help-23

[lone arch]

,w plot (x - 2)^8

[flat frigate]

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[lone arch]

Here, x = 2 is a zero of multiplicity 8

,w plot (x - 2)^9

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[lone arch]

Here, x = 2 is a zero of multiplicity 9

See how zeros of even and odd multiplicites differ visually?

[sly lintel]

yess

so is the multiplicitie in this case 2? since it looks kinda like an parabel?

[lone arch]

Yes!

For x = -1

For x = 0.5, it's 1

[sly lintel]

ohh alright

[lone arch]

So we can construct f already

[sly lintel]

and how do i summarize it in one equation?

[lone arch]

f(x) = (x-a)(x-b)(x-c). We have -1 of multiplicity 2, so we take for example a = b = -1.

Then f(x) = (x+1)^2(x - c)

Now we take c = 0.5, because that's the other zeor

So f(x) = (x+1)^2(x - 0.5)

[sly lintel]

thank you so much!! ur a life saver

[lone arch]

Now there is one thing we might consider if we are very strict and careful

,w plot (x - 2)^2

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[lone arch]

,w plot (x - 2)^4

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[lone arch]

Those almost look the same, right?

[sly lintel]

yeah

[lone arch]

So really, we would also argue that it's f(x) = (x +1)^2(x - 0.5) and not f(x) = (x + 1)^4(x - 0.5) (or any higher even exponent)

To do that, just pick some value

[sly lintel]

any?

like 3

[lone arch]

At x = 1, it looks like it's at -4

[sly lintel]

yeah

[lone arch]

Or really, I guess 0 would be the easiest value

f(0) = 1^2(0 - 0.5) = -0.5

Well, but if we look at the graph, it's supposed to be 2 there!

So we notice that we actually need a factor too

Namely a factor of -4

So what we get is $f(x) = -4(x + 1)^2(x - 0.5)$

[flat frigate]

Kepe

[sly lintel]

ohh alright

i get it now

[lone arch]

Ok it's really small, wait a second

[sly lintel]

okok

[lone arch]

There

[sly lintel]

perfecttt

thanks alot!!

[lone arch]

Np

[sly lintel]

.close

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[lone arch]

(Regarding this, I don't think you need to further discuss it in your working, only if you want to be really really precise, which I really think is not necessary here)

[sly lintel]

okok got it, thanks :DD

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[hollow stream]

Hello so i was working on this problem earlier and I just wanted to make sure I understand the process correctly. The correct thing to do here is to set u=1+9x and then to find the new bounds I plug in the old bounds into my value for u.

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[lean otter]

how to work out a

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@lean otter Has your question been resolved?

[lean otter]

no

how do u get the end number

what is the method

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[lean otter]

no

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[opaque brook]

is this question badly worded or am I missing something

[opaque brook]

part (b), unsure what to do

[icy lance]

ii took a second but otherwise its fine
ii: find the equation of the circle with center P, which passes through point T

[opaque brook]

but what is point T

[icy lance]

there is a line joining the circles edge and P
said line is tangent to the circle C
T is where the line intersects the circle

[opaque brook]

its tangent?

ahh ‘touches the circle’

the wording threw me off

anyway cheers

.close

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[glossy mica]

is this right so far?

[drowsy moss]

looks like you've assumed the thing is already true.

[gusty trench]

if you're proving by induction you probably want to be very careful about how you present it

base case, inductive hypothesis, then inductive step

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@glossy mica Has your question been resolved?

[glossy mica]

so base case would be like... determining it's true for n=5 (because n=5 isn't shown)

inductive hypothesis would be the "p(k)" where i replace n with k

and then the inductive step is k+1?

@gusty trench

Base case: n=5
$1^2+2^2+3^2+4^2+5^2=((5(5+1))(2(5)+1))/6$
$55=55$
Inductive hypothesis:
$1^2+2^2+3^2+4^2+\cdots+k^2=\frac{k(k+1)(2k+1)}6$

[flat frigate]

AWACS Sky Eye

[glossy mica]

<@&286206848099549185>
i'm just stuck now.

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@glossy mica Has your question been resolved?

[gusty trench]

just show that it’s true for n = 1, then assume it’s true for n = k and show that it must be true for n = k+1

[proper moth]

Yeah it's the inductive hypothesis

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[heavy venture]

not sure where to start with part b

[merry swift]

how about the hint

[heavy venture]

i thought that was for determining linear independence

for part a

[merry swift]

No, I think for a) you've done a lot of extra fluff. Linearly independent -> basis in R^2 is justification enough if you hvae that from class

the hint is definitely for part b)

To understand why the hint applies for part b), by the definition of basis the coordinates of w in a basis {x_1,x_2} are just a and b s.t. ax_1+bx_2=w

[heavy venture]

what is the notation on top of w?

[merry swift]

cool squiggly line

[heavy venture]

lol yeah

[merry swift]

I think what they mean is

w_1 with a squiggle

is the first coordinate of w in the basis

w_2 with a squiggle is the second coordinate of w in the basis

and w_1 and w_2 are just the coordinates in the standard basis

[heavy venture]

so

[merry swift]

by "the basis" i mean {[1,2], [2,5]}

[heavy venture]

1 2 and 2 5

1 2 woudl be the first coordinate?

anything that is a linear combination of 1 2 and 2 5?

[merry swift]

When you express w in that basis that means ur expressing it as some number of copies $\tilde{w_1}$ of $[1,2]$ + some number of copies $\tilde{w_2}$ of $[2,5]$, i.e. you're finding $\tilde{w_1}$ and $\tilde{w_2}$ such that $$[w_1,w_2]=\tilde{w_1} [1,2] + \tilde{w_2} [2,5]$$

well that hasnt worked has it

[flat frigate]

Damian

[heavy venture]

is that really what the question is asking?

