help-23

[white umbra]

If we can successfully do that, then it is. If we can't, then maybe we start to reconsider.

[mild bridge]

ok

.

[white umbra]

Do you remember what the definition of continuous is?

Using epsilons and deltas

[mild bridge]

um

hold up

[mild bridge]

i don't

i thought it had something to do with limits from both sides

and epsilon delta is the tool for it

[white umbra]

Okay we need to start there then

That's right

[calm violet]

@white umbra considering his current foundations I think we should work out my example g first

[mild bridge]

the function with the one higher point?

[white umbra]

You can step in if you want

[calm violet]

yeah

umm sorry now i really gtg

[white umbra]

Oh alright

Yeah let's talk about that example function g

So let's say g(x) = 1 when x=0 and 0 when x≠0.

Where is it continuous?

[mild bridge]

everywhere except for x = 0

right

?

[white umbra]

Yup

So intuitively that's because if you move a bit, the y-value doesn't change a lot, right

[mild bridge]

yeah

when approaching any point from both directions, we get at the same value

[white umbra]

Yeah

Careful though, that's not sufficient

Because if you think about x=0

[mild bridge]

but in the case of x = 0, when we apprach it from both directions, we actually got to 0 instead of 1

[white umbra]

Yeah exactly

When approaching from both directions, we get the same value

But it also has to match the actual value for the function to be continuous

[mild bridge]

the limit from sides has to be equal to the image of that value

[mild bridge]

ok got it

and how does show it mathematically?

like lim x - > 0 (g(x)) = 0 = / = 1

?

[white umbra]

Yeah you would write it like
[\lim_{x\to 0}g(x) = g(0)]
if (g) is continuous at (0)

[flat frigate]

pERICyclic_reaction

[white umbra]

But in this case, the left-hand side is 0 and the right-hand side is 1

So it wouldn't be continuous at x=0

Does that make sense?

[mild bridge]

okk

yes

[white umbra]

Okay cool

But then there's the question: what does a "limit" actually mean, rigorously?

It basically means I can get as close as I want, right?

[mild bridge]

something with sequences

[mild bridge]

yes

[white umbra]

So to write that in rigorous mathematical notation

I like to think of it as a game

If [\lim_{x\to0} g(x) = 0]

[flat frigate]

pERICyclic_reaction

[white umbra]

Then that means I can get g as close as I want to 0, so if I tell you I want to make g(x) really close to 0, like between -0.1 and 0.1 for example

Then you should be able to tell me how close I need to make x to 0 to make that happen

So like for example "if 0<|x|<1, then |g(x)|<0.1"

Does that make sense?

[mild bridge]

yeah?

so

i keep plugin in x values from the domain that make g(x) get closer to the 0

are we still talking about the same g function?

[white umbra]

Yeah

In this case it's pretty trivial because g(x) is always 0 (except when x=0)

[mild bridge]

yeah ok

[white umbra]

So you could really tell me any value of "closeness" for x

Like for example you could equally say

"if 0 < |x| < 10000, then |g(x)| < 0.1"

and that would still be valid

But in general, if g wasn't constant, then you might have to make that closeness value for x really small

In order to get g(x) into the desired range

Does that make sense?

[mild bridge]

But in general, if g wasn't constant, then you might have to make that closeness value for x really small

why?

[white umbra]

Let's think about another example, g(x) = x^2

[mild bridge]

ok

[white umbra]

Let's say I tell you, I want |g(x)| < 0.1

And I want to know what x values are acceptable

If you then told me "0 < |x| < 10000"

That wouldn't work, right?

[mild bridge]

but it would no?

[white umbra]

How come?

[mild bridge]

like x = 100000 lets say

[white umbra]

Yeah

[mild bridge]

that squared is smaller than 0.1

[white umbra]

Huh?

100000^2 is a lot bigger than 0.1

[mild bridge]

omg

for some reasing i thought that's a decimal number

0.00001

like

[white umbra]

xD

[mild bridge]

sorry

ive been awake for a while

[white umbra]

Yeah so you have to make the bound on x pretty small right

like for example you could make it

[mild bridge]

yes

[white umbra]

"0 < |x| < 0.3"

and then when you square that you get under 0.1

[mild bridge]

yup

[white umbra]

And then if I told you like

I want precision to within 10^-6, so |g(x)| < 10^-6

What's some bound on the x-value that you could choose to guarantee that?

[mild bridge]

0<|x|<10^-4

[white umbra]

yup

So now the point is

No matter what precision value I pick, whether it's |g(x)| < 0.1 or |g(x)| < 10^-6 or whatever

You can find an appropriate range of x which guarantees the precision I want

That's the definition of a limit

[mild bridge]

ookkk

[white umbra]

So we define
[\lim_{x\to a}f(x) = L]
to mean that for any (\epsilon>0) (the precision I want), there's some (\delta>0) (the range on (x) that you pick for me) so that (0<|x-a|<\delta) guarantees that (|f(x)-L|<\epsilon).

[flat frigate]

pERICyclic_reaction

[white umbra]

Or to make it more concise,
[\forall\epsilon>0,,\exists\delta>0,,(0<|x-a|<\delta)\implies(|f(x)-L|<\epsilon).]

[flat frigate]

pERICyclic_reaction

[mild bridge]

ok

got it

what is this a

x - a

why not just x

[white umbra]

|x-a| is the distance between x and a

in the previous example, I had x→0

so it was the distance between x and 0, or |x-0|, or just |x|

but in general, I can have x approaching any value a

and then I would want the distance between x and a, |x-a|, to get really small

does that make sense?

[mild bridge]

yes

[white umbra]

awesome

[white umbra]

So now your challenge is, consider again that function g which was like g(x) = 1 when x=0 and 0 when x≠0, prove to me using this definition that g is not continuous at x=0.

