help-36

[wary helm]

16+9=25, now you take the sqrt and you get 5

In the end, for the first option you get 7=5, which is obviously not correct, so the first option is not a correct answer

Now you need to repeat the same for the other answers

Btw this procedure is exactly what you used to find that the third option was correct

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@brisk bane Has your question been resolved?

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[ancient violet]

does anyone know what should be the approach?

[ancient violet]

what i did was

a<b<c<d<e

so

c - (a+b+c+d+e)/5

[ancient violet]

we want maximum of that

i’m stuck on it

the solution is very not solution

this is what i mean

this is one of exam olympiad question

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@ancient violet Has your question been resolved?

[final saddle]

@ancient violet Has your question been resolved?

[smoky shore]

i think the best way to approach this solution is to first leverage it on the assumption that the median has to be either 100 or 0 to get the maximum difference between the median and mean, because that maximizes the range of the dataset and the difference between the median and one lower/upper bound

[ancient violet]

yes the solution is bit intuitive, but how do i prove it algebraically

i get your point

[smoky shore]

i mean, after the assumption that the median has to be 0 or 100, the rest is proved easily. but i'm not sure there would be a straightforward way of proving algebraically that the median has to be 0 or 100 (at least, i'm not coming up with any)

The best approach I can think of is:
-Isolate the lowest three values: a,b, and c. c is our median, and a<b<c.
-We know that the maximum difference between mean and median will scale proportionally with the size of our range, so a=0 and e=100, utilizing the maximum possible range (e is the largest value from the dataset). this means our mean must be between 20 and 80.
-Then, the maximum possible difference between the mean and median will be achieved if and only if b=c,d=e or b=a,d=c (giving the highest weights to the mean). Using b=c and d=e, our dataset is now [0,c,c,100,100].
-The difference between mean and median with respect to c can now be represented by the linear expression c-(2c+200)/5. Simplifying gives us 3c/5-40. In this linear equation, the extrema are at c=100 and c=0, so the range of possible differences is from 3(100)/5-40=20 to 3(0)/5-40=-40.
-Because -40 has the greatest magnitude, we can conclude 40 is the greatest possible difference between mean and median.

[smoky shore]

It's not the most clear or intuitive solution (nor is it that well written) but i think it gets the solution concisely and, more importantly, algebraically

[tiny gorge]

a perturbation argument might work...

suppose you have five sorted numbers a,b,c,d,e

and suppose c is not 100

then I can increase the median by delta if I increase c by delta (and if necessary, increase d and e to keep the numbers sorted; at most you would have to increase them by delta)

and that only increases the mean by at most (3/5)delta

so it increases the difference

similar argument if c is not zero, you can make it smaller (decrease a and b if necessary to maintain the sorting) and thereby increase the difference
therefore the best solution must have either c=0 (and hence a=b=0), or c=100 (and hence d=e=100)

[shell hare]

Yo can I get help

For b(I) and b(ii)

[tiny gorge]

#❓how-to-get-help

[shell hare]

alr

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@ancient violet Has your question been resolved?

[ancient violet]

ohhh

i understand it now ig

thanks @smoky shore and @tiny gorge for your remarks

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[lean dawn]

How did the third equation get to the last one

Oh nvm i got it

Lol

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[tranquil pine]

I have to determine the pairs (n,m) for which f is holomorphic

[tired walrus]

throw cauchy riemann equations at it?

[tranquil pine]

This is what i have done so far

but i dont now how to continue further

[tired walrus]

you want $nx^{n-1} = my^{m-1}$ to hold for all $x, y \in \bR$

[soft zealot]

Ann

[tranquil pine]

yes

[tired walrus]

this means that the exponents on x and y have to be zero

otherwise one side is not constant wrt x while the other is, or vice versa

[tranquil pine]

u can do that???

[tired walrus]

of course you can reason like this. why couldn't you?

[tranquil pine]

sry I'm studying physics and I always think that something complicated has to come out of something like that

Thankss

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[lament mirage]

What's the highest power of 18 that divides 672! . I thought about using this formula but I think it's only for P prime

[sturdy cypress]

it's half of that

for p = 3

[final saddle]

@lament mirage Has your question been resolved?

[lament mirage]

I don't understand

with p=3, α=54

with 18 α = 27?

[sturdy cypress]

yeah that's what i meant

[lament mirage]

but why

[sturdy cypress]

how did you get 54 though

I get 332

[lament mirage]

A friend sent me that, sorry. got the same

[sturdy cypress]

so that's how many 3s the factorial has

you need two 3s

for one 18

[lament mirage]

I don't understand

sry

[sturdy cypress]

no problem i don't know how to explain though

like the factorial of 5 is (2)(3)(2×2)(5)

it can be divided by 2³

or 6^1, or 5^1

there is enough 2s to make three 2s

there's only enough 3s to make 1 six

the formula tells you how many of the prime there is

and then 18 requires two 3s and one 2

if there is enough 3s there's def enough 2s, so we focus on 3s

[lament mirage]

Ok, got that part

[sturdy cypress]

ok

[mortal field]

but how you get the 18?

[lament mirage]

but how do you know it's the half? it's like a pattern?

[sturdy cypress]

I know I need two 3s per 18

because it's 2×3×3

[mortal field]

yeah

[lament mirage]

I think i got it. If you need 332 for 672! and every 2 (3) you get 18, then you need 332 (3) for 166 (18) ?