[merry swift]

Yeah! Then the final step is just to write that as a vector [\tilde{w_1},\tilde{w_2}]

[heavy venture]

isn't expressiong them in terms of eachother what i did in problem 1?

[merry swift]

You're not expressing them in terms of each other

You're expressing some arbitrary vector

w

in terms of the both of them

[heavy venture]

okay

so just rewrite in terms of w1 and w2

easy

almost too easy i think

thanks though

close

[merry swift]

just comes down to solving a linear system with some arbitrary constants so not too complex

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[open spade]

Can someone help me learn optimization I have a test tomorrow, and I don't understand any of this.

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[frail mountain]

need some

[frail mountain]

help with calc 3

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[covert gorge]

how do i solve this problem

[covert gorge]

where do i start learning math

i've found an answer but i need to know how to solve it

[charred saffron]

multiplying the numerator and denominator by the conjugate

[proper moth]

It comes by practicing. Solve lots and lots of examples

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[tranquil geyser]

I cant tell if im being stupid or not

is showing 4k + 3 not as a sum of squares super trivial

[tranquil geyser]

if its not ill keep trying and figure out something

[fleet condor]

What is the actual thing you are trying to prove

[obsidian oracle]

show that a square is either 0 or 1 mod 4

[tranquil geyser]

damn i was gonna use mod then convert it to a divisibility problem

[obsidian oracle]

oh you can stay with mod

[tranquil geyser]

ah

alright ill probably just change approach then

just out of curiousity

is showing 4 | x^2 + y^2 - 3 even possible?

without reverting back to modulo

[obsidian oracle]

you mean show it's false

like I said

modulo definition is made to make your life easier

there always is a way through 'divisibility'

since there is a way through mod and mod is defined using divisibility

[tranquil geyser]

ah okay i see

[tranquil geyser]

how did u get this idea so fast btw

my brain never even came pu with this thought

[obsidian oracle]

well apart from the fact that I already did this problem

[tranquil geyser]

oh 💀 idk did u ever struggle with stuff like this for ur first time because i certainly am 😃

[obsidian oracle]

showing that a sum of squares is never 4k+3 is the same as showing a sum of squares is always 0,1 or 2 mod 4

[tranquil geyser]

it is my first proof class outside of discrete so alot of these questions are killing me

[tranquil geyser]

ah

[cold relic]

Totally normal

[tranquil geyser]

does this end up being quadractic residues

🤔

im trying to avoid that rn

[obsidian oracle]

this is literally mod 4

there aren't many options

(there are 4 options)

[tranquil geyser]

wait

im being dumb

0 and 1 are the only possible residues here

🤦‍♂️

bruh

is this just a bunch of cases

im goign to kms

😭

ok got it ty @obsidian oracle

.close

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[vague pewter]

ive tried so many answers but i cant seem to do it (polar coords calculus 2)

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[knotty jungle]

If they say that 0 isn’t an element of f([a,b]) how can f(x) be 0 then

Like obviously if the integral is zero f(x) could be 0 therefore the integral will still be 0

But they just said it can’t be 0

[robust river]

Looks like poor wording, yeah

[knotty jungle]

lol

[robust river]

If a continuous function doesn't change the sign, then the integral is either strictly positive or strictly negative

Ok I think we can fix their statement

Let f be a continuous function such that f(x) >= 0 on [a, b]

Then we have this

Same for the case when f(x) <= 0 on [a, b]

[heady flare]

This property states that if the integral of a continuous and sign-definite function over an interval [a,b] equals zero, then the function itself must be identically zero on that interval. Intuitively, for a continuous function that doesn’t change sign (either always non-negative or always non-positive) to have a zero integral, it would imply that the function has no "area" under its curve across the interval, hence f (x) = 0 for all x ∈ [a,b].

[knotty jungle]

Okay thx

.close

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[knotty jungle]

.reopen

[safe radish]

✅

[knotty jungle]

actually if y=x^3 and we integrate it over -1,1 then it’ll be 0 too

But in this case y wasn’t 0

[robust river]

However x^3 is not sign-definite on [-1, 1]

[knotty jungle]

Hmm

Do u know an example

So this makes sense haha

[robust river]

Sign-definite continuous function f(x) on [a,b] integrates to zero if and only if f(x)=0 on [a,b]

[knotty jungle]

like the only example is y=0 itself id think

Because if you have y=3 which means its sign indefinite because its always positive

The integral over a certain interval won’t be 0

So it literally only makes sense for y=0 but that seems like such a useless statement

[robust river]

Well, we're working with this statement

And it's useful, I assure you

For example in variational calculus that's one of the key steps in deriving the Euler-Lagrange equations

[robust river]

Let's prove it, actually

Let f(x) be a sign-definite function on [a, b]. Without loss of generality, we can just say f(x) >=0 on [a, b]

So we want to prove that for such $f(x)$ we have
$$\int_a^b f(x)dx = 0 \Leftrightarrow f(x)\equiv 0$$

[knotty jungle]

Crazy

[flat frigate]

EQUENOS

[robust river]

Implication $\Leftarrow$ is obvious, right?

[flat frigate]

EQUENOS

[robust river]

Now let's prove $\Rightarrow$, using continuity of $f(x)$ and the fact that $f(x) \ge 0$ on $[a,b]$

[flat frigate]

EQUENOS

[robust river]

Suppose $f(x) > 0$ at some $x_0 \in [a,b]$. Then, due to continuity there exists $\delta > 0$ such that $f(x) > 0$ on $[x_0-\delta, x_0+delta]$. Hence
$$\int_a^b f(x)dx \ge \int_{x_0-\delta}^{x_0+\delta} f(x)dx > 0$$

[flat frigate]

EQUENOS

[robust river]

So the integral can't be zero for non-zero non-negative continuous functions

[knotty jungle]

Ok let me read

Crazy

Ok thx

.close

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[noble wing]

Hello, I would like some help for this question (specifically the one on the right).

