help-33

[ebon depot]

On b) it says:

if $\lambda^4 = 1$ then $v_1 = [1, \lambda, \lambda^2, \lambda^4]^T$ Which should hold true since one of the eigenvalues is 1 and the corresponding eigenvector is $[1,1,1,1]^T$

Am i thinking correct?

[elfin berry]

Merineth

[ebon depot]

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[urban bobcat]

can someone help me solve this i have already solved a. "Let F : R^4 → R^4 cyclically map the basis vectors e1F→ e4F→ e3F→ e2F→ e1.
(a) Derive F's mapping matrix A, such that F(x) = Ax. Show that A4 = I4 and that AT = A−1 = A3.
(b) Show that if λ^4 = 1, then λ is an eigenvalue of A with associated eigenvector (1, λ, λ^2, λ^4)T.
(c) Brilliant and diagonalize: Find a diagonal matrix D and an invertible P such that A = P DP ^−1

[marsh citrus]

@urban bobcat Has your question been resolved?

[urban bobcat]

Can I use the formula A*e = lambda *e?

[marsh citrus]

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@urban bobcat Has your question been resolved?

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[slow holly]

How would I approach this? I’ve already tried reversing the summation and converting it all to exponential functions but neither of those seem to lead anywhere

[slow holly]

Dn is the dirichlet kernel btw

[upbeat sable]

Looks like they multiplied the numerator and denominator by sin(x/2) and used the product to sum formula

[slow holly]

Hmmm ill try that

I need to look at the product to sum formulas I guess because all I’ve been doing is converting to exponents, doing some algebra, and converting back lol

I’ll keep the channel open for now in case I fail miserably

Wait this is a telescoping series now

Thanks!!

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[upbeat sable]

Yes

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[lilac scaffold]

would anyone like to verify my work?

[lilac scaffold]

this is the original question

part b

I'm adding 2 vectors and using cosine and sine law to find the magnitude and bearing

I believe my cosine law is 100% correct, I just want someone to double check the sine law

[marsh citrus]

@lilac scaffold Has your question been resolved?

[lilac scaffold]

<@&286206848099549185>

[marsh citrus]

@lilac scaffold Has your question been resolved?

[slim spire]

I think you used the wrong numbers for the sine law, it should always be the angle over the side directly opposite of it

[lilac scaffold]

wdym

[slim spire]

which sin(B)/7 isn't

[lilac scaffold]

oh thats my bad

[lilac scaffold]

it should be sin(b)/2 right?

[slim spire]

if you're trying to find that angle on the left yea

[lilac scaffold]

yeah exactly

cause when i find that interior angle, i can add the angle between the line a and 90 degrees

then subtract it by 90 degrees

and that would be the distance between 0 degrees and the dotted line

ykwim?

which should be 45.4

...if i got evertything right

[slim spire]

I get 20 degrees, let me check it with sine law

[lilac scaffold]

ok

[slim spire]

was this what you had?

[lilac scaffold]

i got the answer in a decimal

and no

the top part is 2sin133

which is the same as 2sin47 i believe but whatever

either way there should be that 2 in front

[slim spire]

oh I have the wrong angle mb

uhh what numbers did you get for the sine law angle then the final answer?

[lilac scaffold]

the sine law angle was about 14.6

then considering theres 60 degrees above that, the final answer (the heading of the dotted line) should be 45.4

...i think

[slim spire]

I guess it's the cosine law magnitude giving the discrepancy hmm

maybe a calculator typo?

ok yea you have to extend this square root over the whole thing to get 8.49

[lilac scaffold]

am I supposed to get 8.49?

[lilac scaffold]

oh wait

sorry thats also a mistake

its supposed to be over the whole thing

[slim spire]

yea if you fix the square root it gets 8.49 and that'll fix everything

[lilac scaffold]

oh shit wait

cos133 is different than cos47

i originally used cos47 in my calculation

and theyre not the same at all

uh ohhh

okay gimme a couple minutes let me redo it

okay it says the interior angle is 9.9

that seems quite small..

thats possible though right?

like it doesnt seem unrealistic?

[slim spire]

9.9 is right

[lilac scaffold]

and i just did it for the other angle

itll add up to 180

[slim spire]

probably because the side length 2 is so small

[lilac scaffold]

or like.. 179.9

[lilac scaffold]

yeah exactly

my drawing isnt quite to scale loll

okay i guess thats the end of it

[slim spire]

just to check what do you get after 9.9? for the final angle bearing

[lilac scaffold]

50.1

so

therefore statement

[slim spire]

oh 50.1 is off let me draw it

[lilac scaffold]

a+b-> is 8.5m/s [050.1 degrees]

[lilac scaffold]

wdym

[slim spire]

like this

and you want the angle under the red

[lilac scaffold]

yeah

but nono

hold on

oooh ya i even did that wrong

jeez

it should be 30 - 9.9 right?

[slim spire]

right

[lilac scaffold]

i was doing 60-9.9 like from the top

[slim spire]

the other way should've worked but there's a lot of minus signs to mess up

[lilac scaffold]

hm

okok

so

therefore statement

a+b-> is 8.5m/s [020.1 degrees]

[slim spire]

yup I confirm

[lilac scaffold]

awesome

jeez i wouldve messed that up so badly

thanks a lot

[slim spire]

👍 np

[lilac scaffold]

situations like these helps me understand how important it is to check eevery possible mistake

🙏

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[hexed pilot]

Really can't think of much, tried substituting some variables in hope but got nowhere

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[vestal mist]

i tried using ptolemies but now im stuck

[marsh citrus]

@vestal mist Has your question been resolved?