[sturdy cypress]

i guess

i don't know how to solve it in general, if it was something more complicated than 2×3×3

here it's obvious that you can make 18 out of two 3s

so half the exponent of 3 in the factorization of the factorial

[lament mirage]

Ok. I understand it know. Tysm

It was kinda difficult to figure it out by myself

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[formal flicker]

If $A\in M_n(\mathbb{C})$, prove that there is an upper triangular matrix $T$ and an invertible matrix $P$ such that $P^-1AP=T.$

[soft zealot]

ohNoiAmHere

[formal flicker]

I was thinking to do some sort of induction or maybe try 'n mimic a proof of the spectral theorem

but i just don't know how to start it

any tips?

[gleaming bramble]

Yes induction

Use induction on size of A_n*n

[final saddle]

@formal flicker Has your question been resolved?

[gleaming bramble]

Let x = v bar , v / ||v bar , v||
Take u = x - e_1
Let Q be the householder matrix associated with U

[formal flicker]

yea i think i can come up with something

[gleaming bramble]

By above
X is the first column of Q
And thus Q is hermitian and unitary

Ok if u need further guide I can pm u the whole thing once I’m done

[formal flicker]

oki thanks

i was just struggling with the start

which happens all too often

once i get into it usually it works out

[gleaming bramble]

The first sentence here
The denom is “ || v bar , v || “

Zzz

Double lines

| | v bar , v | |

[formal flicker]

norm

right

[gleaming bramble]

Ok

[formal flicker]

wait so i showed that the eigenvalues of A are in the diagonal of the upper triangular matrix

that is correct right?

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[wraith crater]

What’s the formula for the area of a rectangle?

[grave tundra]

ye gimme 1 sec to send what ive done so far

[lean dawn]

Oops

[lean dawn]

2x^2 +x -3

2x^2 +x -3 = 30

2x^2 +x -33=0

[lean dawn]

And u put it in the formular

[grave tundra]

ye

but which one is it is what im stuck on

[lean dawn]

Wdym

[wraith crater]

Should be -4.31971…

[grave tundra]

usually its one minus and one plus but both are positive

[lean dawn]

Ye

[grave tundra]

alr lemme try

[wraith crater]

,w solve 2x^2 -x -33

[lean dawn]

And the one that is positive is the right one

[soft zealot]

Results provided by WolframAlpha

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

Requested by Pure#9389

[final tangle]

+x

[grave tundra]

it was wrong when i rounded it

and it gave me a new question D;

[final tangle]

you had an issue with signs

[lean dawn]

D:

[final tangle]

doesn't hurt to solve this one properly

[grave tundra]

this is the new one i just got

idk why this website is so shit

[lean dawn]

Lol

[final tangle]

initial approach is alright, just be careful with signs this time

[grave tundra]

ok gimme a minute to work it out

so its just 5.1?

since u round to nearest 1 dp and thats the positive one

ok it was right

[final tangle]

,W (2x+7)(x-3) where x= 5.141

[grave tundra]

ty guys

cya

uh actually, if i have another question can i just ask it here

cos this next one is harder than the previous

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[ornate sun]

I understand if given a point that I can substitute f(a) and a in this equation. How do I manage this if not given any point?

[ornate sun]

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[steady trout]

question 23

can someone help with what i need to do in the beginning so i can solve it?

[spiral quail]

Can u plz translate into English?

[steady trout]

The ratio of the large base to the small base of a trapezoid is p. The ratio of the largest upright side to the smallest upright side is q. If one draws the two diagonals, then the ratio A/B of the area of the two triangles is which do not certain the bases (see figure) equal to

[spiral quail]

Thank you!

What does this part mean “which do net certain the bases (see figure) equal to”?

[steady trout]

not*

i misstyped my bad

u understand the question?

[spiral quail]

Yes. Just having trouble solving it lol

[steady trout]

I thought if we use q = 1 we have a isosceles trapezoid so A = B

[spiral quail]

Idk sorry

<@&286206848099549185>

[steady trout]

i feel like that doesnt really help

they never answer

[civic sparrow]

the ratio should be 1

[steady trout]

can u justify ur answer?

[civic sparrow]

first off, what can you say about the altitude of A and B?

[steady trout]

idk

[civic sparrow]

draw the altitudes for them and see what it could be

[steady trout]

uh i really dont know

[civic sparrow]

if you look at this image, the two altitudes would be h1 and h2, what can you say about them?

[steady trout]

idk how to say it in english

[steady trout]

is this q?

[civic sparrow]

there's no need to involve q or p into this

[steady trout]

oh

[civic sparrow]

it turns out that both h1 and h2 are the same

[steady trout]

yeah

[civic sparrow]

so using that info and the base DC, what can you say about the areas of ADC and BCD?

[steady trout]

they are = ?

[civic sparrow]

yep, and we can see they have an overlap area of DPC, right?

[steady trout]

yeah

[civic sparrow]

so what can you say about the areas of APD and BPC?

[steady trout]

equal too? idk sorry

[civic sparrow]

yep, since area(ADC) = area(BCD), and they have DPC in common, we can see that area(ADC)-area(DPC) = area(APD) and area(BCD)-area(DPC) = area(BPC) and so area(APD) = area(BPC)

so going back to your question, what does that mean for triangle A and triangle B?

