help-13

[near quiver]

On the set of real numbers, this associative law of composition defines itself as: x * y = 2xy - 6(x+y) + 21

I need help at b)

Prove that N = that monstrosity is a natural number

[past wave]

WHAT?

[near quiver]

It's like creating new mathematics

[past wave]

Oh

[near quiver]

Just like addition and multiplication

[past wave]

I get it

x*y is now a function in x, y

Basically f(x, y)

[near quiver]

We can create any function for x and y belonging to a set of numbers whose outputs also admit to the same set of numbers

Yes.

For this specific question, my f(x, y) is this

[past wave]

For us * means multiplication

That's why I was confused

[near quiver]

I get it, but the problem gave it away

[past wave]

In computer programming this is widely used

[near quiver]

Obviously, there's no dot or just )(

I was beginning to question whether you were trolling

[past wave]

Ok I understand the problem now

Thanks

[near quiver]

Dw

[past wave]

Lol

[near quiver]

Well uhm

Do you have any idea?

[past wave]

Lemme rethink

[near quiver]

Alright

.

<@&286206848099549185>

[near quiver]

English translation.

[past wave]

What did a*b*c mean then

[near quiver]

a) b) ?

[steep badge]

@past wave padhle bhai

[past wave]

Abbe mai iss liye puch rha hu kyuki

a*b = (a-3)(b-3) + 3
a*b*c = (c-3)((a-3)(b-3) + 3

Wait this is useful

[past wave]

@near quiver

[near quiver]

Okay so first of all

a * b would be 2(a-3)(b-3) + 3

Then that * c

[past wave]

Yes just multiply the first term by 4

But the constant term just disappeares

[near quiver]

Which is... 2(2(a-3)(b-3) - 3)(c-3) + 3

[past wave]

2(2(a-3)(b-3) + 3 - 3)(c-3) + 3

So 3 cancels

[near quiver]

Okay true true

[past wave]

I think it's doable now

[near quiver]

4(a-3)(b-3)(c-3) + 3 you say?

[past wave]

Bhai tujhe kaise pata chala mai hindi bolta hu?

[past wave]

Yes

[near quiver]

Then 8(a-3)(b-3)(c-3)(d-3) + 3

OH

[past wave]

Bata bro

[near quiver]

And there should be a cbrt of 27

[past wave]

But I don't see how we solve it now

[near quiver]

Well it's easy

[past wave]

a²-b²?

[near quiver]

It'd be 2^4047(a-3)(b-3)...(z-3) + 3

And since one of those is cbrt of 27

Cbrt of 27 - 3 is 0

Thus making the whole thing fall apart

[past wave]

Ohh

So my ans was correct lol

[past wave]

Why don't such things happen during exams

[near quiver]

:))

Always.

WOW

That was genious ngl

Thanks a lot!

[past wave]

Nice question

[near quiver]

True, we all love maths ^^

Have a nice day or evening, and thanks for sticking out

Patience is key!

/close

!close

How do I close?

[steep badge]

.close

[near quiver]

Thank you

.close

[cedar kiln]

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[crimson sedge]

So I've been trying to solve this for a while (I failed) but I want to understand it more deeper, I don't understand why the first question's answer is -2 and the second question's answer is 3, can someone help me?

[exotic sphinx]

do you see where the minimum point is

[crimson sedge]

No, i've been trying to figure it out, is the minimum point where the line points to (3, 2) ?

[exotic sphinx]

yes

[crimson sedge]

Oh

[exotic sphinx]

so if you want to move the point vertically so that it is on the x-axis you will have to make the y cordinate 0

[crimson sedge]

So if the y = 0 then the extreme point at the top will point at 3, and the minimum point will be -2, so I do understand it now

[exotic sphinx]

yea that

[crimson sedge]

Ohhhhh

Thx so much

[exotic sphinx]

you're welcome

[cedar kiln]

@crimson sedge Has your question been resolved?

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[fathom star]

I used the formula for acceleration to find the missing velocity.

oh wait

thats the accelerations magnitude

Guessing I need to do this to it

Ayy

.close

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[glass spruce]

why is it that if f is differentiable at a poin then its continuous at this point?

[ancient lodge]

https://math.stackexchange.com/questions/1048539/why-differentiability-implies-continuity-but-continuity-does-not-imply-differen

[glass spruce]

if we take f(x) = (x^2 if x < 0 else x^2 + 1)

then the derivative of f(x) at 0 from the left is the same as from the right, so its differentiable at 0

but clearly its not continuous at 0

[dire geode]

your claim is circular. you assumed the derivative of f(x) is 2x at x=0

use the limit definition of derivative on your function to see why it's not differentiable at 0
https://tutorial.math.lamar.edu/classes/calcI/DefnOfDerivative.aspx

[glass spruce]

wdym

how did i assume?

the definition tells us

oh wait

oh

[dire geode]

prove this

[glass spruce]

oh yeah the definition from the left is not defined

oh wait

does that mean its not defined?

that means its infinite

[dire geode]

this f(x) is not differentiable at 0

[glass spruce]

right yeah

thanks

.close

[cedar kiln]

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[cedar kiln]

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[crimson sedge]

How can I calculate the second limit?

[hollow trail]

do you know derivatives yet?

[crimson sedge]

No

[dire geode]

use squeeze theorem

[crimson sedge]

What's that

?

