help-49

[last slate]

well i know the answer is not equialteral because if it is fully contained inside DEF it is equilateral and i think some "rules" are probably broke if it isn't fully confined inside DEF

anyway how do i even draw this diagram for b)?

[midnight plank]

@last slate Has your question been resolved?

[midnight plank]

@last slate Has your question been resolved?

[hybrid crane]

You can definitely construct one that isn't fully confined

[last slate]

oh yeah nice, what "software" did you use?

[hybrid crane]

desmos

[last slate]

wtf? 😭

desmos geometry?

[hybrid crane]

https://www.desmos.com/calculator/isvcjgakaf

nah, the graphing calculator

[last slate]

ohh

idk how to do that :(

how'd you learn?

[hybrid crane]

I just read the documentation whenever I needed a certain feature

[last slate]

what's this?

i mean how is this relevant

[hybrid crane]

It's a function that returns a point. If you're familiar with complex numbers, it makes more sense

It basically gives n equal spaces points on a circle of radius s

[last slate]

right i follow that, but is this just boiler plate code?

[hybrid crane]

v is just the "point number" like point 0, point 1

[hybrid crane]

yeah it is

[last slate]

mmmmm interesting

[hybrid crane]

On the topic at hand, I think it is necessarily equilateral

I'm not entirely sure but lemme explain my reasoning

[last slate]

if it's contained inside it is equilateral

but otherwise it's not

[hybrid crane]

the diagram I sent shows a case where it is equilateral when it's not confined to EFD

[last slate]

hmmm wait are we sure that's how you would draw the diagram?

okay wait your case is a special case,no?

[hybrid crane]

It's one case

[last slate]

well the answer says

something else tbh

iirc

[hybrid crane]

Idk then tbh

[last slate]

wait i'll find you it

idr either

okay here

[hybrid crane]

Ohhhhhhhhhh

I didn't think about the triangles not necessarily oriented is a similar way

[last slate]

i still don't get how to draw it tbh

😭

i suck at geometry arghhh

how did they draw that

[hybrid crane]

is this any help?

[last slate]

no i mean probably dumb question but it's late for me

how do you draw that on paper

or whatever

[hybrid crane]

oops, error

[last slate]

what's the motivation to draw it

[hybrid crane]

Well if you really wanted to, you could use a ruler and a protractor to draw the EFD.
The angle DEA, BFE, CDF are the same so there's that

[last slate]

no no

how do i make a sketch of it lol

for this i first drew ABC

then i just drew outward triangles BCF, ACD, and ABE

then connected the vertex of each of those outward triangles and got DEF

i don't care about the exact measurement ofc

i just have to draw a sketch

but i can't think of how to draw a sketch of however the ABC thing will look like after it's out of DEF

[last slate]

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[midnight plank]

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[hybrid crane]

just reflect A in line ED and join up the points

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[bleak wren]

cat, nut, not, act, art, bat

this is an unique arithmetic sequence, where a letter corresponds to a number.

determine the next word in this sequence

[bleak wren]

hard stuck

last letter is t obv but then I'm stuck

[fossil knot]

hm

so c, n, a, and b are consecutive

The tens digits also jump from c to r to a

so r=n

cat nut not act ant bat

you also go from a back to a in five numbers

So the common difference is a multiple of 20

probably either 40 or 60

[midnight plank]

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[normal citrus]

f is a function of R^3, how can i show this?

[deep vine]

do you now the domain and codomain?

[normal citrus]

hum i guess yes

it is just write that it's a function of R^3 so i guess the domain and codomain are R^3

[hard umbra]

it doesn't look like f is just any function

perhaps you mean to specify that it is linear?

[normal citrus]

it's not write but we can assume it, what do we gain with the linearity

[hard umbra]

well nobody talks about kernel and image for non-linear stuff

[normal citrus]

ye that's true

[rose trout]

For the problem, start by showing that ker(f) and im(f) do indeed make a direct sum (show that their intersection is {0}).
Then you can use a really nice theorem about the dimensions of kernels and images :p

[normal citrus]

that's the part i'm strugling with

x€Ker so f(x) = 0
and f^3(x) + f(x) = 0

idk what to do with that

[rose trout]

Take some element y in both ker(f) and in im(f). You want to show that this has to be 0.

If it's in ker(f), then f(y) = 0.
If it's in im(f), then there is some x such that f(x) = y.

You'll want to use both of those, along with f^3(x) + f(x) = 0 to get to y = 0.

[normal citrus]

ok i made it

for the second part we're using the fact that we are in finished dimension

[rose trout]

Yeah. The direct sum is a subspace of R^3, and you can get its dimension from what you constructed.

[normal citrus]

alr thx

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[sharp wave]

hello

[sharp wave]

these are the options

[midnight plank]

@sharp wave Has your question been resolved?