[vestal mist]

anyoneeeee

<@&286206848099549185>

[marsh citrus]

@vestal mist Has your question been resolved?

[vestal mist]

Anyoneeee

What

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[vestal mist]

.reopen

[marsh citrus]

✅

[vestal mist]

You guys really hate geo

[cobalt sentinel]

who doesnt

[void elm]

I personally don't know any geometry, so I couldn't help if I wanted to

not that I want to do geometry anyways lol

I suppose you should just be patient and wait for someone

or you can post in #geometry-and-trigonometry

[still temple]

me.

[still temple]

nvm i hate it

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[vestal mist]

nooooooo

.close

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[open sentinel]

can some help me show that these are equal by hand?

[open sentinel]

(1/3) k (k+1) (k+2) + (k+1) (k+2) = (1/3) (k+1) (k+2) (k+3)

I put it into the computer and it told me there are equal but i cant figure out how to do it by hand

[whole sleet]

~~Expand the left side
Expand the right side

You should get the same thing~~

Actually nvm. Take the left side and common factor (k + 1)(k + 2)

[open sentinel]

factored it out and got (1/3)(k+1)(k+2)(k+1)

im not sure where the k+3 is coming from

[whole sleet]

Watch out for that 1/3. You should have got:
(k + 1)(k + 2)(1/3 k + 1)

[open sentinel]

okay I see

[whole sleet]

If you then proceed to factor that 1/3 out, you will get the right side exactly

[open sentinel]

okay thank you

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[eternal meteor]

anyone how to approach this problem?

[static quarry]

well first find the "orthogonal complement of the column space of (1,1)"

which is a rather obtuse way of saying something much simpler in this case

[eternal meteor]

right, how do I find an orthogonal complement of a vector

[static quarry]

well in this case you're in 2 dimensional space

so what is the dimension of the orthogonal complement of a vector?

[eternal meteor]

wait, vector orthogonal to [1,1] is either [-1,1] or [1,-1] right?

[static quarry]

sure

[eternal meteor]

cool thnks

that's neat

and then project the 2 other vector onto [1,-1] right?

[static quarry]

yep

[eternal meteor]

aight thnks

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[tall sun]

the only limitation i see is like

its not an actual graph or whatever

what do they want??

[haughty portal]

limitations refer to domain and range of function

[tall sun]

oh my god

hello again

[haughty portal]

yeah lmao

[tall sun]

how embarassing

anyways-

that makes this significantly easier

[haughty portal]

yep

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[normal knot]

Been studying for too long and officially braindead. I'm sure it's obvious, but I can't think.

Why would it be nonsensical or invalid to do:

P=P(1+r/e)^et

Yes, I know this is incorrect. I know that e represents continuous growth. I know I'm plugging e into the discrete compound interest formula. I'm just asking for clarification on specifically why we can't just say "e times over the course of a year," etc.

[normal knot]

Like, if we were just some crazy bankers looking to have fun, and we just decided to get wonky and do interest with this formula (whenever the interest rate is supposed to be continuous), what would happen to the money?

[cobalt sentinel]

,w eulers number

[elfin berry]

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Requested by northsteve

[cobalt sentinel]

e is approximately 2.718

meaning you'd compound 2.718 times a year

which would work just like how youd compound 2 times a year

or 4 times a year

or 10 times a year

if u want to compound continuously, use A(t) = P*e^(rt)

[normal knot]

Why do we raise e to the rate? Maybe I need to sleep lol

[cobalt sentinel]

im sure theres some proof of continuous compounding out there somewhere

[fervent rampart]

https://www.khanacademy.org/economics-finance-domain/core-finance/interest-tutorial/cont-comp-int-and-e/v/continuously-compounding-interest-formula-e

[cobalt sentinel]

(1 + 1/x)^n as n to infty

[normal knot]

Thanks folks, I'll take a look at that and maybe some proofs.

[marsh citrus]

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[slim brook]

I don’t know if I’m on the right track or if I did this correctly.

[slim brook]

I’m so confused

[calm harbor]

Nope, I don’t think you’re on the right track

[calm harbor]

You’re not supposed to get x when solving for p

[slim brook]

Idk how to approach the question

[calm harbor]

I suggest you set f(x) = g(x)Q(x)+r(x)

Q stands for quotient

[slim brook]

Oh that’s the division statement?

[calm harbor]

yep

[slim brook]

I did that and then I rearranged to isolate for p

[calm harbor]

yes

[slim brook]

Should I do the division statement for my long division or for the info in the question? I mean with -13 and stuff

[calm harbor]

Just set every unknown coefficient about Q as a b c d

[slim brook]

Oh ok

[calm harbor]

Ping me when you’re still confused after trying

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[sudden mauve]

where is (4)(24) coming from, since we are not given a function?

[void elm]

It’s a triangle!

base is 4, height is 24

[sudden mauve]

why are we just ignoring the intgrals then?

[void elm]

wdym

the definite integral from 0 to 4 is just the area of a triangle

[sudden mauve]

like youre not integrating anything youre just using the forumla for the area

[sudden mauve]

so youre using it in a more indirect way?

[void elm]

the thing is, there is a defined function you can take the integral over

It’s f(x) = 6x when 0 leq x leq 4, and f(x) = 24 when 4 leq x leq 8

if you wish, you may integrate that as you like

But you’ll see that you get the same result as just adding the areas of the triangle and rectangle components

that’s why it’s far quicker (and probably preferable) to just use formulas for the area of a rectangle and triangle

[sudden mauve]

ohhh i see

thank you!!!!

