help-36

[rigid cedar]

this was the gx graph

this is fx

[sinful fjord]

yes

[rigid cedar]

for the second graph how can i exclude {x < 1}, {x > 13/7}

nvm

i think

\left{x<1,\ x>13/7\right}

{x < 1, x > 13/7}

liek that worked

lets go

.close

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[lone fog]

for part a, I need to use normal distrubution

for part b, i need to use binomial expansion

the issue is, how on earth do i know how to identify that

chatgpt says this

[spare summit]

!noai please.

[final saddle]

Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).

[lone fog]

i know thats why i put it out there as it seems correct

idk how else to spot it

[opal pelican]

That you need to use the normal distribution for Part a) is literally stated in the problem

States X~N(239,25)

[topaz kite]

lengths, and by extension heights, can take on any range of values. for instance, someone can be 140cm tall, or 140.01cm tall, or 140.001cm tall, or 150cm tall, etc.
the idea is, measurements are not discrete because of the possibility of these in-between values (depending on the precision of the measuring device, of course).

on the contrary, part b) asks for heights greater than 145cm. the opposite of this scenario would be heights less than or equal to 145cm.
there are only two categories here: either a height is > 145cm, or <= 145cm, hence the use of a binomial expansion.

(and of course, if the question says so, then...)

[opal pelican]

As for part b), the binomial distribution gives you the probability that, in a group of n people, something labeled as "success" happens k times

Identifying being over 145 cm as "success" and using the probability you computed above allows you to think the problem in terms of a binomial distribution

Essentially: the use of the normal distribution should be somewhat easy to see, as it's stated in the problem that you should use it

The binomial distribution is a handy tool when dealing with dichotomies, i.e "this happens or doesn't happen"

In this case, you either have height greater than or equal to 145, or height lesser than 145

[lone fog]

im gonna process this give me a sec

okay so

can i just approach it being normal vs binomial regarding the number of people being counted?

like for a) it is one person and b) it is three people

[opal pelican]

I don't get what you mean

[lone fog]

for part a) it mentions "a ten year old" so one person

for part b) it mentions "that 3 or more" so it is at least 3 people

[grizzled tusk]

that's not the point, think about different variables

first, there is the height of children

the exercise says to you that it's normally distributed

if H is the variable for the height, H ~ N(139, 5), and a) asks the probability that it's > 145

the b) is asking something else, it says you're taking 20 children height, under the law of H

repeating an experiment n times and counting the number of successes is the definition of a binomial law

your variable for the number of success, S, is S~B(20, p)

where p is the probability of a children above 145cm, you calculated in a)

those are two different problems about two different things

[lone fog]

is that the part that says "3 or more"

[grizzled tusk]

it could be 5 or more, or 2 or less

or exactly 8

that's not the point

it's about counting successes

when you repeat an experience that has a probability p of success, n times

the probability of a given number of successes is given by B(n, p)

[lone fog]

okay so like

[grizzled tusk]

using these laws to solve these problems is a different thing to think about, the fact that these are yours laws are consequence of the nature of the experiment

calculating things with that will come in a second time

you have an experiment, it follows a certain law

then you use this law to find the probabilities you're looking for

[lone fog]

i see

ill need to break it down myself more

i sort of get it

[grizzled tusk]

my advice would be to take your time to realise the two questions are very different, maybe only thinking about solving a) first

once it seems clear enough, moving onto b

[lone fog]

okay thanks

okay so

i went away

and i was able to conclude that

part a is just talking about a continuous measurement

part b is talking about the number of successes regarding a continuous measurement

[grizzled tusk]

yep, more precisely, b is about repeating measures 20 times, and counting successes

[lone fog]

is that what gives it away to use binomial expansion

the fact it mentions "20 times"

[grizzled tusk]

if we want to be very clear, it's because of what a success is

a success is binary, either you fail, and it's 0, or you succeed, and it's one success

a variable about 1 success follows what we call a Bernouilli law, of probability of success p

then, when you repeat such an experiment n times, it's the sum of n identical Bernouilli law

and it follows what we call a binomial law of parameters n and p

the clearest way to understand binomial law: if you repeat n times an experiment and you succeed one with probability p, or fail with proba 1-p

then it's a binomial law by definition

[lone fog]

i see okay

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[left trail]

what can I do about the inductive step? If I say $f^{(n)}(0) = c_0 + nc_1 +n(n-1)c_2$. then we have $f^{(n+1)}(0) = d/dx f^{(n)}(0) = d/dx (c_0 +nc_1 + n(n+1)c_2)$ but then our $c_0$ will disappear. I was trying to find another way to represent the nth deriviatve but I don't know how I would represent that sum. Any suggestions?

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Please don't occupy multiple help channels.

[vital crag]

Close your other channel

[soft zealot]

BigBen

[drowsy epoch]

That's wrong, f^(n+1)(0) is some constant so you cant really write it as d/dx [f^n(0)]

You can use the general leibniz rule

[left trail]

ok I will look into this. Thank you

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[frank ginkgo]

$\int^{+\infty}_{-\infty} \frac{\cos(x)}{x^2+2x+2} \mathrm dx$

[soft zealot]

Custos Caeli Invictus

[frank ginkgo]

I rewrote it as

$\int^{+\infty}_{-\infty} \frac{\cos(x)}{(x+1)^2+1} \mathrm dx$

[soft zealot]

Custos Caeli Invictus

[frank ginkgo]

Using t = x+1

$\int^{+\infty}_{-\infty} \frac{\cos(t-1)}{t^2+1} \mathrm dt$

[soft zealot]

Custos Caeli Invictus

[frank ginkgo]

I used that cos(t-1) = cos t cos 1 + sin t sin 1

$\cos 1 \int^{+\infty}{-\infty} \frac{\cos(t)}{t^2+1} \mathrm dt + \sin 1 \int^{+\infty}{-\infty} \frac{\sin(t)}{t^2+1} \mathrm dt$

[soft zealot]

Custos Caeli Invictus

[drowsy epoch]

Maybe you could try residues, I don't see where this could be going

[frank ginkgo]

The second integral is 0 since it is odd and is also integrable on R

$I = \cos 1 \int^{+\infty}_{-\infty} \frac{\cos(t)}{t^2+1} \mathrm dt$

[soft zealot]

Custos Caeli Invictus

[drowsy epoch]

I feel I was wrong

[frank ginkgo]

If i cosider

$f(t) = \frac{1}{t^2+1}$

[soft zealot]

Custos Caeli Invictus

[vital crag]

i'm sure your way would work too

[frank ginkgo]

$\widehat f(\xi)=\pi e^{-|\xi|}$

[soft zealot]

Custos Caeli Invictus

[frank ginkgo]

