help-27

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[restive river]

kind of having a brain fart moment

[restive river]

ratio test is inconclusive, leads me to believe this needs some application of a comparison test

not sure what to compare it with though

[zenith jacinth]

you want to use stirling ?

[zenith jacinth]

double factorial, woaw

[restive river]

What even is that

Lmfao

if the number is odd

then the double factorial of an odd number

is that you multiply ever odd number until 1

[restive river]

same with even until the even number

[restive river]

wowzers

talk about maths shenanigans i never thought were a thing

its used for when you do

[arctic field]

bessel functions

[restive river]

[ \prod_{n=1}^{\infty} n ]

[woven radish]

muk

[restive river]

you can use double factorials

to simplify this

[arctic field]

,w maclaurin series BesselJ[0, x]

[restive river]

very interesting

[woven radish]

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[arctic field]

this expansion is in terms of double factorials

[arctic field]

iirc

[restive river]

i am not sure if our teacher wanted us to do this in a different way tho since this was like our fourth question in the slides 2 months ago when we first got introduced to those concepts

if they didnt introduce it, they want you to do it a different way, unless they told you to learn it

[arctic field]

the fact that it can be written in double factorial notation doesnt preclude you from using the idea of double factorials

all it is is separating the factorial into its odd/even terms

[restive river]

makes sense, really interesting

will research more about it later

thanks

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[thorny pawn]

For this integral what’s the rationale behind making the substitution tan(theta)=sqrt(e^x -1)

[thorny pawn]

<@&286206848099549185>

[prime egret]

no rationale whatsoever

[thorny pawn]

Bruh

Fine

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[proven kite]

I have no idea how to get rid of inf/inf.

I tried some algebraic manipulations, but to no avail. I cannot use L'Hospital and neither Taylor series

[torn wave]

yk the limit for ln x to infinity is infinity

[prime egret]

Hint Squeeze theorem

[torn wave]

yeah

[proven kite]

Yes, but isnt inf^inf also undefined?

[torn wave]

lim of 1/x to infinity is 0

infinity times 0

rudimentary explaination you should do squeeze theroem

but like

[proven kite]

Ohhh, hmm i didnt get squeeze theorem that good, but ill try that then. Thank you

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[viscid crown]

If there is a cycle containing edge e and f and another cycle containing edge f and g, then there is a cycle containing edge e and g

[viscid crown]

So let u, v, w be the vertices such that uv=e, vw=f and uw=g

then our cycle1 would be written as (uv, vv_1, ...., v_kv, vw, ww_1, ..., w_kw, wu)

wait that doesnt look right

okay instead cycle1 will be (e, something, f, something-that-ends-with u)

[wild creek]

i think its true if you dont require elementary cycle

[viscid crown]

there is no restricitions

so it should be true

but im struggling on proving it lol

[wild creek]

you should consider the concatenation

and remove double edges maybe

in some way

if theres is backtracking

[viscid crown]

so C3 cycle must pass at least 1 vertex in C1 and C2

otherwise it wont exist

[devout snow]

@viscid crown Has your question been resolved?

[viscid crown]

it cant just be cycle3 = (e, some walk, f, some walk, g, some walk, f, some walk-ending-with u)

[devout snow]

@viscid crown Has your question been resolved?

[devout snow]

@viscid crown Has your question been resolved?

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[deep sentinel]

I'm stuck on this, I know it's not A. I'm pretty sure it's C?

Or B

Nvm it's b

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✅

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[azure yew]

how am i supposed to proceed here?

[devout snow]

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[azure yew]

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[shut crown]

Q 3

[devout snow]

@shut crown Has your question been resolved?

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[gloomy fox]

Describe as simply as possible the set of all real numbers x, for
which applies (the solution must be specified):
x − 3| < 8

[devout snow]

Please don't occupy multiple help channels.

[gloomy fox]

sry

[gloomy fox]

| is missing

is the answer x > -5?

[amber valley]

You can read that as, the distance of x to 3 is less than eight

What numbers are 8 units away from 3?

[gloomy fox]

11 and -5

-5 < x < 11?

[amber valley]

No, outside that range

Wait no, my bad lol

You're right

[gloomy fox]

xD

but i need a answer way

what do i have to calculate

-x+3 < 8?

[amber valley]

|a| < b is the same as a<b and -a<b

[gloomy fox]

so i have to calculate -x+3 < 8 and x+3 <8?

-x + 3 <8 is x > -5

[amber valley]

Don't forget parentheses!

The second inequality should be x-3>8

[gloomy fox]

oh ok

so i could write first case : -x+3 < 8 ... and second case: x-3>8 ...

[dark dawn]

can u also write this as a piecewise function?

[real grail]

or you could say :
-8 <x - 3< 8

to make it simple

[gloomy fox]

yeah i know but i need a solve way i have to do it step by step

Describe as simply as possible the set of all real numbers x, for
which applies (the solution must be specified):

solution must be specified

[dark dawn]

this is that

[gloomy fox]

i think to write down only that isnt enough

[dark dawn]

your solution is just -5<x<11

well

you will be writing for an infinite amount of time then if its for all reals

[amber valley]

^

[gloomy fox]

they have to see how i get to this solution

[dark dawn]

you did it

[gloomy fox]

hmm ok

[dark dawn]

using property ral said

[dark dawn]

this one

[real grail]

you can add 3 in all the side of -8 <x - 3< 8

-8 +3 <x - 3 + 3< 8 + 3

[gloomy fox]

ah ok

[real grail]

that will explain to him how did you get it

[gloomy fox]

ok thanks guys ❤️

im closing

.close

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[amber valley]

[devout snow]

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[bleak sun]

Use $Area = \sqrt{(s-a)(s-b)(s-c)(s-d)-abcd * cos^2(\theta / 2)}$

[woven radish]

cantprogram

[quartz ice]

what the fuck

[bleak sun]

where theta is the sum of the opposite angles and s is the sum of all the sides divided by 2

[quartz ice]

I haven't gotten near this bro

can I not use sine or cosine rule for it?

