help-27

[errant sparrow]

wdym add a constant?

[topaz beacon]

for example if you get f(1) = -4, what do you add to get 2?

[errant sparrow]

-2?

[topaz beacon]

-4+_=2

[errant sparrow]

6?

[topaz beacon]

yes

that is the constant you add

[errant sparrow]

for the second term here i get 1/4=c and n=4?

[topaz beacon]

dont forget about the negative sign

[errant sparrow]

so its -1/4?

[topaz beacon]

c = -1/4

yeah

[errant sparrow]

so my equation shouldve been -x^3 = cnx^n-1 ?

[topaz beacon]

yeah

or you could remember that there is a minus sign

[errant sparrow]

the last term is just x?

[topaz beacon]

but i prefer keeping the minus sign to not forget

yes, the last term is just x

[errant sparrow]

i got
f(x) = (1/2) x^2 - (1/4) x^4 +x

[topaz beacon]

right, but we also have a really important +C

[errant sparrow]

wdym +c?

i just add +c to the equation and then sub in f(1)=2?

[topaz beacon]

remember how 3x+5 and 3x+7 have the same derivative?

[errant sparrow]

yes

[topaz beacon]

we need a function value to find what C is

plug the value in and solve for C

[errant sparrow]

so my equation is
f(x) = (1/2) x^2 - (1/4) x^4 +x +c
then i sub in f(1)=2?

[topaz beacon]

yes

many people forget the +C when doing antiderivatives

[errant sparrow]

for this one theres they give me two to plug in
do i do that the same way?

[topaz beacon]

notice that you have f''

you have to be a little careful with this one

first we find f' by finding the function that has the derivative equal to f''

what function has a derivative e^x

[errant sparrow]

part A i got
f(x)= (1/2)x^2 - (1/4)x^4 +x +3
c=3?

[errant sparrow]

e^x?

[topaz beacon]

yes

[topaz beacon]

3 doesnt sound right

[errant sparrow]

-3

[topaz beacon]

still doesnt sound right

how did you get that?

[topaz beacon]

when you have f(1) = 2, you replace all instance of x with 1

on the left, you have f(1), which you can replace with 2
and on the right, you have (1/2)(1)^2-...

[topaz beacon]

thats better

[errant sparrow]

f(x)= (1/2)x^2 - (1/4)x^4 +x +(3/4)
thats my final equation?

[topaz beacon]

yup

[errant sparrow]

this one i got
f(x) = x + 2x^(3/2) +c
and then i solve for c?

[topaz beacon]

yup

[errant sparrow]

the second term is ok ?

[topaz beacon]

yup

you can check by doing power rule

[errant sparrow]

yes it looks correct

f(x) = x + 2x^(3/2) +5
i get c=5?

[topaz beacon]

looks right

[errant sparrow]

and then for the last one , is f'' the same way to solve?

[topaz beacon]

you have to be a little careful

get to f', then ill show you what to do

[errant sparrow]

i have to do it twice?

[topaz beacon]

yes

[errant sparrow]

the -2sinx do i have to use the product rule?

or can it just be 2cosX

[topaz beacon]

just 2cosx

you only need to worry about product rule if the two factors are both functions

if one is a constant, you dont need to worry

[errant sparrow]

f'(x) = e^x - 2cosX

?

[topaz beacon]

dont forget the +C

[errant sparrow]

i dont have to solve for f(x) first before i do the +c?

[topaz beacon]

but i want you to write it as $C_1$

[woven radish]

me when the gbf

[topaz beacon]

you dont know f(x), so you cant solve for C

C1, i should say

[errant sparrow]

so i sub it in twice with f(0) for c1 and then f(pi/2) for x2?

[topaz beacon]

not yet

first off i want you to be careful with the -2sinx part

derivative of cosx = -sinx, so the minus sign turns into a plus sign

[errant sparrow]

so it would be +2cosx?

[topaz beacon]

yes

so our f' is e^x+2cosx+C1

[errant sparrow]

and then i sub in f(0)=3?

[topaz beacon]

no not yet

we have our f', we need f

can you find the first two terms of f?

[errant sparrow]

f(x)= e^x -2sinx

the c1 becomes a 0?

[topaz beacon]

we are working backwards, so the derivative of what becomes a constant?

[errant sparrow]

wdym?

[topaz beacon]

using the reverse power rule method or any other method, the derivative of what becomes C1?

[errant sparrow]

cx?

[topaz beacon]

yeah C1*x

[errant sparrow]

f(x)= e^x -2sinx +c1x ?

[topaz beacon]

watch your signs on the trig function again

derivative of what function becomes +cosx? is it +sinx or -sinx?

[errant sparrow]

+2sinx

[topaz beacon]

yes

so our function is f(x) = e^x+2sinx+C1*x, but we are msising something

[errant sparrow]

another constant?

[topaz beacon]

yes

C2

[errant sparrow]

we add c2 to the end of that equation?

[topaz beacon]

yup

then you can plug in your function values to get a system of equations to solve for C1 and C2

[errant sparrow]

one at a time?>

[errant sparrow]

wait wdym

[topaz beacon]

plug in x=0 first

this will get rid of C1, because C1*x = C1*0 = 0

no system of equations needed

just some algebra

[errant sparrow]

c2=2?

[topaz beacon]

yup

[errant sparrow]

f(x)=e^x +2sinx +c1x + 2

how did i know to plug in f(0) =3 first?

