help-27

[little totem]

1+z+z^2+z^3

1+z+z^2+1

no way this is 0 anymore

[spring oasis]

we need multiples of 3 to sum to 0

[little totem]

bingo!

we have to add a number of terms that is a multiple of 3

if we add 8 terms, not 0

if we add 30 terms, thats 0

because every 3 terms cycles around the circle

ok so finally we can recap

we need 2n+1 a multiple of 11

we need n+4 to be 2 more than a multiple of 3

find all n that satisfy this, and that's your answer

the rest of the steps are just modular arithmetic

you good with that?

[spring oasis]

what?

[little totem]

so remember

when we summed from j=0 to 2

we got 0

so the upper limit being 2 works

we can add 3 terms to that and it will still be 0, in other words, upper limit can also be 5

similarly, it can be 8, 11, etc

the upper limit is n+4

so n+4 must be 2, 5, 8, 11, etc

[spring oasis]

can we go through that again

[little totem]

sure

what part

[spring oasis]

the fuck is going on?

0 <=j <= n + 4

[little totem]

yes

[spring oasis]

which is

1 <= j <= n + 5

[little totem]

uhhhh thats not the same thing

oh

wait why are you doing that

[spring oasis]

n +5 = 0 (mod 3)

n = -5 = 1 (mod 3)

[little totem]

ahhh i see

uh

[spring oasis]

n = 3k + 1

[little totem]

clearly thats wrong because

we explicitly demonstrated that when the upper limit is 2

we get the zero that we want

but that must mean n=6 works and it doesn't

[little totem]

this is the problematic step

[spring oasis]

what do i do

[little totem]

you notice that upper limit being 2, 5, 8, etc works

so far so good?

[spring oasis]

yes

[little totem]

so thats 2 mod 3

which is what n+4 should be

oh wait

wait i think youre right i did n-4 instead of n+4 sorry

my bad

yeah its 3k+1

[spring oasis]

z0 + z1 + z2 + z3 + z4 + z5
z0 + z1 + z2 + z0 + z1 + z2

[little totem]

upper limit there is 5, which means n+4 is 5, n is 1

so you were right, 3k+1

[spring oasis]

so what is the answer

n >= 2, n = 3k + 1 with k in Z

ah i am getting confused again

[little totem]

thats the restriction of what n can be based on the sum

but remember we also had a restriction based on the condition

11 must divide 2n+1

you have to combine the two and simplify

[spring oasis]

2(3k+1) = 0 (mod 11)

6k = -2 (mod 11)

6k = 9 (mod 11)

[little totem]

so the nice thing about having one of these conditions mod 3 is you can just check the pattern

[spring oasis]

k = 18 (mod 11)

[little totem]

you dont have to do fancy mod math

watch:
n = 1 mod 3 <- we focus on mod 3
2n+1 = 0 mod 11 <- while checking these values

odd multiples of 11: 11, 33, 55, 77, etc

11 + 22k

2n+1 = 11+22k

n = 5+11k

these n values are:

5, 16, 27, 38, 49, etc

mod 3 they are

2, 1, 0, 2, 1, 0, etc

so we want the 16, the 49, etc

16 + 33k

16 mod 33 are all of your answers

you want to get comfortable with just brute force playing with numbers and experimenting, looking for patterns

dont just rely entirely on symbolic manipulation

ok i gotta go, i hope ive been helpful

[spring oasis]

reminds of crt

[devout snow]

@spring oasis Has your question been resolved?

[spring oasis]

I appreciate the help

@little totem

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[devout snow]

@languid escarp Has your question been resolved?

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[thin fern]

Let $f : \mathbb{R}^n \to \mathbb{R}^m$ be $C^1$. Show that $df$ is constant if and only if $f$ is an affine function.

[woven radish]

Ari

[thin fern]

My idea was to let $df = A$ where $A$ is a constant matrix, and show that $g(x) = f(x) - A(x)$ is constant

[woven radish]

Ari

[thin fern]

Ok I just noticed what I was missing lol

I was struggling also with the converse direction though

If we assume $f(x) = Ax + b$, I'm not really sure where to from there to show that $df$ is constant

[woven radish]

Ari

[devout snow]

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[thin fern]

nvm I think I got it

.close

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[stoic relic]

is this correct

[stoic relic]

correctly simplified

[winter patrol]

no

[stoic relic]

where did i go wrong

[winter patrol]

getting a common denominator, Ur multiplying the wrong things

let's use a simpler example

$$\frac23 - \frac17$$
how would you simplify this?

[woven radish]

ραμOmeganato5

[stoic relic]

uhh

21 below

14-3 / 21

i think is what youre asking

[winter patrol]

() around the (14-3)
when communicating in plain text, but yes

now the issue with what you did,
is that instead of doing that,
you were multiplying the numerator and denominator of the first fraction together etc

here that would be doing 2 * 3
instead of the 2 * 7 = 14 you just did

[stoic relic]

so is it

this maybe

nvm

[winter patrol]

and denominator was also expanded incorrectly as well

[stoic relic]

yeah i just realized that

like a minute ago

[winter patrol]

it's also quite hard to read, just to be clear, is the initial expression
$$\frac{\sqrt{a}}{\sqrt{ae}+e}-\frac{\sqrt{e}}{a+\sqrt{ae}}$$

[woven radish]

ραμOmeganato5

[stoic relic]

its a b but yes, it is

sorry for the ass handwriting

[winter patrol]

in future always post an image of the original if able to

id recommend just leaving the denominator in factored form

focus on getting the correct numerator

[devout snow]

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[stoic relic]

.close

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[devout snow]

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[novel lichen]

I know the answer is x = 0 but how? this is AP Calc AB

[novel lichen]

<@&286206848099549185>

[devout snow]

@novel lichen Has your question been resolved?

