help-17

[burnt temple]

If K = ceil(epsilon) + 1 , then since N >= K , N > ceil(epsilon) >= epsilon

actually ik how to proceed with this

[thin vale]

Okay

[burnt temple]

But ima do it my own

[thin vale]

I am going to interrupt quickly here

[burnt temple]

Iw ont botheryou anym9re

[thin vale]

let me show my final work

[burnt temple]

oh

Yes

Do it

[thin vale]

First we state a few theorems we will use

if lim n->infinity |x_n-L| ->0 then x_n->L

if a_n -> a then sqrt(a_n) -> sqrt(a)

also, {1/k} -> 0

the epsilon n proof is easy for that

let k>1/epsilon

[burnt temple]

What

Wtf

[thin vale]

?

[burnt temple]

$\lim( |x_n - L|) = 0$ , then $x_n \to L$

[twin meteor]

Bishop

[burnt temple]

????????????

[thin vale]

yea

L in R

[burnt temple]

Oh

Interesting

I never thought of it that way

Neither did i ever see that theorem ever

[thin vale]

lim xn -> x if and only if lim |xn-x|=0

anyways

using those theorems

this is |x_k-L|

by the theorem about sqrt(a_n)->sqrt(a) and 1/k -> 0

we get that the numerator -> sqrt(2)

using just that 1/k->0

we get that the denominator -> 2

using the theorem about if a_n -> a and b_n -> b and b!=0 then a_n/b_n -> a/b

we get here that the left hand fraction (x_k) goes to sqrt(2)/2

so the limit of |x_k-L| is sqrt(2)/2-sqrt(2)/2 is 0

so we get that x_k -> L=sqrt(2)/2

that seems valid to me

[dull bear]

Works though of course if you do it the algebra of limits way you can cut down a bit

[thin vale]

awsum

[dull bear]

From there you'd be fine

[thin vale]

ok cool ty chartbit

sorry it took me so long

I was trying to choose a k for too long

finally gave up on that

[dull bear]

Mind you, if you did choose the epsilon-N way, then provided you could show this that way

You could manipulate the upper bound to be that and you're happy

[thin vale]

yes let k>1/eps

[thin vale]

hmmm

[dull bear]

"making denom smaller" and all

Either way algebra of limits is much nicer to do these with honestly, saves so much time

(it also isn't too bad to prove the algebra of limits either )

[thin vale]

mhm

[vocal sleet]

@thin vale Has your question been resolved?

[sly sierra]

you got your name back

[thin vale]

yes I got an i earlier

I am working on this now

I am fairly confident I have the (=> ) direction

but I get stuck working in the (<= ) direction

We let ${x_k}$ be a sequence of points of $S\setminus {\vec{a}}$ s.t ${x_k}\to \vec{a}$ and we want to show that this implies that $\vec{a}$ is an accumulation point of $S$

[twin meteor]

Austin

[thin vale]

${x_k}\to \vec{a}$ tells us for all $\varepsilon >0$ we can choose a $k\in \mathbb{N}$ s.t $$|x_k-\vec{a}|<\varepsilon$$

[twin meteor]

Austin

[thin vale]

which means that no matter how small we shrink the radius of our ball around a, (in S\{a}) there are still infinitely many x_k around a?

and x_k are in S so?

[sly sierra]

yes, the tail of the sequence is any epsilon ball

[thin vale]

I am having trouble putting this into words for Rn spaces Bungo

[sly sierra]

you have to be somewhat cautious though, as the elements of the sequence aren't necessarily distinct from each other (so if you knew nothing else, the sequence might only assume finitely many values, and then your argument doesn't work)

so you have to use that none of them are a, yet the sequence converges to a

[thin vale]

if none of them are a, then the sequence can't be a constant sequence b, because that would converge to b, but b not equal to a

so the sequence is not a constant sequence

so the elements are distinct

[sly sierra]

they don't have to all be distinct

just rule out the possibility that there are only finitely many values

[thin vale]

can I ask you a different question actually Bungo

[sly sierra]

sure

[thin vale]

I think it'd be better I sort out my confusions with the other part first

because I have the idea but I have trouble making arguments with balls

so

let $\vec{a}$ be an accumulation point of $S\subset \mathbb{R}^{n}$

[twin meteor]

Austin

[thin vale]

then $\forall \varepsilon >0$ the set of $\vec{x}$ within a ball of radius $\varepsilon$ of $\vec{a}$ interesected with $S$ disjoint $\vec{a}$ is nonempty

[twin meteor]

Austin

[sly sierra]

yep, not only nonempty but infinite

[thin vale]

${|\vec{x}-\vec{a}|<\varepsilon} \cap S\setminus {\vec{a}}\neq \emptyset$

[twin meteor]

Austin

[thin vale]

I can write this statement for all epsilon>0 correct?

and like you said it is also infinite

[sly sierra]

yes

[thin vale]

$${|\vec{x}-\vec{a}|<1} \cap S\setminus {\vec{a}}\neq \emptyset$$
$${|\vec{x}-\vec{a}|<\frac{1}{2}} \cap S\setminus {\vec{a}}\neq \emptyset$$
$${|\vec{x}-\vec{a}|<\frac{1}{3}} \cap S\setminus {\vec{a}}\neq \emptyset$$
Continuing in this fashion
$${|\vec{x}-\vec{a}|<\frac{1}{n}} \cap S\setminus {\vec{a}}\neq \emptyset$$

[sly sierra]

sure

[thin vale]

lost my brackets there

[twin meteor]