[dire geode]

...

what HAVE you learned

[crimson sedge]

I can send u the whole ppt

of the class

[dire geode]

just show the main results

not every little step is important

[crimson sedge]

That's it

[crimson sedge]

I think I saw it

[dire geode]

🤔

[crimson sedge]

when I searched the proof by myself

but in spanish

so I didn't know the name

but I wasn't taught that

[crimson sedge]

is this helpful?

there is no x in the exponent

[thick heath]

puede ser que lo pasaran como teorema del sandwich, sino puede que sea con otro nombre

[crimson sedge]

Lo vi en yt

pero no en clase

no me lo han enseñado

lo vi al buscar la demostración the lim x-> 0 = senx/x = 1

pero no sé qué hacer con esto

idk what's that

Yeah

[thick heath]

para el primer caso por lo menos puedes utilizar esto creo

[crimson sedge]

ah eso sí sé

ya lo había hecho

grc igualmente

<@&286206848099549185>

Is there some way to calculate this without calculus?

[dire geode]

you can also set x = log(y) and and x goes to 0, y goes to 1

write the limit in terms of y

[crimson sedge]

mm

idk

like x = lny

?

and I replace x with that

[crimson sedge]

I'm sorry, but could u walk me through it, I haven't done this before

[dire geode]

just do this

e^(log(y)) = y

[crimson sedge]

ok, why?

[crimson sedge]

how is this true?

like why

[dire geode]

wot

have you learned logs before

[crimson sedge]

yes

[dire geode]

what do you think e^(log(y)) equals?

[crimson sedge]

well if log(y) = x then e^(10^x)

but maybe you're confusing it w ln

[full rover]

he means ln(x)

the natural log

[dire geode]

log = log base e in math

,w log(e)

[wraith dagger]

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Requested by rie.mann

[full rover]

after a certain level of math we just consider log to be ln by default

[crimson sedge]

I was taught that ln is log base e

and log is base 10

default

[full rover]

yeah i know the terminology switches

[crimson sedge]

I

[full rover]

when we're talking about calculus, log usually means log base e, or lnx

[crimson sedge]

I'll use ln

[full rover]

that works fine

[crimson sedge]

m ok

x = lny

@dire geode I did it, what u said

[dire geode]

take the limit

[dire geode]

.

if x is zero, what's 0 = log(y) ?

[crimson sedge]

mm y = 1

but for the limit

it would be like when y -> 1?

like how

do I change that part

pls idk what to do

[dire geode]

that's right

did you learn any limits about log

[crimson sedge]

no

[dire geode]

and remember any properties of logs

$\log(a^b) = b \log(a)$

[wraith dagger]

riemann

[crimson sedge]

mm how do I use that here?

[dire geode]

you have something like the right side

use uhhh $\frac{a}{b} = \frac{1}{b/a}$

[wraith dagger]

riemann

[bitter perch]

lmao

[crimson sedge]

done

[dire geode]

can you take the limit as y goes to 1 now?

[crimson sedge]

is the same

1/0/0

which is 0/0

[dire geode]

ah shit i missed a - 1

[full rover]

this makes me realize just how much i take lhopital's for granted damn

[dire geode]

it's more convenient to set x = log(y) - 1

the numerator will just be y now

and your limit as y was going to 1 before

[full rover]

im curious what your teacher intended you to do tho because this limt is mildly annoying without calculus

[dire geode]

but now with the new x, y will go to 0

[bitter perch]

yeah i've been looking at this for a bit and i have no idea

[dire geode]

top should just be y, and denominator should be ln(y+1)

[bitter perch]

can you graph this? or does your teacher want it done without

[crimson sedge]

m isn't it y/e - 1

bc it's e^(lny-1) - 1

[dire geode]

sorry i messed up again

[dire geode]

x = ln(y + 1)

pretty sure that works now

x -> 0 so y -> 0 again

e^x - 1 = e^(ln(y+1)) - 1 = y

[crimson sedge]

mm idk, perhaps I could, but I haven't solved any exercise of limits like that, neither the teachers

[dire geode]

you'll need one of these next

[bitter perch]

your teachers don't show you how to solve a limit, by looking at a graph?

[dire geode]

you're not really helping can you stop

[crimson sedge]

mm how

[crimson sedge]

Juust showed the meaning of it

[dire geode]

use this

[crimson sedge]

here?

[dire geode]

almost there. this step again

[full rover]

ohhhh

[crimson sedge]

sorry to say this, but I really don't see it

[dire geode]

and this

[dire geode]

can you not spoil it

[full rover]

sure lmfao

[dire geode]

once you bring y to the denom, what do you get

[crimson sedge]

1/ln(y+1)/y

[dire geode]

you have something like this in the denominator

[crimson sedge]

oh

m

like 1/y is the b

[dire geode]

yea

[crimson sedge]

do I put it in e or y+1

?

[dire geode]

i don't know what you mean by "put it in e"

you should only have ln and y

[crimson sedge]

there is a property

[dire geode]

yea if you're jumping ahead go for it

[crimson sedge]

then to the other one

like a root

[crimson sedge]

for what

[crimson sedge]

mm how

like for multplication

do I separate them

[dire geode]

i have no idea what you're doing

[crimson sedge]

this is the property

but I'll asume you mean to put it in x

[dire geode]

there's no need to change base

[dire geode]

did you use this yet

[crimson sedge]

yes

[dire geode]

show

[dire geode]

yea now pass the limit into the log

$\lim \log(...) = \log(\lim(...) )$

[wraith dagger]

riemann

[crimson sedge]

mm idk that, but ig it's like taking the inside part to the limit?

and the log outside?

[dire geode]

yes

similar to this

[crimson sedge]

how do I deal with the 1/

[dire geode]

the lim swapped with exponential

lim 1 /(stuff) = 1 / lim (stuff)

[crimson sedge]

?