[glacial tide]

How do you know this

[novel lichen]

cuz I got it wrong

[glacial tide]

okay

have you heard of candidate's test?

[novel lichen]

yeah

[glacial tide]

so just apply that

that's literally it

[novel lichen]

How would i do that? I just learned it today so I'm kinda sped on it

[glacial tide]

wait I take that back

the way you reason this is just that

from 0 to 6 the function is increasing

so every point after f(0) on that interval is greater than f(0)

[novel lichen]

mhm

[glacial tide]

and then the function decreases again

but not enough to make up for the massive increase it saw earlier

so the later decrease from 6 to 8 won't drop the value lower than what it was at f(0)

since the area under the graph is far below the area above it

[novel lichen]

OHHHHH

WAIT THAT MAKES SENSE

I never thought abouit it like that

alr tyty

[glacial tide]

yeah that's not super rigorous but it'll do for ab

[novel lichen]

alr alr

[glacial tide]

but you can use

ftoc

too

[novel lichen]

what that

[glacial tide]

which is basically what I implicitly did

fundamental theorem of calculus

did u learn that yet?

[novel lichen]

no 😭

I'm on unit 5

of calc

[glacial tide]

oh

all good

[novel lichen]

yeah lol

but what you said makese sene

sense

[glacial tide]

it might make some sense

but the area is what matters

just saying

[novel lichen]

alr

[glacial tide]

if that dip from 6 to 8

was like super super deep

then the function decreases really fast

[novel lichen]

then there would be a min from then right

[glacial tide]

possibly yes

[novel lichen]

icic

alr ty Holo

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[normal bolt]

[
\sin^2(6^\circ) + \sin^2(12^\circ) + \sin^2(18^\circ) + \cdots +
\sin^2(54^\circ) + \cdots + \sin^2(72^\circ) + \sin^2(84^\circ) + \sin^2(90^\circ)
]

[
= \left( \sin^2(6^\circ) + \sin^2(84^\circ) \right)

• \left( \sin^2(12^\circ) + \sin^2(78^\circ) \right)
• \cdots
• \left( , ____ ; + \sin^2(90^\circ) \right)
  ]

[woven radish]

BlackidoZΣ

[hoary pawn]

Can someone help me on BEDMAS

[normal bolt]

i can't identify in the place of blank with sin^2 (90 ) which angle to put there which is multiple of 6 and make pair with 90 degree

[devout snow]

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

Please do not advertise your help channel or thread in other parts of the server. There are many people who need help, so advertising can quickly turn into spam.

[cinder bobcat]

sin^2(0)?

[normal bolt]

yes that will make 1 with sin^2 90

your right

now how will i proceed in next step?

[grand edge]

Try to exploit the pythagorean identity

$\sin^{2}(x) + \cos^{2}(x) = 1$

[normal bolt]

i mean i will count how many ones ill have

[woven radish]

Roy

[normal bolt]

but i need one step back angle from sin^(0) + sin^2 (90)

how do i get that

so that i can count how many ones are there from sin^ (6) to sin^2 (84) to that one step back angle from sin^(0) + sin^2 (90)

[normal bolt]

not clear exactly what you wanna say

[slate peak]

do you really need one for sin^2(90)?

its just (sin(90))^2

just match all the other ones and use the identity for switching sin to cos

no need to pair

[normal bolt]

can you show it

[restive river]

could this help

[normal bolt]

i know that formula cos(90-x) = sinx

[slate peak]

yup

thus, $\sin^2(6)+\sin^2(84)=\sin^2(6)+\cos^2(6)$

[woven radish]

4-aminopyrimidin-2(1H)-one

[slate peak]

repeat for every pair

[normal bolt]

okay got 8

answer

[slate peak]

correct

[devout snow]

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[devout snow]

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[devout snow]

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[olive osprey]

can sm1 help me with A)

[errant harbor]

We have 3 size categories, and we want 6 blocks
That means we want 2 age categories, we can call them "young" and "old"

In the first paragraph, it has been told that dogs aged over 8 years commonly exhibit joint problems

So we can make the age categories as so :
Young -> <8
Old -> >=8

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[rancid wharf]

idk what im doing for 2 ngl

[rancid wharf]

i dont even have an intuitive idea of why 2 is true

oh shoot nvm i do

[devout snow]

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[bold pier]

Can someone help me pls

[bold pier]

<@&286206848099549185>

I need help

This is due tmrw and I don’t get this

[devout snow]

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[bold pier]

Oh mb

<@&286206848099549185>

[calm blade]

What is q?

[bold pier]

Wdym?

[calm blade]

Question?

[bold pier]

Like what is the question

Ooh mb

Idk what to do after

Like I’m lost

@calm blade did you get it, sorry for pinging you

[calm blade]

You got two sides and angle

[bold pier]

So SSA?

Wait now

[calm blade]

SAS

[bold pier]

Ysah

Wait

How do you find the order?

[calm blade]

Angle is between sides

[bold pier]

Like in between?

[calm blade]

Yes

[bold pier]

Oh so like I should say Triangle ECD =~ Triangle BAF for the statement?

[calm blade]

Nah

As it is asked

ACD =~ BCE

[calm blade]

Those are different

[bold pier]

Wait I’m sped I was looking at the wrong question

Yeah you right

Tysm my good man

.stop

.close

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[warm plinth]

how did the answers get 2/sqrt21?

