help-49

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

so $n=dk$

[grand pond]

ƒ(Why am. I here)=I don't Know

[soft token]

Well, you are almost done.

[twilit field]

yeah, I just like working on discord, as the server is usualy less active now, I figured there's no harm in working on discord

[viral dagger]

agree

[twilit field]

$\frac{2^{dk}-1}{2^d-1}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[soft token]

Oh no. I just meant that you have almost done it. I didn't mean to object to you doing it here. Lol

[twilit field]

now let $2^d=x$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

then (x-1) cancels out

is that it?

[hollow oyster]

yes

[viral dagger]

i wanna know how do you cancel out x^k-1/x-1

[hollow oyster]

long div

[viral dagger]

ok

o ye

[twilit field]

That was a hint in my book lol, they gave the factorisation

[hollow oyster]

,w (x^k-1)/(x-1)

[grand pond]

Results provided by WolframAlpha

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

Requested by convergence0091

[zealous schooner]

well the factor theorem does that for you

[hollow oyster]

shit not this

[soft token]

x in denominator cancels out the one in numerator. 1 cancels 1. You are left with k. /s

[zealous schooner]

that's wild

[nova yoke]

just plug x=1 into x^k - 1 to see that it's a root, so x-1 has to be a factor, easy peasy

[zealous schooner]

yeah, factor theorem

[viral dagger]

i think you can also use syntethic

[zealous schooner]

I mean, you can use whatever you want

[viral dagger]

fayr

btw is this what people call number theory?

[nova yoke]

but yea the factorization is also nice,
(x-1)(x^(k-1) + x^(k-2) + ... + x + 1) = x^k - 1

[zealous schooner]

yeah it's introductory number theory

[viral dagger]

i like it :3

[hollow oyster]

yes bungo

[viral dagger]

maybe ill search a free course online

[twilit field]

on Number theory ?

[viral dagger]

ye

[hollow oyster]

there one done by Richard E boress

[zealous schooner]

gotta be the best math field

[twilit field]

I can suggest an affordable book

David Burton was around 10 USD where I'm at

[zealous schooner]

wai what are you using for NT?

[hollow oyster]

nah the best is SOLs

[twilit field]

david M burton

[zealous schooner]

I don't think indian prices compare to other countries lmao

[zealous schooner]

oooh nice

[twilit field]

the server fav was 20,000 INR

[zealous schooner]

what's SOLs?

[hollow oyster]

second order logic

[zealous schooner]

yeah... although, you can probably find pdf's online

[hollow oyster]

try it its fun

[zealous schooner]

oh what does that entail

[viral dagger]

i dont know if im at that level yet lol

[hollow oyster]

which one idr

[viral dagger]

do you have link? i cant find it on google

[soft token]

This is the one I used too. I like the way he tells stories of how theorems came to be and so on.

[twilit field]

Niven, Zuckerman, Montogomery

[hollow oyster]

oh i see

[twilit field]

yeah, and the best part is unlike other books, it'a affordable

[hollow oyster]

it is there on yt

[twilit field]

.close

[midnight plank]

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[viral dagger]

ok

[midnight plank]

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[twilit field]

Let $t_n$ denote the nth triangular number. for what value of n, does $t_n$ divide the sum $\sum t_i$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

so $t_i= \frac{(i)(i+1}{2}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

so the sum is given by $\frac{(n)(n+1)(2n+1)}{12}+ \frac{n(n+1)}{4}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

now this is always divisible by n

is that right?

[hollow oyster]

wait so let me get the question right
it asks find the value of n: $t_n|\sum t_n$ right?

[grand pond]

convergence

[twilit field]

yeah

oh

wait

my bad

yeah, nvm, true for all n

[frank wolf]

Nope!

It’s not true for all n

I think

[twilit field]

but this is the sum

right

[frank wolf]

Yes

Consider n = 2

[hollow oyster]

here the sum does it mean t_1+t_2+....+t_n right

[frank wolf]

Yeah i think so

[twilit field]

and $t_n= n(n+1)/2$

[grand pond]

ƒ(Why am. I here)=I don't Know

[frank wolf]

Yep

[hollow oyster]

yes

[twilit field]

so true for all n

[frank wolf]

Nope!

Again, consider n = 2

[hollow oyster]

@twilit field clarify this

[twilit field]

yes

[hollow oyster]

oh ok

[twilit field]

hmm,ok

so $t_2=3$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

the sum is 4

oh

[frank wolf]

You see?

It can’t be true for all n

[twilit field]

yeah

but why is my working wrong?

[frank wolf]

So, what is your working?

[twilit field]

this is S_n

t_n= n(n+1)/2

[frank wolf]

That is correct! As far as i can tell

Both are correct

[twilit field]

so I find S_n/t_n at an arbitrary n

[frank wolf]

Which is?

[twilit field]

(2n+1)/12+1/4

[hollow oyster]

this means $\sum t_n= \farc {\left(1(2)+2(3)+\dots +n(n+1)\right)}{2}$

[frank wolf]

No

[grand pond]

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

[twilit field]

2n+1/6+1/42

[frank wolf]

Still no

[twilit field]

huh

[frank wolf]

It’s just algebra mistakes you’re making

Try factoring out $\frac{n (n + 1)}{2}$ from the expression for $\sum t_i$

[grand pond]

Pseudonium

[twilit field]

I did

[frank wolf]

Show me your working?