[topaz beacon]

because it gets rid of the c1x term

[errant sparrow]

would it be wrong if i pluged in f(pi/2)=0?

[topaz beacon]

if you plugged in pi/2, then both variables are there and you cant solve anything

i should say unknown constants, not variables

[errant sparrow]

do i always plug in the first one or i have to know which one to plug in?

[topaz beacon]

you could plug in either first, and then youll find out which one is more useful

[errant sparrow]

got it

so now i have to find c1?

[topaz beacon]

you may find both are not too useful, but together they form a system of equations to solve them

[topaz beacon]

yes

[errant sparrow]

i plug in the f(pi/2)?

[topaz beacon]

yeah, plug in x= pi/2

[errant sparrow]

How would I solve this?

[topaz beacon]

you can leave it in exact form or approximate it by plugging it into a calculator

it has no clean answer

[errant sparrow]

I got this?

[topaz beacon]

,calc (-e^(pi/2)-4)/(pi/2)

[woven radish]

Result:

-5.6089241047199

[topaz beacon]

looks right

[errant sparrow]

my final equation
f(x) = e^x + 2sinx -5.61x +2

[topaz beacon]

looks good

[errant sparrow]

ok thats the last question

tysm for ur help

[topaz beacon]

youre welcome

[errant sparrow]

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[restive river]

simple question, for part c what does it mean by initial depth?

[restive river]

Would that be 0?

oh ok got it

.close

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[eager cliff]

Two rectangles have areas in the ratio 36:121. If the length of the smaller rectangle is 15cm, what is the length of the bigger rectangle?

[eager cliff]

Just wondering if I’ve gone wrong here

Since the answer is irrational

<@&286206848099549185>

<@&286206848099549185>

[opal cloak]

all caps when you spell the mans name

[eager cliff]

Is my Spotify status still visible?

[opal cloak]

no

[eager cliff]

Oh

[opal cloak]

ur name is a dead giveaway

[eager cliff]

Yeah

So just checking I haven’t done a step wrong

I feel like I have but I can’t tell why

[opal cloak]

the ratio between the sides and their areas arent the same i believe

[eager cliff]

My answer is irrational and I feel like it shouldn’t be so I’m running around in circles trying to figure out what I did wrong

Well if the ratio is 36:121

And the smaller side is 15cm

Wouldn’t it make sense to cross multiply

?

[opal cloak]

the answer being irrational shouldnt be a problem

just let me check something

[eager cliff]

K

[opal cloak]

is the question assuming that they have the same heights?

[eager cliff]

Well that’s all the info given in the question

If the height on the smaller triangle was 36cm (hypothetical) than the height on the bigger rectangle would be 121cm since that’s the ratio

[opal cloak]

thats not what im asking

[eager cliff]

Oh

[opal cloak]

the height of the smaller rectangle is 2.4cm so are we assuming that is also the height of the larger rectangle?

[eager cliff]

There’s no given height?

And the heights of the two rectangles would correlate to the ratio 36:121

Wait I just realised

[opal cloak]

yeah but thats my problem, the areas are in a ratio

[eager cliff]

Areas

Oh my days I need to square root it

Ok I’m all good

[opal cloak]

uhh ok

[eager cliff]

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[stray ravine]

im not sure about how i would start question D

[devout snow]

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[rancid pelican]

this is rational function and i am not sure about the graph im just posting here just to check the graph

[past bough]

not quite

look at values close to -1 and see what happens

and you're y-axis isn't consistent

[glossy dew]

the y-markings are not uniform

meaning having the same difference

for a normal graph the difference in nearby markings would be same

you would get a function like this if plotted properly

[rancid pelican]

okok i see

thanks as always

[glossy dew]

just one more thing

[rancid pelican]

yeah wahts up

[glossy dew]

maybe try marking the graph with y axis marked as -1, -0.5, 0, 0.5, 1... and then drawing the graph

[rancid pelican]

gotcha gotcha im noting all this down

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[vast quiver]

Let the family set Ai with i ∈ N ∶ Ai =] - ∞, i]. Show that ∩i∈N Ai = ⋯

[vast quiver]

is ∩i∈N Ai = (-inf,0]?

[winter torrent]

seems right to me

assuming 0 is a natural number

[vast quiver]

yeah

and are there any problems while trying to prove it?

Is this fine

[devout snow]

@vast quiver Has your question been resolved?

[devout snow]

@vast quiver Has your question been resolved?

[vast quiver]

@winter torrent

[winter torrent]

i think that's probably fine, although I'm not sure I believe that line 3 implies line 2

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[radiant rune]

how do i solve these two quesitons?

[radiant rune]

im confused

[hushed wraith]

f is a simple enuf function

I think u can sketch what the integral of that looks like

[coral violet]

do you know the libenez theorem

i think that’s what its called..