[sharp wave]

<@&286206848099549185>

[hidden stone]

what step do you need help with

like where are you

[sharp wave]

i substituted the values of f(x) and f(pi/2-x) but then im stuck on how to simplify

[tidal turret]

prolly we need tfc1,no? but i cant manage to put all the pieces together

[sharp wave]

tfc1?

should i write it down?

and show you guys

we have been taught a property but that dosent help

over here

[tidal turret]

double angle formula

sin(2t)=2sin(t)cos(t)

mmm

[sharp wave]

okay let me try

how does that work

youre stuck with a sin2x term right

wait should we sub 2x as u maybe thatll work

since the lim its then come in the asnwers format

[neat hamlet]

no

do partial fraction decomposition first

that way you can separate out the x and pi/2 - x

[sharp wave]

but there are still two terms in the numerator no

wait let me try i like this

yep that worked @neat hamlet

i still had to sub 2x as u though

thank you so much both of you

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[twilit field]

prove that $a^n|b^n \implies a|b$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

without prime factorisation

I would like to use EDA if possible

eulclds division algorithm that is

so I have $b^n=da^n$

[grand pond]

ƒ(Why am. I here)=I don't Know

[frank wolf]

hm

the main proof of this I know does use prime factorisation

[twilit field]

I was thinking of equating b^n/a^n to some integer k^n

then take the nth root of both sides

[frank wolf]

well

with divisibility

it is usually a bad idea to introduce fractions…

[twilit field]

yeah, I know

[frank wolf]

I feel like you’ve been told this many times but for some reason you insist on using fractions

[twilit field]

sorry

[frank wolf]

hm, im not sure how Euclidean division could help..

[twilit field]

alternatively bezout's lemma may help

[frank wolf]

it would if we were talking about gcd

but we’re not

[twilit field]

my book does have a hint,just as minute

put d=gcd(a,b) and write a=rd, b=sd where gcd(r,s)=1. my part (a) gcd(r^n,s^n)=1 . show that r=1 whence a=d

part (a) is if gcd(a,b)=1 then gcd(a^n,b^n)=1

[frank wolf]

oh huh, interesting

[twilit field]

problem a is if gcd(a,b)=1 then gcd(a^n, b^n)=1

[worthy kestrel]

cant u show the contra positive easily

[twilit field]

that is if a|b \implies a^n|b^n

[worthy kestrel]

no

a implies b
contrapositive is
not b implies not a

[frank wolf]

hmm how would you show the contrapositive?

[frank wolf]

Do you know how to do this?

[twilit field]

yeah, bezout's lemma

[frank wolf]

ok cool

can you show that r = 1?

[twilit field]

let me try

[worthy kestrel]

ok wait this is hard too :(

[twilit field]

I'm not too sure, but I guess bézout's lemma

[frank wolf]

so try to use a^n | b^n

[twilit field]

ok, no

so I have $(rd)^n x+(sd)^n y=d^n$

[grand pond]

ƒ(Why am. I here)=I don't Know

[worthy kestrel]

assume a doesnt divide b
so there exists some p^w that divides a that doesnt divide b. the p power in b is <w. looking at the p power of a^n is >=wn. while p power of b^n is <wn. thus a^n doesnt divide b^n

that works?

[twilit field]

what?

I don;t follow

I feel like I should try to understand EDA once more I thought solving problems would help, but doesn't seem like it's helping

can I close this for now?

[pearl idol]

sure

[twilit field]

.close

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[midnight plank]

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[twilit field]

https://math.stackexchange.com/a/696537/879009 could someone explain this?

[twilit field]

how is$ab=-a^2?$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

hmm, if gcd(a,b)=1

[worthy kestrel]

gcd(A,B)=gcd(A,(kA+B))

in this case k is -a

[twilit field]

no, I get that, but how is $ab=-a^2$

[grand pond]

ƒ(Why am. I here)=I don't Know

[worthy kestrel]

it isnt

[visual tiger]

They never said that

ab - a(a+b) = -a²

[worthy kestrel]

its just with gcd there are some properties that allow u to get from one part to the next

[twilit field]

yes

but we want gcd(a+b,ab)

[visual tiger]

Yes

gcd(x,y) = gcd(x,y+kx)

replace x,y by a+b and ab

And k = -a

[twilit field]

ah

ok

How am I supposed to know this identity if I've never seen it before, is it supposed to be obvious?

[visual tiger]

Yes, bezout

ux + vy = 1

this implies

(u-kv)x +v(y+kx) = 1

[twilit field]

wait

okay

yeah, makes sense

[visual tiger]

Another way to see it

Is the euclid algorithm

To compute gcd(x,y) you compute the remainder of y/x

And so on until you reach 0

[twilit field]

yes

[visual tiger]

Well if y = ax+b

Then y+kx = (a+k)x + b

Same remainder

[twilit field]

ah, makes sense

thanks!

.close

[midnight plank]

Channel closed

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[twilit field]

.reopen

[midnight plank]

✅

[twilit field]

I'm having a hard time understanding this

so the only elements of the field are -\infty, 0 and \infy?

[pearl idol]

no

it's all real numbers, as well as ∞ and -∞

[twilit field]

ah

[pearl idol]

and it's not a field

[twilit field]

so it's. a vector space?

[pearl idol]

what makes you think that?