[hardy trail]

Hi! Firstly, do you understand the logic of the question? I.e do you understand why the minimum value of arg(z) is at the tangent of the circle?

[warm plinth]

not really 😭

[hardy trail]

Well

It may help if you draw the circle :)

Anyway that's also how you get 2/sqrt(21)

But I'll wait before we get there

[warm plinth]

where should i go from here?

[hardy trail]

Nice. Now, show me the point on this circle which has the minimum argument.

Remember - the argument is the angle a complex number makes with the x axis. I wonder visually where this would be the smallest?

[hardy trail]

Good! And in a way it kinda makes sense - if you rotate the complex number any lower, you will fly off the circle.

Sweet, so that's the intuition behind the problem, now you have to calculate it.

[hardy trail]

Sooo any ideas on how to calculate the angle of this line?

[warm plinth]

using the point where the tangent intersects the circle and then making a right-angle triangle?

wait also quick question: why do we need to find the argument of the center first?

[hardy trail]

That's a very good question! Let me give you a hint

[hardy trail]

Let's say I drew the diagram, as per what you said.

Now, the argument of the centre is represented by this red angle, right?

[warm plinth]

yes

[hardy trail]

I wonder how this helps us find the argument of the blue line?

[warm plinth]

ohh so it helps us find the angle between the lines

okay thank you so much!

.close

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[atomic idol]

I need to decide whether these are subspaces or not and I’m not quite sure what the notation here means or how to go about this
I know about the subspace criterion but not how to apply it

I did that for vector spaces with real numbers already but I’m not sure if that was correct either

[atomic idol]

The specific thing that I’m struggling with notation wise is this
—- ——
w1 ; w2 …

What does that mean for a vector?

[stone stump]

the w_i are complex numbers

the dash is conjugation

[atomic idol]

Oh I see

So it’d be like

(0,3+i, i)—> (0,3-i, -i)
?

[stone stump]

w is some vector

maybe (0,3+i,i)

and then in S are all vectors with 0 z_1 + (3-i)z_2 + (-i) z_3 = 0

[atomic idol]

So i need to decide whether picking a random vector w will lead to a subspace of this format?

[stone stump]

basically, yeah

[atomic idol]

Do I just start by showing the the zero-vector thing (i don’t know the English term) exists within said subspace?

[stone stump]

yes

you check the subspace criterion

[atomic idol]

Is z_1 = z_2 = […] = z_N = 0 valid for this?

[stone stump]

thats the zero vector, yes

[atomic idol]

How exactly would i prove tjat vector x + a * vector y = an element from S

[stone stump]

you need to check that the equation holds for the new vector

I would recommend to do it in two steps

show that x+y is an element from S

and show that a*x is an element from S

[atomic idol]

For this part, do I just say x and y are both complex thus x+y must be complex too? I honestly don’t understand how this „proof“ works but i feel like ive seen it in use multiple times before

Technically whole numbers can lead to complex numbers depending on the operation (sqrt(x) where x<0 e.g.)

Does it depend on the operation? Am I allowed to assume addition and multiplication won’t change the type of number we’re looking at? If so why do i need to prove x+y? Wouldn’t that be trivial?

[stone stump]

you need to show that the vector x+y still satisfies the equation

yes you can assume that x+y still has complex numbers as entries

[atomic idol]

Can’t I just say that x_1 + y_1 = z_1 and so on to use the fact that z_i is a Part of the subspace anyway then?

Just substituting all z_1 and saying the conditions still apply because they’re equivalent

[stone stump]

x is an element from S

so x satisfies the equation

so $\overline{w_1}x_1 + \overline{w_2}x_2 +\ldots+\overline{w_N}x_N=0$

[woven radish]

Denascite

[stone stump]

similarly, $\overline{w_1}y_1+\overline{w_2}y_2+\ldots+\overline{w_N}y_N=0$

[woven radish]

Denascite

[stone stump]

you have to show that also $\overline{w_1}z_1+\overline{w_2}z_2+\ldots+\overline{w_N}z_N=0$

[woven radish]

Denascite

[stone stump]

using that z_i=x_i+y_i and what you know about x and y

[atomic idol]

I see

I think I’m struggling to explain my thought process mathematically
Can you help me express it better if the thought is correct at least?

The idea was:
we know it applies for a vector like x with a bunch of complex numbers even if we have no information about those complex numbers

So it’s true for all complex numbers

z is just x and y together
And both x and y are complex
Leading to the conclusion that z is complex

Which is the requirement to satisfy the equation, no?

[stone stump]

no

the equation does not hold if you just pick all the x_i randomly

in your previous example, 0x_1+(3-i)x_2+(-i)x_3=0 does not hold for example x=(0,1+i,0)

[atomic idol]

Oh wait you’re right
I confused w and z i think

ChatGPT decided to just show that you could put 0 everywhere and it’d give you the 0 vector, which is a part of the set because it contains the 0 vector

That feels like an incomplete proof tho, is it fine as it is?
To me it seems like showing that the case of everything being 0 working out doesn’t give you any information on the other cases

[devout snow]

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[bitter marsh]

prove that in a sequence of n^2 + 1 distinct integers, there is either an increasing subsequence of length n + 1 or a decreasing subsequence of length n + 1

[bitter marsh]

i do not know how to start with it

[vital edge]

Oh god this problem

[ocean gale]

Yea gl on that bro

[vital edge]

Yeah there really isn't a way to do this without like, knowing how to do it

[bitter marsh]

so there's a standard approach for it?