[bleak sun]

yeah thats derived from cosine rule and herons formula

[quartz ice]

herons formula? bro u talkin bout greke phiolosophers nd shit

[bleak sun]

nvm i guess u can split into triangles and solve that way then

[quartz ice]

yea thats what i was doing

so a 1/2 15.8 x 9.2 x sin 72 for the first one

cause I split it along where 220 is

can I use simple trig for the 2nd triangle

cause its perpendicular

Okay lets use that formula

@bleak sunwhat would a b c and d be

[bleak sun]

they are just the side lengths

[quartz ice]

can they be in any order

[bleak sun]

yes

[quartz ice]

did that

was fuckin

wrong

damn im an idiot

@bleak sunman

can you put it in t

the formula every time i do it

wrong

[bleak sun]

huh

show your working out

[quartz ice]

alr

New questionright

[bleak sun]

ok

[quartz ice]

so s =38

correct?

[bleak sun]

,w (16.6 + 31 + 10 + 18.4) / 2

[woven radish]

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[bleak sun]

no.

[quartz ice]

???

How

all sides

[bleak sun]

oh

yeah i thought u wrote 28

[quartz ice]

divided by two?

Man

[bleak sun]

ok yeah s=38

[quartz ice]

imma stop tiring ur eyes i promise

[bleak sun]

theta = 206+63

then just plug in

[bleak sun]

oh it should be cos^2(theta/2)

$Area = \sqrt{(s-a)(s-b)(s-c)(s-d)-abcd * cos^2(\theta / 2)}$

[woven radish]

cantprogram

[quartz ice]

on my calc

it does cos(^2 only

not cos^2

can i do cos(cos) ?

[bleak sun]

no

cos(x)^2

is the same as

cos^2(x)

[quartz ice]

what should x be then

or just close the bracket

[bleak sun]

what

in the formula its

cos(theta/2)^2

,w cos^2((206+63)/2)

[woven radish]

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[quartz ice]

Man the answer is STILL WRONG I GOT 134 PUT IT IN WRONG

Please can we just walk through splitting the triangles please i dont wanna waste any more of your time

So far got this

.close

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[winter topaz]

Hello, recently I tried to solve this question, but the teacher told me the bounds on the integral were wrong

I understand that this is wrong because if x is less than 2, then y shouldn't be integrated from 0 to 2

but, I'm not sure how to fix these bounds

[zenith jacinth]

should be -infty right ?

🤔

[winter topaz]

but the bounds for the joint integral itself are 0 < y < x < 4

[zenith jacinth]

thats the expression of f(x,y) when y and x are between 0 and 4

elsewhere its 0

it doesnt always mean that f(x,y) is defined only on 0<y<x<4

[winter topaz]

but if it's 0 elsewhere

if integrated isn't the same as being defined only on 0<y<x<4

since the rest doesn't contribute to the integral

[zenith jacinth]

yea I know but you should still write the correct bound

[winter topaz]

that still wouldn't be the answer tho right

wait you mean (-infty, 2) and (-infty, 3)?

for y and x respectively

[zenith jacinth]

yes thats the correct bound

[winter topaz]

but doesn't y depend on x for its bound?

[zenith jacinth]

hmmm maybe

[winter topaz]

if it does, then (-infty, 2) and (-infty, 3) wouldn't be correct I think

the (-infty, 3) I think is right, it's just the y bounds

and (0, 3) still means the same thing for the x bound as well

but my question is if I let's say put (0, x) for the y bound

and x goes to 3

then y also goes to 3

when I'm supposed to find (X < 3, Y < 2)

so what should the y bound be?

[zenith jacinth]

$\int_{-\infty}^3 \left(\int_{-\infty}^x f(x,y) dy \right) dx$

[woven radish]

Ｈｅｒｅｌｓ

[zenith jacinth]

🤔

[winter topaz]

ohhh

wait

y first?

I think

I think this might work

YES

idk why I was so into doing x first

ty ty

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[restive river]

Tried doing a comparison test but idk how to start with it, or if I should do direct or limit comparison, any help? I asked about this a bit ago but I’m still lost T_T

[bleak sun]

whats maximum value of cos^2(x)

[restive river]

1 I think ;-;

[bleak sun]

yes

so its less than or equal to what ?

[restive river]

E^n?

[bleak sun]

not quite

close though

[restive river]

uhhhh, 1/e^n?

[bleak sun]

yup

so our series is less than or equal to 1/e^n

and we can find is 1/e^n diverges or not easily because its just a geometric series

[restive river]

it converges right?

[bleak sun]

how'd you get that

its a geometric series so find r

and see if the absolute value of r is less than 1

[restive river]

yes and since the value is less than 1

[bleak sun]

yup

what did u get r as ?

[restive river]

this might sound stupid but (1/e)/[1-(1/e)], im so sorry T_T

[bleak sun]

yea r = 1/e lol

and abs(1/e) < 1

so 1/e^n converges

and by comparison test our series converges

[restive river]

ohhh okay

[restive river]

i just realized you just asked me for r

[bleak sun]

yeah lol

[restive river]

man.

what happens when you're gone from class bc of a flu and have to learn inf series in 3 days before the final exam that's worth half the semester grade

[bleak sun]

damn good luck

[restive river]

thank you, still have alternating series along with maclaurin and taylor series :), and ratio test

i now try my best with the other problems, thank you T_T

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[wanton storm]

hii can someone help me with this?