[steady trout]

that they are equal too so the ratio of A/B is 1?

[civic sparrow]

there you go

note that the only property we used in this entire thing is that the top and bottom bases are parallel, which is one of the properties of all trapezoids, so this will apply to any trapezoid

[steady trout]

should i give the corners a name on the figure?

[civic sparrow]

if you need to prove it, then yeah it could be helpful to give them names

[steady trout]

yep thanks for your help!

have a nice day

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[mellow fox]

I dont know the steps to solving

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[unique tapir]

do you know calc?

[unique tapir]

this answer key doesnt make snese to me

because my first though would be to use product rule but that answer doesnt make sense to me

[fossil geyser]

That's the answer key? Or that's your work?

If that's the answer key, it's wrong

[unique tapir]

answer key

i thought so

[fossil geyser]

Using the product rule is right, which is what they tried to do here, but incorrectly

[unique tapir]

can i show you my work to check real quick?

[fossil geyser]

sure

[unique tapir]

ok give me a sec sorry dont wait up for me if you have others to help

[fossil geyser]

It's all good I'll see it

[unique tapir]

what do i do next?

question l btw

[fossil geyser]

Looks good

[unique tapir]

photomath gives me some weird shit

[fossil geyser]

The goal is just to find the derivative?

[unique tapir]

ye

[fossil geyser]

Then I'd say you're done

I mean, you could distribute the 5, or factor out a sec(theta)

But neither form is better than what you've already got

[unique tapir]

alright thanks

[fossil geyser]

Sure thing

[unique tapir]

im all good now

[fossil geyser]

You can use .close to close the channel if you want

Otherwise a bot will bug you about it later lol

[unique tapir]

.close

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[worthy briar]

I don't even know how to start here... seems like there is missing some important information

[dusty harbor]

it might help to know that the sample is hypergeometrically distributed

since its sampling without replacement

[worthy briar]

I've never heard about that...
But lets assume i'd know what that is, what would the parameter K be in my case?

[dusty harbor]

alright so

this is an example from wikipedia

wait are youre x_i's bernoulli?

i might have made a mistake

[worthy briar]

idk maybe...

[dusty harbor]

oh theyre definitely not actually my bad

subset of $\real$

[soft zealot]

Gijs

[worthy briar]

Is there a simple definition of "one dimensional marginal distribution"?

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@worthy briar Has your question been resolved?

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@worthy briar Has your question been resolved?

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@worthy briar Has your question been resolved?

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[quaint crescent]

Hello, i just need clarification on quadratic equations, any trick so i cant forget the quadratic formula?

[summer vale]

aye aye

wassup

practice

secret formula

is that

or

wanna listen

to a song?

I got a catchy one

[quaint crescent]

No?

[summer vale]

sorta

ah then ye

practice

[quaint crescent]

Oh ok

[spare mirage]

https://www.youtube.com/watch?v=E2eVZFy9yzk

[quaint crescent]

Woah

[summer vale]

^

ye

I was gonna

recc that

[quaint crescent]

Thanks a lot

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[pastel roost]

can somene tell me if I'm missing something. is this factor incorrect?

[grim badger]

It's correct

Why do you think it's wrong?

[pastel roost]

because -2 * 4 = -8

but -2 + 4 does not equal -2

[grim badger]

Why are you doing -2 + 4?

[pastel roost]

to factor

find 2 number that multiply to be -8 and add to be -2

[grim badger]

When you expand, it's 3x * -2 + 4 * x = -2x

[pastel roost]

what is the $?

[grim badger]

Supposed to be x, opps

[pastel roost]

and how did a -2x get on the other side

[grim badger]

By simplifying the left side

https://youtube.com/watch?v=4v-EQIxqpMQ

[pastel roost]

oh I see, is their a different way of facotirng when a does not equal 0?

[grim badger]

Here's a good video to watch for factoring when a is not 1

[pastel roost]

ok ty

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[fickle remnant]

so if the denominator is neg and im trying to equal equations on both side would i need to m8ultiply by a neg denominator both sides?

[fickle remnant]

example 2= 2/-4

[fathom walrus]

Show problem

Well that’s just not true lol

U mean like x = 2/-4

[fickle remnant]

yea

[fathom walrus]

Actually here’s a better example

x/-6 = 3

To solve for x, multiply both sides by -6

[fickle remnant]

so yea, muti by a neg denominator right

[fathom walrus]

Ye

Or u could just multi by pos denom then in the next step, multi both sides by -1 but that’s the same thing

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[eager finch]

are tangent lines intersecting?

[summer vale]

aye aye

wassup

wdym

[eager finch]

helloooo

[summer vale]

ayeee

wassup

[eager finch]

not much

correcting some hw

[summer vale]

alr

cools

so

what’s the qn

[eager finch]

so if two parabolas are touching at their vertexes
are they considered intersecting

it's another question but I was just wondering

[summer vale]

ye they’re touching

at one point

ask intersect at one

point

ye

[round fern]

Yo

I need help!!!!!!!

[summer vale]

wassup

well

this channel

sadly taken

maybe

try

#help-0

[eager finch]

I can close it and it'll reopen

or you can use another free channel

thanks @summer vale

.close

[final saddle]

Channel closed

Closed by @eager finch

Use .reopen if this was a mistake.