Austin

[thin vale]

so this tells us that there exists $\vec{x_1},\vec{x_2},\vec{x_3}\dots \vec{x_n}\in S\setminus {a}$

[twin meteor]

Austin

[thin vale]

that are within any arbitrarily small radius of $\vec{a}$

[twin meteor]

Austin

[sly sierra]

sure, you can pick x1 in the first set, x2 in the second set, x3 in the third, etc

[thin vale]

yeah

[sly sierra]

notice that you need the axiom of countable choice for this

[thin vale]

and this sequence of xn goes to a

yet none of the xn are a

[sly sierra]

correct

on both counts

[thin vale]

is that enough justification you would say?

because I feel like

[thin vale]

I lose my mind right here

[sly sierra]

yes, that construction is fine

[thin vale]

I can no longer formally state what I am trying to say

after that point

It frustrates me

[sly sierra]

you don't need to say "arbitrarily small radius", you can say:

the distance between x_n and a is less than 1/n

(and greater than 0)

so for each n you have the inequality

0 < |x_n - a| < 1/n

[untold surge]

The other direction is very easy. It’s just that suppose not, you have a r, such that the neighbor of a, a ball radius r, center at a, contains finitely many x_k: k from J, then those distance(x_j,a) >r for any j that isn’t in J, so for all great enough j, distance(x_j,a)>r. Then x_j impossible to converge to a, contradiction

[sly sierra]

then |x_n - a| is sandwiched between 0 and 1/n, both of which go to 0 as n goes to infinity

hence |x_n - a| must go to 0

and hence...

[thin vale]

x_n -> a

[sly sierra]

yep

[thin vale]

and all of the x_n are not equal to a

[sly sierra]

correct

[thin vale]

and they are a sequence in S

so perfect

[sly sierra]

indeed

[thin vale]

Okay ty bungo I appreciate that

Idk if you have ever related to that kind of feeling but gosh it is just like everything I just studied in R being taken to Rn and I have the ideas but I just completely suck at trying to say them

[sly sierra]

yea, things can get a bit confusing every time you generalize a bit

but you won't lose a ton of intuition if you consider that balls in R^n are just intervals in R

[thin vale]

with intervals this is so easy

[sly sierra]

the main thing that breaks down in R^n but works in R is that in R^n, you can't necessarily express every open set as a countable union of disjoint open balls

whereas in R you can

[sly sierra]

but that doesn't affect you here

[thin vale]

this is for in R... I had no trouble with it whatsoever

cluster point = accumulation point

[sly sierra]

yea, to translate this sort of argument from R to R^n it's helpful to translate things of the form
x - epsilon < x_n < x + epsilon
to
-epsilon < x_n - x < epsilon
to
|x_n - x| < epsilon
all of which are equivalent in R, but only the last one can be used in R^n

[thin vale]

so now I want to translate the (<= ) argument into Rn

which started by supposing the existence of a sequence xn in S disjoint a that converges to a

and from this we have to show a is an accumulation point

[sly sierra]

right

[thin vale]

since the definition of the sequence converging gives for all eps>0 |xn-a|<eps we can form a ball around a of radius epsilon which contains elements of the sequence xn?

for any radius we'd like

elements of xn are in S yet not equal to a

[sly sierra]

right

[thin vale]

hmmm

so

any ball around a

of any size

contains elements of S disjoint a

these elements being the xn in our sequence

but how do we know there are infinitely many?

[sly sierra]

suppose for contradiction that there are only finitely many

so the sequence xn assumes only finitely many values

[thin vale]

oh right

the thing about

[sly sierra]

that means there exists a subsequence which is constant

[thin vale]

yes

[sly sierra]

and not equal to a

[thin vale]

right

[sly sierra]

that subsequence converges to a

[thin vale]

which isn't possible

[sly sierra]

because all subsequences must converge to a

right

[thin vale]

Okay so let me restate

---

Because we have a sequence of elements of S disjoint a that converges to a, from the definition of sequence convergence we get that for all balls centered around a of radius epsilon,
|x_n-a|<epsilon
That is, any ball around a contains elements of our sequence, which are elements of S disjoint a.

Suppose that our sequence contains finitely many elements, then there must exist a constant subsequence that converges to a. But, for a constant sequence to converge to a value it must take on that value. But, this subsequence of our original sequence cannot take on a, and thus cannot converge to a. This contradicts the existence of a constant subsequence converging to a, which contradicts our sequence containing finitely many elements.

Is this proper?

[sly sierra]

"from the definition of sequence convergence we get that for all balls centered around a of radius epsilon,
|x_n-a|<epsilon"
This part holds for all n past some N, not necessarily for all n

(where the N depends on epsilon)

[thin vale]

yes

does that hurt a part of my argument by leaving that out

[sly sierra]

the rest looks ok

[sly sierra]

nope, just mentioning it for completeness

[thin vale]

I have one final piece of confusion about what I just said

any ball contains elements of our sequence yes

and our sequence doesn't have finitely many elements yes

but how do we know that a certain ball doesn't have finitely many elements from our sequence?

