help-0

[azure needle]

It's vowel and consonants

[junior kernel]

like sum of the positions of each letter

[azure needle]

@junior kernel

[junior kernel]

like a = 1, b = 2, c = 3 etc

[junior kernel]

oh

[azure needle]

Count vowel and consonants

[junior kernel]

ill check

one min

[azure needle]

[junior kernel]

oh god

it is

i hate this exam

[azure needle]

Yay!!!

[junior kernel]

thank you

[azure needle]

Exam?
Are you asking for help on an exam?

@junior kernel

Answer or you get modded

[junior kernel]

no, it was for an exam i gave before

i skipped this one

i can send u a pic of the answer sheet

[azure needle]

Oh ok 😂

[junior kernel]

and the time right now

[azure needle]

Nah nvm

[lone heart]

If you are done with this channel, please mark your problem as solved by typing .close

[junior kernel]

anyway cya

.close

[lone heart]

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[shadow saffron]

wait what was the answer, how to solve it?

[proven depot]

count the vowel and consonants

[junior kernel]

the nubmer of vowels is the first digit

number of consonents is 2nd

[shadow saffron]

oh goddd

[azure needle]

Yeah lol

Has to be the worst code ever

[junior kernel]

on god

and we have 1.5 min per question

[azure needle]

Bag=dab

Decoding will be impossible

[junior kernel]

yeah

[lone heart]

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[fervent ferry]

confused about notation. do we use NABLA only when it's a vector of partials? why r' is not expressed as nabla as well?

[edgy vine]

can anybody teach me integration

[cedar juniper]

#❓how-to-get-help

[fervent ferry]

nabla, not delta sorry. i've corrected above

[left wharf]

nabla is used to denote operations that concern functions that take in several inputs

For a univariate function the derivative at a point, if it exists, is defined as the limit (f(x+h)-f(x))/h

note that that definition makes sense even if the function takes on vectors as values

but as soon as the function takes in several inputs it's no longer obvious how you'd extend that definition

And nabla is basically an attempt to define certain "derivative-like" operations for functions that take in several variables

You write r' instead of nabla r because r takes in a single variable (namely time)

Although technically the gradient of a single variable function is more or less just the derivative

(it's a 1x1 matrix whose sole entry is the derivative function)

(or, more precisely, it's a function mapping each input to a 1x1 matrix corresponding to the value of the derivative at that point)

[lone heart]

@fervent ferry Has your question been resolved?

[fervent ferry]

can i see nabla as derivative operator ?

@left wharf ?

[fervent ferry]

sorry this part is unclear.

can you make an example? how would i do vector+h ?

is h a vector?

[left wharf]

If f(t) = (g(t), h(t)) for high school functions g and h then f'(t) (by the high school definition) is (g'(t), h'(t))

So f is vector-valued even though its derivative could be computed by a high schooler

However if f(s, t) = s + t then "the derivative of f" no longer makes any sense

However the gradient of f (denoted nabla f) is the function that maps (x, y) to (1, 1)

[fervent ferry]

yea so the derivative of f is the gradient

[left wharf]

kind of yeah

[fervent ferry]

i'ts a vector whose components are its partials

[left wharf]

yeah exactly

[fervent ferry]

what you mean vectors as values then?

as inputs?

f(t) and t is a vector?

or f(t) outputs a vector?

i think i understood but i want to be sure i get what you were saying here

[left wharf]

as outputs

[fervent ferry]

ok fine

[left wharf]

the value of a function is its output

[fervent ferry]

the "takes" confused me

[left wharf]

takes on

[fervent ferry]

english is not my language, i tend to associate with takes IN

ok

thanks, all clear

[left wharf]

no worries ^^

[fervent ferry]

.close

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[austere meteor]

hi! is anyone able to help?

[hot dawn]

yeah wassup

[lone heart]

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[grizzled falcon]

Hello! I was wondering how i could solve this question..

So I want to find the limit, is that what they are asking for ?

[daring parcel]

how

[grizzled falcon]

?

[lavish cave]

no

do you know what the condition is for a geometric series to converge?

[grizzled falcon]

That the partial sum tends to a limit?

[lavish cave]

no, that's true for any series to converge

what condition must there be on the common ratio?

[grizzled falcon]

Absolute value of r is less than one?

i dont know

[lavish cave]

yeah that one

[grizzled falcon]

hmm

But how do i find the R in this geometric sum ?

[lavish cave]

it's the 2nd term divided by the 1st term for instance

$\frac{-10e^{-x}}{5} = -2e^{-x}$

[ocean seal]

south's secret twin brother

[grizzled falcon]

Okay and how di know if the absolute value of this is less than one ?

but this will never happen?

there is no such x that will make it converge

[lavish cave]

well $|-2e^{-x}| < 1$ implies $e^{-x} < 1/2$

[ocean seal]

south's secret twin brother

[lavish cave]

it's definitely possible

[grizzled falcon]

ohhhhh

[lavish cave]

$|ab| = |a| |b|$ so you can take a $|-2| = 2$ out

then divide by 2

[ocean seal]

south's secret twin brother

[grizzled falcon]

so 2< e^x ?

and ln(2) = x ?

