dat omdat het niet (4^x)^2 is

[severe stone]

h'(x)=f'(g(x)) * g'(x) dus de afgeleide van f'(x) met g(x) erin dus afgeleide van 4^x maar dan op de x zit x²

maar wat ik dus altijd zo doen met exponentiele functies met afgeleides

altijd overnemen

dus e^cos(x) de afgeleide hiervan

bevat sws iets met e^cos(x)

en 4^x de afgeleide bevat altijd iets met 4^x

dus altijd beginnen met overschrijven

[burnt lily]

tering ik weet die ook niet denk ik

[severe stone]

probeer het

[burnt lily]

ok

ik heb alleen die e nooit gesnapt bij afleiden

die blijft hetzelfde ofzo

e^x is nogsteeds e^x

echt vaag

[severe stone]

ja klopt

aangezien je eigenlijk doet

[burnt lily]

-sin(x)*e^cos(x)*e^cos(x) is het enigste wat ik kan bedenken

maar dat zal wel fout zijn

[severe stone]

functie overschrijven * ln(grondgetal, in dit geval e) dus e^x * ln(e) = e^x * 1 = e^x

[severe stone]

hoezo 2x de e^cos(x)?

[burnt lily]

oh ok dat is wel logisch

[severe stone]

wat is jouw reden daarachter?

[burnt lily]

omdat je ketting regel doet dus een keer om e^cos(x) te differentieren en 1 keer als bijproduct van cos(x) te differentieren als f(x)

maar miss niet dan

[severe stone]

de ketting regel is die -sin(x)

[burnt lily]

ja klopt

maar je moet toch ook nog g(x) laten staan

of is dat dan gewoon e

[severe stone]

nou wat ik dus doe bij exponentiele

is dus gewoon het volgende

e^cos(x) afgeleide : eerst overschrijven, dus e^cos(x) keer ln(grondtal) dus ln(e) * afgeleide in de exponent dus -sin(x)

dus dit wordt e^cos(x) * -sin(x) (ln(e)) is 1

[burnt lily]

ik zat wel in de buurt

[severe stone]

absoluut

dit is hoe chatgpt het uitlegt

misschien is dit fijner voor je

[burnt lily]

ik denk dat ik dit snap

ik denk dat ike r nu nog een zou kunnen

met iets in exponent

[severe stone]

2^ln(x³+x²) ?

als je dit kan kan je ze allemaal eigk wel

[burnt lily]

oh shit ln was iets met delen vgm ik weet niet of ik die nog kan hahaaha

ok ik ga proberen

[severe stone]

is goed

succes

[burnt lily]

2^ln(x^3+x^2) * ln(2) * (1/(x^3+x^2)) denk ik

ben alleen niet helemaal zeker van ln(x^3+x^2) gedifferentieerd

[severe stone]

eerste deel is indd goed

je vergeet 1 ding

nog

want nu is het niet h(x)=f(g(x)) maar nu heb je

l(x)=f(g(h(x)))

aangezien in de ln ook een functie zit

[burnt lily]

ah

ofc

dus is ook nog eens * 3x^2+2x in dat laatste deel

[severe stone]

precies

en dan heb je hem helemaal

[burnt lily]

damn ik heb dat ln goed gedaan/

[severe stone]

jup

[burnt lily]

dacht echt dat ik dat vergeten was hahaa

[severe stone]

neejohh

welk jaar zit je nu?

[burnt lily]

1e jaar hbo

[severe stone]

krijg je veel wiskunde op je studie?

[burnt lily]

redelijk wat volgensmij

maar vooral lineare algebra

maar nog niet

dit is zegmaar voorbereidend

[severe stone]

hahaha succes :p

[burnt lily]

ik ben goed met vectoren tho

[severe stone]

ja weet ni hoe diep ze er op in gaan op hbo

[severe stone]

er is trouwens nog een handig trucje als je differentieren echt lastig vindt

dat is differentieren door gebruik te maken van de ln

[burnt lily]

nou normaal heb ik er geen problemen mee maar het is nogal een tijdje geleden

[severe stone]

aaah oke

[burnt lily]

en ik moet nu in hele korte tijd

compleet vwo wiskunde doen

dus het blijft moeilijk hangen

[severe stone]

dalijk dus lekker integralen enzo ook

[burnt lily]

jep

;-;

[severe stone]

dan moet je extra goed differentieren

[burnt lily]

ik weet

[severe stone]

dus vooral goed je differentieren trainen

[burnt lily]

maar vgm gaat het niet heel diep in op integralen

ik vind gewoon al die regels zo kut

[severe stone]