[sly sierra]

because all terms of the sequence must be in that ball, for n >= N

[thin vale]

aha!

so it does matter

[sly sierra]

i.e. the tail of the sequence must be in the ball

[thin vale]

hahaha

you're awesome Bungo

[sly sierra]

if the tail only takes on finitely many values then so does the entire sequence

[main ferry]

why are you guys so smart

[thin vale]

Bungo is the only smart one here

[sly sierra]

since the whole sequence is the first part (with finitely many n) followed by the tail

[main ferry]

idek what u guys are talking abt

[sly sierra]

not smart, just more experienced with this material

[main ferry]

btw lmk if u guys wanna help me solve a probability dice problem whenver u guys have time:D

[thin vale]

the compliment was self-serving

I see

hahaha

[main ferry]

HAHAHAH

[sly sierra]

is it this one? #help-3

[main ferry]

but i meant it

yes

[sly sierra]

i can take a look, if austin is all set now

[thin vale]

Let me have 30 more seconds to confirm that

[sly sierra]

sure

[thin vale]

yes

[main ferry]

oke tysm<3

[thin vale]

I am all set

[sly sierra]

cool

[thin vale]

thank you very much Bungo

[sly sierra]

yw

[thin vale]

you're great

[sly sierra]

cheers

[thin vale]

seriously

cya!

Since the closure of S is defined to be the S union Boundary of S

I need to show that boundary of S equals the set of accumulation points?

that is, the accumulation points are a subset of the boundary points,
and the boundary points are a subset of the accumulation points?

Is that true or am I missing something here

[sly sierra]

no, an accumulation point need not be on the boundary, and a point can be in the boundary without being an accumulation point

example 1: in R, let S = (0,1), then 1/2 is an accumulation point not on the boundary

example 2: in R, let S = {0}, then 0 is on the boundary but not an accumulation point

what you need to show is that if you have a point in the boundary of S that is not in S itself, then that point is an accumulation point... that gives you the containment closure(S) ⊆ (S U accumulation points of S)

then you need to show the reverse containment too

[vocal sleet]

@thin vale Has your question been resolved?

[thin vale]

Okay TY Bungo I will work on that

[vocal sleet]

@thin vale Has your question been resolved?

[vocal sleet]

@thin vale Has your question been resolved?

[vocal sleet]

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[civic geyser]

Im having a problem on my exam review. You’re supposed to solve a multi step question, which i have, but then I have to check it and it always stumps me. The first photo is an example of what the checking looks like (done by my friend) and the other picture is my work and what it looks like. I would like someone to guide me, i would appreciate it very much
how do i check my answer?

[karmic imp]

Just plug in the value you got into the equation

So you got x = -1, plug that into the equation

[civic geyser]

like that?

[karmic imp]

Yes, simplify that

See if the left side equals the right side

[civic geyser]

okay got it hold on

[karmic imp]

Also it's -7x

You wrote 7

[civic geyser]

oh oops thank you

i think i did something wrong

[karmic imp]

What's -6 * -1

[civic geyser]

it’s just 6

oh wait

thank you so much im like rlly slow with math i need like a guide when i do it sometimes but thank you

[karmic imp]

That's all you do to check your answer, you just plug it back in

[civic geyser]

alright!

.close

[vocal sleet]

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• After 15 minutes, feel free to ping <@&286206848099549185>.
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[civic geyser]

another question. gave this distance question a shot and i’m still so confused. i need some help

[thin vale]

Draw a right triangle connecting the points

Where the hypotenuse of the triangle is the straight line shortest distance between the two points

And then use Pythagorean theorem to solve for the length

[civic geyser]

well it’s distance like the distance formula i was taught this way to do it

but that was with my teacher

[thin vale]

The distance formula comes from the method I described

[civic geyser]

oh

[thin vale]

They are the same thing

You will see

Try it out

[toxic kettle]

It’s asking me to identify the function family to which f belongs but how am I supposed to identify that?

[thin vale]

This channel is occupied

!help

[vocal sleet]

Please read #❓how-to-get-help

[thin vale]

Open an available help channel. This one is taken

[civic geyser]

okay so would it look like this? sorry it’s hard to read

[thin vale]

Yes like that

[civic geyser]

okay okay i’m gonna use the pythagorean theorem formula real quick

so the distance is 5?

[thin vale]

Yes

[civic geyser]

thank you sm!

[thin vale]

Do you see how that is the same as the distance formula

[civic geyser]

yeah i think so

[thin vale]

Great

[civic geyser]

thank you again, have a good day or night

.close

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[vast shale]

How do I use desmos to graph vector functions

[vast shale]

like how would i graph this situation?

[thin vale]

desmos can graph the parametrization of the vector valued function

so if you are able to find the position vector

it can graph this

but the accerlation vector is constant, there is really no function to graph in that case

[vocal sleet]

@vast shale Has your question been resolved?

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[fervent reef]

can someone help with the last two questions of part b? part a answer is 1/2

[ebon rapids]

Write it out on a paper

It’s nothing more than taking your answer from a and solving for cos(x) step-by-step

[fervent reef]

i’m setting 1-cosx equal to 1/2?

[ebon rapids]

Not quite

What’s your equation from part a?

[fervent reef]

sin^2x/x^2 multiplied by 1/1+cosx

[ebon rapids]

No no, skip to the end

[twin meteor]

CST

[fervent reef]

yea

[ebon rapids]

Part b is essentially saying “take this and solve for cos(x)”

It’s a step-by-step

[fervent reef]

so 1 - cosx = 1/2?

[ebon rapids]

Where did the x^2 go?

[fervent reef]

in part b the question is 1 - cosx

[ebon rapids]

What’s your answer to box 1 of part b?

[fervent reef]

1/2

[ebon rapids]

Ok

Now from box 1 to box 2, how did the left-hand side change?

[fervent reef]

there’s no x^2

[ebon rapids]

Yes

So how did the x^2 “disappear”?

[fervent reef]

multiply by x^2??

[ebon rapids]

Yep

So what will you do to the RHS to maintain equality?