[lavish cave]

the sign flips when you take the reciprocal

wait so e^x > 2 yes that's correct

and x > ln(2)

cause e^x is a 1-to-1 increasing function

yeah the beauty of this approach is from 2 < 3, you actually have e^(-2) > e^(-3)

the sign flips cause e^(-x) is decreasing

or just solve for x and note the shape of the graph of e^(-x), also works

[lone heart]

@grizzled falcon Has your question been resolved?

[grizzled falcon]

Thank you !

For your help:D

[lone heart]

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[hot crescent]

Hello, the question is to descide wether the statements are true or not. The first one is:
∀x ∈ R ∃y ∈ R : x2 + y2 ≥ 4

[hot crescent]

I understand this is

For all X in the Real numbers there exists a y in the real number so that X^2+y^2>=4

[summer dirge]

that’s correct

do you think it’s true?

[hot crescent]

Thats the thing Im unsure

if x were to be lets say 1 then 1^2+y^2>=4

so y^2>=3

[summer dirge]

you don’t have to pick the best y, just one y

[hot crescent]

cant y be any possible number?

real number

[summer dirge]

indeed

we get to pick y

[hot crescent]

So is it correct?

[summer dirge]

I’ll let you think about it some more

[hot crescent]

hahah damn okay

Ohhhh

wait nvm

[summer dirge]

for any given x, what y would suffice so that the inequality x^2 + y^2 >= 4 is true?

there are multiple different values of y that’ll work, but you only need to find one

[hot crescent]

y=2?

[summer dirge]

yeah!

y = 2 works perfectly

so does y = 10, or y = 472648203748

but y = 2 is sufficient

[modern sedge]

note that this y actually proves a stronger statement:

∃y ∈ R : ∀x ∈ R (x^2 + y^2 ≥ 4)

[hidden monolith]

bro did everything but use latex 😭

[hot crescent]

I just copied

[modern sedge]

me too

[hot crescent]

:D

so to finish of the statement is correct.

[lone heart]

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[hot crescent]

.reopen

[lone heart]

✅

[hot crescent]

,, \sum_{k=0}^{100}\begin{pmatrix}
100 \k
\end{pmatrix} = 2^{100}

[ocean seal]

vrsth

[hot crescent]

So i know that

,, \begin{pmatrix}
100 \k
\end{pmatrix} = \frac{100!}{(100-k)!\times k!}

[ocean seal]

vrsth

[hot crescent]

but after that I am not sure what I shall do

,, \sum_{k=0}^{100}\frac{100!}{(100-k)!\times k!}

[ocean seal]

vrsth

[hot crescent]

Should i rewrite it like this?

[jade moat]

hey guys. im confused as to why in the second picture (uppgift 15), we are not needed to do a divide by n.
(screenshot from gemini is actually a translation of the answer from the professor)

[hot crescent]

this channel was quite occupied already

Nvm can have this channel i solved the question

.close

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[ocean hound]

Please I have to show that with the recurrence

[mystic swallow]

,rccw

[alpine sable]

start by substituting U_n-1 and U

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[austere nexus]

How would I do BII?

[tight pier]

you can resolve the absolute value

f(x-2) for x ≥ 0
f(-(x+2)) for x < 0

[austere nexus]

can u explain why its f(-(x+2)) for x<0

[ocean seal]

bacc (unhelpful)

[tight pier]

so for x < 0

f(-x-2) = f(-(x+2))

[austere nexus]

ok

[austere nexus]

im kinda confused how i would draw that

[tight pier]

- means reflection at the y-axis and x+2 means shift by 2 units to the left

[austere nexus]

Like how do you know the order of translations and reflections?

[tight pier]

the blue is the first transformation

you flip it at the y-axis

then from blue to green you translate it by 2 units to the left

[austere nexus]

ok

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[paper trellis]

How much should pablo invest at 5.75% rate compounded semi-annually for 5 years if he wants to accumulate 45,000 dollars

[vast rapids]

A=P(1+r/n)^(n×t)

r is your rate as a decimal
t is time
n is number of times compounded
A is the amount you have after t years

[paper trellis]

r = 0.0575 is the annual interest rate of 5.75%
n = 2
t = 5

PV = 45,000/1.32769549

I did that

[tacit arch]

,calc 45000/1.32769549 * (1+0.0575/2)^(2*5)

[ocean seal]

Result:

45000.000226313

[lone heart]

@paper trellis Has your question been resolved?

[paper trellis]

is that correct

my final answer is 33,893.31

[lone heart]

@paper trellis Has your question been resolved?

[tender heart]

Y’all I need help with my ixl please help

Here is a pic

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@paper trellis Has your question been resolved?

[frosty venture]

Help with finding y and x please

[lone heart]

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[umbral arrow]

Any explanation?

[normal bone]

A is a linear operator, you can think of it as taking linear combinations of x to get y

differentiating Ax gives you a constant

namely, A itself

[umbral arrow]

Doesn't A multiplied by x give a column vector

Then by differentiating that it gives a column vector, not A

But the column vector with sums

[normal bone]

yes, but you are taking the partial with respect to every single entry in x

so the derivative will really be an m by n matrix

[umbral arrow]

So kinda like this? How would we know when we do it this way?