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[velvet sedge]

.close

[potent token]

why is 0^0 equal to one? if i use the same argument for 0/0 being indeterminant (approaching negative infinity, positive infinity, and zero), 0^x equals zero for all x values up to 0₊, and on the right it equals 1/0^x, which is also undefined. meanwhile, x^0 always equals 1. is there something i'm misunderstanding?

[desert dirge]

this is just one of those things
some will say its 1, some undefined, some 0, though usually either of the first two

saying its 1 is just from the definition that any number ^0 is 1

its context dependent

[potent token]

like i understand how it can be seen as indeterminate for calculus, but if we can just set 0^0 to be defined as 1 in some cases, why cant we do 0/0 = 1 because any number x/x = 1?

[limber lintel]

Because you’re dividing by 0

[desert dirge]

if x=0/0 then x*0=0, x could be anything for this to hold so x doesnt really have a defined value, thats perhaps an overly simplisitic look at it

[marsh citrus]

@potent token Has your question been resolved?

[zenith knoll]

if we define 0/0 to be 1 everything will break

0/0 is undefined

[potent token]

im not confused rn im just testing something out

one moment

[potent token]

okay wait

this might be horribly wrong but

is this a valid way to define 0^0?

[desert dirge]

that works i suppose

actually

no

[potent token]

it's not really rigorous since im multiplying by ln(0) and just setting it to zero

that's the problem

[desert dirge]

yeah its dodgy

its more a definition than something to be proven

[potent token]

im aware of that but if its a definition then it has to make sense somewhere

[desert dirge]

definitions are just things we accept as true

[potent token]

but i dont like that :(

[desert dirge]

0^0 is defined differently based on the context

[desert dirge]

sorry, thats just the way it is

[potent token]

welp

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[foggy atlas]

Which of these is a map/illustration (not sure what the correct term of it is in english) and why (not)?

z : IR → IR with z(x) = { ⌊x⌋ for x ≥ 5
−x + 9 for x ≤ 5}

w : IR → IR with w(x) = { 4x² for x ≥ 2
⌊9x − 3⌋ for x ≤ 2}

[foggy atlas]

my approach is: both aren't maps because the in the definition area are values that are connected with values more than one time. If I put 5 in ⌊x⌋, I get 5. If I put 5 in −x + 9, I get 4. So 4 != 5

[slim spire]

I agree, they're both not functions/maps because of the multivalued parts you said

[foggy atlas]

and is this a function? f : IN → Z with f(x) = 3x/sin²(x)+4

I would say it is again not a function

[slim spire]

uhh how does you class define IN because everything is fine besides x=0

[foggy atlas]

oh IN is like {1.....100000+}

1,2,3,4,5

all natural numbers from 1

I thought that sin²(x) is (sin(x))² and then sin(x)

so 3x/sin(x)+4, when I put in 1 I get 3/0,017+4
3/4,017 = 0,74

and 0,74 is not a natural number

but maybe I'm wrong

[slim spire]

oh right makes sense, not a valid function

[foggy atlas]

okok

.close

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[still temple]

Find any horizontal or vertical tangents to:

[marsh citrus]

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[still temple]

<@&286206848099549185>

[still temple]

<@&286206848099549185>

<@&286206848099549185>

[marsh citrus]

@still temple Has your question been resolved?

[grave sequoia]

you can find the equations for x and y in terms of theta (x = r * cos(theta) and y = r * sin(theta))

then take the derivative of each and equate to 0

[still temple]

i know that already

[grave sequoia]

okay

what were you unsure about then

[still temple]

the rest of the process

[grave sequoia]

okay

if you start with x

you can multiply the expression for r by cos(theta)

so 3(1 - cos(theta))(cos(theta))

then you'll end up with an expression for x in terms of theta

and if you take the derivative with respect to theta

then equate that to zero

then you can solve for the angle at which the rate of change of x is zero

which would be a vertical tangent

then you can take the angle you get and substitute it back into the x = r * cos(theta)

and you should get an x = ___ line as your tangent

then you can do the same for y

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[wary yew]

Can you please check if the proof is correct

[wary yew]

Prove $(k,m)=1 \implies (m-k,m)= 1$\

Proof: Suppose $(m-k,m)=d \neq 1. \
d |(m-k)$ and $d|m
\implies d|m$ and $d|k$

but this contradicts (m,k)=1

[elfin berry]

Dubs

[wary yew]

i think i should add little more on how d|(m-k) implies d|m and d|k

Improved version, please check

[marsh citrus]

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[marsh citrus]

@mint wind Has your question been resolved?

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[shadow thorn]

Does this limit exist (as x approaches 0)?

[charred pelican]

yes

[fathom ridge]

does it?

[shadow thorn]

Because from both sides they approach 5

[fathom ridge]

yup

[shadow thorn]

but at x = 0 its -2

[fathom ridge]

the actual value of he function doesnt matter

[charred pelican]

your question should be "why does this limit exist"

[fathom ridge]

you just need to see what it approaches

[shadow thorn]

Ohhh

so limit is different from equalling?

basically

is equalling a word

i meant equality

[charred pelican]

equalling ?