[fervent reef]

multiply by x^2

[ebon rapids]

Yep

So what’s in box 2 now?

[fervent reef]

1/2x^2

[ebon rapids]

please use parentheses so that it’s clear

[fervent reef]

(1/2)(x^2)

[ebon rapids]

Good

Now onto box 3

[twin meteor]

CST

[ebon rapids]

Isolate cos(x) now

[fervent reef]

1 - (1/2)x^2

[ebon rapids]

Yep

So you see how we got to this?

[fervent reef]

no

[ebon rapids]

Where are you confused, the transition from box 1 to 2 or box 2 to 3?

[fervent reef]

2 to 3

[ebon rapids]

Ok, let’s walk through this

[ebon rapids]

So from here

Do you have any idea how we can proceed?

[fervent reef]

wait i get it ty

[ebon rapids]

Nice, no problem!

[fervent reef]

.close

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[craggy acorn]

hey is anyone good at fraction

[craggy acorn]

i need help i forgot it

[gritty zenith]

Send the q

[civic otter]

"good at fraction" means nothing by the way

[vocal sleet]

@craggy acorn Has your question been resolved?

[craggy acorn]

The first 1 is rewrite fractions into mixed numbers the 2 is rewrite to a spurious one and the 3 is reduce as much as possible, submit as a real fraction or mixed number

[craggy acorn]

im bad at math

<@&286206848099549185>

[visual drift]

what question

[craggy acorn]

all

nah im done with number 1 and number 2

i just need 3

[visual drift]

your elementary school?

[tired nebula]

Lmao

[craggy acorn]

?

im in 8 grade

ye

[visual drift]

bro thats kinda easy

[craggy acorn]

ik

i just forgot

[tired nebula]

Kinda?

[visual drift]

yes, why?

[tired nebula]

It's too ez

[craggy acorn]

...

[visual drift]

BRO GOT DAMMIT, i will teach you

[craggy acorn]

Thx

[visual drift]

opgave 3 right?

[craggy acorn]

yes

or

wait

nvm

[visual drift]

dont make me to a asian mom

[craggy acorn]

i need to finnish it now and send it to my teacher

jsut 1 question then

number 2 and

this one

imma just not do opgave 3

[visual drift]

ok

[civic otter]

How old are you?

Because Discord is for people with age ≥ 13

[craggy acorn]

14

[burnt temple]

💀 🐒

[craggy acorn]

shut up

[civic otter]

Ok no prob then 👍

[burnt temple]

Sorry 🥺

Plz no hit me daddy

[visual drift]

be my boyfriend

jk

[burnt temple]

Yes

Noooo

[visual drift]

hell nah

[burnt temple]

😭

[visual drift]

Pretend to be a couple then

[burnt temple]

There's no need to pretend

When it's the real deal 😏

[visual drift]

maybe i wanna die, i will chat you soon, cuz i wanna turn off this laptop and wanna go to the summer

i already add you

[burnt temple]

😳 🍴

👍

[paper depot]

@visual drift @burnt temple this is not the place.

@craggy acorn is there anythin you still need help with?

[vocal sleet]

@craggy acorn Has your question been resolved?

[vocal sleet]

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[boreal path]

what does complex exponentiation actually mean

[boreal path]

ho do you multiply a number by itself i times

[hard atlas]

you dont

you take a function exp(x) which for real numbers corresponds to repeated multiplication, and then extend that function to complex numbers

[vast shale]

I always thought it's more intuitive to think of multiplication as operations on a vector rather than the repeated addition idea schools give you

The idea itself breaks apart for negative numbers

[hard atlas]

even something like x^0.5 challenges the idea of repeated multiplication. you can kind of imagine multiplying a number half times but it doesnt really work

[boreal path]

but weve defined those

half times means srud

[vast shale]

That's not the point

[boreal path]

then hat is

[hard atlas]

do you really think of x^0.5 as repeated multiplication or do you think of it as something else

whatever that something else might be

[boreal path]

well

it is repreated

but you cant repeat half

so we multiply half the amount of times we normally do

[round plover]

it's kinda the same thing we've done to multiplication to extend it from natural numbers to real numbers

like, if you view 4 * 3 as adding 3 copies of 4

something like 3.2 * 5 doesn't make sense, so we extend the definition in a way that is back-compatible with our previous notion

and note that it really is a definition at that point

the taylor series of e^x is a theorem

but the extension of e^x for complex x, or other inputs, by means of plugging them into this series can be taken to be a definition

[boreal path]

so its just undefined?

[round plover]

the question "how do you multiply a number by itself i times" is undefined

because that's not what we mean by exponentiating by a complex or imaginary number

[boreal path]

so hwo do we define

2^i

[round plover]

we can extend e^x to complex x

so we just combine it with (complex) logarithms

and we can transform the e base to any number, like 2

or more simply, write 2 in polar form and combine it with i

[hard atlas]

another perspective: if we step really far back, multiplication is repeated addition. so if x^0.5 is multiplying by itself half times, then it is also in some sense repeated addition. but by this point you really have to see that this is not any useful way to view what is going on

somewhere along the line we stop viewing multiplication as repeated addition and see it as its own thing

and this is also now happening with exponentiation

[vocal sleet]

@boreal path Has your question been resolved?

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[twin meteor]

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

[quick estuary]

Binomial

Approximation

[half rover]

I can help

Give me 1 min

[cursive turret]

write an x insteadof 38 and solve it for x

[half rover]

It does not say whether interest is compund or simple

So I guess, too - can you be sure?