Since f is even, using $e^{i\xi t}$ also yields the same real value

[soft zealot]

Custos Caeli Invictus

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

[vital crag]

do you need to prove this or do you have a table with formulas

[drowsy epoch]

I actually thought they would resub t->t+1 and do that trig thing again, but maybe it's circular

[frank ginkgo]

Im using notable Fourier transform

[frank ginkgo]

this Is πe^(-1)

$\int_{-\infty}^{+\infty}\frac{\cos t}{t^2+1} \mathrm dt
=\widehat{\left(\frac{1}{1+t^2}\right)}(1)
=\pi e^{-1}$

[soft zealot]

Custos Caeli Invictus

[frank ginkgo]

so $I = \frac{\pi}{e} \cos 1$

[soft zealot]

Custos Caeli Invictus

[vital crag]

yea see basically you use residue theorem to prove the fourier transform identity they're using

[vital crag]

,w \int^\infty_{-\infty} \frac{\cos(x)}{x^2+2x+2} dx

[soft zealot]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by rie.mann

[vital crag]

looks right

[frank ginkgo]

Should this be the fastest method?

[vital crag]

who cares

[frank ginkgo]

I should see with the residues

But I think it takes more time

[frank ginkgo]

I like to solve problems with the fastest methods

[drowsy epoch]

people with little lifespan

[vital crag]

apply linear operator with eigenvalue > 1

[frank ginkgo]

.close

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[real cove]

I am making a quiz/setwork and was wondering if this is a good answer key to process.

[deep condor]

looks good to me

[real cove]

I am worried I am just using the differentiation formula for e^x only.

[final tangle]

du/dx, not just du

[real cove]

I am using power rule for #2 but I kinda feel like I should include more formulas.

[final tangle]

whether you feel the need to include depends on how much they've been taught

the rule is essentially implied when applying it

or do you mean make it more complicated

[real cove]

(I am a student teacher and I do want to go easy on them since the exam will be a bit harder.)

[real cove]

Yeah.

This is a seatwork/quiz

[final tangle]

i suppose you can chuck another expression on the side
or have a denom
so they'd have to apply product or quotient rule

[real cove]

That was what I was thinking. Maybe do that for a number 2 somehow.

[final tangle]

sounds good

also don't forget to fix those du,dv in your answer key

[real cove]

I did. du/dx and dv/du

[final tangle]

actually other notation issues as well

those should still be x

[real cove]

No, because we are still substitution u into the equation, right? x should be there when we replace u with the original value of u.

[final tangle]

no

if you're going to replace, you'd need to replace with something equivalent

e.g. note that with your sub, u = x^2 - 3x

you'll have
f(x) = e^u
and just because the right side has the variable u
does NOT mean that you can just replace x in the left side with u

writing f(u) = e^u implies
f(x) = e^x
which is clearly different from what you started with

[real cove]

Gotcha

[final tangle]

you "could" express the x in f(x) in terms of u
but that's tedious and unnecessary since you have access to chain rule

[real cove]

This is long but uses the formula for differentiation of sin and also uses product rule.

[real cove]

Yeah. I see my issue. Will correct that and my lesson soon.

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[hushed marten]

can anyone help me with this problem?

[gritty chasm]

hint: square both sides and see what happens

and make use of the fact that $(|x|)^2 = x^2$

[soft zealot]

Annie Maqionde

[hushed marten]

okay ty, i'll try

ccould you please remind me how to expand the identity (a+b+c)^2?

I forgot.

[coral zenith]

(a+b+c)^2 = a^2 + b^2 + c^2 + 2(ab+bc+ca)

[hushed marten]

oh
I solved it.

tysm

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[coral zenith]

<@&268886789983436800>

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[patent steeple]

How was factorization of d concluded?

[slow wolf]

we know that the gcd(c, d) = 54, since 54 is the greatest common divisor of c and d, what does this tell us about c and d?

[patent steeple]

D is a multiple or 2×3³ yeah correct idk why I stared at this for so long

[slow wolf]

Yes! good job. In other words, d = 2 x 3^3 * t

[patent steeple]

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[open musk]

Hello everyone, I’m new to this server and hope we can help each other here. If you need any help, feel free to reach out to me.

[slow wolf]

Sorry, this a help channel.
If you want to talk with others, check out #discussion . To help someone with math, look at other help channels where someone is still looking for help!

[timber leaf]

W

[open musk]

could you elaborate, i didn't get that

I know this is a help channel

[slow wolf]

This persons question has already been answered. Help channels treat one person at a time

[glossy zephyr]

https://discord.com/channels/268882317391429632/488120190538743810
Consider reading this ig. Wrong reason to open a help channel

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[stray notch]

Is there any particular reason why we need to substitute x with t at the end?

[stray notch]

like, cant we just use x?

[versed crater]

well it gives you a function f(t) = (x(t), y(t), z(t)) that traces the line of intersection

[opaque ember]

its not strictly needed. for some ppl it helps them see the solution parametrized that way

[stray notch]

ohhhh

i see

[gritty chasm]

There's no mathematical reason actually, some textbooks write it that way.

[stray notch]

so it's just there to make the function of the line easier to understand right?

alright

thanks yall, appreciate it

[opaque ember]

for example if the solution depends on two parameters then we might replace them with s and t in the final answer

[lilac moat]

Another way to find the line of intersection is to cross the two plane normals

[stray notch]

damn

ye i will def try that. that's an interesting way

thanks

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[lilac bison]

you're welcome

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[tired iron]

js wanna ask for graphing a linear equation x and yintercept and using the slope r enough/?

[tired iron]

found the slope by comparing with y=mx+c

and xand y intercept r the point of intersections( of cordinate axis) for it?

[rugged merlin]

You just need two points to form a line actually

So you dont even need to be given the slope

[tired iron]

ye x and y doesthe work too

[tired iron]

also this

[tired iron]

the thing with making table n then graphing was such a pain

[rugged merlin]

Sorry, there might be a language barrier because I have a hard time understanding what you are saying

[tired iron]

i meant putting x values in equation to get y values then graphing those points

instead of jst getting x and y intercept

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[spice hull]

hi, i'm doing a report for a linear algebra 2 class on how linear algebra applies to facial recognition, PCA etc... I'm just confused/stuck in thinking that maybe i've got something horribly wrong in this report and i could really use someone to double check it 😭 🙏

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[spice hull]

^^

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@spice hull Has your question been resolved?

[gritty chasm]

<@&268886789983436800>

[final saddle]

@spice hull Has your question been resolved?

[spice hull]

<@&286206848099549185>

[signal vector]

which part are you confused about?