Aaah dat scheelt

[severe stone]

succes met lineair algebra :p

[burnt lily]

lol

[severe stone]

gewoon veel oefenen en dan lukt het je sws

[burnt lily]

heb hier bijvoorbeeld totaal geen moeite mee

[severe stone]

kijk das top

[burnt lily]

het is gewoon kut dat het in een keer moet in deze oefeningen

je kan zegmaar niet ding voor ding differentieren en dan [x]' erin zetten

[severe stone]

nee maar dat gaat wel wennen

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[finite hatch]

how did they get (4x^3 - 5x)^3

[finite hatch]

on the denominator

[empty frigate]

the u^-3 becomes u^3 in the denominator, and then you replace u with 4x^3-5x because that's what we originally said it is

[finite hatch]

also i dont get when its dy/du or du/dx or dx/du

[finite hatch]

oh

[bitter pilot]

that's basically chain rule

dy/dx = dy/du * du/dx

[finite hatch]

what does dy/du mean

like whenever i see smt like that dx is on the top

[atomic jasper]

its the derivative of y with respect to u

[bitter pilot]

like instead when you do u = 4x³-5x then y = (4x³-5x)^-2 becomes y = u^-2

and dy/du means you differentiate y = u^-2 with respect to u first (the outer function) and afterwards multiply by the derivative of u (inner function) which is a function of x

[finite hatch]

im kind of confused

why cant it be dy/dx

@bitter pilot

oh

dy/dx is the gradient thingy

but why can the other one be du/dx

but y cant be dy/dx

[bitter pilot]

you basically differentiate y not directly with respect to x because that's kinda meh, how would u do that other than trying to expand or simplify (4x³-5x)^(-2)

so the idea is to cover the complicated thing for a moment and call it u for example

so y = u^(-2) now

that is now easy to differentiate

so it becomes dy/du = -2u^(-3)

now if you want to get dy/dx

[finite hatch]

wait did they substitute u with 4x^3-5x

[bitter pilot]

you have to multiply by the inner derivative of u

which is du/dx

[bitter pilot]

yes

[finite hatch]

oh

they remove expanding

is that what this is for

[bitter pilot]

basically

like how do you expand and simplify that in the first place

so you try to rather think of your current function as a composite function

[finite hatch]

tbh my school hasnt taught me function notation yet

so weird

bcoz i need it now

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[civic otter]

Can anyone help me, please?

[civic otter]

In particular, I don't think it is solvable with the classical methods

Because it's neither linear nor separable.

And even if I wanted to make a substitution, I can't think of which one

[bitter pilot]

maybe you can think of it as a bernoulli differential equation

no nvm

maybe polar substitution?

https://math.stackexchange.com/questions/446926/riccati-differential-equation-y-x2y2

[civic otter]

I've never done Riccati actually

[civic otter]

Even if I understood this I don't think my professor wants me to use that

But true, I don't see other ways to do it

[bitter pilot]

i sent it because that person also asked for a method with classical tools and there might be a hint

[civic otter]

But how can I obtain this crazy thing? 😅 I've never seen these "J" functions

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@civic otter Has your question been resolved?

[sudden compass]

Those j functions are called bessel functions

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[gusty mortar]

how do i differentiate this my solution was (x-1)^2/2rootx + 2rootx(x-1)

[reef agate]

✅

[gusty mortar]

it is incorrect the marking scheme says so atleast

[reef agate]

Your marking scheme not this is the correct answer?

[gusty mortar]

this is the answer in the ms

hello ?

<@&286206848099549185>

solved it

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[bleak prawn]

I think I can solve this but im just stuck on one thing

I know when I convert it into the form with u and du, when integrating I need to also do something with the limits of integration, no?

[bitter pilot]

yes

[bleak prawn]

I dont remember specifically what I have to do

[bitter pilot]

if you dont want to resubstitute in the end you need to change the bounds

[sudden compass]

Do you know what to substitute though?

[bitter pilot]

if you have a substitution u = u(x) then you simply do u_1 = u(x_1) and u_2 = u(x_2)

if x_1 was the lower bound then u_1 becomes the new lower bound, and if x_2 was the upper bound then u_2 becomes the new upper bound

[bleak prawn]

So here u = sqrt(x)+4, so my new bounds become
sqrt(1)+4 = 5
sqrt(0)+4 = 4

Right?

Hmm I got a negative number which cant be right, unless I made a mistake somewhere else...

[bleak prawn]

Is this bit right at least?

[bitter pilot]

sqrt(0)+4 = 4

can u show you work

integrand is positive so you should be get positive result

[bleak prawn]

u = sqrt(x) + 4
du = 1/2sqrt(x) dx

[bitter pilot]

dx

,, \int_4^5 \frac{1}{u^4} \cdot 2\sqrt{x} : du = : ?