[vital edge]

I'm gonna try and motivate this as well as I can

[vital edge]

Yes

[bitter marsh]

all right

[vital edge]

It's dynamic programming esque

So might be more digestible to you than it was to me

[bitter marsh]

probably not haha

[vital edge]

Nah, i hate combi stuff so I actively refuse to digest this proof

Anyway

You have a sequence with n² + 1 distinct integers

Let's index it

[woven radish]

Xavier 🌺

$x_1, x_2, \ldots, x_{n^2+1}$

Xavier 🌺

Now for every index, define $I_k$ and $D_k$, which are the lengths of the longest strictly increasing and decreasing sequences starting at $x_k$

[vital edge]

With me so far?

[bitter marsh]

yeah

[vital edge]

Well now what do we want to prove

[bitter marsh]

that there exists an Ik of length n+1 or a Dk of the same length?

[vital edge]

Well ≥ n+1

But yea

[violet wind]

well if there exists one longer than n+1 there's another one of exactly n+1

[woven radish]

Xavier 🌺

There exists some $k$ such that either $I_k \geq n+1$ or $D_k \geq n+1$

[vital edge]

I will bite you

On a serious note, yes. But we want to phrase it as ≥ to make the next step make sense

It's fine for the second best proof method

@bitter marsh any guesses?

[bitter marsh]

i am not really sure

[vital edge]

Well proof by contradiction

[bitter marsh]

i was thinking of

pigeonhole principle

does that work here?

or no?

[vital edge]

It will eventually

But yes, good intuition

We just need to hit a point where we can use pigeon hole principle

[vital edge]

So to prove this by contradiction, what do we assume

[bitter marsh]

this but with <=

[vital edge]

Nope

[bitter marsh]

Ik <= n and Dk <= n for every k

no?

[vital edge]

<

[vital edge]

Oh with n

Yea thats correct

[bitter marsh]

makes sense

[vital edge]

Okay now is the step you guessed

How do we pigeon hole this

[bitter marsh]

not sure if this is correct but

pairs (Ik, Dk) are all distinct

so

[vital edge]

Well not exactly but you're on the right trac

[bitter marsh]

we suppose i < j and (Ii, Di) < (Ij, Dj). because the terms are distinct either xi < xj or xi > xj

[bitter marsh]

what is wrong?

[vital edge]

Correct idea, slightly imprecise

Essentially we have $(I_k,D_k)$

[woven radish]

Xavier 🌺

[vital edge]

With n² possible values

But we have n²+1 possible values for k

[vital edge]

So two of these tuples must be identical

[bitter marsh]

makes sense

[vital edge]

And then back to what you said

[bitter marsh]

so for case one (xi < xj), we take a longest increasing subsequence starting at xj of length Ij?
​

[vital edge]

Yes

And now what happens

[young spade]

Hope this doesnt interrupt
Totally out of my actual knowledge since I dont really know much about it, but this is also called the Erdős–Szekeres theorem (more like a re-statement of it)

And there are a few theorems that basically have the truth of it as corollary.

[vital edge]

Yup, I was planning to drop the wikipedia link after we were done with the problem

Well more appropriately, Dreyuk asked me to

[young spade]

One of them is intuitive enough and is also basically a sort of Pigeonhole reasoning

[vital edge]

Yeye

[young spade]

gl

[vital edge]

I wanted to walk her through it rather than slapping a link on her face

[bitter marsh]

not really sure

[vital edge]

Well

[woven radish]

Xavier 🌺

Let's say $i < j \newline
I_i = I_j \text{ and } D_i = D_j \newline
\text{Now assume } x_i < x_j$

[bitter marsh]

wait

we are going to get a contradiction here aren't we

because

[vital edge]

I mean I started this by saying we do my second favourite proof technique

[bitter marsh]

Prepending xi produces an increasing subsequence starting at xi of length Ij + 1

[vital edge]

Yup

[bitter marsh]

and the first is?

[vital edge]

"proof by it's obvious"

[bitter marsh]

haha, fair enough

[bitter marsh]

so this contradicts Ii = Ij

[vital edge]

And when x_i > x_j we get the same, just with the D this time

[vital edge]

Yes

Thus we can't limit all the I's and D's to be <= n

[bitter marsh]

so no two indices have the same (Ik, Dk)

[bitter marsh]

right, makes sense

[vital edge]

Yup

And now you're done

And as promised

https://en.wikipedia.org/wiki/Erdős–Szekeres_theorem

[bitter marsh]

all right, thanks a lot!

.close

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[devout snow]

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[upbeat breach]

Ive done a and b, just confused on c)

[autumn girder]

!status

[devout snow]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[hollow ice]

do you know how z1 and z2 are related?

[upbeat breach]

No

[hollow ice]

ok, a broader question then, what do you know about the roots of unity then?

[upbeat breach]

We did it in class

But i forgot about it because i thought it wasnt important

😭

[hollow ice]

💀

open your notes or textbook then

[upbeat breach]

Zk = cos(2pik/n) + isin(2pik/n)

Right?

[hollow ice]

yes

so the basic idea is that you divide the angle between the real line (argument 0 line) and the z complex number into n equal parts

for 5th roots of +1, that angle is gonna be 0

but you need 5th roots of -1 here

so can you adjust that accordingly?

[upbeat breach]

No 💀

Im looking at my textbook rn

It never did anything w -1

[hollow ice]

well, I just told you what to do here...