[charred pelican]

uh

Basically if there is a point which exists when the limit approaches some "x" then the limit exists

given that the LHL and RHL are the same

[shadow thorn]

soo

[charred pelican]

yes

it is different

[shadow thorn]

the limit is not always the same as the value of the function at that point

[charred pelican]

yep

[shadow thorn]

got it

thx

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[zealous glacier]

how to do part f

[oblique grove]

in the numerator

[cloud field]

do the prime factorization for each of those numbers

[oblique grove]

24^(x+1)

can be written as (2^3 * 3)^(x+1)

[zealous glacier]

oh

[oblique grove]

the 8^-1

goes to the denominator

as a^-b = 1/a^b

so the Dr is (6^2x*8)

the 6 is written ad 2*3

and the 8 as 2^3

so the Denominator is (2*3)^2x X2^3

using the laws of exponents it can be finally written as

2^x*3^(1-x)

[zealous glacier]

cheers mate

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[dire pulsar]

I need help doing calculus

[dire pulsar]

Could someone provide me basic calculus worksheets that will help me understand because I am learning.

[coral cloak]

are you in calc 1?

[marsh citrus]

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[twilit musk]

In an alien world, there’s a peculiar plant known as the Starlight Creeper. This plant has a unique growth pattern; from each of its stems, three new stems emerge every year. This process starts with a single stem and continues for 10 years. Determine the total number of stems that the Starlight Creeper will have at the end of the 10th year, including the original stem.

[twilit musk]

So I understand that in the beginning theres one stem, then 1+3, then 1+3+(12), so basically its f(n) = f(n-1) + f(n-1)*3 right where f(0) = 1

is that correct?

in any case, how do I proceed?

i asked chat gpt and it formed a geometric series which I kind of understand where it comes from but Im hoping for a detailed explanation for an understanding so that I can solve similar questions

[charred pelican]

We have 1 stem

in 1 year we have +3 stems

so we have 4 stems

[twilit musk]

yes

[charred pelican]

In the next year, we get 3 stems from each stem

so we get 12

so we have 16

in the third year,
we have 3 stems growing from one and we have 16 stems
so we get 48 stems

Can you see a pattern?

[twilit musk]

but 16 right cause the one in the beginning

[charred pelican]

ah yes my bad

[twilit musk]

so the sequence is 1 + 3 + 12 + 48 and so on

[charred pelican]

ah

[twilit musk]

take the previous addition multiply by 3

?

[charred pelican]

at the origin of year, we have 1 stem

at the end of one year, we have 4

at the end of second year, we have 16

at the end of third year we, have 48

[charred pelican]

also, why are you adding these

[twilit musk]

ah yes

[charred pelican]

is it obvious now

[brittle dust]

there’s an easier way to do this

what the sequence is is an arithmetic sequence

this is the formula for it

[twilit musk]

my bad, it asked the total number of stems at the end of 10th year, i got confused and I was already adding them in the sequence

[brittle dust]

a sub n = the number term you are trying to find

[charred pelican]

how is it an arithmetic series

[brittle dust]

a sub 1 = the first term

[brittle dust]

there is a common difference of 3

[left wadi]

isnt it geometric series?

[twilit musk]

if anything it should probably be a geometric series

right

[brittle dust]

geometric is when there is a common ratio

[twilit musk]

I wanted to understand why

[left wadi]

a_10 gives you 28

[charred pelican]

..?

[left wadi]

whats way too low

[charred pelican]

uhm

[twilit musk]

I think we messed up here

[charred pelican]

am i being dumb

[brittle dust]

the difference between each term is 3

when we add by increments of the same number in a sequence it’s called arithmetic

geometric is when we multiply each term by the same quantity

which is not the case here

[charred pelican]

For, T0 =1

for T1 = 4

for T2 = 16

[twilit musk]

its 1, 4 and then 16. clearly theres a factor and not a difference

[charred pelican]

no

[brittle dust]

oh EACH stem grows 3

ok then this is geometric

my bad

[charred pelican]

Find T10

which will be your answer

[brittle dust]

you can use this for geometric

[cunning jackal]

They should look for T11

Because T1 is at the start of the first year

[twilit musk]

no I meant that T4 should have been 64

[charred pelican]

ah yes

[brittle dust]

a sub 1 = 1 (because we start with 1 stem)

and we let n = 11

[charred pelican]

the end of third year is T4 so yes

Both of them mean the same stuff

[left wadi]

if a_1=1 then a_2=3* a_1 then a_3=3* (a_1+a_2) and so on then a_n=3*(a_1+a_2+...+a_(n-1))

[brittle dust]

so it’s just 3^10

[twilit musk]

oh a makes sense

[brittle dust]

59,049 stems

[left wadi]

nah if that would be correct then T_0=1 and T_1=3 but its T_1=4

wouldnt it?

[brittle dust]

oh true

[left wadi]

and T_2=9 but its T_2=16

[brittle dust]

ok im out of my league here lol

[left wadi]

its recursive as i sent here

[twilit musk]

common ratio is 4 right

[left wadi]

why?

16-9=7

[cunning jackal]

Ratio*

[left wadi]

and 16/9 =/= 4

[twilit musk]

my bad yes

[cunning jackal]

Idk why this is being complicated here when you had the answer in your first message

[cunning jackal]

Here

You said it yourself f(n) = f(n-1)+3f(n-1)

=4f(n-1)

[twilit musk]

idkk I just wanted an understanding, which I guess i got. You check for first year, second year, third year, find a pattern, see if its a common series and work with it

[cunning jackal]

Depends on what you call T1.

[twilit musk]

oh yeah

[cunning jackal]

1,4,16,64..