[dire geode]

yea that's right

[crimson sedge]

oh

ok

mmm I get -1

[dire geode]

my bad this is wrong again

this is right

this is not

$\lim \frac{1}{\log(...)} = \frac{1}{\lim \log(...)}$

[wraith dagger]

riemann

[dire geode]

this basically

[dire geode]

once the limit and log are in the denominator, then use this

[crimson sedge]

is 1

[dire geode]

,w lim x to 0 (e^x - 1)/x

[wraith dagger]

Results provided by WolframAlpha

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Requested by rie.mann

[dire geode]

yea

long problem

[crimson sedge]

thx 🙂 yeah
can I post other mm

[dire geode]

https://socratic.org/questions/how-do-you-solve-the-following-limit-e-x-1-x-as-x-approaches-zero

probably should just close and reopen with a new one unless you need everything in this channel

[crimson sedge]

thx

I'll read it

[crimson sedge]

ok yes

.close

.close

[cedar kiln]

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[astral drum]

is my answer for question B correct since aparently its suppose to be a minimiser but i think i found a point that contradicts it

[mellow haven]

youre correct

[cedar kiln]

@astral drum Has your question been resolved?

[astral drum]

oh thanks

[mellow haven]

how do you know that its supposed to be a minimiser?

[astral drum]

uhhh GPT lol

thats what they said

but ig thats why u dont trust it 😂

[mellow haven]

yes exactly, its not that good "yet" at solving problems :d

theyre asking you to check whether or not the condition is satisfied

so it doesnt really have to be satisfied

maybe it is, maybe not :p

[astral drum]

can u help me with this tho? please

im like stuck where to rly begin

[mellow haven]

me not study this lol, but ill try to look

[dire geode]

don't use gpt for math

[mellow haven]

post this in another help channel, you may get better/faster help

[cedar kiln]

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[novel magnet]

hey i have been banging my head trying to solve it ,any idea ?

[dire geode]

have you learned $\lim_{t \to 0} \frac{\sin(t)}{t} = 1?$

[wraith dagger]

riemann

[novel magnet]

oooh swapping the variable ! i didnt think about it !

[dreamy void]

3/3 screams

[novel magnet]

ig it was really simple after all

ty all :D

.close

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[remote trail]

erm

[vagrant pewter]

<@&268886789983436800>

This is not it

[cedar kiln]

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[celest seal]

very funny scam

dropshipping has to be like, top 3 funniest scams

"just act as a middleman for shitty chinese goods bro, youre the real scammer not me"

"buy my guide for $500 thatll teach you how to do this"

[vagrant pewter]

Fr

That guy is capping

Mlm type shit

[cedar kiln]

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[wheat sphinx]

Can anyone help with this Calculus volume problem please

[wheat sphinx]

am I using the right volume formula? I feel like I have my upper/lower bounds incorrect

[carmine bronze]

The question states that you are to revolve the region about the x-axis.

[remote trail]

why do you feel like your bounds are incorrect?

[wheat sphinx]

so my bounds are from 0 to 20

[remote trail]

didn't you use from 0 to 5 in your working?

what's wrong with using the bounds from 0 to 5

[wheat sphinx]

since the rectangle is horizontal and not vertical

I'm supposed to be integrating for y so the formula should be dy right

?

[carmine bronze]

Do you understand which method the question is asking you to use?

[remote trail]

I'm pretty sure you should be integrating wrt x

[wheat sphinx]

I think it's the disk method

[remote trail]

via the disk method

[carmine bronze]

No.

[wheat sphinx]

shell method?

[carmine bronze]

A horizontal rectangle that revolves around the x-axis is the Shell method.

[wheat sphinx]

ok i kinda figured something was off

[remote trail]

doesn't the question say revolving around the x-axis?

given we are rotating the region y=4x from 0 to 5 wouldn't it be logical to use the disk method

[wheat sphinx]

Yes but they want me to draw HORIZONTAL rectangle

[remote trail]

or am I reading the question incorrectly

[wheat sphinx]

in another problem i did the same for VERTICLE rectangle

see

did i do problem 5 correct atleast LOL

[carmine bronze]

The second part of the question states to draw a horizontal rectangle.

[wheat sphinx]

@carmine bronze

nvm I know i did it right

[carmine bronze]

Do you know the general formula for the Shell method?

[wheat sphinx]

no

[carmine bronze]

The Shell method uses the formula for a cylinder to calculate the volume.

What is the formula for the volume of a cylinder?

[wheat sphinx]

this is the formula

[carmine bronze]

You are not integrating with respect to y, wry, though.

It's usually better to write it out as

Obviously use the appropriate "with respect to x" or "with respect to y".

[wheat sphinx]

👍

[carmine bronze]

This form makes it more apparent what you are evaluating.

r(y) * h(y)

The equation for r(y) is hopefully obvious.

h(y) requires a bit of manipulation.

[cedar kiln]

@wheat sphinx Has your question been resolved?

[wheat sphinx]

@carmine bronze so for r(y) I have y/4 but how do i get h(y) Is it 5 - 4x?

[carmine bronze]

No, you will need to solve for the height in terms of y.

[wheat sphinx]

?

[carmine bronze]

Almost.

These are the two endpoints in which you are calculating the height.

If you use the distance formula, what would the length/height of that segment be?

[wheat sphinx]

5-y/4 ?

[carmine bronze]

Yes.

And what would the equation for the radius be?

[wheat sphinx]

thats what i have for my formula for volume

[carmine bronze]

The radius is incorrect.

[wheat sphinx]

oh wait i think i know

[wheat sphinx]

it's just y

?

[carmine bronze]

Correct.

[wheat sphinx]

those diagrams you showed helped alot

[cedar kiln]

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[olive dirge]

hi

[olive dirge]

just had a small question

[bleak viper]

which is?