[radiant rune]

0_0

[coral violet]

Here

Ill give you the proof, hold on..

here

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[somber shore]

hi

[somber shore]

could you please give me a hint for this

here is how far i could go

btw

i made a mistake

[strong tiger]

why did you substitue x = 1 - e^4

[somber shore]

e^ u

bad handwriting

[strong tiger]

you should follow a rule

that substitute a function whose derivative is present

try again using that again

[somber shore]

i tried that but nothing came to my mind

1/x is deriv of ln (x)

the only thing that might work is this ig

[strong tiger]

do you know what is the derivative of ln(1+x)

[somber shore]

1/(x+1)

[strong tiger]

oh wait

ah damn im dumb

[somber shore]

all good

[strong tiger]

sorry

[somber shore]

the ln on the top

is so annoying

[snow raptor]

you can use series

ln(1-x)

find the series of that

[strong tiger]

uhh

listen

[somber shore]

idk how to use it ):
if its the only way then i should move on

[strong tiger]

try ln(1-x) = u

that will work maybe

[snow raptor]

I'm not sure

bc this can be expressed into special integral of log

[strong tiger]

huh

[snow raptor]

Li(x)

[strong tiger]

im getting something

lemme cook

[snow raptor]

https://en.wikipedia.org/wiki/Polylogarithm#Dilogarithm

[somber shore]

this is so cool

[snow raptor]

I don't know which sub to take but I only know using series

[strong tiger]

😭 i tried

i got the same integral again

wasted yall time sorry

[somber shore]

should we use series to get this ?

[somber shore]

its all good man tysm

[snow raptor]

I don't know lol, I just shared you a source assuming you know it

to match that form

[somber shore]

LOL tysm
this is way beyond my knowledge

[snow raptor]

multiply by -1

I will show you wait

$-\int(-\frac{ln(1-x)}{x})$

[woven radish]

Alaska

[somber shore]

ok ok

[snow raptor]

so its -Li_2 (x)

I don't really know how to evaluate limits

this is higher math

[somber shore]

tysm then i should wait till i understand series

.close

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[snow raptor]

https://study.com/academy/lesson/representing-the-ln-1-x-power-series-how-to-steps.html

then you divide by x

[devout snow]

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[gloomy moat]

How can I quickly tell if this matrix has independent columns?

[gloomy moat]

and how can i see which columns are not independent?

[acoustic leaf]

consider whether 4 vectors in R^3 can be linearly independent, generally

[gloomy moat]

so i can just chose any column?

[acoustic leaf]

it's not necessarily obvious whether any column could be removed to make the columns linearly independent. a systematic way would be to compute the REF

[gloomy moat]

hm okay thx

do you know how i could do the "The Strang Decomposition A = CR" on this?

[acoustic leaf]

i believe you start by taking the RREF of the matrix in question

[gloomy moat]

idk i never heard of this

RREF

i prob wont have enough time to do all these steps anyway in the exam lol

there has to be a trick

[acoustic leaf]

you haven't heard of reduced row echelon form?

[gloomy moat]

this ?

[acoustic leaf]

yes

[gloomy moat]

yeah im confident u can do this without using RREF

the prof also said that we wont have time to do calculations

so yeah

[acoustic leaf]

the "R" part of the A = CR decomposition is the rref of A (0 rows removed) so i'm not sure how you would do that

[devout snow]

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[plucky maple]

Do they mean probability of 1st success of 7 MORE trials after the 5 failed trials(same for 2?)?

[faint zinc]

No, 2 more trials for a total of 7.

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[plucky maple]

Wait lemme process

[plucky maple]

Ohh my god

Of course yeah that makes sense

Ty so much

.close

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[covert umbra]

!status

[devout snow]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[covert umbra]

1......

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@covert umbra Has your question been resolved?

[faint gorge]

Start by multiplying by $\dd x$ you would get the form
[ \underbrace{\left ( 2y^{a+2}+3\sin x \right )}{M} : \dd x + \underbrace{(10y^4x+e^{ay})}{N} : \dd y = 0. ]
Now recall, the differential equation is exact if and only if
[ M_y = N_x ]

[woven radish]

𝔸dωn𝓲²s

[covert umbra]

what does My mean

[faint gorge]

partial derivative wrt y

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[mossy rapids]

i solved and got 2 for the value of k

now orthocentre is equidistant from all the vertices

so i took ortho centre as (x', y') and did using distance method

but unable to get the ans using that

[uncut imp]

Maybe just compute where 2 altitudes intersect?

[mossy rapids]

yea that is another way, but isnt this logical too??

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[marble otter]

Hello!
I have the following "table" and want to calculate this:

[marble otter]

I know that the solution would be this:
I don´t know where to start

[upper schooner]

Do you have any additional context?

[marble otter]

no, sadly not

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[warped prism]

Consider the infinite series
[
\sum_{n=2}^\infty n^{-1 - \frac{1}{(\log_2 n)^2}}.
]
Determine whether the series converges or diverges and prove your answer. Use a dyadic partition.

[woven radish]

ZaoWuZhu

[warped prism]

Broke it into the dyadic partitions, but struggle to see what to do after

[random helm]

I dont know what a dyadic partition is but I see that the series is greater than 1/(n log n) which is divergent. That's because 1/log n < loglog n.

[devout snow]

@warped prism Has your question been resolved?

[warped prism]

Yeah, I know it diverges

but really not understanding how to procede from the the partition

[
\sum_{n=2}^\infty n^{-1 - \frac{1}{(\log_2 n)^2}} = \sum_{k=1}^\infty \sum_{n=2^k}^{2^{k+1} - 1} n^{-1 - \frac{1}{(\log_2 n)^2}}.
]

[woven radish]

ZaoWuZhu

[warped prism]

Let ( S_k ) denote the sum over the ( k )-th dyadic interval:
[
S_k = \sum_{n=2^k}^{2^{k+1} - 1} n^{-1 - \frac{1}{(\log_2 n)^2}}.
]

[woven radish]

ZaoWuZhu

[magic wind]

I think, if you substitute n with 2^k, you can a lower bound for S_k. Like S_k > 2^k * (2^k)^(-1-1/k^2)), the first 2^k is the number of summands in the sum.