[bitter pilot]

how did you solve for sqrt(x)

[bleak prawn]

I calculated x first

u = sqrt(x) + 4

(u - 4)^2 = x

[bitter pilot]

yikes

[bleak prawn]

Then sub in that for x gives

2(u-4)/u^4

[bitter pilot]

it's not really wrong but very complicated

u = sqrt(x) + 4
just move the 4 to the left

besides that you'd introduce with your method unnecessary absolute values

[bleak prawn]

Ah yeah I didnt notice it was square root x I couldve just subbed in directly

[bitter pilot]

yea

[bleak prawn]

But yeah I got this

(2u-8)/u^4

2/u^3 - 8/u^4

2u^(-3) - 8u^(-4)

integrates to
-1/u^(2) + 8/3u^(3)

[bitter pilot]

you forgot to divide by -2

the left term

[bleak prawn]

Yepppp that gave me the right answer

[bitter pilot]

,w (-1/5^2 + 8/(3(5^3)) - (-1/4^2 + 8/(3(4^3))

[twin meteor]

Results provided by WolframAlpha

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

Requested by adoniszgks

[bleak prawn]

Noice

[bitter pilot]

seems you messed up a sign before

[bleak prawn]

I messed up a few things 🥲

But my method was right thats the most important I guess

Thank you!!

❤️

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[lime gorge]

how do i even start this?

[lime gorge]

maybe is it finding if f_r,\theta = f_r * f_theta ?

We know x^2 + y^2 = r^2

.clos

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[simple kraken]

I’ve been trying and k can’t seem to remember how to find the equation

[reef agate]

$y = a(x - h)^2 + k$

[twin meteor]

King Leo

[reef agate]

vertex: $(h, k)$

[twin meteor]

King Leo

[simple kraken]

I’ve gotten to 5=a(0-3)^2-4

[reef agate]

how did you get - 3 and -4

[simple kraken]

I got 3 from the y axis and don’t know how I got -4, I think it should be +5 instead of -4

[reef agate]

can you first label the vertex shown, and write its coordinate as an ordered pair

i think youre getting confused

[simple kraken]

V(-1,3)

[reef agate]

so what are h and k

[simple kraken]

-1 and 3?

[reef agate]

now can you apply $h = -1$ and $k = 3$ to $$y = a(x - h)^2 + k$$

[twin meteor]

King Leo

[simple kraken]

I’ll try that right now

Would it turn to a positive 1?

Would it be y=a(x+1)^2 + 3

[reef agate]

now pick any non-vertex point, plug it into your vertex-form parabola, and solve for a

[simple kraken]

-2,5

5=-a+3

[reef agate]

what is (-2 + 1)^2

[simple kraken]

Ohh it’s 1

So just 5=a+3?

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[reef agate]

✅

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[heady timber]

I cannot figure out how to find f with like 3 variables

[vocal sleet]

@heady timber Has your question been resolved?

[heady timber]

<@&286206848099549185>

[gaunt sparrow]

I've not done this in a while tbh.
Isn't the wave speed something like $\frac{\omega}{k}?$

[twin meteor]

Azyrashacorki

[gaunt sparrow]

Where your function is expressed as $y = A \sin(kx - \omega t)$

[twin meteor]

Azyrashacorki

[gaunt sparrow]

Then all you really need is to solve for $f$ in $\frac{\omega}{k} = 36$

[twin meteor]

Azyrashacorki

[heady timber]

thank you

[heady timber]

can you tell me what omega is ?

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[inland flower]

am i plugging in -1,4,9 into T(x1,x2)

[inland flower]

..

[balmy salmon]

You must create a system of equations

[proud terrace]

What is the expression of $T(x_1,x_2)$? And what is it equal to according to question b) ?

[twin meteor]

ss

[proud terrace]

then, you will have a system

[twin meteor]

ss

[sacred flicker]

yo since yall know what yall talking about

can someone help

[inland flower]

??

[sacred flicker]

this is grade 11 math btw

[inland flower]

that is not how this server works

[proud terrace]

its busy

go to help #11

@inland flower you understood ?