[twilit field]

oh right

[prime hornet]

check the vsp axioms

[twilit field]

that's the question

[pearl idol]

what's the question?

it doesn't seem like you included it in the screenshot

[twilit field]

Yeah,just relaised

[pearl idol]

ah

[twilit field]

OK, I think I have a rough idea of how to do this now, let me TeX a proof

so we'll have to prove teh following 1) Commutativity, associativity, additive identity, aaditive inverse, multiplicative idenrity and the distibutive property

[pearl idol]

lots of things to prove

[twilit field]

1. commutativity
   \
   $a+b=b+a$ , from the defn it follows that $\infty+a=a+ \infty$ and the same for negative $\infty$
   2)associativity : $(a+b)+c=a+(b+c)$ over R and as any number + $\infty$ is \infty the same holds true as one or more variables tends to infty
2. 0 is an element by defn
3. as is1

[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]

TeX massacred

oof

1. commutativity
   
   $a+b=b+a$ , from the defn it follows that $\infty+a=a+ \infty$ and the same for negative $\infty$
   \
   2)associativity : $(a+b)+c=a+(b+c)$ over R and as any number + $\infty$
   \is $\infty$ the same holds true as one or more variables tends to infty
   \
2. 0 is an element by defn
3. as is 1

[pearl idol]

you need to be more careful than this

you're missing cases

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

hmm?

oh right

[pearl idol]

for example for commutativity, you have the case where a and b are both real numbers, and the case where a is ∞ or -∞

there's still another case

[twilit field]

where both are infty

of some kind

[pearl idol]

yes

what happens then

[twilit field]

$\pm \infty \pm \infty = \pm \infty$

[grand pond]

ƒ(Why am. I here)=I don't Know

[pearl idol]

I'm not sure what that says

oh wait

I get it nvm

[twilit field]

and $\infty - \infty =0$

[grand pond]

ƒ(Why am. I here)=I don't Know

[pearl idol]

right

so if a and b are equal

then commutativity is obvious: a + b is clearly equal to b + a

so the only case you need to consider is ∞ + (-∞) = (-∞) + ∞

which is fine bc they're both 0

so commutativity is good

what about associativity

[twilit field]

That's true over R and true at $\pm \infty$ as $ a+ \pm \infty = \pm \infty$

[grand pond]

ƒ(Why am. I here)=I don't Know

[pearl idol]

I don't know what you mean by that

there are three different variables in the statement of associativity

(a + b) + c = a + (b + c)

which ones are in R and which ones are infinite, in what you are considering?

[twilit field]

(a+b)+c=a+(b+c) is a property of R

[pearl idol]

okay, so that handles the case where a, b, and c are all part of R

what about the other cases?

[worthy kestrel]

yep

7 cases to go

[twilit field]

wait

7?

oh

right

[worthy kestrel]

actually more

[pearl idol]

well

just keep checking cases

I'll tell you when you've finished checking them all

[twilit field]

I guess I can prove it for one case and say it's simialrly true for all other cases?

[pearl idol]

(you should also be able to tell when yo'ure done, ofc)

[pearl idol]

why?

why would the other cases be similar

[twilit field]

one var is \pm \infty the other two belong to R

in the next case two vars are some form of infty and the other blongs to R

[worthy kestrel]

write it out

[pearl idol]

yeah, let's consider that first case where one var is ±∞ first

[twilit field]

$\infty+(a+b) = \infty = (\infty+a)+b= \infty )$

[grand pond]

ƒ(Why am. I here)=I don't Know

[pearl idol]

I think you wrote it a little weirdly, but good

[worthy kestrel]

uve an extra ) but its goos

[pearl idol]

I would've written something like: [\infty + (b + c) = \infty = \infty + c = (\infty + b) + c]

[grand pond]

biased_estimatERIC

[pearl idol]

and then the same proof works for a = -∞

what about if b is ±∞, and a and c are real? does it also work

[twilit field]

yes

[pearl idol]

okay

and then what about if c is ±∞, and a and b are real? does it also work

[twilit field]

yeah, it works simialry for all cases

[worthy kestrel]

write it

[pearl idol]

yeah, it's good to practice writing it out

even if it may seem tedious

associativity is often hard to check because there are three variables involved!

[twilit field]

yeah, Wrote it in my notebook

[pearl idol]

okay can you show me

[twilit field]

uh, it's kind of messy, let me re-write it

[pearl idol]

alright

[pearl idol]

okay yeah that looks good

[worthy kestrel]

btw isn't this not exactly exhaustive

u need more details no?

[pearl idol]

there's still more cases, yeah, if that's what you mean

[worthy kestrel]

yea

[twilit field]

yeah, I realise there are 3 possibilities for each variable, so 27 possible cases

[pearl idol]

I'm assuming b, c are real for that case

yeah, a lot of them are similar, but not all of them are similar

what about the options where both a and b are infinite now? (and c is a real number)

this is where it might get tricky

[twilit field]

that's still \infty

[worthy kestrel]

i mean inf in the middle or last position

[pearl idol]

I'm assuming a = ±∞ for that case; I made him do the other two positions in the picture he sent (b = ±∞ and c = ±∞)

[worthy kestrel]

ah i see thats good

[twilit field]

now to prove the distributive property for the additive identity and multiplicative identity are trivially ture, right?