[olive dirge]

one sec ill send a pic

[olive dirge]

ok when doing limits with square root and multiplying it

why does the sign on the rightneeds to be positive?

[bleak viper]

because $(a-b)(a+b) = a^2-b^2$

[wraith dagger]

Astar777

[bleak viper]

its to remove the square root

[olive dirge]

OHH

ok lets goo your the goat

.close

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[brittle terrace]

how come the limit is equal to 0 and not undefined?

heres my reasoning
x is slightly larger than 3, so x^2 slightly larger than 3, 9-x^2 slightly less than 0 = negative, sqrt negative = undefined

[hollow trail]

you are correct that the limit as written is not defined. they probably meant 3^-

[brittle terrace]

oh alr, thanks

.close

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[hallow quartz]

Need help computing a basis for the kernel of a linear transformation

[hallow quartz]

i know that the kernel are the vectors in the V space such that when transformed will equal 0. However we have done examples of this using matrices before but struggling using polynomials. If someone could explain what steps i need to take or how to do it would be appreciated.

[hollow trail]

post the full question

[hallow quartz]

<@&286206848099549185>

[cedar kiln]

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[cedar kiln]

@hallow quartz Has your question been resolved?

[cedar kiln]

@hallow quartz Has your question been resolved?

[cedar kiln]

@hallow quartz Has your question been resolved?

[cedar kiln]

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[mental trail]

!status

[cedar kiln]

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

[sharp talon]

my bad

i'd say i dont know where to begin

[mental trail]

ok

maybe compute what the product of those 2 matrices on the left should be

you can keep g outside for now

[sharp talon]

okay

46 0
7h-28 h-24

like that?

[mental trail]

not sure about the bottom right

you should check it again

[sharp talon]

isnt it (h x 1)+(-4 x 6)?

[mental trail]

-4 and 6 are on the same matrix...

[sharp talon]

damn careless mistake

mb

46 0
7h-28 h+42

like that?

[mental trail]

yess

[sharp talon]

yess

im slow at this

[mental trail]

alright

so multiplying every entry by g

should get you to
1 0
0 1

maybe focus on a specific entry

to find the value of g

[sharp talon]

g x (7h-28)?

like all four number?

[mental trail]

yes all 4 numbers

[sharp talon]

46g 0
7(h-4g) h+42g

like this?

[mental trail]

uh

g*[7h - 28] for the bottom left

g*[h+42] bottom right

you multiply the WHOLE entry

[sharp talon]

eh

dont understand

😅

[mental trail]

if you have to multiply some compound expression by g

you multiply the whole expression by g

not the stuff of your choosing

so h+42

multiplied by g

is not h+42g

because you only multiplied 42

when you're supposed to multiply everything by g

instead

it should be g*(h+42)

[sharp talon]

ohh

i see what u mean

i only multiply the numbers and leave h behind

right?

[mental trail]

?

well that is what YOU did

and I'm saying it's wrong

[sharp talon]

my mistake is i left h behind

[mental trail]

yeah

h is also a number

you don't know which number, but it's also a number

[sharp talon]

alright

i see what u mean

okay next step

[mental trail]

ok

now the matrix that you have

is supposed to be
1 0
0 1

so each entry must match, right?

[sharp talon]

yes

so 46 x g = 1?

and the rest follows

right?

[mental trail]

yes it should

[sharp talon]

so 1/46 =0.02173913

like that

[mental trail]

yeah g = 1/46

[sharp talon]

as for the one with h idk how to do it

[sharp talon]

so this is final?

46g 0
g(7h-28) g(h+42)

with only 46g u can find g ?

and then use the g to find h?

[mental trail]

yeah, 46g is the best entry to find g

since it's an entry that only depends on g

[mental trail]

afterwards you will know the value of g so only h is unknown

[sharp talon]

okay

im not good at algebra

g(7h-28) = 7h/46 x -14/23

like that?

[mental trail]

"x"?

[sharp talon]

multiply

[mental trail]

no

[sharp talon]

no?

then whats the symbol

[mental trail]

I don't know, maybe I thought it was a typo from you

2*(2+3) is not equal to 2*2 x 2*3

[sharp talon]

ooo

then what is it suppose to be

[mental trail]

it's +

[sharp talon]

so its 22 + 23

[mental trail]

yes

here you just distribute the 2 on each term

[sharp talon]

ooo

[mental trail]

it doesn't change the fact that you were supposed to add them

[sharp talon]

1/46(h + 42) = h/46 + 21/23

[mental trail]

alright, just quick disclaimer, we don't need to do that if you want to find h

[sharp talon]

eh

[mental trail]

like

take the bottom left entry

g(7h-28)

it's supposed to be 0

[sharp talon]

yews

yes

[mental trail]

since g is not 0

the other term you multiply it with needs to be 0

7h - 28 = 0

[sharp talon]

wait 4h - 28?

not 7h - 28?

[mental trail]

oops

[mental trail]

edited

[sharp talon]

so h = 4

is g(7h-28) still 4?

[mental trail]

no, if h = 4

then 7h - 28 is 0

then multiplying by anything (like g) doesn't change the fact it's 0

[sharp talon]

wait

explain it again

sorry man english is not my main language

nvm

i get it now

if the answer for an equation is 0, no matter how many multiplying to the equation, its still 0

like that

anyways

thank you so much for the help @mental trail

[cedar kiln]

@sharp talon Has your question been resolved?