[twilit field]

one minute,

oh, nvm, I did it in my head( factoring it out), let me try again

sorry

[frank wolf]

It’s okkk

[twilit field]

(2n+1)/3+1/2

[frank wolf]

Still no lol

[twilit field]

(2n+1)/6+1/2

[frank wolf]

There you go! That’s correct

[hollow oyster]

yes

[twilit field]

so why are there any restrictions on n

[frank wolf]

Well

Think about what happens for n = 2

[twilit field]

ah

right

so 2n+1 should be a multiple of 3

[hollow oyster]

does (2n+1)/6 +1/2 always be a natural number

[twilit field]

no

[hollow oyster]

yes phy

[frank wolf]

This isn’t the condition i think

Though hmm hang on

[hollow oyster]

its 6|2n+4

[frank wolf]

Actually this is correct

It’s equivalent to the condition i had in mind

[twilit field]

Thanks! Both of you

[hollow oyster]

;)

[twilit field]

.close

[midnight plank]

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[blissful drum]

For number 2, what error did I make?

[tribal tartan]

you have two different variables

you either express everything in terms of x

or everything in terms of u

you cant have both there

[blissful drum]

Oh I see

Alr thanks

.close

[tribal tartan]

yw!

[midnight plank]

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[viral dagger]

<@&268886789983436800>

bro the audacity of saying "oh come on" 😭

[grim vector]

Kek

[midnight plank]

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[woven solstice]

Help me with integration basic ↓

[woven solstice]

integrate each of the following with respect to x

[tribal tartan]

do you know the inverse power rule?

[woven solstice]

yes

power + 1 things right?

[tribal tartan]

yeah add 1 to the power and divide it by the new power

just do that here

[woven solstice]

wait lemme try

[tribal tartan]

generally $\int x^n dx=\frac{x^{n+1}}{n+1}+C, n\neq -1$

[grand pond]

y0shi

[woven solstice]

@tribal tartan

am I right?

[tribal tartan]

yep

dont forget the +C

[woven solstice]

everything right?

ohh yea

because integrate

[tribal tartan]

yep

the derivative of a constant is just 0

[woven solstice]

(4/3) / 4?

how to divide this

[tribal tartan]

yep divide it by 4

$\frac43\cdot\frac14=\frac13$

[grand pond]

y0shi

[woven solstice]

thanks

so it would be 1/3 x^4 + c?

[tribal tartan]

yep thats it

[woven solstice]

I look at the answer (x^4)/3 + c

it is the same thing right?

[tribal tartan]

thats the same thing

yeah

[woven solstice]

because

1/3 × x⁴ = x⁴/3

[tribal tartan]

mhm

[midnight plank]

@woven solstice Has your question been resolved?

[woven solstice]

@tribal tartan sorry last ques

am I right?

[midnight plank]

@woven solstice Has your question been resolved?

[unborn terrace]

Its sooo much easier if you get rid of the fraction by making so its x^-2

then you would realise if you add 1, it actually becomes x^-1

and you would have to multiply it by something else, not 1/3

If I ever have an x on the bottom of a fraction, i always convert it to a negative power so i don't make mistakes, and its easier to process

[woven solstice]

ohh yea ty

[midnight plank]

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[left cedar]

Is there a way to know if this equation has positive integral solutions for b?
And n and m must also be positive integers
Is there a way to know it without having to evaluate for different values of n and m

[stiff sundial]

need help?

[left cedar]

Obviously

[stiff sundial]

cool

i mean

who doesn't

[jagged saffron]

cut this out stop doing this

[stiff sundial]

ok 😔

[midnight plank]

@left cedar Has your question been resolved?

[gloomy scaffold]

positive integral or positive integer

[left cedar]

Positive integer

[gloomy scaffold]

b(2^m-3^n)=3^n-2^n

wait nvm I'm blind LOL

[left cedar]

Why you first thought that? And what made you change your mind?

[gloomy scaffold]

nono I just flipped something

-b(3^n-2^m)=3^n-2^n

do u think this could help

[left cedar]

Yea maybe
Now we have 3 and 2 in same order
But there isn't any existing mathematical technique for such problems?

[gloomy scaffold]

probably are but too niche for me to know lol

[left cedar]

Ok then
Thanks

[gloomy scaffold]

there's something you can do by the way

[gloomy scaffold]

^

[drifting root]

Doesnt it depend on if m is bigger than n?

[gloomy scaffold]

YES!

[drifting root]

Ah

[gloomy scaffold]

also this is just binomial which is probably overcomplicating I feel

[left cedar]

Yes, m would definitely b larger cuz b is positive

[drifting root]

Yea

[left cedar]

I will try it

[left cedar]

Ok, I'll try this as well

.close

[midnight plank]

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[midnight plank]

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[fallen creek]

ok im even more cooked now

[fallen creek]

<@&286206848099549185>

[fierce saffron]

he's back

[fallen creek]

bruh

its worse

now

[fierce saffron]

what

[fallen creek]

thefolder tried to help

and now im autistic

anyways

the new question is

5

help

please

[fierce saffron]

determinant <= 0

since it's either always positive or it has solutions

[fallen creek]

huh

[fierce saffron]

for both

[fallen creek]

Bro please js gimme working and answers atp

im bout to cry

ive been for 2 hours

[fierce saffron]

i can't send working im on laptop let me solve one sec

[fallen creek]

aigh

[fierce saffron]

is it 25 for the first one?

bro

[fallen creek]

im back

25?

wdym

oh yeah yeah

whats the working

[fierce saffron]

you know

the determinant of the equation right

D = b^2 - 4ac

equate to zero

[fallen creek]

huh

b^2-4ac is out of the quadratic formula

[fierce saffron]

you do know what the determinant of a quadratic equation is right

[fallen creek]

js go through everything for me

yk what

[fierce saffron]

nvm bro ur cooked 🙏

[fallen creek]

lets come back to this question

wait wait wait

[fierce saffron]

what grade are you in

[fallen creek]

help me w 123

and literally call it a day

fuck number 5

[fallen creek]

10

hydronuke?