[devout snow]

Please don't occupy multiple help channels.

[bleak sun]

just plug in x= -1 and see if matches

same for x= 1/2 and x = 1

[wanton storm]

how?

match with what?

[bleak sun]

y = (1/2)^x

y = (1/2)^-1

y = (1/2)^(1/2)

y = (1/2)^1

[devout snow]

@wanton storm Has your question been resolved?

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[fair spruce]

Have I interpreted this correctly for solution?

[brisk totem]

does it logically make sense that given 6 of the 30 tickets are lucky, a particular individual buying 1 ticket has a more than 70% chance to get a lucky ticket?

check your arguments again

[fair spruce]

Yeah... Maybe I should go with Bayes' rule with 3 events?

E1 won, E2 won, E3 won
E1 won, E2 lost, E3 won (because odds for E1 and E2 are stated to be same odds no point in doing E1 lost E2 won, E3 won)
E1 lost, E2 lost, E3 won.

[brisk totem]

bayes works, combinatorical argument will also work in the same fashion if set up correctly

if you want to figure out what's wrong with your probabilities, explaining each fraction in words might help? idk

[fair spruce]

Yeah the binom (24,3) doesn't make much sense actually now that I look at it.

[brisk totem]

there are also nice tricks to skip a lot of the working out if you can argue properly

very easy question: what's the probability of student 1 winning?

slightly harder question: what's the probability of student 2 winning?

which leads to your question

[fair spruce]

well for first its 6/30

and id assume its 5 / 29 for second but then the question states they have the same probability :/

so for third it can be like 6/ 28 or 5 / 28 or 4 / 28

[brisk totem]

think again, does student 2 winning imply student 1 is guaranteed to have won?

[fair spruce]

no their independent

so 6/30 and 6 / 29 ?

[brisk totem]

does student 2 winning imply that student 1 is guaranteed to have lost?

[fair spruce]

no 😄

[brisk totem]

so it probably lies somewhere in between 😛

bayes rule will help just like it would for student 3, then you might be able to find a pattern

[fair spruce]

Im currently writing a way to chain the 3 events

[devout snow]

@fair spruce Has your question been resolved?

[fair spruce]

Idk I still keep arriving to 6/30 * 6 /29 * 6 /28

For both the 1st and 2nd student we have events where they either drew a winning ticket or not and the odds for 3rd student will depend on those events.

[brisk totem]

not sure where you are going wrong, but by your work the probability of student 3 winning is a sum of 4 events right? where students 1 and 2 can win or lose

so determine the probability of those 4 cases individually

[fair spruce]

So the answer will be that there are 4 different outcomes right?

But we just can't determine which one

[brisk totem]

no, by your construction you have 4 cases where student 3 can win:

case 1: s_1 wins, s_2 wins, s_3 wins
case 2: s_1 wins, s_2 loses, s_3 wins
case 3: ..
case 4: ..

since those cases can't simultaneously happen, the probability of s_3 winning is the sum of the probability of those 4 cases

[fair spruce]

ah okay se we may sum the cases, okay.

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[jovial temple]

How to prove this statement?

[devout snow]

@jovial temple Has your question been resolved?

[jovial temple]

<@&286206848099549185>

[devout snow]

@jovial temple Has your question been resolved?

[devout snow]

@jovial temple Has your question been resolved?

[devout snow]

@jovial temple Has your question been resolved?

[supple trench]

there's a conflict of role of the variable n:

1. upper limit of the product
2. instance variable of the product

[jovial temple]

right

I mean upper bound k

The series ```1/23/45/6...

[supple trench]

,,\prod_{k=1}^n \frac{2k-1}{2k} \le \frac{1}{\sqrt{n + 1/2}}?

[woven radish]

vin100

[supple trench]

sorry i dun hav idea. you may kill this problem with Stirling's approximation. Approach0 might give a more elementary solution

[jovial temple]

Ok I will look into it thank you

[jovial temple]

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[magic gust]

hello

[magic gust]

anyone can help me in informatics please

i can't launch my programm

i have an error message

but i don't understand why it's flase

false*

[long kettle]

Y.append(y)

Not Y.append[y]

[magic gust]

ahhhhh okay

thanks !

.close

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[next root]

This is the math book its in Dutch

And i have a question about the argument alpha

So you calculate the argument by tangens = b/a

By the first one after "voorbeelden"

My answer is -1

And out of nothing the answer is -45°

Never mind by explaining my problem i found the problem

[devout snow]

@next root Has your question been resolved?

[next root]

half i got a new question now its one with my calculator

For the argument on the first excersice

Tang alpha = -.577

So alpha is -30

Never mind

Found the problem while explaining it 😅

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[restive river]

How is the interval found exactly, I don’t get it

I understand that these underlined functions must both be continuous in a certain interval and that t=0 is undefined here

But why isn’t it ]-inf, inf[ \ {0}?

I mean how can I tell y(4) is not correct just by looking at this?

Nevermind just had to google this 😛

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[shut crown]

Q 13 can someone help me solve this?

Simultaneously equations

Linear and non linear

[bleak sun]

find what x is in terms of y using first equation

and plug into 2nd equation

[shut crown]

Yes I know that bit what do I do when it's a fraction?

[bleak sun]

why would a fraction make a difference ?

[shut crown]

Browhat

I need help look I turn thr x into ×=

Ok

So then it would be 15-5y over 3x

Squared

The what do I do after I square it

[devout snow]

@shut crown Has your question been resolved?