[inland flower]

im trying too lol hold up

[inland flower]

is this the expression

and is it equal to -1,4,9

[proud terrace]

yes so how can we find x_1 and x_2

[inland flower]

do i write it out like this
V

[proud terrace]

there is an equal but yes

you can also write it like 2 equations

[balmy salmon]

That's not possible

[inland flower]

what

[proud terrace]

its [
\begin{pmatrix} 1 & -2 \ -1 & 3 \ 3 & -2 \end{pmatrix}
\begin{pmatrix} x_1 \ x_2 \end{pmatrix}
= \begin{pmatrix} -1 \ 4 \ 9 \end{pmatrix}
]

[twin meteor]

ss

[proud terrace]

there is an equal

you forgot it but nvm

[proud terrace]

write it like a system of 2 equations

[inland flower]

oh yeah i added but why is the person saying what i wrote isnt possible

[proud terrace]

you need an equal here

i think that’s why

[balmy salmon]

The matrix with x_1 and x_2 cannot multiply with the matrix with -1, 4 and 9

[balmy salmon]

Yes

[inland flower]

but isnt this matrix only with x1 and x2

[inland flower]

how am i supposed to make it so i can multiply it with -1,4,9

[proud terrace]

you can have a 3rd line with 0 on it

[
\begin{pmatrix} x_1 \ x_2 \ 0 \end{pmatrix}
]

[balmy salmon]

Yup

[inland flower]

do i just plug those numbers back into the equation now

and should i write the matrix out like this now

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@inland flower Has your question been resolved?

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[reef agate]

This looks like a test

[atomic jasper]

whats your question??

[thorn spindle]

no its hw

[thorn spindle]

isnt similar triangles r/h

r/x*

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[rigid maple]

someone explain how to answer this please

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[rigid maple]

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[kind fiber]

Can somebody tell me if I'm doing this correctly? I'm having trouble with it

[kind fiber]

Sorry

[reef agate]

No biggie, but in the future, its faster if you ,rotate your own images

[kind fiber]

I didn't even know that was a thing TBF

But yea I'm struggling with this

Any advice?

[reef agate]

You got it right, and you used proper methods

Writing 0/0 on your paper is a BIG no-no

[kind fiber]

Hmmmm. Okay I will take note of that. This is why I did it tho

[reef agate]

This statement is unneeded, because youre evaluating the limit x -> -5, not the value x = -5

[kind fiber]

These are class notes and he does the whole 0/0 (needs more work) thing

Thats why i did it

Should i speak with my professor about it just in case?

[reef agate]

Are you in usa?

And are you taking AP calculus AB

[kind fiber]

Isn't the limit x--> -5?

[kind fiber]

This is college 131

Math 131

Calc with geometry

Usa

[reef agate]

Edited ✅

[reef agate]

Ok, i'm still pretty sure that you should avoid 0/0

[kind fiber]

So would I be correct or wrong about writing that part specifically?

[kind fiber]

No issue and I appreciate you letting me know. He was probably just writing it to show that we need to have it in our heads that it "needs more work"

[reef agate]

Instead, you should write:
$$\lim_{x \to -5} \qty(x^2 + 3x - 10) = 0$$
$$\lim_{x \to -5} \qty(x^3 + 11x^2 + 30x) = 0$$
$$\text{Indeterminate Form}$$

[twin meteor]

King Leo

[kind fiber]

I love that

Thanks buddy

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[viral galleon]

I have a specific question about this problem

so basically I have a model where red dots can only connect to blue dots. And this problem is in the context of axioms of betweenness and incidence by the way

Why does line r1q (q being r2qb1) not intersect pr2

I know its not suppoed to because of my professor saying so but him re explaining it to me didn't make sense

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@viral galleon Has your question been resolved?

[viral galleon]

<@&286206848099549185>

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<@&286206848099549185>

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[snow nimbus]

My teacher told me to use the pythagorean theorem to solve this but I dont know which values to insert 🥹

[snow nimbus]

Dont mind the pencil writings

[snow nimbus]

Also idk if what they said is right, after all they are my science teacher.. 🥲

[lilac sluice]

so using the pythag theorem is correct here. we know XN is 10 units right?

and we know the radius is 12?

[snow nimbus]

Yes

[lilac sluice]

draw a line connecting N and Q and what do you notice about that line

oh wait hold on

i misread gimme a sec

[snow nimbus]

Its a hypotenuse 😛

[lilac sluice]

but it's also the radius

[snow nimbus]

Ok jett 🔥🔥

Ig so

How do I apply this, if this is the right theorem

[lilac sluice]

so i actually dont know how to solve this

my bad

u can ping helpers if you'd like

[snow nimbus]

Oker ☺️

<@&286206848099549185> hi . . . pardon

[dusky light]

Hello

[peak fjord]

Shouldn't QX equal QY if both chords are congruent?

[dusky light]

I was about to say that, yes.

[peak fjord]

Then just connect Q and N

You got two sides already, use Pythagoras to find the third i.e QX

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[rigid torrent]

why do I go from 1/2 to 1/4?