[pearl idol]

hang on

stay with associativity

did you consider the options where a and b are infinite, and c is a real number yet?

[twilit field]

I did , I got $\infty$

[grand pond]

ƒ(Why am. I here)=I don't Know

[pearl idol]

show me what you did

keep in mind there's not just +∞, there's also -∞

[twilit field]

$(\infty + \infty)+c = \infty +c =\infty=\infty+(\infty+c)=\infty$
\
$(- \infty + \infty )+c=0+c=c but - \infty+(\infty +c)= -\infty+\infty =0$
\

[pearl idol]

how does that demonstrate the associative property?

[twilit field]

oh, right

[worthy kestrel]

associativity is
(a+b)+c=a+(b+c)

[grand pond]

ƒ(Why am. I here)=I don't Know

[pearl idol]

so then what can you conclude

[twilit field]

this isn't a vector space

[pearl idol]

good! because associativity doesn't work

intuitively, you can think of adding infinity as "losing information"

if you start with some real number c, then add infinity, then subtract infinity, you lose track of the original number c was

that's not possible in a vector space

[twilit field]

I see. yeah, makes sense

[pearl idol]

any addition operation is perfectly reversible in a vector space

[twilit field]

thanks!

[pearl idol]

(more generally, in what's called an "abelian group" this is also true)

(but don't worry about that too much right now)

[twilit field]

I pressume group theory is a late undergrad course?

[pearl idol]

it's an undergrad course

when you take it in undergrad depends on your interests and school

[twilit field]

👍

thanks

[pearl idol]

you're welcome! you're doing well

[twilit field]

Thanks

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[worthy kestrel]

good job

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[urban prism]

find a condition a,b and c so that one zero of ax^2+bx+c is the square of other

[urban prism]

Let me tell u my side first

basically

it is saying one should be alpha so other should be alpha^2

right?

sum = alpha + alpha^2 = -b/a

product = alpha X alpha^2 = alpha^3 = c/a

but what after that

[zealous schooner]

Yeah that’s right

$\alpha+\alpha^2=-\frac{b}{a}, \alpha^3=\frac{c}{a}$

[grand pond]

kheerii

[zealous schooner]

What variable do you wish to eliminate from this?

[urban prism]

b?

no

a is common

so a

hm?

[worthy wing]

$\alpha^3=\alpha^2 \alpha$

[grand pond]

Samuel

[urban prism]

I already know that

ig

[urban prism]

^^^

$\alpha+\alpha^2=-\frac{b}{a}, \alpha^3=\frac{c}{a}$

[grand pond]

Thor

[midnight plank]

@urban prism Has your question been resolved?

[urban prism]

<@&286206848099549185> 15 mins have passed so

[dim edge]

guys what is a good game rn which is like either a city game or maybe like a vintage old game or vintage old military related stuff in roblox

[urban prism]

??

.close

[midnight plank]

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

I need help

[grim vector]

Distribute it first

[vast bone]

okay so kx^2-56x = -16

[grim vector]

Yes

Put the 16 in the lhs

[vast bone]

so would it be kx^2-56x+16=0?

[grim vector]

Yes it would

[vast bone]

but what is the least possibel

[grim vector]

Now do you know delta ?

[vast bone]

no im in 9th grade

[grim vector]

The discriminant ?

[vast bone]

yeah

b-4ac

[grim vector]

b^2 - 4ac

[vast bone]

ohh

:(

but what do i do after?

[grim vector]

You apply it to the quadratic you got

[vast bone]

but idk k

or like the a value

[grim vector]

Yes so let k as it is

[grim vector]

The a value is k

[vast bone]

ohhh wait i get it

so then i solve for k

after doing the discriminant?

[grim vector]

You solve your discriminant < 0

For k

[vast bone]

wait is < 0 one solution?

[grim vector]

No its no solution

The one solution is when the discriminant = 0

[vast bone]

but why do i have to solve for no solution

[grim vector]

Cuz they ask for

[vast bone]

srry im confused on this part

ohhhh

wait yeah wait icic

but after you set it as less than what do you do

[grim vector]

You have 56^2 -4k*16 < 0

[vast bone]

mhm

[grim vector]

You solve for k

[vast bone]

k<49

but it says least value of k

[grim vector]

I think you forgot to turn the inequality sign no ?

When divided by a negative numbers

[vast bone]

ohh

oops

okay

k>49

[grim vector]

Then you have the least

Its 50

[vast bone]

thank you!!!

i get it now

[grim vector]

There you go

[midnight plank]

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[vast bone]

can someone explain why

[vast bone]

I know the answer is 1/4

but why

...

[unreal fjord]

think about it...

after 4 hours, you have double bacteria, right?

[vast bone]

yeah

[unreal fjord]

so... when t = 0 (initially), you have P(0) = (C)(2)^0. Agree?

[vast bone]

yes

[unreal fjord]

so C = 12000, agree?

[vast bone]

yeah

ohhh

[unreal fjord]

if it is double after 4 hours, then 2^(4r) = 2

but you know that 2^1 = 2

[vast bone]

wait what?

[unreal fjord]

Ok so P(4) = 24000 = 12000(2^(4r)), agree?

