help-39

[vestal tapir]

second guess is 19
the response is "lower", so I guess 18 and I win

[tropic saddle]

well if anything your next guess should be 21 if you wanna trigger the 3x lower rule

[vestal tapir]

i want to guess 19

[devout hamlet]

yeah but in 6 guesses am i guaranteed to guess a number in a range of 1-40

[vestal tapir]

what do you mean

[tropic saddle]

I dont exactly understand the third rule. are you somehow guaranteed to not get 3x higher or lower or are you not allowed to get it

[vestal tapir]

yes, because 3 guesses is enough for any range

[devout hamlet]

i'm not allowed to

so if i do 20 21 22 and they all had a response of "higher," i lose

[vestal tapir]

ah ok you lose if you let it happen

[tropic saddle]

well that basically means after two times higher you have to guess 40 next

[devout hamlet]

yeah in this case the numbers are narrowed to 1-19. i would guess 10, but then if it was "lower" i would have to guess "1"

and then idt it'll be possible

but i still wanna prove it without making it sound messy

[tropic saddle]

well then you still have three guesses for numbers 2-9

or four guesses for 11-19

hmm but if you guess 5 then next you would have to guess 9. so that doenst work

[vestal tapir]

your second guess should be low, like 7 , so you penalize "lower"

[devout hamlet]

yeah but i think this situation would have a recursive formula or some pattern, i just can't find it

[pearl pond]

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[fading ocean]

I cant apply lagrange to this right?

[fading ocean]

how do i deal with the inequality applied to the constraint

[acoustic path]

depends if you can prove whether the maximum occurs during the equality case

[fading ocean]

is that where i should start?

[acoustic path]

sure might be worth looking into

you can consider ||what this setup looks like in the 2d plane||

(it should be noted there are multiple ways to do this without using lagrange, such as ||cauchy-schwarz inequality|| or ||solving a quadratic||)
unless this is specifically a lagrange exercise and you're required to use it ig

[fading ocean]

we technically did not learn lagrange in lecture yet

just double intergrals and implicit

but some of the hmwk qs clearly use lagrange so idk why my prof is doing this

[pearl pond]

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[pearl pond]

@fading ocean Has your question been resolved?

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[storm spruce]

I am trying to solve analytically for the contour integral over C of the function f(z). I am having considerable trouble reproducing the answer the textbook provides. I am using the fact that z^(a-1) = exp[(a-1)Log(z)] to simplify the integral, but keep getting stuck. any advice you may have for solving problems of this type would be greatly appreciated.

[pearl pond]

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@storm spruce Has your question been resolved?

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[tropic zenith]

anyone here can with me with my data management homeowrk?

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[wraith sorrel]

how would I find out if e.) converges or diverges

[regal herald]

write it out for 5 odd numbers and 5 even numbers

the first 5

then youll need to see if theyre moving towards the same thing

(i would split the fraction of the odd n's, it makes it quite clear)

[wraith sorrel]

so are there two different answers?

[regal herald]

no? its one sequence

[wraith sorrel]

so the limit of 1/n would be 0

and the limit of 2n+1/n would be 2

[regal herald]

yeah, so is it convergent

[wraith sorrel]

and it would converge at 2?

or 0

[regal herald]

it just wouldnt converge

at some point the sequence will essentially look like
0,2,0,2,0,2,0,2,...

thats not converging

[wraith sorrel]

so just convergent and the limit is 0 and 2

[regal herald]

no its not convergent

the limit doesnt exist

[wraith sorrel]

ohh

because 2 does not equal 0

I soo

see

[regal herald]

yeah, itll just bounce between them

[wraith sorrel]

I get it

thank you

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[hybrid basin]

Hello

[hybrid basin]

Does det only work for square matrices?

[terse cedar]

yes

[grizzled dust]

yes only square matrices have determinants

[hybrid basin]

They are racist af

But ic ty!

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[hybrid basin]

.reopen

Wait

[pearl pond]

✅

[hybrid basin]

Why tho

Can't you just add 0, 0, 0 in the last column?

[meager trellis]

well i guess you could do that, but then the determinant would be 0 so that's not very interesting

[hybrid basin]

So if the matrix is not in square, bye bye linear transformation?

[meager trellis]

...?
it's still linear, there just isn't a useful definition of determinant

a non-square matrix corresponds to a linear transformation that takes n-dimensional vectors and outputs m-dimensional vectors for some m that isn't the same as n

[hybrid basin]

Oh you're right

It's gonna output a different dimension

What are 2 x 3 matrices for then?

Why do they exist?

Why not all matrices have exact same ranks

Even the identity matrices only apply to square matrices

[pearl pond]

@hybrid basin Has your question been resolved?

[grizzled dust]

because sometimes you wanna work with linear transformations that have a different domain and codomain

like example: projection of points in R3 onto some 2 dimensional plane working in computer graphics

[hybrid basin]

Ohh I see

Ty

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[nova mortar]

why does this function not have an oblique asymptote?