[normal bone]

yes this is what i mean

what do you mean, when to do it this way?

this is the definition of the jacobian

[umbral arrow]

Oh I see

[umbral arrow]

What if x depends on z?

y=Ax

dy/dz = ?

[normal bone]

then you should apply the chain rule

[umbral arrow]

Is it the same as the normal chain rule in calculus or is there a difference when it is linear algebra?

[normal bone]

it is essentially the same, $J_{f\circ g} = (J_f \circ g)\cdot J_g$

[ocean seal]

Bair

[normal bone]

the proof is somewhat more technical

[lone heart]

@umbral arrow Has your question been resolved?

[golden elk]

Help I’m unsure how to even do these problems

[normal bone]

!occupied

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Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

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[marble crown]

A question like this is very common

[marble crown]

Like a chain rule with a product rule

[tall topaz]

Seems like you’re in need of dire help… are you trying to differentiate the expression?

[marble crown]

Yes

[lethal belfry]

||Logrithmatic differentiation ||

[marble crown]

Sorry didn’t specify

[marble crown]

Calc 1

Simplify but that’s chain rule

[tall topaz]

Yeah, seems good - that’s the derivative of (3x-1)^5

[marble crown]

Yes

[tall topaz]

Well, you can just continue using the product rule

[marble crown]

Derivative of the first one multiply second function plus the second function of the first one

OK let me try actually yeah

[tall topaz]

[tall topaz]

Seems right to me

[marble crown]

And I can’t simplify the expression on right because the exponent ^5 has to opened up first right?

[tall topaz]

Yes, but you can pull out a common factor of (3x-1)^4 from both terms

[marble crown]

I see

Not necessary atm for scope

Thank you

[tall topaz]

No problem

[marble crown]

.close

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[marble crown]

.reopen

[lone heart]

✅

[marble crown]

Just did this as well

More in depth

Quotient rule + Chain rule + product rule

Using logarithmic differentiation (I didn’t do it that way) is the one closest to my answer my best is that they factored out (3x-1)^4 ?

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[timber briar]

Can anyone help me with the c part of this question

[royal vine]

how do you solve this Q without using higher than quadratic degree formulas

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@timber briar Has your question been resolved?

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[unreal terrace]

,rotate

[unreal terrace]

So I think that the answer to b should be 0

Btw in the white space it's supposed to say:
1,2,2²,2³,...,2⁸

[unreal terrace]

Because 2⁸ = 2⁷ + 2⁶ + 2⁵ ... + 1

And you can't fit 2⁸ in all sums so there's no way all of them can come out equal

And by a I've found out s = 8097 but I haven't found s = 8104

Can someone help me find it or help me prove that there isn't one

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[lone heart]

@covert sonnet Has your question been resolved?

[tacit arch]

what's the question

[covert sonnet]

i dont get how to answer it

[tacit arch]

what is the quesiton

[covert sonnet]

is this right?

[covert sonnet]

which exponent do i simpify first

[tacit arch]

order doesn't matter

[covert sonnet]

oh

[covert sonnet]

but is this right?

[covert sonnet]

@tacit arch

[covert sonnet]

what do i do next after this?

[tacit arch]

,w simplify (-2x^(-11/3))^6 * (4x^2)^(9/2) / (8x^36)^(1/3)

pain

write 4 = 2^2 and 8 = 2^3 then combine the numbers together as a single power of 2.

then do the same thing for all the powers of x

,tex .exp rules

[ocean seal]

riemann

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[covert sonnet]

huh

.reopen

[lone heart]

✅

[tacit arch]

what part of that was confusing?

[covert sonnet]

then what will happen to their exponents?

[covert sonnet]

the 4^9/70 and 8^1/105

[tacit arch]

use power of power

[tacit arch]

after this

[covert sonnet]

,rotate

[covert sonnet]

this? @tacit arch

[tacit arch]

you still have multiple terms of power of 2.

do you know how to combine them into a single power?

[covert sonnet]

huh?

[covert sonnet]

no

[tacit arch]

$a \cdot b \cdot a = a^2 b$

[ocean seal]

riemann

[tacit arch]

just put some powers on a and they combine the same way

[covert sonnet]

i rly dont get what u mean

[tacit arch]

which part

[tacit arch]

of this?

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[mossy crypt]

I need to find MLE estimates for the exponential and pareto distributions for right-censored data
I think I've found the likelihood functions:
For exp:
klog(lambda) - lambda x sum(obs) - lambda (n - k) M
For pareto:
klog(alpha) + k x alpha x log(lambda) - (alpha + 1) sum(log(obs + lambda)) + alpha (n - k) log(lambda / (M + lambda))
where k is the number of observed data points, n is the total number of data points, M is the censoring point, obs is the vector of all the observed data points
Pareto is defined as the two parameter version, alpha x lambda^alpha/(lambda + x)^(alpha +1)
I hope you can check if these likelihood functions are correct because when I optimise and plot it, it doesn't seem to account for the censored points

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[mossy crypt]

<@&286206848099549185>

[mossy crypt]

<@&286206848099549185>

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[brazen swan]

This question says these forces are in the horizontal plane; doesn't that mean the particle's weight should be into the page in the vertical plane? The worked example seems to ignore that.