[vast bone]

ohhh wait yeah i get what you mean now

so would i just solve for r by doing

[vast bone]

this

[unreal fjord]

yes

[vast bone]

ohhhh okay tyy

so (log base 2)2=4r

[unreal fjord]

you can take log. or you can just realize that 2^(4r) = 2^1, and conclude that 4r = 1

[vast bone]

oh yeah tht works too

tyyy!!!

[midnight plank]

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[midnight plank]

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[inland summit]

Can someone tell me the answer for this

[velvet compass]

Have you tried solving it?

[inland summit]

Ya

I have a doubt in sign

When x< 0

-x^2 -7x-8 =0

[velvet compass]

more precisely when x < -2, but yes

[inland summit]

And if I directly use quadratic formula

I get (7-root 17) /2

[inland summit]

But if I multiply -1 here

[velvet compass]

or (-7 - root(17))/2

[inland summit]

ya

[velvet compass]

but here the assumption is important

since x < -2

[inland summit]

K

[velvet compass]

then x = (-7-root(17))/2

[inland summit]

Ohh

[velvet compass]

is the only solution

[inland summit]

Thx bro

[velvet compass]

Do you know why x < -2?

[inland summit]

Absolute value

[velvet compass]

yep

[inland summit]

But how is it

For the first mod 3 is absolute value

And for the second -1

How is it -2

[velvet compass]

what do you mean by "mod"? Case?

[inland summit]

Modulus

[velvet compass]

(x+3) doesn't have modulus

so we don't even take it into account

[inland summit]

Oh shit

My bad

[velvet compass]

for the 1st one we have |x+2| then x = -2

[inland summit]

I was seeing the 52nd one instead of 53rd

Thx brother

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[twilit field]

let $a<b$, where a and b are positive , show $a^2<b^2$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

I mean this is obviosuly true

[hard shard]

there are a coupld of ways to do this

wat are you thinking

[obtuse void]

let b=a plus some constant k

[pearl hull]

Same question

[twilit field]

Not too sure tbh

am working on it now

just like working on discord

[hard shard]

oh

[hard umbra]

do you have any axioms to work with?

[humble torrent]

I guess you’re allowed to multiply by a positive number by both sides of the inequality?

[dreamy lichen]

k > 0
m < n => km < kn

This should be enough to prove it

[dreamy lichen]

yeah, this

[twilit field]

I was actually thinking of arguing if b>a adding b to it self b times will be more than adding a to itself a times

[dreamy lichen]

that works only for integers

[twilit field]

oh,right

[hard shard]

eh its nt anyway

[hard umbra]

adding b to it self b times
adding a to itself a times

apparently not

lang's calculus is it?

[twilit field]

A first course in calculus

[hard shard]

oh

drat

[hard umbra]

calculus yes

[obtuse void]

couldn’t you use geometry since both a and b are positive and can be used as side lengths?

[humble torrent]

Try and see if you can find this @twilit field in the textbook

[hard shard]

you could

[twilit field]

i could

[hard umbra]

i dont have a copy of the textbook so idk what lang covers

[twilit field]

I'll check

there a few rules

if a>b and b>c then a>c

if a>b and c>0

then ac>bc

I suppsoed I could use this

[dreamy lichen]

those are sufficient actually

[twilit field]

a^2>ab

ab>b^2

so a^2>b^2

[hard shard]

dont flip your signs

[humble torrent]

Yup

[dreamy lichen]

do you mean a^2 < ab?

[twilit field]

yeah

my bad

[dreamy lichen]

this, otherwise it's correct

[twilit field]

is it normal to miss the obvious stuff while just starting college , haven't have had to think in such an elemenatry way in a while

thanks everyon!

[dreamy lichen]

You got it quite quickly after looking at the axioms

[humble torrent]

Yeah I think you’re always doing very well the times I chime into these help channels

[twilit field]

👍

thanks both of you

thanks everyone, once again!

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[twilit field]

if a/b<c/d , show that a/b<(a+c)/(b+d)

[twilit field]

I just cross multiplied and got back the oG inequality

is there any other way to do this?

[humble torrent]

You can reverse those steps

[surreal moon]

Can you assume all a,b,c,d>0?

[twilit field]

oh yeah, that too

[twilit field]

so I start with ad<bc

then I add ab to both sides

and I'm done I think

thanks

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[twilit field]

so to start let $v_1,v_2\in V_c$
where $v_1=a+bi$

$v_2=c+di$

then $v_1+v_2 =(a+c)+(b+d)i$

similarly $v_2+v_1 = (c+a)+(d+b)i$

these two req equivalent due to the commutativity on $\R$

[grand pond]

ƒ(Why am. I here)=I don't Know

[clear void]

Yeah

[twilit field]

just a minute

now to prove associativity

let $v_1 and v_2$ be defined the same way they were above and then we have $v_3 =e+if$

so$ (v_1+v_2)+v_3_= ((a+c)+e+i((b+d)+f)$

and $v_1+(v_2+v_3)=(a+(c+e))+i(b+(d+f))$

which are once again equal due to the laws of associativity on $\R$

[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.)