That is our series

[twilit musk]

yes this

[left wadi]

ok

[twilit musk]

well ty guys

[left wadi]

i just leave it out to you

[twilit musk]

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[hoary veldt]

Hi, can anyone help me with the integral of $\sqrt{tanx}$

[elfin berry]

Aetherfly

[novel juniper]

$\int \sqrt{tan(x)}dx$?

[elfin berry]

Why am. I here

[hoary veldt]

yes

ive heard that its very hard

[novel juniper]

ok, so which method do you think you should use

[hoary veldt]

uh

substitution at fist

[novel juniper]

cool

what sub

[hoary veldt]

sqrt cosx maybe?

or no

sqrt cosx

[novel juniper]

hint tan(x)=u works well

[hoary veldt]

would be better

[hoary veldt]

what

I never thought ofthat

[novel juniper]

you first multiply and divide by $sec^2(x)$

[elfin berry]

Why am. I here

[hoary veldt]

😭 why

like how do I know

[novel juniper]

try it

[hoary veldt]

but

how do i know

to do it

[novel juniper]

as in how to think of it?

[hoary veldt]

(dont say intuition)

[hoary veldt]

yes

[novel juniper]

I won't

I basically want to introduce the derivative into the integrand somehow

to perform a u-sub

as I know that $d(tan(x))=sec^2(x)$

[elfin berry]

Why am. I here

[hoary veldt]

hmm

[novel juniper]

so you get $\frac{\sqrt{u}}{1+u^{2\ }}$

[elfin berry]

Why am. I here

[novel juniper]

right?

wait, this may not work

let me try

[hoary veldt]

oh

[novel juniper]

one min

ok, I was sort of wrong

$tan(x)=u^2$ works better

[elfin berry]

Why am. I here

[novel juniper]

allowes you to elimiate the root too

[hoary veldt]

hmm

[novel juniper]

makes sense?

[hoary veldt]

wait let me try

to di it

do*

[marsh citrus]

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[marsh citrus]

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[dreamy hull]

Hello, does anyone know why this change is legal?

[dreamy hull]

Does it have something to do with complementary angles? I know that sin(x) = cos(y) if and only if x+y=pi/2.

[novel juniper]

yes

[dreamy hull]

but how would it apply in this case?

the other thing I know is that cos and sin are equal at 45°

but I don't quite understand what was done there

[novel juniper]

no, I don't think that would help here

$sin(\alpha+pi/4)=cos(pi/2-(/alpha+pi/4))$

[elfin berry]

Why am. I here

[marsh citrus]

@dreamy hull Has your question been resolved?

[dreamy hull]

but how do I get to the cos(a - pi/4) expression?

ah, by the parity of cosine cos(x) = cos(-x)

thanks

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[inner grove]

Can someone explain me please, what does "df" over equal sign means in this notation?

[inner grove]

In my book about probability it's called upper bound of sequence of events A_n

I think it's some kind of convergence, but what exactly

https://tenor.com/view/i-saw-w-gus-fring-gus-gustavo-deleted-gif-25440636

I see on the internet, there is word "def", rather than "df". I think if that's the same, cause I don't see anyone to use df

but it might be it

I will wait for more answers, cause I am not quite sure

[static trench]

yeah that is why I deleted since I wasnt 100% sure and dont want to mislead you

[wooden elbow]

Defined as

[inner grove]

k, thank you

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[neon imp]

How do I confirm this operation is closed?

[neon imp]

I also have the addition, if needed

Can I do neutral element of addition * vector = neutral element of addition?

[marsh citrus]

@neon imp Has your question been resolved?

[neon imp]

<@&286206848099549185>

[marsh citrus]

@neon imp Has your question been resolved?

[neon imp]

<@&286206848099549185>

[remote granite]

nie wiem

[neon imp]

Damn, I really need to understand this.

Cuz if a scalar times the vector comes to something, it also has to be inside the same subspace so it is closed under scalar multiplication.

[marsh citrus]

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[neon imp]

.close

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[craggy skiff]

help

If I were to add 50% to a number, how could I reverse this? like if I was given a number and was told it had 50 percent added to it, how could I get its original value?

[sleek lake]

times 2 divide by 3

[craggy skiff]

ty

[marsh citrus]

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[past shard]

Does anyone mind explaining why $\frac{d}{dx}\left(\frac{\sqrt{1^2+\left(15-x\right)^2}}{2.5}\right) = \frac{0.4\left(x-15\right)}{\sqrt{x^2-30x+226}}$ and not $\frac{1}{6}+\frac{\frac{2}{5}x-6}{\sqrt{1^2+\left(15-x\right)^2}}$?

[elfin berry]

Punisher

[honest berry]

hey guys

can someone help me

[tiny patrol]

try simplifying the fraction, to 1/2.5(sqrt(16-x^2)) and apply differentiation rules

@honest berry go to a closed help channel and open it

[honest berry]

good idea

[past shard]

Which fraction?

[tiny patrol]

@past shard hint:

this will make things much easier

[tiny patrol]

the initial one, the one you wanted to find the derivative of

[tiny patrol]

like this

now can you find the derivative of that?

[past shard]

Where are you getting the 16 from?

[tiny patrol]

you have 1^2 + (15-x^2) correct?