[noble wing]

Im new so i hope i used this help thing right!

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[noble wing]

<@&286206848099549185>

[proper moth]

You should label some coordinates on the curve like this

[noble wing]

Oh I see thank you for letting me know

[noble wing]

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[proper moth]

.reopen

How do you get the equation if I may ask?

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[mystic junco]

i am currently stuck at iii

[mystic junco]

i have was trying to prove that (1+iz)/(1-iz) = fifth root of unity://

then i selfcontradict myself by assuming z is purely real

💀 i have no idea how to proceed

[obsidian oracle]

well (1+iz)/(1-iz) is a fifth root of unity

it's just that z is not necessarily real

(1+iz)/(1-iz) = e^(itheta)

with the correct values of theta

(1+iz) = (1-iz)(e^(itheta))

isolate z

then z = ...

[mystic junco]

aigh lemme try that

oh 💀

i got it tks

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[inland void]

Hi please can anyone help me with this stats. I need help. The requirement is: I would like to know the potential number of staff (and that number as a percentage), who may have fingerprints eliminated from the total number of fingerprints taken in 2022 and 2023.

This is the stats

Have 33,509 staff as of Sept 2024
Approximately 19,000 fingerprints missing from staff
We only have 14,509 staff fingerprints on database

2022
Fingerprints collected: 19,775
Fingerprints identified to database: 163
Unidentified fingerprints on database: 12,023

2023
Fingerprints collected: 18,095
Fingerprints identified to PPE: 89
Unidentified fingerprints on database: 11,437

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[daring tundra]

how to solve this

[daring tundra]

bprp tried an entirely different approach
idk if its a known approach
would appreciate if someone can name or explain this method

[proper moth]

You know liate?

[daring tundra]

parts? yea

I tried that but you get stuck

[proper moth]

Let me try going through

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[proper moth]

Done.

It's tiring haha

@daring tundra

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[lean otter]

How is this obtained

[lean otter]

The derivative part (x-1)/x

[pine horizon]

derivative of ln(x) is 1/x

and the latter part cuz chain rule

[lean otter]

<@&286206848099549185>

Where did we apply the chain rule

If ln(x) is 1/x

[pine horizon]

its ln(x/(x-1)) in your question

[lean otter]

Yes

[pine horizon]

so first you differentiate the log

then x/(x-1) because chain rule

[lean otter]

Ok

Ah ok

1/x >> (x-1)/x

So we always do the log one and then multiply it with the normal derivative

And how did we get the -1/x^2-x

[pine horizon]

by simplyfying this

[lean otter]

Ok tysm bro!

.close

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[frigid ocean]

I'm trying to do backprop manually like the derivatives etc

But I don't know exactly I'm doing it correct?

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[sinful stratus]

Why did I get marked wrong here? Is my Integral correct and just my solving was off?

[safe radish]

@sinful stratus Has your question been resolved?

[safe radish]

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[safe radish]

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[hollow yacht]

Due to an awareness campaign, the number of sedentary people in a city has been decreasing. It is estimated that, at the beginning of the campaign, 4 out of every 5 inhabitants were sedentary, and that this number has been decreasing by 10% per year. If this trend continues, after how many years from the beginning of the campaign will only 1 in every 5 people be sedentary? If necessary, use log 2 = 0.3 and log 3 = 0.48.

[hollow yacht]

I don't even know where to start

[safe radish]

@hollow yacht Has your question been resolved?

[potent jolt]

so at first it was 4/5 = 80%, then it decreases 10% a year, a decrease of 10% is the same as multiplying by 90%, so if we were to make a formula for what's that number after X amount of years we could do: $f(x) = (4/5) \cdot \underbrace{90% \cdot 90% \cdots 90%}_{\text{x number of 90 percents}} = (4/5) \cdot (90%)^x$

[flat frigate]

Sires

[potent jolt]

ok so we wish to know for what x(the number of years) will that function be equal to 1/5

so let's just plug that number in:
$$
1/5 = (4/5) \cdot (90%)^x = (4/5) \cdot(9/10)^x
$$
$$
\Rightarrow 1/4 = (9/10)^x
$$

[flat frigate]

Sires

[potent jolt]

so now we have this exponential equation to solve, the way you do that is using logarithms since they turn exponentials into multiplications

let's apply the base 10 log on both sides

$$
\log (1/4) = \log (9/10)^x = x \log(9/10)
$$
$$
\Rightarrow x = \frac{\log(1/4)}{\log(9/10)}
$$

[flat frigate]

Sires

[potent jolt]

so we just need to calculate those logarithms, we can do that using a bunch of properties they have(log of 1/x is -log of x, log of x^y is y log of x, etc...)

[hollow yacht]

Why is it 1/4?

[potent jolt]

I divided both sides by 4/5

the 5 cancels out and the 4 goes there dividing

[hollow yacht]

oh ok

[potent jolt]

$$
x = \frac{\log(1/4)}{\log(9/10)} = \frac{-\log(4)}{\log(9) - \log(10)} = \frac{-\log 2^2}{\log 9^2 - 1} = \frac{-2 \log 2}{2 \log 3 - 1}
$$

[flat frigate]

Sires

[potent jolt]

and now we can use the values provided by the question

$$
= \frac{-2 \cdot 0.3}{2 \cdot 0.48 - 1} = \frac{-0.6}{0.96 - 1} = \frac{-0.6}{-0.04} = \frac{60} 4 = 15
$$

[flat frigate]

Sires

[potent jolt]

so yeah, we found x, its gonna shrink to 1/5 in 15 years

[hollow yacht]

Why did log(9) - log(10) turn into log 9^2 - 1?