[spice hull]

just wondering if it's all correct that's all

[signal vector]

if you think that something is wrong, there must be a location in the document where you think you might have written something wrong. please point it out. I dont think anyone will read the entire report to find for mistakes. Also, on reports and stuff like this it is advisable to add references.

[spice hull]

the bit where i wrote 'this maps a matrix to a one-dimensional vector elementwise' and wrote out how phi of X equals to [M_1,1,1,...] etc is this corrct notation? Also i wrote about that the covariance matrix is symetric which means it can be diagonalised, we didn't really go over symmetric matrices in my couse so i'm just curious if that's true. I think that's kinda it. But about the rferences i'm in the process of adding those in the actual report doc i created so that should be fine

[signal vector]

yes, any symmetric matrix can be diagonalised. as for the notation i dont know if its the standard in this field of study but you never use that notation afterwards in the document anyways so i dont think it matters

[spice hull]

ahh ok perfect thank you 🙏

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[gloomy rose]

I discovered something that I don't know if it had been discovered before:

If you multiply 2 numbers (one of one digit and the other of two)

For example: 11x2=22

And then you change the units of the two numbers with each other

For example: 11x2=22 is changed to 12x1=12

The multiplications will have the number of the tens of the difference multiplication.

And I tried it with other numbers to test if it always works and it gave me the same:

13x4=52 and if you change it there is 14x3=42 difference=10

15x4=60 and if you change it there is 14x5=70 difference=10

27x6=162 and if you change it it's 26x7=182 difference=20

31x2=62 and if you change it it is 32x1=32 difference=30

Something I don't understand is because here he gave me something different if in the others if it is fulfilled (I think it's because I multiply it by 5 upwards in the units and by 0):

35x9=315 and if you change it it is 39x5=195 difference=120

43x5=215 and if you change it it is 45x3=135 difference=80

41x9=369 and if you change it there is 49x1=49 difference=320

26x2=52 and if you change it it's 22x6=132 difference=80

37x4=148 and if you change it there is 34x7=238 difference=90

18x3=54 and if you change it there is 13x8=104 difference=50

10x5=50 and if you change it there is 15x0=0 difference= 50
So, is it correct yes or no?

[gritty chasm]

!15m

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[sterile marsh]

Umm

(10a+b) c

Becomes (10a+c) b

[shell condor]

I guess the one's place doesn't change, that's why you always get a difference of a number that is a multiple of 10

[south dirge]

only works when you take mumber which has consecutive digits in ones and tens

[wary juniper]

(10a+b)(c)-(10a+c)b=10(ab-ac)

[south dirge]

as in n n+1

[gloomy rose]

Sorry, I didn’t read that.

[silent mango]

the difference is 10a(c-b)

not just the tens digit

[sterile marsh]

Where the original multiplication was (10a+b)×c

[silent mango]

yup

[gloomy rose]

Thanks.

[silent mango]

no problem :D

[shell condor]

!done?

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[gloomy rose]

.close

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[gloomy rose]

Thank you

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[torn helm]

i was trying to solve a physics problem which boiled down to finding which one of the two expressions is greater than the other. can someone tell me if it is possible to figure it out with the given conditions? if not then i'd probably have to try and solve it in a different way

[torn helm]

my bad both expressions are multiplied by by 0.5

oh wait that doesn't really matter

[wary juniper]

Removing common factors and adding fraction, we can simplify it down to comparing
nb+ms and b-nb+s-ms

[torn helm]

oh i see ya

also i forgot to mention that all the variables here are positive

[wary juniper]

Take the 1st minus 2nd: (2n-1)b +(2m-1)s
At maximum this is s>0
At minimum this is -b<0

Which means we can't compare them

[wary juniper]

Next time you should write >0 instead of R

[torn helm]

how did u figure out what it's gonna be at maximum and at minimum?

[torn helm]

okay

[wary juniper]

b and s are positive so it's max when 2n-1 and 2m-1 is max, so when n and m are max
Same reasoning for min

[torn helm]

oh okay i understand

[torn helm]

and how does it conclude that we can't compare them?

[torn helm]

i mean these two are just two of the given conditions

[wary juniper]

Imagine 2 values comparing a and b
a>b <=> a-b>0
a<b <=> a-b<0
But we cant compare a and b if a-b can be <0 or>0

[wary juniper]

What?

[torn helm]

so if we were given that a-b>0 or a-b<0 then we wouldn't be able to compare a and b, is that correct?

[torn helm]

like s>0 and -b<0 are two of the given conditions

[torn helm]

oh i got it like, if were given a-b>0 or a-b<0 then we find a<b or a>b , which is obviously true and doesnt give us any information about which value is greater or less than the other

[torn helm]

but how do we apply this idea in this case?

[wary juniper]

a is nb+ms, b is b-nb+s-ms

[torn helm]

wait if i say that a = nb+ ms
and c = b - nb + s - ms

then subtract 2nd from 1st to get a - c = (2n-1)b + (2m-1)s

and at maximum a-c =s >0 so a>c
and at minimum a-c=-b<0 so
a>c

so in both cases a>c
so nb+ms > b-nb+s-ms

no?

[torn helm]

yah that's what i did

[torn helm]

awwh wait no..a-c=-b<0 means a<c

thanks a lot man, u were of great help, i appreciate your time

.close

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[runic phoenix]

Exercise 1.1.11: Let 𝐹beanorderedfieldand 𝑥,𝑦,𝑧,𝑤∈𝐹.
a) Prove part (vii) of Proposition1.1.8 . That is, if 𝑥≤𝑦and 𝑧≤𝑤,then 𝑥+𝑧≤𝑦+𝑤.
b) Prove thatif 𝑥 < 𝑦and 𝑧≤𝑤,then 𝑥+𝑧 < 𝑦+𝑤.

[runic phoenix]

That is proof of part a

<@&286206848099549185>

[opal pelican]

I don't know if it's a me problem, but I can't read your handwriting

Well, at least not being confident I'm doing it correctly

[runic phoenix]

Oh

Let me write it in latex

[south dirge]

-# what in the gods abomination is that😭

[south dirge]

-# yeah that would be helpful no lie

[runic phoenix]

$$\text{Let } x,y,z,w \in \mathbb{F} \text{, then } w - z <= (y-x) - (w - z)
$$
$$
\text{ we then rewrite right hand }
w-z <= (y+w) - (x +z) \text{, so we have } w <= (y + w) - x
$$
$$
z <= w \text{ and } w<=(y+w) -x \text{, by transitivity of ordered sets, }
z<=(y+w)-x \implies z+x<=y+w \implies x+z<=y+w
$$

[amber dock]

For part (b), since you know x < y and want to show x + z < y + w, it might be helpful to use something like x + z < y + z

[runic phoenix]

yes

okay

[royal gust]

You likely have very specific rules about what ≤ can do

What's the book?