[past shard]

No. I have sqrt of 1^2 + (15 - x)^2

Oh

[tiny patrol]

that changes things then

[past shard]

That's my bad; I didn't indicate that it was the quantity (15 - x)^2

[tiny patrol]

alright then dont simplify, go ahead and differentiate like this, moving the constant to the outside

and letting the inside be

sqrt(1 + (15-x)^2)

step by step, using the chain rule

and tell me where you will get

[past shard]

d/dx [√1^2 + (15 - x)^2] => 1/2 • d/dx [1/√1^2 + (15 - x)^2]

Am I right so far?

[tiny patrol]

you are missing something, apply the chain rule correctly

you need to multiply the whole fraction by the derivative of the inside of the square root

so (1/2 • d/dx [1/√1^2 + (15 - x)^2]) * d/dx[1 + (15-x)^2]

if you have done it correctly it should be this

dont forget to multiply this whole thing by the constant at the start though!!

1/2.5

[past shard]

Right. Ok, so that would equal (-1)(15 - x)(1/√1 + (15 - x)^2)

[tiny patrol]

not quite, how did you get the -1 at the start?

[past shard]

from d/x[(15 - x)^2]

well, from the interior function

[tiny patrol]

ohhh yes, i forgot to write it down on the final answer!

the minus is there in the last step

[past shard]

Right. So, the final answer should then be (x – 15)/6.25(√1 + (15 - x)^2. Right?

[tiny patrol]

why 6.25?

the constant is 2.5

[past shard]

2.5^2 = 6.25

The denominator must be squared, no?

[tiny patrol]

nono you have placed the constant outside the derivative so you just multiply it at the end

d/dx[c * f(x)]= c * d/dx[f(x)]

are you there? do you need me to explain any further on that?

[past shard]

Sorry. I just re-derived the original function (d/dx[(√1^2 + (15 - x)^2)/2.5]), and I got a very similar answer to my original answer (minus the 1/6).

^ I thought the quotient rule of derivatives stated that the derivative of a fraction is the derivative of the numerator multiplied by the denominator minus the derivative of the denominator multiplied by the numerator, and all divided by the denominator squared.

Wait. I was right this whole time 😆; the answer I looked up was just in decimal form (namely, 0.4(15 - x)/√x^2 -30x +226)

0.4 => 2/5 lmbo

Anyways

Thanks @tiny patrol

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[flat bear]

This is the first exercise in chapter 4 of the book think bayes 2 and i can't understand why they choose the prior for getting a hit as uniform between 0.1 and 0.4? I don't see any motivation for that in the question, imo either you use the info given to make a more accurate prior or you just do a uniform between 0 and 1. Anyone understand this choice? Hopefully this is ok as a statistics question, feel free to remove if its not appropriate to ask that here

[marsh citrus]

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[tawny herald]

How do I find the center on the ellipse based off the plotted points without looking at the graph?

[sand fable]

plug in theta = pi/2 and theta = 3pi/2 to the polar equation

average those 2 points to find center

[tawny herald]

Thank you that works. 🙂

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[balmy vault]

Im trying to find the real and imaginary values of the function g(z) = sin(1/z)

[balmy vault]

Im kinda stuck at (e^(i/z)-e^(-i/z))/2i

[still temple]

euler’s formula for the trig functions?

[balmy vault]

imma try 2 sec

[stray hound]

let z = a+bi first to separate real and imaginary parts

[balmy vault]

wait how would i then spilt the z=a+bi inside the cos and sin

[still temple]

you dont

[stray hound]

1/(a+bi) = (?) + (?)i

[balmy vault]

then every part would be imaginary no?

[still temple]

because e^(a + bi) = e^a(e^(bi))

[balmy vault]

wait i can split fractions like that?

[still temple]

1/(a+bi) = (a - bi)/((a+bi)(a-bi))

simplify that

again the last time i was messing around with trig in C i used the taylor series

[balmy vault]

(a-bi)/(a^2-b^2)

?

[still temple]

equal to a/(a^2-b^2) - bi/(a^2-b^2)

[balmy vault]

yeahh oooh i think i got now ty

[still temple]

now expand into cos x + i sin x in your formula thing

also can you tell me once you have your answer i might need it for a coding project lol

[balmy vault]

Sure give me little time tho something came up i gotta go for 5 ish min

[marsh citrus]

@balmy vault Has your question been resolved?

[balmy vault]

wha

well i found a section in my book after i was stuck

sin(z)=sin(x)cosh(y)+icos(x)sinh(y)

i am assuming i can use this property here no?

[still temple]

where z = x + iy? yes definitely

just gonna screenshot this rq for whenever i need brute force complexexp rq

[balmy vault]

This is correct no? Sorry bout handwriting and Angle

[still temple]

how do x and y relate to z

[still temple]

oh ok

[balmy vault]

Isnt the real part denoted by x and the imaginary by y?

[still temple]

yes

so it would be sin(x + iy) = sin(x)cosh(y) + i cos(x)sinh(y)

[balmy vault]

Ye

But I got sin(1/z)

[still temple]

now substituting that into your formula seems to match that paper

[balmy vault]

^^ yay

[still temple]

yes i think

[balmy vault]

Ty for help ^^ ❤️.close

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[marsh citrus]

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[cobalt garnet]

Suppose that two cards are selected at random from a standard 52-card deck. What is the probability that both cards are less than 10?

Since there are 4 suits and each of them has 8 cards that are less than 10 we get 32 cards that are less than 10 in a 52-card deck. To get the probability we divide 32C2 by 52C2?

[slim spire]

👍 that's correct, there's a way that has a more simplified answer if you do (chance of first <10 card)*(chance of second <10 card given the smaller deck)

[cobalt garnet]

Thanks

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[odd sparrow]

Is my working out correct?