[cursive turret]

!show

[vocal sleet]

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

[half rover]

I think bot give a good idea. Show? Meanwhile, I shall solve it myself and then explain

ok

[half rover]

Where is your formular from? The way I read your problem - it's just like this - GCSE level

[cursive turret]

that is wrong, you missed the 800 are paid every xear

[half rover]

If you are 15 years old - you find n by trial and error, if you are 17-18 year old level - use log

You are right

Wait

[cursive turret]

i guess the -800 at the start are wrong. but anyway. from the last line multiply by 1,025-1 then add 1 then use log

if youe aplly log at the right side - what do you get?

[half rover]

@cursive turret - I'll leave it to you - I overlooked... yes You are on track

Good luck !

[cursive turret]

lets discuss this leter, now for log. ok apply log to the left side?

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

[twin meteor]

ThM

[cursive turret]

x times log (1.025) = log (16.556)

divide by log 1.o25 and you have x

i guess you made the "multiply by 1.025-1" part wrong, this gave me some smaller then 16

yes

something like that

and the -800 is right, as you need the value at the end of the year

lets start with the last year. 800 at the start of the year you need the value at the end -> 800 times 1.025

for the year before: 800 times 1.025 ^2 and so on.

so yo have the sum of 800 tines 1.025 ^i with i from 1 to n.

no. wait a moment. you have now $\sum_{i=1}^{n} (800 \cdot 1.025^i)$

[twin meteor]

ThM

[cursive turret]

this can be written as $800 \cdot \sum_{i=1}^n 1.025^i$

[twin meteor]

ThM

[cursive turret]

and now look at the sum this is the geometric sum, where a formula exists.

but this formula starts with i = 0

$\sum_{i=1}^n 1.025^i=(\sum_{i=0}^n 1.025^i)-1$

[twin meteor]

ThM

[cursive turret]

just wait.

$800 \cdot \sum_{i=1}^n 1.025^i=800 \cdot ((\sum_{i=0}^n 1.025^i)-1) = 800 \cdot(\frac {1.025^{n+1}-1}{1.025-1}-1)=800 \cdot(\frac {1.025^{n+1}-1}{1.025-1})-800$

[twin meteor]

ThM

[cursive turret]

where your solution 38 means n+1

$800 \cdot \sum_{i=1}^n 1.025^i=800 \cdot ((\sum_{i=0}^n 1.025^i)-1)$ is this clear?

[twin meteor]

ThM

[cursive turret]

$800 \cdot ((\sum_{i=0}^n 1.025^i)-1) = 800 \cdot(\frac {1.025^{n+1}-1}{1.025-1}-1)$ is this clear?

[twin meteor]

ThM

[cursive turret]

i did not change the -1 it is on the left side in the bracket, it is on the right side in the bracket. i only replaced the sum by the geometric formula. look at https://mathworld.wolfram.com/GeometricSeries.html for the S_N-Formula, if the sum goes from 0 to n the exponent in the fraction is n+1

because i didnt evalute the brackets at this moment, this is the last step.
$800 \cdot(\frac {1.025^{n+1}-1}{1.025-1}-1)=800 \cdot(\frac {1.025^{n+1}-1}{1.025-1})-800$

[twin meteor]

ThM

[cursive turret]

we started with this. which gave as a sum from 1 to n. where n is the number of year.s. then we added the i = 0 term and subtracted him again (the minus 1 part), we got a formula with n+1 which we solved and got 38. so n+1 = 38 -> so the answer is 37 years.

just drop the -800 is a wrong solution. there is a way without the -800 part, but then the term to solve is a little bit different. a different way, which gave you the same (!) solution, not only in a rounded way.

if you are interested we can do this other way

lets start with the same sum: $800 \cdot \sum_{i=1}^n 1.025^i$

[twin meteor]

ThM

[cursive turret]

we write this now as:

$800 \cdot 1.025 \cdot \sum_{i=1}^n 1.025^{i-1}$ ok?

[twin meteor]

ThM

[cursive turret]

now replace the sum:

$800 \cdot 1.025 \cdot \frac{1.025^n-1}{1.025-1}$

[twin meteor]

ThM

[cursive turret]

you said ok to this.

ok

lets look on this

$\sum_{i=1}^n 1.025^{i}=\sum_{i=1}^n 1.025 \cdot 1.025^{i-1}$

[twin meteor]

ThM

[cursive turret]

and now take the 1.025 before the sum

we want a sum starting from 0 to use the formula for the geometric series.

$800 \cdot(\frac {1.025^{n+1}-1}{1.025-1})-800 = 48980$

[twin meteor]

ThM

[cursive turret]

that is the starting point.

add 800

we did this step by step, and this was the last step: #help-17 message

if it is needed we can start from the beginning again.

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[forest crater]

hey guys I think the solution I am given is wrong but I need to reassure myself so I'd really appreciate someone helping me out by confirming that this is wrong or otherwise tell me that I'm dumb

so I have the integer set 1 through 8 and B:={at most 7} and C :={all odd numbers}
Now the book says that BuC would be {1,3,5,7} but I believe it should definitely be [7] so B, right ?

[blissful sentinel]

Yeah it should be {1,2,...,7}

It looks like they calculated intersection instead of union

which would be (B \cap C)

[twin meteor]

fréchet_filter

[vocal sleet]

@forest crater Has your question been resolved?

[elder totem]

What is your neuromanifold.

[elder totem]

What is your neuromanifold.

Give me your neuromanifold

[blissful sentinel]

stop trolling

[forest crater]

alright, thank you very much!!

