help-17

[outer warren]

if you wanted to expand, consider binomial theorem
or applying the definition of squaring,
(x-50)^2 = (x-50)(x-50)
and expand using distributive property (or the foil system which i have a disdain for)

also its not ideal to expand that here

[wise oriole]

Ohhh I gotta factor for this

[outer warren]

its already in factored form

[wise oriole]

so does it turn into 4=0.0048(x^2-100x+2500)

[outer warren]

that would be valid.

but like i said

also its not ideal to expand that here

[wise oriole]

Yeah I dont like that way either but idk what you mean

[outer warren]

first isolate
(x-50)^2

then take square root of both sides

[wise oriole]

2=0.0048(x-50)

[outer warren]

no

[wise oriole]

IIm getting more lost now

[outer warren]

please read my instructions carefully

[wise oriole]

Alright I'll try

[outer warren]

that's not the message you should be focussing on

that was a description of the invalid math you did

first isolate
(x-50)^2

[wise oriole]

4/0.0048=(x-50)^2

nvm thats wrong

I've been trying this for the past 10 minutes and I dont think I am getting it right, could I get an addition hint

[outer warren]

well it wasn't,
that was exactly what i was asking for

[wise oriole]

Wait really

[outer warren]

yes, you've done the valid manipulation and
(x-50)^2 has been isolated

now take the square root of both sides,
don't forget about the - case

[wise oriole]

4/0.0048=(x-50)^2
833.33333=(x-50)^2
28.86751346=(x-50)
78.86751346=x

[outer warren]

that's one of the values

don't forget about the - case

[wise oriole]

78.86751346-50=21.13248654=x

Could you remind me why this one has two solutions?

[outer warren]

whare are the solutions to
p^2 = 4

[wise oriole]

-2 and 2

oh I see

[outer warren]

yes

[wise oriole]

its because of square rt

Alright thank you, but I have one more question

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[outer panther]

im so lost

[outer panther]

i found the derivative to be

$\frac{-.8abe^(-.8t)}{(1+be^(-.8t))^2}$

[twin meteor]

MWB117

[outer panther]

set equal to 14

then set $\frac{a}{1+b} = 20$

[twin meteor]

MWB117

[outer panther]

am i going in the right path?

from there i set a = 20b+20 and plugged into derivative

and from there i got b = -7/15 and -1

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@outer panther Has your question been resolved?

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@outer panther Has your question been resolved?

[wraith python]

I believe you mathed wrong somewhere. The logic is sound but I get a different answer for b.

[outer panther]

i can't figure it out

[wraith python]

Show your work for how you calculated b.

[outer panther]

$\frac{-.8(20b+20)b*1}{(1+b)^2}$

[twin meteor]

MWB117

[outer panther]

set that equal to 20

solving for b

[wraith python]

No.

[outer panther]

oof

[outer panther]

i dont have my work on me so im trying to figure it out again

well if u set it to 14

$-16b^2-16b=14+28b+14b^2$

[twin meteor]

MWB117

[outer panther]

right?

so $-30b^2-44b-14=0$

[twin meteor]

MWB117

[wraith python]

The LHS is incorrect. The negative sign should have cancelled out.

[outer panther]

LHS?

[wraith python]

Left-hand side

[outer panther]

hm

$2b^2-12b-14=0$?

[twin meteor]

MWB117

[outer panther]

b= -1,7

is that what u got

[outer panther]

$16b^2+16b = 14+28b+14b^2$

[twin meteor]

MWB117

[wraith python]

You can reduce the fraction on the LHS by eliminating (b+1)/(b+1)

[outer panther]

oh

16b = 14

[wraith python]

16b/(b+1) = 14

[outer panther]

ohh

$16b=14b+14$

[wraith python]

b+1 in the numerator is reduced to 1 and (b+1)^2 is reduced to b+1 in the denominator.

[twin meteor]

MWB117

[outer panther]

7?

[wraith python]

👍

[outer panther]

ok thanks!

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[vast shale]

how is the first line true? im missing something fs

[vast shale]

there’s a factor of sin^2(x) in both terms

So bring out as a factor from both

same way you could factorise x - zx as x(1-z)

oh okay i got it. it makes more sense when i factor it in. thx

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[faint frigate]

hi

[sacred remnant]

What is the question you need help with?

[faint frigate]

one second

heres the problem

i was wondering if theres a method to approach this besides counting in cases

there was this pset i was working on and this is one of the three questions i cant figure out

.end

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[sharp comet]

the whole diagonal is 6srt2 long

[hushed pewter]

looks good

[sharp comet]

ok

thanks

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[novel warren]

How do i solve question no 14

[novel warren]

I can get to $$AP+BP=2b$$

[twin meteor]

neon

[novel warren]

And solving that i get to $$x^2+y^2+a^2=2b^2$$

[twin meteor]

neon

[novel warren]

But i dont know what to do after that

I tried doing $$\frac{x^2+y^2+a^2}{2b^2}=1$$ but that doesnt seem to work

[twin meteor]

neon

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[covert python]

convert the roots to powers of 1/2 and divide both sides by one of the roots maybe?