[upbeat breach]

I understand the 1 part though

[upbeat breach]

Yeah

And im confused on the wording

[hollow ice]

what is the argument gonna be for a number like -1?

[upbeat breach]

-pi?

No

Pi

Right?

[hollow ice]

yea

both are equivalent tbh, tho one of them is principal argument

anyways, you know how the complex numbers multiply right?

so by de moivre theorem, you have the cos(5 * theta_2) = -1 and sin(5 * theta_2) = 0, for which you have to find solutions

from your basic trig, can you find the solutions?

[upbeat breach]

Does _2 just mean /2?

[hollow ice]

no, its a subscript

theta_2 is the angle that corresponds to z_2

[upbeat breach]

Oh

Alr

[upbeat breach]

Can you put this in texit

I can barely read it on text

[hollow ice]

$\cos(5\theta_2) = -1$ and $\sin(5\theta_2) = 0$

[upbeat breach]

So trig equations?

[woven radish]

βαχτϵρ10Φρ4γ

[hollow ice]

yea

[upbeat breach]

Just 5theta = pi

So theta = pi/5?

[hollow ice]

pi/5 is just one of the answers

[upbeat breach]

Oh

[hollow ice]

thats z_1 btw

[upbeat breach]

5theta = pi + 2pi

[hollow ice]

yea

[upbeat breach]

Ahhh

So theta = 3pi/5

[hollow ice]

that would be the argument for z_2 since its gonna land in 2nd quadrant

[upbeat breach]

Alr thanks

That made sense

.close

[devout snow]

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[hollow ice]

yea, same idea for part d too

[devout snow]

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[wet haven]

Need help with this because I don't know how to do this without grouping

Given the polynomial ( W(x) = 3x^3 - 14x^2 + 95x - 58 ).
Determine all rational numbers among which one should look for the roots of this polynomial.
Factor the polynomial ( W(x) ) over the real and complex numbers without using grouping.

[fossil locust]

you need the rational root theorem

go search it up

[wet haven]

alr

[fossil locust]

it's the "guess a root r then divide by (x - r) strategy"

synthetic division is nice if you know it

[wet haven]

I don't so I am gonna grind this stuff now

I will ping you when I do it or fail to if that is okay

[devout snow]

@wet haven Has your question been resolved?

[fossil locust]

pls don't ping me

[wet haven]

oh ok

[fossil locust]

but then other people will come

or if your channel goes to the bottom of the list, you can close and reopen it

[wet haven]

is it okay if I stay till I resolve it or do I close?

ok

[fossil locust]

you can stay

[wet haven]

alright, thanks

[fossil locust]

no worries!

[devout snow]

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[hard sable]

for a

[hard sable]

i got

f'(x) = 3 + (5)/(2x+1)^2

and then, idk how to do part b

i tried doing y - y1 = m(x - x1)

but it just seems like way too much expansion and rearrangement for 3 marks

there has to be another way

[winter patrol]

show your attempt

at this level basic manipulation is worth very little

[hard sable]

it would just take long

[hard sable]

ok wait lemme go on my phone

i aint got discord on my phone hold up

[winter patrol]

not longer than part a)

[hard sable]

part a was straightforward

[winter patrol]

ok, that's wrong

[hard sable]

yeah

[winter patrol]

you should first get the numerical value of the desired slope at P

[hard sable]

yeah thats it

idk why i didnt do that for this question

brain moment

thanks

.close

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[thin fern]

I'm not really sure where to start here tbh

[sand dove]

Can you show it for f a polynomial?

[thin fern]

let me try

you could just integrate it directly then

giving you some linear combination of the coefficients of f(x) equal 0

but you can do this for lots of values of n

[sand dove]

Well you could try that, though you would need to argue that you have enough linearly independent equations to say 0 is the only solution

Here's a hint: show integral(f²) = 0

[thin fern]

Ohhhh

because you can split integral of f^2 into a bunch of a_n * int(f * t^n)

which are all 0

So if f is not a polynomial, is this something to due with the weierstrass approx theorem

that polynomials are dense

[sand dove]

Exactly

Because f : [0,1]-> R

[thin fern]

why is the [0,1] of special importance here?

[sand dove]

Because Weierstrass only works on closed intervals

Or compact sets in general

[thin fern]

Oh right

So the argument would be

any continuous function from [0,1] to R by the weierastrasss approx thm can be written as the sum of polynomials

and we've shown that all polynomials are 0 under the given assumption

[sand dove]

Sequence of polynomials*

It's an approximation with respect to sup norm

[thin fern]

Oh right

Okay I think I've got it now thanks!

.close

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[sand dove]

So, to be more precise that's not exactly it

[thin fern]

.reopen

[devout snow]

✅ Original question: #help-27 message

[sand dove]

You have to use that if $p_n \to f$, then $\int_0^1fp_n\to \int_0^1f^2$

[woven radish]

Raphaelisius Maximus MMIII

[sand dove]

But the integral of fp_n is 0 for all n because of the assumption

So integral of f² is 0 by limit

Same argument as before

[thin fern]

Let (p_n) be a sequence of polynomials that converges to f. We've shown already that integral of f*p_n from 0 to 1 must be 0. Therefore, it follows the integral from 0 to 1 of f^2 is also 0. Hence f=0.