[mild bridge]

ok

when trying to apply the definition of limit to a function

do i use actual values for my epsilon and delta?

[white umbra]

So when you want to prove a limit holds, you need to show that it works for any possible value of epsilon

So you won't use actual values

But when you want to prove a limit doesn't hold, you just need to find one value of epsilon that it doesn't work for, and then you need to show that no value of delta works

So that's a case where you'd want to pick an actual value for epsilon and then show that no possible value of delta works

[mild bridge]

and L would be the limit that I'd be testing

in this case 1

[white umbra]

Yup

[mild bridge]

and

i guess I can't use 1 which is the limit itself to check

sorry I mean 0

I have to check the function at all values except for the value of the limit that I'm testing

[white umbra]

Right now I just want you to show that
[\lim_{x\to0}g(x) = g(0)]
is false

[flat frigate]

pERICyclic_reaction

[white umbra]

yeah, the limit doesn't depend on g(0)

[mild bridge]

so

I choose and epsilon value

= 0.1

[white umbra]

right

[mild bridge]

then I identify the limit I'm testing

which is 1

L = 1

[white umbra]

and you want to prove that there's no value of delta so that 0<|x|<delta guarantees that g(x) is within 0.1 of 1

So for any value of delta, there's at least one value of x where 0<|x|<delta so that g(x) is farther than 0.1 from 1

[mild bridge]

and

i just take a look at the function

I see that for no value of x regardless of delta can the function every be closer than 1 to the limit

[white umbra]

Yeah. That's right.

So that proves the function isn't continuous at 0.

Let's now try this with the original function

[mild bridge]

ok

[white umbra]

This one

How would you prove this function isn't continuous at x=0?

[mild bridge]

I would first

set an epsilon value as an irrational number

a really small one

and then

do

| f(x) | < ε

wait nvm

I select a really small irrational delta

wait im confused

say ε = 0.1

and we wanna show that no value for x fulfills | f(x) | < 0.1

there will always be a value that fulfills this tho

[white umbra]

Okay lemme reorganize your thoughts here

So we set epsilon = 0.1

[mild bridge]

ok

[white umbra]

And we want to show that for any delta, there's some x satisfying 0 < |x| < delta so that |f(x)| ≥ 0.1

[mild bridge]

ok

[white umbra]

how would we go about that?

[mild bridge]

is it possible to say that delta is an infinitely small irrational number?

[white umbra]

No, there's no such thing as infinitely small.

[mild bridge]

wait

[white umbra]

And also you don't get to choose delta.

I choose delta and I give it to you.

[mild bridge]

we want to show the opposite tho

that there isn't such a delta value

[white umbra]

And then you have to choose a value of x according to the delta that I gave you.

[white umbra]

Yes, showing that there isn't a delta value that works is equivalent to showing that all delta values don't work.

[mild bridge]

:D

[white umbra]

So for every delta value I give you, you need to demonstrate that you can show me a value of x that's a counterexample

i.e. some x where 0 < |x| < delta but |f(x)| ≥ 0.1

Does that make sense?

[mild bridge]

yes

and regardless of the delta, we can always subtract a really small irrational number from delta

and land on an x value, where when we take it and plug it in f(x), we are not in our epsilon value range

[white umbra]

That's the right type of reasoning. You need to be a little more careful though because if my delta is irrational and you subtract an irrational from it, the result might end up being rational.

But it's true that there's always some irrational number x between 0 and delta.

And then how do you know that f(x) is not in the range, i.e. |f(x)| ≥ 0.1?

[mild bridge]

because it is more than at least 1 away from f(0)

[white umbra]

why?

[mild bridge]

as it is irrational?

and 0 rational

[white umbra]

Well let's do an example right

Let's say I pick the delta value 0.1

and you pick an irrational number between 0 and 0.1

[mild bridge]

okj

ok

[white umbra]

call that number x

and then f(x) = 1-x, which is somewhere between 0.9 and 1

[mild bridge]

ok

[white umbra]

And you know that's more than 0.1

It's not more than 1 though, like you said

Let's say I give you a value for delta

Can you pick x to be any irrational number between 0 and delta?

Or is there some sort of restriction?

[mild bridge]

yes why not

[white umbra]

Well let's say for example I pick delta = 1

[mild bridge]

ok

[white umbra]

And you say let x be any irrational number between 0 and 1

[mild bridge]

ok

[white umbra]

So I say okay pick an irrational number which is really really close to 0.99

Then what's f(x)?

[mild bridge]

really close to 0

[white umbra]

yeah, really really close to 0.01, and that's less than 0.1

but your claim was that it was more than 0.1

So we need to do some sort of fix to make sure x can't be chosen like that

Does that make sense?

[mild bridge]

wait im confused

when you were writing out the definition of lim

[white umbra]

Yeah

[mild bridge]

u wrote |f(x) - a| < ε

[mild bridge]

but here you wrote the opposite

[white umbra]

Yeah that's because we're trying to prove the opposite

We're trying to show that the limit doesn't exist

[mild bridge]

ok ok

[white umbra]

Or at least, that the limit is not equal to f(0)

[white umbra]

So this is saying that the limit is L

To say that the limit is not L, we would have to write this

[\exists\epsilon>0,,\forall\delta>0,,\exists x,,(0<|x-a|<\delta)\text{ and }(|f(x)-L|\ge\epsilon).]

Make sure you understand where that comes from

[flat frigate]

pERICyclic_reaction

[white umbra]

(Btw I actually forgot the "forall x" in the original, so lemme retype the definition of the limit being equal to L)

[\lim_{x\to a}f(x) = L]
means that
[\forall\epsilon>0,,\exists\delta>0,,\forall x,,(0<|x-a|<\delta)\implies(|f(x)-L|<\epsilon).]