[vast shale]

be po be po be po

[mellow oyster]

multiply both sides by sqrt(x2 / x1)

this will cancel the square root on the RHS and simplify the square root on the LHS

[covert python]

$$\sqrt{\frac{x_2}{x_1}}=\frac{x_2^{\frac{1}{2}}}{x_1^{\frac{1}{2}}}$$

[twin meteor]

CrEpasPmkinPie

[mellow oyster]

well consider the LHS, you'll have sqrt(x2 / x1) * sqrt(x2 / x1) so that just becomes x2/x1, right?

and on the RHS, you'll have sqrt(x1 / x2) * sqrt(x2 / x1) = sqrt(x1 * x2 / (x2 * x1)) = sqrt(1) = 1

so after that multiplication you'll have 1/(2p1) * x2 / x1 = 1/(2p2 + t)

it's a bit easier to solve for x2 now

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[undone crane]

my brain is refusing to work today

[gaunt sparrow]

What is the definition of (n+1)! ?

[undone crane]

what do you mean

[undone crane]

that dancing pumpkin guy is such nostalgia, I'd jumpscare people in videogames with it when I was a kid

[gaunt sparrow]

You have to use the definition of the factorial.

$n! = n\cdot (n-1) \cdot \ldots 3\cdot 2 \cdot 1$

[twin meteor]

Azyrashacorki

[gaunt sparrow]

Heheh

[undone crane]

I have not been introduced to this one yet

"by definition of the factorial"

this is the schoolbook answer

I don't understand what is this

[gaunt sparrow]

Yes, they use the definition I just showed you

[undone crane]

but that definition has not been taught to me

[gaunt sparrow]

It's how you define factorials in the first place, how did you learn it?

[undone crane]

in the previous page before that I was only introduced to factorials with this

[gaunt sparrow]

Well that's exactly the definition I gave you

[undone crane]

where does the (n-1) come from

[gaunt sparrow]

It's just the previous integer.

6! = (6)(6-1)(6-2)...

You can write it starting from 1 as well it doesn't matter.

[undone crane]

I see

[undone crane]

that makes total sense, what doesn't make sense to me is how this proves anything

they pretty much start the proof with (n+1)! = n!*(n+1) and end the proof with the same thing

[gaunt sparrow]

They start the proof just by writing (n+1)! with this definition

Then they take the first n factors and realize that this is just n!

[undone crane]

oh

[gaunt sparrow]

The goal is to show how you can define factorials recursively essentially

It's baked into the very definition, so it's fairly straight to the point

[undone crane]

I'm also curious, how would you write (n+1)! in this form?

[gaunt sparrow]

Just write n+1 at the top of Pi

All it means is multiply all integers from 1 to n+1

[undone crane]

I see thanks

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[vast shale]

My mistake??

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[vital wave]

I thought I was doing this right but I got the wrong answer

[mellow void]

the algebra is goofed starting on line 4

[vital wave]

I think I found my problem

[vital wave]

Yeah

Thy

Am I allowed to simplify this? Or do I leave it with the polynomial on top too?

@mellow void

Nevermind it’s the same to solve for x

I think I messed up my algebra again

Damn I messed up the f(l)

Is this right?

[mellow void]

it looks plausible

I don't check arithmetic but the algebra looks fine

by the way, quotient rule is complete overkill

your denominator is a constant

$\frac d{dx}\left(\frac{f(x)}c\right)=\frac d{dx}\left(\frac1c f(x)\right)=\frac1c\frac d{dx}f(x)=\frac1cf'(x)$

[twin meteor]

Flip

[mellow void]

1/c is a harmless constant

it did nothing wrong

[vital wave]

Ok

[mellow void]

it doesn't need to be annihilated in the derivation process

[vital wave]

I’m going to start a new chat o solved this

⛅️

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[barren kindle]

how do you prove that the nine point circle is tangent to the incircle

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[barren kindle]

how do you prove that the nine point circle is tangent to the incircle

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[barren kindle]

how do you prove that the nine point circle is tangent to the incircle

[barren kindle]

<@&286206848099549185>

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[bleak prawn]

Not sure what I did wrong here

Sin^2(x)cos^2(x)

Wait I just saw

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[hybrid fossil]

suppose i have a function F'(x - ct) such that F'(x - ct) = h(t)

[hybrid fossil]

and i wish to find F(x - ct)

i am trying to integrate both sides with respect to t but its just not happening lol

$\int_0^t F'(x - cs)\dd s = \int_0^t h(s)ds$

[twin meteor]

heavy

[hybrid fossil]

is this wrong already?

[lethal plank]

This is a bit weird

[hybrid fossil]

yeah right

[lethal plank]

Because F'(x-ct)=h(t) means that the derivative is independent of x and c

And if you are independent of c

You are independent of t

Due to the symmetry

So h=constant

[hybrid fossil]

well i mean no

[hidden bramble]

I don't think anything is wrong with it

[lethal plank]

Could you elaborate ?