[sand dove]

We haven't really "shown it already"

It's just that p_n is a linear combination of some t^k

[thin fern]

and by assumption the integral of each f*t^k is 0

[sand dove]

the "f polynomial" case was more of a hint as to how to do it in the general case

[devout snow]

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[bronze urchin]

Anyone whos into finance, if i have a stock A whats the probability that stock will be above its original price in 24h or any other timeframe. What equation would you use here to get that so that you get a function as of time remaining and current price

[lusty sapphire]

Too many real world factors to say reliably. Otherwise, we would all be rich

[bronze urchin]

Well I mean there must be something it doesnt have to be perfect ofc

[lusty sapphire]

Realistically, the probability would be zero if you consider finances to be zero sum

[bronze urchin]

I don’t think thats the right answer

[harsh sierra]

the expected change of the stock would be 0

and there is no reason to expect the stock to not be equally likely to go up/down

[bronze urchin]

Ok but say theres like 12h left and stock is up 1% from original price whats the probability it will stay above the original price

If you get me

[harsh sierra]

this would depend on how you model a stock

[supple knot]

^

this isn't a math question, you're just speculating

[lusty sapphire]

If you had a good answer for this, you would be very rich

[bronze urchin]

I mean id assume there are some methods or models

Like im not that advanced but stoll thanks for help

[lusty sapphire]

There's really no way to know reliably. We would all be millionaires if we could predict stock prices

[bronze urchin]

Its not about predicting its price its abt predicting if it will stay above a certain price

[lusty sapphire]

That is the same thing

[bronze urchin]

Is it

[lusty sapphire]

Yes

[bronze urchin]

If i say what will stock a be tomorrow thats a guess

But if i say it wont be down 90% tomorrow

Thats like more certain

Bc it never goes down 90%

[lusty sapphire]

There's no way to say definitively

[bronze urchin]

Ig you’re right but i thought u could at least somewhat predict it stays above something ig not

[lusty sapphire]

You can measure on past actions, but there are far too many factors to say anything with reliability beyond guessing

[bronze urchin]

Thanks

[lusty sapphire]

There is a tolerance. You're right, it won't drop to zero or 10x overnight. The best you can do is make a trendline. But on such a short time period by such a small amount, it's almost entirely random

[devout snow]

@bronze urchin Has your question been resolved?

[devout snow]

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[rocky marlin]

I cant find a its giving me a number with e in it

[supple knot]

what is a supposed to be

[rocky marlin]

The starting value

Y=ab^x

[supple knot]

do you know the y value when x = 0?

oh i see

you should think of x as years since 1995 instead of years after 0

so the first value is simply 690 = a * b^(years since 1995)

[rocky marlin]

So its 690 in year 1995 actually the starting value?

[supple knot]

yes

so 1999 has x value equal to ?

[rocky marlin]

4?

[supple knot]

yes

[rocky marlin]

Ok thanks

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[neon crystal]

i dont understand d...

The picture shows the graph of the function ( y = f(x) ).
Find, by reading from the graph:

a) the coordinates of points A and B

b) ( f(4) )

c) the value(s) of ( x ) such that ( f(x) = -4 )

d) the value(s) of ( x ) such that ( f(f(x)) = -1 )

[neon crystal]

f(0) = -1

[devout snow]

@neon crystal Has your question been resolved?

[neon crystal]

@tranquil iron

[neon crystal]

.close

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[warm plinth]

why did they add pi/2?

[rare kernel]

angle is measured from positive real axis

check what quadrant the complex number is in

[devout snow]

@warm plinth Has your question been resolved?

[warm plinth]

why would doing π - angle be different?

[devout snow]

@warm plinth Has your question been resolved?

[devout snow]

@warm plinth Has your question been resolved?

[devout snow]

@warm plinth Has your question been resolved?

[lunar harbor]

note that your values of theta are different

$\frac{\pi}{2}+\arctan \frac{4}{5}=\frac{\pi}{2}+\left(\frac{\pi}{2}-\arctan \frac{5}{4} \right)=\pi-\arctan \frac{5}{4}$

[woven radish]

Civil Service Pigeon

[devout snow]

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[cinder plover]

Can someone tell me how Im suppose to mathc and match

[cinder plover]

by just looking at the equation

i dont understand bro

cn someone guide me please

likestepby step and stuff

[devout snow]

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[knotty sapphire]

g(x) = ((x)^2 + 1) * e ^ ( -x^2) - 1/2
What does x equal to when g(x) = 0

[supple knot]

,w roots ((x)^2 + 1) * e ^ ( -x^2) - 1/2

[woven radish]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by rie.mann

[supple knot]

do you know https://en.wikipedia.org/wiki/Lambert_W_function ?

[knotty sapphire]

no actually? this is new

[supple knot]

well then you might have messed up earlier in your work then

[knotty sapphire]

oh well i guess i will be looking into that

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[lime harbor]

F(x) =
-x when x <= 0
x+1 when x > 0
is the point 0, 0 consider a local min

[lime harbor]

Because for any small x range around 0, (0, 0) is the smallest point

[vital edge]

Yes

I misread the question first lol

[timber pebble]

its the global minimum

so it should probably be a local minimum too

thats as much as youd have to justify it, i think

[lime harbor]

Global min does not imply local min

[timber pebble]

i guess not it depends on the neighborhood sure

[vital edge]

Global min not at extremes of the domain does usually imply local min

[lime harbor]

For example if the min is on a closed interval its not local, by what my prog said at least

[timber pebble]

sure

what class is this for

you have calculus?

[lime harbor]

ya

[timber pebble]

so is this a proof?