[potent jolt]

my bad the 9^2 is a typo

should have been 3^2

and log(10) = 1

[hollow yacht]

okk

[potent jolt]

very much by definition

which power of 10 = 10? its 1

[hollow yacht]

You make it look easy

The equation

But thank you

Now I understood more

[potent jolt]

because it is, just practice and it will come naturally

[hollow yacht]

I tried, but I suck at log

[potent jolt]

keep in mind all those properties and you're good to go

[hollow yacht]

I can't see those things

[potent jolt]

Its a matter of knowing the log properties by heart and knowing what log means

what does logarithm actually mean?

[hollow yacht]

It's like log b = c, then 10^c = b, no? Something like that

[potent jolt]

yeah

it asks the question "What power of my base(in this case 10) equals that number?"

so say I know log 3, I know what power of 10 equals 3

if I need to know 9, you see its 3², then what power of 10 will equal 3²? its gonna be whatever the last one was but times 2

$$10^{2\log 3} = (10^{log 3})^2 = 3^2 = 9$$

[flat frigate]

Sires

[potent jolt]

tbh the way I remember those things(other than just endless practice) is thinking the logarithms turns operations down a notch: exponentiation becomes multiplication, multiplication becomes addition

similarly, radiciation(idk if thats the word) becomes division, division becomes subtraction

[hollow yacht]

Yeah

I can remember the properties of log, but I just can't have that "vision" of when to "decompose" (I don't know if that's the word, I'm using the translator) the numbers

[potent jolt]

the question gave you the log of 3 and log of 2, so ideally you want to decompose it all down to log 3 and log 2, you see 9 and 4 which are 3² and 2²

if it gave you log 5 you'd probably want to decompose into that

[hollow yacht]

Yeah. I should try to think more like this lol

But anyways, thank you!

.close

[safe radish]

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[brazen helm]

Find the smallest possible number c such that for big enough n, [ \int_{\pi}^{n\pi} \frac{\sin x}{ x} , dx < c.]

[flat frigate]

bigpufik

[brazen helm]

Very odd problem for me, any help would be appreciated

<@&286206848099549185>

[safe radish]

@brazen helm Has your question been resolved?

[safe radish]

@brazen helm Has your question been resolved?

[brazen helm]

.close

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[willow jungle]

We draw a 2024 × 2024 grid of unit squares. We call the vertices of the unit squares in the
grid lattice points; there are 2025^2 of these in our grid. Someone chose 10 of these lattice
points, and drew all the line segments that connect any two of them. Show that at least
one of these drawn line segments will contain at least two more lattice points besides its end
points.

[willow jungle]

Use the formula: for any two points (a,b) and (c,d), the number of lattice points INCLUDING THE ENDPOINTS is: gcd(a-c,b-d) +1

[safe radish]

@willow jungle Has your question been resolved?

[willow jungle]

<@&286206848099549185>

<@&286206848099549185> is this the right ping?

<@&286206848099549185> if so pls help me thx :D

[lean otter]

Yes

[willow jungle]

ok great

:)

[lean otter]

So whats going on

[willow jungle]

ok so we have 10 lattice points

randomly picked

[lean otter]

Damn probability

[willow jungle]

prove that when you connect all points to each other that at least one segment has 2 lattice points excluding endpoints

not probability its pigeonhole principle

[lean otter]

Oooooh

[willow jungle]

and sort of number parity

[lean otter]

Okay

[willow jungle]

yeah

btw 2024 by 2024 points doesnt matter at all

just pretend that its an infinite griid

[lean otter]

Yes aight

[willow jungle]

k

[lean otter]

We call it
Principle drawers

Hhhh

Okay

[willow jungle]

here's a very helpful property: for any two points (a,b) and (c,d), the number of lattice points INCLUDING THE ENDPOINTS is: gcd(a-c,b-d) +1

[lean otter]

Hmmmm

That's hard fr

[lean otter]

Okay i got an idea

Lets list a variable that explain the mouvement of the lattices

t between 0 and 1

put (x,y)=((1-t)x1+x2t,(1-t)y1+ty2)

[willow jungle]

mhm

[lean otter]

And puting a variation x and a variation of y and d=gcd(vara,varb)

Its like

[willow jungle]

ok

i see

[lean otter]

Difference coordinates between the two lattices

We note by delta yeah

[willow jungle]

yeah

[lean otter]

Then we have d|both of var x and b
Var y
Var x=d.a and y=d.b

So if d>1

It finished
We already prove it
We need only to work on d=1

[willow jungle]

alr

my teacher half showed me how to do it

it involves parity with 3x

let 0,0 be (3x,3y)

[lean otter]

Yes

[willow jungle]

0,1 = (3x,3y+1)

so on

we need gcd(a-c,b-d) to be 3 or greater

since gcd + 1 >= 4 right?

gcd >= 3

4 because we count two endpoints and 2 extra points on the line

so our proof is this: Prove that if we have ten points and we connect all of them to another point (45 total line) gcd(a-c,b-d)>=3

[lean otter]

Yes

Combination of 10

[willow jungle]

where a,b and c,d are points

[willow jungle]

yeah 10c2

btw what grade u in?

[lean otter]

Here we dont have grades
We have prepa system
So u can say thats my third year in engineering

[willow jungle]

damn

[lean otter]

But we called student that's why i named my tag undergraduate hh

S*

[willow jungle]

im in 8th grade im only 13 yo 💀

gotcha

[lean otter]

Hhhhhhhhhhhh yes difference type of studying

[willow jungle]

lol

[lean otter]

Were focusing now only on topology and analyse

[willow jungle]

cool

u do chess?

whats ur elo

[lean otter]

Yeah

[lean otter]

I donot play for like 2 years but u can say appro 1700elo

My name in chesscom is

Micheallordzada

[willow jungle]

k

im 2100

lichess

[lean otter]

Ooh thats cool

[willow jungle]

well its blitz

[lean otter]

Ur good for ur age tho

[willow jungle]

everything else like 1900 1800

rapid 1900

[willow jungle]

thx i play competitively

like uscf/fide

[lean otter]

Hmmm wanna play ?