[hybrid fossil]

i obtain $F(x - ct)= -c \int_0^t h(s) \dd s$

but thats wrong

[twin meteor]

heavy

[hybrid fossil]

well you can find nonconstant functions such that it is true

[hidden bramble]

how do you obtain it?

[cyan talon]

looks like some diffusion equation stuff, how did you arrive to that problem ?

[hybrid fossil]

in a sense F' is just a shifted and stretched version of h(t)

[hybrid fossil]

method of characteristics for the wave equation

[lethal plank]

Can you give an example

[hybrid fossil]

yeah suure hold on

[hidden bramble]

I think what you obtain is $F(x-ct) - F(x) = \int_0^t h(s)ds$

[twin meteor]

MetuMortis

[hybrid fossil]

oh yes youre right, i left it out because here F(x) = 0 for x > 0

sorry lol

[hidden bramble]

how do you get -c in front of the int?

[hybrid fossil]

well i did a u sub with u = x - cs

wrong reply

[hybrid fossil]

working on it

[cyan talon]

wait why would that be the case

[hybrid fossil]

remember here that x is constant so instead lets use y as dummy variable and you could have $F'(y) = \left(\frac{y - x}{c}\right)^2$

[twin meteor]

heavy

[cyan talon]

isn't F supposed to be an arbitrary function

or is it a condition of your question you haven't told us about

[hybrid fossil]

then $F'(x - ct) = \left(\frac{x - ct - x}{c}\right)^2 = t^2 = h(t)$

[twin meteor]

heavy

[hidden bramble]

So you get $\int_0^{-ct} F'(u)\dd u = \int_0^t h(s)ds$ right?

[hybrid fossil]

well because of the initial conditions we necessarily have that u(x,y) = F(x) = 0 whenever x < ct

hmm uhm wait am i saying it right

point is i didnt want to make it too complicated as i didnt that context was relevant

this should just be a calc 1 problem

[cyan talon]

you subbed the bounds incorrectly

[hybrid fossil]

wouldnt you get x and x - ct in your bounds?

remember its du/ds not du/dx for u sub

[lethal plank]

If you treat both x and c as constants which you are allowed to use outside the equation then yeah sure

[hidden bramble]

oh yes sry

[lethal plank]

(x-(x-ct))/c=t

[cyan talon]

and you forgot the -1/c factor again

[lethal plank]

But this seems like abuse of notation

[hybrid fossil]

haha maybe

[cyan talon]

if u = x-cs, then du = -c ds, so ds = -1/c du

[cyan talon]

that's the problem when you don't provide context, ppl are like wtf is going on

you're green, you should know better lol

$$F(x-ct) - F(x) = -c \int_0^t h(s) \dd{s}$$

[hybrid fossil]

yeah fair enough

[twin meteor]

aPlatypus

[cyan talon]

and so what's your goal actually here ?

if you have one that's not just screwing around with wave equations

[lethal plank]

Yeah this is the objection I was raising

See the derivative is evaluated at (x=0,t)

[hybrid fossil]

wait ohg my lord

[lethal plank]

Trolling

It's ok it happens

[hybrid fossil]

well you actually make a good point

x = 0

cant believe i actually caused an XY problem myself lol

[hybrid flicker]

So do you know the general solution form to dtt u = c² dxx u?

[hybrid fossil]

F(x - ct) + G(x + ct)

[hybrid flicker]

Ok great

[hybrid fossil]

with F and G defined by the intial conditions

[hybrid flicker]

Yep

Meaning F(x) = -G(x) and etc...

[hybrid fossil]

yep

[lethal plank]

Unfortunately this is inhomogenous

[hybrid flicker]

Wdym and how is it a problem

[lethal plank]

Not sure trying to pull D'alembert's solution would work

[hybrid flicker]

u / t² is the same dimension as (velocity)² * u/x²

[lethal plank]

h(t) on the boundary

[hybrid flicker]

?

[hybrid fossil]

i figured it out

[lethal plank]

It is not a free propagating wave

[hybrid fossil]

$F(x - ct) = -c\int_0^{t - \frac{x}{c}} h(s) \dd s$

[twin meteor]

heavy

[hybrid fossil]

alright sorry for wasting everyones time

!XYproblem

[vocal sleet]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[hybrid fossil]

midnight math got me like

thanks though

.close

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[hybrid flicker]

As long as the equation ruling on the interior of the domain is the d'Alembert one, we use the d'Alembert solution on the interior, at least for C² solutions

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[ornate solstice]

can someone explain me

what does the -B means?

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[covert flint]

what exactly are you confused about?

[spiral needle]

What is meant by Vertical angle

here

[covert flint]

it's prob just a isosceles triangle with the 120 angle on top and base on bottom

[spiral needle]

ok

then one angle is 120 other 2 are 30 right

[covert flint]

yeah

[spiral needle]

how to find then altitude

[covert flint]

use some trig

[spiral needle]

i divided the triangle in to right angle triangle

[covert flint]

thats a good first step

[spiral needle]

then using Pythagorus therom

[covert flint]

yes

[spiral needle]

Thanks

Can i ask one more question

[covert flint]

yes

[spiral needle]

If alpha is the measure of an angle of triangle ABC then tan( alpha / 2 ) in always

This is new question

A) Negative B) Zero C) Positive D) Undefined

[covert flint]

is there any more context?