[bleak sun]

no

x = (15-5y)/3

not 3x

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[mint mortar]

what is meant by fifth root in this question?

[bleak sun]

x^(1/5)

[celest depot]

The fifth root of a number x is the number that must be multiplied five times by itself to get x.

[mint mortar]

oh

thank you

btw, how can i close this thread?

[bleak sun]

.close

[celest depot]

.close

[mint mortar]

.close

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[copper arrow]

Does anyone know how to draw a complex plane in LaTex? I'm using Tikz trying to plot 1 + i and 1 - i and don't know what I'm doing.

[bleak sun]

#latex-help

[devout snow]

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[slow trench]

quick question, what is this weird a supposed to mean?

[prime egret]

Partial derivative

(It’s del)

[slow trench]

thanks!!

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[restive river]

Oh is it pronounced as del?

Never knew

So like, del of y with respect to x?

[prime egret]

I read it as partial y with respect to x

Don’t think anyone actually says del

[devout snow]

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[wintry rapids]

Hi

[wintry rapids]

Basically my teacher is really bad and I didn’t learn much from him.

So I need help

It’s task 1

[dapper reef]

I believe what you can do is sub in 7

and you have the value of f(x)

and then to find the intrest rate

multiply 0.93 by 7 and that should be your intrest rate don't trust me on that tho i'm not sure

[devout snow]

@wintry rapids Has your question been resolved?

[wintry rapids]

So basically f(7)?

[dapper reef]

yeah

I got 150425.2177

I got -6.9999999 for the intrest rate so

[wintry rapids]

Damn

[wintry rapids]

You’d swear to my teacher

My teacher is really bad at teaching

But can you show me how I am supposed to write the answer?

With proof of how I got there

[devout snow]

@wintry rapids Has your question been resolved?

[devout snow]

@wintry rapids Has your question been resolved?

[devout snow]

@wintry rapids Has your question been resolved?

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@wintry rapids Has your question been resolved?

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[devout snow]

@restive river Has your question been resolved?

[devout snow]

@restive river Has your question been resolved?

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[plush osprey]

how would i start these problems

[devout snow]

@plush osprey Has your question been resolved?

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<@&286206848099549185>

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[late surge]

hi guys

[devout snow]

Please don't occupy multiple help channels.

[late surge]

can someonr please help ne with either one of these questions 🙏🙏

[main gull]

You already have a channel open

Don't open multiple channels

.close

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[main gull]

#help-11 is the channel you have

[devout snow]

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[topaz iron]

So I solved it and got the correct answer

[topaz iron]

but idk if the process is correct

would this be a correct way to use the CRT to compute the question?

[devout snow]

@topaz iron Has your question been resolved?

[topaz iron]

<@&286206848099549185>

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.close

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[finite inlet]

When you have something like $|x|$, you need to differentiate between 2 cases, $x < 1$ or $x \geq 1$, right?

[woven radish]

madmike

[rich summit]

you mean x < 0 and x >= 0?

[finite inlet]

oh yeah sorry

[rich summit]

most of time you probably do

[finite inlet]

However I get the wrong result if I do $x \geq 0$, and I get the correct result if I do $x > 0$.

[woven radish]

madmike

[rich summit]

what was the original problem?

[long kestrel]

Yoo jelle

[long kestrel]

Can I ask u a favor

[finite inlet]

no

[rich summit]

#❓how-to-get-help

[rich summit]

you considered 2 cases? or 3?

[finite inlet]

4 cases

at the end I get this

[finite inlet]

which is wrong because try putting 2 in here

you get 1 and 1 > 1 is false

[rich summit]

yes

[finite inlet]

so I wonder why this is wrong

I changed my cases from x > 0 to x >= 0, I assume the latter is correct...

and then I get a wrong result

I can send the whole thing but it's a lot

[rich summit]

you should split it up into 3 cases: x < 1, 1 <= x <= 2, x > 2

where you put the = sign doesn't matter

[finite inlet]

huh

why 3 cases

[rich summit]

because the 4th case: x > 2 and x < 1 doesn't exist

[finite inlet]

either x-1 is positive and x-2 is positive, or x-1 is positive and x-2 is negative, or x-1 is negative and x-2 is positive, or x-1 is negative and x-2 is negative

or not?

[rich summit]

or x-1 is negative and x-2 is positive, this case can never happen

[finite inlet]

thats right, I assume that's why I get gibberish from that case

apparently also x-1 >= 0 and x-2 < 0 can never happen

or 🤔

[rich summit]

that would imply: x >= 1 and x < 2, which can happen

x = 1.5 for example

[finite inlet]

then idk maybe my process is all wrong

because I got 2 > 2 I just ignored this case

[rich summit]

that's always false

you should probably be careful with the difference between >= and >

[finite inlet]

actually if you insert 1.5 or whatever into |x-1| + |x-2| > 1 then you get 1>1

so I think you're wrong about this case?

[rich summit]

Oh, I though you meant no x existed that satisfied those inequalities

you're correct, 1 <= x <= 2 implies 1 > 1

[finite inlet]

yeah maybe I made a tiny mistake somewhere trying to change it up

I'll look over it again

[rich summit]

maybe what you did wrong is assume x >= 2

$x\ge2:\ x-1+x-2>1\to2x-3>1\to x>2,\ true\ if\ x\ \neq\ 2$

[woven radish]

Jelle

[real grail]

there is a simple way to prove it

since |x| + |y| >= |x + y|

[finite inlet]

I think I found my mistake here

x>=1 and x>=2 and x>2

if you simply it then you get x>2 , right?

not x>=2

[rich summit]

I think so, because the original question is a strict inequality

[rich summit]

you also need to prove x < 1 and x > 2 are in fact solutions

[real grail]

no no need to

[finite inlet]

this is my whole thing I had to do 9 of those

and that's just exercise 3, we have 4 exercises

god

[rich summit]

jeez

[finite inlet]

your tag is pre-university

are you in high school?