This simplifies to S_k > 2^(-1/k), and the sum over k diverges for k-> infinity.

[warped prism]

The series can then be written as
[
\sum_{n=2}^\infty n^{-1 - \frac{1}{(\log_2 n)^2}} = \sum_{k=1}^\infty \sum_{n=2^k}^{2^{k+1} - 1} n^{-1 - \frac{1}{(\log_2 n)^2}}.
]

Let ( S_k ) denote the sum over the ( k )-th dyadic interval:
[
S_k = \sum_{n=2^k}^{2^{k+1} - 1} n^{-1 - \frac{1}{(\log_2 n)^2}}.
]

For ( n \in [2^k, 2^{k+1}) ), note that ( \log_2 n \geq k ). Thus, ( \frac{1}{(\log_2 n)^2} \leq \frac{1}{k^2} ), and the terms in the interval satisfy:
[
n^{-1 - \frac{1}{(\log_2 n)^2}} \geq n^{-1 - \frac{1}{k^2}}.
]

This gives a lower bound for ( S_k ):
[
S_k \geq \sum_{n=2^k}^{2^{k+1} - 1} n^{-1 - \frac{1}{k^2}}.
]

Since ( n \geq 2^k ) within the interval, we further bound each term by ( (2^k)^{-1 - \frac{1}{k^2}} ), giving:
[
S_k \geq (2^k)^{-1 - \frac{1}{k^2}} \cdot \left( 2^k - 1 \right).
]

Simplify:
[
S_k \geq (2^k)^{-1 - \frac{1}{k^2}} \cdot 2^k = (2^k)^{-\frac{1}{k^2}}.
]

To simplify ( (2^k)^{-\frac{1}{k^2}} ), use the identity ( a^x = e^{x \ln a} ):
[
(2^k)^{-\frac{1}{k^2}} = e^{-\frac{k \ln 2}{k^2}} = e^{-\frac{\ln 2}{k}}.
]

As ( k \to \infty ), ( e^{-\frac{\ln 2}{k}} \to 1 ), so:
[
S_k \geq C,
]
where ( C > 0 ) is a constant independent of ( k ).

To prove divergence, we compute the total sum of the first (N) dyadic intervals:
[
\text{Partial sum: } \sum_{k=1}^N S_k.
]

Using the lower bound ( S_k \geq C > 0 ), we find:
[
\sum_{k=1}^N S_k \geq C + C + \cdots + C \quad \text{(a total of (N) terms)}.
]

Simplify:
[
\sum_{k=1}^N S_k \geq C \cdot N.
]

Since ( C > 0 ) is constant, the partial sums grow with ( N ). Letting ( N \to \infty ), we see:
[
\sum_{k=1}^N S_k \to \infty.
]

This shows that the sequence of partial sums is unbounded, and the series diverges.

The series
[
\sum_{n=2}^\infty n^{-1 - \frac{1}{(\log_2 n)^2}}
]
diverges because the partial sums grow without bound,

[woven radish]

ZaoWuZhu

[warped prism]

I tried to flesh it out a bit

[magic wind]

Looks good.

[warped prism]

appreciate man 🙂

that hint really helped out

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[lapis sparrow]

I feel like I have an idea of where to start; however, I would like some assistance

[supple knot]

explain your idea

[lapis sparrow]

So my thought is firstly, finding the Eigenvalues of A, then finding the Generalized Eigenvalues to then use the General Solution form

So, then you would have eigen-values of -1 with multiplicity three

And then you have eigenvector (1, 1, 1)?

Wait wait

Nope, I have no idea

Wait

Yeah I have no clue just because of the repeated Eigen Values

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[green vortex]

Just need quick and simple help, i understand the question however the only thin i dont understand is how where 2x-1=ln(50), the -1 moves to the other side, dont you have to add 1 to each side meaning it would be 2x=ln(50) +1 ?? sorry if this is very simple just going over my basics

[green vortex]

nevermind, the teacher made an error it was +1

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[wind root]

Hallo I need an explanation why these are the formula when finding FV and PV in compound interest.

[wind root]

I'm confused how we will get PV if we need PV to get FV

Nvm there will be a given for FV

I just need an explanation why division and multiplication

NVM I got it... I just need to divide it both sides by the parenthesis

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[prime whale]

I want to redefine this dot product using the definition involving the magnitued of the vectors and the ange between them

should I keep or discard the negative?

I imagine discard? Like it's basically just flipping the mu vector

[acoustic leaf]

the negative should still be there

if you flip the mu vector that changes the angle

[prime whale]

so instead of |a||b| cos its

-|a||b| cos

?

[acoustic leaf]

yes

alternatively you can bring the - inside the cos as a change to the angle

[prime whale]

hmm ok

thank you

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[unreal lynx]

what is the technique called that allows you to do this kind of math by hand without a calculator? 11^5 mod 13

[faint gorge]

11 is congruent to -2

[unreal lynx]

that's great but how does that answer the question?

[faint gorge]

there is no techniuqe

it's just logic and knowing some modular arithmetic

[unreal lynx]

I think it might be "square and multiply" @faint gorge

[fossil locust]

sounds correct

there's also Hensel's lifting lemma if you are interested

[unreal lynx]

I am interested in the simplest and quickest solution

which with a name like that I think I will stick to the first

[fossil locust]

if you want to find say 11^5 mod 13^2 or 13^3

but yeah the simplest is just (-2)^5 = -32 = -32 + 39 = 7 mod 13

[unreal lynx]

thanks

[fossil locust]

no worries!