[spiral needle]

nope

[covert flint]

oh wait

have u tried this question?

[spiral needle]

no

but i try to understand it

so i did not know what they are saying

[covert flint]

alpha is just a variable

it's commonly used when u need a variable for an angle

[spiral needle]

alpha is just any angle

Variable or angle

[covert flint]

it's a variable for an angle

[spiral needle]

ya

[covert flint]

so what are the possible alpha values

[spiral needle]

it domain

[covert flint]

i wouldnt really call it the domain

[spiral needle]

ok?

[covert flint]

so what possible alpha values are there

like a range

[spiral needle]

infinty

• infinity , + infinity

[covert flint]

i was more of going for (0,180)

because it's describing the angle of a triangle

[spiral needle]

ok

Sorry i miss that

[covert flint]

do u understand that

[spiral needle]

ya i understand know

[covert flint]

do u know how the tangent function works

[spiral needle]

ya

i think

[covert flint]

so if the range of alpha is (0,180), then what is the range of alpha/2

[spiral needle]

0 , 90

[covert flint]

correct

[spiral needle]

here tan is positive

[covert flint]

yes

[spiral needle]

Thanks bro

[covert flint]

and it's important that u have () and not []

or blank

so specify that it's non-inclusive (tan 90 degrees is undefined)

[spiral needle]

( 0 - 90 )

[covert flint]

also that looks like 0 minus 90

(0 , 90)

[spiral needle]

ok

Thanks for the help

.close

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[covert flint]

np

[spiral needle]

bye

[vocal sleet]

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[naive heron]

In triangle BC, BC = 2r. Denote the midpoint of BC to be M. From point M, a circle of radius r is drawn, and intersects the triangle at 4 spots: B, P, Q, and C, in order. PB = 20, and QC = 22, and PQ = 22. If possible, find the side lengths of the triangle, or prove why there is not enough information to solve the problem.

[naive heron]

this took me a long time to word formally, and ive arrived on this step from a bigger problem

nvm, this problem does not help with my entire problem what am i thinking

.close

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[vocal sleet]

@vital wave Has your question been resolved?

[vital wave]

<@&286206848099549185>

[solar valley]

do you know relative velocity?

[vital wave]

I k now related rates and we just started optimization problems

[vital wave]

I’m not sure

[solar valley]

what do you think the relative velocity of first boat about second one is?

[vital wave]

Actually we never learned relative velocity

[solar valley]

okay

[vital wave]

Would you mind teaching it to me?

[solar valley]

if you assume the second boat is stopped, how does the first one move?

[vital wave]

How does it move?

[solar valley]

okay you really don't know about it...

[vital wave]

You mean how fast?

[solar valley]

then,

do you know about the vector

[vital wave]

They are in the water

If one stops the other can keep going

They are not even in each others way

I know what a vector is

[solar valley]

okay.

if we assume that you are in the second boat,

you feel that that second one is stable, not moving

then the first one moves in different velocity from the problem

that is vector v1 - vector v2

[vital wave]

Ok can you just tell me what the related velocity is

It doesn’t say anywhere that one boat relies on another to move

So that doesn’t apply here

You are getting too complex

[solar valley]

this is what the problem says

[vital wave]

Ok

Yes

[solar valley]

this is what when we assume the second one is not moving

and thats because of this

[vital wave]

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[hard silo]

Hi, I have a question about the Euler-Lagrange equation in case of a contstraint, where J = C for example. So generally problems like the hanging chain problem under gravity. Why do we suddenly introduce the Lagrange multiplier in there? how can I interpret this? How is this connected to gradients, or is it even connected to gradients at all

[vocal sleet]

@hard silo Has your question been resolved?

[hard silo]

<@&286206848099549185>

[blissful sentinel]

the lagrange multiplier is actually interpretable as the constraint force

I think a standard analytical mechanics textbook like Taylor should talk about this, if you're looking for places to read about it

[hard silo]

it's a few paragraphs in my Calc 2/3 book under the chapter optimalisation and extremumproblems

they give the equations, and a partial proof of how they get to it

but there's little to no explaining why certain steps are taken

from what I remember from previous chapters, the lagrange multiplier gave the sensitivity of the extremum to changes to contraints, essentially telling us how stable the extremum will remain for changes in the contraints

can I interpret the lambda part of the equations simply as a contraint function which is equal to a constant ?

simply changing the gradients of the functions to the euler-lagrange equation for eacht of the contraints/goal function

so something like ;

$$F_{y} - \frac{d}{dx}F_{y'} + \lambda\sum_{i=1}^{n} \phi_{y_i} - \frac{d}{dx}\phi_{y'_i}= 0$$

[twin meteor]

Mephisto

[hard silo]

with phi being the contraint functional and F the goal functional

[vocal sleet]

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[plucky trout]

nevermind

im dumb

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[lean otter]

my brain is melting from discrete maths, i need help with probably very simple problem i'm just new to it so i'm having a lot of contradicting thoughts that drain me a lot, problem goes how many ways can we put 6 same towers on the chess grid 8x8 that they aren't attacking eachother? i got to an answer 564480 if im using 2 times 8 over 6 * 6! because of the rows and columns and 20160 if i'm using 8 over 6 once * 6! i'm confused on which one is the right answer(564480 feels kinda too much to me but i'm not sure)(6! represents the permutation of 6 figures which is just n!).