[quick iris]

!status

[pearl pond]

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

[nova mortar]

7

it's just that the correction says that there's no oblique asymptote

but the exponent of the numerator = +1 the degree of the denominator

[quick iris]

do you know how to find oblique asymptotes

[nova mortar]

so idk why not

yea

Euclidean division

[quick iris]

long division

perform the long division

[nova mortar]

ok

2sec

[grizzled dust]

or just factor

no need for long division if it can be manually factored easily

numerator = -2(x^2 - x - 12)
maybe now you can see more easily how to factor it

[nova mortar]

with long division i get 2x/3+4/3

[nova mortar]

if theres no more x in the denominator i cant do long division?

[grizzled dust]

basically that cancels out, as long as $x\neq -3$

[jolly parrot]

∫ooshⁱˣ

[grizzled dust]

so it's more or less the graph of a line but with a pinprick hole at x = -3

[nova mortar]

ohhh alrr

so no oblique asymptote

[grizzled dust]

right, so it's an indication that the numerator and denominator of the rational function have a common factor

if you graph it and see that

[pearl pond]

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[heady bone]

can someone help please

[heady bone]

i'm not really sure how to do this question

[west sapphire]

the second line

yea

scooped by doaby haha

[brittle tinsel]

lol, sniped

but fix your Algebra and you've got the right idea

[heady bone]

i'm not really sure where i went wrong

[west sapphire]

you made a sign error on the second line

and you also lost a term

[heady bone]

the term magically reappeared later on after i realised 💀 but regarding the sign error, is it with (k-2)?

[west sapphire]

yes

[heady bone]

so is it x^2 - (k-2)x - k - 3x = 0?

or do i have to flip the sign in the bracket into a +? i'm not really sure

[west sapphire]

it was (k-2)x on the left side

so if you move it to the right side, it becomes -(k-2)x

[heady bone]

ohh okay

does it turn into this?

oh wait oops

kx

[west sapphire]

yea one of the -k's should be -kx

but otherwise correct

[west sapphire]

another sign issue on line 5

[heady bone]

oh it should be + 1?

[west sapphire]

yea

and then on line 6, b would just be -(k+1)

the x is not part of b

[heady bone]

oh

after i expand b^2, would it look like (k+1)^2?

[west sapphire]

so far so good

yea, when you square -(k+1), the - goes away

because -1 times -1 is 1

[heady bone]

i'm not really sure how to factorise the last line

[west sapphire]

use the quadratic formula?

[heady bone]

ohhh

true

[west sapphire]

it's not going to have nice solutions

[heady bone]

so k = -3 + or - 2 x square root 2

how do i determine if it's tangent?

[west sapphire]

looks right

(assuming all the other algebra is right)

[heady bone]

is the rule that if it doesn't equal 0, then it's not tangent?

[west sapphire]

if what doesn't equal zero?

[heady bone]

k

the term]

how do i know if it's tangent to the parabola y = x^2 - 3x?

if the discriminant = 0 right

[west sapphire]

yes

discriminant = 0 means there's exactly one solution to your original equation, which means that the two curves intersect at exactly one point, which means they are tangent to each other

[heady bone]

which means that the question is false

thank you!!

[west sapphire]

wait, why is it false?

you just found two values of k that make it true, didn't you?

[heady bone]

huh?

wait what

ohh I found the values which satisfy the rule?

which means the question is true?

tbh idek what i'm doing i just plug it into the formula they tell us

i don't really understand it, i just do it

[west sapphire]

it's not a true/false question
they tell you to find the values of k that work

and you did that

[heady bone]

oh

What about this question?

[pearl pond]

@heady bone Has your question been resolved?

[pearl pond]

@heady bone Has your question been resolved?

[pearl pond]

@heady bone Has your question been resolved?

[vivid grove]

Use long division

refer to youtube if need be

[pearl pond]

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[scarlet vector]

hello

[midnight haven]

hello

do you have a question?

[pearl pond]

@scarlet vector Has your question been resolved?

[torpid mountain]

@scarlet vector

!done

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[mental verge]

can someone help me with a true or false statement?

the question is

True or False? If c is a real zero of the polynomial $$P$$, then all the other zeros of $$P$$ are zeros of $$\frac{P(x)}{x-c}$$

[jolly parrot]

ates

[mental verge]

okay so basically all i know is P(x) over x-c means P(c) = the remainder

from the remainder theorem

but im not rlly sure whether this is related to the remaider theorem or not

i mean i think this is false though because it doesnt rlly makes sense but idk whether there's a specific rule for this or not

<@&286206848099549185>

[empty needle]

hm yeah true I think

[mental verge]

why

[empty needle]

c is a zero of P right?

[mental verge]

yeah it says c is a real zero

[empty needle]

uhuh alr

P(x) = 0

say that P(x)/(x-c) = Q(x)

which implies

P(x) = Q(x)(x-c)

to find the zeroes

we have

Q(x)(x-c) =0

we know c is a zero of P(x) already

so to find the other zeroes we equate Q(x) =0, which would contain the zeroes of Q(x)

[mental verge]

but it would contain the zeroes of Q(x) not P(x)?

i mean as i look more into it it seems true but i just cand state a reason lol

[empty needle]

okay

Q(x)(x-c) =0

[mental verge]

yes

[empty needle]

I mean

correct?

so if Q(x) = 0

P(x) is also 0

correct?

[mental verge]

yeah

[empty needle]

so the value of x for which

Q is 0

is also a value of x for which P is 0 right

[mental verge]

yea

[empty needle]

doesn't that imply a zero of Q is a zero of P?