@fierce saffron

[fierce saffron]

yeah im here

all these questions can be solved using the determinant for each eq

or inequation whatever

[fallen creek]

whats the determinant

js zooom through it

[fierce saffron]

b^2-4ac

[fallen creek]

tell me what i need to know

[fierce saffron]

just

remember this

[fallen creek]

yeah

[fierce saffron]

ok

now

if a quadratic expression is always positive or always negative

[fallen creek]

wait what question are we doing

[fierce saffron]

all of them???

theyre like

the same

[fallen creek]

ok start w 1

alr go on

[fierce saffron]

for 1 right

so since the coefficient of x^2 is positive it opens upwards

and if you find out b^2-4ac and that comes out to be positive then it also has roots

so if it satisfies both these conditions the expression is always >= 0

[fallen creek]

mate yk what i gtg

im too cooked

thanks so much

[fierce saffron]

🙏

yeah np

[fallen creek]

its over for me

i think i cooked some bs that looks like working ngl @fierce saffron

hopefully prof js looks at the answer and goes "hm good job"

[fierce saffron]

you could just

use the formula i gave you

[fierce saffron]

cool

[fallen creek]

nah i looked at the textbook

its says

write it in the form

completed square form

and then you get -b as x

and c is y

so then you have the turning point

and you can look at the graph

look at x=-5 for this curve

and y= 25

c=y

25=y

you see?

@fierce saffron

[fierce saffron]

oh

yeah that works too

[midnight plank]

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[clear nacelle]

How can i find the common perpendicular line of AB and CD, where A(1,3,1), B(1,5,1) C(-3,2,1) D(0,3,-1)

[midnight plank]

@clear nacelle Has your question been resolved?

[midnight plank]

@clear nacelle Has your question been resolved?

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[waxen silo]

could soemone guide me through on how to do this?

[waxen silo]

I kind of know where to go by moving x/100 to the LHS and then intergrating it

but idk if I'm doing it right bc there isn't a t inside which feels like they're should be

[midnight plank]

@waxen silo Has your question been resolved?

[waxen silo]

.close

[midnight plank]

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[hollow thicket]

Find the radius of the circle circumscribed about the triangle for which A = 50°, B = 20°, and a = 35.

My answer is r = 22.8.
Am I right on this? I just want to make sure🙂

[hollow thicket]

,rccw

[rose cobalt]

Looks fine

[midnight plank]

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[hollow pike]

I forgot how to do (x-3)^2

[tawdry sapphire]

(a- b)² = a² + b² -2ab

[hollow pike]

A is X right?

X^2 + 3^2 - 2X*3
X^2 + 9 - 6X

[obtuse void]

yeah

[hollow pike]

Aight nice

.close

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[elder socket]

How is this wrong?

[elder socket]

Basically just need help with order of operations? In this instance shouldnt it just be left to right

[last slate]

You wrote 77/8 on the right side od the equal sign

which will be 8/77

[dreamy lichen]

Why?

[last slate]

other than that it's all correct!

oh wait

[elder socket]

how does it become 8/77 though?

[last slate]

actually no

Ye

sorry

sprry

my brain laggin

[dreamy lichen]

,calc (77/8) / (4/3) / (24/13) / (6/11)

[grand pond]

Result:

7.1686197916667

[elder socket]

thats how I read it

solution says 11011/1536

[dreamy lichen]

,calc (77/8) * (3/4) * (13/24) * (11/6)

[grand pond]

Result:

7.1686197916667

[dreamy lichen]

huh

[last slate]

mhm

[dreamy lichen]

oh

its right

[last slate]

its decimal

[dreamy lichen]

you just need to further simplify that

[last slate]

mhm

sorry for that random thing I just told ya

[dreamy lichen]

dawg

thatll be painful

there is nothing that cancels

[elder socket]

I get 273/128

[last slate]

my brain aint braining

[dreamy lichen]

so just multiply all the numerators together and denominators together

[dreamy lichen]

How?

oh wait, 3 cancels. But other than that nothing

,calc 77 * 13 * 11

,calc 8 * 4 * 24 * 2

[grand pond]

Result:

1536

Result:

11011

[dreamy lichen]

yeah you probably did some mistake here

[elder socket]

for me I get the same on both

g and h

looking for mistakes rn

[dreamy lichen]

i think they are not the same

!show

[midnight plank]

Show your work, and if possible, explain where you are stuck.

[elder socket]

nvm I think I got correct now also

tfx solved

thx

.close

[midnight plank]

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[spice scroll]

Is R a function?

[dreamy lichen]

No, but its inverse is

[midnight plank]

@spice scroll Has your question been resolved?

[spice scroll]

how do i prove it isn't?