[final saddle]

@runic phoenix Has your question been resolved?

[runic phoenix]

Basic Analysis I. Jirǐ Lebl

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@runic phoenix Has your question been resolved?

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[livid sky]

I have a question about Graph Powers. I am currently taking Discrete Structures and this is what I know:
"Let G be a digraph. Let u,v be any two vertices in G. There is an edge between u,v in G^k if and only if there is a walk of length k from u to v in G."

Also, G^+ = union of all G^k = G^1 U G^2 U G^3 U...G^k.

"Let G be a diagram with n vertices and let A be the adjacency matric of G. Then for k>=1, A^k is the adjacency matrix for G^k where boolean addition and multiplication are used to compute A^k." Effectively, G^k is the digraph of the adjacency matrix A^k.

My question is how do I tell what k is given a digraph G.

[livid sky]

To add a tad to my question, I am confused if the largest k is 3 or 4. I presume the largest possible walk would have to be from 4 to 1. But I can think of two walks that are reasonable options, <4,2,3,1> (k=3), and <4,2,2,3,1> (k=4).

I guess I'm tempted to go with the smaller option, since you could go infinite with it, but I am also not sure if I should exhaust all edges at least once.

If no one can answer, that's fine. The answer is a, I believe, since the out-degree is the same for A^3 and A^4. Though, I would still like to know if I should have used A^3 or A^4 for exam sake.

And if anyone knows whats going on and sees any mistakes, I'd appreciate a notice. These matrices are using boolean algebra.

[whole halo]

for G^+, k is taken to be a limit?

also, dont forget that you can go 3 -> 1 -> 3, so its possible to have a walk from 3 to itself

[livid sky]

This is what I’ve got in my notes.

[whole halo]

gonna be a yes then

think of it this way

starting at 3, which possible vertices can you eventually reach?

[livid sky]

1 and 3?

[whole halo]

how many vertices is that?

[whole halo]

thats all

you get 1 from G and 3 from G^2

its not possible to reach 2 or 4 from a walk starting at 3

so its just 1 and 3 from G U G^2, the rest of the Gs would alternate between 1 and 3 from a walk starting at G

see here youve circled the 3rd row, and its 1 0 0 0 on odd powers and 0 0 1 0 on even powers

[livid sky]

Interesting.

[whole halo]

this comes from the definition of A^k being the adjacency matrix of G^k, so we dont necessarily need to compute A^k's powers to confirm this

[livid sky]

I see, so if I were to do it this way, I would only need to go to the largest/minimum k for a given vertex? It's hard to phrase, but how from vertex 3 could only take a largest walk of 2 to get back to vertex 3.

[whole halo]

keep in mind they want G^+, not for you to go through G^1, G^2, ...

think broadly here

can you, or cant you reach there from here

thats all for G^+

it would waste time to have to think:
"can I reach there from here in 1 move?"
"can I reach there from here in 2 moves?"
"can I [...]

[livid sky]

I'm not sure I get what you mean.

What's the shortcut you are suggesting?

[whole halo]

this, presumably

[livid sky]

I don't get it. What is G^+?

[whole halo]

take a look at this definition

now note here that theyre using "union"

notice there in the definition that they have defined each of G^1 through G^k

[livid sky]

Okay.

[whole halo]

now G^1 through G^k are graphs or matrices?

[livid sky]

Graphs.

[whole halo]

what about A^1 through A^k?

[livid sky]

Matrices.

[whole halo]

so when they use "union" for graphs, what does that mean

[livid sky]

You add the corresponding matrices together. Or add their relations to each other?

[whole halo]

you merge them together / combine them

[livid sky]

Wait.

[whole halo]

for example if you had A -> B and A <- B
they union to A <-> B

[livid sky]

Okay.

OHHH

Oopsies

[whole halo]

for the matrices, itd be doing logical OR

same idea

[livid sky]

I forgot to add them

[whole halo]

I dont recommend that

again you dont need to multiply matrices to figure this question out, I told you that

G^1 shows where you can reach in 1 move
G^k shows where you can reach in k moves

so if you union all of them

you just get a graph that shows where you can reach in any amount of moves

so just a graph of where you can be in general

[livid sky]

I see.

That's cool.

[whole halo]

they just want the out-degree for G^+'s 3

so you only need to do this much

[livid sky]

So by nature it can only possibly reach 2 vertices.

Alrighty.

[whole halo]

if for example they wanted a picture of G^+, you use similar thinking and can draw G^+ just based on where you can reach based on your starting place

[livid sky]

Fancy pancy.

[whole halo]

there you wouldnt need to calculate the adjacency matrix either

[livid sky]

So, just to test my knowledge, I'm gonna send a picture of question 7 and see if I get it. Is that fine?

[livid sky]

Actually, this might be harder.

a.

1 can never reach 4.

Eh?

Maybe?

[final saddle]

@livid sky Has your question been resolved?

[livid sky]

I would just like to know if I am understanding correctly.

[livid sky]

I will assume yes. I don't want to hog the channel.

.close

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[strange sparrow]

no vertex other than 4 can reach 4 because it has in degree 0, yes

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[solemn osprey]

I got the answer as (a)

but as g(x) is a function why are there two values of g(x) for same x??

[runic needle]

yeah that's not a function

[solemn osprey]

the question is just wrong?

[runic needle]

maybe option A means that options B and C are both correct

[runic needle]

it's ambiguous

[solemn osprey]

it's a single correct type but alright ty❤️

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[runic needle]

i mean option B is option A with + and option C is option A with -

B and C are technically also correct

dumb question phrasing

[solemn osprey]

yh got it man ty🩷

[runic needle]

aight

[solemn osprey]

they be framing anything just to get their job done

[runic needle]

the core of the questions is pretty cool tho

[solemn osprey]

they probly felt like a genius after framing the initial part and just didn't think after😭

[runic needle]

lol

<@&268886789983436800>

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[left trail]

For 21 what can I do for 2x/pi< sinx. I hade the function sinx-2x/pi. And then I want to see where it was strictly increasing. I had cosx-2/pi>0. But it doesn't seem like it would hold up to pi/2

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@left trail Has your question been resolved?

[left trail]

<@&286206848099549185>

[keen delta]

24 or 21?

[left trail]

Sorry 21

[keen delta]

Try looking at the second derivative, f''(x)

[left trail]

Ok so we have sinx<0.