.close

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[long flame]

i keep getting stuck trying to find limits. Am I supposed to just plug in 1.000001 as values and take that answer? Or am I factoring wrongly?

[void latch]

x^2 + x - 2 does not equal (x+2)(x-1)

[vast shale]

Why?

[void latch]

shit i was thinking of someone else's problem

thats right

[vast shale]

Aye aye 😂😂

[void latch]

What does a factor in the denominator imply about rational functions

If it doesn't cancel out

[long flame]

not sure?

[void latch]

Think about the function 1/x

[shut heart]

Well it Doesn't not exist

[long flame]

so theres an asymptote at 0

[void latch]

yeah

[shut heart]

Yeah

[long flame]

how does it help me

[shut heart]

U can draw a rough graph with it

[void latch]

now think about where the vertical asymptote is if the factors in the denominator are (x-1)(x+2)

[long flame]

-2 and 1

[void latch]

Yep

and If there's a vertical asymptote, then that means that the limit...?

[long flame]

approaches 1?

[void latch]

Does not exist

[long flame]

huh

[void latch]

Think about 1/x again

[long flame]

so you're saying the limit doesn't exist

[void latch]

Yep

[long flame]

when it approaches from up to down

[void latch]

The left and right sided limits give different answers

[long flame]

this is right sided

[void latch]

it doesnt say so in the question

[long flame]

yes top left

at least my course is using this notation

[void latch]

Is that downwards arrow?

huh

interesting

[long flame]

i was used to left and right arrows too but

getting used to this now for my exam

[void latch]

So the limit has to be infinity then

[long flame]

brother are u sure about this

[void latch]

Check it by plugging in values close to but greater than 1

[long flame]

so 1.000001 etc which i stated initially

[shut heart]

Ooo this is a new notation to me

[void latch]

yeah or 1.1 or 1.01

[void latch]

I have never seen that either

[shut heart]

Yeah we use 1^+ for RHL and 1^- for LHL

x -> 1 like this

[void latch]

same

[shut heart]

So it's quite interesting

[long flame]

okay so the answer was infinity

[shut heart]

What does the down arrow means

[long flame]

right limit

[shut heart]

Oo then it is +∞

[long flame]

how can u see that

i can't understand why

when i plug in values close to but greater than 1 i get -5.5

for my factored equation at least

[shut heart]

Is u put 1 in (x-7)(x-5) u get positive ryt

The x+2 for 1 is +ve

So just x-1 is left

Now if u approach it from right side which means a little greater than 1

Therefore x-1 for x= something a little greater than 1 should be positive

[long flame]

so you think of it as positive values divided by a veeeery small but positive value. So naturally it must be a v high answer

[shut heart]

Yeah

[long flame]

because (x+2)(x-1) must be v small

[shut heart]

Yeah

[long flame]

and for negative infinity, it would be opposite, so it would be a veeeery small but negative number

division by that

[shut heart]

Yes

[long flame]

what about the nominator

what if the nominator is zero

[shut heart]

U put 1

[long flame]

i guess it will never be zero

hmm

[shut heart]

When if it is 0/0 u gotta do work

If it is 0/something finite then it should be 0

It doesn't matter if it is +ve or -ve since it is very close to 0

[long flame]

for these types of problems, should we always test initially with a lim value or the actual value

plug in 1 or 1.00001 so to speak

[shut heart]

Just do 1

[hollow orbit]

YES

[shut heart]

Always be on look out for indeterminate forms

[hollow orbit]

you should work out the left and right hand limits

[shut heart]

The question is RHL

Tho

[hollow orbit]

IF IT IS NOT A FINITE #

[shut heart]

If it is not mentioned then check both sides

[hollow orbit]

Im seeing a 1 not 1^+

[long flame]

ok so i plug in the value. Now I can get many different types of results. What should I look out for? 0/0 means more work

[hollow orbit]

oh so its mentioned?

[round plover]

the downarrow is notation for right hand limit

[shut heart]

Yeah I was confused too the down arrow supposed to represent 1^+

[hollow orbit]

oh new concept for me

[shut heart]

Me too

[round plover]

yeah, ^+ and ^- is definitely more usual

[long flame]

when i plug in values, 0/0 case means more work. 0/finite means 0, what other cases are there

[hollow orbit]

in this its gon be +INFINITY

[shut heart]

Yeah 0/0 ,∞/∞,0⁰ , etc

1^∞

[hollow orbit]

0*infinity

[shut heart]

∞⁰

[hollow orbit]

1^infinity

[shut heart]

Oops my bad

[long flame]

? i don't understand these

okay infinity/infinity means also more work?

[shut heart]

Which one

[shut heart]

Nice

[hollow orbit]

yeah

these means indeterminate forms which are a non defined result

[long flame]

oh okay so i was only working with fraction limits so far

[hollow orbit]

means more work

[long flame]

so i see these are other forms

[hollow orbit]

might be simplifications

[hollow orbit]

you might not have taken them yet but you will soon regarding limits

[shut heart]

Why?

[long flame]

wow, this feels like a lot to remember

[hollow orbit]

not really

[shut heart]

Nah bro just practice

[long flame]

and these indeterminate forms

we can determine already from its initial form? before any simplifying?