[dreamy lichen]

show that there are some f, (g, h), (g', h') such that f R (g, h) and f R (g', h')

meaning that both f and (g, h) are related through R, and f and (g', h') are related through R

which comes to showing that f = g ∘ h and f = g' ∘ h'

for some f, g, h, g', h'

[spice scroll]

i dont follow, do i prove by contradiction with specific h=... and g=....

?

[dreamy lichen]

yeah, you would have to show counterexample

[spice scroll]

was that what you were saying?

[dreamy lichen]

yeah

basically you need to show that for 1 input f, the "function" R would have to have 2 outputs: (g, h) and (g', h')

that would show that it's not a function

[spice scroll]

wait how's that possible?

[dreamy lichen]

if R was a function, it wouldnt be possible. That's why it proves that R is not a function

[spice scroll]

i mean if i get an argument f=g ball h, how do i even get <g, h>?

[dreamy lichen]

by (g, h) i meant <g, h>

[spice scroll]

i know

[dreamy lichen]

f = g ∘ h iff (f, <g, h>) ∈ R

by definition of R

[spice scroll]

im confused

ill try something and compare

[dreamy lichen]

this notation might be better for you?

idk which notation are you using

[spice scroll]

we use the <a,b>, but i can follow both

i just dont understand how to disprove formally

[dreamy lichen]

For that we need the definition of function

[spice scroll]

whole and injective

[dreamy lichen]

not sure whether injective is the right term, but i think the idea is correct

[spice scroll]

i mean if we get function f as an argument, we are to return a function, but this doesnt make sense

[spice scroll]

one-to-one

[dreamy lichen]

its hard to grasp, but it makes sense

[spice scroll]

u get it

[dreamy lichen]

i get what you mean

you mean that it can have only one output for one input

right?

[midnight plank]

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[spice scroll]

no it doesnt, because f is supposed to return an integer

[dreamy lichen]

.reopne

.reopen

[midnight plank]

✅

[spice scroll]

yes

[dreamy lichen]

yes, f is supposed to return integer

[spice scroll]

it's contradicting, but i dont know how to formalize the proof

[dreamy lichen]

but if R was a function, it would take function as an input, and return pair of functions as output

[dreamy lichen]

that's not the contradiction there

e.g. R^-1 would be a completely valid function

[spice scroll]

i dont understand then

[spice scroll]

yes i agree

[dreamy lichen]

okay

so let's analyze R

[spice scroll]

two arguments that are functions (double lambda) and you get a returned value

an integer*

[dreamy lichen]

not really, R^-1 would take 2 functions as input and return their composition (also a function) as output

[spice scroll]

oh right

[dreamy lichen]

If R was a function, then (by what you call "injectivity") if <f, <g, h>> ∈ R and <f, <g', h'>> ∈ R, then g = g' and h = h'

right?

[spice scroll]

yes

[dreamy lichen]

okay cool

[spice scroll]

but then not neccarily we have more than 1 pair in R though?

[dreamy lichen]

so to show that R is not a function, we will have to find specific f, <g,h> and <g', h'>, such that g and g' and h and h' aren't same

[dreamy lichen]

yeah, but we dont need to contradict that

it suffices to get one contradiction

[spice scroll]

alright

[dreamy lichen]

to show that R is not a function

and in this case, it actually is "whole". Because for every possible input function f, you can take f = f ∘ I and so <f, <f, I>> ∈ R

where I is identity function

[spice scroll]

true

so how do we disprove injectivity? as in i understand we need to find 2 pairs for the same f

but how?

[dreamy lichen]

take any concrete example

[spice scroll]

do we try to pick literal example?

[dreamy lichen]

e.g. f(x) = (2x)^2

yeah

[spice scroll]

so g(x) = x+1
h(x) = x-1

h(x) = i_Z
g`(x) = i_Z

[dreamy lichen]

hmm

you need to choose g and h, such that f = g composition h

okay

[dreamy lichen]

but this is good

f(x) = i_Z

[spice scroll]

lets goo

yes

[dreamy lichen]

this basically shows that for i_Z, R has at least 2 outputs, and from this it follows that R is not a function

[spice scroll]

so i can say then "let f_1(x)=... and let f_2(x)=..."?

[spice scroll]

yes exactly

[dreamy lichen]

to avoid confusion, i would use
f(x) = i_Z
g(x) = x+1
h(x) = x-1
h'(x) = i_Z
g`(x) = i_Z

this preserves the names of functions they use

[spice scroll]

yes true

[dreamy lichen]

then f = g ∘ h and f = g' ∘ h'

[spice scroll]

yes

[dreamy lichen]

meaning <f, <g, h>> in R, and <f, <g', h'>> in R

hence R cant be a function

since <g, h> doesnt equal <g', h'>

[spice scroll]

yes exactly

[dreamy lichen]

the precise wording would depend on precise definition of function that's given in your book

but that's the general outline

[spice scroll]

we dont have a book

[dreamy lichen]

oh

[spice scroll]

my lecturer those a 10-page scribble of the subject

[dreamy lichen]

i see

[spice scroll]

the horrors

[spice scroll]

thank you very much for the help mate!

[dreamy lichen]

np

[spice scroll]

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[simple frost]

x^2(1-x)y'' - (1+x)y' + 2xy = 0

x=0 is irregular singular point and x =1 is regular singular point and everything else is ordinary points
Am I correct

[midnight plank]

@simple frost Has your question been resolved?

[midnight plank]

@simple frost Has your question been resolved?

[midnight plank]

@simple frost Has your question been resolved?