[cedar kiln]

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[elder jasper]

I’m not sure if I’m going the right way with this

[dreamy void]

no

you need product rule if you wanna continue this way

[elder jasper]

Ahh your right

[toxic sky]

I think use chain rule is easier

[elder jasper]

^2

[toxic sky]

You forgot power 2?

[tepid ginkgo]

Hy i am a new here can anyone tell me how this group works

[elder jasper]

I’m confused how to get domain

[elder jasper]

Go to help rooms and they will come after you ask a question

[tepid ginkgo]

i am a 11th grade student can i find school students here or only graduates or postgraduates

[elder jasper]

i think they use all kinds of tutors here

[tepid ginkgo]

okayy what are you pursuing by the way

[elder jasper]

civil engineering

wbu

[tepid ginkgo]

what is your nationalty

[tepid ginkgo]

a 11th grader student

[elder jasper]

@gleaming path

[cedar kiln]

@elder jasper Has your question been resolved?

[elder jasper]

nope

[cedar kiln]

@elder jasper Has your question been resolved?

[elder jasper]

.close

[cedar kiln]

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[cedar kiln]

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[dapper raven]

Does anyone know what they mean with $S_t \subseteq S$?

[wraith dagger]

Katharine

[dapper raven]

I don't mean the notation i know that i just don't get what they're asking.

I don't get how anything is provable in their question

[crimson delta]

for example if t=1 then S_t={1}. because all the elements in S_t add upto t

for every number t you can pick some of the numbers from 1 to 24 such that the sum of those is t

[dapper raven]

I don't see how that could possibly translate to a proof by induction, there is not regularity

[crimson delta]

if you have some numbers which add to t, how could you slightly change them so that they add up to t+1

[crimson sedge]

i think its like since the numbers in S are distinct and range from 1 to 24, there are combinations of these numbers that can sum to every value in the range 1 to 300. so basically you can use induction to show that each step can be reduced to the next by either adding a small number or shifting around subset elements

[dapper raven]

proof by induction is horrible when it's anything but a problem where i can easily show that case k+1 uses case k to be shown to be true

:(

i don't understand how i can show that if it holds for k then it holds for k+1

because if it holds for 24

then 25 is completely different case

the subset is a different size

and similarly for those cases that have subsets of any size bigger than 1

[crimson delta]

what do you do for the step 16->17

or lets say 27=25+2 -> 28

[cedar kiln]

@dapper raven Has your question been resolved?

[cedar kiln]

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[sand cradle]

I want to ask if there is generally a systematic way to do these or if you just try to visualize it in your head
I.e. think of more complex figures, not a cuboid

[cedar kiln]

@sand cradle Has your question been resolved?

[cedar kiln]

@sand cradle Has your question been resolved?

[cedar kiln]

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[tawdry plaza]

Idk how to do 2b

[tawdry plaza]

Can someone pls hell

Help

[daring karma]

hi

is this year 11? GCSE mock?

Alright,

[tawdry plaza]

Nah this is my a level class

💀

You guys did this at gcse?

😭

[daring karma]

ah I am in year 13

[tawdry plaza]

Probably further then

[daring karma]

yep

[tawdry plaza]

Fair enough

[daring karma]

any ideas to the answer?

[tawdry plaza]

Yeah so I think you gotta put discriminant to 0

But idk how to find k from that

[digital cliff]

i detected some fellow tea drinkers

[tawdry plaza]

🤝

[digital cliff]

your discriminant should never have xs in it (for a quadratic of x)

[daring karma]

(1) To find the discriminate of a quadratic equation you do f (x) = x2 - ( k + 8)x + (8k + 1) Use the formula (Triangle) = b2 - 4ac. Where a = 1. b = - (k + 8) and c = 8k + 1

[tawdry plaza]

Oh so I did the 1st party wrong then it seems

If a = 1

That's probably why I can't do the 2nd bit then

[daring karma]

(2) Substitute the values of a. b. and c into the discriminate formula to get
(Triangle) = { - ( k + 8). and c = 8k + 1

Do you understand?

[tawdry plaza]

Yeah

[daring karma]

So the question which you are stuck on is B, correct?

[tawdry plaza]

Yeah

I think I did a wrong so I'll have redo that

[daring karma]

It'll be k = 6 or k = 10

There is the answer.

[tawdry plaza]

Yo what other subjects do you take?

[daring karma]

For GCSE?

[tawdry plaza]

A level

[daring karma]

Geography, Maths

Also Physics

[tawdry plaza]

I'm.takjng physixs aswell

[daring karma]

I am going to try and get a PHD in physics

[tawdry plaza]

Are the rumors true that it becomes hell?

Someone's lurking lmao

[daring karma]

I done a mock paper, the examination is pretty hard. But I got a good grade in it.

[tawdry plaza]

Fair enough then

Well imma vet on with my work.ig

.close

[cedar kiln]

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[cedar kiln]

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[fleet canyon]

Integrate the following where b is a real positive number :

[fleet canyon]

I’ve tried solving this, but I fail to compute the integral at the end. I’m getting smth along the lines of e^(i(inf)), even though that does not exist right?

Here’s my work so far :

I’ll get e^((iω + b)(inf)) which doesn’t exist

[dawn junco]

you typo'd while copying on the 2nd page

[fleet canyon]

Whoops that’s true thanks

[dawn junco]

but anyway are you so sure that limit to infty doesn't exist

what do you think $|e^{(i \omega-b)\eta}|$ is ?

[wraith dagger]

aPlatypus

[fleet canyon]

I have the intuition that this limit doesn’t exist, because it seems as though it could give any value on the unitary circle of the complex plane right?

[dawn junco]

true but you're neglecting the real part inside the exponential

[fleet canyon]

The norm of a complex number? I’m not sure

[fleet canyon]

Hmm ok yeah

Separating the two should help you mean?