[flat frigate]

pERICyclic_reaction

[mild bridge]

ok

[mild bridge]

in order to prove that a limit doesn't exist

can we change the for all x to there doesn't exist an x

and keep everything else the same

[white umbra]

So when you negate something

You flip all the quantifiers

So "for all" becomes "there exists"

That's because to disprove that something is true for all x, you only need to find one counterexample, so there exists an x so that it's false

Similarly "there exists" becomes "for all"

Hopefully that makes sense

[mild bridge]

yes it does

[white umbra]

I have to go now, but I hope this was helpful. These steps here are also very well thought out ^

[mild bridge]

ah ok

thanks a lot for helping me out

really appreciate it

[white umbra]

But to solve the problem we were facing, try considering an irrational x which is between 0 and delta, and also less than some value that will make sure I can't do something sneaky like pick x to be an irrational value close to 0.99.

No problem!

[white umbra]

And then once you've figured out how to prove the function is discontinuous at 0, try proving it's discontinuous at other points too. What you really want to prove is that it's discontinuous at every point except for 0.5.

And then finally, prove it's continuous at x=0.5.

Good luck.

[mild bridge]

ok thanks for the rundown

babye

[safe radish]

@mild bridge Has your question been resolved?

[safe radish]

@mild bridge Has your question been resolved?

[safe radish]

@mild bridge Has your question been resolved?

[safe radish]

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[storm ice]

Am I doing this right?

And what happens to the y(x) that's being integrated? Would it become Y(x)..?

if I want to do partial decomposition(?) how would that look with the y(x)? or is it "illegal"?

[desert pasture]

for simplicity, let y(x)=y, this requires something called an integrating factor to be solved

[storm ice]

ohhhh

yes its the thing where i do

okokok

[desert pasture]

https://youtu.be/20x2dNVztlU?si=X6y8NCgHoma7Vx0I

[storm ice]

i dont exactly remember the steps though but it's uhh

[desert pasture]

this may help

[storm ice]

With that said - is the approach I was going for impossible?

[desert pasture]

I think so

I could be wrong though

[storm ice]

and what the frick would the integral of 2x/1+x^2 be?

[desert pasture]

ln(1+x^2)

[storm ice]

man do i feel dumb

[desert pasture]

let 1+x^2=u

you can take it from here

[storm ice]

i was thinking i had to do something with arctan

because the derivative of arctan is 1/1+x^2 right?

[desert pasture]

yes

[storm ice]

so to dumb it down for me

if the nominator is the derivative of the denominator

you just do ln(denominator)

[desert pasture]

yes

[storm ice]

perfect

what an angel you are

[desert pasture]

to generalise the f(x) be a function

and f'(x) its derivative

[storm ice]

i have nothing but the utmost love for u <33

[desert pasture]

then $\int \frac{f'(x)dx}{f(x)}$ is $ln|f(x)| +C$

[flat frigate]

Why am. I here

[storm ice]

<33333333323333

.close

[safe radish]

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[tender hill]

i have no idea how to even start this question:(

[lean otter]

,rccw

[lean otter]

so you know m = 3/2 for both right? @tender hill

[tender hill]

mhm!

[lean otter]

if the lines are tangent to the circle, so the perpendicular line from those lines (at the point of tangency) would be the radius

[tender hill]

i see

yeah?

[lean otter]

lets call the point of tangency for the first line to be $(a,b)$, then the equation of the circle would be [
y = \f32 x + b
]
and the radius from the point $(3,-3)$ (the origin) to the point of tangency can be given by [
d = \f{\abs{ax + by +c}}{\s{a^2 + b^2}}
]

[lean otter]

i think what what you can do here is recognise that you have [
y - \f32 x - b = 0
]
and then substitute to get [
\f{\abs{-\f32(3) + 1(-3) - b}}{\s{(\ff32)^2 + b^2}}
]

[lean otter]

is this fine? @tender hill

[tender hill]

ive never seen this before :(

[lean otter]

its the equation of like the distance from a point to a line

was it not covered in ur syllabus

[tender hill]

it wasnt

[lean otter]

that sucks i think you get a quadratic to solve and then using that equation u can recover the point

and that should be it

[tender hill]

ah a;rghty then

[lean otter]

i might have fucked up though because i dont think there are any real solutions for b

let me scan

[
\s{52} = \t{Radius} = \f{\abs{-\f32(3) + 1(-3) - b}}{\s{(\ff32)^2 + 1^2}}
]

[lean otter]

okay ig that should be fine

anyways @tender hill

we can use this to verify our answer

but what have u learnt so far?

maybe i can gauge a different method

[safe radish]

@tender hill Has your question been resolved?

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[royal sorrel]

Ummm helo?

[royal sorrel]

Simplify

I have zero idea about this

[spice grove]

There is no question.

[royal sorrel]

And ive tried

Simplify

[royal sorrel]

Simplify

Now, im no thought head empty

[hearty egret]

try some easier problem before attempting harder ones mate

[royal sorrel]

Yes i had done it

Now about the b guy

Multiplies are easy

But when they meet add

They say "im gonna end this man whole career"

[safe radish]

@royal sorrel Has your question been resolved?

[safe radish]

@royal sorrel Has your question been resolved?

[royal sorrel]

<@&286206848099549185> please Q_Q

[royal sorrel]

Ive tried, and the answer i made doesnt make sense

[daring crescent]

You don't need to multiply them, there is a general rule to dividing numbers that have the same base but different powers

[cinder lily]

b^(-3x+3y)?