[willow jungle]

sure bullet mayb

[lean otter]

U have fide id?

[willow jungle]

uscf

im from usa

there are rarely fide tourneys

[lean otter]

I got 1900 in elo in my fide id hhh i was playing tournaments but i got nerfed

[willow jungle]

dammmmmmnn

[lean otter]

Nice i like uscf matches

[willow jungle]

thats insane

whats ur uscf rating

[lean otter]

Im a french guy hh i have fide and national par elo

[willow jungle]

k

[lean otter]

Which is an intertournaments but ill try when im free

[willow jungle]

cool

[lean otter]

To make it to usa hh

[willow jungle]

k

[lean otter]

Anyway i cant give exact answer for ur question

But may sum get a percise one

[willow jungle]

k

[lean otter]

<@&286206848099549185>

[willow jungle]

do you know anybody rlly good at math

like reallllly good at math

[lean otter]

Hmmm maybe profs?

[willow jungle]

yeah sure

[lean otter]

Like doctorate guy is good

[willow jungle]

who?

maybe since ur already a helper you could help me reach him?

cuz ive asked this question on this discord server multiple times and nobody has solveed it

[lean otter]

?

[lean otter]

Hmmmm

Listen

This question needs to be answered as ur level

[willow jungle]

meaning?

[lean otter]

I was going to do it with variation but no way ur gonna understand this

Derivative and terrorying

[willow jungle]

well im doing precalc in 8thh grade

actually i finished precalc

soo

[lean otter]

U understand derivative?

[willow jungle]

sort of

i dont think its applicable

to this question

[lean otter]

In this question

I need u to try to imagine

There is between the distance of two lattice

[willow jungle]

yeah this question involves parity and pigeonhole

im 100% sure

that you dont need deririvatives

to solve it

[lean otter]

A really quiet distance so u gotta need to make the division between tiroir 1 and tiroir two diffrent than one

Cause they are different it cannot have 1 in pgcd
It need to be 2 or idk how

Cause if it one then the distance is not prime and the division will say that we have diffrent tiroirs

We need to catch 1 to apply it into the announcement

[willow jungle]

k

[lean otter]

So were gonna count the distincts line segments

Which is 45

[willow jungle]

yes

[lean otter]

It’s impossible to arrange 10 lattice points in such a way that all 45 segments connecting pairs of points would have
d = 1
d=1 (which would mean no internal lattice points), due to the density and distribution of lattice points
By what u called pingeonshole

U got it?

Thus

[willow jungle]

mhm

[lean otter]

The PP at least one of the segments must have
d>1
d>1 meaning it contains at least one additional lattice point along the segment excluding endpoints

[willow jungle]

yes

[lean otter]

So qed

Thats how it works

[willow jungle]

ok

i need to condense all this into a neat proof

[lean otter]

Yes it quiet easy

[willow jungle]

could you help me with that cuz i 75% understood what u said

[lean otter]

Okay what u need

[willow jungle]

just a step by step proof

be specific plz

[lean otter]

Hmmmmm

Okay listen

Mathematically

A lattice point is a point in the plane with integer coordinates such as (x,y)
Right?

[willow jungle]

yes

[lean otter]

When we pick any two lattice points and connect them

the line segment between them will have integer endpoints

[willow jungle]

duh

[lean otter]

U got it?

[willow jungle]

yes

[lean otter]

We want to prove that at least one of these segments will contain another lattice point along the segment not counting the endpoints

[willow jungle]

2

not 1

"will contain TWO lattice points"

[lean otter]

Yes thats what it mean contain another lattice

Its 1+1=2

Hhhh

Like a hole contain 1
Will contain extra one

[willow jungle]

wdym

ok whatever keep going

[lean otter]

A segment contain 2
Means it contain extra ine

Cause already it contain 1

[willow jungle]

its 4 total

[lean otter]

Yes

[willow jungle]

we need one to have at least 4 total lattice points

[lean otter]

2+2

[willow jungle]

ok got it

yes

keep going

[lean otter]

Now if u understand what next its great

[willow jungle]

ok

keep going

[lean otter]

If we connect two lattice points, say we can describe any point on the line segment AB using the parameter t where 0<t<1
And A (x1,y1) , B(x1,y1)
Thus
(x,y)=((1-t)x1+x2t,(1-t)y1+y2t)

Its describe the mouvement

[willow jungle]

hmmmm

[lean otter]

Pffft there is no way u can get that

hmmm

[willow jungle]

my teacher gave me a different way to solve it

[lean otter]

Show me

[willow jungle]

so remember that we need to prove that gcd(a-c,b-d)>=3

for at least 1 out of the 45 line segments

if you dont remmeber i can show you why

oh yeah lets get another person to help us

<@&286206848099549185>

SOMEONE HELP USSSSSSSSSSSSSSSSSSSSSSSSS HElPERS PlS

<@&286206848099549185>

[safe radish]

@willow jungle Has your question been resolved?

[safe radish]

@willow jungle Has your question been resolved?

[safe radish]

@willow jungle Has your question been resolved?

[safe radish]

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[modern bloom]

Can we use the weight function stuff to prove it?

The question is to prove this equation

[eternal carbon]

||u = (2n - 1)pi - x||

https://brilliant.org/wiki/integration-tricks/#reflections

[modern bloom]

Ty!