[mental verge]

oh ok

[empty needle]

yeah

you get it right?

[mental verge]

okay yea kinda like Q(x) implies quotient right

[empty needle]

ye

[mental verge]

like it makes sense but it doesnt lol

[empty needle]

damn

[mental verge]

i mean P(x) / x-c means P(c) = 0 if c is a zero

[empty needle]

what part doesn't make sense

[mental verge]

right

[empty needle]

?

yes

hm

no wait

no

I mean

not exactly? i guess

okay just run me through your thought process

[mental verge]

uhm it says zeros of (P(x)) / x-c firslty what is that

what are zeroes of this?

[empty needle]

ah okay

I think You'll understand if I give you an expression

alright

say you have

P(x) = x^2-5x+6

you know that 3 is a zero of this right?

[mental verge]

okay

[empty needle]

so what would be P(x)/x-3?

[mental verge]

0

[empty needle]

really?

say x is not equals 3

[mental verge]

oh okay

so idk like the other zero of P?

oh wait

okay

it would be equal to the other zero of P(x) right?

[empty needle]

yes

[mental verge]

oh okay now i got it

[empty needle]

yeye

[mental verge]

thank youu

[empty needle]

nws

[mental verge]

anyway im closing the chat have a great day

.close

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[winged haven]

For this question, does the answer $\frac{1}{4} ln((2 + x^4)^2)$ also work?

[jolly parrot]

Kai

[eternal reef]

what answer are you getting right now?

[eternal reef]

and a constant

[winged haven]

sorry yes + c

but does that work?

cuz the organic chem tutor did the reverse power rule

[eternal reef]

need to check it hold on

[eternal reef]

uh I thing that's wrong

[winged haven]

Oh fr?

[eternal reef]

what I am getting is $\frac{-1}{4(2+x^4)} + C$

[jolly parrot]

Wither

[eternal reef]

what have you tried?

[winged haven]

Well, I got till $\frac{1}{4} \int \frac{1}{u^2} du$ so I treated it like it's anti derivative of a 1/x function and just said $ln(u^2) \cdot \frac{1}{4}$

[jolly parrot]

Kai

[eternal reef]

ohh

that's the mistake you've made sir

[winged haven]

Oh really? We're not allowed to do this?

[eternal reef]

the antiderivative of 1/x is ln(x)

you cannot use this fact for 1/x^2

because power rule is valid here

and the derivative of ln(x^2) is not 1/x^2

[winged haven]

Interesting. What if I counter it? derivative of ln(x^2) is 2x / x^2 yes?

so I could say the answer is ln(u^2) * 1/8x, that'll cancel out the power rule

[eternal reef]

ah

only if that worked

[winged haven]

💀 Do none of my methods work

[eternal reef]

see what you're doing is using substitution in a wrong way

so the derivative of $ln(x^2) is \frac{2x}{x^2}$

[jolly parrot]

Wither

[winged haven]

Yes

[eternal reef]

but

lets say we want to find the antiderivative of 1/x^2

what you're doing is trying to divide the 2x

which is not possible because only constants can move out of the integral sign

and not variables

[winged haven]

ohhhh

ok I see what I've done

[eternal reef]

yeah

if that would be possible then integrals would be too easy

dont worry if you have such kind of doubts

[winged haven]

Got it. Thanks I'll work on other problems 👍

[eternal reef]

cos its good to have them, since you're actually getting these topics

[eternal reef]

welcome :)

[winged haven]

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[forest vine]

need to prove whether or not this sequence is convergent and if yes, find it's limit value.

[near echo]

probably want to multiply by the conjugate

@forest vine

[pearl pond]

@forest vine Has your question been resolved?

[forest vine]

what is a conjugate?

[near echo]

the conjugate of sqrt(n+1) - sqrt(n) is sqrt(n+1) + sqrt(n)

[forest vine]

ahh ok ok

[near echo]

so write
[
c_n = \sqrt{n}(\sqrt{n+1} - \sqrt{n}) = \sqrt{n}\frac{(\sqrt{n+1} - \sqrt{n})(\sqrt{n+1} + \sqrt{n+1})}{\sqrt{n+1}+\sqrt{n}}
]

then distribute on the top

oh sorry

[jolly parrot]

maximo

[forest vine]

and in this form, it will be easier to analyse right?

[near echo]

yes. the square roots on the top will cancel, and you can factor a sqrt(n) from the denominator to cancel the one up front

[forest vine]

sqrt(n)/(sqrt(n+1)+sqrt(n)

[forest vine]

I didn't get that part

do I just multiply ot again by sqrt(n)/sqrt(n)?

[near echo]

no

the denominator can be written as

\begin{align*}
\sqrt{n+1}+\sqrt{n} &= \sqrt{n\left(1 + \frac{1}{n}\right)} + \sqrt{n}
\&= \sqrt{n}\sqrt{1+\frac{1}{n}} + \sqrt{n}
\&= \sqrt{n}\left(\sqrt{1 + \frac{1}{n}} + 1\right)
\end{align*}

oh ran out of space sorry

[jolly parrot]

maximo

[near echo]

so we factor an n from inside the first root

then factor a sqrt(n) from the whole thing

so you could replace the denominator with that

and you'll have a sqrt(n) in the bottom, which will cancel with the sqrt(n) up front

[forest vine]

ok ok

I couldn't think of that

thank you @near echo

@near echo and then can I rationalise it using the conjugate of the denominator?