[daring crescent]

You take the numerators power and subtract the denominators power. Also I would keep the x and y on separate powers @royal sorrel

No

[cinder lily]

oh wait i didnt see it properly

[daring crescent]

a^(3x - 2x) * a^(-y - 3y)

[cinder lily]

b^(-2x-x+2y) + b^(3y-x+2y)
b^(-3x+2y) + b^(5y-x)

[cinder lily]

when they add

split them

[daring crescent]

I would keep for simplicty the x and y seperate or you can do power rules

[daring crescent]

a^(3x - y)/a^(2x + 3y) and then

a^(3x - y - 2x - 3y) then combine like terms

[cinder lily]

best way is to take log

and solve

[royal sorrel]

but this question act as preface for the log...

i must answer the b guy first

[royal sorrel]

i tried, and the answer didnt make any sense

[cinder lily]

similar to how l hopital rule doesnt make any sense when studying limits and after learning derivatives we only use that

[daring crescent]

Ah there is more above

You can split that into two separate parts

$\frac{b^{-2x}}{b^{x} * b^{-2y}} + \frac{b^{3y}}{b^{x} * b^{-2y}}$

[flat frigate]

dragonbreath

[royal sorrel]

so the answer ends with +?

[daring crescent]

My connection was so spotty for and so weird, it sent my messages but didn't recieve all of the messages just bits of the conversation so sorry for the untimely interjections

[daring crescent]

If you want it too

In order to add they would have to have the same power and you would add the bases

[royal sorrel]

((b^2y)/(b^3x))+((b^y)/(b^x))

?

does it ends there?

[daring crescent]

Nvm, was thinking of sig figs I think

[daring crescent]

For the most part yes

[royal sorrel]

hmmmm

let me think again

[daring crescent]

You can try and combine the fractions but you'll need equal denominators

[royal sorrel]

hmmmm

[daring crescent]

It'll probably be an uglier equation

[royal sorrel]

if it ends with + , then it cannot be really called simplified...

[daring crescent]

I don't think there is really a fun way to get rid of the +, if there is I am unaware of it.

[royal sorrel]

okie

we must meet again sum times

thx very much

.close

[safe radish]

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[stiff vine]

Is this graph answer [0,inf)U (3,3] U (2,inf)?

[quasi bison]

"answer"?

[fiery merlin]

What about that graph are you saying fits those ranges combined?

[stiff vine]

Interval notation

[fiery merlin]

Oh, the range?

[quasi bison]

this doesn't tell us much

[stiff vine]

Oh my bad I meant the range

[stiff vine]

Yeah

[fiery merlin]

It looks like it's the ranges of each piece of the piecewise function.

[stiff vine]

So is this the correct range answer?

[quasi bison]

(3, 3]?

[fiery merlin]

Yeah, just fix the (3, 3]. (3 means it's higher than 3.

[quasi bison]

you are technically right but unsimplified!

[0, +infty) already includes both {3} and (2, +infty) as subsets. so no need to state them twice.

[fiery merlin]

You'd write [3, 3] or {3}.

[stiff vine]

But the first point isn’t included

[fiery merlin]

Yes, but it's not about the first point. Like your quadratic there doesn't start with 0, but it reaches it.

So, you'd have [0 because 0 is in it.

Same with the [3, 3].

It's about the collection of outputs, which goes as low as exactly 3.

So, exactly 3 is [3.

It's [lowest, highest] or (lowest, highest] or whatever.

The lowest is exactly 3.

So, [3.

It's not [leftmost, rightmost].

[stiff vine]

So I just change the (3,3] to [3,3] and my answer is right?

[fiery merlin]

Yes, but it's not simplified.

Like with [0, infinity), [3, 3] is inside that.

So, you can just do [0, infinity) by itself because it handles [3, 3] for you.

Same with [2, infinity).

Those numbers are already in [0, infinity).

So, you can just say [0, infinity) as the final answer, because it contains every single output.

And it doesn't include anything that's not an output.

So, it's exactly the numbers that you can get as outputs.

Nothing added in, nothing removed.

Does that make sense?

The only reason you'd use U in the final answer is if there's gaps between two intervals.

[stiff vine]

Ok so a graph like this is [0,5]U(2,5]? For the range

[fiery merlin]

Yes, and then there's no gap between the intervals.

Like you don't stop at 5 for one and then start at 6 for another.

Where there's a gap.

[stiff vine]

Yeah so [0,5]U(2,5] would be correct for this graph right

[fiery merlin]

It wouldn't be simplified. I'm drawing something related.

[stiff vine]

So the answer is just [0,5]??

[fiery merlin]

See how they overlap?

Like both of them handle 2 to 5, right?

They overlap there.

[stiff vine]

Ok so it’s just [0,5]?

[fiery merlin]

Yes, but there's a trick.

You have two lower endpoints.

Like (a, b] and [c, d) have (a and [c as the lower endpoints.

[stiff vine]

Is this interval notation still a right answer?

[fiery merlin]

No, you want to simplify it.

It's correct, but it's not what you should submit.

So, what you do when they overlap is that you see which starting endpoint is lower.

Like if it's (3, 10] and [4, 25), you have (3 and [4. (3 is lower.

So, you write (3 in your final answer.

On the ending endpoint, you look at which is higher.

25. is higher than 10].

[stiff vine]

Damn that’s what I submitted my answer as

[fiery merlin]

So, you have (3, 25).

Does that make sense?

[stiff vine]

Kinda

[stiff vine]

But I should still get full points?

[fiery merlin]

It's possible, but you might get a point or two taken for not simplifying it.

Knowing how to combine overlapping intervals into one is a good thing to know for these problems.

Like you have [0, infinity) and [3, 3]. The overlap. The lower starting number is [0. The higher ending number is infinity), so [0, infinity) is their combined interval.

[stiff vine]

Ok so this would be [0,inf)?

[fiery merlin]

Then, you have (2, inf). It overlaps [0, inf).

Yes, that's right.

The intervals overlap.