[unreal lynx]

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[proven grove]

help please I am getting 0 as an inflection point and as a critical point and when I test with the first derivative, there is a change in sign which means it should be a critical point and at the same time I also got it as an inflection point. However in my answer key it says that the answer is E

[devout snow]

@proven grove Has your question been resolved?

[proven grove]

<@&286206848099549185>

[dense badge]

find where the first derivative is equal to 0

then find where the second derivative is also equal to 0

if they are at the same point than count it

[proven grove]

i did that

first deriv = 0 @ x = 0,1/3

second deriv = 0 @ x = 0,1/4

and i tested 0 and it changes

from pos to neg

around the point

but the answer key i have says that no critical points are the same as inflection points

[devout snow]

@proven grove Has your question been resolved?

[charred zodiac]

Test this for second deriv. If it's an inflection point, then there is a change in the sign of f''

[devout snow]

@proven grove Has your question been resolved?

[proven grove]

i tested it and it does change

but the answer key says that there are none inflection points that are also extrema

here is the graph in desmos

[charred zodiac]

An inflection point must show a change in concavity, what do you see in the graph? 🤔

[charred zodiac]

also what I meant is that you should check if the sign of second deriv changes

[proven grove]

I mean there’s a slight dip but idk

also it’s a no calc problem i wasn’t even supposed to be graphing it

[proven grove]

ye I did that

it changes from concave down to up

[charred zodiac]

1/3 is past another point where f'' is 0

try a smaller interval

I mean, the points where f'' = 0 are x=0 and 1/4 as you calculated

If you want to check change in sign for x=0, don't check a value greater that 1/4

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[rancid spire]

Hey! How to multiply

[stone stump]

how to multiply what

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[fierce matrix]

https://ncatlab.org/nlab/show/cartesian+product

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[upbeat pawn]

why does labda become 1

[fringe meadow]

what does the eigenvalue show

like what does it actually mean

[upbeat pawn]

the scalar right

[fringe meadow]

Yes

How much everything gets scaled by

If you look at the first line

There is no scaling happening

So its lambda = 1

Remember Ax=Lamdax?

[upbeat pawn]

ya

[fringe meadow]

Well theres the Ax on the right hand side

and the lambda x on the other side

And as you can see there is no scalar value

so lambda is just one

you follow me?

[upbeat pawn]

uh not rly

why dont they write lambda in the first place

Ax= lambda x

i understand this much

[fringe meadow]

Because it says the number of visitors is constant, so if lambda was there it would mean that its not constant

[upbeat pawn]

and when lambda =1 it becomes Ax= x

[fringe meadow]

as every transformation would scale up or down the number of visitors

[fringe meadow]

x = the vector with GIM, and do you see a constant in front of the vector on the left hand side of the side equation?

[upbeat pawn]

no

[fringe meadow]

So what must the scalar be?

[upbeat pawn]

1

[fringe meadow]

There you go

[upbeat pawn]

wait

lemme slowly digest

[fringe meadow]

Let me explain like this

imagine you have kid, adult and elderly visitors and you use the vector they gave to represent it

and imagine the transformation aka matrix-vector multiplication with A is applied every day to get the next days number of visitors (kid, adult and elderly)

And it tells you that the number of visitors is constant

so there cannot be any scaling

The eigenvalue shows how much the eigenvector gets scaled by

therefore the eigenvalue must be 1

Is this clearer?

[upbeat pawn]

only the total (kid , adult, elderly visitors) remain the same but the individual visitors of each category changes ?

[fringe meadow]

No, every single one remains constant

[upbeat pawn]

this part multiples with identity matrix to factorise [G I M]?

[fringe meadow]

Yes because you can't subtract a matrix and a scalar so it can't just be A-1 you need the identity

[upbeat pawn]

ok thx

[fringe meadow]

All clear?

[upbeat pawn]

i kindda get the basic concept

this chapter is so ass

[fringe meadow]

watch 3b1bs essense of LA it clears up alot of stuff

[fringe meadow]

Whats the chapter

[upbeat pawn]

eigenvector and eigenvalues

[fringe meadow]

you take a levels by any chance?

[upbeat pawn]

this was the question

[fringe meadow]

I see

[upbeat pawn]

yes

but without further maths

[fringe meadow]

Then why you doing this, its FP2?

[upbeat pawn]

im in uni

[fringe meadow]

do some research on eigenvector convergence

[upbeat pawn]

egineering maths shi

[fringe meadow]

Oh

you had a pre uni maths

role

[upbeat pawn]

so this chap is basically new to me

college math basically

only standard a level maths

not the further maths

[fringe meadow]

same tbh, I learnt it a couple months ago, 3b1b's videos helped so much

[upbeat pawn]

youtue?

[fringe meadow]

would highly highly recommend

[fringe meadow]

yes, watch his playlists on the essense of linear algebra and calculus

what uni you in?

if you dont mind sharing

[upbeat pawn]

monash

wbu

[fringe meadow]

Im in y12

[upbeat pawn]

h

oh

[fringe meadow]

still got a while

[upbeat pawn]

college at?

[fringe meadow]

Eyeing trinity maths

[upbeat pawn]

u took further maths?