[near echo]

no need

that would undo your work sort of

the idea is now that the sqrt(n)'s cancelled

the denominator goes to 1

[forest vine]

,rotate

[jolly parrot]

Couldn't find an attached image in the last 10 messages.

Mar 25, 2024, 12:45 PM

[forest vine]

goes to 2 right?

[near echo]

oh yes

2 my bad

[forest vine]

I know what I did here is probably useless

but have I made a mistake maybe?

,rotate?

[forest vine]

nvm I figured it out

you were right it makes it worse

[forest vine]

@near echo any ideas about this one?

[near echo]

factor an n from the bottom

it's a similar principle

[forest vine]

I know that it's approaching 1 from both directions

ok ok

lemme try

can it be considered a limit, if the sequence os approaching it from two different directions?

[pearl pond]

@forest vine Has your question been resolved?

[compact ridge]

Yes, you can use something called the squeeze theorem

If $a_n \le b_n \le c_n$ for all $n$ and $a_n \to L, c_n \to L$, then $b_n \to L$ also

[jolly parrot]

south

[pearl pond]

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[midnight haven]

Can anyone give me a concise solution for this

[pearl pond]

@midnight haven Has your question been resolved?

[earnest stratus]

We don't hand out solutions here.

Do you have anything you've tried?

[midnight haven]

i tried doing triple scalar product but it is coming so long calculation

[earnest stratus]

Are you saying that because you think the calculation is tedious you just assumed it's wrong?

You can post your work and we'll help you proceed from there.

[midnight haven]

ok wait

[earnest stratus]

,rcw

[midnight haven]

See it's so humongous

[earnest stratus]

You could just do something like

(B + C) × (C + A) = B × C + B × A + C × A

And try to generate the scalar product [A B C]

Instead of brute forcing like that.

[pearl pond]

@midnight haven Has your question been resolved?

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[sharp smelt]

$\frac{12}{m}=\frac{n}{12}$ how many ordered pairs of integers(m,n) satisfy this

[jolly parrot]

ƒ(Why am. I here)=I don't know

[sharp smelt]

so mn=144

so I suppose i start by prime factorising 144

,w factorise 144

[jolly parrot]

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[sharp smelt]

hmm

so 15 ways?

[midnight haven]

yeah

5.3=15 so yeah

[sharp smelt]

got it

thanks!

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[midnight haven]

Please help

[midnight haven]

<@&286206848099549185>

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[exotic idol]

Determine the characteristic polynomial pa of A

the answer should be z(z-2)² according to the problem sheet, but i get -z(z-2)². i have used several online calculators and get both results.

[prime bramble]

it’s a matter of convention usually

some people and calculators use det(zI - A) to be the characteristic polynomial, some people use det(A - zI)

if you use the former, you won’t get the (-1)^n

in any case, it doesn’t really matter, because we almost always only care about the roots of the char poly, which isn’t affected by the leading sign

[exotic idol]

ahh ok thx

.close

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[pearl pond]

@restive pike Has your question been resolved?

[pearl pond]

@restive pike Has your question been resolved?

[subtle shard]

@restive pike try to see parallelograms

[restive pike]

In what way?

[subtle shard]

For the lower and upper edges of the trapezoid

For example QT and RS are parallel

QR and TS are parallel also

then QRST is paralellogram

opposite sides equal

lengths

[restive pike]

Opposite sides of a parallelogram are equal?

oh aight

[subtle shard]

yes

and there are too many parallelogram here

[restive pike]

So ig that's the ans to the first qstn?

there's 2 marks though

[subtle shard]

yes

[restive pike]

PQDU is a parallelogram
QRTS is a parallelogram

[subtle shard]

if this answer is valid you can write this

yes

[restive pike]

It's a past paper but no marking scheme which is annoying

[subtle shard]

hm

[restive pike]

For the 3rd qstn I was thinking of total area - white box area, but I couldn't find the length of the 2nd white box

[subtle shard]

just find the shaded areas ig you want

directly

[restive pike]

dont understand where to find the last shaded boxs length

first two shaded boxes are fine

[subtle shard]

area of ​​parallelogram = base times height

area of a triangle = base times height divided by 2

use them

find how many cm is AP

PQ = 3cm

QR = 2cm

RB = 3cm

AP = 9cm-3cm-2cm-3cm = 1cm

The height of the APD triangle is 7cm

1cm × 7cm/2 = 7cm²/2

[restive pike]

found that and the second

[subtle shard]

you can continue like this

[restive pike]

this is my problem

[subtle shard]

base times height

21cm²

btw i mentioned about area of RBCS

[restive pike]

where did you get 7cm ?

[subtle shard]

7cm is given

by the question

AD = 7cm

that means the distance between line AB and line DC is 7cm

[shy imp]

Yea so base is up-down, not side length of something else

[pearl pond]

@restive pike Has your question been resolved?