[stiff vine]

Instead of [0,inf)U[4,4]

Ok

[fiery merlin]

Right, you graph them and see if they overlap.

Then, if so, you take the lowest start and the highest end for the combined interval.

And [3 counts as lower than (3. 5] counts as higher than 5).

That's because 3 is lower than the numbers just above 3, and 5 is higher than the numbers just before 5.

[stiff vine]

Which problem u talking about

[fiery merlin]

Oh, the example.

The question is, if you have [3 and (3, which one wins?

The square bracket one wins.

Same with 5] and 5). The square bracket wins.

If the numbers are the same, but you have both square and round brackets, pick the square bracket.

It wasn't about a problem you posted.

Just rules for how to combine them.

[stiff vine]

Why

Can u show an example

[fiery merlin]

Because [3 includes 3, right?

(3 doesn't include 3.

It's numbers above 3.

[stiff vine]

( means it gets close to 3 but never reaches it?

[fiery merlin]

Right.

Anything just above 3 is fine.

But 3 isn't.

So, [3 can go a little bit lower because it can actually become 3.

So, if you're combining two overlapping intervals, and the number on the left is the same for both intervals, you want to pick the one with a square bracket if one of the intervals has one.

If it's (3 and (3, you just do (3.

[stiff vine]

If 3 isn’t fine then why would u include the [

[fiery merlin]

3 is fine when you do [3.

[fiery merlin]

I was responding to this ^

(3 means anything just above 3 is fine.

(3 means 3 isn't fine.

But [3 means both 3 and anything just above 3 are fine, so it can get a little bit lower.

Does that make sense?

[stiff vine]

Ok so in this I just make it [0,2] instead of (0,2]

Is that what u mean

[fiery merlin]

Well, I've said a lot of things. Which one are you asking the meaning of?

[stiff vine]

That u always use [ over ( if they’re same spot

[fiery merlin]

Oh, I see.

Yes, that's right.

They overlap, which is very important, since you need that to combine them.

Then, you have [0 and (0. The numbers are the same, so you pick the [0 if it's there.

Like if it was (0 and (0, you just have (0.

But if it's mixed like that, you do [0.

[stiff vine]

👍

[safe radish]

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[safe radish]

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[narrow vault]

how to find the center of gravity?

[narrow vault]

yall hate physics dont you

did noone take any physics courses here?

[safe radish]

@narrow vault Has your question been resolved?

[safe radish]

@narrow vault Has your question been resolved?

[plucky elk]

#old-network for physics server

Or go hire a physics tutor

[narrow vault]

im definetly not hiring anyone

hiring a tutor is for people who are too lazy to study

.close

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[deep dagger]

I'll probably need more help soon, but this is my biggest source of confusion. The whole page has me nervous, and I just cannot work out practice 2. I've tried -40 + 360 and -40 - 360, but it's never between the given numbers T~T

[light shoal]

well subtracting 360 isn't going to help, because that'll make it more negative

you added 360 once and didn't get there

how about doing it a second time?

[deep dagger]

I know, I just didn't know what to do D:

ooo

I forgot you could do that, omg xD

you're amazing

680 would work, right?

[light shoal]

yep that's correct

[deep dagger]

yesss!!

and do you know about practice 3?

I'm not sure what I'm looking at, tbh

[light shoal]

it's basically the same idea

take the first number on the left, 30 degrees

match it with a number on the right that is coterminal with it

meaning, a number that differs from 30 by a multiple of 360

[deep dagger]

like just the -360/+360 stuff again or would stuff like 12 be involved since it's a multiple of 360

if that makes sense

[light shoal]

nope, just -360/+360 stuff

[deep dagger]

oki cool cool

[light shoal]

"coterminal" means they agree to within a multiple of 360

[deep dagger]

ooo wait this is a lot simpler than I thought xD

with your help, of course

[light shoal]

yea, it shouldn't be too bad, and it gets easier as you go along because there are fewer options to check

[deep dagger]

yeh :D

I can definitely do this on my own time, are you cool to help me with 4 b? I can do a, I hope xD

[light shoal]

sure, do you know in general how to convert from radians to degrees?

[deep dagger]

I took notes today, lemme grab that rq

probs should've done that beforehand, sorry abt that

[light shoal]

nw

[deep dagger]

r = 180/pi ?

ah wait hm

uh T~T

managed to lose my original notes, sob

[light shoal]

180/pi is involved for sure!

[deep dagger]

well that's good :D

[light shoal]

it's:
angle in degrees = (180/pi) x angle in radians

[deep dagger]

oooo right

I'll rewrite that down xD

this is probably very wrong, but would that result in (1260pi)/(9pi)?

[light shoal]

yes

[deep dagger]

:D

[light shoal]

but that can be simplified quite a bit

[deep dagger]

ofc ofc

[light shoal]

oh, i see they want decimal digits anyway

so you'll get to break out the calculator or software

[deep dagger]

:D

1 moment then

[light shoal]

(actually it turns out you don't need a calculator, the numbers are nice)

[deep dagger]

(I'm not that smart, sob)

[light shoal]

since 9 divides evenly into 180

[deep dagger]

o

why is 180 still in play?