[dawn junco]

that can help yes

[fleet canyon]

Ok I’ll try that and see how it goes

[dawn junco]

you still forgot the **-**b eta

quite essential here

[fleet canyon]

Yea mb

[dawn junco]

I assume b is positive?

[fleet canyon]

I’ll put it back everywhere one second

[fleet canyon]

Oh yes srry

I mentioned it earlier

[fleet canyon]

.

[dawn junco]

ah yeah

[fleet canyon]

Hmm so e^(-bM) would give 0

And even if e^(iωM) diverges

Can I assume that it would still give 0 when multiplied by zero?

[fleet canyon]

Wait no that doesn’t diverge right?

Rather it just

[dawn junco]

well 0 * something that diverges is quite uncertain

[fleet canyon]

Gives an infinite amount of possible values

[fleet canyon]

Yes true

[fleet canyon]

So it’s rather this right?

[dawn junco]

that's why I mentioned the norm earlier

[fleet canyon]

Whatever value it will give

That times 0 will give 0 still

Right?

[dawn junco]

if you can show the norm goes to 0, the thing itself also goes to 0

with that split you made, the norm is easier to compute

[fleet canyon]

Oh is e^(-bM) the norm?

[dawn junco]

yeah

gg

[fleet canyon]

Wait wait

Even if it has a complex number?

In it?

Because it’s not in exponential form yet right?

[dawn junco]

what has a complex number

e^(-bM) is real

M is real

[fleet canyon]

Idk I assumed ω was a complex number

The exponential form is re^(iθ)

So I would need to get rid of the ω somehow if I want this exact form right?

If ω is a complex number

(Tbh idk if it’s a complex number)

[dawn junco]

i doubt it's a complex number yeah

anything is said about w ?

[fleet canyon]

At other places in the book ω denotes a complex number

The question right under that one actually defines ω as

Actually no

Idk

But right under in the next question

ω(z) is given

Where ω(z) = arccos(z)

And z obv is a complex number

[dawn junco]

well w might as well be an angular frequency for all we know

[fleet canyon]

But anyways that’s for the next question

[fleet canyon]

True

It’s ill defined

So maybe whether it’s complex or not doesn’t matter

So perhaps we should interpret it as a complex number anyway

[dawn junco]

let's just say it's real here

[fleet canyon]

Since the set of* all real numbers is* a subset of complex numbers

[fleet canyon]

Hm Alr

[dawn junco]

the problem if it's complex is that the integral might not converge for all w

so like you can look at the conditions of convergence in that case if you want

but I'm lazy

[fleet canyon]

Oh yeah we didn’t learn that

[fleet canyon]

Haha dw u helped so much

But yeah since we didn’t learn that

Then I’m guessing I shouldn’t do it?

Right?

[dawn junco]

well it's just doing what we're doing here

but more carefully

[fleet canyon]

Hmm

What?

[dawn junco]

cause w complex will influence that norm we found

[fleet canyon]

Yuh that’s what I was thinking

[dawn junco]

but let's do real first

[fleet canyon]

If I expand w in Cartesian form

Then maybe I can do smth?

[dawn junco]

essentially yes

[fleet canyon]

I’ll just do it it’s just a few extra lines

Ok lemme try

Wait did I make a mistake?

On line 6?

[fleet canyon]

When I took out the exponent on line 6 I left a minus there

And so now I have positive e^(bM) which diverges

[dawn junco]

[fleet canyon]

We’re left with bM right?

Yeah I think it was a mistake

Help

<@&286206848099549185>

[cedar kiln]

@fleet canyon Has your question been resolved?

[cedar kiln]

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[idle crystal]

similar triangles is all ive found😭

[limber dawn]

you could just like break everything into x, y components

that's a very non-mathematician way to do it tho

[idle crystal]

huh

[limber dawn]

like define (x, y) coordinates on this x.x

[idle crystal]

oh

[limber dawn]

very engineer-ish way to do it lol

[idle crystal]

howd i find where point d and e are though

[limber dawn]

you can use just sin and cos

well

you have all the info for triangle ABD for example

enough to get the (x, y) coordinate of point D with not too much work

[idle crystal]

oh

could anyone tell me the "geometrical" way😭

<@&286206848099549185>

[limber dawn]

i want to know too

[cedar kiln]

@idle crystal Has your question been resolved?

[reef halo]

im too lazy to solve this but just plug in everything into the bottom formula

[limber dawn]

ah so you found z the angle?

that makes sense

[reef halo]

hold on use this one

[idle crystal]

z is 45

.close

[cedar kiln]

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[cedar kiln]

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[blazing zephyr]

find all positive integers (a,b) such that ab | a^2024+b

[blazing zephyr]

status 1

[vague rapids]

I think starting off with odd/even could help

It is pretty clear that a and b both need to be even

[blazing zephyr]

mhm

hmm i guess we can deduce b is a multiple of 4

wait huh

[blazing zephyr]

(1,1) works?

[vague rapids]

that's a trivial solution

[blazing zephyr]

i dont see how you can rule out both odd

[vague rapids]

make cases

[blazing zephyr]

wdym

[undone star]

if both odd, then ab is still odd, but rhs u have odd + odd

i dont know how that rules it out tho

[upper laurel]

3 | 6?

??????