[vocal sleet]

@lean otter Has your question been resolved?

[hidden bramble]

are we talking about rooks?

[lean otter]

wym

[lean otter]

just checked what it means yeah it's rooks

[hidden bramble]

I also find 564480 doing:
I select 6 columns: C(8,6)
I select 6 rows: C(8,6)
I have to create pairs using those rows and columns so: 6!

If I do C(8,6) * C(8,6) * 6! I get 564480

[lean otter]

yeah exactly what i did, are you 100% sure that's the way?

[hidden bramble]

I don't think anything is wrong with that

[lean otter]

i saw teachers video from my university, with 8 rooks instead of 6 and he did C(8,8) * 8! which lead me to assumption 564480 is faulty

[hidden bramble]

I think it is actually C(8,8) * C(8,8) * 8!
with the same logic we use

[lean otter]

im not sure

40320 is for 8

564k for 6??

[hidden bramble]

I mean it is definitely not a linear function

[lean otter]

i mean that's true

but you'd expect bit more alike numbers no?

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[upbeat jungle]

I got stuck in this

[vocal sleet]

@upbeat jungle Has your question been resolved?

[bold quail]

u-sub

[upbeat jungle]

Omg, thanks

[bold quail]

its complicated u-sub but it should still work

[upbeat jungle]

I didn't know where to start, it's already a step forward

[bold quail]

tell me what u get when u evaluate it

[upbeat jungle]

results in a common integral

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[half imp]

A little weird that 13 precedes 14 here

[vast shale]

is the answer of 17th is 1 and answer of 20 is m + 1?

[half imp]

17 is not 1

[vast shale]

WAIT

there is empty set then there is single set of m

number

[sly sierra]

(unless A is the empty set)

[vast shale]

m + 1?

[half imp]

Yes, m+1

[vast shale]

damn

i see

and 20 is correct?

[half imp]

No

[vast shale]

let me think

okay 20 is actually

1

[half imp]

No

[vast shale]

WAIT

ughhh

{phi, {x1}, {x2}....{xm}....}

thats m + 1?

[half imp]

{x1} isn't a subset of P(X)

[vast shale]

OH WAIT its about oh damn im dumb okay so like every set in that power set can be put in single set

[half imp]

ye

[vast shale]

there is

2^m

sets

so there is 2^m

elements

OH YEAH

i think that should be it

[sly sierra]

don't forget the empty set

[vast shale]

AH

yeah

2^m + 1

[half imp]

ye

[vast shale]

{{}}

i was just taking this

but

{}

itself is too

is one

[sly sierra]

yep

[half imp]

Ye

[vast shale]

Cool Cool thank you guys!

[half imp]

Np

[vast shale]

.close

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[hybrid fossil]

why are they asking to find $L * \text{such that} \int_a^b\left[u L^*(v)-v L(u)\right] d x=\left.H(x)\right|_a ^b$

[twin meteor]

heavy

[hybrid fossil]

wouldnt the left side be 0 since it just means $<u, L^{*}v> - <v, Lu> = <Lu, v> - <v, Lu> = 0$ by definition of the inner product and adjoint

[twin meteor]

heavy

[hybrid fossil]

so $H(b) - H(a) = 0$

[twin meteor]

heavy

[hybrid fossil]

am I understanding the question correcttly?

[vocal sleet]

@hybrid fossil Has your question been resolved?

[obsidian stream]

I haven't done any differential equations for a while but when L = L*, L is called a Sturm-Liouville operator, I assume the purpose of the exercise is to help motivate the study of what is to follow.

I also haven't attempted this exercise, but I assume the goal is to start with <v, L(u)> and apply integration by parts until it becomes something of the form H + <Mv, u> so that you can get an expression of the form <v, Lu> - <Mv,u> = F and study M since, if M is "nice", i.e., M = L* and L = L*, then H = 0.

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[hybrid fossil]

oh yeah actually that makes a lot of sense, thanks

[hybrid fossil]

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[raw trout]

how do I find the suitable value of x

[lucid crypt]

you can express $\frac{3}{4}$ as $1 - \frac{1}{4}$

[twin meteor]

AlphaNull

[raw trout]

didn't get it

[lucid crypt]

find an expression for $\frac{48^{\tfrac{1}{3}}}{4}$, i.e. $\left({\frac{3}{4}}\right)^{\tfrac{1}{3}}$ via part b

[twin meteor]

AlphaNull

[lucid crypt]

then using my hint from earlier, we have $\left({\frac{3}{4}}\right)^{\tfrac{1}{3}} = \left(1 - {\frac{1}{4}}\right)^{\tfrac{1}{3}}$

[twin meteor]

AlphaNull

[lucid crypt]

if you want it to look like $(1-5x)^{\tfrac{1}{3}}$, what should $x$ be?