[vernal pike]

Yes

[marsh citrus]

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[balmy vault]

im trying to sketch f(z)= zi where Re(z) > 0

[balmy vault]

so let z = x+yi so f(z) = xi-y

x would still be the real part of the equation so would this be the functional equivalent of x*1

?

which would mean that this would essentially be the same sketching z with Re(z)>0

feel free to ask questions if i didnt properly explain myself

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[sharp mauve]

0,3 <= π^2/8-0.5arctan(N) how can i find n without trail and error?

[sharp mauve]

This was my attempt

[zenith dragon]

i'd proly do it by trial and error

[sharp mauve]

why?

[marsh citrus]

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[tidal wasp]

help

[tidal wasp]

hello

i need help

on this question

[winter nest]

you have only considered one possibility, the magnets could be UDU or UUD

[tidal wasp]

yes

added them up and i get 9/16

i tried tht but it doesnt work sadly

[winter nest]

is there a chance that the quesetion is written wrong and wants you to do at least 2 pointing upwards?

[tidal wasp]

idk

its unlikely

but i could try the answer for what u said

if so i would have to include UUU

[winter nest]

yeah

[tidal wasp]

yh it isn't 10/16 either

[winter nest]

yeah you have done the probabilities wrong

[tidal wasp]

oh

[winter nest]

the answer i got was 3/4, but working out P(UD)xP(DU) + P(UU)xP(UD)

[tidal wasp]

yes thats correct

could u draw a ven diagram

[winter nest]

so the probabilities you are wanting is 3/4 x 3/4 + 1/4 x 3/4

[tidal wasp]

yeah

what about these two questios

[winter nest]

you want to work out P(UDD) as that is the only configuration that will have only one magnet pointing upwards

[tidal wasp]

so would that be 3/4 *1/4

[winter nest]

yeah thats correct

[tidal wasp]

P(DDD)

would it be 3/16

[winter nest]

you will have to read the question carefully for that one

[tidal wasp]

tbh

i need to work on how i read that question

but would part D u just need to find P(DDD)

[winter nest]

yeah you need to find P(DDD) but when you read the question you can see that the left most magnet is always U, so the probability is 0

[tidal wasp]

oh

so 0 * 1/4

[winter nest]

yyeah exactly

[tidal wasp]

oh

can i also get help on a kinematics question

[winter nest]

i can try, i havent really done any kinematics for about 3 or 4 years but i can have a look and point you in the right direction if i dont understand it

[tidal wasp]

alright

what about a vectors questin

[winter nest]

yeah i can help with that

[winter nest]

with this one, if you have the speed of the yacht u = v then they are both heading in the same direction, so the least value of u would have to be going in the opposite direction to v

[tidal wasp]

alright

so u = -v

[winter nest]

thats what i think yeah

[tidal wasp]

nope its wrong

i think it needs to include a and b

ah don't worry anyway

thank u for your help

may your pillows be forever cold to have a good sleep

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[wintry moth]

Help

[wintry moth]

(╯°□°)╯︵ ┻━┻

.flip

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@wintry moth Has your question been resolved?

[celest geyser]

quadratic formula

[marsh citrus]

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[urban grove]

How much weight of lead do I need for a 52 cm sphere made out of concrete, if I want the desired weight to be 220 kg?

The sphere diameter cannot change, only the lead and concrete volume.
Calculated with 2200 kg/m3 concrete volume and 11 340kg/m3 with lead volume.

52 cm sphere = +- 162 kg, that would simply mean I need 58 kg lead core of a 21,4 cm diameter.
However, while inserting the core, there needs to be taken in consideration, that also the same volume of concrete needs to be removed. Which is around 11,3 kg with that diameter and concrete volume. And THAT again... would mean I need to add another 11,3 kg of lead to the core ?
Which would mean -> 22,7 cm diameter of the lead core, which again removes roughly 13,5 kg of concrete and again calculation of core... I dont think I am going the right way... I would like to reach the exact weight of lead needed, which doesnt get me there by this approach.

(I am long gone from math classes, sorry, be nice to me.)

[marsh citrus]

@urban grove Has your question been resolved?

[marsh citrus]

@urban grove Has your question been resolved?

[urban grove]

<@&286206848099549185>

[marsh citrus]

@urban grove Has your question been resolved?

[molten moss]

Quick question, do you want the lead inside the sphere to be any specific shape? In terms of calculations one easy solution would be having it as a sphere

Then, the only unknown is the radius/diameter of the lead sphere inside. Let me try and walk you through the theory

Let's fix some notation: $m$ is your desired total mass (220kg), $\rho_c$ is the density of concrete (2200 kg/m3), $\rho_{l}$ is the density of lead (11340kg/m3), $R$ is the outer radius (52/2 cm = 26 cm) and $r$ is the radius of the inner sphere.