[fair parcel]

No that would not be considered

[brazen swan]

But there would be a weight into the page? I just wanna know if I interpreted it correctly.

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[tacit arch]

try multiplying the inequality by f(x) and squeeze theorem

[sterile trench]

aha squeeze theorem?

[tacit arch]

also called sandwich theorem in some places

[sterile trench]

actually my sir doesnt tell methods/names, directly teaches what to do

[sterile trench]

aha gotch

a

[sterile trench]

yup

[tacit arch]

unfortunately squeeze theorem wouldn't rule out (A) if the left and right limits are different

might help if you bounded f(x) in terms of x first

x in [1/(n+1), 1/n) implies 1/(n+1) <= x < 1/n implies 1/x > n

there's another bound with the first inequality

[sterile trench]

ahaa

yeah nvm, its hard for me to understand over messages

will ask someone tmrw

as i am sleepy rn

thanks tho

cya gn

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[sterile trench]

it is ques from one of toughest exam in world

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[elfin rapids]

Math book won't give an answer so

[elfin rapids]

Is this correct

[tacit arch]

is this the question?

showing m choose 6 equals m choose 3?

[lone heart]

@elfin rapids Has your question been resolved?

[elfin rapids]

The question was.

Find m value

[tacit arch]

i see. binomial coefficients satisfy symmetry
$\binom{m}{m-k} = \binom{m}{k}$
apply that with $k=3$ and solve for m

[ocean seal]

riemann

[elfin rapids]

Ah, my bad. I noticed that picture look awful.

Here, a screenshot.

@tacit arch

So, for the first one...

(m n) = ( m m-n)

[tacit arch]

Yea your n is my k

[elfin rapids]

therefore

[tacit arch]

Just apply

[elfin rapids]

9, right?

As for B :

My answer is 11.

For C: 5

[elfin rapids]

Is my approach correct in question D and E?

I tried Pascal.

(m n) + (m n-1) = m+1 n

[tacit arch]

,w 11 choose 2

[ocean seal]

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[tacit arch]

,w 5 choose 3

[ocean seal]

Results provided by WolframAlpha

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[elfin rapids]

FUCK YEAH

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[elfin rapids]

.reopen

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✅

[elfin rapids]

@tacit arch what about e and d

I do not know how to use WolframAlpha

[tacit arch]

There's an adjacent sum identity you can use.

[elfin rapids]

Pascal, right?

[tacit arch]

https://present5.com/presentation/97d2593349ed3134dbd46a627e66e8cb/image-48.jpg

[elfin rapids]

My answer is (16 9)

Exactly!

[tacit arch]

I think it works for both d and e

[elfin rapids]

Yes, that's what I did.

So, 16 9 it seems

So, I did it flawless. = )

I'm gonna pass my test.

and be a math king.

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[robust crypt]

I am asked to prove that a bijection F : R2 -> R2, whose images on lines are lines and F(0) = 0 is a linear function.
I've only managed to prove that F preserves parallelity, I don't know where to go from here.

[robust crypt]

Of course, we are talking about R2 as a vector space, not just a set of real pairs without any other structure

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[untold lake]

Given the regular expression $\newline P \to PBP \newline B \to BB \newline$ How do I get the [n, m] interval of the Nth line?
3rd line would be PBPBBPBP, so [3,6] would be PBBP

[ocean seal]

Megu

[untold lake]

The starting seed is P

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@untold lake Has your question been resolved?

[bitter tundra]

In the Nth line you have 2^(N-1) 'P's and clusters of 'B's in between them. So there are 2^(N-1)-1 of these B-clusters. The number of B-s in k-th cluster equals the binary number formed by 1 and trailing zeroes in binary expansion of k. For example in 5th line you have 4 B in 12th cluster as 12=1100 and 4=100. I think this is enough to make fast algorithm to get [n,m] segment.

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@untold lake Has your question been resolved?

[untold lake]

Could you run me through using this way to find number of P's in [27,41] of 5th line? I'm having some trouble visualizing it

[bitter tundra]

I'm not yet ready to give a complete algorithm. Here is the rule for forming this sequence.

[untold lake]

How do I know what cluster 27 and 41 fall in though

[bitter tundra]

yeah, I agree, it needs further thinking 🙂

[untold lake]

You can get the starting index of the n-th cluster by summing n+1 (number of P's) + sum of number of B's up till that point, I guess

but that doesn't help much
and when you're working on the 1000th line, the numbers add up quick

[bitter tundra]

You see, if we take some number n, then we know how many ones, how many 2s and so on are there before that n. I think this should work fast

so lets take that 27

oops, I forgot there are P in between.

I think the number of pees can be predicted too.

[untold lake]

cluster n has n P's behind it

[bitter tundra]

Do you need an algorithm/program or just the formula? What are the limits for N?

[untold lake]

just the formula

lines go up to 1000, positions up to 10^16

so in [n,m] if n lies in the i-th cluster and m in j-th cluster, you'd have j - i P's?

[bitter tundra]

Any time limit?

[bitter tundra]

what is the maximum length of [m,n] or maybe you give a link to this?