But that doesn't satisfy the x in our interval

[keen delta]

Check the values at the endpoints 0 and π/2; since f''(x) = -sin x < 0, the function is concave and must have a maximum in (0,π/2)

[left trail]

Ok so the endpoints give us 0 and -1. The max would be x such that cosx=2/pi

[keen delta]

the endpoints of f(x)

[left trail]

0 and 1

[keen delta]

f(x)=sinx-2x/pi

0 and 0

[left trail]

Yes

Also are you going to say we are going to look at the strictly increasing part of our function

[keen delta]

It’s not strictly increasing everywhere, but since it’s concave and both endpoints are zero, the entire curve must stay above the x-axis.

[left trail]

Oh I see so the difference is always positive. I was just so focused on trying to make it strictly increasing

Thanks

.solved

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[empty grotto]

How to solve this one 😅 can someone explain to me 👉👈

[gritty chasm]

Well do you know something called vieta's relations?

https://en.wikipedia.org/wiki/Vieta's_formulas

[ivory sierra]

Let the roots of P(x) be x, y, z. Then, we have the roots of Q(x) as 3x, 3y, 3z.
Use vieta's relations on both polynomials, and determine b, c, d by factoring out the multiples of 3, substituting the values got from the coefficient of P(x) and multiply.
Thats the outline

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[gentle zephyr]

can I get some help proving this three

.close

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[tropic crest]

what.

insta close 😭 🙏

[gentle zephyr]

I realized this exercise is kinda hard

and not so relevant to my course

even though is in the homework exercises I dont think I will get so much out of it in terms of practice for the exam

It will take me a long time and also will consume much of my time while I would be better of practicing for the exam

[proper raptor]

it’s kinda short to prove these relations

[gentle zephyr]

double inclusion right

[proper raptor]

ah hell nah it’s shorter than that

you can do it with iff

[gentle zephyr]

how?

[proper raptor]

for the first one, let $x \in B\setminus \cup_{i \in I} A_i \iff x \in B \text{ and } \forall i \in I \ x \notin A_i \iff \forall i \in I (x \in B \text{ and } x \notin A_i)…$

[soft zealot]

tm

[gentle zephyr]

you make it seem easy

[proper raptor]

lol

try the second one

It’s the same line of reasoning

[gentle zephyr]

.reopen

[final saddle]

✅ Original question: #help-36 message

[gentle zephyr]

suppose $x \in B \setminus \bigcup_{i \in I} A_i \iff x \in B \land x \not\in A_i \forall i \in I$

[soft zealot]

Renato

[proper raptor]

we traduce it by $x \in B \setminus \bigcup_{i \in I} A_i \iff x \in B \land x \not\in \cap_{i \in I} A_i$

[soft zealot]

tm

[proper raptor]

first, what is the « formal expression » of $x \in \cap_{i \in I} A_i$

[soft zealot]

tm

[gentle zephyr]

dude even if its short this is not easy

[proper raptor]

bro you can do it 😭

[modest sequoia]

instead of saying "this is not easy" or "you make it seem easy" it might be better to say what you don't understand. in that case, helpers know where you need help or if there are any fundamentals you are missing.

[gentle zephyr]

,, x \in B \setminus \bigcup_{i \in I} A_i \ \iff x \in B \land x \not\in A_i \quad , \forall i \in I \ \iff x \in B \land x \not\in \bigcap_{i \in I} A_i \quad , \forall i \in I \ \iff x \in B \setminus \bigcap_{i \in I} A_i \quad , \forall i \in I \ \iff x \in \bigcap_{i \in I} \left(B \setminus A_i \right)

[proper raptor]

it’s for the first one ?

[proper raptor]

look what i did here

[soft zealot]

Renato

[proper raptor]

i put everything under the « I » index

[proper raptor]

i see the error

lemme write

[gentle zephyr]

,align x \in B\setminus \cup_{i \in I} A_i \ &\iff x \in B \text{ and } \forall i \in I \ x \notin A_i \ &\iff \forall i \in I (x \in B \text{ and } x \notin A_i) \ &\iff \forall i \in I (x \in B \setminus A_i) \ &\iff \forall i \in I (x \in B \setminus \cap_{i \in I} A_i)

[soft zealot]

Renato

[proper raptor]

$$x \notin \bigcup_{i \in I} A_i \iff x \notin A_i \ \forall i \in I$$
is correct, but

$$x \notin A_i \ \forall i \in I \implies x \notin \bigcap_{i \in I} A_i$$
is false

[soft zealot]

tm

[proper raptor]

yes bwo this one is good 👍

[gentle zephyr]

you mean if x is not in every Ai it is still possible for x to be in the intersection of Ai?

[proper raptor]

so now try the same method on the second

[kindred mortar]

not the last line

[kindred mortar]

here

[proper raptor]

yeah oops

im blind

[gentle zephyr]

suppose I = [1,2] with i in N

x is not in A1 and x is not in A2

how can it be in the intersection of A1 and A2?

[gentle zephyr]

u sure?

[gentle zephyr]

yeah the last line its not an IFF but its an implication

[kindred mortar]

that's not what you want to prove

but you are almost done

[gentle zephyr]

,align x \in B\setminus \cup_{i \in I} A_i \ &\iff x \in B \text{ and } \forall i \in I \ x \notin A_i \ &\iff \forall i \in I (x \in B \text{ and } x \notin A_i) \ &\iff \forall i \in I (x \in B \setminus A_i) \ &\iff x \in \cap_{i \in I} B \setminus A_i

[proper raptor]

you don’t need to put \forall i \in I at the end

the indexation is already here with the \cap

[gentle zephyr]

where?

[proper raptor]

last line

[soft zealot]

Renato

[proper raptor]

now that’s good

[gentle zephyr]

you just joined the server?

I thought you were doing prepa, how did it went

[gentle zephyr]

is this iffs or implications ?

[proper raptor]

i left in december i think i’m not new here

[proper raptor]

im in uni since 2 years i only did 1 year of prepa lol

[proper raptor]

its iffs

much better than double inclusions 😶‍🌫️

[gentle zephyr]

let me try to do the second one

[proper raptor]

yea only the quantifiers change but its quite the same

[gentle zephyr]

,align x \in B \setminus \cap_{i \in I} A_i \ &\iff x \in B \text{ and } \exists i \in I \text{ s.t. } x \not\in A_i \ &\iff \exists i \in I (x \in B \text{ and } x \not\in A_i) \ &\iff \exists i \in I (x \in B \setminus A_i)

[proper raptor]

absolutely

[gentle zephyr]

is hard to continue from here, any hint

[proper raptor]

do the same as the first one, put everything under the indexation of I

yes bwo

[soft zealot]

Renato

[proper raptor]

perfect

[gentle zephyr]

I feel like I am going in the wrong direction, perhaps?

how to introduce CUP

[kindred mortar]

no good direction

[proper raptor]

well it’s under your eyes 😭

$x \in \cup B_i \iff \exists i \in I \text{ such that } x \in B_i$

[gentle zephyr]