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[merry stirrup]

EX 16

How to do it 🥲

I Approached this question like this
3^x =y
9y²+(a²-4a-2)y+1>0
y(9y+1/y)+(a-2)²>0

Now I don't know what to do

[glacial sequoia]

the third line was unecessary

[merry stirrup]

Yep I don't understand the concept

[glacial sequoia]

You can use the discriminant of a quadratic from the second line

[merry stirrup]

Why a≠2?

[merry stirrup]

For ex 16

[glacial sequoia]

oh wait you didn't complete the square properly

I thought something looked off

[glacial sequoia]

yes indeed

[merry stirrup]

So like
let's just say y is the variable of this quadratic eq

[jolly parrot]

²⁰Ne

[merry stirrup]

Ogey

(a-2)⁴-4(9)(1)>0
(a-2)⁴-2²3³>0
{(a-2)²-6}{(a-2)²+6}>0
Uggghhh

It just turned ugly

[glacial sequoia]

what are you doing

power 4?

[merry stirrup]

(a-2)²=b right

b²=(a-2)⁴

[jolly parrot]

²⁰Ne

[merry stirrup]

Why ?

[glacial sequoia]

uh

it just isn't

😭

[jolly parrot]

²⁰Ne

[merry stirrup]

If this is equal then the roots are equal too right ?

[merry stirrup]

If this is true

Hen

Then*

Ohhhhh

,w factor a²-4a-2

[jolly parrot]

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[merry stirrup]

Ohhh

[glacial sequoia]

tough luck

[merry stirrup]

🥲🥲🥲

That's why I was fked

[merry stirrup]

Different steps but I don't understand the last step

Again the quadratic determinant thing?

[pearl pond]

@merry stirrup Has your question been resolved?

[merry stirrup]

I understand that
Putting a=2
In
(a²-4a-2)<36 doesn't satisfy the eq so a≠2
But I wanna know this method

[merry stirrup]

Typo (a²-4a-2)²<36

<@&286206848099549185>

[merry stirrup]

.close

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[near echo]

sorry for the vague question, but i had the original expression (w - 27)sqrt(w) = 2t * sqrt(13a), a > 0 fixed, and solved for w by squaring and finding the real cubic solution
im curious why i suddenly lost the domain of definition though. the original expression is defined for t >= 0 pretty trivially, but the cubic solution gives what i show in the picture

[near echo]

which is only defined for t >= 27/sqrt(13a) or t <= -27/sqrt(13a)

ah ok never mind i see why taking the roots might remove solutions

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[topaz valve]

needed some help on this calculus problem.. i understand that i need to use the product rule, but how do i read f'(1) and g'(1) from the given graph?

[sharp vigil]

they look like piecewise lines, so find the slope assuming they're lines

[topaz valve]

ahh okay, thanks so much

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[silent parcel]

Not sure how to get the graph started here

Oh nvm it’s easy

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[sand robin]

How to determine whether this converges

pls help

[glacial scroll]

first take out 1/7 of the sum

[sand robin]

Yees and then?

[glacial scroll]

you also know -1 <= cos(u) <= 1. Either way -1 <= $\cos^3u$ <= 1

[jolly parrot]

sssssssssssvvvvvvvvvvccccccccccc

[sand robin]

Hm maybe it’s not clear but it’s cos(3n)

[glacial scroll]

oh

[sand robin]

Sorry

[glacial scroll]

maybe ratio test?

[sand robin]

I tried but wolfram does not want to do it

[glacial scroll]

,w lim u-->inf [(abs((cos(3u)/u)((u+1)/(cos(3u+3))))]

oops

[sand robin]

,w lim n to inf ((|cos(3n)/(7n)|)/((|cos(3n+3)/(7n+7)|)

[jolly parrot]

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[glacial scroll]

lol

[jolly parrot]

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[sand robin]

Yes it sucks

[glacial scroll]

ok $|\cos (3x)| \le 1$, or in general for any angle inside the cosine

[jolly parrot]

sssssssssssvvvvvvvvvvccccccccccc

[sand robin]

Yes

[glacial scroll]

because $x \ge 1$, you have $|\cos (3x)| \le 1 \le x$

[jolly parrot]

sssssssssssvvvvvvvvvvccccccccccc

[glacial scroll]

(i meant 1)

[subtle ginkgo]

think wisely and use it in your case

[glacial scroll]

know what prob that's better i think mines is a bad choice

[sand robin]

hm

[sand robin]

could you please give me another hint on how to use it because unfortunately i dont see it

[pearl pond]

@sand robin Has your question been resolved?

[pearl pond]

@sand robin Has your question been resolved?

[pearl pond]

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[pearl pond]

Please don't occupy multiple help channels.

[hybrid haven]

can someone tell me what im supposed to do after this step

[wet elbow]

divide both sides by -ln(2)?

[eternal reef]

please close this channel

you already have an active help channel

[hybrid haven]

ok mb

noones responding on the otehr oen tho

[hybrid haven]

can i just equate it

.close

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[hybrid haven]

is m a constant in this scenario

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[gusty prism]

U

Uh

f(x)=a^x?

[regal herald]

huh?

its asking you to find a essentially

[kindred geyser]

You have x and y

[gusty prism]

yes

[kindred geyser]

then u find a

[pseudo oxide]

okay hang on, this is what we have

[jolly parrot]

ren

[pseudo oxide]

we need to find "a"

now at x = -1, y = 1/3

so we get

[jolly parrot]

ren

[pseudo oxide]

i think you can find "a" yourself now

@gusty prism

[pearl pond]

@gusty prism Has your question been resolved?