[clear void]

Yea

[twilit field]

next let $v_1= a+ib ; v_2=0+0i$ so we have $v_1+v_2=a+ib$, but this doesn't prove the existance of an additive identity

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

[jagged saffron]

yes it does you just wrote it down

[twilit field]

huh, it does?

ok,thanks

[jagged saffron]

you just showed that there is some vector such that you add it to any other vector you get that vector back

also i should note that everything you are doing is essentially identical to how you (presumably) showed C is a vector space

[twilit field]

yeah, that's essentially the question

for the existance of an additive inverse let $v_1 =a +ib ; v_2 =-a+-ib$ so $v_1+v_2=0$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

now to prove the existence of a multiplicative identity if $v_1=a+bi$ and $v_2=1+0i$ then $v_1 v_2=v_1 proving the existance of the same$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

the distributive property follows similarly

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[twilit field]

thanks everyone!

[midnight plank]

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[chrome root]

How do I ask for help

[chrome root]

How do I solve this

[fleet flume]

(pi(11 square)(22.5))/3

[chrome root]

ahuh

I thionk I got it

Cuz

im currently doing my onine math test

did I get this right?

and dis

[chrome root]

this is volume

I need sa

[sage olive]

it's against server rules to ask for help for a graded assessment

[chrome root]

Oh

mb

[midnight plank]

@chrome root Has your question been resolved?

[violet storm]

.close

[midnight plank]

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[twilit field]

$\lim_{x\rightarrow a}\left(f\left(x\right)+g\left(x\right)\right)=\ \lim_{x\rightarrow a}f\left(x\right)+\lim_{x\ \rightarrow a}g\left(x\right)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

I want ot prove this

could I have a hint

[carmine void]

Look at the proof for sequences

[twilit field]

without epsilon-delta limits

[twilit field]

👍

[zealous schooner]

Hey wai

[twilit field]

what exactly should I look up??

[twilit field]

hey

[carmine void]

The proof that you can split the limit of a sum of two sequences if both limits exist

[twilit field]

oh, ok

thanks

everything seems to be using epsilon delta limits

[zealous schooner]

This is only true if both limits exist, which is important

[twilit field]

I know that

[zealous schooner]

Yeah that’s what I was thinking too

[twilit field]

so for now I just accept it as a fact?

I mean it's obvious

I could probably learn epsilon delta limits

thanks

[zealous schooner]

Honestly it’s like 2 lines lmao

[twilit field]

I know, but I've never done epsilon delta limits before

thanks again everyone

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[zealous schooner]

Hmm, the sequence one is also using epsilon delta?

[twilit field]

yeah

[zealous schooner]

Can you show me

[twilit field]

https://www.quora.com/How-do-you-prove-that-the-limit-of-the-sum-product-of-two-sequences-is-equal-to-the-sum-product-of-the-individual-limits

[sage helm]

there's no delta to talk of

[zealous schooner]

Oh, hmm

I guess it’s just the same as a normal limit

[zealous schooner]

What

[sage helm]

when i proved that stuff there was no delta

I think epsilon-delta is only for limits of functions

[zealous schooner]

Huh?

[sage helm]

???

def of convergence for sequences

nvm

[zealous schooner]

Oh, for limits at infinity there’s no delta yeah

You usually use N for that

[twilit field]

ok, I don't know how to solve epsilon limits either

[sage helm]

yeah

[zealous schooner]

Or there isn’t an epsilon rather

[twilit field]

what

[zealous schooner]

If you’re taking a limit to infinity you don’t use |x-a|<epsilon, you use x>N

[twilit field]

I have no idea what that means

[zealous schooner]

When you study epsilon delta proofs you’ll understand

[twilit field]

Thanks a lot though

[midnight plank]

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[last slate]

Please solve it

I might be offline so please solve fully. I tried solving it but could only work out 2 ordered pairs, one is (1,2) and second is (2,1).

[potent veldt]

Against the rules bruv

if u want a hint ask for that but no sols

[last slate]

Well, I did get solutions so-

[last slate]

And it's only against the rules if you haven't fully solved the question, but if you have then you might. Still, I'll try to work with a hint if that's the best I can get.

[worthy wing]

!show

[midnight plank]

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

[potent veldt]

The show ur work instead

[last slate]

This is my work ;-;

I got (1,2) and (2,1)

[potent veldt]

So u didn't solve it

[last slate]

But optipns dont have 2 ordered pairs

Bruh

[potent veldt]

look ill just give u the hint

Empty set

[last slate]

Wait

How

[worthy wing]

He literally gave you the answer

Work a bit

[last slate]

But then it will be (phi,1) or (phi,2) and A union B won't be {1,2}

It's gonna be either {1} or {2}

I guess

I'm confused

[potent veldt]

why's B gotta be {1} or {2}

[last slate]

Wait

Can it have more than 1 element?

Wait a sec

[potent veldt]

A and B are sets

[last slate]

OH

My bad

[potent veldt]

np

[last slate]

I got 9 elements

Combinations of A and B*

Oh

Wait

4 combos

Because others have phi as an element which will not satisfy the condition

Lemme fheck the answer

Oh ;-;

It's 9 for some reason

[worthy wing]

Read the question again

[last slate]

Nvm

.close

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[sage helm]

Have I proved it the wrong way around

[frozen talon]

is def'n 2.2.3 just that of a sequence converging to its limit?