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[untold lake]

.reopen

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✅

[untold lake]

1<=m<=n<=10^16

it's in a university page, so locked behind a login

[bitter tundra]

well, if m is 1 and n=10^16 do you still need to output all the string?

[untold lake]

I just need to output the number of P's

[bitter tundra]

oh, thats easier

[untold lake]

I didn't ask correctly, yeah

[bitter tundra]

Ok, I'll take a break. Possibly figure it out tomorrow, if it's still relevant

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[crystal valley]

so I set f(x)=0?

what is the best way to go about this problem

[tacit arch]

inverse function theorem

[crystal valley]

so am I using
$(f^{-1})' (x)=\frac{1}{f'(f^{-1}(x))$

so this?

[tacit arch]

right

[crystal valley]

I cant find f inverse of 0

that is my problem

[tacit arch]

use 0 + 0 = 0

[crystal valley]

thank you

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[craggy remnant]

where did i go wrong?

[tacit arch]

this is not correct

[craggy remnant]

so what would v equal?

[alpine sable]

😔

Try u•dv for x•xe^-x²

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[dense garden]

how do i do part b?

[loud warren]

Idk what theorems you know to show that a function like this dont have an existing limit but what you could and also should do is try to "go to the limit" from different directions if you know what i mean

@dense garden

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[dense garden]

how do i show the epsilon delta proof

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[sacred haven]

how would I go forward with this?

proving

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@sacred haven Has your question been resolved?

[acoustic glen]

@sacred haven here you go:

[sacred haven]

danm i was so clsoe

thank you

[acoustic glen]

sure, you're welcome

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[twilit solstice]

im stuck which is wrong or right in this equation for rational function mines is the solution 1 and the solution 2 is my friends im quite weary on the graph as well

[buoyant saddle]

solution to what

[twilit solstice]

for this equation

[buoyant saddle]

what are you trying to find

brother

stating an equation is not a question

there is nothing to solve

do you mean the roots

do you mean the asymptotes

[twilit solstice]

the asymptotes

[buoyant saddle]

well what do you know about vertical asymptotes

start there

[twilit solstice]

A vertical line that guides the graph?

[buoyant saddle]

...

analytically speaking

[spring mica]

solution two finds the domain of the function, but as @buoyant saddle said, “solution” means nothing here

you both need to specify what these steps are for

but yes, solution 2 solves for the domain and therefore vertical asymptote. looks like they solved for the vertical intercept as well. the graph looks okay

solution 1, i’m sorry, is very misguided

there is no reason you should equate the numerator and denominator of a rational function, like, ever

@twilit solstice

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[coral night]

Log(X²-7x-30)-Log(x+3). Log base 6 but that doesn't really matter.

In this question, you factor the quadratic (x+3)(x-10) and simplify to log(x-10). I get that. Now when it comes to stating restriction I get confused. I know that it has to be positive because 6 to whatever power will never be negative or 0. So what my teacher does I'd (x+3)(x-10)>0. x>10, x<-3. I agree with this, but the method he uses for the 10 is x-10>0 -> add 10 to the other side x>10. Now for the 3 I have no idea why he flips the sign. I know that if you graph the quadratic it's positive before x=-3, so I agree with the restriction I just don't know how he got the restriction without Graphing it. Am I just missing something? Any help would be appreciated

[coral night]

Also my final statement in this case would be log(x-10) x>10, x<-3 right?

The way I've been imagining it is just whether the graph opens up or down and where the intercepts are, but idk how to show my work for that

[buoyant saddle]

he flips the sign becuase becuase if x>-3 then sure x+3>0 but x-10 isnt, hence their product is negative

but thats not really how you solve it

it isnt just about flipping a sign whenever

i recommend you search up sign chart for inequalities on youtube

@coral night hop on clash bro

[coral night]

Yea so he's been doing sign charts, but he hasn't been plugging in values to test he's just been filling then out, and a sub said testing values takes too much time. I'm about to go to bed so I just skimmed through a video, but it seems they just plug in values to test

[buoyant saddle]

i mean you can just do it in your head eventually

[coral night]

And it's x<-3 isn't it?

[buoyant saddle]

you know its positive if and only if the two factors have the same sign

so you just look for where they are both negative or both posiitve

[coral night]

Wait hold on, so my final statement for this would just be x>10? But isn't 10 negative when x<-3 or do I put them in like -3>x>10?

Because like if I do a sign chart for the factors, it's positive in the red areas but our answer is only on the right

Oh wait I think I get it, so if we just have the function log(x²-7x-30), we do get an answer before x<-3, but because we have the restriction from the second part of the question where the restriction is x>-3, it leaves only x>10 as the restriction where all are positive

But that leave the question, why does the factor (x+3) and the second part of the question (x+3) have different restrictions?