,align x \in B \setminus \cap_{i \in I} A_i \ &\iff x \in B \text{ and } \exists i \in I \text{ s.t. } x \not\in A_i \ &\iff \exists i \in I (x \in B \text{ and } x \not\in A_i) \ &\iff \exists i \in I (x \in B \setminus A_i) \ &\iff x \in \cup_{ i \in I} (B \setminus A_i)

[soft zealot]

tm

Renato

[proper raptor]

https://tenor.com/view/homelander-the-boys-gif-18572939

[gentle zephyr]

,align x \in \cup_{i \in I} (A_i \cap B) \ &\iff \exists i \in I ( x \in A_i \text{ and } x \in B) \ &\iff x \in B \text{ and } \exists i \in I (x \in A_i) \ &\iff x \in B \text{ and } x \in \cup_{i \in I} A_i \ &\iff x \in B \cap \left(\bigcup_{i \in I} A_i \right)

[proper raptor]

arg not perfect this time

reread yourself

u will find the little error

lol now it’s good

the last line is a bit messy but i get what u mean

good 👍

[soft zealot]

Renato

[gentle zephyr]

I appreciate the help

[proper raptor]

np u welcome

[gentle zephyr]

if I would have done this using double inclusion it would have taken me ages

I appreciate it really

.solved

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[cedar obsidian]

I have to find domain of f(x)

But at last step it looks weird

[strange pelican]

Go on

What does that tell you

[strange pelican]

[vital crag]

denominator doesn't ever look zero

[strange pelican]

-1<=sin(x)<=1

[vital crag]

you basically have it here

[cedar obsidian]

Answer?

[vital crag]

does sin(thing) ever equal sqrt(2) ?

[cedar obsidian]

Nope it's greater than 1

[vital crag]

sqrt(2) ~ 1.4 > 1 yes

[cedar obsidian]

then how do I conclude domain in inequality from this statement

Is the domain x E R?

[vital crag]

the domain is thus {all x such that sin(2x) is not equal to sqrt(2)}. what elements x are in that set?

[vital crag]

oh right. you answered already

[cedar obsidian]

so from definition x is input means x belongs real

Can I say (-infinity,infinity)

Ok thx

[tiny gorge]

just a little side note, sin^4(x) + cos^4(x) is the sum of two nonnegative numbers, so the only way it can be zero is if both of those numbers are zero, that that would imply sin(x) = 0 and cos(x) = 0; you can easily check that this never happens

(one way to check that it never happens is that it would violate sin^2(x) + cos^2(x) = 1)

[cedar obsidian]

Yes by saying that there is no common angle where both are 0

right?

[tiny gorge]

yep, using sin^2 + cos^2 = 1 is one way to demonstrate that

your way is fine, just pointing out that this is a shortcut

[cedar obsidian]

How can I generalise this reason for such functions or similar questions

[vital crag]

do more problems of such functions and similar questions

[tiny gorge]

well the only generalization i can see is that if your denominator is the sum of things that are nonnegative, then the only way it can be zero is if all of those things are zero

[cedar obsidian]

Yes doing but still I asked before looking to another ones

Alr thx

[vital crag]

you don't even know what you're generalizing to

[cedar obsidian]

nvm I gotta do more questions for that matters

[final saddle]

@cedar obsidian Has your question been resolved?

[cedar obsidian]

For this how do I check for what x that inequality is satisfied

@vital crag

[strange pelican]

For the square root you should write >=0

[strange pelican]

sin2x>=0
When is sin (theta)>=0?

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@cedar obsidian Has your question been resolved?

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[sharp jacinth]

Yoo wsup

[sharp jacinth]

Just made this and need help

Question:

A cylinder has a height of 15 cm and a volume of 11,304 cm³. Calculate the radius.

Now i am on the last step where i got 753,6

Is it 753,6cm³ now ? If yes how do make it normal on that calc

[honest gust]

seems way too big

[sharp jacinth]

Ohh shooot

[sharp jacinth]

Oh not the end btw if you think that

[honest gust]

a good rule of thumb is that the radius will be less than or equal to the volume

well in this case i can see that happening

[sharp jacinth]

So you mean i gotta decrease it since its ³rn?

[honest gust]

nah something might've happened

[sharp jacinth]

Tell me

[honest gust]

ok you js need to do some algebra

$V=h\pi r^{2}$

[soft zealot]

chudcel

[sharp jacinth]

Yep i did it dif tho may i?

[honest gust]

like you want r on its own

[sharp jacinth]

Yes

[honest gust]

did you sqrt at the end

maybe that's the reason

[sharp jacinth]

Nope i wanted to ask if i have to do that

[honest gust]

you need to

like after you get r^2 on its own

sqrt

[sharp jacinth]

So if i have ³ do i use normal square root?

[honest gust]

you have to use like

the normal sqrt

[sharp jacinth]

If i used the normal i'd get 27,5

Cm

For radius

[honest gust]

ok i'll check

[sharp jacinth]

Okay

[honest gust]

it seems like given allat that 0 < r < 1

because the volume < height

the radius has to be very thin for this to happen

[sharp jacinth]

I dont get what you mean rn tbh

[honest gust]

and yes this did happen

[honest gust]

it's good to visualize it

[sharp jacinth]

So you think mine was right or wrong?

[honest gust]

let me see your work

we'll see what happened

[sharp jacinth]

Aight its kinda messy tho

[honest gust]

it's fine

my handwriting sucks so i'll be able to read it eventually

[sharp jacinth]

If you cant read it dont try just tell me

G is the main side

Hk is just height

[honest gust]

yeah it seems like

this solves for the volume

but the problem wants you to solve for the radius

[sharp jacinth]

Ik i need to explain how i did it one sec

So first i got my volume and did it ÷ the height so i have the main sides idk how to call em in english math.

And now i see my mistake

Lemme redo

[honest gust]

you also need to divide by π

the main sides don't seem to be very nice to work with here

[sharp jacinth]

So

Now that 239,8

Is that ² or ³?

Do i just do square root and thats r then?

[honest gust]

ok can you explain the steps to get there

[sharp jacinth]

I have volume divided by height so i get main sides

Then main side divided by pi

Now i get that 239,8 which i think i have to square root

[sharp jacinth]

Which should be 15,5 or 15,48 at the end

[honest gust]

radius still seems pretty big

did you do like volume divided by (radius multiplied by π)

[sharp jacinth]

No my formula different

I use main formula and make it like back

If yk

I'm really sure its 15,5

If we round it up

[honest gust]

waht is the formula

oh wait are your units all in centimeters

[sharp jacinth]

Yes

U tought 15m ?