[rich summit]

yeah

[finite inlet]

how do you know this stuff

😂

[rich summit]

absolute values do get taught in my country in high school

[finite inlet]

nice

[rich summit]

but I also just watch videos on youtube about other stuff

[real grail]

there is a simple way to prove it:

we know that |x| = |-x|
so |x-1| = |1-x|

|x-1| + |x - 2| > 1 <=> |1-x| + |x - 2| > 1

since |x| + |y| >= |x + y|
so |1-x| + |x-2| >= |1-x + x-2|
|1-x| + |x-2| >= |1-2|
|1-x| + |x-2| >= 1
so |x-1| + |x-2| > 1 true if |x-1| + |x - 2| not equal to 1

and that when x not equal to 1
so the solution is x not equal to x

simple.

[finite inlet]

I am confused

what is this for

[rich summit]

but why would 1 <= x <= 2 imply |x - 1| + |x - 2| = 1

[real grail]

you want to prove that |x-1| + |x-2| > 1 right ?

[rich summit]

yeah

[real grail]

I will explain step by step just follow me

there is a rule in math said that :
|x| + |y| is always bigger or equal than |x+y|

so we will use it

okay ?

[rich summit]

yes, I understood everything, up until "so |x-1| + |x-2| > 1 true if |x-1| + |x - 2| not equal to 1 "

I understood that line

but what followed was a bit weird

[real grail]

what i find is that "what ever x is , |x-1| + |x-2| is bigger or equal than 1"

what I want is "|x-1| + |x-2| is bigger than 1"

[rich summit]

yes

but your proof kinda stopped at the conclusion: |x-1| + |x - 2| not equal to 1

[real grail]

|x -1 | + |x - 2| = 1 when x = 1

[rich summit]

and x = 1.5 and x = 1.1 and ....

[real grail]

don't forget that |x| is always bigger or equal to 0

|1.5 - 1| + |1.5 - 2| = |0.5| + |-0.5|
= 0.5 + 0.5
= 1

|-0.5| = 0.5

let solve |x-1| + |x-2| = 1

[old rover]

Take three cases

[real grail]

|x-1| + |x - 2| = 1 when :

1. (|x-1| = 0 and |x-2| = 1)
   or
2. (|x-1| = 1 and |x-2| = 0)

[old rover]

x>2

[finite inlet]

guys I have a different question 👀 😂

[old rover]

1<=x<=2

x<1

[woven radish]

madmike

[finite inlet]

is the summary down there legit?

or do I need to include more conditions

[old rover]

yes it's legit

And it's using way too many cases!

three cases are sufficient

x>1
x<-1
-1<=x<=1

[real grail]

you don't need many cases since :
|x - 1| = 0 when x - 1 = 0 so x = 1
|x + 1| = 0 when x + 1 = 0 so x = -1
so solution are {-1,1}

[old rover]

even better!

[finite inlet]

oh

I don't need to do the case where x-1 < 0 ?

[real grail]

I thought that your question is |x-1| + |x - 2| > 1 😆

[finite inlet]

I solved that one, I have 9 like these lol

now I'm finally done

[real grail]

nope , you don't actually

just because |x-1| = 0

that will change if you have a number different than 0

[old rover]

You don't even need cases (kinda) $|x^2-1|=|(x-1)(x+1)|=|x-1||x+1|=0 \leadsto |x^2-1|=0 \leadsto x^2-1=0 \leadsto x=\pm 1$

[woven radish]

Mohammad

[finite inlet]

🤯

[real grail]

yeap , that's right

[finite inlet]

anyway, thank you guys

[devout snow]

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[bold venture]

“An object experiences an acceleration of -6.8 m/s^2. As a result, it accelerates from 54 m/s to a complete stop. How much distance did it travel during that acceleration?”

[proud perch]

2aΔx = v² - v_0²

[bold venture]

??

[proud perch]

This is the time independent acceleration eqn

Have you seen it before?

[bold venture]

No

[proud perch]

Well now you have

Gl!

[jaunty acorn]

if anyone is free to help please go to #help-6

[bold venture]

No

[bold venture]

It’s (vf^2 - vo^2)/2a?

[proud perch]

2aΔx = v² - v_0²

don't forget the a

[bold venture]

Ok is it correct now

[proud perch]

Yea

[bold venture]

Yay thanks

What lmao

Sorry for saying thanks I guess

[devout snow]

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[devout snow]

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[feral pine]

how i solve this , i dont have any idea and exercises

[feral pine]

this is the base but in a exercise i dont know where start or end

for example the 5

i get stuck here

[restive river]

It's really hard to know what the question is to begin with and what's in those pictures

[feral pine]

sorry for the problems the english is not my mother lenguaje

[restive river]

Hablas español? puedo hablar un poco de español

[feral pine]

si si puedo hablar español

[restive river]

Está bien, mi español es un poco flojo, pero lo intentaré jaja

qué problema tienes con la pregunta?

[feral pine]

que entiendo el concepto de lo que hay que hacer con esos ejercicios pero en la practica no tengo idea

[restive river]

Hmm okey

cual es el mayor problema que tienes con el?