[lime harbor]

No just trying to understand what local min and max is

[timber pebble]

or what kind of argument are you expected to make

[timber pebble]

it usually means how it sounds, they give you a point, so it means if there is an immediate (probably infinitesimally close) point which is higher or lower respectively

if you have calculus and a nice function, usually that means like

for a local min

showing that the function is increasing in a region of interest around that point

idk, is that helpful?

here I think you have the benefit that the function is always positive, so you could argue that, then show its 0 at (0,0), so it must be a local min

[lime harbor]

So from what i understand, local min is equivalent to ithere exists an interval around that point st f(x) is bigger or equal then the point for every x

And f(x) has to be defined in the interval

[timber pebble]

IDK if it has to be defined

but, yea

[lime harbor]

Im just saying that cause of the case where its on the edge

[timber pebble]

like if you had $f(x) = x$ for $x\geq 0$

[woven radish]

jan Niku

[timber pebble]

then $f(0)=0$ is a local min

[woven radish]

jan Niku

[lime harbor]

Wait really

I think our definition is different then

[timber pebble]

its possible, yea

i cant think of any reason to require it to be defined

[lime harbor]

Idk prof said if its on the edge of a function it can be global min/max but not local

Ok then

[timber pebble]

I guess you have to follow your prof, then

[lime harbor]

Lol yea
Thanks, you did help :)

[timber pebble]

[lime harbor]

.close

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[vital edge]

Yeah but if it's not on the edge then global implies local no?

[lime harbor]

.reopen

[devout snow]

✅ Original question: #help-27 message

[vital edge]

No don't reopen, I just had that one comment lol

[lime harbor]

.close

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[lime harbor]

:p

But ya I think so

[vital edge]

Ask your prof if that's enough justification ig

[timber pebble]

theres nothing to do here but go back to definitions tbh

which could vary

or at least, youre compelled to use your profs definitions

[lime harbor]

Shouldn't rly matter tho for the questions we get, I just was curious exactly what it means

[woven radish]

jan Niku

[timber pebble]

@lime harbor

usually we'd use something like this

[lime harbor]

Yeah that makes sense
I figured ours is probably the same thing except f(x) has to be defined around the interval

[devout snow]

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[lunar cloud]

This question regards graphing and how to dissect a standard form quadratic. If given the equation f(x) = x^2 - 8x + 12 how do I use that to get the vertex, axis of symmetry, min/max, andd the x and y intercepts? All i remember is my teacher mentioning something about -b/2a

[misty crest]

complete the square

[lunar cloud]

?

[mighty thicket]

x intercept is the values for where f(x) = 0

y intercept is ur idependent term ( 12 in this case )

axis of simetry is ur -b/2a which is ur X pos of ur vertex

[mighty thicket]

to get the Y of ur vertex u can get by using -delta/4a or plug in -b/2a into ur function

[lunar cloud]

thanks, let me try out and report back

[mighty thicket]

doesnt that get a bit complicated when the function isnt close to a perfect square trinomial*

like y= 3x^2 -3/5x + 17

[lunar cloud]

also where do you get the 36 from in the chart above?

[misty crest]

why is this any different, you just factor the 3 out first

[sour osprey]

itll be good practice with fractions but ya

[mighty thicket]

ur making it into a form of (x + y)^2

which gives u x^2 + 2xy + y^2

u can see 2xy = 12x

so y is 6

[mighty thicket]

still, a lot easier to just -b/2a and plug that value in the function no?

[lunar cloud]

how do I know if the graph opens upwards or downwards?

[sour osprey]

if a in ax^2 is positive i think it opens upwards, and downwards for negative

[lunar cloud]

So if I did all this right, I got y intercept as 12, x intercept also as 12, line of symmetry as x = 4, vertex as (4,-4), and min as -4, does that look good?

[sour osprey]

you only got one x-intercept ?

[lunar cloud]

well it would be -12 and 12 no? My apologies

[sour osprey]

hmmm i dont think so

was the function

$y=x^2-8x+12$ ?

[woven radish]

exoulos

[lunar cloud]

then I make y = 0, so then 8+-√-8^2 - 4(1)(12)/2

so then 8+-4/2
then 4+-2, so x = 6 and x = 2?

i see what i did wrong

are the other parts of my solution correct otherwise?

[sour osprey]

yea it looks pretty much right

[lunar cloud]

thanks so much for the help in that case

im going to continue doing problems and if I run into any more trouble ill open another ticket. Much appreciated!!!!

[sour osprey]

of course ! best of luck, you got it

[lunar cloud]

.close

[devout snow]

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[devout snow]

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[remote copper]

My answer is B

[misty crest]

why?

do you know the definition of local max?

[devout snow]

@remote copper Has your question been resolved?

[misty crest]

@remote copper brother

you hit the X and didn't respond

😭

[remote copper]

Why B is wrong?

[misty crest]

why do you think it isn't a local max

[remote copper]

Because graph will looks like

[misty crest]

just because it isn't continuous?

[misty crest]

thats not how you're supposed to reason

[remote copper]

No

[misty crest]

so how does that not look like a local max to you then

you don't know the definition of a local max do you

[remote copper]

if it was x=0 f(x)=0 then local max

[misty crest]

🤔

no

that would be a local min

[remote copper]

Local minima

yeah

[misty crest]

i'll give you the actual definition since you don't know it

[remote copper]

Definition is simple

All the values around the interval will be less or max for local minima or maxima

At that particular point

[misty crest]

f has a local max at x = a if there exists delta > 0 such that for all x in (a - delta, a + delta), f(a) >= f(x)

[remote copper]

at x=0 we have value 1 which is middle of the graph

[misty crest]

this is certainly true

we can find some interval around x = 0 so that for all x in the interval, f(x) <= f(0) = 1

just take delta = 1

or anything smaller

[remote copper]

So you meant x=0 would be local maxima

@misty crest

[misty crest]

f has a local max at x = 0 yes

[remote copper]

How much longer should the interval have to be?