[viral copper]

2 was factored out

you can multiply the 2 back in and see it's the same

[foggy rune]

U could do the same thing without factoring out right
(1/sqrt(2) x+sqrt(2))(1/sqrt(2) x-sqrt(2)=1/2 x^2 - 1
I guess it’s just cleaner to not have sqrt of 2 in there

[quiet echo]

2*1/4=1/2

[rigid torrent]

so can I say, 1/2x^2-2=2(1/2)x^2-2(1)?

and from that if do different of squires?

[quiet echo]

Ah. Not necessarily.

ab+ac=a(b+c)

[rigid torrent]

Sry I still dont understand it.

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[tough lily]

stupid question but how do I solve this.. I feel like I'm not understanding the question or I'm missing some key information. The screenshot is translated from finnish to english so it might be a little bit scuffed..

[hearty jungle]

I think it just wants you to subtract the numbers

like for a) you just put sqrt(3)

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[pliant rose]

i need help

[hearty jungle]

did you evaluate u2

to check if it increases or decreases

[pliant rose]

it decreases

[hearty jungle]

doesn't that kind of mean it has to have a finite limit

ok not really

idk

[pliant rose]

🐧

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@pliant rose Has your question been resolved?

[leaden ingot]

Prove that $f(x)=\frac{x\left( x^2+3 \right)}{3x^2+1}$ is strictly increasing

[twin meteor]

TargetVN

[pliant rose]

.close

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[jagged cargo]

how is my proof here?

[vocal sleet]

@jagged cargo Has your question been resolved?

[cobalt crypt]

yeah its fine

also

this is a special case of the lemma i got you to prove before

with the exercise you gave up on

[jagged cargo]

wait rlly?

[cobalt crypt]

yes

a basis of F is any non-zero number, so here you can take phi(u)

and u will be a preimage

[jagged cargo]

oh i see

[cobalt crypt]

X = span{u}

[jagged cargo]

yeah makes sense

alright thanks

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[trail moss]

Hello

[night glen]

hey crazy

[trail moss]

How can i solve ex 23

[full summit]

,rccw

[silk osprey]

pascal

[trail moss]

Binomial theorem?

Ok thx

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[fluid flare]

Hello

[fluid flare]

i am unable to reach an answer

not in options

please ping me

<@&286206848099549185>

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@fluid flare Has your question been resolved?

[fluid flare]

<@&286206848099549185>

[kindred hearth]

hello

[fluid flare]

hello

[uneven flicker]

@fluid flare still need help?

[fluid flare]

yes

[uneven flicker]

What have you tried?

[fluid flare]

I constructed AE

then got BOA+COD=135

so BOC 90

but it is wrong

and i am unable to find another solution

[uneven flicker]

I think 90 is correct too

My way is draw OM

M is midpoint of AD

CDO congruent to MDO

Therefore DOC + AOB = 135

[uneven flicker]

RHS

[fluid flare]

Also how can i get better at geometry

this question took me way too long

any suggestions

[uneven flicker]

Practice

Try to think of multiple solutions of questions you do

[fluid flare]

Thanks

.close

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[uneven flicker]

Npp

[vocal sleet]

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[inland flower]

is thiis being done right and if it is im stuck now idk what the next step is

[bitter pilot]

how did you get rid of the 2 in the last row

0020 -> 0000

up before that it looks good

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[inland flower]

divided r3 by 2.. is dat allowed

[inland flower]

.reopen

[bitter pilot]

and since when is 2/2 = 0

[inland flower]

oh wait

yeah brain wasnt braining i did that last night

i just realized

thanks

[bitter pilot]

u r welcome

[inland flower]

so would this be righrt

do i turn it into an equation now

...

[left portal]

x3 = 0

x1 -4x2 -5x4 = 0
x2 + 3x4 = 0

[inland flower]

would that be the final answer

[left portal]

no you need to find a basis

[inland flower]

whats that

[bitter pilot]

he means you need to determine the solution

[bitter pilot]

since x_3 = 0 you can put that into equation 2 and then you will notice you have one free variable

[left portal]

1. x1 = 4x2 + 5x4
2. x2 = -3x4
3. x3 = 0

1.1) x1 = 4(-3x4) + 5x4
1.2) x1 = -12x4 + 5x4
1.3) x1 = -7x4
2) x2 = -3x4
3) x3 = 0

[inland flower]

ohh ok

[left portal]

(x1,x2,x3,x4) = (-7x4, -3x4, 0, x4)
(x1,x2,x3,x4) = x4(-7,-3,0,1)
x4 ∈ R, x4 is free variable

[vocal sleet]

@inland flower Has your question been resolved?

[inland flower]

where did the x4 come from in the first (x1,x2,x3,x4)=
ik the second one u took out x4 as the common factor right?...