.close

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[onyx narwhal]

im sorry that this probably isnt the right channel for this, but im not sure where else. Are higher order differential equations more accurate in general for modeling systems or do they just allow for more complex systems?

[onyx narwhal]

.close

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[safe cypress]

.

[safe cypress]

Show that: $\cos ^{2}75\degree = \frac{2-\sqrt{ 3 }}{4}$

[flat frigate]

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

[lapis prism]

Could you not just use compound angle formula?

[safe cypress]

Wdym

[lapis prism]

Split it so it's (cos(45+30)^2

And then expand it out

[lapis prism]

This triangle also makes it pretty simple, cos75 equals 1 over root (8+4root3). Square that and rationalise the roots and you should get the answer

[safe radish]

@safe cypress Has your question been resolved?

[safe cypress]

This is as far as I got

and then i dont know how to simplify it more

[keen nexus]

I was thinking, dont simplify it

Just square it

And formulate an equation with the other value

Then show that both sides are equal

like 1/(8+4√3) = (2-√3)/4

Something like that

[safe cypress]

Nvm

.solved

[safe radish]

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[paper mantle]

could i have some help understanding the division of an expresion by another expresion when it comes to quadratics, i have a worked example here relating to the conjugate root therom done by my teacher but it doesnt make any sense to me, help would be appriciated 😭

[lapis shadow]

u want to understand how to carry out the division?

[paper mantle]

yeah 😭 😭 😭

it sounds so silly im sorry

i just wanted to understand how my teacher does it with short division

i havent had to divide polynomial expressions since aos 1 of methods

could you help me with that please and thank you🙏

[lapis shadow]

by short division do u mean the thing on the paper?

[paper mantle]

yess

[lapis shadow]

okay so u write the expression z^4 - 4z^3 + 7z^2 - 4z + 6

this is what u want to divide

u want to divide it by z^2 + 1

[paper mantle]

mhm

[lapis shadow]

u look at the largest power term in each expression

u must divide z^4 by z^2

[paper mantle]

which would be z^2??

[lapis shadow]

yeah

write that on top

now multiply what u just wrote on the top by z^2 + 1

and write it below

z^2 * (z^2 + 1) = z^4 + z^2

[paper mantle]

z^4+z^2

[lapis shadow]

yeah

subtract that from z^4 - 4z^3 + 7z^2 - 4z + 6

what does that give

[paper mantle]

-4z^3+6z^2-4z+6???

[lapis shadow]

yes

[paper mantle]

yay

[lapis shadow]

now repeat using this

keep going until u can no longer divide

so for example z^3 / z^2 is okay, and ull get z
z^2 / z^2 is also okay and u get 1
but z / z^2 is not okay, this is when u must stop

get it?

[paper mantle]

yess i think so

yes omg that makes so much more sense

[lapis shadow]

so either that, or u stop when it becomes 0

like stop as soon as the degree of the thing u r dividing is less than the degree of the divisor (which in ur case is z^2 + 1, so stop as soon as the degree becomes 1 or less)

[paper mantle]

mmmmm

okay

okay okay yeah i see how it works now

just like another side not

i look at what we were being taught in aos 1 with polynomials and we were using some methods/theorums instead of just doing division like that

ill pop the theorums up but would you recomend doing division like that or using the theorums

remainder and factor theorums

[lapis shadow]

the remainder and factor theorems are useful when u r trying to find linear factors

[paper mantle]

hmm so because we are working with quadratic factors then division is better?

[lapis shadow]

u cud say that yes. coz i dont know of any other way to divide something by an expression of degree 2 or more

[paper mantle]

it seems to be that

im looking at the rest of the lesson and all the questions say to use long division

here i think it is a lianr factor though right

since its just z-2

so its just the paramaters of the question enforcing long division there ig??

sorry if i have a bunch of typos and wrong lingo, im really tired😭

[lapis shadow]

yeah. like u can use the remainder and factor theorems to see if something is a factor or not. and "guess" factors. but if u want to find the quotient upon dividing an expression by a linear factor, ull still need long division ig...u can like try to be a little "clever" in some cases

like so u know z - 2 is a factor

so u want (z - 2)( something of degree 2 ) = z^3 + 2z^2 - 3z - 10

something of degree must have a z^2 in order that it forms z^3

and it must have a constant of +5 in order that it forms -10

so its (z - 2)(z^2 + z term + 5)

now a z term results from 5z - 2*(z term), and u want that to be -3z, so u can find the z term

in any case, division always works

when in doubt use that ig..

[paper mantle]

i think for this we had a formula like (z-2)(az^2 + bz + c) and then you like equate coefficients

[lapis shadow]

yeah thats the same thing im doing but u dont really need to write all that out

i mean u might need to if its an exam

and it requires u to

but this is kinda how u cud do it quickly if the actual steps of factorization don't matter to ur answer

[paper mantle]

cool cool cool, yeah, i think the factor and remainder theorums would come in more in methods since thats where i learnt it

but im doing specalists as well and there its more about the aplications of like complex numbers or tig functions or whatever

[paper mantle]

probably why my specalists teacher uses long division so much 💀

[paper mantle]

i hated the guess and check aspect when it came to factors of polynomials, especially since theres like more sure ways it ig 😭

i feel like i sound dumb oml, (i have exams in a week im so cooked), thanks for the help though this really helped me understand 🙏

[lapis shadow]

hope u do well on ur exams 🙂

[paper mantle]

thanksss, its only units 1 and 2 for most of my subjects so as long i just dont miss the whole exam i can just lock in next year

sorry i dragged this out so much, do i just close now 😭

.close

[safe radish]