[light shoal]

well if you left it as 180 x 7 in the numerator instead of multiplying it out

then the 9 in the denominator

divides evenly into the 180

[deep dagger]

ahhh

that makes sense

[light shoal]

but if you didn't notice that, no big deal, just plug it into your calculator

[deep dagger]

I'll just plug it in so I don't hafta deal with pi as much xD

[light shoal]

actually the pi's cancel

in the num and denom

[deep dagger]

o h

[light shoal]

so you can immediately simplify it to just 1260/9

[deep dagger]

I thought abt that, but assumed it wouldn't work here cause circles Ig

okie that is wayyy easier :D

[light shoal]

it's just a ratio of two numbers, you can cancel any common factors the way you usually do

even if the factors are pi

[deep dagger]

in my defense, I am working on very little sleep :')

oo okie

that makes sense

should I be worried that I don't have any decimal places-

[light shoal]

nope, because the numbers divide evenly

1260 is an integer multiple of 9

[deep dagger]

oh true

normally if they say that, it means there's gonna be decimals, but maybe it was a mistake or just a deviation from how it normally is

in the manuals I mean

okie that makes so much sense :o

[light shoal]

yea in general there will be fractional digits

you just got lucky in this one

[deep dagger]

gross

[light shoal]

or technically depending on what they expect, maybe you should write .000000 after your integer answer

[deep dagger]

oh true-

I doubt she'll care, but better safe than sorry

[light shoal]

yea

[deep dagger]

... ok now I'm scared of a >.>

Idrk what they mean by "in terms of pi"

[light shoal]

oh, they just mean "don't plug it into your calculator, leave it as a fraction involving pi"

[deep dagger]

ooo snazzy ok

ok that was what I was super stuck on (for now >.>), but do you mind checking some of my answers for earlier review problems?

there's like 5

[light shoal]

ah maybe open a new channel so you can find another helper, i'm gonna take off and play video games haha

[deep dagger]

ooo very fun :D

best choice tbh xD

[light shoal]

yea

[deep dagger]

have a good day!

[light shoal]

you too, enjoy

[deep dagger]

.close

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[broken sphinx]

is it true that $a^n \mod b = a\mod b$

[flat frigate]

kungulus

[plucky elk]

Try finding a counter example

[broken sphinx]

got it

.close

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[ionic spire]

is this answer correct?

[raven heart]

@ionic spire yep

[ionic spire]

thank you very much

.close

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[gilded marsh]

.reopen

Hello, im in quite need of these 3 problems, theyre all about parallelograms.

[main mural]

one by one

[gilded marsh]

Any help would be appreciated

[ruby bolt]

This is easy tbh

[main mural]

crazy right

[gilded marsh]

Im overthinking too much, so i dont have any progress at all

[ruby bolt]

Damn yeah

[main mural]

which one do you want to start with

[gilded marsh]

Start with b and c actually

Im good with d

[main mural]

in B b)

IJK is 97°

so IHK is 97° too

right?

[gilded marsh]

Well yea opposite sides are equal

[ruby bolt]

angles*

[gilded marsh]

Thats honestly the one i was wondering

[main mural]

yes

[gilded marsh]

x+y=180

[main mural]

huh?

[gilded marsh]

Wait let me give spme references

[main mural]

yes that's what i was gonna talk about regarding JIH

[main mural]

yeah so

IJK and JIH are consecutive

hence supplementary

[gilded marsh]

Oh thanks

[main mural]

so 97° + JIH = 180°

[gilded marsh]

What about C. Thou? Do you have answers to that?

[main mural]

there's a lot in C

i won't be solving everything

[gilded marsh]

Its ok

[ruby bolt]

Which question is c?

[gilded marsh]

Wait ill send the problem again

Wait actually on second thought i dont need help at c anymore?

Are you guys solving thou?

Cuz i can close this now

.close

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[quartz wasp]

triangles sides will apply to each other by 1 . 2. 3 the smallest side is 3cm whats the biggest side

[potent seal]

Apply to each other? Wdym?

[hardy lion]

If youre talking ratios, a 1:2:3 side ratio is not a triangle

[eternal carbon]

i'm assuming that's referring to .. ^

[lean otter]

Not a healthy triangle

[ruby bolt]

degenerate triangle

[lean otter]

Hey, that's mean! His mother said he was just a special boy :(

[ruby bolt]

@quartz wasp

[quartz wasp]

no the problem says its a triangle

[ruby bolt]

huh?

[lean otter]

Okay, what does "apply to each other" mean?

[ruby bolt]

@quartz wasp

[quartz wasp]

ratio

[potent seal]

Oh

[lean otter]

Right, then it should be pretty simple

How much bigger is the biggest side than the smallest side?

[quartz wasp]

3 times?

[lean otter]

Yes

[ruby bolt]

It'll be a degenerate triangle

[lean otter]

So, if the smallest side is 3cm, how long is the longest side?

[quartz wasp]

thats wrong tho you mean the biggest triangle is 9 right there is no answer 9

so therefore theres some other way to do it

[lean otter]

Please send us a screenshot or picture of the problem exactly as it is written

Including all context

[quartz wasp]

its in anothr language

[lean otter]

What language?

[quartz wasp]

the ratio is 1 2 3 and the smallest side is 3 need to find the biggest side

[lean otter]

As your question is stated, the answer here is 9

If that's not the right answer then something is wrong/missing

[past birch]

let the sides be k,2k,3k respectively, if smallest is 3 then k=3 and the longest is 3(3) = 9

[lean otter]

So we need to see the problem exactly as it is written with context

@quartz wasp Could you take a picture/screenshot?

[quartz wasp]

its in another language

[lean otter]

Yes I understood that

As I've said though, with how you stated the question, the answer is 9

So if that's not the right answer then we need more context one way or another

[quartz wasp]

There is no more context and th answer 9 is not on the answers

Omg

Im so stupid

It ment the angles ratio is 1. 2. 3.

Not the sides

[lean otter]

So there was more context

Or rather the current context was incorrect

[lean otter]

As stated here

[quartz wasp]

It would be a right triangle right

[lean otter]

@quartz wasp Anyways, unless you have any more questions you can now .close this channel

[quartz wasp]

30 60 90

[lean otter]

Yeah that seems correct

[quartz wasp]

Yea thanks

[lean otter]

@quartz wasp Don't forget the .close

It's a command

[safe radish]

@quartz wasp Has your question been resolved?