[summer vale]

npnp

[final saddle]

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[nimble frost]

ik this is right but why? according to discrete math rules

[final saddle]

@nimble frost Has your question been resolved?

[final saddle]

@nimble frost Has your question been resolved?

[final saddle]

@nimble frost Has your question been resolved?

[muted prairie]

https://en.m.wikipedia.org/wiki/Hypothetical_syllogism

[final saddle]

@nimble frost Has your question been resolved?

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[tranquil pine]

Whats the asymptote of the blue function?

[south cosmos]

what are your thoughts on it?

[tranquil pine]

sth. like -0.5x+0.5

[true escarp]

.

do you know what a asymptote is

[final tangle]

didn't you get the equation of the graph earlier?

[true escarp]

oh the diagonal one

[tranquil pine]

a line that the function never touches, it can be upside down, left to right and diagonal

[true escarp]

yea

[final tangle]

misonception

[true escarp]

some what

the line can touch it

graph*

[final tangle]

function can touch/intersect with stuff like horizontal and slant asymptotes infinitely many times

anyway

didn't you get the equation of the graph earlier?

[tranquil pine]

yes but I didn't got the explanation of the equation

[final tangle]

were you told the equation in the book?

[tranquil pine]

What would be a function that touches an asymptote 10 times?

[tranquil pine]

no from a forum

[final tangle]

,w graph sin(x)/x

[soft zealot]

Results provided by WolframAlpha

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

Requested by ramonov#0921

[tranquil pine]

whats the asymptote here?

[final tangle]

has an asymptote of y=0 but the graph crosses the x-axis infinitenly many times

[tranquil pine]

got it

apparently the asymptote is -0.75x + 0.5

[true escarp]

yeah it is

its easy to find with polynomial division

if u do polynomial division on f(x)

youll get some linear function + remainder

and the linear function will be the asymptote

works for this where the degree of the denominator is one less than the numerator one

[tranquil pine]

whats with the rest?

\polyset{%
style=C,
delims={\big(}{\big)},
div=:
}
[
\polylongdiv{6x^2-7x+6}{4-8x}
]

[true escarp]

whats with what rest

[soft zealot]

/\

[true escarp]

you leave the remainder

the asymptote will be the quotient without remainder

so see how the linear function formed by the first two terms there

[true escarp]

is this

just with fractions

[tranquil pine]

and how do I get from -0.75x + 0.5

plus the informations that f(0) = 1.5
and 1 vertical asymptote at y = 0.5

to this:

[true escarp]

what have you managed to gather from that info

[tranquil pine]

I don't understand that function

wait I know that that the demoninator for x = 0.5 has to be 0

[true escarp]

yeah

[tranquil pine]

and the function has to be f(0) = 1.5

6/4 = 1.5

or 3/2 = 1.5

but why does this function not work?

[true escarp]

the line isnt an asymptote

[smoky shore]

because it has to tend to -0.75x+0.5 as x tends to infinity

[tranquil pine]

so how to I get from

to this

[smoky shore]

what is this

oh nvm

[tranquil pine]

two informations
f(0) = 1.5
and vertical asymptote at y = 0.5

third information is -0.75x+0.5

[final tangle]

considering

its relatively the intended asymptote of the blue graph will be y= -3x/4 + 0.5

[smoky shore]

the greatest power terms on the numerator and denominator of this function have to have a ratio of -3x/4

[tranquil pine]

now they have, but why -7x ?

[final tangle]

<@&268886789983436800>

from drawing the asymptote the slant asymptote on the graph, it would be safe to assume that the slant asymptote will be $y= -\frac{3x}{4} + 0.5$ \
from the vertical asymptote, you function can be expressed in the form
$$y = -\frac{3x}{4} + 0.5 + \frac{c}{2x - 1}$$
from the $y$ intercept of 1.5, the value of $c$ can be determine to be $-1$, hence
$$y = -\frac{3x}{4} + 0.5 - \frac{1}{2x - 1}$$

[soft zealot]

ℝamonov

[final tangle]

its fine to just leave this as it is

combining these three terms into a single fraction will result in what they had

[tranquil pine]

no its not fine, i need to know how to get from this

to this

[final tangle]

combining these three terms into a single fraction will result in what they had

[true escarp]

very simple

[final tangle]

basic fraction addition, lcd etc

[tranquil pine]

i see

[smoky shore]

yeah, i can confirm that these terms combine to get the f(x) there.

seems to be about it for the blue function

[tranquil pine]

now I understand, thanks

.close

[final saddle]

Channel closed

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[tranquil pine]

how do I do this

[tranquil pine]

i first tried using expansion fo the cubic on the top

and I got (sinx + cosx) ( 1 - sinxcosx) all over sinx + cosx

but that's no tthe answer

i need to prove the identity

[final saddle]

@tranquil pine Has your question been resolved?

[inland bison]

That is indeed the answer...

Assuming that sinx+cosx!=0

[tranquil pine]

im trying to prove the identity

how can I prove this?

[inland bison]

Yes, i know, $\frac{(\sin x +\cos x)(1-\sin x\cos x)}{\sin x +\cos x}=1-\sin x +\cos x$ is true for $\sin x + \cos x\neq 0$

[soft zealot]

waler

[inland bison]

You were heading in the right path home, but you suddenly decided to stop right at the door of your house and ran away

[tranquil pine]

ok

look

it becomes 1 - (1-sinxcosx)/(sinx + cosx)

right

what does the left side simplify to

[inland bison]

No, of course not

[tranquil pine]

$ test

what dose the left side simplify to then

[inland bison]

To the right side.....