[twin meteor]

AlphaNull

[raw trout]

uh

-1/5?

[lucid crypt]

can you explain how you got that?

its not right

[raw trout]

I don't think if I understood tbh

[lucid crypt]

did you do a and b?

[raw trout]

yes

[lucid crypt]

48^(1/3) = 4(3/4)^(1/3), divide 4 on both sides

[lucid crypt]

then you can follow from here

[raw trout]

wbu this

how can we express this

[vocal sleet]

@raw trout Has your question been resolved?

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[next burrow]

need help understanding this

[flint idol]

well so lets go bit by bit

the patient can only be sick or healthy

so the probability of a sick would be?

[next burrow]

0.01?

[flint idol]

yep

now do the same for the others

[flint idol]

mhm yeah thats it

they all look good to me

[next burrow]

can you double check C

[flint idol]

yeah it looks good

[next burrow]

okay

thank you, even though this was very much a short session

[jade anvil]

Paradigm paradox

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@next burrow Has your question been resolved?

[next burrow]

forgot to close my bad

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[uneven sage]

single fraction in jts simplest form

[lusty snow]

Common denominator?

[uneven sage]

lcm abc

[lusty snow]

Wha

Where you got l and m

Oh i see

Least common multiple

yeah it's abc

[uneven sage]

what do i do next

[lusty snow]

Just multiply each term with unsufficent variables

1/ab×c/c and etc

[uneven sage]

i don’t get it

[uneven sage]

why is it like that

[lusty snow]

c/c is equal 1

If you multiply by 1 nothing will change

We multiple by c/c to get abc in the denominator

[uneven sage]

no i mean like why 1/ab x c/c

[lusty snow]

🙂

[warm vine]

to complete the abc

you will do a/a for the second term

so completing abc again in denominator

basically make common denominators for all the terms

and now since the denominator is common

you can just sum up numerator

[uneven sage]

ohh

thanks

one more

[lusty snow]

x=100%
28=70%

Proportions

[uneven sage]

ohh thanks

[uneven sage]

40

[lusty snow]

Ye

[uneven sage]

last one

[warm vine]

split into 2 inequalities

[uneven sage]

ya i tried that but

7< x + 3

just seems weird

[hushed pewter]

why

[warm vine]

wot

its the same as x+3>7

[uneven sage]

for x+1<7 i got x<6

[uneven sage]

ya but then the sign is <x 3-7

[hushed pewter]

<x 3-7?

[warm vine]

so thats 0 < x-4

add 4 on both sides you get 4 < x

so x>4

[hushed pewter]

Or, you can subtract 3 on both sides here

If you think it's weird because x is on the RHS, you can flip the inequality sign to swap the equations. If 4<x, that means the same thing as x>4

[uneven sage]

i got -4< x

[warm vine]

how

[uneven sage]

ohh thanks

[lusty snow]

So the answer is xЄ...?

[uneven sage]

4 > x > 6?

[warm vine]

4 < x < 6*

[uneven sage]

ohh thanks

[uneven sage]

but the sign was switched for 4<x to x>4

[warm vine]

4<x is the same as x>4

[uneven sage]

oh thanks

okay thanks

got it

.close

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[last canyon]

can i have help with c and d please

[last canyon]

on B2

[vocal sleet]

@last canyon Has your question been resolved?

[dapper condor]

just a quick thought, you could try something like working backwards

so you have your inequality

try maybe diffing both sides

and see if you can get a logically sound statement

then, you can start from this logically sound statement, and use your working from before to arrive at your original inequality

[last canyon]

do u mean differentiate aha sorry

ok so

differentiate both sides of this statement

[dapper condor]

also just realized

you don't need to do this lmao

[last canyon]

ok ahahah what do i do

[dapper condor]

so you know from the first step that In+1 < 1/2In

[last canyon]

wait how

wait ahah yes

[dapper condor]

you know what In+1 is

[last canyon]

sorry i thought your In was an Ln

[dapper condor]

yeah my bad

dont know latex

[last canyon]

same

[dapper condor]

so sub In+1 into your inequality

and after doing some simple evaluation, i believe you get your answer

[last canyon]

wait where do i sub In+1

@dapper condor

as in rearrange for In?

[last canyon]

as i see

[dapper condor]

the rest is just rearranging and stuff

[last canyon]

got it

can u help with d?