[misty crest]

what?

it doesn't matter the delta you choose so long as its <= 1

[remote copper]

I meant after some points our condition will fail

[misty crest]

if delta > 1

[remote copper]

Yeah

[remote copper]

I meant do we have to look the whole graph?

[misty crest]

what does that mean

[calm blade]

Is it

[misty crest]

🤔

yea

[calm blade]

Nvm

Mb

[devout snow]

@remote copper Has your question been resolved?

[devout snow]

Channel closed

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[devout snow]

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[placid mauve]

hey guys how do u become better at algebra and indices?

[pseudo basin]

practice, practice and practice

[placid mauve]

when i do questions, sometimes i get answers and redo them and get different answers

[silk terrace]

Check your reasoning and where you went wrong in it

[pseudo basin]

well not just reasoning tbh

[placid mauve]

okayy

[pseudo basin]

execution also matters

cause like if you fuck up arithmetic for example that means you need to practice arithmetic more

[placid mauve]

ooh okay

wait what do u mean by reasoning

can u give an example pls

[silk terrace]

All math problems are in some way a process of deducing "new" true statements from given data/statements

[abstract plover]

"why am I doing this? why am I allowed to do this? what happens if I do this?"

[silk terrace]

Which is done by deduction

And deduction goes thru steps.
From statements 1,2,3, you in step 1 deduce statement 4, then with 1,2,3,4 deduce statement 5, and so on, till you get your desired statement

<@&268886789983436800>

[placid mauve]

u can contineu

continue*

[silk terrace]

If you haven't found your desired statement it means you went wrong in one of these steps

[long adder]

it was deleted

[placid mauve]

so like

rechecking?

each process?

[silk terrace]

Yes

Each step

[placid mauve]

okay

[abstract plover]

not just that, but also being able to justify each step

[placid mauve]

what if we dont have time for that in the exam

[silk terrace]

Well, this is why I hate time trouble in exams

One should be given enough time to think through

But to answer your question, you just need to get used to the process of

Checking

[placid mauve]

checking while answering?

[silk terrace]

And also to the process of deduction, so that you don't make mistakes

[silk terrace]

Yeah

[placid mauve]

okayy

[abstract plover]

on top of that, make a habit to write the justification of each step as you do them

[placid mauve]

like the workings?

or why we did that

[abstract plover]

the why, yes

[silk terrace]

Yeah make sure you know exactly why you've written what you've written

[abstract plover]

suppose we have an equation to solve

[silk terrace]

How it follows from the data

[placid mauve]

oooh okay

[silk terrace]

Etc

[placid mauve]

thank you guyyss

hope u have a great day

[silk terrace]

You too

[placid mauve]

.close

[devout snow]

Channel closed

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[abstract plover]

ah ok

wanted to give an example, but all good

[devout snow]

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[restive river]

Hey, I'm a bit confused about the "b" in the function. Does this mean the function is undefined for x<0, since we cannot get real values rooting negative numbers?

[misty crest]

you have a smaller domain than that

[restive river]

oh the domain is when b>=0

oops

x>=b

[misty crest]

it’s true that x cant be < 0 but it also can’t be < b

[restive river]

Ok thanks!

.close

[devout snow]

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[glass canopy]

Am I correct that the contrapositive statement is equivalent to:

$f(\tau) := G_{4}^{3}(\tau) - 27G_{6}^{2}(\tau)$ has $f(\mathbb{H}) \supset \mathbb{C} \setminus {0}$

[woven radish]

Varixiuqlhfbgraijbzjnqghppxnqmvw

[thorny bloom]

u should ask this in advance channels

[glass canopy]

Ah, apologies

[thorny bloom]

which subject is this related to?

[glass canopy]

Modular forms

[thorny bloom]

ah no u would get better answer there

[glass canopy]

Thank you very much!

.close

[devout snow]

Channel closed

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[glass canopy]

.reopen

[devout snow]

✅ Original question: #help-27 message

[glass canopy]

You're typing, I'll let you finish 😊

[thorny bloom]

omg no i was just giving u the channel name 😭😭😭

#real-complex-analysis

[glass canopy]

Ohhh that'd actually be really helpful

Thank you so much!

.close

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[harsh sierra]

#advanced-number-theory

[devout snow]

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[sour spire]

guys why am i wrong?

[pseudo basin]

setting aside potential arithmetic issues i don't see any issue with the process at the moment

why are you so certain you're wrong

,calc exp(138/55)

[woven radish]

Result:

12.293748843597

[calm blade]

why'd you say its wrong

[pseudo basin]

actually yeah let's check your values on desmos...

,calc (6.46-3.7)/(2.2-1)

[woven radish]

Result:

2.3

[pseudo basin]

,calc 138/55

[woven radish]

Result:

2.5090909090909

[pseudo basin]

@sour spire

i do wonder how on earth you could end up with 138/55 here.

like what was the arithmetical reasoning that led to 55 of all things ending up in your denominator

let's wait for op to get back online...

[sour spire]

the mark scheme says im wrong

[pseudo basin]

read what i wrote later on

[sour spire]

i calculated it w calculator

[pseudo basin]

can you show exact calculator input and output

[sour spire]

just know i did (6.46-3.7) then i divided by 1.2

it was 138/55

[pseudo basin]

do you have your calculator with you

[sour spire]

yess

[pseudo basin]

can you reproduce this

do the same calculation again