[feral pine]

pongamos el primer ejemplo

[feral pine]

de aqui para ser especifico

[restive river]

Okey

[feral pine]

imaginemos que n es 3 y m es 12

[restive river]

ok lo hare paso a paso

[feral pine]

ok

[restive river]

$a^{\frac{12}{3}}$

[woven radish]

♡LexQa♡

[restive river]

Por que: $\sqrt[n]{a^{m}} \to a^{\frac{m}{n}}, m = 12, n = 3$

[woven radish]

♡LexQa♡

[restive river]

@feral pine

$a^{\frac{12}{3}} \to a^4$

[woven radish]

♡LexQa♡

[restive river]

Entiendes amigo?

[feral pine]

eso intento

creo que ya entiendo

[restive river]

Está bien. No te preocupes

[feral pine]

aqui tengo una duda se divide 12/3 o 3/12 ?

[restive river]

Muy bueno!

[restive river]

12/3

sería 3/12 cuando: $\sqrt[12]{a^{3}}$

[woven radish]

♡LexQa♡

[feral pine]

ahora tengo otra duda si a fuera aqui un 3 se multiplicaria 3.3.3.3

?

[restive river]

Sí. los exponentes son multiplicaciones repetidas

Igual que la multiplicación es una suma repetida

[feral pine]

ahora probemos el 2

supongamos que n es 2 a 5 y b 7

[restive river]

Okey

[restive river]

$\sqrt[2]{5 \cdot 7} \to \sqrt[2]{5} \cdot \sqrt[2]{7} \to 5^{\frac{1}{2}} \cdot 7^{\frac{1}{2}}$

[woven radish]

♡LexQa♡

[feral pine]

pero de donde salen los 1/2 del final ?

o mejor dicho por que

?

[restive river]

Por que: $\sqrt[n]{a^{m}} \to a^{\frac{m}{n}}$ sentido: $\sqrt[2]{7^1} \to 7^{\frac{1}{2}}$

[woven radish]

♡LexQa♡

[restive river]

además: $7^1 \iff 7$

[woven radish]

♡LexQa♡

[feral pine]

osea que siempre hay un 1 aunque no se vea

[restive river]

Sí, eso es verdad

[feral pine]

y que se hace si 5 y 7 tienen exponentes ?

@restive river

[restive river]

$a^{5^{8}}$?

[woven radish]

♡LexQa♡

[feral pine]

a eso me refiero

[restive river]

Ahh

[feral pine]

sí en este ejercicio 5 y 7 como se resuelve ?

[restive river]

Siento haberme ido de repente, pero se está haciendo tarde y tengo que irme a dormir :(( @feral pine

podemos continuar mañana?

[feral pine]

si no hay problema yo practicare los dos ejercicios que vimos hoy

buenas noches :)

[restive river]

¡por supuesto! Buena suerte amigo

[restive river]

tú también <3

perdon por mi mal español jajaja

Another day failing Spanish less go 😎

[feral pine]

lo mismos digo por mi ingles :]

another day not knowing talk with people less go :}

[restive river]

😭

It's okay really, learning languages are very hard

Especially English

Tienes este amigo :)

[feral pine]

and you have me :)

bye :>

[restive river]

Byee :>>

[devout snow]

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[restive river]

Help

[devout snow]

Please don't occupy multiple help channels.

[long kettle]

Don't open a second channel

[restive river]

i wakt someone else to help me

so i opned this one

[long kettle]

Then wait in your original channel

[restive river]

no

+ratio

someone help me

pls

that id not gamer dio

[winter patrol]

.close

[devout snow]

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[vestal dirge]

Is this correct

[vestal dirge]

*3rd binomial formula

weil = because

[devout snow]

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[muted sierra]

did I do this right?

[devout snow]

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[muted sierra]

.close

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[fading ether]

Can someone pls explain why the answer isn’t 2..?

[proud perch]

Does it cross the x axis?

[fading ether]

The graph? No, but the equation f(x)=0 supposedly means that it does

[proud perch]

Thats not what that means

[fading ether]

And I also know that if you put an equation to equal 0, there should be a real answer, not an unequal one
It wouldn’t be irrational, right..?

[proud perch]

The equation f(x) = 0 means that its asking you to find x values that make f(x)=0

If there are no such x values over the reals, then the graph will not cross the x axis

Because f(x) will never be 0 for any real value of x

[fading ether]

So what yourse saying is, this current graph can’t cross the x-axis, so this equation would give irrational roots?

[proud perch]

Would it even give real roots?

Are there any solutions over the reals?

[fading ether]

Sorry I have a smooth brain
When you say the reals, are you referring to the solutions/roots?

[proud perch]

Yeah. Are there real solutions to f(x) = 0?

[fading ether]

Not to this graph

[proud perch]

Then what do we call numbers that aren't real?

[fading ether]

Oh I gotcha
Sorry for having to break it down so far

So it would be real, irrational and unequal?

(Still figuring out the unequal part but I think it means a real and an irrational root)

[proud perch]

Not quite

Didn't we just say that there's no real solutions

As the graph never touches the x axis

[fading ether]

Yup, so that makes irrational roots!

[proud perch]

Maybe you should relearn some of these definitions

Irrational numbers can be real

[fading ether]

Yeah ik that

[proud perch]

Consider the square root of 2

[fading ether]

The only part that’s confusing me is the equal/unequal parts
But Ik that most numbers are real

[proud perch]

You misunderstand completely. Let's define all these terms before you confuse yourself

What is a rational number?