[left portal]

I dont know if its easy to explain but
we know

1. x1 = -7x4
2. x2 = -3x4
3. x3 = 0
   since x4 is free, x4 = t
   x1 = -7t
   x2 = -3t
   x3 = 0
   x4 = t

t ∈ R

maybe I am not making things more clear

[inland flower]

oh so just the fact that we know that x4 is a free variable

[left portal]

x2 and x1 are constrainted to the value of x4

and x4 is constrainted to t which is a real number

maybe I am making things more confusing, I am sorry

[inland flower]

i think i understand

so when its saying to find all the vectors

do i turn
x1 = -7t
x2 = -3t
x3 = 0
x4 = t

into a vector

like

...

[left portal]

but most likely we messed up in the row reduction step because

,w Nullspace {{1,-4,7,-5},{0,1,-4,3},{2,-6,6,4}}

[twin meteor]

Results provided by WolframAlpha

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

Requested by 938c2cc0dcc05f2b68c4287040cfcf71

[left portal]

wolfram Is giving span{(9,4,1,0)}

[inland flower]

damn

so i gotta restart

[left portal]

well go through the row reduction and see where u messed up

but you can check with wolfram

[inland flower]

either im blind or there shouldnt be anything wrong

[left portal]

after you divide row 3 by 2

you confused the 3 with a 5

thats my 2 cents, maybe I am wrong haha

[left portal]

@inland flower

[inland flower]

checking rq

hmm

OH

waittt

[inland flower]

yeah

thats what im looking at rn

lol

[left portal]

third row third col

for the second matrix

[inland flower]

it was upposed to be -4

right

[left portal]

yea

[inland flower]

i did see my own handwriting as a 5

damn

😭

thats crazy

how did i go that wrong

all i gotta do now is subtract r3-r2

wait

yeah

[inland flower]

do i do the same plugging in thing i did before

this not seeming right now

[vocal sleet]

@inland flower Has your question been resolved?

[bitter pilot]

still here?

[inland flower]

yeah

[bitter pilot]

whats the issue any updates

[inland flower]

I went back and redid my matrix because it was wrong and now i have the last row 0000 and i made an equation but im not sure what to do next

because when i first did it i just had x3= 0 so i plugged than in and went from that

but now the second equation is long

and i made x2=4xx3-3x4 and tried plugging it in but idk if it seems right

so idk what the next step is

[bitter pilot]

ok yes

ok so you got basically two equations

it will have two free variables if you solve for one

[bitter pilot]

so far it's correct you just need to finish

figure x_1

x_2 is already given

and then your solution is of the form

you can write it as a vector

I will write it as a spoiler so you can check up

[inland flower]

am i doing this riight

[bitter pilot]

it's 4x_3 - 3x_4

not +

[inland flower]

oh shoot

[bitter pilot]

forgot the = 0

but yes you are going right

[left portal]

the problem is that when I checked with wolfram, the nullspace of the matrix A has only one free variable and not 2, maybe there is another misstep in the row reduction

[bitter pilot]

where

i checked as well and it's got two

[bitter pilot]

you put a 4 instead -4

[bitter pilot]

and we would write this in set notation

so we express all solutions

[inland flower]

the x1 -9x3 7x4? = 0

[bitter pilot]

x_1 will be replaced with 9x_3 - 7x_4

[inland flower]

do i plug that back in

[bitter pilot]

yes you replace x_1 with that

your solutions only depend on x_3 and x_4

since they are free variables

[inland flower]

do i plug x2 back in as well

in here?

[bitter pilot]

wait

what are you doing rn

[inland flower]

i thought i was replacing x1

[bitter pilot]

yes but with 9x_3 - 7x_4 only

they are the free variables!

x_ 1 = 9x_3 - 7x_4

[inland flower]

in here right

[bitter pilot]

we just derived that

no

[bitter pilot]

here

[inland flower]

oh

[bitter pilot]

when you solve a system of equations you can write the solutions in vector form

[inland flower]

can i put x2 as 4x3-3x4

[bitter pilot]

yes and you have!

[inland flower]

um

idk if im understanding but

[bitter pilot]

yes

and now x_3 and x_4 too

you got 4 coordinates

[inland flower]

how do i get x3 and x4

[bitter pilot]

😄

[inland flower]

or do i already have that

[bitter pilot]

you already have that

they are free

so they remain themselves

[inland flower]

ohh ok

[bitter pilot]

yep

Now if you want

you can but you dont have to replace x_3 and x_4 with variables like s and t

to make it look nicer

[inland flower]

is that all

for the question

thats the vector?

[bitter pilot]

you would write it into a set

let me show

[bitter pilot]

like this

can pick either of these two

this set contains all possible solutions of your system

[inland flower]

so writing either one of these ways
would be right?