[vague rapids]

oh mb

[upper laurel]

what is happening with yall

here I was thinking you somehow figured this out instantly

[undone star]

sry we are nuckleheads

[upper laurel]

the best Ive found rn is that
ab | a^2024 + b
abk = a^2024 + b
abk - a^2024 = b
(bk - a^2023) a = b
bk - a^2023 | b
which is certainly something, if you have a and b be relatively similar in size then k has to be incredibly large

[blazing zephyr]

hm ;-;

[upper laurel]

then again ab is already much smaller than a^2024 + b so that wouldve been clear anyways

[loud gyro]

One of the answers is b is of form a^n where n<2024

[upper laurel]

knucklehead moment

[loud gyro]

but that may not be the only possible b

[blazing zephyr]

n=0 :3

[loud gyro]

n = 0 works

oh no it doesnt

mb

[upper laurel]

ab | a^2024 + b
b | 1 + b
(1, 2) is a trivial solution

(its not)

whoops

[loud gyro]

Welcome to the thread where everyone is making dumb mistakes

[upper laurel]

you missed the term "knucklehead"

[blazing zephyr]

divisibility is so ass 🔥

[loud gyro]

its not even thread 😭

[upper laurel]

you need to use the proper terminology

gain the proper Knucklehead status

[loud gyro]

I am a knucklehead fr

[upper laurel]

we found (1, 1) as the only solution for a = 1 this is such a discovery

[loud gyro]

imma suggest in meta to rename helpful as knucklehead

[upper laurel]

a = 2
2b | 2^2024 + b
2bk = 2^2024 + b
2bk - b = 2^2024
(2k - 1) b = 2^2024
2k - 1 | 2^2024
2k - 1 is odd, so 2k - 1 = (1 or -1)
2k = (2 or 0)
k = (1 or 0)

(2(0) - 1) b = 2^2024 would mean b is negative
(2(1) - 1) b = 2^2024
b = 2^2024

only solution for a = 2 is (2, 2^2024)

[loud gyro]

what about 2,2^3

that also works

a^n for all 1 < n < 2024 should work

[upper laurel]

(2)(2^3) | 2^2024 + 2^3
(2)(2^3)k = 2^2024 + 2^3
k = (2^2024 + 2^3)/((2)(2^3))
k = 2^2020 + 1/2 is not an integer

[loud gyro]

how did you get "2k - 1 is odd, so 2k - 1 = (1 or -1)"

2k-1 is odd for any integer

[blazing zephyr]

the only odd number thats divides 2^2024 is 1 or -1

[upper laurel]

2k - 1 | 2^2024
2k - 1 is a positive or negative factor of 2^2024
the factors of 2^2024 are: -2^2024, -2^2023, ..., -8, -4, -2, -1, 1, 2, 4, 8, ..., 2^2023, 2^2024
therefore 2k - 1 is 1 or -1

[loud gyro]

another knucklehead moment smh

[upper laurel]

you missed the Knucklehead warning sign on the road, it had the right of way

(a)(a^n) | a^2024 + a^n
(a)(a^n)k = a^2024 + a^n
k = (a^2024 + a^n)/((a)(a^n))
k = a^(2024 - n - 1) + 1/a
if 2024 - n - 1 > 0, then k is not an integer
2024 > n + 1
2023 > n

I think you managed to find the form of b = a^n that arent solutions

[crimson sedge]

can someone do my math homework please

[blazing zephyr]

!occupied

[cedar kiln]

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).

[loud gyro]

you know what, im gonna close discord and come back in like an hour when my brain has got working

[upper laurel]

save the rest of us knuckleheads when you come back

[blazing zephyr]

dw my teacher gave me 12 questions that get progressively harder and this is q9 :(((

[upper laurel]

"gave me 12 questions but we ignore 8 of them so really its 4 questions"

[blazing zephyr]

q11 q12 are fucking imo questions im so cooked

[blazing zephyr]

i already did q1-8 much help to you guys

[upper laurel]

assume that a > 1 and 2024 - n - 1 < 0 instead
then 0 < a^(2024 - n - 1) < 1
k = a^(2024 - n - 1) + 1/a
k - 1/a = a^(2024 - n - 1)
0 < k - 1/a = a^(2024 - n - 1) < 1
0 < k - 1/a < 1
1/a < k < 1 + 1/a
0 < 1/a < k < 1 + 1/a < 2
k = 1
1 - 1/a = a^(2024 - n - 1)
a - 1 = a^(2024 - n)
a^(2024 - n) is an integer
2024 - n > 0
2024 > n
since 2024 - n - 1 < 0,
2023 < n
n = 2024
b = a^2024 might work?

a(a^2024) | a^2024 + a^2024
a | 1 + 1
a | 2

uh oh

so other than (1, 1) and (2, 2^2024) we gotta find a new form to use

b = 1
a | a^2024 + 1
ak = a^2024 + 1
ak - a^2024 = 1
a(k - a^2023) = 1
a is positive, and a = 1 leads to (1, 1)
no new solutions

b is prime p
ap | a^2024 + p
apk = a^2024 + p
apk - a^2024 = p
a (pk - a^2023) = p
a is positive, and a = 1 leads to (1, 1)
a = p
(a, b) = (p, p)
it was already found earlier that solutions of the form (a, a^n) for n > 1 only fit (1, 1) and (2, 2^2024)
no new solutions

[blazing zephyr]

so those are the only ones?