[hollow orbit]

cannot

[shut heart]

U need to do more work if u see these

[hollow orbit]

you mean determine if its indeterminate then yes

[long flame]

yes

i mean to determine if its indeterminate

so we dont need to first simplify to see what type of limit it is

[shut heart]

No just plug the value in the function

[hollow orbit]

yeah you should first in limits plug the value to see if its a number then we dont need any simplification

[hollow orbit]

indeterminate forms need

[long flame]

okay good

so first plug in values. Determine form. Then try to simplify. After simplifying, i understood for fractions, if the denominator is infinitely positively small it is positive infinite limit, if infinitely negatively small it is negative infinite limit

[hollow orbit]

yes but

watch the numerator too

[shut heart]

The small factor might even cancel out

[hollow orbit]

its sign might be negative or there is cancelling out process

[long flame]

if the small factor cancels out, then it is ... 1 ?

or negative 1?

or i guess it depends ?

[round plover]

it depends

[long flame]

i'm still confused as to know when a simplification is really "finished" (if i did it right in the first place), and then what to do, and then how to analyze the results of what i did

for the "what to do" step, is it just simply to plug in values again?

[hollow orbit]

plug -> the result is a finite number? you are done, the result is an indeterminate form? do some work.

[shut heart]

Maybe explaining with an example question might help

[hollow orbit]

from marty?

[long flame]

i can get another one

[shut heart]

Yeah

[long flame]

lets go through this one

0/12

[shut heart]

This ones not indeterminate...

[hollow orbit]

0 i believe

for just plugging 5

[long flame]

so since its 0/12 (finite), its 0

lemme get one thats indeterminate

[hollow orbit]

yes

[shut heart]

I have one u can try

A easy one

Should I give it?

[hollow orbit]

let em try it

[shut heart]

1st one

U can check by plugging the value

[long flame]

okay great hold on

i tried getting an indeterminate one for 2 min all were determinate -_-

lemme look at yours

[shut heart]

-2^+ is just the downward arrow to -2

[long flame]

3/0, meaning indeterminate ?

[shut heart]

No

Shit

[long flame]

more work

i mean

[shut heart]

My bad

[hollow orbit]

try this

[shut heart]

I messed up bad lol

[long flame]

0/0 indeterminate

[hollow orbit]

yeah

[shut heart]

Bro I was trying to find one all the questions I have have log or a trigo function

[hollow orbit]

i got plenty of poly

[shut heart]

All the poly I have u need to take log both sides 💀

[hollow orbit]

1^inf situation

[shut heart]

Yeah 0^0 and ∞^0 too

[hollow orbit]

these are the best

and using lhospital also

my game now lol

[hollow orbit]

more simplification and you get 3/5

great

correct one

[shut heart]

We were forbidden to use L'Hospital by our teacher 🥲

[shut heart]

Nice

[hollow orbit]

me too at first but then he allowed it when we took the rule

check this one

[long flame]

yeah true

[shut heart]

As long as it's not log 0 somewhere we don't use L'Hospital

[hollow orbit]

i only got 0/0

[long flame]

surely here you would use a different method, because plugging in q gives weird eq

[hollow orbit]

just a simplification

[shut heart]

Its 9 not q

[long flame]

oh lol

lol

[shut heart]

😂

[long flame]

that makes sense

[hollow orbit]

i didnt realize lol

[shut heart]

Can u give me some tough ones it's been a while I did limits

[hollow orbit]

allowing lhospital?

[shut heart]

I have been doing indefinite Integration for quite a while

[shut heart]

Hmm u can give any

[hollow orbit]

i have done the course of integration, was greatttt

[shut heart]

Yeah it's good

[dry lynx]

i have a limit if anyone wants it

number 4

[hollow orbit]

my neck

lol

[dry lynx]

,rotate

[dry lynx]

i dont know how to do it without LH

and using LH gets messy

[shut heart]

I am trying

U know expansion of cosx and sinx

[dry lynx]

like taylor series

[shut heart]

Yeah

[dry lynx]

yeah

[shut heart]

It always works if L hospital is to hard to apply

[dry lynx]

but that gets messy too

i have to use up to 2 terms for both

[shut heart]

U need to factorise first

[dry lynx]

ok

[shut heart]

a²-b²

To (a+b)(a-b)

[dry lynx]

yeah

[shut heart]

U can take cos(sinx) + cosx out of lim

[dry lynx]

why?

[shut heart]

Bez it's finite

It's 2

[dry lynx]

oh

i see it now

thanks

[shut heart]

No problem

But why did u use this channel 💀

[dry lynx]

u asked for a hard limit

and i had this somewhere

[shut heart]

U could have dmed

[dry lynx]

sorry

[shut heart]

@long flame u here?

[long flame]

yeah

[shut heart]

It's ok

[long flame]

i was working on the sqrt lim

[shut heart]

U tried the limit?

[long flame]

i simplified it to

$(x+9)*(sqrt(x)-3)$

[twin meteor]

marty

[long flame]

which still gives zero

[dry lynx]

isnt it sqrt(x)+3

[long flame]

oh yeah

[hollow orbit]

no

[long flame]

shit

wrong sign

[hollow orbit]

$(x+9)*(sqrt(x)+3)$

[twin meteor]

Sherlock

[long flame]

i wrote plus in my calc even im getting too tired lol

so 18 * 6

[hollow orbit]

happens a lot

yeah

[long flame]

108

[hollow orbit]

my man

[shut heart]

Correct

[long flame]

cool

[shut heart]

U understand it now ryt

[long flame]

well the steps are getting clearer

here i felt confident the simplification was done because i was no longer dividing by zero

and i was always trying to cancel that term

[shut heart]

Nice

[long flame]

and finite division by zero is still workable right

[hollow orbit]

yeah its DNE

[shut heart]

Don't exist

[hollow orbit]

specifically +- infinity

[shut heart]

Yeah

[long flame]

i mean for the initial stage before simplification

[hollow orbit]

yeah

no need for further process

[long flame]

oh

[hollow orbit]

unless its an indeterminate form

[long flame]

okay so it is determined, and its either + or - infinity

[hollow orbit]

yeah

[shut heart]

Save this

[long flame]

and zero division by finite is again

i saved it

also determinate , infinite ?