[bitter pilot]

yes

it means for every vector in R^4 it must have the form of this vector inside

because then it is a solution 😄

expanding it like in the right can give you an idea of how it looks

and yea it's a plane because it's two independent vectors and they always span a plane

and that plane is in 4dimensional space

kinda hard to visualize

[inland flower]

this stuff feels like it should be easy but im struggling🥲

[bitter pilot]

nooo

it's okay

take your time to understand

[inland flower]

my two other questions are like

half started

cuz i got stuck

like

for this question

are the numbers in green being plugged back into the equations

[bitter pilot]

plugged back into the equations?

i dont understand the question

[inland flower]

like

[bitter pilot]

are you trying to write down the equations?

oh

no

no

no

not atll

else it would be

T(-1,4,9)

[inland flower]

this seems like the first question but the vector is already given to us

so working backwards or sum

?

[bitter pilot]

but you want to find x such that T(x) = (-1,4,9)

yes

[inland flower]

so whats the first step i take

[bitter pilot]

to solve T(x) = (-1,4,9)?

[inland flower]

yes

assuming everything else i did so far is right

[bitter pilot]

yes splendid

next step is you can actually multiply the matrix

and derive the system

and the equations

does it make sense?

[inland flower]

i think so

[bitter pilot]

I realized a little mistake you did

[x_1, x_2] just

3 components then the matrix multiplication is invalid

(3x2) (2x1) -> (3x1)

so let me fix something

[inland flower]

ok

[bitter pilot]

can you write down the equations?

[inland flower]

like x1-2x2 -x1+3x2...

[bitter pilot]

a bit clearer please

the first one you started right

1 * x_1 + (-2) * x_2 + 0 * 0 = -1
x_1 -2x_2 = -1

[bitter pilot]

these would be equations if they had a = too

but yes you are on the right track

[inland flower]

are u saying like this

[bitter pilot]

oh simpler

x_1 - 2x_2 = -1 right

-1x_1 + 3x_2 = 4

3x_1 - 2x_2 = 9

these are our equations

[inland flower]

ohh

[bitter pilot]

and we can write them into a matrix!

i do this so thoroughly so you understand

So we can solve the following
[ \left [ \begin{array}{cc|c} ,1 & -2 & -1 \ -1 & ,3 & ,4 \ 3 & -2 & 9 \end{array} \right ] ]

[inland flower]

lemme guess

[inland flower]

perform row operations ?

[bitter pilot]

yes!

so you get the solution

for x_1 and x_2

u get it?

[inland flower]

i think so

lemme do this rq

[bitter pilot]

that's all we are looking for

find x_1 and x_2 such that T(x_1,x_2) = (-1,4,9)

[inland flower]

uh does this seem right

and if so when turning it into an equation do i do x1-2_x2=-1 or is it x1-2_x2-x_3=0

[bitter pilot]

last row operation R_3 - 2 is illegal

instead do -2 * R_1 + R_3

and also it's better to utilize the diagonal elements when row reducing

like after R_3 - R_1

do immediately -3R_1 + R_3 to also make the 3 a 0

then move to the entry where -2 was

i gtg for now, i hope you can somehow manage on your own for now, for reference what to do you try to reread how we approached the previous system

and how we wrote down the solution

[inland flower]

after i do this

do i equal the equation to 0

and do what i did in the first question

[bitter pilot]

not to 0!

[bitter pilot]

I did put the vertical line for a reason

[inland flower]

oh ok

just checking

[bitter pilot]

you equate it to the numbers in the 3rd column

the vertical line means basically what come after the = sign

[inland flower]

thanks

[vocal sleet]

@inland flower Has your question been resolved?

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[lyric shadow]

So I have an electron, and the probrability of finding it at n state is:
p_e = (196/(pi^4n^4) ) * sin(pin/2)
Thus the commulative distribution is the sum of the p_e's to N.
Now, how can I find the smallest N such that the added probrabilities are greater than 0.999?

[lyric shadow]

So how can I find the minimum amount of probrabilities, added from n=1, so that they add up to being greater than 0.999?

Like I could just brute force, trial and error this but it seems, unscientific

This probrability distribution is multinomial (to pretty much infinity) so it's hard to work around

[spiral turtle]

Your probability distribution goes negative.

Did this a probability distribution or amplitude?

@lyric shadow

[lyric shadow]

it shouldn't, the probrabilities should just be getting exponentially smaller as n increases

[spiral turtle]

also the value for n = 1 is larger than 2

have you tried actually computing these values?