[simple frost]

<@&286206848099549185>

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@simple frost Has your question been resolved?

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[twilit field]

if $a|bc$ prove $a|gcd(a,b)gcd(a,c)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

hints please

like I know how to do this using prime factorisation

but without that?

well, I guss using euclid's lemma I could argue that a|bc \implies a|b or a|c

[frank wolf]

Bezout

[frank wolf]

No unless you know a is prime

[twilit field]

ah

ok

thanks!

$ax+by=d_1;az+ct=d_2$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

hmm

so I have

$\frac{(ax+by)(az+ct)}{a}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

yeah, that works!

Thank you so much !

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[twilit field]

what exactly is the difference. between -f(x) and (-f)(x)

[twilit field]

context

[hard shard]

(-f)(x) is the additive inverse of f(x) while -f(x) is the negative of f(x)

[twilit field]

hmm, ok

thanks!

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[shut fossil]

No problem @twilit field

[hard shard]

best helper fr

[viral dagger]

reel

[frank wolf]

Wait how are these different

[hard shard]

additive inverse and negative could mean different things

[frank wolf]

When

[hard shard]

definitely not intuitive

um good question

[frank wolf]

Can you give an example?

[hard shard]

not that i can think of right now

[worthy kestrel]

isnt negative normally defined as the additive inverse

[midnight plank]

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[twilit field]

so this is my working

--v-v+v= -v(-1+1)+v=v

[hollow oyster]

It's best done in eq format

[twilit field]

eq?

equation form ?

so I was thinking I'll first add 0 to both sides

[quiet hinge]

Not complete. You have only proven that --v-v+v=v

[subtle blaze]

remember that -b has the property that b + (-b) = 0

let b = (-v)

[twilit field]

oh, congrats on getting helpful

[quiet hinge]

|| use the associative property to get --v-v+v=--v, finally giving --v=v||

[hollow oyster]

What frosst said

Use that

[twilit field]

hmm

ok, let me try

ok, adding (-v+v), which is 0

[subtle blaze]

no

add b to both sides!

[twilit field]

to both sides as in both sides of -(-v)=v?

[subtle blaze]

-(-v) has the property that (-v) + -(-v) = 0

[subtle blaze]

that is exactly what i wrote here

[twilit field]

hmm

ok

[subtle blaze]

$-(-v) = v \iff (-v) + -(-v) = (-v) + v$

[grand pond]

frosst

[twilit field]

I don't follow

$-(-v) =v\iff -(-v)+-v=-v+v$

[subtle blaze]

how can an expression iff an equation

[grand pond]

ƒ(Why am. I here)=I don't Know

[subtle blaze]

that's the same thing as what i wrote

[twilit field]

yeah, I don't follow

this just means --v= v is -v+v=0

[subtle blaze]

$a=b \iff a+c = b + c$

[grand pond]

frosst

[twilit field]

hmm, ok

makes sense

thanks

but why is what I did wrong

[subtle blaze]

you then chain another iff line into 0 = 0

by the fact that you have a thing adding its own inverse (on both sides of the equality sign)

then you go oh, 0=0 is true, so then -(-v) = v is also true

[subtle blaze]

i dont know what you're doing making up the +1-1 stuff

you dont really need it

[twilit field]

so from this I get -(-v)=v iff 0=0

thus our assumption is true

that's what you're saying, right?

[subtle blaze]

why do you have b here

[hidden stone]

assume -(-v)=b

[twilit field]

I meant to type v

[hidden stone]

and then add -v on both sides

[subtle blaze]

that is what we're doing

[hidden stone]

oh i didnt see this part my bad

[quiet hinge]

Ypu can read this later if you like (completely verbal arguement, no equatipons)||When we attach a '-' to something, we are denoting its additive inverse. The additive inverse of -v, from above can be denoted as -(-v) but it is also v, since -v+v=0. Therefore -(-v)=v||

[subtle blaze]

and i've already used b for -v so... let's use a different letter if you want to do this

but it is also v you're packing the entire problem into this phrase lol

we don't know that it's also v

that's the problem we're trying to solve

[quiet hinge]

We actually do. The additive inverse of -v is v...

[subtle blaze]

we dont know that the additive inverse of -v is --v

[twilit field]

we only know the additive inverse of v is -v though, not the other way round

[subtle blaze]

yeah

[twilit field]

so this would be the last line of the proof, right?

[subtle blaze]

i would have after what i've written

\iff 0 = 0

hence the first statement is true

if that's what you meant then yes

[twilit field]

got it, thanks!

[hidden stone]

$$(a^{-1})^{-1}=b \Longrightarrow a^{-1}(a^{-1})^{-1}=ba^{-1} \Longrightarrow e=ba^{-1} \Longrightarrow b=a^{-1}$$

[grand pond]

aZxc

[quiet hinge]

How? If ypu accept that v+-v=0, by commutativity, -v+v=0. So v is the additive inverse of -v

I always think of attaching a - sign to something as writing its additive inverse

[hidden stone]

this would be the approach for most algebraic structures u can use it for a vector space

[subtle blaze]

how have you gotten from a^-1^-1 = b to b = a^-1

you've messed something in between

[hidden stone]

a^-1 is an element

(a^-1)^-1 is the inverse of said element by definition

[subtle blaze]

shouldn't it say a = b at the end

[quiet hinge]