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[brazen parrot]

guys

[brazen parrot]

for my math exam i had this question

5^(x+1) = 15 what is value of 5^(x-1)

i got 5 x 5^x = 15

5^x = 3

and then 5^x x 5^-1

= 3 x 5^-1

= 3/5

but when i plug 5^(3/5) back in

i dont get 3

btw logs are not in my course yet

we do next year

[pseudo scroll]

you can't really solve this without logarithms

Oh wait

mb

[oblique crypt]

I was about to say

[pseudo scroll]

how did you do this

[brazen parrot]

i checked it when i got home

i hate the test makers

):

[oblique crypt]

how did you cancel out X???

without log

[brazen parrot]

5^x = 3

[brazen parrot]

ghnfewoihjgewgfew

[pseudo scroll]

3/5 is correct

[oblique crypt]

wait how

am getting dumber by the min

[pseudo scroll]

5^x = 3 so 5^(x-1) = (5^x)/5 = 3/5

[oblique crypt]

ahhh[

okay

[safe radish]

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[sweet nexus]

Can someone help me solve these please will send photos

[hearty egret]

they are supposed to be done by yourself...if there is some particular thing that you cannot figure out, we can help with it

[sweet nexus]

Oh

[hearty egret]

I tell you this as the problems seem very different from each other

it is like you've just posted here the work that was given to you without even trying

[sweet nexus]

Yes it’s like our practice test for school that we get

[granite flower]

@sweet nexus
the objective of the help channel is to help you solve problems that you got stuck in while solving it or help you check your work.
when posting you should send what indicates that you did try a little in the problem.
Even if you posted we will not answer your question immediatly, we are going to help you get to the correct answers by giving you directions, explinations, or hints till you reach to the answer by yourself.

[sweet nexus]

Oh ok

[granite flower]

That is kinda the basics of learning. you learn from the channel to be able to solve other related problems

[sweet nexus]

So if I go try to solve them all I can come back with the ones I had problems with?

[granite flower]

yes

and post one at a time,
check #❓how-to-get-help for more information

close the channel and open a new one when you want to ask

[safe radish]

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

[sweet nexus]

.close

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[desert pewter]

can someone help me

[safe radish]

@desert pewter Has your question been resolved?

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[whole forge]

In an abelian grp, if $H,N \le G$, then $HN \le G$ and has order $\frac{|H||N|}{|H\cap N|}$

[flat frigate]

somethingwrong

[whole forge]

Is the statement about the order true? and any hints on how to prove it?

[obsidian oracle]

by "<=" you mean "subset"?

[whole forge]

subgrp

[obsidian oracle]

ok

well find a group morphism

to apply isomorphism theorem

not necessarily kernel equal to what you deleted, only isomorphic to it

reminder: cartesian product of groups is a group with its induced operation

[whole forge]

Let $\phi: H\times N \rightarrow NH$ be defined by $\phi(h,n)=h+n$. We then show that this is a surjective homomorphism(omitted). If $\phi(h,n)=0$, then $h+n=0$. This implies that $h$ is the inverse of $-n$ and thus $h\in N$. By symmetry we get that $n \in H$ and thus $n,h \in H \cap N$

[flat frigate]

somethingwrong

[whole forge]

hm im not sure how to proceed from here, to show that the kernel is isomorphic to H \cap N

oh wait

The above would mean that $\phi(h,n)=0$ implies $(h,n)=(h,-h)$ where $h\in H \cap N$

[flat frigate]

somethingwrong

[obsidian oracle]

yes

[whole forge]

Then right now we have that $\ker(\phi)\subseteq {(h,-h): h \in H\cap N}$

[flat frigate]

somethingwrong

[whole forge]

the other direction should be very straightforward and then it is easy to see that the RHS Is isomorphic to H\cap N

first isomorphism theorem then gives us the result at the very very top right?

[obsidian oracle]

yes

[whole forge]

okay thanks for your help, i get it now

.close

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[mystic junco]

hi im currently stuck at the cos(ktheta) part 💀

[mystic junco]

first part ive shown by using geometric series.

then the second part i attempted

my idea is to find the real part of the right side only.-.

yet i have no idea how to deal with the denominator of the right side

🫠

[misty talon]

ez

wait

[robust river]

That's a good idea

To deal with denominator just use that fact that $\frac{1}{x+iy} = \frac{x-iy}{(x+iy)(x-iy)} = \frac{x-iy}{x^2 + y^2}$

[flat frigate]

EQUENOS

[mystic junco]

aight lemme reattempt

tysm

got it 😅

.close

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[fading lodge]

My brain isn't working properly rn can someone help with this question on coordinate geometry of the line?

[fading lodge]

Idk how they get the coordinates when it doesn't have numbers on the chart

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[mighty haven]

For every natural n, this expression is real. True or false

With demonstration

Please I’m stuck with it I can’t continue

[quiet plume]

You can use DeMoivre and the definition of sin(x) in terms of complex exponentiels

[safe radish]

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[graceful pike]

guys how do i solve this implicit derivatives

[teal thicket]

i just applied a small change on both sides!

[graceful pike]

thank youuu

can you explain in details

im a bit lost

@teal thicket

[teal thicket]

sure! Where are you lost?

[graceful pike]

all the steps tbh

can you clarify the steps please

[teal thicket]

aleight!

umm.. so i applied a small change on both sodes

sides*

[graceful pike]

why arent we using y prime?

[teal thicket]

well, right!

we can do that too

i just thought youd be more comfortable with the eroor and approximation process

alright, lets derive it!

[graceful pike]

thats the thing im stuck on

[teal thicket]

here

what term did you not understand the derivation of?

[graceful pike]

the step after this

[teal thicket]

alright

here

any misclarifications?

well

And do you know the partial derivative rule?

this thing

you can use this too!

[graceful pike]

ermmm i havent taken that

[teal thicket]

oh alright!

lets focuz

ont eh first method then

where are you lost?