My bad i didnt say cm

[honest gust]

yeah

it's fine

[sharp jacinth]

Sorry sorry

[honest gust]

it just seems like a very big number

because if we like plug in r = 15

we should get like a way bigger volume than 11

[sharp jacinth]

11304cm³ volume

[honest gust]

yeah that's like really big

the volume given was like

11.304 cm^3

[sharp jacinth]

I just checked 15,5cm is gud

[honest gust]

didn't it give you this

[sharp jacinth]

Idk but i checked and it was correct

Thanks for help i need to go to sleep now

[honest gust]

ok ion think it works

alright good night

[sharp jacinth]

LOL

[sharp jacinth]

Thx

May i close?

[honest gust]

<@&268886789983436800> he is copying my method

[honest gust]

sure

[sharp jacinth]

.close

[final saddle]

Channel closed

Closed by @sharp jacinth

Use .reopen if this was a mistake.

[opaque ember]

wdym?

[honest gust]

someone posted scam images

[opaque ember]

oh i banned em already

[honest gust]

https://cdn.discordapp.com/attachments/268882317391429632/1495736762268520568/d70c5dceba3e23da.gif?ex=69e754e9&is=69e60369&hm=ed081dfc6262b2be41e6b89c5d138c43af50fbf5b29469120823d5c045bef985&

looked like this

[final saddle]

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
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[gritty flume]

Number f

Erm for the cos one, is the whole fraction negative

[vital crag]

what does "the cos one" refer to

[gritty flume]

The cos b part of the question

[vital crag]

right -12/13 is negative

[gritty flume]

No like is the whole fraction negative

Does the sign affect the whole thing?

[vital crag]

$-\frac{12}{13} = \frac{-12}{13} = \frac{12}{-13}$ if that's what you're asking

[soft zealot]

riemann

[gritty flume]

Yes that

Tysm

[vital crag]

yea asking with math is more clear

[gritty flume]

Sorry

I’m just super bad at explaining

Thanks again

.close

[final saddle]

Channel closed

Closed by @gritty flume

Use .reopen if this was a mistake.

[gritty flume]

.reopen

[final saddle]

✅ Original question: #help-36 message

[gritty flume]

Can someone help me w the negatives for questions a

[grizzled tusk]

either you guess or you use trig formulas

like cos(x-y) = cos(x)cos(y) + sin(x)sin(y) gives you an answer for 6

[gritty flume]

Huh

[grizzled tusk]

but you could also notice that for theta = 90°, cos(theta - 90°) = cos(0) = 1, so it's kinda "clear" it's the shift to get sin(theta)

[modest sequoia]

some of these identities may help

[bold turtle]

(the question's in degrees fyi)

[modest sequoia]

well yes but you can convert pi/2 to 90

[gritty flume]

Idk the pi way

[bold turtle]

(the implication being that you can't necessarily assume she's seen thos-

CALLED IT

[bold turtle]

Similarly I'm guessing you haven't seen these? (e.g. addition rules for cosine)

@gritty flume^

[gritty flume]

I’m not sure

I may have seen it but in a different way

I can’t easily read stuff like that

[vital crag]

do you not have notes from class or a book you can look up

[gritty flume]

Kind of

[bold turtle]

(it's more that I would imagine this exercise could appear before those sorts of thing)

[gritty flume]

Ik how to do the one w the degree in frog if the teta

Idk how to do these

[bold turtle]

Well, apply your arithmetic skills to rearrange some stuff

e.g. 90 - theta = -theta + 90

that is, you negate theta, then add 90

(hopefully all the other rules you need are right above that exercise)

[gritty flume]

So I can switch it like this?

[bold turtle]

That one's valid specifically because (theta - 90) = minus (90 - theta), and cos(-x) = cos(+x)

[gritty flume]

The second one is the one Ik how yo solve

I don’t understand that but Ty

[bold turtle]

As in, you couldn't say the same thing if that said sin(...) = sin(...)

[gritty flume]

Could you maybe write it out

I can’t see the picture

[bold turtle]

cosine of a negative value is the same as cosine of the positive value

the negative of theta-90 is 90-theta

[gritty flume]

Oh

[bold turtle]

(cf. how the negative of 3-5 is 5-3)

[gritty flume]

What’s cf

[bold turtle]

Ergo, [the thing you've just written]

[gritty flume]

Ohhh

[bold turtle]

confer, Latin, roughly "compare how/with"

[gritty flume]

Wouldn’t it be the opposite if each other

[bold turtle]

...?

3-5 is -2

[gritty flume]

Yes

[bold turtle]

The negative of -2 is (+)2

[gritty flume]

And 5-3 is 2

[bold turtle]

which is what 5-3 is

[gritty flume]

Ooooo

Wow that’s smart

[bold turtle]

It's a good algebra thing to bear in mind - you'll see that shit crop up everywhere when you realise it

Similarly, but distinctly, sine and tangent work slightly differently -
-sin(x) = sin(-x) [i.e. negative of sine is sine of negative]
-tan(x) = tan(-x) [i.e. negative of tangent is tangent of negative]

These, along with what you've claimed to know, should hopefully get you through that question

[gritty flume]

Can I solve one w what ik now

And show u

[bold turtle]

sure

[gritty flume]

Ty

[bold turtle]

Careful

How'd you get to that?

[gritty flume]

370 changes the uh idk wht it’s called

270

Mb

[bold turtle]

Show the steps you take to get to that equivalence...?

[gritty flume]

yeah I’m confused

Initially I used the formula u gave

That negative makes cos positive

[vital crag]

show waes how you used his formula

[gritty flume]

Oh

Idk how

I just assumed

[vital crag]

you can assume his statement is correct. but you should show how you applied his formulas

[bold turtle]

(I mean, technically she did apply it correctly)

(it's just not really a useful one to apply for this question)

[gritty flume]

Oh ok

[bold turtle]

Since the theta there is already positive, what we should try and do first is manage the -270, because it's a bit massive

[gritty flume]

Oh

[bold turtle]

We can shift cosine by how much before it repeats itself?

[gritty flume]

I don’t understand that

Y teacher said if a number is big I should minus by 360

[bold turtle]

Sure

[gritty flume]

Sorry if it’s contradictory to what u said

I’m not trying to be rude

[bold turtle]

But here it's "big"-ly negative

So instead of subtracting by 360, what could we do instead?

[gritty flume]

I’m not sure

I did 360-270

[bold turtle]

ye

You're adding 360 to it

[gritty flume]

Ohh

[bold turtle]

But that leaves us with a (+)90 in the cosine's argument

[gritty flume]

I thought that makes it bigger

[bold turtle]

ye

So we have cos(x + 90)

Erm, if you've seen the graphs of cosine (and of sine) this might make finishing this off a bit easier...?