Because (a^-1)^-1 is the inverse of a^-1, so multiplying by a^-1 gives you e

[hidden stone]

by inverse axiom the left hand simplifies to the identity element

[twilit field]

I don't follow the algebric proof at all, sorry

[hidden stone]

yeah yeah mb

[subtle blaze]

doesn't $(a^{-1})^{-1} = b \implies a = b$

[hidden stone]

but you get the point

[grand pond]

frosst

[subtle blaze]

lol yeah

[hidden stone]

i added an extra -1 oop

[subtle blaze]

something in the middle went wrong

[hidden stone]

no i think the proof is good up until the last point

just a typing error

[quiet hinge]

The last iff is incorrect

Right multiply by a to get b=a

[hidden stone]

right

as i said

[subtle blaze]

i think because at this stage you dont know about uniqueness of inverses yet

think that's the problem

[hidden stone]

$$(a^{-1})^{-1}=b \Longrightarrow a^{-1}(a^{-1})^{-1}=ba^{-1} \Longrightarrow e=ba^{-1} \Longrightarrow b=a$$

[grand pond]

aZxc

[hidden stone]

there

[subtle blaze]

this argument works for stuff like groups and rings as well it's good

[hidden stone]

if the symbols are intimidating just think

[quiet hinge]

If x and y are two seperare inverses, v+x=v+y=0. Giving x=y.

[twilit field]

what's e here though

[quiet hinge]

Identity element (0)

[subtle blaze]

we just havne't proven it yet

[hidden stone]

since we're dealing with very abstract structures we can't assume anything. so even tho it seems obvious that the inverse of an inverse is just itself we first assume its any number. then we use some axioms (fundamental laws) to simplify the equation

[hidden stone]

we have to prove that v+x=v+y=0 essentially

i dont think its an axiom

[twilit field]

I get that bit, but what's the $e$

[grand pond]

ƒ(Why am. I here)=I don't Know

[subtle blaze]

you know how in the axioms

we said the addition operation has an identity

[hidden stone]

usually in abstract algebra, e stands for the identity of a structure

so like 1 for multiplication

or 0 for addition

[twilit field]

oh,ok

[subtle blaze]

maybe $e_+$

[grand pond]

frosst

[quiet hinge]

X and y being inverses of v, by definition, mean that v+x=0 and v+y=0

[subtle blaze]

but not really necessary it should be obvious which identity you're talking about

[twilit field]

got it

so I follow until the last step

$1=ba^{-1}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[hidden stone]

great

[quiet hinge]

Now right multiply, by a

[twilit field]

now you multiply both sides by a, right?

[hidden stone]

yes

[twilit field]

cool

thanks!

[hidden stone]

mhm :)

np

[twilit field]

let me go though all the proofs again now

ok

got it

thanks everyone

But one thing

why is adding 0 to prove this wrong though

$-(-v)+(-v)+v=-(v)(-1+1)+v \impllies -(-v)=v$

[grand pond]

ƒ(Why am. I here)=I don't Know
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[twilit field]

I still don't get why this is wrong

[hidden stone]

wait can u fix the formatting

[twilit field]

ok

$-(-v)+(-v)+v = (-v)(-1+1)+v = v$

[grand pond]

ƒ(Why am. I here)=I don't Know

[hidden stone]

no thats fine

[hidden stone]

its actually equivalent to this i think

[quiet hinge]

read this

[hidden stone]

ye

[twilit field]

but -v+v=0

[hidden stone]

bc -(-v)+(-v)=0

and -v+v=0

[quiet hinge]

Yeah, so write it

[twilit field]

oh

I have to mention that?

ok

[quiet hinge]

--v-v+v=--v, so finally ypu have --v=v

[twilit field]

got it

thanks everyone!

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[twilit field]

what do I say here other than if a=0, av=0 and if v=0, av=0?

[wary thorn]

you're tasked with proving that av=0 => a=0 or v=0
not a=0 => av=0 and v=0 => av=0

[twilit field]

hmm

ok, can I assume multiplication by 1/a is defined ?

if so I multiply both sides by 1/a

thus getting v=0

[wary thorn]

well, when do you have multiplicative inverses in a field?

[twilit field]

when $a\neq 0$

[grand pond]

ƒ(Why am. I here)=I don't Know

[wary thorn]

You're assuming by writing "mutliplication that 1/a" that 1/a is the multiplicative inverse of a

yes so

if a!=0 then you can multiply both sides by the multiplicative inverse of a

which implies v=0

[twilit field]

got it

thanks

[wary thorn]

Now what's the other case

[twilit field]

a=0

when a=0, av=0

[wary thorn]

well actually, that is only if (1/a)*0 = 0

justify that aswell

[wary thorn]

this is not needed

you're given av=0
either a=0, and you're done
or a!=0, and you just proved v=0 and you're done again

[twilit field]

got it!

thanks so much !

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[twilit field]

so $x =\frac{w-v}{3}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

now I know w-v is unique

thus x is unique?

or do I have to prove w-v is unqiue too?

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[blissful drum]

why is it wrong to do f(0) = 7 then plug it in getting y = ln7 then doing the dy/dx of ln(7) to get the gradient?

[blissful drum]

is it because its basically like a coordinate?

ln7 would be the y coordnate when x = 0?

(0, ln7)

[midnight plank]

@blissful drum Has your question been resolved?

[prime leaf]

hi

If you do dy/dx of y = ln7 is equals to 0

bc ln7 is a constant

I think you have to do:

dy/dx of y = ln(f(x))

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[midnight plank]

@foggy tinsel Has your question been resolved?