[fading ether]

Any number that isn’t constantly repeating, like i or e (I can’t make the symbol but yeah)

[proud perch]

Bad definition, let's use a better one

A rational number is one that can be expressed as the ratio of two integers

[fading ether]

Ah ok- I wasn’t taught that part haha

[proud perch]

For example:
0/1, -3/4, 7/3

Etc

[fading ether]

Aight 👌

[topaz axle]

it's not just bad it's the opposite

of true

[fading ether]

Good to know 👍

[topaz axle]

e is not repeating, it's always something new

a rational number is repeating

[fading ether]

Ok but pi is

And pi isn’t really rational

[topaz axle]

pi is also not repeating

[fading ether]

my teachers lied to me..
I was told it was basically infinity

[topaz axle]

it's always some new "unpredictable" digit

[fading ether]

Aight so a rational number is everything but irrational
Negatives, 0, repeating decimals and fractions

[topaz axle]

mostly yeah

[hybrid snow]

Pretty much

[fading ether]

so going back to the main problem, the roots of that specific graph cannot touch 0, meaning they’re…not real

Or imaginary..? That’s one of the options

||Again, I apologize that simple stuff like this has gotta be constantly repeated, I’m just weird like that||

Ok that was the right answer

[dim cliff]

Technically yes, a rational number is an exact value that can be expressed as a fraction, ie. p/q where p and q are integers. An irrational number is a number that cannot be expressed as a fraction where p and q are integers. So, the square root of 2 is an example of a non-rational number, since there is no way to represent this number as a fraction. This is something many have tried. Note that it's also the case that this is not a root (as a fraction) of any integer either. The only way to precisely calculate this value is to use the method from calculus known as Newton's Method. Newton's Method is a process for approximating the value of a function based on the values of its derivatives. This allows us a method of calculating the root of functions. With Newton's method, you start with an initial guess and an initial value of the function's derivatives. Then, you can repeat this process until you get close enough to the actual value. This method can be utilized for any function, not just square roots.

[wooden veldt]

Newtons method is definitely not the only way to calculate square roots to arbitrary precision

[fading ether]

I’m gonna loose all my brain cells soon:j

[dim cliff]

I agree, it is just one such method. There is another, much faster, which is known as the Babylonian Method. In this method, we compute the square root as a limit. In this method we can get the square root of 2 to 15 digits in about 10 steps, and a much larger number of significant digits in a few more steps. This is a far faster method than Newton's method.

[wooden veldt]

Babylonians schmabylonians

[fading ether]

Aight so most all numbers/ square roots can be rational if you try hard enough

[dim cliff]

∞ + ∞ = 16

[wooden veldt]

How have you come to this conclusion lmao

[fading ether]

Y’all litterally just took an irrational number, and used two methods to simplify it into a rational one
And if it’s still irrational don’t mind me idk what I’m talking about lmao

[dim cliff]

Hold tight 🏎️ !

The problem is there are more irrational numbers than rational numbers. For example let's look at all real numbers between -1 and 1. There are an infinite number of rational numbers, and also an infinite number of irrational numbers. However there are many more numbers between -1 and 1 that are irrational. This is due to the fact that we cannot simply enumerate all irrational numbers on some sort of sequence. But we can enumerate all rational numbers. The fundamental fact behind this is that we can describe rational numbers by using fractions. Any rational number can be described exactly by some sequence of fractions. On the other hand, we have no way of describing irrational numbers with fractions. All the irrational numbers are simply non-terminating decimals. And we can never represent all the non-terminating decimals with a sequence of fractions. You can prove that there are more irrational numbers than rational numbers by using a proof called Cantor's Diagonal Argument.

[wooden veldt]

Stop, this poor person doesn't need walls of text thrown at them when they're trying to grasp the concept of a rational number

[dim cliff]

But irrational numbers are not rational numbers. The closest rational number we can get to the constant pi is 355/113, and even this is still incorrect. You cannot represent irrational numbers with fractions because these numbers are non-terminating decimals, ie. they are infinitely long and have a non-repeating decimal digits. The point is, this representation that we get out of these methods is not rational. It will be a decimal that is a more accurate approximation than we had before, but still infinitely long and non-terminating, thus we cannot represent these numbers with fractions.

[wooden veldt]

Try having a conversation rather than being an information machine

[fading ether]

Nah it’s okay it’s pretty interesting to learn about
But yeah I mainly just forgot what I learned 2 years ago, it’s not that I completely don’t understand math 😅

[dim cliff]

I try.

[wooden veldt]

The closest rational number we can get to the constant pi is 355/113

This is just not true

[fading ether]

Afk for a minute

[topaz axle]

why reply to scriptod?

[dim cliff]

Gregory-Leibniz series 🤦‍♂️

[dim cliff]

We all forget math, so I want you to consider the length of a curve like the curve of a string. The exact length is impossible to measure since we are constrained in our ability to measure something with an infinite number of degrees of freedom in the real world. However, we can estimate the length of the curve by measuring the length of short segments of the curve. These short segments will always be an irrational number. Therefore, we measure irrational numbers every day in the real world.

[dim cliff]

I would like to apologize for any misconceptions or misunderstanding we have had before, I'm merely a knowledge seeker and there is no knowledge without questions.

[topaz axle]

i don't remember us interacting this is the first time

i decided to try it too

[fading ether]

Aight !!
By curve you mean a parabola, or litterally a single curve? Like an absolute value graph? (I think..)

[dim cliff]

This is correct, and the fact that these lengths can only be approximated is due to the fact they are irrational numbers. When it comes to mathematical objects like curves, we can use the idea of calculus to describe this. A curve like a circle or a parabola will have length, and the value of that length will be given as a definite integral. However if we only measure small parts of these curves, then we will only get approximations of this value. The length of a curve is a non-trivial concept to measure. In mathematical terms, we would say that the length of the curve is the integral of the derivative, ie. its arc length. For some basic curves this can be calculated rather easily (like the case of a sphere or a cylinder). However, the curve for the string requires a bit more effort than you might think. ||A good way of thinking about the string is that we can define it to be the zero curve of the following differential equation: y'' - y = 0 (where y is the curve).||

[fading ether]

Okay that should do for the question, I’ll stay in the channel in case another one pops up

[devout snow]

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[glass bison]

just a quick question

[glass bison]

how do I find the overestimate and underestimate?

they tried explaining but it still doesn't make sense to me

[supple knot]

This is not a quick question

Depends on what method approximation you're using

[glass bison]

approximation?