The weight of the inner lead sphere is then the volume of it, times the density: $\frac{4}{3} \pi r^3 \rho_l$, the weight of the concrete hull around it is $\frac{4}{3} \pi (R^3 - r^3) \rho_c$, so we have that the total mass is $m = \frac{4}{3} \pi r^3 \rho_l + \frac{4}{3} \pi (R^3 - r^3) \rho_c$. Since the only unknown is $r$, we should be able to solve this equation for $r$, let's do that

[elfin berry]

Lartomato

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[molten moss]

Just reshuffling terms gets us $\frac{m - \frac{4}{3} \pi R^3 \rho_c}{\frac{4}{3} \pi (\rho_l - \rho_c)} = r^3$

[elfin berry]

Lartomato

[molten moss]

goddamnit i'm not asking a question i'm responding to one

Simplifying that and taking the third root, we get

$\sqrt[3]{ \frac{3m}{4 \pi (\rho_l - \rho_c)} - \frac{R^3 \rho_c}{\rho_l - \rho_c}} = r$

[elfin berry]

Lartomato

[molten moss]

If you trust this, then the radius of your inner lead sphere should be 11.487cm

[molten moss]

And 11.487 * 2 is 22.974cm, so your estimate of 22.7cm diameter seems pretty good!

and the double-check does pretty well

@urban grove let me know if that helped you! good luck with your extremely heavy spheres

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[urban shard]

Looking at a quick proof for symmetry in A^T A

[urban shard]

It is the dot product of row i of AT (column i of A) with column j of A. The (j, i) entry
is the same dot product, column j with column i.```

😭 Lol why is this the case? I don't see it

,, (row i){A^T} \cdot (col j)A = (row i){A} \cdot (col_j){A^T}

[elfin berry]

nyxie9151

[urban shard]

they're equal?

[iron meadow]

not quite like this, row i of A becomes column i of AT, column j of A becomes row j of AT

1st column = 1st row of transpse

[marsh citrus]

@urban shard Has your question been resolved?

[urban shard]

yes i know how tranposes work

i mean like

i'm talking about symmetry in A^T A

What does this mean? Maybe i'm a bit dense now but yeah

Could you explain that bit

[urban shard]

They're saying this right?

actually i got it

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[half portal]

hello, i'd appreciate some help with this problem, i know there's probably a formula but i would like to know if i could integrate shape (i) 👍

[half portal]

• if i put it in a cartesian plane, maybe try to get a function for the horizontal crossection, i assume it would be y=(1/2rx+r), because of the gradient and the fact that the integration would start from y=r, is this the correct approach? can't seem to get it right

[stray hound]

it's the right idea, but I think your line is wrong since it should depend on h in some way

connect the points (0,r) and (h,3r/2)

then find the volume

[half portal]

r/2h right?

then the line should be y=(rx/2h)+r?

got it, it was (19pi/12)(r^2)(h) thanks

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[shut stone]

find the real number m so that the equation has at least two real solutions

[shut stone]

!status

[marsh citrus]

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

[shut stone]

1

<@&286206848099549185>

[spark siren]

just try some values. assume x = 0. does the equation hold?

[shut stone]

it does not hold

[spark siren]

sorry, not x = 0. m = 0.

[shut stone]

it does

[spark siren]

for m = 0 you get 2 soluions for x?

[shut stone]

yes

wait no

yes

x=+-sqrt(e)

[spark siren]

now look at your solutions a) to f) which contains 0?

[shut stone]

did you mean a and c?

[spark siren]

a and c does not contain 0, so they cant be right.

[shut stone]

yes

I also took the derivative and I got 2x(1/x^2-m)=0, and for m=-e^(-2) we have 1/x^2=-e^(2) which is false. So we also eliminate d

and e and f

so it must be b

right?

[spark siren]

why not f?

[shut stone]

we have 1/x^2=m in the derivative so m cant be negative

and it need to have at least two points in which is 0

except for x=0

[spark siren]

why do you use the derivative?

[shut stone]

I thought it would be easier

[spark siren]

easier for what?

[shut stone]

if it changes sign at a point then it has at least one solution in that interval

[spark siren]

m=-1 has to solutions:

[shut stone]

oh well

what would be the better approach

[spark siren]

just try some appropriate values.

[shut stone]

ok ill try

so there is m=e^(-2)

so we check now only d and e

and since we also have m=-1 the answer is f

thank you for help

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[urban cradle]

quick question, does e stand for joules or p stand for joules

[frozen delta]

J stands for joules

but

[urban cradle]

J?

yeah ik

[frozen delta]

Joules are a measure of energy

so

[urban cradle]

but like u have E P T

aahh

so E

[frozen delta]

E

yeah

:)

[urban cradle]

thx

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[rare pollen]

You want to buy a new sports car from Muscle Motors for $44,300. The contract is in the form of a 72-month annuity due at an APR of 6.05 percent. What will your monthly payment be?

[storm fern]

so

[rare pollen]

hi

[storm fern]

you could use this calculation to find the total price

[rare pollen]

uh

i have not seen that equation before

[storm fern]

n is expressed for the n-th value of a certain period which is annually increased based on the question

so you have 72 months right?

[rare pollen]

correct

[storm fern]

how many years would it be

[rare pollen]

6

[storm fern]

the n would be the exponent you use on this equation

and the i would be the annual percentage rate

which is the interest

[rare pollen]

right

[storm fern]

and Mo is the price offered by Muscle Motors

which is 44.300

[rare pollen]

right

[storm fern]

so what's the total price would be by using this formula?

[rare pollen]

well i have to first fit everything into the equation

[storm fern]

after you found the total price, divide it by 72 as you have to find the amount to pay monthly

[rare pollen]

72 = 44,300 (1 + 6.05) 6 ?

that might be wrong

[storm fern]

no

72 is not the total price of it

[rare pollen]

o the left is what we are finding out

[storm fern]

right

[rare pollen]

44,300 (1 + 6.05) 72. ?

[storm fern]

$44.300(1+6.05)^6$

[elfin berry]

sho

[rare pollen]

ah i see

5,439.242.58