[buoyant saddle]

because we cant use the simplified function at values for which the original function isnt defined

yo saladman98

[coral night]

Well yea but if we just look at the 2 (x+3)s, they have different restrictions on them. I get that the restriction comes from the very start of the question I'm just confused in why one gets the sign flipped

[buoyant saddle]

your pfp is so heat bro

[coral night]

Thanks king

[buoyant saddle]

i thought i explained this already

[buoyant saddle]

^

we need (x+3)(x-10) > 0 and (x+3) > 0

now, since x + 3 > 0 we can omit values of x such that x <= -3

and if x+3 > 0 then the only way (x+3)(x-10) > 0 is if (x-10) is also > 0

[coral night]

Yea I get that but we have to state every restriction, and I'm not sure how to show my work for the x<-3

[buoyant saddle]

hence x > 10

[buoyant saddle]

you dont need to

its useless

because you already know x > -3

[buoyant saddle]

so who cares about

but again i can show you how he got that if youd like

despite being utterly useless

[coral night]

Yes please, I understand that it's useless but he wants us to state all restrictions and then do the unit chart to find the one that works

But then I have to show how I get to x<-3

[buoyant saddle]

ok, so ill reiterate, a product with two factors is positive if and only if both factors have the same sign (i.e both positive or both negative)

[coral night]

Like he said he's gonna take marks off if I go straight to x>-3 from the equation I gotta write x+3>0 lmao

[buoyant saddle]

since we need to solve (x+3)(x-10) > 0 we need either (x + 3 > 0 and x - 10 > 0) or (x + 3 < 0 and x - 10 < 0)

so that they have the same sign

hence for the first part where both factors are positive, we take the intersection of the two intervals, namely x > -3 and x > 10 ; thus the intersection is x > 10

then for the second part where both factors are negative (meaning their product is positive) we take the intersection of the two intervals (x + 3 < 0 and x - 10 < 0) hence x < -3 and x < 10. the intersection of these intervals is just x < -3

yo thesaladman98, i just wanted to say that youre my fucking guy bro

[coral night]

Omg I think I get it now

You're the goat I love you bro

[buoyant saddle]

i love you too bro

[coral night]

Aight I gotta head to bed I'm so cooked today

[buoyant saddle]

🐐

[coral night]

How do I mark this as answered

[buoyant saddle]

nah youre chillin youll ace it

type .close

[coral night]

.close

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[warped sail]

can someone pls explain me this part

[buoyant saddle]

!help

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To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.

[warped sail]

didnt get it

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[molten arch]

<@&286206848099549185>

[molten arch]

Shouldn't it be 34.6

[keen mason]

yea

[reef wren]

Yea

[molten arch]

So the answer in book is wrong

[molten arch]

Thanks

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[pallid ferry]

I’m tryna do this invertible matrix question, where A,B, and X are invertible square matrices of the same size. Solve for X in terms of A and B and simply result as much as possible. What I did here is it correct?

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[pallid ferry]

<@&286206848099549185>

[alpine sable]

yes

[pallid ferry]

🙏

[tardy stag]

yep

[pallid ferry]

.

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[pallid ferry]

<@&286206848099549185>

Ohhh

You guys were saying yes to the answer

Lmao

.close

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[copper willow]

My approach would be to use a piecewise function of asymptotes or tan

I assume you might not be able to use the latter though lol

[pliant terrace]

like arctan ? 😅

[copper willow]

Not quite. Think of how you can create a map which covers all of R^2 using tan

[pliant terrace]

if im being entirely honest i haven’t revisited geometry in a long time so give me a second

nah i think im a little lost

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[keen saffron]

Computer organization help

[keen saffron]

how do i do this

(mock exam question)

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[arctic tulip]

Why is the answer infinity

If I rationalize the expression and then evaluate the limit, I get -3

[alpine sable]

rationalize the expression?

[alpine sable]

not completely sure

but i think your second line is correct

your third line is not because square roots are always positive so you're not allowed to simply factor out the x from under the root

[alpine sable]

from here, top becomes +infinity and denominator approaches 0

[alpine sable]

this makes sense as well if the limit was x->+infinity it would approach -3 (i graphed in desmos)

uhm, i hope that makes some sense

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[arctic tulip]

Can u explain this part again??

[alpine sable]

it's kind of a weird convention

the square root symbol as used above essentially refers to the principal (positive) square root of the radicand

although mathematically speaking, there are two square roots but conventionally the specific symbol only outputs the positive version

thus, to perform the operation in this way, you would need to do x in the numerator and |x| in the denominator
edit: careful of how you factor the constant 1 though since it will be become negative 1

[arctic tulip]

Ahhh I seee it now

It does make sense

Also I just noticed that

If I plug infinity into the limit directly

It becomes infinity+infinity

Which is also just infinity

Thanks mate 👍

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[alpine sable]

yes but

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[alpine sable]

technically it's an "indeterminate form" though so we can't say exactly say that

[alpine sable]

but that only matters more when we have expressions like infinity/infinity when each infinity is a different "size"

[alpine sable]

your welcome :)

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.close

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[outer lark]

hey, im going through my notes from the beginning on fourier series, and having now gone through a textbook chapter, course notes, and a couple videos, each of them introduce the idea of representing f(x) as an infinite sum of sines and cosines as something along the lines of 'suppose we can do this, then later we'll prove we can do this' , is there a good argument for choosing to represent functions this way, does it just come later?

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[ocean ravine]

how do i simplfy this to match the inductive hypothesis

[ocean ravine]

I tried expanding it out, but nothing was simplfying

[hidden knoll]

what does 1^2+2^2+3^2...+k^2 equal to

not 1^2+2^2+3^2...+k^2+(k+1)^2

in our assumption

[ocean ravine]

oh

you mean it should have k+1 appended to it?