[pearl pond]

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[waxen talon]

How can I calculate $\sum_{n=1}^{\infty} \f{n}{4^n}$

[jolly parrot]

ColdTe²

[shy imp]

so work out the pattern maybe?

lemme see, 1/4 then 2/16 = 1/8 then 3/64 then 4/256 = 1/64

[sharp vigil]

since it looks quite similar, you may want to start with the geometric power series (what sort of calculus could you do to make it look like that?)

[west sapphire]

do you know about differentiating power series term by term?

[waxen talon]

I don't

Let me show you the original question

I divided this into two parts

3 + 3/4 + 3/4² + ... \infty

[west sapphire]

ugh, it's not even clear from the first three terms if the numerator of the next term is gonna be 3+4p or 3+3p

poorly worded question imo

[waxen talon]

p/4 + 2p/4² + .... \infty

[waxen talon]

This turned out to be 4

[spiral sierra]

this can be easily converted to geometric series

[waxen talon]

$4 = p\sum_{n=1}^{\infty} \frac{n}{4^n}$

[jolly parrot]

ColdTe²

[spiral sierra]

multiply the sum with 4

and subtract the the 4*sum by original sum

you'll end up with a geometric series

[waxen talon]

Let me try that but other than that

[waxen talon]

@spiral sierra are you talking about this

[spiral sierra]

yes

[waxen talon]

Alright

Let me try

I don't quite get it $\f34 + \f{6}{4^2} + \f{9}{4^3} + \dots \infty$

[jolly parrot]

ColdTe²

[waxen talon]

Maybe I didn't do it the way you expected me to

[spiral sierra]

where did you get 3, 6 and 9 from

[spiral sierra]

yeah i think so

[waxen talon]

$\f14 + \f{2}{4^2} + \f{3}{4^3} + \dots \infty$

[jolly parrot]

ColdTe²

[waxen talon]

Multiplying by 4

$1 + \f24 + \f{3}{4^2} + \dots \infty$

[jolly parrot]

ColdTe²

[spiral sierra]

now subtract the ones with same denominator

[waxen talon]

What about 1

[spiral sierra]

leave the 1 out of the subtraction process

[waxen talon]

Ahh

I see

$\f14 + \f1{4^2} + \f1{4^3} + \dots \infty$

[jolly parrot]

ColdTe²

[spiral sierra]

geometric series

[waxen talon]

So that's

1/3

$\f13$

[jolly parrot]

ColdTe²

[spiral sierra]

so 3S = 1/3 + 1

[waxen talon]

4/3

4/9

I get it

So that'll be

$4 = \f{4p}{9}$

[jolly parrot]

ColdTe²

[spiral sierra]

[waxen talon]

Cool

[waxen talon]

Sir Bungo

I'm interested in this way

As well

What do I search

[waxen talon]

Thank you

[spiral sierra]

you're welcome :)

[west sapphire]

brief summary:
$$\frac{d}{dx}\sum_{n=0}^\infty x^n = \sum_{n=1}^\infty nx^{n-1}$$
for $|x|<1$

[jolly parrot]

Bungo

[west sapphire]

you can apply that here with x = 1/4

the advantage is that you already know a closed form expression for the sum on the left, and it's easy to differentiate that closed form expression

[waxen talon]

Alright let me try this

$\sum_{n=0}^{\infty} \frac{1}{4^n} = \sum_{n=1}^{\infty} \frac{n}{4^{n-1}}$

[jolly parrot]

ColdTe²

[waxen talon]

$4 = p\sum_{n=1}^{\infty} \frac{n}{4^n}$

[jolly parrot]

ColdTe²

[waxen talon]

Damn

$\f13 = \f14 \sum_{n=1}^{\infty} \frac{n}{4^n}$

[jolly parrot]

ColdTe²

[waxen talon]

$\sum_{n=1}^{\infty} \frac{n}{4^n} = \f43$

[jolly parrot]

ColdTe²

[waxen talon]

Thank you Sir

I'll look more into this

.close

[pearl pond]

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[waxen talon]

.reopen

[pearl pond]

✅

[waxen talon]

$\f43 = 4 \sum_{n=1}^{\infty} \frac{n}{4^n}$

I'm making a mistake idk where

[jolly parrot]

ColdTe²

[west sapphire]

how did you get that?

[waxen talon]

$\frac{d}{dx}\sum_{n=0}^{\infty} \frac{1}{4^n} = \sum_{n=1}^{\infty} \frac{n}{4^{n-1}}$

[jolly parrot]

ColdTe²

[waxen talon]

Is this wrong

[west sapphire]

yes, you can't plug in x=1/4 until after you do the derivative of the LHS

$$\sum_{n=0}^\infty x^n = \frac{1}{1-x}$$
take the derivative of that

[jolly parrot]

Bungo

[west sapphire]

then plug in x=1/4

[waxen talon]

Ohhh

Wouldn't that be 4/3

Let me rethink this

[west sapphire]

let me check...