[safe radish]

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[dire escarp]

dont know if this is relevant but this is on propositional form and i missed the class so im confused on what the a d c rows mean, can someone explain what they are to me?

[safe radish]

@dire escarp Has your question been resolved?

[dire escarp]

<@&286206848099549185>

[lean otter]

Looks like A is just what is written on top

It checks whether a implies b and b are both true

[dire escarp]

what about d and c?

[dire escarp]

wait before that, if this is the case, then why isnt row 2 true since there are both false

[lean otter]

Probably not possible to figure out from the context you gave

But I don't know much about whatever you're studying

I'm only answering because no one else is

[dire escarp]

i just started discrete math for context so i guess this is like the most basic topic cause its topic 1

[lean otter]

Because... They need to be both true?

[lean otter]

I meant the context above the question

[lean otter]

Look at how A is defined above

(a => b) n b is what I'm reading

Might be a weirdly written upside down U though

[dire escarp]

isnt it both true or both false = true

[lean otter]

No?

Plus that doesn't change the fact that both false does not mean both true

I see they need to be both true

If they're both false then they're not both true

[dire escarp]

i think i mixed that up with biconditional

[lean otter]

And the symbol they used, whatever it actually is, means "and"

[dire escarp]

like v right

[lean otter]

No, that's "or"

[dire escarp]

upside down is and?

[lean otter]

Yep

[dire escarp]

could the d and c mean dnf/cnf?

[lean otter]

¯_(ツ)_/¯

gotta go

[dire escarp]

i see

thanks for the clarification

.close

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[hybrid plover]

Suppose $p$ is a prime and $3\mid p-1$. Prove that $$p\mid (p-1)! \sum_{k=1}^{\frac{p-1}3} \frac{(-1)^{k-1}}{(3k-1)(3k-2)}.$$

[flat frigate]

moriaritie

[hybrid plover]

can i get a hint for this? plz

nvm, the question seems incorrect lol. ill ask for help later when i get the right thing

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[hybrid plover]

.reopen

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✅

[hybrid plover]

Suppose $p$ is a prime and $3\mid p-1$. Prove that $$p\mid (p-1)! \sum_{k=1}^{\frac{p-1}3} \frac{(-1)^{k-1}(2k-1)}{(3k-2)(3k-1)}.$$

[flat frigate]

moriaritie

[hybrid plover]

alright, this seems to be right...

how do i approach this problem? partial fractions will give 2(p-1)/3 fractions only, so we can't apply wolstenholmes.

[safe radish]

@hybrid plover Has your question been resolved?

[hybrid plover]

ok ill figure it myself

[safe radish]

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[buoyant pond]

.close

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[keen pecan]

so guys i came across a problem which state that asuming |a| <= |b| then : min(|a - b|, |a + b|) <= |a|, |b| <= max(|a - b|, |a + b|) iff : |b| <= 2 * |a|
i know to proff this i need to proof :p -> q and q -> p but the thing is the idea that |a - b|, |a + b| can be swaped together confuse me alittle what is the right way to approach this problem?
btw the domain of a, b are Integers set

[covert yoke]

@keen pecan abs everywhere.

So as a hint, consider first the values |a + b| and |a - b| alongside the values |a + c| and |a - c| where c = -b

What can we say about these two pairs of values?

What about |d + b| and |d - b| where d = -a?

[keen pecan]

for (|a + b|, |a - b|), (|a + c|, |a - c|) they both the same pair?

no wait sorry

the first one is (max, min), second pair is (min, max) or the oposite

right?

[covert yoke]

Yup

So, the smaller of the two values is going to be whatever the positive differences in the magnitudes are

And the larger of the two values is the sum of the two magnitudes.

[keen pecan]

im sorry i think i got lost here do u mean that the max will be |a| + |b| and min |a - b| ?

[covert yoke]

The max will be |a| + |b|

And the min will be ||a| - |b||

[keen pecan]

i think i got it ok so next we will repharse the problem as this?
||a| - |b|| <= |x|, |y| <= |a| + |b| <-> |b| <= 2 * |a|

x, y _> a, b sorry a typo

[covert yoke]

Yup, because if |b| were larger than 2|a| then ||a|-|b|| would be larger than |a|

As an example consider a = 1 and b = 10, so our two values of max and min would be 11 and 9 respectively

[keen pecan]

oh got it its a proof by contradiction
thnx alot for the help🙏

[covert yoke]

No problem

[keen pecan]

.close

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[lavish wharf]

Hello

[lavish wharf]

I got majorly two questions to ask

The question is written on paper, and I wonder how to transform it into the remainder of the question required to prove, or did I approach the wrong method

Second question is that what should I do now if it seems to be 1 by ratio test

<@&286206848099549185>

[lavish wharf]

( It is 100% that we gonna use ratio test btw )

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[lavish wharf]

.reopen

[safe radish]

✅

[lavish wharf]

<@&286206848099549185>

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@lavish wharf Has your question been resolved?

[safe radish]

@lavish wharf Has your question been resolved?