[dapper condor]

yeah so i was looking at it

and it looks like a doozy

there are some things that we can start with tho

[dapper condor]

look at the summand

[last canyon]

ahaah yh

[dapper condor]

1/(s*2^(s))

looks really close to the expression in part (c)

[last canyon]

yeah

[dapper condor]

all you have to do is take out a 1/2 from the sum

and you get it exactly

initally i thought this was a riemann sum thing cause of the bounds

but maybe not

[last canyon]

hm yeah

[dapper condor]

anyways

we get the sum from s = 1 to n of I(s)

where I(s) refers to an integral

so this is like I1 + I2 + .... In

[last canyon]

ok

[dapper condor]

also here i am talking about integrals

[last canyon]

so the 1/s2^s is the integral part

[dapper condor]

yeah

but since we replaced it with the integral, we need to do something else important

[last canyon]

what about the In+1

[dapper condor]

that will come later

[last canyon]

okie

[dapper condor]

the main thing is that ln(2) > 1/2 * sum of integrals + In+1

[last canyon]

ok

[dapper condor]

cause of the inequality

since there is a 1/2 outside of this sum and you can think of this like being multiplied with every integral

[last canyon]

wait so how would i clearly write down where the half came from

[dapper condor]

this

2^n = 2^(n-1) *2 right

[last canyon]

yup

[dapper condor]

proceeding on, we can now use a fact we have from the other parts

namely part(a)

[last canyon]

just so we are on the same page

i can write those

this

[dapper condor]

you can't put the inequality sign yet

thats only when you use the inequality

[last canyon]

ok

[dapper condor]

ok, so in order to get a particular value, we need to see how we can get this infinite series to cancel out

well its not really infinite but you get the gist

[last canyon]

ok yeah

[last canyon]

the integral that’s added on is smaller than what’s in the sum? does that help

[dapper condor]

i was thinking about using that

but doesn't seem to help that much

i think the essence of the question is to use these different inequalities to derive the upper and lower bounds

[last canyon]

ok

[dapper condor]

anyways the statement in part (b)

seems interesting

[last canyon]

hm yeah ur right

[dapper condor]

after subbing it and doing some stuff, you get the sum of integrals I from 1 to n

now we just need to compute this series

[last canyon]

so we sub what we’ve been told in b into In+1

[dapper condor]

yeah

[last canyon]

becsue it’s using ns, does this mean it’s taking away the highest this series inside the sum can go up to?

or something like that

[dapper condor]

something like that

[last canyon]

ok so we have this

ok so did we rlly need to take the half out cos if we didn’t it looks the same with S as the one we are taking away with n

[dapper condor]

yeah we don't take out the half yet

keep it in the standard form

do some stuff and then remove the half

[last canyon]

like what?

[dapper condor]

you can remove the last term from the series

[last canyon]

ok

@dapper condor this look ok?

[dapper condor]

what did you do for the last step?

[last canyon]

wrote out the series

then did …

till one before the the last term

(n-1) term

then added on the rest

[dapper condor]

why is there still the -1/(n*2^n)

[last canyon]

that’s from subbing in for In+1

that’s just added on to the series

ahhhhhh wait

ur right

forgot that’s what’s cancelling out

ok got it

what next?

[dapper condor]

hmm well from here you could do the 1/2 trick

and you would get 1/2*(I1 + I2 + ... + In-1) + In

[last canyon]

ok

[dapper condor]

and using In+1 < 1/2 * In

we get this weird summation of integrals

which ig you could try to see if they converged but that doesn't look like what they want you to do

[last canyon]

hm yeah ok

it’s a bit odd aha

[dapper condor]

not an easy question that is for sure

did you get any more information

like a table of values

or something like that

[last canyon]

nope

my friend has done this

but i’m not sure what’s going on

can u understand that?

[dapper condor]

huh

they proved the statement

well 'proved' to be more exact

this is not working out for bounds

[last canyon]

so is it not correct?

[dapper condor]

i mean do they prove the bounds

unless there is more working out, it doens't look like it

oops

i didn't read the whole question

they want to prove the statement and then prove for bounds

my bad

okay still

this doesn't work

this is not a proof at all

this is just showing it works

i assume they are looking for a proof

cause the latter is elementary

[last canyon]

hm yeah ur right

why can i just show it works tho?

@dapper condor any more ideas?

it’s does just say “show that”

[vocal sleet]

@last canyon Has your question been resolved?

[vocal sleet]

@last canyon Has your question been resolved?

[vocal sleet]

@last canyon Has your question been resolved?

[tawny hollow]

@last canyon Have you done part c) ?

[last canyon]

yes

im very stuck on part d however

also i am on a boat rn so wifi is very slow

i have tried lots of algebraic ways to do d, do u have any clue? @tawny hollow

[tawny hollow]

so with d) the idea is just to iterate the recursive formula from b)

first off notice how I_1 = ln 2

[last canyon]

what does that mean

ahhhhhh

[tawny hollow]

So you have that I_n+1 = -1/n2^n + I_n

correct?

let me tex this up so the formula is clearer

$I_{n+1} = -\frac{1}{n2^n} + I_n$

[twin meteor]

Aslan

[tawny hollow]

we have this right?

[last canyon]

yup

wait im trying to write down

how ln(2)=

[tawny hollow]

alright, did you notice how $I_1 = \ln 2$

ah gotcha

[twin meteor]

Aslan

[last canyon]

I1

[tawny hollow]

lmk if you got it

[last canyon]

not yet

[tawny hollow]

take ur time!