[hidden knoll]

i am asking you what information can we assume

[ocean ravine]

well this is equal to

[hidden knoll]

correct

different channel

[hidden knoll]

so would you agree that $1^2+2^2...+(k+1)^2=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$?

[ocean seal]

the breadwave

[hidden knoll]

because $1^2+2^2...+k^2=\frac{k(k+1)(2k+1)}{6}$

[ocean seal]

the breadwave

[ocean ravine]

hmmm yes, that would be another way to write that, because you take the total of the previous term, and add the (k+1)^2 to it

did i just set it up wrong?

[hidden knoll]

yes

[hidden knoll]

no

you just didn't use the assumption that's given in the inductive hypothesis

[ocean ravine]

i thought this was the assumption

[hidden knoll]

yes, you were supposed to use this information

[ocean ravine]

then I replaced k, with k+1

[hidden knoll]

so we did here

[hidden knoll]

you're trying to prove the equality for k+1 using the equality we assume to be true for k

[ocean ravine]

and I cant do that?

[hidden knoll]

you can

[hidden knoll]

we did here

[ocean ravine]

im not completely following, but I can try to simplify this?

[hidden knoll]

correct

we are using the assumption that $1^2+2^2...+k^2=\frac{k(k+1)(2k+1)}{6}$ is true for $k$ to show that $1^2+2^2...+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}$ is true (which proves for $k+1$)

[ocean seal]

the breadwave

[ocean ravine]

shouldnt we be showing that it's equal to this instead?

[hidden knoll]

maybe try showing me $\frac{(k+1)(k+2)(2k+3)}{6}=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$

[ocean seal]

the breadwave

[ocean ravine]

im ngl this is making it more confusing

[hidden knoll]

try expanding $k(k+1)(2k+1)+6(k+1)^2$ out

[ocean seal]

the breadwave

[hidden knoll]

and $(k+1)(k+2)(2k+3)$

[ocean seal]

the breadwave

[ocean ravine]

the whole thing?

[hidden knoll]

the same way you did with (k+1)(k+2)(2k+3)

[ocean ravine]

like distribute the k, distirbute the 6((k+1) * (k+1)),

[hidden knoll]

ya so you just end up with a polynomial

[ocean ravine]

ok this is going to take a while

is there a point to this?

i feel like its over complicating it

[hidden knoll]

you will find something

interesting

[placid trellis]

not a gf

[ocean ravine]

they are equal

?

[hidden knoll]

yes

correct

[ocean ravine]

yay

[hidden knoll]

so we know that $1^2+2^2...+(k+1)^2=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$ and that $\frac{k(k+1)(2k+1)}{6}+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}$

[ocean ravine]

and that concludes the proof?

[ocean seal]

the breadwave

[hidden knoll]

so what can we conclude

therefore

[ocean ravine]

well I see you can solve this in two ways

one by substituing k+1 directly in the equation

and the other way by appending it to the end

[hidden knoll]

no

we used the appending to the end to arrive at the equality for k+1

[ocean ravine]

OH WAIT

[hidden knoll]

do you get what we're doing now

[ocean ravine]

so is the +(k+1)^2 part of the hypotheis?

OHHHHHHHHHHHHHHHH

[hidden knoll]

yeah we were using the fact that we know $1^2+2^2...+k^2=\frac{k(k+1)(2k+1)}{6}$ is true

[ocean ravine]

so we just proved that because its equal, it means that the ladder holds true

[ocean seal]

the breadwave

[hidden knoll]

so therefore $1^2+2^2...+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}$ is true (as we demonstrated $\frac{k(k+1)(2k+1)}{6}+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}$

[ocean seal]

the breadwave

[ocean ravine]

hmmmm ok

but why does this take 100 steps to prove

i feel like there has to be a simpler way to prove they are equal

[hidden knoll]

once you have the idea it's just basic algebra

[ocean ravine]

not the distributing part

its making the question's answer take up 3 pages lol

[hidden knoll]

i usually dont show my steps because it's just basic algebra but if your teacher wants that

[ocean ravine]

right but there probably some laws that can show they are equal faster than distribtuing everything to find a one-liner polynomial

right?

[hidden knoll]

yeah you can use whatever method you can to show those 2 are equal

but that's not the point of the exercise

the point of the exercise is to see whether you can set up induction correctly

[ocean ravine]

well in that case, did I? because I see two inductive hypothesis

or wait

[hidden knoll]

you are not seeing two inductive hypotheses

[ocean ravine]

I guess the hypothesis is appending (k+1)^2

and then the inductive steps are substituing k+1 in

and arriving at the same conclusion

[hidden knoll]

our hypothesis is that $1^2+2^2...+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}$

[ocean seal]

the breadwave

[hidden knoll]

our proof was using the fact that $1^2+2^2...+(k+1)^2=\frac{k(k+1)(2k+1)}{6}+(k+1)^2$ to demonstrate $1^2+2^2...+(k+1)^2=\frac{(k+1)(k+2)(2k+3)}{6}$

[ocean seal]

the breadwave

[ocean ravine]

woah

ok

i sort of get it

my teacher might take off points for doubling its complexity

but

ig it still holds true

i just cant figure out the algebraic shortcuts

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[brittle lake]

Find second derivative with respect to y

[brittle lake]

Problem: $x^2+4y^2=7$

[ocean seal]

Light

[brittle lake]

Did I do this correct?