$\frac{ab}{a}=b$ for $a\neq0$

Where a and b can be any kind of expression in any terms

Not just numbers

[soft zealot]

waler

[tranquil pine]

ok

[inland bison]

Typo there, be careful

[tranquil pine]

but i get rid of the denominator

and the left side of the numerator

im left with 1 + sinxcosx != 1 + sinx + cosx

[inland bison]

Huh?

[inland bison]

Take a really careful look at what im writing here

[inland bison]

Bruh, and a typo

What i meant is 1-sinxcosx

[tranquil pine]

bro ok

i see

[inland bison]

$\frac{(\sin x +\cos x)(1-\sin x\cos x)}{\sin x +\cos x}=1-\sin x \cos x$ is true for $\sin x +\cos x\neq 0$

[tranquil pine]

I get it now

[inland bison]

This is what I meant, excuse me

[tranquil pine]

still wrong

-_-

[inland bison]

Oh wait i fixed the wrong bit

[soft zealot]

waler

[inland bison]

Right, excuse my attempt to latex on phone

[tranquil pine]

wait i used the expansion of squares

is a^3 + b^3 the same as (a+b)^3?

[inland bison]

No

You factor a^3+b^3

And this can be called the sum of cubes identity

[tranquil pine]

ohh

how can I prove cos^4x - sin^2x = cos2x?

[void crest]

idk prob start with the sin^2 +cos^2 =1

[tranquil pine]

where do u get the sin^2x + cos^2x

[inland bison]

Did you mean cos^2(x)?

[void crest]

u prob manipulate a bit

so like

[inland bison]

Because that equation is not true in general

[void crest]

isit idk i always assume its true when ppl said to prove

do u memorise trig stuff T.T

[inland bison]

Well not everything, that's a common identity to remember

[tranquil pine]

yes

how do I do it

or is there no way

i can only manipulate one side of the equation

im not allowed to do both

[inland bison]

Like i said, was there a typo?

If not you can not prove that it is true in general

God my head is not working right

[void crest]

just check with wolfie

,w cos^4x - sin^2x = cos2x

[soft zealot]

Results provided by WolframAlpha

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

Requested by Circle#0025

[void crest]

yea its prob not always true

[inland bison]

It is not, like i said

[tranquil pine]

i don't understand this

it jumped from cos^2x - sin^2x to 2cos^2x - 1

[inland bison]

Ah, indeed so, it is not cos^4(x)

[void crest]

ahh i see where u got ur value from now hahhaha

literal double angle formula

[tranquil pine]

wait i get it

ok it's here on my class notes

[inland bison]

Yes everything looks good

(Note that you wrote sin^4(x) here, and not sin^2(x) like above)

[tranquil pine]

I need help on (cotx - cscx)(cosx+1) = sin(-x)

I got -sin^2x/sinx ultimately

but that equals to 1, not sin(-x)

[inland bison]

a^2/a=a assuming a!=0

[tranquil pine]

where

are u talking about

[inland bison]

Sin^2(x)/sin(x) is sin(x)

[tranquil pine]

bri

bRH

but it's -sinx tho

that's not even

even function is f(-x) = f(x)

not f(-x) = -f(x)

nvm

[inland bison]

There is a - infront of that....

[tranquil pine]

why did this skip to pi(n)

[inland bison]

Are you familiar with the general solutions form of trig equations?

[tranquil pine]

yes

ik only tan does pi

rest is 2pi

[inland bison]

Yes, now for sin(x)=0, this is quite a special case

So technically, our two general solutions is x=2kpi and pi+2kpi, where k is in Z

[tranquil pine]

ok

[inland bison]

But, we can also simplify this, as in, rewrite the set of solutions

Note that 2kpi is just a full rotation around the circle, starting at 0 rad, as for pi+2kpi, its pretty much the same thing but starting from pi rads

So this is technically just us starting from 0 rad and doing half a circle rotation

Which means, we can write it as npi, where n is another parameter in Z

Also note that it does not matter what we use, k, n, a, b, c, just as long as you remember to say they are integers

[tranquil pine]

can u explain how this equals to the bottom?

[inland bison]

sin^2(theta)=1-cos^2(theta)

[final saddle]

@tranquil pine Has your question been resolved?

[tranquil pine]

i don't understand symbolab's explanation

why would 1-cos(2x)/2 be equal to sin^2x

which identity is this

[loud sundial]

Double angle formula and then Pythagorean identity

[tranquil pine]

my brain is dying on this problem

how?

[cerulean bay]

whats csc

cosec?

[green river]

I think so

[tranquil pine]

yes

[tranquil pine]

how else do u write it

[cerulean bay]

i havent seen csc

:/

[green river]

If you use the definitions csc(x) = 1/sin(x), cot(x) = cos(x)/sin(x) and sec(x) = 1/cos(x), you can plug them into the LHS of the equation, rearrange terms and easily get to the RHS

Do you want me to show you the calculations?