[fringe meadow]

Im taking right now

[upbeat pawn]

oh nice

[fringe meadow]

Exam is next year

[upbeat pawn]

i dropped it after a few months lmao

[fringe meadow]

I have to take it cause I wanna study maths 💀

but our school picked further stat and mech

which I hate

[upbeat pawn]

well if u like maths den go ahead

[fringe meadow]

hate hate hate

mechanics is just yuck

[upbeat pawn]

thats the requirement for everyone who took fm

¯_(ツ)_/¯

[fringe meadow]

No its optional in FM

[upbeat pawn]

huh

[fringe meadow]

school chooses 2 from 8

there are 8 papers

FM1 and 2, FS 1 and 2, further pure 1 and 2 and decision 1 and 2

school picks 2 for you

[upbeat pawn]

my college set it for us, basically AS do further pure and further stats, A2 do further mechanics and furher pre

same ig

u in A2?

[fringe meadow]

We re doing AS topics rn in school but Im doing A2

[upbeat pawn]

smart guy

prodigy type shi

[fringe meadow]

below average for cambridge applicants

lol

[upbeat pawn]

thats a high aim

[fringe meadow]

they did this in primary

[upbeat pawn]

nah

no way

[fringe meadow]

Im obviously joking but they're still really cracked

anyways i gotta get some work done

good luck with your eigenvalues and vectors chapter

[upbeat pawn]

kk thx for help

bye

[fringe meadow]

cya

[upbeat pawn]

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[restive river]

Can some pls explain example 38.1

I want the example explaines not the exercise

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[restive river]

Can anyone pls explain example 38.1 pls

[restive river]

<@&286206848099549185>

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[left palm]

you have two right triangles and they're congruent by hypotenuse-leg

[restive river]

Oki

[left palm]

then you compare congruent parts

that should be it

[restive river]

In the book if zayz

AB is a chord of the circle,centre O
X is the midpoint of AB
OA and OB are radii,

So truangle OAB is iosceles

OX is a line of symmetry for triangle OAB

Therefore OX is perpendicular to AB

This shows that

The perpendicular from the centre to a chord bisects the chord

How does that apply in the example

[left palm]

well that perpendicular part gives you the right triangles

[restive river]

I get that AB is a chord
OA and OB are the same length

But OBA

[left palm]

the chord bisection doesn't matter so much since they told you it's the midpoint already

what about OAB?

[restive river]

Sorry OBA

[left palm]

right, we have to right triangles

[restive river]

How is OAB and AOB same

[left palm]

they share the hypotenuse (radii) and a leg (OX)

[restive river]

Mhm

[left palm]

there's a hypotenuse-leg theorem for right triangles which says they're congruent

[restive river]

Ok

[left palm]

then you can use CPCTC (corresponding parts of congruent triangles are congruent)

[restive river]

Ok

[left palm]

that should be it

[restive river]

Ok.thank u very much.have a nice dayy

[left palm]

thanks you too!

[restive river]

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[oblique mulch]

I did the ratio test to find the convergence for this power series but I’m not really sure at this point?

[devout snow]

@oblique mulch Has your question been resolved?

[faint gorge]

you calculated 1/R = 1/2 so it's actually R = 2

also check end points

[sage burrow]

assume x = 2 and look what series you would get ...

[oblique mulch]

why 1/R?

[oblique mulch]

Interesting

so I should be taking the ratio test upside down with an above an+1?

[faint gorge]

either that or just consider the reciprocal as you see, a_n/a_(n+1)

https://en.wikipedia.org/wiki/Radius_of_convergence

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[grim ginkgo]

is the answer b =2?

thats what i got

if it isnt imma need some help cantlie

[supple knot]

explain your work

[grim ginkgo]

give me a sec

this should be really easy to double check

oh wait

its 3

hahaha im stupid

wait no

nah ill respond back in a sec

no

ok now im even more stupid then stupid

the answer is 2

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[blazing crest]

3b anyone?

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[supple knot]

did you try proving it when I = [a, b]

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[blazing crest]

idrk

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[cerulean wharf]

Good morning, if we have two uniform distributions D1 and D2 bounded by B1 and B2 respectivelly, can we have a relation between the bounds such that the two distributions are indistinguishable?

[cerulean wharf]

Is there any theorem or lemma about this?

Maybe leftover hash lemma 🤔

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[cerulean wharf]

Hmm B1 and B2 must be very close because the min entropy differs by an exponential 😔

[cerulean wharf]

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[hexed holly]

Have I done this right?

[hexed holly]

i think i need to add b + a = RHS, but there's no answer so i'm not sure.

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[rose ingot]

Seems about right. The + law is commutative in the real or complex ring so when adding two 3 dimensional vectors it boils down to doing 3 normal additions like you did.

[hexed holly]

okay, thank you!

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[north pivot]

Hey ! Can anyone help with finding the ways to resolve the first integral ? I am kind of stuck

[proud perch]

if you solve for y (or x) in the conditions you can turn this into a calc 1 problem

at least for part a

x^2 = 2y
x^2 = 1 + y^2

1+y^2 = 2y
y^2 - 2y + 1 = 0
(y-1)^2 = 0
y = 1

x^2 = 2 -> x = +-sqrt(2)

[devout snow]

@north pivot Has your question been resolved?