[gritty flume]

Cos is negative?

[bold turtle]

?

[gritty flume]

Bc it’ll be on the 2ndquard

[bold turtle]

The CAST diagram isn't gonna really be practical for transformations like these

[gritty flume]

Oh

[bold turtle]

not least of which because the values are on the axes themselves, and not in any quadrant

[gritty flume]

Aww

[bold turtle]

you familiar with these graphs?

[gritty flume]

Yes

I not good at them tho

[bold turtle]

Right - so we can try and deal with a graph transformation here

we want to handle (y = ) cos(theta + 90)

It's gonna look like cos(theta) i.e. the blue graph, but we're going to shift it in which direction, and by how much?

[gritty flume]

Erm the left

I think

[bold turtle]

ye...

by how much?

[gritty flume]

90

[bold turtle]

yeee

[gritty flume]

I have to relearn this

[bold turtle]

You'll get there

[gritty flume]

Ty

[bold turtle]

If we do shift that graph left 90 degrees, we thus get this thing:

Now, in a sense, that kinda looks like sine ig?

Except...?

[gritty flume]

Idk..

[bold turtle]

well if I plonk sine on (in orange) here?

It's almost the same but flipped

How can we flip a function upside down?

[gritty flume]

Oo

[gritty flume]

Idk

[bold turtle]

(think of a graph transformation)

(can you recall what y = -f(x) does...?)

[gritty flume]

Ik it flips

I just don’t know which direction

[bold turtle]

up-down flip

[gritty flume]

I think there are two flip formula

[bold turtle]

because you're taking the y-value from f(x) and then negating it

So you're flipping the y-values, meaning your graph is flipping upside down

[gritty flume]

Oh

[bold turtle]

Given that - can you figure out what this blue graph should be then?

[gritty flume]

-x?

F(-x)

[bold turtle]

If the orange one is sin(x), then...?

(negate the whole function)

[gritty flume]

-sin

[bold turtle]

ye

So our blue function (which remember is the one we wanted to find) is -sin(theta)

[gritty flume]

YES

[bold turtle]

That is,
cos(theta - 270) = -sin(theta)

okay let's try another one just to get a sure footing - say (9)?

tan(720 - theta)

Walk me through it

[spice flicker]

I don't think she's online

[gritty flume]

I am

I’m suppose to be off dc but I need help w math

[spice flicker]

OK nevermind

what grade is this math

[gritty flume]

I can’t give a correct answer for deducting the 720

I got 90

[bold turtle]

Well what numbers can we deduct?

[gritty flume]

360 and 90

[bold turtle]

We can't strictly deduct 90

[gritty flume]

Idk but I’m in grade 11

[gritty flume]

Yes

[bold turtle]

We can deduct 360s, though, that's not a problem

Notice anything about 360 and 720 tho?

[gritty flume]

It’s in 360

Likeadoubke

[bold turtle]

ye

So we can just do away with that 720

[gritty flume]

So we have 360?

[bold turtle]

...yh

Though again that's another 360 we can remove

[gritty flume]

Uh

[spice flicker]

sorry accident

[bold turtle]

@spice flicker did you really 💀 "grade 11"?

[bold turtle]

So far we've said that tan(720 -theta) = tan(360 -theta)

[spice flicker]

im not up to that grade bro

[gritty flume]

Oh

[gritty flume]

Yes

[bold turtle]

Right, but then that's irrelevant to the discussion

[bold turtle]

and then for the same reason, we can just deduct 360 again

(careful with the sign - what then do we have inside the tan()?)

[gritty flume]

Wdym

Won’t that make it 0

[bold turtle]

careful

We have 360-theta inside our tan()

[gritty flume]

Oh

[bold turtle]

we've still got something left over when we subtract the 360: ...?

[gritty flume]

360 from 720?

[bold turtle]

okay so

tan(720 - theta) = tan(360 - theta), that bit's done

But we can subtract 360 again from the tan's argument there

[gritty flume]

Yes

I so sorry it’s my idk how to say it but it’s my bed time

[bold turtle]

ah no worries

[gritty flume]

Tysm

[bold turtle]

(if you like, .close it, and pick it up later )

[gritty flume]

Yeah

.closr

Omds

.close

[final saddle]

Channel closed

Closed by @gritty flume

Use .reopen if this was a mistake.

[final saddle]

Available help channel!

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• Show your work, and if possible, explain where you are stuck.
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[final saddle]

@warped herald Has your question been resolved?

[edgy mauve]

Hey, which part do you need help with? What have you tried?

[final saddle]

Channel closed

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Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[spiral abyss]

Guys

[spiral abyss]

pls help me

[orchid coral]

!D2DK

[spiral abyss]

huh?

[orchid coral]

Don't ask to ask

[spiral abyss]

oh

[tropic crest]

!da2a

[final saddle]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[spiral abyss]

Guys, I'm having a problem with a math equation. If someone can help me, I appreciate it!
Problem:
A = (1-(1/1+2))(1-(1/1+2+3))...(1-(1/1+2+3+...+2021))

[orchid coral]

Use the 1 + 2 + ..... + n = n(n+1)/2 formula, and then manipulate the general term into a telescoping product.

[final saddle]

@spiral abyss Has your question been resolved?

[spiral abyss]

Yes

[final saddle]

Channel closed

Closed by @spiral abyss

Use .reopen if this was a mistake.

Available help channel!

Send your question here to claim the channel.

Remember:
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• Show your work, and if possible, explain where you are stuck.
• Do not immediately ping people or roles. After 15 minutes, feel free to ping <@&286206848099549185> once.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

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[left trail]

Any suggestions on the substitution for 28d? I know how to get it in the form of 28c. But I don't see what substitution we could make to have the bounds and the integrand match 28d

[gritty chasm]

28d is kinda hidden

[left trail]

Oh sorry one second

[scarlet sequoia]

So "Express $\int_c^x \frac{e^{2t}}{t-1}, dt$ in terms of $Li$, where $c = 1 + \frac 12 \log 2$"?

[soft zealot]

Rafilouyear2026

[scarlet sequoia]

Is what I could read

[scarlet sequoia]

Also talking about 28c, what is the constant b you found

[left trail]

Log2

I feel as though there is a connection between the two two parts but I don't see it

[hasty inlet]

Hello everyone im new here

[scarlet sequoia]

How about trying to find a change of variable

[left trail]

You mean a u sub from part c to d?

[scarlet sequoia]

If you have a math question, open an available channel of your own and state your problem there, follow #❓how-to-get-help for more info. Otherwise go to #discussion if you just wanna chat

[scarlet sequoia]

well rather part d to c

Starting from the integral given in d, we need to go back to something of the form c

[left trail]

Interesting. Let me see what I can do