[graceful pike]

i have the solving

but

idk why we wrote the highlighted part instead of writing 4y’+3y^2 y’

[wet merlin]

because of product rule

[graceful pike]

so 4x alone and the y alone?

like the 4x is f

and y is g

[wet merlin]

yeah
(y)d(4x)/dx+4xd(y)/dx

[graceful pike]

i feel like i get it

but im still confused a bit

i dont get the concept ill try solving another question

[safe radish]

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[brittle storm]

How do I solve equations like this?

<@&286206848099549185>

[frozen marlin]

!15m

[safe radish]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[left hound]

What’s the original question

[brittle storm]

Solve the equation

[frozen marlin]

presumably, that is the original question

[left hound]

That is?

[brittle storm]

Yes

[sharp cliff]

Solve for x?

[brittle storm]

Yes

[left hound]

Can u once type it out for me

[brittle storm]

|-2x-5| - |-3x| = -5

so do I rewrite it as 2x - 3x = -5 - 5?

The question says there are 2 x values 🤔

[junior smelt]

Be very careful with absolute value equations

[charred saffron]

the absolute value of something is always positive, but that doesnt imply whats between the absolute value lines is positive

| -3 | = 3
| 3 | = 3

[junior smelt]

(on that note, are you aware of the piecewise definition of the absolute value function?)

[brittle storm]

|x-3| = 1

solution 1: x-3=1
x = 4
solution 2: x-3=-1
x = 2

This is how I solved previous question

[junior smelt]

In this case, you're doing something similar - to make things a bit easier, I'm gonna rewrite it as |2x + 5| - |3x| = -5

[brittle storm]

x if x > 0
-x if x<=0

[junior smelt]

Then you wanna consider the cases where

• both 2x + 5 and 3x are positive (so when x >= 0),
• when 3x is negative but 2x + 5 is positive (so when x is between -5/2 and 0), and
• where both 2x + 5 and 3x are negative (so x <= -5/2)

[junior smelt]

In each of those cases, you can then rewrite the absolute value equation using this, and then solve (and check the values you find correspond to the interval you should be in, e.g. if you take the first case, you shouldn't get something like -43 as a solution)

[brittle storm]

I don't understand how I am supposed to use it, can you show example?

[junior smelt]

E.g. for the first case, you know that |2x + 5| = 2x + 5 and |3x| = 3x, so put those into the equation and (try to) solve (2x + 5) - (3x) = -5: do you get a solution x that's nonnegative?

Second case is solving (2x + 5) - (-3x) = -5 and checking if any solution you find is between -5/2 and 0, third is solving -(2x + 5) - (-3x) = -5 and checking if the solution you find is at most -5/2

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[mystic vale]

hello, here is the problem for context

[mystic vale]

this is solution for 1

and here is solution to 3. where i don't understand what is going on.

so first of all, i think there is a missing negative sign in the result of dy

then they are substituting the dy in the first differential. but i don't get why some stuff becomes 0

the thing i've underlined

the way i think about solving it is
partial dw/dx is
= ze^y +zxe^y dy/dx +e^z + e^z dy/dx
(zxe^y + e^z ) dy/dx + ze^y + e^z

and for the constraint function if i use chain rule i get
(x^2 +2xy)dy/dx + 2xy + y^2 = 0
dy/dx = -(2xy + y^2) / (x^2 +2xy)

is it because z is a constant? but why they set it = 0 ?

[brittle storm]

@junior smelt thanks for the help

[safe radish]

@mystic vale Has your question been resolved?

[mystic vale]

<@&286206848099549185>

[safe radish]

@mystic vale Has your question been resolved?

[mystic vale]

https://tenor.com/view/sad-sigh-monkey-gif-2049020038486032791

going to sleep, if anyone can help out understanding why they seem to set z=0 ping me please so I'll see it tomorrow

[safe radish]

@mystic vale Has your question been resolved?

[safe radish]

@mystic vale Has your question been resolved?

[safe radish]

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[soft halo]

Can someone help me understand the inequality, specifically where the -1 in the denom come from in the ans key?

[fathom jewel]

Basically one way to look at it is, you can increase a fraction by making the denominator smaller, e.g z²-1 is smaller than z²+1

[soft halo]

Ohh I see, would the statement hold if I just get rid of the 1? like z^2-1 smaller than z^2?

[fathom jewel]

z^2+1 > z^2 > z^2 -1

[soft halo]

alr, thx i got it now

.close

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[junior smelt]

.(reverse triangle inequality for the "smaller denom bigger fraction", as those are complex numbers)

[soft halo]

sorry how to apply that here? I googled reverse triangle inequality and it's a difference term

[junior smelt]

.reopen

[safe radish]

✅

[sand leaf]

After you're done with him can i get some help then please?

[safe radish]

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.

[junior smelt]

Reverse triangle inequality says $\abs{\abs{z} - \abs{w}} \leq \abs{z + w}$, if you've seen it, so that implies $\frac1{\abs{\abs{z} - \abs{w}}} \geq \frac1{\abs{z + w}}$ (provided no dividing by zero etc)

[flat frigate]

@junior smelt

[soft halo]

oh I was looking at this one

unless they are the same thing

[junior smelt]

In this case, replace z with z^2 and w with 1 (and on C, you have |z| = 5, so you know |z^2| = |z|^2 = 25 and the "included absolute value" is positive)

[junior smelt]

(and they are, you could e.g. replace b with -b and then you get the form I stated)

[soft halo]

I see, that makes more sense. Thank you so much

[junior smelt]

If you're happy you can close the channel again if you want (reopened it so it didn't get locked/recycled)

[soft halo]

Alr will do 🙏

.close

[safe radish]

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[safe radish]

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[flat frigate]

gmdn