[supple knot]

Yes.

Approximate area implies a method of approximation

[glass bison]

if ur talking about /\ x, then I use this:

.close

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[devout snow]

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[ebon wyvern]

Lmk if u can help i can also make my work a bit neater if needed

[devout snow]

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[ebon wyvern]

.close

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[left robin]

.

hello

[left robin]

i have a problem about lagrange interpolation

let x1,...,xn be in R and each being different

let y1,...,yn also be in R

show that there exists exactly one polynomial with degree n-1 such that all points p(xk)=yk are in p for all k in 1<=k<=n

i dont really know where to start here

[stone stump]

Suppose there are two such polynomials p,q. What can you tell me about the difference p-q?

And for existence, either solve the linear system of equations or use lagrange base polynomials

[devout snow]

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[devout snow]

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[quaint anchor]

Hey guys

[quaint anchor]

My question is so easy

What's the antiderivative from 4/x?

[sonic smelt]

4ln|x| + c

For some constant c

[devout snow]

@quaint anchor Has your question been resolved?

[devout snow]

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[supple flax]

A coffee factory mixes 3 different mixtures in order
produce special coffee blends. Mixture A contains 5 grams
Coffee, 2 grams of sugar and 3 grams of creamer. Mixture B contains 3 grams
Coffee, 7 grams of sugar and ? gram creamer. Mixture C consists only
of 2 grams of coffee and 3 grams of sugar. How many hunchbacks each -_
each of these mixtures must be combined in order to produce
a special coffee blend containing 21 grams of coffee, 10 grams of sugar and
2? grams of creamer?

[devout snow]

Please don't occupy multiple help channels.

[supple flax]

help

[torn crane]

did you try anything?

[supple flax]

i got the answer

but i didnt know the way

[torn crane]

there seems to be errors in the task, like it was OCR'd

can you make sure it's correct

[supple flax]

wdym ocr?

[rapid atlas]

Optical character recognition, aka maybe it was scanned and translated by a computer.

[supple flax]

oo

wait lemme fix that

[rapid atlas]

It often results in some mistranslation like missing characters and things.

[supple flax]

A coffee factory mixes 3 different mixtures in order
produce special coffee blends. Mixture A contains 5 grams
Coffee, 2 grams of sugar and 3 grams of creamer. Mixture B contains 3 grams
Coffee, 7 grams of sugar and 7 gram creamer. Mixture C consists only
of 2 grams of coffee and 3 grams of sugar. How many mixtures must be combined in order to produce
a special coffee blend containing 21 grams of coffee, 10 grams of sugar and
27 grams of creamer

[rapid atlas]

What about the hunchback thing

[supple flax]

oo wait

[rapid atlas]

I have a feeling that's not right but

[supple flax]

is that right?

[rapid atlas]

Yeah looks fine

I think I'll let vfio finish the explanation though

[supple flax]

okie

you can explain it if you want

if you dont mind

@rapid atlas @torn crane help pls

idk whos the helper

[torn crane]

sorry, I'm done for today. I will be taking my leave. but whoever comes next will enjoy the fixed text

[supple flax]

okie

[devout snow]

@supple flax Has your question been resolved?

[devout snow]

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[devout snow]

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[pulsar ermine]

Can any basis be transformed into the standard basis?

[zenith jacinth]

if its possible to find a matrix that transform the basis into the standard basis, why not

[torn vessel]

if you're talking about a basis for R^n, then yes

if it's a basis for a subspace of R^n, then no

[pulsar ermine]

Thanks, another question I have

They want to prove whether this set is a linearly independent or not

So they assumed that it isn’t independent?

However if a set of vectors is linearly dependent that doesn’t mean that all the coefficients are non zero though

[torn vessel]

it means some of the coefficients can be non-zero, yeah

also that last equation is wrong. the last vector should be subtracted, not added.
The first assumption was delta = -1, not 1

at least I think.

[pulsar ermine]

So how can they assume that delta is -1

Couldn’t it possibly be 0?

[torn vessel]

possibly, but they're using this theorem

since there's 4 vectors you know they're dependent (there's just too many of them), and this theorem says you can write the last one as a linear combination of the preceding ones

[pulsar ermine]

Does this apply to every vector in the set?

[torn vessel]

no, this just says there is some vector you can write as a linear combination of the previous ones.

so, if you have v1,v2,v3,v4 then you can write v4 as a linear combination of v1,v2,v3

You can move things around, and reapply the theorem, like
v4,v1,v2,v3 lets you write v3 as a linear combination of v4,v1,v2

But the way the theorem is stated you can't, for example, necessarily write v2 as a linear combination of v1 from your list v1,v2,v3,v4

[pulsar ermine]

Wait if I’m understanding this correctly, it doesn’t matter what vectors you choose for the order. So the theorem is just saying let any v1,v2,…,vj-1 then you can write Vj as a linear combination if they’re linearly dependent

[torn vessel]

yes

order doesn't matter, but you'll need the j-1 vectors to write the j-th one.

[pulsar ermine]

I see thanks

[torn vessel]

so, in your problem, you need 3 vectors to write the 4th one. whichever one you choose doesn't really matter, but you need 3 of them.

[devout snow]

@pulsar ermine Has your question been resolved?