[hollow orbit]

determinate but it is zero

[long flame]

ah

ah ofc

[hollow orbit]

like 0/2 ?

[long flame]

yes

that makes sense

so thats the symmetry

0/finite = 0, finite/0 = +- infinity

[hollow orbit]

yessir

[long flame]

excellent

thank you all very much

[hollow orbit]

no worries bro

[shut heart]

No problem

[hollow orbit]

any question just send a mention

[long flame]

i will close for now 😄 🙏

[long flame]

🙏

.close

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

how is this undefined at f(4). It is 0 over 1 which is 0, not undefined

[sand lodge]

denominator becomes 1/4 - 1/4

[long flame]

omg

too quick sorry

i was thinking it was adding for some reason

now makes sense

.close

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[bleak prawn]

Trying to spot where they are haha

Is it like this?

The one on the right is obvious but I wasnt so sure about the one on the left

I wouldve guessed there was only one point of inflection if they didnt tell me

[silent folio]

Yeah it's quite hard to see but I would agree

[bleak prawn]

Okok nice

Points of inflection are kinda new to me so I just wanted to be sure haha

Thank you! Thats all I Wanted to ask

❤️

.close

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[glacial meadow]

how did i do?

[vocal sleet]

@glacial meadow Has your question been resolved?

[vocal sleet]

@glacial meadow Has your question been resolved?

[vocal sleet]

@glacial meadow Has your question been resolved?

[vocal sleet]

@glacial meadow Has your question been resolved?

[glacial meadow]

.close

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[timid leaf]

Why might it be useful to convert radians to degrees?

[timid leaf]

can someone help me answer this

like

what is a radian really

[iron flame]

a radian is the angle where the arclength of a circle is equal to its radius

[thin vale]

https://youtu.be/opQhR1vX1U4

[timid leaf]

ohhkay

[civic otter]

Because radians are unitless, since they arise from ratio of lengths

[timid leaf]

so like indefinable?

[civic otter]

And unitless quantities mean numbers which we can put inside functions

[civic otter]

No no, indeed you have been given the definition of radians above

Unitless means that there's no unit, it's not in meters, neither in kilometres, nor in seconds, nor in kilograms and so on

Simply it is a number, hence a quantity with no unit

[vocal sleet]

@timid leaf Has your question been resolved?

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[potent orchid]

Hello

[vocal sleet]

Please don't occupy multiple help channels.

[thin vale]

.close

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[thin vale]

Don't open multiple channels

use #help-9

[vocal sleet]

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[potent orchid]

.close

Hello I am having trouble comparing graphs of a few functions

[opal obsidian]

Please post your question

[potent orchid]

For the first value of a i think it’s either a horizontal stretch

But the amplitude is different also

<@&286206848099549185>

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@potent orchid Has your question been resolved?

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.close

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[sly sierra]

!status

[vocal sleet]

What step are you on?
1. I don't know where to begin
2. I have begun but got stuck midway
3. I got an answer but I'm told it's wrong
4. I got an answer and would like my work checked
5. I have a question about someone else's worked solution
6. None of the above

[mint nebula]

4

[vocal sleet]

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

[sly sierra]

cool, can you show what you did?

[mint nebula]

SOrry i mean 1

I don't kknow where to begin

is it simultaneous equations?

wait nvm

someone else gonna hel me

thansk

.close

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[long flame]

i didn't feel satisified in this simplification, is it rly enough to leave it like this? I just started using trig. addition rule

[pallid forge]

it could be simplified a little bit by adding the fractions together.

[long flame]

oh yeah fair enough

[pallid forge]

you could also factor out sqrt(2), but i'm not sure that it's "simpler" after that

[long flame]

$\frac{sqrt(2)+sqrt(3)*sqrt(2)}{4}$

[twin meteor]

marty

[pallid forge]

yeah but also sqrt(3)*sqrt(2)=sqrt(6)

[long flame]

wait

[pallid forge]

so you could have $\frac{\sqrt(2)+\sqrt(6)}{4}$

[twin meteor]

Wumbo

[long flame]

i forgot apparently how sqrt works. so sqrt(x) * sqrt(y) = sqrt(x*y)

[pallid forge]

yep!

[long flame]

v good to know

yeah this feels simple enough

[pallid forge]

so like sqrt(8)=sqrt(4)*sqrt(2)

so sqrt(8)=2*sqrt(2)

[long flame]

no rules for addition right

like we have here

[pallid forge]

not really no

[long flame]

okay great

thanks again

.close

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[long flame]

cos120 is the same as -cos30 right?

[twin meteor]

Results provided by WolframAlpha

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

Requested by neonperseus

[quick estuary]

No

[gritty zenith]

Odd question

[quick estuary]

Cos 90+x is

• sin x

• sin x

Bro

[viral copper]

F Discord

[quick estuary]

-sin x

[long flame]

i was thinking in terms of the unit circle

[quick estuary]

Even then

[long flame]

that we've gone past 90, and now we're in negative space for cos, and then 30

how about, cos120 = -cos60?

that should be correct right?

[round plover]

ah wait mb, yes looks good

[long flame]

how do u do the wolfram query

[round plover]

,w (query here)

[long flame]

it wouldnt be 30 because the angle is measured from the horizontal

i keep forgetting that