[cerulean bay]

i got it

@tranquil pine

make all the terms in sin and cos

u will get ur answer in 1 sec

[green river]

I think the difficulty of this problem is knowing the definitions of csc, cot and sec, but just ask WolframAlpha for the definitions (I did so too)

[tranquil pine]

but how do u write it

[cerulean bay]

wdym

cosec theta

lol?

[tranquil pine]

bruh

[green river]

csc(x) = 1/sin(x)

[cerulean bay]

yes

[tranquil pine]

for A, why does it include pi/2 + pin

why isn't it pi/2 + 2pin AND 3pi/2 + 2pin

[cerulean bay]

pi means its in radians

[tranquil pine]

bro i know

[cerulean bay]

then?

[tranquil pine]

that's not the question

why is it just pi/2 + 2pin

there's another one at 3pi(n)

it didn't include that in part A

[green river]

because that's equivalent to what you said

[cerulean bay]

yes

[tranquil pine]

pi/2 + 2pi(n) is the same as pi/2 + pi(n)?

[cerulean bay]

its hit and try bruh

[green river]

No like every term of the form pi/2 + pi(n) is either expressable as pi/2+2pi(n) or 3pi/2+2pi(n)

[tranquil pine]

ok

For the previous problem

I’m stuck here

[green river]

It's right so far. Now notice the nominator to be (1-cos(x))/sin(x) and the denominator to be (1-cos(x))/cos(x). Then you can just divide the two fractions and arrive at the answer

Sorry that I didn't TeX it, I was too lazy to type backslash and dollar symbols at my mobile keyboard

[tranquil pine]

how do i divide that

idk how to divide fraction fraction

[green river]

Swap nominator and denominator of the second fraction and then multiply

[tranquil pine]

got it ez

EZ

[green river]

nice

[tranquil pine]

how do I make a triangle for this

i need to use this to solve for sin(2x) so i first need to make a triangle

which quadrant do i kniw this will be in

[green river]

The first one because both cos(x)>0 and sin(x)>0

[tranquil pine]

what are these called and where acn I learn this?

[green river]

You know the definition of sin and cos on the unit circle, right? Then you can just picture it

If you want to do trig graphically, always have the picture of the unit circle in mind

But I don't know whether this has a name

[green river]

But if you want to solve this equation for sin(2x), you don't need to picture it. You can just use double angle formula and trigonometric pythagorean theorem

[tranquil pine]

yes

[tranquil pine]

ik

i just need to know where the quadrant is

is there a pattern that I can memorize?

i forgot what it's called

it's like sin > 0 cos > 0 = 1, sin <0 cos <1 = 4

or something

[green river]

sin>0, cos>0 => 1st quadrant

sin>0, cos<0 => 2nd quadrant

sin< 0, cos< 0 => 3rd quadrant

sin<0, cos>0 => 4th quadrant

[tranquil pine]

what's tan?

would that just be 3rd and first?

how do I do this?

[tranquil pine]

what is the sine and cosine in this case?

[green river]

What do you mean?

[tranquil pine]

im stuck on cscx = 5/2

if tanx < 0

like

where's the cosine and sine

there's only 1/sine

where's the cosine

i don't get it

what's this type of problem called so I can go watch Youtube?

[green river]

Do you mean this problem?

[tranquil pine]

yes

solving stuff that's not on the unit circle

[green river]

If $\frac{1}{\sin x} > 0$, then $\sin x > 0$

[soft zealot]

Mathemaddict

[tranquil pine]

ok

so sinx is greather than 0

is cosine x greater than 0?

[green river]

Yes because it's $\frac{1}{7}$, that's greater than zero

[soft zealot]

Mathemaddict

[tranquil pine]

OH

what if it's tanx >0

what do I do

[green river]

Then it's either the first or the third quadrant because tan is sin over cos

But to solve the problem from above, you don't need this

[tranquil pine]

so cscx = 5/2 and tanx >0

it's 1st quadrant

because cos >0 and sin >0

[green river]

Right

[tranquil pine]

what if it's in terms of pi?

the restriction

#help-36 message as shown here

[green river]

It gives you a unique solution. Normally you have infinitely many solutions because of periodicity

[tranquil pine]

how do I draw my triangle

[green river]

Why would you do this with a triangle?

[tranquil pine]

which quadrant is the triangle?

[green river]

third

Because first quadrant is $x \in [0,\frac]$, second is $x \in [

[tranquil pine]

how do I tell??

[soft zealot]

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

[tranquil pine]

im so confused

[green river]

Sorry

Wait, let me TeX this, I'm so bad at typing

[tranquil pine]

ok this is 3rd quadrant but how do i put the points

tanx = 1/5

it becomes sinx/cosx = 1/5

how do I put that on the triangle to get the 3rd side?

[green river]

If you have angles $x \in [0,2\pi]$, then the first quadrant is $x \in [0, \frac{\pi}{2}]$, the second is $x \in [\frac{\pi}{2}, \pi]$, third is $x \in [\pi, \frac{3\pi}{2}],$ and the fourth is $x \in [\frac{3\pi}{2}, 2\pi]$

[tranquil pine]

bruh lol

[green river]

Sorry, I always accidentally hit enter

[tranquil pine]

ohhhh

i see

so in 3rd quadrant how do I put in the side lengths

[green river]

You have $\sin^2 x + \cos^2 x = 1$ and $\cos x = 5\sin x$. You can plug in the second equation into the first one to get the sine.