[waxen talon]

Wait I think I got it

[west sapphire]

i get -16/9

i think you didn't actually take the derivative

[waxen talon]

Yeah mb

,w diff (1-x)^{-1}, x= 1/4

[west sapphire]

ah i screwed up too

i think it's just 16/9

i forgot another -1 factor due to the chain rule

[jolly parrot]

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[waxen talon]

Alright this works

Now

So it's

$\f{16}{9} = \sum_{n=1}^{\infty} \frac{n}{4^{n-1}}$

[jolly parrot]

ColdTe²

[waxen talon]

$\f{16}{9} = 4\sum_{n=1}^{\infty} \frac{n}{4^{n}}$

[jolly parrot]

ColdTe²

[waxen talon]

So I just write the sum then take derivative of the sum

Which would then equal

Alright I kinda get it

Thank you Again

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[west sapphire]

yea this is right

[pearl pond]

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[mental verge]

can someone help me factorize $$x^4-x^3-23x^2-3x+90$$

[jolly parrot]

ates

[mental verge]

!status

[pearl pond]

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

[mental verge]

im in stat 3

well, i find $$(x-3)^2(x+5)(x-2)$$

[jolly parrot]

ates

[mental verge]

using root theorem

when i check this using desmos it gives me the graph of another polynom so this is incorrect

firstly, i found 2 and -3 as roots. then the quotient when i divided the p(x) by 2 was $$x^3-x^2-21x-45$$

[jolly parrot]

ates

[mental verge]

i factorized this one using rational root theorem as well and found the root -5 and to find another root i factorized the quotient of the polynom this time (the polynom is the quotient of the previous polynom) and got $$x^2-6x+9$$ as quotient which makes (x-3)^2

[jolly parrot]

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

[mental verge]

ik this is long but this is my solution

i did like 13 questions like this

and this one gave me an error

<@&286206848099549185>

[alpine elbow]

u will have to find one of solutions for it to be equal to zero and then devide it by that

so u will go on and on till u get it done

[alpine elbow]

well for start this can be devided by (x-5)

[mental verge]

i did that

[alpine elbow]

so what did u get

[mental verge]

and at the end this is what i got but idk where i did this mistake

i got

[alpine elbow]

lemmy calculate

[mental verge]

this

i check the accuracy of the answer using desmos to see whether this equals to the same polynomial but it wasnt equal

[alpine elbow]

ok so when u devide it by x-5 u get
x^3+4x^2-3x-18

ok now put in 2

so devide by x-2

[mental verge]

i get 0 as remainder

and $$x^3-x^2-21x-45$$ as quotient

[jolly parrot]

ates

[alpine elbow]

wait ima write it on my pc and send calculation if u want

[mental verge]

alright

i divided this by -5 and got

$$x^2-6x+9$$

[jolly parrot]

ates

[alpine elbow]

this is for second part

[mental verge]

which makes $$(x-3)^2$$

[jolly parrot]

ates

[alpine elbow]

ima write first one rn

[mental verge]

ok

[mental verge]

hold on aint we dividing it by -5? also the polynomial is not x^3+4x^2-3x-28

[alpine elbow]

this is the first one

[alpine elbow]

this is first dividation

[alpine elbow]

that one is second one

[alpine elbow]

well this is equal to (x+3)^2

so this equation will be equal to
(x-5)(x-2)(x+3)^2

[mental verge]

this is the calculation i did

we're doing two different things w ya

u get remainder i use rational root theorem

[alpine elbow]

idk i do those like this

[mental verge]

okay so lemme understand ur pov

[alpine elbow]

its called Bézout's theorem

[mental verge]

why did u divide it by -5 at first

[mental verge]

ur calculations?

[alpine elbow]

no

like this uses Bézout's theorem

u know

like if it can be unfolded

and if u equate it to 0 then all of those like we got here (x-5) (x-2)

and stuff

[mental verge]

wait okay

[alpine elbow]

then we will get x for this example 5 2 and 3

-3*

[mental verge]

no

uhm

i got my calcs wrong with -5

its +5

[alpine elbow]

ye

[mental verge]

wait its not 5 either

still cant get the graph correct

[alpine elbow]

wait lemmy explane it through translator

i dont know english math termines that good

[mental verge]

okay so the answer is$$ \left(x+3\right)^{2}\left(x-5\right)\left(x-2\right)$$

[jolly parrot]

ates

[mental verge]

according to desmos

but there are two things i dont get

[alpine elbow]

If you want to expand a polynomial into a product of monomials, then you have to equate this polynomial to zero and find one of the solutions, as here we have 5 and divide by (x-5) because one of the terms of the final product (if it exists) will be x-5

[alpine elbow]

ye thats it

[mental verge]

when i divide $$x^3-x^2-21x-45$$ by 5 i get

$$x^2-4x-1$$ as quotient and -50 as remainder

[jolly parrot]

ates

[alpine elbow]

no u shouldnt devide it by 5 again

like u only got one (x-5)

[mental verge]

i divide it by 2 at first

[alpine elbow]

you should start finding new solution for next ones

like see if 1 works then 2 and etc

[mental verge]

i did

[alpine elbow]

but it also may logically accuire

[mental verge]

only 2 and -3 worked

[alpine elbow]

ye so devide it either by x-2 or x+3

[mental verge]

okay i divided it by 2 using synthetic division at first

its the same

[alpine elbow]

ok i can explain in easier way

[mental verge]

alr

[alpine elbow]

I mean, look here
x^4-x^3 -23x^2 - 3x+90 is solved as
(ax^3 +b x^2 +cx+d) (ex + g)

now u gind that ex+g

(for now e =1)

[mental verge]

okay how do i find g

[alpine elbow]

u start to put things for x

like so that equation will be equal to 0

[mental verge]

okay like trying to find zeros just like we do using rational root theorem

[alpine elbow]

YESSSSSSSSSSSSS

[mental verge]

alr i did the same now what

[pearl pond]

@mental verge Has your question been resolved?