[twilit jetty]

this one is partly a trick question

they tell you that ZX is a diameter

that means Y is just an angle from one side to the other side of a circle

does that give you any hints about what Y is

@foggy tinsel

[foggy tinsel]

Mo

No

@twilit jetty

[twilit jetty]

@foggy tinsel have you heard of thale's theorem

[midnight plank]

@foggy tinsel Has your question been resolved?

[foggy tinsel]

No

Is he european or black

[twilit jetty]

@foggy tinsel Thale is a greek name

thale's theorem says that if you draw a diameter

then connect it to a triangle

the angle up there is a right angle

[twilit jetty]

so here, with diameter XZ, thale's theorem would say that Y is a right angle

[midnight plank]

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[lunar dome]

Is b right

[rose cobalt]

Incorrect

Why is y=0

[green marlin]

Nope the key words that give you the information to create an equation are "initially" and "decreases at the rate of 0.08grams/hour"

[lunar dome]

Oh

Ok

[lunar dome]

Yeah I’m confused

[green marlin]

What does initially tell us?

[midnight plank]

@lunar dome Has your question been resolved?

[midnight plank]

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[left cedar]

It is obvious for collatz conjecture that if we only prove odd numbers to follow the conjecture, the whole conjecture would be true
Has someone yet classified it further and reduced it to a further sequence?
I heard of the sequence, 5,9,13,17.. before
Is it another simplification series of the conjecture? If yes then how is it so? What is the reason behind it?
Also are there further sequences inside this sequence that if proved true, would prove the whole conjecture?

[left cedar]

I mean is it simplified to more sequence (than the one I have mentioned) that would represent all positive integers? And if proven true would prove the whole conjecture true?

[midnight plank]

@left cedar Has your question been resolved?

[west iron]

if you study the residues mod n you can try to show that every number must eventually hit certain residues

it's easy to show every number will hit something that's 1 mod 4 this way

1,3,5 mod 6 -> 4 mod 6

So you can say the same for 4,10,16,22,28,...

[green marlin]

Easy?

[west iron]

Ah nvm

4k+1 -> 12k+4 -> 3k+1

Which hits every number that's 4 mod 6 so works

but I can't show every number hits something 1 mod 4

[green marlin]

Where did 2k+3 come from?

[west iron]

dividing 12k+4 by 2 twice

[green marlin]

Surely it'll become 3k+1

[west iron]

oh lol

oops

[green marlin]

Even so like when you consider other mod 4s it's not simple to see where they end up

Like 4n, goes to n, which completely relies on n

[west iron]

yeah

[green marlin]

Wait

Ur right

If a number is 0 or 2 mod 4, it's even and so eventually it'll become 1 or 3 mod 4

[west iron]

ah it's eventually true

when you find the first 0 in the binary expansion of the number reading backwards

still not an easy mod argument

So something like 31 will take a while to be 1 mod 4

31 94 47 142 71 214 107 322 161

161 is 1 mod 4

[midnight plank]

@left cedar Has your question been resolved?

[green marlin]

Yeah eventually true but hard to exactly prove

[left cedar]

So there is or isn't another simplified sequence known?

[green marlin]

Look up papers on the collatz conjecture, I doubt people in this server will be experts on it, the thing with problems like this is: the probability of actually making meaningful progress is so low that most people don't study or look at the collatz conjecture much

Need to find specific research then look at papers "collatz conjecture mod 4 papers PDF"

[left cedar]

Ok thanks

[midnight plank]

@left cedar Has your question been resolved?

[midnight plank]

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[fluid island]

Hello everyone :)
kinda struggling with the following question

a) Sketch the region area B
b) Calculate the area F of B
c) Calculate the coordinates of the center of gravity S(x_s, y_s) of B

[fluid island]

i ll provide the task images in a sec

That would be my approach, but I don't know which integration limits I have to use for the integral 😅

[midnight plank]

@fluid island Has your question been resolved?

[fluid island]

<@&286206848099549185>

[last slate]

what ya need

[fluid island]

I dont know how to find the integral limits

from c)

[midnight plank]

@fluid island Has your question been resolved?

[silk saddle]

Kannst du vielleicht die gesamte Aufgabenstellung bereitstellen

[silk saddle]

Also das B unter dem Integral sagt dir, dass du den Bereich von B verwenden sollst

[fluid island]

also bei b) hab ich die Grenzen 0 und 1 verwendet aber ich weiß nicht wie ich die Grenzen bei zwei Integralen machen soll

[silk saddle]

Also: $ \int_0^1 \int_0^{2\sqrt{x}}$

$\int_0^1$

[grand pond]

NulledOutChicken

[silk saddle]

\int_0^{2 \sqrt{x}}

$\int_0^{2 \sqrt{x}}$

[grand pond]

NulledOutChicken

[fluid island]

also setz ich die Funktion von y als Grenze ein?

[silk saddle]

$\int_0^1 \int_0^{2 \sqrt{x}}$

[grand pond]

NulledOutChicken

[silk saddle]

Yapp, ziemlich meta, oder?