[north pivot]

This is what I found

but when i write the integral down it seems unclear on the limits

[proud perch]

wdym

[proud perch]

i'd say the easiest thing here is to use the symmetry of the region

[devout snow]

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[proud perch]

.reopen

[devout snow]

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[slate rose]

sure

[devout snow]

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[vale pawn]

Let V be the volume of the solid obtained by rotating the region bounded by the curves y=(tan(x))^3, x=π/6, x=0 and y = 0 about the line y = 2

[vale pawn]

great i forgot how to use latex after years uh

this

i got a and b correct but i dont know why is this wrong

please enlighten me

thx

oh, nvm

.close

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[restive river]

just want some confirmation on this

[restive river]

I did evreything now for the helicopter off the ground

would I just do 661sin 28 degrees 11 and 445sin 44.5 degrees

[devout snow]

@restive river Has your question been resolved?

[restive river]

<@&286206848099549185>

[atomic sparrow]

There are three interior angles in a triangle

U know 2 of them

Find the top angle by doing 180-first angle - second angle

Then use the sine law to find the lengths of the legs of the triangle

[devout snow]

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[hot hamlet]

how do i solve this

[hot hamlet]

i need to draw something too

pls help\

[supple knot]

solve sin(t) = sqrt(3) / 2 for t first. then solve t = x/4 + pi/3 for x

definitely more than 1 solution for x

[hot hamlet]

sorry, what do you mean by sin(t) = sqrt(3) / 2 for t first

By t what do you mean?

[supple knot]

t is another variable

[hot hamlet]

ok and what about this:

[supple knot]

,tex .shift trig

[woven radish]

riemann

[supple knot]

either top left or one beneathe it

[hot hamlet]

beside this any other method?

[supple knot]

why

[hot hamlet]

my teacher asks me various method, he might ask me to do another way

and the test is tomorrow

😭

[hot hamlet]

ofc if there IS another way.

[supple knot]

go through your other trig identities then

i don't know what you learned

[hot hamlet]

alr

.close

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[sonic igloo]

help

[sonic igloo]

can sum1 tell me if the power point is ok

[cold bough]

sus

[sonic igloo]

and be honest

[cold bough]

it could easily be a virus

[sonic igloo]

it could also easily real..

[cold bough]

oh I expected it was about math

though

[sonic igloo]

nope

computer sci

but i need sum1 to rate it

[supple knot]

wtf

<@&268886789983436800>

[sonic igloo]

what happen

oh

can sum1 dm me saying rate nd then go thru and tell me if theres anything i could do better

[modest dagger]

We generally discourage people moving to dms for this kind of thing.

It can get kinda sketchy

[sonic igloo]

well can i send it here

[modest dagger]

How long is the presentation?

If it's not too long just taking screenshots might be viable.

The whole "download my file and check it" thing is also sorta sketchy

Idk if people are going to want to do that.

[sonic igloo]

ok

[left palm]

if this is going to be presented on a large screen, some of the text might be quite small. i didn't look at the content, but the presentation is quite sleek. also you censored your name page 1 but your full name is signed on each page

also this isn't really math feedback lmao

[devout snow]

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[hollow jay]

I want help on Q6 of BMO1 2021, doing the olympiad on Wednesday

[hollow jay]

Haven't really got anywhere

[violet wind]

so it's $(a_1+1)(a_2+1)\cdots(a_n+1)$ where $\sum a_i = 70$

[woven radish]

Dreyuk

[violet wind]

36+12+10+6+2+2+2 = 70 so it's just the number of ways to rearrange the factors of 999,999

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[modest pagoda]

I am at a complete loss for trying to solve for k. I’ve tried setting the area of the triangle formed by g(x) equal to the integral the upper region plus the lower region to isolate k, and got k = 3, which makes zero sense in the context of this problem. I got the same result by trying the same method but with the trapezoidal shape. One of my buddies set up a system of equations and got k as 6.61, and another one got it as exactly 7. We are all at a loss.

[supple knot]

show all your work

[modest pagoda]

mine:

my friend’s system:

and i do not have access to the 3rd person’s work

[supple knot]

why are you calculating this area

oh are you calculating the area of the triangle connecting A to (k, 0) and (k, g(k))?

yea you can do that and then subtract the area under f(x) between 5 and k

you don't need an integral to find the area of a triangle

but you would for $\int_5^k \frac{1}{x-4} + 1 dx$. so evaluate that and then do the subtraction

[woven radish]

riemann

[devout snow]

@modest pagoda Has your question been resolved?

[modest pagoda]

won’t that leave this or am i misinterpreting?

[supple knot]

oh yes you have to subtract off another triangle

oh that gives you the unshaded region

[modest pagoda]

the full triangle minus the integral you described gives the yellow region

[supple knot]

yea it's probably just easier to find the small triangle with AB hypotenuse + the second integral i mentioned.

[modest pagoda]

setting all that up causes k to cancel out

[supple knot]

what is this supposed to be

[devout snow]

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[supple knot]

also why does your definite integral still have an x

[modest pagoda]

left side is the full triangle, right side is the sum of the areas making up the triangle to make them equivalent

[supple knot]

.reopen

[devout snow]

✅

[supple knot]

that's circular

[modest pagoda]

typo, meant to be k. still canceled out in my work

[supple knot]

after this, you set it equal to log(p) + 8

then solve for p

and set the non log stuff equal to 8

[modest pagoda]

i cannot leave it in terms of k

[supple knot]

this gives you k

or at least an equation for k

then you have 2 eqns 2 unknowns

[modest pagoda]

why equal to 8?