[sage helm]

yeah

[frozen talon]

then yeah, you did it the other way around. you used that which you wanted to prove as an assumption

also you never declared a positive epsilon

so we don't know who this epsilon character is on mathy line 2

[sage helm]

Does this fix all of that?

[frozen talon]

we ignore the first line you write because it assumes the conclusion. also, the algebraic step between 2 and 3 may be unclear

[sage helm]

Oh yeah the first line

[sage helm]

multiplying 2/3?

[frozen talon]

I guess it depends lol

[sage helm]

Yeah I've done worse before

I think I'll let it slide

[frozen talon]

that's fair

there's a better procedure that doesn't incur the handwaviness at the end

[sage helm]

That being?

Define a new sequence

?

[frozen talon]

you just massage the inequality of (2) a bit to look like (2x_n - 1)/3 converges to 1

[sage helm]

I guess I could do some triangle inequality stuff

[frozen talon]

that's not necessary either

[sage helm]

Huh

[frozen talon]

2x/3 - 4/3 = (2x-4)/3 = (2x-1)/3 - 1

and abs((2x-1)/3 - 1) < epsilon for all n >= N, hence (2x-1)/3 converges to 1

[sage helm]

Ohhh

Yeah

That makes sense

Okay yeah, fixed

[frozen talon]

rad

[frozen talon]

I was typing up a comment about using a colon after the N on the first line but I rescind it lol

I think the colon is good and fixes a problem that I have

[sage helm]

colon where?

you mean replacing "such that"

?

[frozen talon]

wherein me using a comma in place of a colon on the first line (such that for all n >= N{: -> ,}) makes me want to treat the math mode stuff like a sentence and thus put a period on the last math line in (3)

[sage helm]

Oh

Well

English is weird

Does it really matter though?

[frozen talon]

probably not but I figured I'd share my thought anyway in case you had any more insight about the usage of the colon lol

[sage helm]

¯_(ツ)_/¯

If that's it, then I'll close the channel

Thanks a bunch :))

[frozen talon]

pog, np

[sage helm]

.close

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[agile carbon]

Help for d and e pls

[agile carbon]

its tech active so i can use my cas to find the discriminant, but i need some guidance

[midnight plank]

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[agile carbon]

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[slow moon]

someone tell me where i went wrong

.close

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[waxen silo]

how do I solve the last question?

[waxen silo]

I tried doing it by subbing 0.004 as s and find for t but it gives me a decimal like 0.0666687... and that is clearly not correct

how are you suppose to do this?

[midnight plank]

@waxen silo Has your question been resolved?

[waxen silo]

<@&286206848099549185>

[waxen silo]

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[twilit field]

prove that $x-\frac{x^3}{6} \leq sin(x)$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

my first idea was to say that x-x^3/6 is only a part of sin(x)'s taylor expansion

or alternatively, first find the derivative of both sides

to obtain $1-cos(x) - \frac{x^2}{2}$

[grand pond]

ƒ(Why am. I here)=I don't Know

[last slate]

are you allowed to use taylor series?

[twilit field]

I know taylor series, but haven't covered it yet in the book so idts

this is the same as $2sin^2(x/2) -x^2/2$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

hmm

let x/2=t

[frank wolf]

This is only for $x \geq 0$?

[grand pond]

Pseudonium

[twilit field]

yes

[frank wolf]

Otherwise im pretty sure its false

Right

[twilit field]

so this is the samee as 2sin^2(t)-2t^2<0

or $sin(t)<t$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

which is true

that's it i guess?

[frank wolf]

Uh

I don’t quite follow your argument

[frank wolf]

Also this is false for $t \geq 0$

[grand pond]

Pseudonium

[last slate]

i think

[twilit field]

,w is x>sin(x)

[last slate]

just using a counter example would be fine

[frank wolf]

It’s true for x > 0

[grand pond]

Results provided by WolframAlpha

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Requested by physicsrocks

[twilit field]

yeah

so t>0

[frank wolf]

But i assume you need it to work for x >= 0

[twilit field]

as t=x/2

[twilit field]

that's why it's $\leq$ here

[grand pond]

ƒ(Why am. I here)=I don't Know

[frank wolf]

Yeah so why not just do $\sin x \leq x$

[grand pond]

Pseudonium

[twilit field]

from here, to move foward , I need to sub x/2=t

right

[twilit field]

to bring it to this form

[frank wolf]

Oh yeah sure

But I don’t see how you get from that back to your original thing

Like I totally agree that $2 \sin^2(x / 2) - x^2 / 2 \leq 0$

[grand pond]

Pseudonium

[frank wolf]

And I agree with your proof strategy of setting $t = \frac x2$ to do this

[grand pond]

Pseudonium

[frank wolf]

Actually

Wait

What is your strategy exactly

[twilit field]

$2sin^2(t)-2t^2<0$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

or $sin^2(t) \leq t^2$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

or sin(t)<t or -sin(t)<t

which is always true

for t>0

[subtle blaze]

But how do we define sin(x) if not with Taylor series

[frank wolf]

You seem to keep introducing strict inequalities when they’re unnecessary

[twilit field]

lang has defined it as $\frac{y}{\sqrt{x^2+y^2}}$

[frank wolf]

Right, so you do need the second one

[twilit field]

or using a triangle

[frank wolf]

And you know why both are true?