[pearl pond]

@mental verge Has your question been resolved?

[pearl pond]

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[drowsy gust]

Hey all,

Linear algebra question: Tried to find the eigenvector using eigenvalue lambda = -1. The teacher got something completely different, how did he get to those steps?

[drowsy gust]

<@&286206848099549185>

[thorny sleet]

I need your help brother.

[unreal thistle]

#1021175428326633542

[vivid grove]

which step

[drowsy gust]

The first one im already confused by, howd he go from

4 -2 0
-2 3 -2
0 -2 -2

to

0 4 4
-2 3 -2
0 -2 2

etc

@thorny sleet go to #help-21｜아리스킨충1 or #help-36 or smt, find an open channel to request in

[vivid grove]

It is just row operations. The one in the slides is correct. Your post has arithmetic error

[drowsy gust]

Oh, really? I just did R2 = R2 - R1 and then R3 = R3 - 3R1

[vivid grove]

It is similar to Gaussian elimination which you probably learned earlier

[drowsy gust]

then switched R2 and R3

[vivid grove]

There are multiple ways of performing calculations. What is most commonly done is upper triangular matrix is constructed

[drowsy gust]

i see, I dont think i'[ve heard of thaht. Is that what my teacher did there?

[thorny sleet]

Can anyone solve that step by step

[vivid grove]

The row operation your teacher did was R1 <- R1 + 2R2

[drowsy gust]

Whyd he do that instead of R2 = R2-R1?

is it preference or is it based on smt?

[vivid grove]

Your teacher didn't construct an upper triangular matrix. But there calculation seems correct.

[drowsy gust]

ur right yeah he did do that, but then howd he get from there to the matrix after that?

[vivid grove]

This will result in the second row becoming [6 -5 1]. which is not what we want. We want find the null vectors, i.e. the basis that form the nullspace. To do that we need to make use of the free variables.

To put it simply we want to find a vector X such that (A-lambda*I)X = 0

where X is a non-trivial solution

[drowsy gust]

Ohhh alright, I see. So you want to do row operations until the top row is fully 0? Or just until x1 y1 is zero?

[vivid grove]

You can make the element a11 0 or make the element a21= 0 both options are correct

[drowsy gust]

Alright and when those elements are 0 you can start calculating the x1 x2 x3's ?

Tysm for all your help, Im sorry im a bit slow. Just trying to understand it

[vivid grove]

When the matrix is the correct form we can find the general solution.

[drowsy gust]

I see, but when would I know when the matrix is the correct form?

Is it the correct form if a11 and a21 = 0?

[vivid grove]

It would be row-reduced echleon form. So you would try to construct an upper triangular matrix.

[vivid grove]

It is not possible to have both a21 and a11 equal 0 using row operations

[drowsy gust]

Alright, I can continue with this info, again thank u so much for your help and your time.

[vivid grove]

remember to close if you are done

[drowsy gust]

Sorry!

.close

[pearl pond]

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[pearl pond]

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[midnight haven]

1. Idk where to start

Do I use V=Lwh?

Idk what the 30,40,50 are

V=(30)(40)(50)?

And divide them by 2.8?

Yep

Divide?

Since its per

For every

I honestly don’t know

But I did get the answer

Ohh

So for every

Cubic cm

Which are 60,000cubic cm

U plug in 2.8

In the cubic cm

And get letter D

Thanks!

.close

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[stark osprey]

Can someone help me solve this for all values of x, the defined range is simple but I don't know how to proceed..?

[white juniper]

maybe start with trying to solve =0?

would you know how to do that?

[stark osprey]

I'll try

[white juniper]

if you can find roots of an equation its should be easy to find inequality from there

[stark osprey]

ok so -3, -7 and 0

ok I got it

tysm 🙂

@white juniper

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[raven spade]

how do I solve this question?

[raven spade]

there are multiple correct options to the question

!status

[pearl pond]

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

[raven spade]

1

[ornate lagoon]

you have to find all the values of p_1, p_2, p_3 ... till p_5
i dont think theres any other way

[acoustic path]

u would be surprised

[ornate lagoon]

theres another way?

[acoustic path]

there’s a reason all p1 p2 p3 etc is an integer yk

[ornate lagoon]

?

[raven spade]

how do we know for sure they're all integers though

[marble sigil]

there's a thing called newton's identities

[raven spade]

oh yeah right on

wasn't aware of it though

[raven spade]

how do I use that in this question though

[marble sigil]

you need to find S0 and S1, then you can just build up all the S's

like a^0+b^0 and a^1+b^1