[supple knot]

that's what the question gives

use the information given

[modest pagoda]

why is the 8 the only part of that equation being used in the part you described

[supple knot]

you see your log here right?

[modest pagoda]

yes

[supple knot]

that argument is equal to p

[modest pagoda]

that’s where my disconnect was. thank you for your help

[devout snow]

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[raven swift]

if im given tangent on a unit circle, how do i find secant?

[coral heron]

$\tan^2(x)=\sec^2(x)-1$

[woven radish]

fish

[acoustic leaf]

you can then use the appropriate quadrant to figure out the sign

[devout snow]

@raven swift Has your question been resolved?

[raven swift]

suppose im giving tangent as -3/4

ik tangent is sin/cosine

and secant is 1/cos

so 1/4?

[coral heron]

no. that would be assuming that sine is -3, which is outside the range of the function

[devout snow]

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[lone imp]

I'm doing line integrals this week, I know the basics of them, I'm just not sure how to set this one up.

[lone imp]

What would the integral bounds and r(t) be?

[devout snow]

@lone imp Has your question been resolved?

[lone imp]

No

[devout snow]

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[faint gorge]

ok

[misty crest]

oh

it was 8x btw

you forgot the x

[whole raven]

yeah i realised, i just entered it again with the x and still got marked wrong 😭

[faint gorge]

(4x²)^4?

[misty crest]

yea

that too

[whole raven]

oh right

thanks guys

[faint gorge]

thank you too

[devout snow]

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[old carbon]

How am I supposed to set this up?

[old carbon]

Our teacher just gave us our first actual worksheet/packet on optimization and I feel kinda lost. I figured there was a relationship between the triangles (similar) and I can use that w/ pythagorean formula?

[tame spear]

Work in terms of z^2, not z. Minimising z^2 is the same as minimising z.

I would start afresh on new paper from "x/(x+1) = 8/y". The form "(x+1)/x = y/8" is preferable, as you can simplify (x+1)/x to 1 + 1/x. Eliminate one of the variables in the pythagorean formula and minimise what you get

Have you done differentiation yet?

[old carbon]

yes, we've covered differentiation

[tame spear]

Just expand that, differentiate it, set it equal to zero and solve for x

[old carbon]

gotcha; what was wrong w/ what I did then 😦

[tame spear]

if you get multiple solutions check what values of z they give you and see which is smallest

[tame spear]

let me read it

• respectfully, I would always try and work downwards rather than forcing yourself to stay on the same page. It's far far better to use more paper and make your work clearer

• Best to minimise z^2 rather than z. It's far better to think of it that way if you can see why - no confusing square roots lying around

[old carbon]

i understand; I've ran out of paper though. I'll try to see if I have a sheet that's blank in the back somewhere

[old carbon]

yeah I see what you mean

I'll give it a shot and see how far I get, thanks mate

before I dive head first into that, do you mind helping me with setting up this problem:

so what I did was I made a similar triangle relationship then I did all the math only to realize there's no max to the relationship I made

[tame spear]

The way I would think about this is a circle centred on the corner

[old carbon]

Chat GPT said something about making two similar triangles in the hallways and adding up their hypoteneous values to form the ladder, but when I tried that I got crit pt at x=0 but ig I could've done the math wrong

[tame spear]

Don't use chatgpt for maths help, it's dumb

[old carbon]

hmmmm. How does that work 🤔

[old carbon]

yeah it was rly annoying

[tame spear]

If you have the best ladder length, you'll put the ladder centre at the corner and rotate it about the centre. You want the arc traced out by the ladder edge (the arc of a circle) to fit inside the corridor

that may be wrong actually

yeah that's only correct when the hallways are the same width, ignore me

[old carbon]

I barely understood enough to read 😭

wait, that's actually really really neat

def will use that if they r the same width

[tame spear]

I don't think 5/7 = (5+y)/(7+x) is correct

Because as you move the ladder through, your triangle will dynamically change what angles it's at, you're rotating it

[old carbon]

it's not

i tried doing it that way but the similar triangle relationship I'm making is always increasing (you can see the remnants of my work in the eraser marks)

[tame spear]

May be a good observation that these ones are similar

[old carbon]

yeah that's what GPT said it just went no where with it

[tame spear]

ladder length = L

as you move the ladder, the lengths x and y change

but you'll have L^2 = (5+x)^2 + (7+y)^2 at all times

may be best to swap x and y actually, since x usually means horizontal distance

[old carbon]

the x's and y's in pencil, right?

[tame spear]

What do you mean?

[old carbon]

red x and blue y r constant

i could be mis-interpreting keep that in mind

but if you were to move the ladder

[tame spear]

not constant

[old carbon]

it would reach a point where y=0, and another where x=0

pencil x and y

[old carbon]

why not

oh wait

i see

cuz the ladder moving is the hypotenous

[tame spear]

this only holds if your ladder is touching against both walls

if x=0, your ladder is horizontal. It can't touch both the corner and the bottom line.

[tame spear]

yeah

[old carbon]

yeah

it's a thinker

[tame spear]

and then your similarity constraint is 5/y = x/7

[tame spear]

So, "maximise (5+x)^2 + (7+y)^2 with constraint 5/y = x/7"

[old carbon]

where'd u get this constraint

[tame spear]

Can you see from that diagram that they're similar triangles?

[old carbon]

the two similar triangles?

[tame spear]

yeah

when you rotate the ladder to a different angle, they stay similar

[old carbon]

that I do not get