[subtle blaze]

Missing } at the end

[twilit field]

yes

[frank wolf]

Ok cool

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

in a right triangle

[humble torrent]

Lol

[subtle blaze]

And what’s y

[twilit field]

side length of opposite side

[subtle blaze]

But surely then x is not the input

You’d need it to be the angle

[twilit field]

right

let me check once more

just a min

[frank wolf]

You need to define a function $\text{sin} : \mathbb{R} \to \mathbb{R}$

[grand pond]

Pseudonium

[twilit field]

x isn't the input here

it's the adjacent side

[subtle blaze]

geometry

I feel like I’d want to approach questions with sine using the definition of sine given

[subtle blaze]

Like how do you know this is true

You’d want to refer to the definition of sine

[twilit field]

I proved that earlier

so I can use it

[subtle blaze]

Or at least I’d want to

Ah ok

[twilit field]

hmm?

[frank wolf]

As in

[hard umbra]

then that raises the question

[humble torrent]

Yeah in an earlier question they said it was given

[hard umbra]

how did you prove that result

[frank wolf]

Given a real number t, you need to be able to give me the output sin(t)

[twilit field]

derivatives

[frank wolf]

Otherwise im not convinced you have a function

[subtle blaze]

Hey I was gonna sweep it under the rug and not look

But now you’ve opened the can of worms

[frank wolf]

Because, after all, a function is supposed to send inputs to outputs

[twilit field]

sin(x) can't really be defined wrt to its input without taylor series afaik

[humble torrent]

well

yes it can

[subtle blaze]

Well I mean idk dude you want to use sine in a question you better give me a definition

[twilit field]

this

[subtle blaze]

Otherwise we’re gonna just pull random facts out of our ass

[frank wolf]

This definition doesn’t meet my requirement

How do I use this definition to find sin(1)?

[humble torrent]

Im pretty sure sinx < x is a given fact, they stated that in an eariler question

[subtle blaze]

“Given fact”

[frank wolf]

Yeah that’s fine

[twilit field]

no, I proved it, I can prove it again if needed

[humble torrent]

damn

[twilit field]

you can't, I'd probably use the unit circle in that case

[subtle blaze]

If you can’t tell me what sin(1) is, you haven’t prescribed to me a function

In order to define a function you must tell me where every input goes

If I have a particular input I must be able to compute the corresponding output

[twilit field]

I'm not sure how I'd do that without the taylor series

[frank wolf]

Yeah that’s reasonable

I mean there are other ways to define sin than Taylor series

[hard umbra]

except when you dont due to uniqueness theorems

[frank wolf]

But Taylor series are the most convenient

[twilit field]

yeah, but they are geometric , right?

[frank wolf]

Oh not necessarily

You can look at the differential equation $y’’ =- y$

[subtle blaze]

But don’t uniqueness theorems tell us where things go in a round about way

[frank wolf]

And define $\sin(x)$ to be the solution satisfying $y(0) = 0, y’(0) = 1$

[grand pond]

Pseudonium

[hard umbra]

are you a constructivist

[frank wolf]

This is a perfectly legitimate definition of sin

[twilit field]

true

[grand pond]

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

[hard umbra]

cant we just assume the inequalities and move on

[frank wolf]

^

This is kinda a distraction

[twilit field]

anyway, should I prove sin(x)<x first?

[frank wolf]

If you say you’ve proved it it’s probably fine

[twilit field]

ok, so in that case is my proof fine?

[frank wolf]

No

Cause ok, I now agree that you’ve shown $2 \sin^2(x/2) - x^2 /2 \leq 0$

[grand pond]

Pseudonium

[frank wolf]

However I don’t see how you get back from this to your original statement

[twilit field]

$2sin^2(t)-2t^2\leq 0$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

so $sin(t) \leq t$

[grand pond]

ƒ(Why am. I here)=I don't Know

[twilit field]

or $-sin(t) \leq t$

[grand pond]

ƒ(Why am. I here)=I don't Know

[frank wolf]

Yeah no I totally agree you’ve proved this (I think)

Although the way you phrase it is a little annoying…

Really, you should convince me that, starting from $\sin(t) \leq t$ and $- \sin(t) \leq t$, we can deduce $2 \sin^2(t) - 2 t^2 \leq 0$

Atm you seem to be proving things backwards

Which, I’m not sure if you’re good enough at proofs for me to trust that you know what you’re doing

[twilit field]

$\sin(t) \leq t$ and $- \sin(t) \leq t$, we can deduce $2 \sin^2(t) - 2 t^2 \leq 0$

[frank wolf]

Ok, why?

[twilit field]

the TeX it bot is down?

[frank wolf]

Yeah I was wondering

[twilit field]

so from $sin(t)\leq t$