[warped marlin]

you mean with respect to x, but answer in terms of y, right

[brittle lake]

Yes

Implicit differentiation

[warped marlin]

nope sadly, i got a different result

why do you make things into fractions before differentiating, it makes it complicated :d

why dont you differentiate like straight forward instead!

try again, you should get $\qquad \ddot{y}=-\frac{7}{16, y^3}$

[ocean seal]

Milena

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@brittle lake Has your question been resolved?

[tight pier]

physicist

[warped marlin]

hahaha

and engineer :d

ngl that notation makes it less complicated and compact

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[alpine sable]

hello

[alpine sable]

how is angle a

angle b

=

<@&286206848099549185>

=

to angle b

[tardy shuttle]

so

is there any other informatino?

*information

[alpine sable]

no

it's just the figure

i just need to know how its equal

[tardy shuttle]

Pongruence probably

Do you know congruence?

[alpine sable]

Yeah i realise it's SAS congruency

[tardy shuttle]

which triangles

[alpine sable]

The Congurency of the figure is SAS

[fair parcel]

yes

[tardy shuttle]

yes that is how you solve

but which triangles are congruent

[alpine sable]

Both of them are equal

[tardy shuttle]

state the name of the triangles

[alpine sable]

Triangle ABD is congurent to Trange BAC

[tardy shuttle]

oh wait

[tardy shuttle]

correct

so what two angles are equal

[alpine sable]

Angle a and b

but how tho

i didn't understand it

[tardy shuttle]

once we prove congruence

[tardy shuttle]

we can use CPCTC(corresponding parts of congruent triangles are congruent)

so angle BAC and ABD are equal

Now, we just need to find another congruence to show that triangles BCD and ADC are equal

[tardy shuttle]

can you do that?

[alpine sable]

Oh no it's proved

i understood

thanks

.close

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[tardy shuttle]

ok

yw

np

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[mellow frigate]

am i doing something wrong here?

[mellow frigate]

.close

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[keen wasp]

hello

[keen wasp]

i need to compute the integral above, given a>3/2

these are the steps i’ve taken so far but i feel like i’ve reached a dead end so i was wondering if anyone could guide me

thanks

[wary crest]

1+1

[wary crest]

.

[keen wasp]

hi

[keen wasp]

can you help

thanks

[wary crest]

? Wdym

[keen wasp]

<@&286206848099549185> :)

[wary crest]

You watch football?

[keen wasp]

yes

no

[wary crest]

👍

[fair parcel]

@wary crest chat in #chill

[keen wasp]

flexo would you like to take a look

<@&286206848099549185> i need help with the problem 😭

[fair parcel]

I am sorry but I am not really good with these types of questions. Others will help.

[molten plover]

What is the name if the aplication to use plz??

[keen wasp]

please

oh

good notes

it costs money

[lone heart]

@keen wasp Has your question been resolved?

[sweet orbit]

@keen wasp trig sub

have u learned that ?

[sweet orbit]

tangent one specifically, try setting x = tan(theta) and use the 1 + tan(theta)^2 = sec(theta)^2 identity

idk wat the a is for?

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[alpine sable]

what does “/{x}”mean

[hushed locust]

could you show the context?

[alpine sable]

yeah one sec

[haughty wigeon]

f(x)?

[alpine sable]

i’m reading my lecture notes and it says

i jus don’t know what the curly brackets is there for

[high cargo]

\

Not /

[sweet orbit]

set difference

[hushed locust]

so {x_0} is the set containing only the number x_0, and (a,b) \ {x_0} is the set difference, which means the set that has all the elements of (a,b) except x_0

[alpine sable]

Okay thanks

.close

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[umbral arrow]

Tips?

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@umbral arrow Has your question been resolved?

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[broken violet]

how do i solve this?

[broken violet]

never mind I didn't read the question rpoperly

.close

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[alpine sable]

.:O help 0

[alpine sable]

For excercise 7-8

,rccw

[alpine sable]

Is this how we are supposed to do it?

The book didn't mention chain for 2 or more independent variables so...

[crystal berry]

when you do partial differentiation, all other variables become constants so it follows the same rule

[crystal berry]

can you walk me through your thinking here

[alpine sable]

Used the dependency deiagram

Assuming I can use it😅

[sly magnet]

looks correct to me

[alpine sable]

Alr

Thank you

.close

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[alpine sable]

What do they mean by reflect the figure over the line and the y=-1?

[neat folio]

reflect over the line y=-1

[limpid heart]

y = -1 is the line

[neat folio]

that you reflect over

[alpine sable]

I did that I just gave the opposite sign

[neat folio]

wdym by just giving the opposite sign

as your answer?

[alpine sable]

I’ll show u

Like this

[limpid heart]

that is not a reflection

not a correct reflection

[patent vale]

its the same as reflecting over the x axis just bring all the points down 1

[alpine sable]

Ohh