[fluid island]

idk xd

aber danke dir :)

[silk saddle]

"aber"

ich weine gleich

[fluid island]

ich tu so als würd mich das nicht treffen :,)

[silk saddle]

Aber keine Sorge, das mit dem Doppel und Trippelintegral macht man nur im ersten Semester explizit

[fluid island]

Das Integral an sich ist gar nicht so das Problem, ich struggle immer die Grenzbereiche zu finden von wo bis wo man integriert

[silk saddle]

genau das mein ich ja auch;)

[fluid island]

😅

[silk saddle]

Später ist dass dan eher $\int \int_{\Omega}$

[grand pond]

NulledOutChicken

[fluid island]

oh man, hoffe das ich da nicht hinkomme o-o

.close

[midnight plank]

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[midnight plank]

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[humble torrent]

,rotate

[humble torrent]

what have you tried so far?

try to recall the definition of differentiability at a point x=a

[wanton shore]

My sir told us a shortcut for this type of problem that we first differentiate the given function and then going in reverse to check if it is differentiable or not

[wanton shore]

Now pls see my doubt

My qstn is that it will only work if f'(x) is continuous at that point?

Am i right?

[humble torrent]

What exactly do you mean by this?
If i'm interpreting this correctly, yes in most cases you can try to differentiate for each case separately and if it's discontinuous, the original function can't be differentiable (at say that point)

[humble torrent]

Yes

However in general, if i recall correctly you have to be careful when doing this

[wanton shore]

What are the limitations?

[humble torrent]

That's what im not sure of, i just remember being careful about this but maybe im deceiving myself

[wanton shore]

i know that f'(x) should be continuous for applying shortcut

[wanton shore]

<@&286206848099549185>

[humble torrent]

Oh, okay so this is a classic example

$$
f(x)=
\begin{cases}
x^{2}\sin(1/x) & \text{if } x\neq 0\
0 & \text{otherwise. }
\end{cases}
$$

[grand pond]

Aslan

[wanton shore]

Yes

Same example used by my teacher

[humble torrent]

it's differentiable, but the derivative is not continuous

Oh, so you were aware of the limitations of your shortcut?

[wanton shore]

I meant other than this are there any more limitations?

[humble torrent]

Well, what do you mean by other than this? Isnt this precisely the limitation you should be aware of?

Since it's not directly obvious

[wanton shore]

f= x^(⅔) suppose if we want to find if this is differentiable at x=0 or not

We can use shortcut even though derivative is not continuous at x=0

[marsh delta]

The limitation that I can think of is

1. The derivative should be continous as said in you example
   2)f(x) should be continous
   for eg f(x) = x^2 for x<1
   2x for x>=1
   in this x^2 derivative at 1 is 2
   and so is the derivative of 2x but as f(x) is not continous f(x) cant be differentiable at x=1

[wanton shore]

Ok

@marsh delta what about f=x^(⅔)

[humble torrent]

Well as we just saw, this is subtle

However i conjecture that the critera we want is that we can use this shortcut whenver the derivative is not oscillating in a wild manner

[marsh delta]

well in f=x^2/3 using the standard limit we get
limh->0+ (0+h)^2/3-0/h
= h^2/3/h
=1/h^1/3
= + infinity

[humble torrent]

Yup

though for somereason they want to use a shortcut

[marsh delta]

thus x^2/3 isnt differentiable at 0

[humble torrent]

I'd say its better to use the definition here directly

[marsh delta]

true

but some question become harder if we use first pricinple

@wanton shore while using the shortcut see if the derivative is defined at the point you are seeing the derivative at
for eg in this question f'=2/3 1/x^1/3 which is not defined at 0
thus try using the first principle

[wanton shore]

Haven't you come across any other func which is differentiable and whise derivative is not continuous other than sin(1/x) types

[last slate]

check if curve is continuous at 1 and if it is then directly check lhd and rhd by differentiating

if not continuous then obviously not differentiable

[marsh delta]

I think this example also fits the description you are asking

|x| could be another function

[humble torrent]

Well yes and no, theyre pathological in some sense so you wont often see them (if any at all) in pratice, but i think its best if you get used to the definintion way of showing something; instead of relying on shortcuts with no well known criteria (maybe you can find one?)

[wanton shore]

I can apply shortcut on |x|

[marsh delta]

try once

[wanton shore]

I will try finding

Thanks for the help

.close

[midnight plank]

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[midnight plank]

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[uneven sandal]

I have been stuck on this problem for 2-3 days and I got nowhere. Any help is appreciated and if you know a source for this proof pls feel free to send it

[uneven sandal]

Given \(0 < a < b\), show that for \(x \geq 0\) and \(x \neq 1\) the inequality
\[
(b-a)^x \neq b^x - a^x
\]
holds, and specifically, the following statements are true:

1. For \(x \in [0, 1)\):
\[
(b-a)^x > b^x - a^x
\]

2. For \(x \in (1, \infty)\):
\[
(b-a)^x < b^x - a^x
\]

[grand pond]

tobi

[midnight plank]

@uneven sandal Has your question been resolved?

[midnight plank]

@uneven sandal Has your question been resolved?

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[last slate]

https://cdn.discordapp.com/attachments/387447700234174464/1258149300370935951/image.png?ex=6686fe58&is=6685acd8&hm=5e5b587df71755545b6d690cffc0c04b4d1c35778f70092670f760f57aa988ea&

[last slate]

https://cdn.discordapp.com/attachments/387447700234174464/1258149300702023680/image.png?ex=6686fe58&is=6685acd8&hm=48e11ccea29d3476d43f1efbab70cf13b17400e6ae131eba608fe31ee6e4b4b8&

Probably me being dumb lol but i can't draw the diagram