[soft zealot]

Mathemaddict

Mathemaddict

[tranquil pine]

wdym 2nd equation

where's the 2nd equation

[green river]

I mean $\cos x = 5\sin x$. You get it from $\cot x = 5$.

[soft zealot]

Mathemaddict

[tranquil pine]

oh

wait that doesn't make sense

i got 26sin^2x = 1

[green river]

And where's the problem? It's correct

[tranquil pine]

do i need to do one for cosine too?

[green river]

If you have $26 \sin^2 x = 1$, then you have the sine, you can get the cosine by taking 5 times the sine, and then you can just use double angle formula for getting $\sin 2x$ and $\cos 2x$.

[soft zealot]

Mathemaddict

[green river]

I think you should try it yourself and I should not give away a complete solution

[tranquil pine]

How tf did I get 10…

That’s not the answer 😭

[green river]

Weird, I don't see the mistake

Oh, I didn't write anything up, but it might be because you did not consider the fact that we are in the third quadrant, so sine and cosine are both negative

[tranquil pine]

Oh

[green river]

Anyways, I don't have time now, need to leave

[tranquil pine]

thank you

you were very helpful

without you i'd have failed my retake

[final saddle]

@tranquil pine Has your question been resolved?

Channel closed

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[final saddle]

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[summer vale]

eh

are u

just suppose

to minus then

them

subtract

and simplify

[zinc dune]

Yeah

[summer vale]

ah

[zinc dune]

Don't know why, or how

[summer vale]

nice nice

alr

have u

started

like an attempt

or nah

[zinc dune]

nah

[summer vale]

alr so

ig u just subtract them

u get

for starters

I THINK

$-x^2 +3x -(2x^3 -\frac{1}{4}x^2)$

[soft zealot]

Springsskateboard

[summer vale]

then

u just gotta

expand the bracket

and simplify

ye

[zinc dune]

Yeah, I think I can do that

[summer vale]

nice nice

see u got this

[zinc dune]

but how'd you get this is what i'm confused about

[summer vale]

so

I just took the first polynomial

and subtracted it using the

second one

they didn’t exactly

specify

so ye

[zinc dune]

I see okay

Thank you

[summer vale]

npnp

[final saddle]

@zinc dune Has your question been resolved?

[final saddle]

Channel closed

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[tranquil pine]

hello, can someone help with this? I've been learning about it but this problem is kinda different from the other ones I have completed and I'm a little confused on how to start it

[tranquil pine]

nvm i got it

.close

[final saddle]

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[patent flame]

Where did I go wrong here?

[patent flame]

Brb going to bathroom

[fathom walrus]

at the start

where u got dy/du

check the sign of that derivative right there

[final saddle]

@patent flame Has your question been resolved?

[patent flame]

Oh

[final saddle]

@patent flame Has your question been resolved?

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

How do I show that this converges?

[unique jolt]

I'm guessing that it's with monotone-bounded, but idk how to show that it's monotone

a1=1/2

obviously its bounded but idk how to get it to be monotone

[strange shore]

study this :
$$f(x) = \sin\left(\frac{x\pi}{3} + \frac{\pi}{6} \right)$$

[soft zealot]

Ｈｅｒｅｌｓ

[strange shore]

derivative, variation, blabla

[unique jolt]

huh

[strange shore]

?

well, I forgot that this thing doesnt work for non-explicit sequence

[unique jolt]

I guess I can use monotonicity of sin and prove it by induction

sin monotone on x in [0,pi/2]

But that feels like cheating, since we haven’t defined trig functions yet

Any ideas outside of that?

[final saddle]

@unique jolt Has your question been resolved?

[unique jolt]

.close

[final saddle]

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[fathom walrus]

Does anyone have any good resources to understand vector spaces? Just reached that section in the textbook and not rlly understanding much

[sweet summit]

if I recall correctly a vector space just needs to satisfy scalar multiplication and vector addiction, yeah? And they have to be closed under those operations.

So for example, you would need to be able to do a in V, b in V, and show that a + b in V and for a in V, then ca in V for some real number c

[final saddle]

@fathom walrus Has your question been resolved?

[fathom walrus]

Hmm, that’s what I didn’t rlly understand. “a in V” and “b in V” and the other highlighted texts

It also had some 10 rules for vector spaces or smth that was kinda confusing

[sweet summit]

the 10 rules stem from those 2 properties I believe:

[fathom walrus]

Yep that’s it

[sweet summit]

yeah so they just stem from those addition and scalar multiplication properties

So let's take an example here

let's say we set our space to be the following:

$V = {<x, y> : x \geq 0, y\geq 0}$

[soft zealot]

MellowDramaLlama

[sweet summit]

here <x, y> is a 2 by 1 matrix like so:

| x |
| y |

so would this be a vector space?

well no

we can tell because scalar multiplicaton is not closed for all c in the reals

since, let's say that c= -1, then our vector would be <-x, -y> ,but x and y have to be positive or 0

[fathom walrus]

Wait a second, sorry to interrupt but I haven’t don’t anything with vectors being matrixes

We just started the vector unit

[sweet summit]

well a vector is a 2 x 1 matrix 🙂

[fathom walrus]

I see

Interesting

[sweet summit]

or it can be a 3 x 1

some n x 1

[fathom walrus]

O

[sweet summit]

but 2 x 1 is easy to play with

[fathom walrus]

So what differentiates a vector from a vector space

Is a vector space like a plane or smth

Where many vectors can exist

Or am I totally off

[sweet summit]

so a vector space is where vectors can "live" if that makes sense

[fathom walrus]

So like a 2d plane?