[lyric shadow]

yeah, that's the point, the probrability of finding the electron in higher states decreases exponentially

the only mistake i've made is putting 196 instead of 96

thus p_e = (96/(pi^4n^4) ) sin(pi*n/2) actually

[spiral turtle]

if it's a probability mass function, then the probabilities should be strictly between 0 and 1

ok, that brings the n=1 case to below 1. This is progress

[lyric shadow]

with 96 nwo as 196 it is so

[spiral turtle]

n=2 yields 0 naturally.

n=3 is negative though

because sin(3pi/2) = -1

[dense eagle]

did you forget to square ur coefficients?

[lyric shadow]

oh fck i made another type sorry

the sin is squared

p_e = (196/(pi^4n^4) ) sin(pi*n/2)^2

so sorry

[dense eagle]

yh lol that's what i suspected

[dense eagle]

nw

anyway if this is in the context of some problem sheet question, probably just compute the first few values

[spiral turtle]

Honestly, I would just compute by hand as well

[lyric shadow]

is there no other way?

[spiral turtle]

there are only a few

[dense eagle]

it decays like 1/n^4

[lyric shadow]

like what if it became greater than 0.999 at N=43?

[spiral turtle]

and the alternatives to bounding the area are, like, integration estimates.

[dense eagle]

so basically you have very fast decay so it should hit 0.999 very quickly

[spiral turtle]

It will become greater than 0.999 in like n=5 or n=7, and you only need to test the odds.

[dense eagle]

if this isn't like as a problem sheet question and there's a chance that N might be very high, you might want to bound N to get some approximate value for N

or find asymptotics

i mean effectively, you want to understand the sum of 1/n^4

there's no closed form formula for that, but the Riemann zeta function is very well understood

(although if ur doing QM i don't think you need to know asymptotics for sum 1/n^4 lol)

[lyric shadow]

yeah, it is part of a problem sheet but it's like 6 marks for the question which seems like a lot for just a brute force question

i guess the only thing that could give extra marks is just saying that it decays very fast due to 1/n^4

but i was also just curious about how you could do it if the value for N was actually really large

thanks

[spiral turtle]

I mean, honestly I'd just write a computer program to do it in that case.

and if it's titanically large, I'd consider thinking of smarter ways to approach it, such as with an integral approximation.

[lyric shadow]

yeah, that's a good idea

thanks again

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[dapper lake]

i want help double checking this answer, i think it might be DNE

[paper depot]

as far as i can tell this limit does exist. show your work for how you got DNE?

[dapper lake]

hmm okay, that was my initial answer too before i second guessed and thought it was dne

my first answer was -5/9

[paper depot]

show work for that one too?

won't hurt to check for arithmetic errors and the like

[dapper lake]

sorry took a bit, i had to rewrite since i erased it

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[urban zinc]

hey

[urban zinc]

So im trying to understand how is d the correct answer

how is 3ln2/ ln4+5 equal to 3ln2/ln2+ln3+ln5, or maybe i have understood this wrong

But how do i actually solve this problem?

[quiet echo]

This is wrong?

$T=\frac{\ln 8}{\ln 30}$, not $\ln 20$.

[twin meteor]

;(

[urban zinc]

Sorry i should have been more clear, first pic is the problem

second is the EXAM answer

3rd is my work

[quiet echo]

I can tell

[quiet echo]

But still

[urban zinc]

yeaah i dont understand what do i do wrong

[quiet echo]

The crazy part is that the answer key is incorrect

[urban zinc]

woah

💀

So whats the correct answer

[quiet echo]

Let me see if I can work it out.

a=ln5, b=ln10, T=ln8/ln30

So...

Legit there are no factors of 3 here. I don't think this is solvable using regular methods.

Valiant effort by you, though.

[urban zinc]

Damn, could that 30 be a typo from 20?

[quiet echo]

Maybe it's vice versa, I don't know.

[urban zinc]

Hmm nvm, its weird cause ln2+3+5

gives 30

[quiet echo]

Well, we can add in a ln3 if we want to.

I was just thinking we were only allowed to express it in a and b, no constants besides the coefficients.

You know what I mean?

[urban zinc]

Hmm not rlly

Im just preparing for an exam idk a lot of math which is wild cause

this is from the university website

Wild because it might be a typo

i was thinking im just bad tho

[finite owl]

probably comes from the 8 being 2^3 but havent calculated anything

[quiet echo]

Probably. Anyways, if you want to solve it by adding an ln3, go for it.

[urban zinc]

we can just do that?

out of nowhere

[quiet echo]

Otherwise, the exam key here does not match with the question

[finite owl]

ln2 = b - a

[quiet echo]

I don't think you should. I'm only saying this if you want to do it for fun.

[urban zinc]

ooh

[quiet echo]

Since it's just an invalid anyways.

[finite owl]

so ln8 = 3(b-a)

prob down below some stuff similar

[urban zinc]

yes