[last canyon]

wait ah yeah

got it

[tawny hollow]

cool!

[tawny hollow]

alright, that will be imporant, because we will basically start from I_1 and iterate using this

[last canyon]

so whats next

[tawny hollow]

so since we only know I_1, lets solve for the "previous term".

[last canyon]

ok

[tawny hollow]

like this $I_n = \frac{1}{n2^n} + I_{n+1}$

[twin meteor]

Aslan

[tawny hollow]

So by setting n=1

what do we get?

[last canyon]

1/2+I2

[tawny hollow]

yeah

[last canyon]

cool

[tawny hollow]

let me write it like this

$\ln 2=I_1 = \frac{1}{1\cdot2^1} + I_2$

[twin meteor]

Aslan

[last canyon]

nice

[tawny hollow]

now let's use the same iteration formula, but for I_2

what do we get if use it for I_2?

[last canyon]

ok

like this

hold on wifi very slow

[tawny hollow]

i see, but let's use the recursion formula the one i wrote above!

[tawny hollow]

this one

[last canyon]

ahhhh ok

[tawny hollow]

you could just write it with text here if youd like

like I_2 = ..

[last canyon]

I_2=1/8+I_3

[tawny hollow]

yup!

let me write it like this, youll see why soon

$I_2 = \frac{1}{2\cdot2^2} + I_3$

[twin meteor]

Aslan

[last canyon]

cool

[tawny hollow]

now, notice

we had this

[tawny hollow]

let's now substitute the I_2 from this!

Then we get something like

$$\ln 2= \frac{1}{1\cdot2^1}+\frac{1}{2\cdot2^2} + I_3$$

[last canyon]

ahhhhhh

[twin meteor]

Aslan

[last canyon]

this will just keep going

[tawny hollow]

yup!

[last canyon]

if we do the same for I_3 and so on

[tawny hollow]

yeah!

so by doing this n-times

what would you get?

a notation here which is helpful is the + ... +, before we use sigma sum

unless youre comfortable with it

[last canyon]

ln(2)=I_1=1/2+I_n ?

maybe

i get where this is going tho im struggling to write it down tho

[tawny hollow]

not quite, let's maybe think this through first by looking at the pattern

oh i see

say if you want to sum the natural numbers up to n, you could also just write it as 1 + 2 + ... + n

[last canyon]

so n number of times

yup that makes sense

[tawny hollow]

yeah so the pattern so far, was how many times?

[last canyon]

3

but we want n

[tawny hollow]

the I_3 here is actually deceiving!

we only did it twice!

We started with I_1

then did I_2

[last canyon]

im sorry aha ive literally just been on this for so long its hard to look at it a different way

[tawny hollow]

the I_3 just comes for free, since we have I_n+1 in the recursion formula

[last canyon]

ahhhhhh

[tawny hollow]

ur doing great, dw

[last canyon]

so we did it twicw

[tawny hollow]

yeah!

[last canyon]

and we want to do it n times

[tawny hollow]

notice how the denominators of the fractions tells us this, that is why i explicitly wrote 1 * 2^1, 2 * 2^2

etc

[last canyon]

yes

so

ln(2)= I1= 1/12^1+1/22^2+...1/n*2^n+I_n+1

?

[tawny hollow]

yeah exactly!

let me tex this up for you, to maybe make it more readable

[last canyon]

cool

[tawny hollow]

$\ln 2 = \frac{1}{1\cdot 2^1}+ \frac{1}{2\cdot 2^2} +\cdots + \frac{1}{n\cdot 2^n} + I_{n+1}$

[twin meteor]

Aslan

[last canyon]

nice!

[tawny hollow]

now we're about done actually, we can if we want just rewrite the sum with sigma notation to get it to look more neater

[last canyon]

so can we put this straight into sigma notation

[tawny hollow]

yeah!

[last canyon]

nice!

[tawny hollow]

we have that $\sum_{s=1}^n \frac{1}{s2^s} =\frac{1}{1\cdot 2^1}+ \frac{1}{2\cdot 2^2} +\cdots + \frac{1}{n\cdot 2^n}$

[twin meteor]

Aslan

[tawny hollow]

so $\ln 2 =\sum_{s=1}^n \frac{1}{s2^s} + I_{n+1}$

[twin meteor]

Aslan

[tawny hollow]

Was this what we wanted to show?

[last canyon]

yup

[tawny hollow]

then youre done with that part of d) !

now we have something left however

the inequality

[last canyon]

yup

[tawny hollow]

try and see if you can show that one by your own for the moment, and come back if you feel like youre stuck or done!

Is that cool?

[last canyon]

yup sounds good

thanks that was very helpful

got a new way to go about these questions now rather than starting with the sigma nottion

[vocal sleet]

@last canyon Has your question been resolved?

[vocal sleet]

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[desert edge]

I don’t get what logic I should use to solve problems like 12.

If b=2a then it’s infinite solutions. What next though?