[upper laurel]

...we didnt rule out b being composite

[blazing zephyr]

gock

[upper laurel]

all Im doing is going through the easy cases

ab | a^2024 + b
abk = a^2024 + b

abk - a^2024 = b
a (bk - a^2023) = b
a | b

abk - b = a^2024
b (ak - 1) = a^2024
b | a^2024

idk why it took me this long to notice this

[blazing zephyr]

what if bk-a^2023 | b

actually er

[blazing zephyr]

yea this

[upper laurel]

the only requirement to m | n is that mk = n for an integer k
bk - a^2023 is an integer, so a | b

[blazing zephyr]

a is also an integer, bk-a^2023 | b works also? or does it like not matter

[upper laurel]

bk - a^2023 | b is useless because it depends on k

it would mean (bk - a^2023) l = b for some integers k, l

[blazing zephyr]

aw

[upper laurel]

we cant use that

similarly ak - 1 | a^2024

[blazing zephyr]

maybe im stupid but what do i do with a | b and b | a^2024

[upper laurel]

not really much of anything

Idk how to really squeeze meaning out of all this, it just provides bounds that a < b < a^2024

at least it would mean that if a is prime, then p | b and b | p^2024, so b must take on the form p^n for 1 < n < 2024, so (2, 2^2024) only

ngl that b = a^n idea is doing wonders for us

[blazing zephyr]

ok makes sense

[upper laurel]

this does mean I didnt have to do the a = 2 case manually

we've done the (a, a^n for n > 0) case and found it to be (1, 1) or (2, 2^2024) only

wait no

it only ended with a | 2, so a=1 and a=2 were both done manually as a result

nvm

[blazing zephyr]

what if a=2 b=2^2

[upper laurel]

Im hoping this works

[blazing zephyr]

it fits these 2 bounds

[upper laurel]

replace the 3s here with 2s, no it doesnt work

to recap, we've done the (a, a^n for n > 0) case and found it to be (1, 1) or (2, 2^2024) only

we did that by first doing a=1, then a=2, then b=a^n

b=a^n reduced down to a|2, cases which were already done

[upper laurel]

the manual work of a=1 and a=2 could not be avoided, that is what I am saying here

I think youre getting confused

[blazing zephyr]

very

[upper laurel]

how are you already lost

I barely did anything

do you need me to repeat

[blazing zephyr]

lemme reread

[upper laurel]

you werent reading?

[blazing zephyr]

im still kinda stuck here on why there are so many solutions thst fit the 2 divisibility but only a single one really works

[upper laurel]

you werent reading

lets recap

[blazing zephyr]

can i be honest

i fell asleep im so tired 😭

[upper laurel]

that makes sense, thats ok

[blazing zephyr]

soo how do we check other b thats not in the form of a^n

[upper laurel]

youre saying this like I know the answer

[blazing zephyr]

ok fair

[upper laurel]

ab | a^2024 + b
abk = a^2024 + b

abk - a^2024 = b
a (bk - a^2023) = b
a | b

abk - b = a^2024
b (ak - 1) = a^2024
b | a^2024
so all solutions must fit a | b | a^2024

now if b = a^n for n > 0,

a | a^n | a^2024
1 < n < 2024

(a)(a^n) | a^2024 + a^n
(a)(a^n)k = a^2024 + a^n
k = (a^2024 + a^n)/((a)(a^n))
k = a^(2024 - n - 1) + 1/a
k - 1/a = a^(2024 - n - 1)
a^(2024 - n - 1) is not an integer
2024 - n - 1 < 0
2023 - n < 0
2023 < n
2024 < n

n = 2024

k - 1/a = a^(2024 - n - 1)
k - 1/a = a^(2024 - 2024 - 1)
k - 1/a = 1/a
k = 2/a
ak = 2
a | 2

a = 1 means b = 1^2024 = 1
a = 2 means b = 2^2024
so if b = a^n for n > 0, then the only solutions are (1, 1) and (2, 2^2024)

this above is the recap

[cedar kiln]

@blazing zephyr Has your question been resolved?

[blazing zephyr]

lemme readbt

fell asleep again sorry

[upper laurel]

nw

eventually the channels gonna close when it asks you for resolved while youre sleeping

[blazing zephyr]

what does a | b| a^2024 mean

[upper laurel]

it means a | b and b | a^2024

I was thinking I wouldnt need to explicitly state that

wont come up with new notation next time

[blazing zephyr]

anyways, do you have the solution for b not in a^n yet?

[upper laurel]

bit of a loaded question there, no I dont

[blazing zephyr]

hm ok

[upper laurel]

youll just have me do all of this?

puts a dampener on things

[blazing zephyr]

wdym

[upper laurel]

think about how you asked this

does that sound like youre trying to solve the problem

[blazing zephyr]

i was thinking if you havent gotten it i could close it for now and take a nap and make a new one when i wake up, but i realize i cant take one rn

[upper laurel]

what in the world are these kinds of shenanigans

[blazing zephyr]

huh

[upper laurel]

ab | a^2024 + b
abk = a^2024 + b
k = (a^2024 + b)/(ab)
k = a^2024/(ab) + b/(ab)
k = a^2023/b + 1/a

a = p^m, b = p^n for m > 1, n > 1?

a | b and b | a^2024
p^m | p^n and p^n | (p^m)^2024
p^m | p^n and p^n | p^(2024m)

m < n < 2024m

k = (p^m)^2023 / (p^n) + 1 / (p^m)
k = p^(2023m - n) + p^(-m)
k - p^(-m) = p^(2023m - n)
p^(2023m - n) is not an integer
2023m - n is negative

k = p^(2023m - n) + p^(-m)
the numbers involved here are:
integer = p^(negative) + p^(negative)

the biggest the RHS can be is 2^-1 + 2^-1 which is 1
for any other p or negative number, it will fall between 0 and 1 and so wont work
so p = 2, 2023m - n = -1, -m = -1

-1 = -m or m = 1

2023m - n = -1
2023(1) - n = -1
2024 - n = 0
n = 2024

so if generalizing (a, b) to both be prime powers, again (2, 2^2024) is the only answer

I think I'm getting closer and closer to the ideal solution each time