[rigid scaffold]

Your intuition is good

[sweet summit]

yep!

[fathom walrus]

Ah i c

Coolio

[sweet summit]

it's not as bad as it seems huh? The vector space will be defined in some way, and you have to ensure that it's closed under vector addition and scalar multiplication

oh and it must contain the 0 vector

[fathom walrus]

<0,0> ?

And im not trying to ask too much, but wdym by defined and closed

[sweet summit]

or <0, 0, .... , 0> depending on how big your vectors are 🙂

[fathom walrus]

Got it

[sweet summit]

so closed means that you do something to two vectors and that something is still a vector

for example

the integers are closed under addition

meaning

if you take two integers, say a and b and you add them together

[final saddle]

Channel closed

Closed due to timeout

Use .reopen if this was a mistake.

[sweet summit]

....okay lol

[fathom walrus]

.reopen

[final saddle]

✅

[sweet summit]

do you get a non-integer or an integer?

[fathom walrus]

Integer

[sweet summit]

bingo!

so integers are closed under addition

there's no way to add two integers and get a non-integer

[fathom walrus]

So what would be a case in which u would do something to 2 vectors and it’s not a vector anymore?

[sweet summit]

so, let's say for example we had something like this:

$V = {<x, 2> : x \in \mathbb{R}}$

how take two vectors in this space and add them together

what do you get?

[soft zealot]

MellowDramaLlama

[sweet summit]

so something like

$<x_1, 2> + <x_2, 2> = ?$

[soft zealot]

MellowDramaLlama

[fathom walrus]

<x1 + x2, 4>

[sweet summit]

bingo!

now is that in V?

[fathom walrus]

Yes, bc still a vector

[sweet summit]

no

[fathom walrus]

O

[sweet summit]

because vectors of V are of hte form <x, 2>

we can't have 4 in the second row

it has to be 2

so this isn't a vector space since we can't add two vectors and be closed under addition in V

[fathom walrus]

Wait a sec

[fathom walrus]

Then what abt this

There’s a 4 in the second row

[sweet summit]

yep! And a 4 is not a 2

[fathom walrus]

O wait that’s not a vector

[sweet summit]

no it is a vector

it's just not in the space of V

so V is not a vector space

[fathom walrus]

O

[sweet summit]

it's still a space, just not a vector space

[fathom walrus]

Why must this be true

[sweet summit]

because we defined it so here.

[fathom walrus]

Ohh that’s what that meant

[sweet summit]

yep!

[fathom walrus]

I thought that was u accidentally not completing the problem lol, mb

[sweet summit]

yeah no worries

V can be anything that involves playing with vectors

but it just has to hold the addition and scalar multiplication and those to be closed to be a vector space

also, <0,0> is not in V

[fathom walrus]

Because 0 is not 2

[sweet summit]

bingo

now if we had something like

$V = {<x, y> : x,y \in \mathbb{R}}$

[fathom walrus]

If we had <x,y> XER would this be in our vector space

[soft zealot]

MellowDramaLlama

[sweet summit]

Then this would be a vector space

yep exactly

but this is trivial

[fathom walrus]

Because it’s not rlly defined abt anything

It’s too broad?

Just a guess*

[sweet summit]

since this is R^2

[fathom walrus]

R^2 means 2D right

[sweet summit]

yep 🙂

[fathom walrus]

So why isn’t <x,y> 2D

[sweet summit]

aka dimension

[sweet summit]

it is

it spans the entire xy-plane

[fathom walrus]

Ah ok

[fathom walrus]

So then what’s the reason for this

Was it what I said abt being too broad and not specific enough?

[sweet summit]

nah it's not so much broad, it's just trivial because it's a natural vector space to work with. You've been doing it this whole time whether you realize it or not lol

trivial in this case means kind of "oh well duh yeah of course that's a vector space"

[fathom walrus]

Oh gotcha

[sweet summit]

<0, 0> is in it, <x, y> + <a, b> = <x + a, y + b>, and c<x, y> = <cx, cy> for some real number c

[fathom walrus]

That’s why it’s allowed to be a vector space?

The rules have been met?

[sweet summit]

yep! it meets the 10 conditions above

[fathom walrus]

Hmm wow I see

Ok final question then I’ll stop bothering u

[sweet summit]

no worries!

what's up?

[fathom walrus]

So what would be a case in which u would do something to 2 vectors in the Vector space and it’s not a vector anymore?

Or is it solely that vectors will stay vectors, but vector spaces may not stay vector spaces phylactery

[sweet summit]

hmm

good question. I can't think of anything that would kick you out of being a vector from just addition and multiplication

so I'll say no

vectors will stay vectors in their spaces

now if they go through some kind of function/transformation, then yes they can turn from vectors to just plain matricies and change rank

which you'll learn about later

like you can turn a 2 x 1 matrix into a 2 x 2 matrix with a function, but that's outside of the scope for now

[fathom walrus]

I see, yea that sounds pretty interesting

[sweet summit]

which is just a linear combination of vectors at that point