[safe radish]

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[fallow aurora]

Was solving like this.
But then saw a solution online that took tan^(-1)(tan θ) as θ.
But i remember learning, cause tanθ domain is bigger than tan^(-1)(x) thus tan^(-1)(tan θ) cannot be equal to θ

[median vigil]

[\tan-1 = \theta + \pi k ] for some integer (k), which is covered by the (+C) in integration

[flat frigate]

cloud

[median vigil]

you could proceed like that and get an integral that's valid only in that particular range, then

[fallow aurora]

[fallow aurora]

Should I keep this in mind and use it or not—

[median vigil]

the only other possibility that i can see would be to integrate by parts from the start, which would involve a pretty hairy derivative but would be valid everywhere

[fallow aurora]

what's a hairy derivative

[median vigil]

just a bit tedious/complicated

[fallow aurora]

I thought it's some shortcut like Heavyside coverup method 😅

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[fallow aurora]

.reopen

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✅

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[fallow aurora]

.reopen

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✅

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[wooden garnet]

my workings are:

it suffices to prove $\forall (\epsilon:R),\epsilon > 0, \exists (N:N), \forall(n:N),n>N \to \mid a_f(n) - a\mid < \epsilon$ and suppose $h:\forall(\epsilon:R), \epsilon > 0,\exists(m:N),\forall(k:N),k>m, \mid a_k - a\mid < \epsilon$

first assume $\epsilon:R$ and assume $h:\epsilon > 0$ take N=m, where m is sufficiently large. to make {$a_n$} converge. it suffices to show
$\forall (n:N) \to n > N \to \mid a_{f(n)} - a\mid < \epsilon$
assume n:N,assume $h_1 : n>m$ it suffices to show $\mid a_{f(n)} - a\mid <\varepsilon$

split the choices of n into n being odd and n being even.

so take n to be odd, so take n to be 2k + 1, where k>m. $f(2k+1) = 2k+1$ since 2k+1 is odd. it suffices to show $\mid a_{2k+1} -a\mid <\varepsilon$. since m<n, and n = 2k+1, 2k+1>m so it converges for odd n

take n to be even, so n = 2k where k>m. $f(2k) = k$ since 2k is even. it suffices to show that $\mid a_k -a \mid <\varepsilon $ since k>m, the sequence converegs for even n by $h$

[wooden garnet]

this doesnt seem correct to me, but I'm not really sure why. Can anyone tell me if this is correct or if I'm miles off

[flat frigate]

lewis_f04

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[fluid wren]

Why am i wrong!

[fluid wren]

?*

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[fluid wren]

Oh my god its just a simple error

.close

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[timid escarp]

Hello

[timid escarp]

A few possible questions

first off im graphing the following (2x+4)/(2x-2) for example im wondering how I can use the horizontal asymptote/ vertical asymptote to graph

[drowsy karma]

analyse it

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[strange plaza]

Feeling a bit silly right now

how did the 1/2 turn into square root

[junior smelt]

Log and exponent rules

notably $k\ln(a) = \ln(a^k)$ and $a^{1/n} = \sqrt[n]{a}$

[flat frigate]

@junior smelt

[strange plaza]

ohhh yep

thank you, that just flew over my head

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[jovial notch]

how do you factor and graph a polynomial with a degree of 3 while using synthetic? for example f(x) = x^3 -x^2 -14x +24

[thin bridge]

first consider rational root theorem to identify candiates to use for your division

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[cobalt tapir]

where does pi/4 come from next to r^3/3?

[past birch]

are you doing r^2 dr dtheta

[kind tinsel]

that is from the outside integral, note that $\int_0^{\sqrt{2}} r^2 dr$ is independent of $\theta$, so the outside integral is like integrating a constant from 0 to $\pi/4$

[flat frigate]

smay

[cobalt tapir]

so the constant is 1?

[kind tinsel]

the constant is the inside integral, like whatever the result is from that (which will just be a single number at the end) you’re integrating it from 0 to pi/4

[cobalt tapir]

so i got r^3/r | sqrt2 to 0 from the first intergral

[kind tinsel]

yep

[cobalt tapir]

which is 2 sqrt2 )/3

and then multiply it with pi/4?

[kind tinsel]

okay so then you integrate this “function” (it’s a constant function with respect to theta) from 0 to pi/4

well the antiderivative would be 2sqrt(2)^3/3 * theta, and you would evaluate that at pi/4 and 0

which makes it look like you’re just multiplying by pi/4

[cobalt tapir]

im confused why the theta comes from

if you intergrate over dr

oh wait

isnt the order wrong?

nvm

i got confused

[kind tinsel]

well, okay we can go over what the general idea is, we evaluate the inside integral first

[cobalt tapir]

i got it now

[kind tinsel]

oh good!

[cobalt tapir]

ty !

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[desert pasture]

I feel some information is missing

[midnight dragon]

that's it man

[desert pasture]

btw, you aren't cheating, are you?

[midnight dragon]

i concluded that this question is solvable by quadratic eqqn

as in the value is prob found through quadratic

[midnight dragon]

??

[desert pasture]

is this from an exam your writing righnt now

[midnight dragon]

no LOL

it's my tutorial

either way i gotta use lockdown browser for online tests

[desert pasture]

you don't know what adot b is, neither axb, you can't find the answer afaik

[forest gust]

use the dot product?

[upper mulch]

Think of a formula you may use.

[midnight dragon]

yep that's what i did

[forest gust]

$b\cdot a=\left|a\right|\left|b\right|\cos\left(\frac{\pi}{4}\right)=4\lambda+3$

[midnight dragon]

maybe there was some calculation error

lemme retry

[flat frigate]

Combustion

[desert pasture]

yes but neither a cross b , nor a dot b is given, so how is lambda even determinable

[forest gust]

hold up let me try

[torpid compass]

= cos theta

[desert pasture]

ah yeah

[midnight dragon]

got an imaginary num

yikes

[forest gust]

yeah it can be solved

$b\cdot a=\left|a\right|\left|b\right|\cos\left(\frac{\pi}{4}\right)=4\lambda+3\to\frac{5\sqrt{\lambda^{2}+1}}{\sqrt{2}}=4\lambda+3$

[flat frigate]

Combustion

[midnight dragon]

ah wait i think i see what i did wrong

where did sqrt2 come from

[forest gust]

cos(pi/4)

[midnight dragon]

oh

must've been because i didn't convert cos pi/4 to fraction

that would make life easier

oh my god

i saw where i did wrong

this is embarrassing ima close this 🥲

.close

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