help-10

[obtuse pebble]

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[cyan tree]

I was trying to do this excercise but I can't think of any other function then sin or cos to have this property

[cyan tree]

can I just say f(x) = sin(x) and then I know the series of it which is

and radius of convergence is infinity

[thin stone]

no, the nth coefficient of the sine expansion doesnt match the n+4th coefficient

[cyan tree]

oh wait u are right its' because of the n!

oh wait is it a geometric series?

oh nvm

[thin stone]

you can try expanding and see what has a common factor of $c_0$, $c_1$, etc

[warm shale]

anonymemes

[cyan tree]

well when i write it out it's c0 + c1x1 + c2x2 + c3x3+ c0x4 .. etc

[thin stone]

okay

[cyan tree]

ooohh

wait

u mean

c0( x1 + x4 + .. )

[thin stone]

yup

[cyan tree]

and also for c1 .. c3

[thin stone]

yup

[cyan tree]

but will that help me

[thin stone]

yes it will

you can get a very useful common factor in the end

[cyan tree]

I have written it out but I still don't know the general formula

[thin stone]

okay. similar to $c_0$, what did you get next to $c_1$?

[warm shale]

anonymemes

[cyan tree]

x^1 + x^5 ...

[thin stone]

okay, try taking an x out

[cyan tree]

for c2 it's x^2 + x^6 ...

[thin stone]

try taking an x^2 out, etc

[cyan tree]

oooh that's so smart

[cyan tree]

the common factorr is that (1 + x^4 + ...)

[thin stone]

exactly

[cyan tree]

is it allowed to say that (c_0 + c_1 x + c_2 x^2 + c_3 * x^3) ( 1 + x^4 + ...) is my general formula?

[thin stone]

no you're not at your general formula yet

[cyan tree]

oh

[thin stone]

the common factor can be made into a sum right?

[cyan tree]

let me see what the sum is

[cyan tree]

x^4 n?

n = 0 to infinity

[thin stone]

yup

and you know the radius of convergence for that

[cyan tree]

uhm let me see

it's just 1

[thin stone]

(and what it equates to)

[thin stone]

yup

[cyan tree]

uhm

oh it equates to geometric series right

(x^4)^n

[thin stone]

yes

[cyan tree]

but then | x^4 | < 1 is my radius of convergence?

oh

and thats 1

[thin stone]

yeah

[cyan tree]

ye

wait so the question is answered right?

[thin stone]

yeah so thats ur radius of convergence

the infinite series becomes a function in this radius right?

[cyan tree]

interrval is just (-1,1 ) btw

[cyan tree]

oh wait ye

1 / (1 - (x^4)

and that I do times the c_0 part

(c_0 + c_1 x + c_2 x^2 + c_3 * x^3) / (1 - (x^4))

is this my general formula?

[thin stone]

exactly

[cyan tree]

it's a little bit long tho

but it's not infinity long atleast

[thin stone]

yeah because of the constants

but infinitely shorter than what it was

[cyan tree]

I can call it my general formula since it's not infinite right?

[thin stone]

yup

[cyan tree]

and uhm this part:
(c_0 + c_1 x + c_2 x^2 + c_3 * x^3)
radius of convergence is just infinity or don't I have to look at that?

[thin stone]

its a finite polynomial, so it has an infinite radius of convergence

[cyan tree]

ye so I take the intersection of radius of convergence right?

which is 1

[thin stone]

yup

[cyan tree]

and the interval is just (-1,1) since thats also for the gemoetirc series

[thin stone]

yes

[cyan tree]

thanks alot for helping

it made it so much easier lol

[thin stone]

you're welcome

[cyan tree]

if u have time can I ask 1 short question which I am confused about

I was doing excercise 9

and I found the power seires representation

which is summation of ((-1)^n * x^(4n+2)) / 2^(4n + 4)

so basically I learned to radius of convergence to do it 2 ways

one is with ratio test

other one is this

which is basically faster way

but it was not working this time

the ratio test was working tho

do u know why?

[thin stone]

one sec

[cyan tree]

ofc

[cyan tree]

to clarify u take x^2 / 16 out to get a geometric series and basically simplify after that

[thin stone]

so why did that one not work exactly?

[thin stone]

yeah i understand, your formula is correct

[cyan tree]

uhm I forgot I did this one like a couple hours ago and I accepted that it didn't worked lol but now that u can help me I thought I might aswell ask 1 sec I will try it again and see why it failed

[cyan tree]

ye so c_n = (-1)^n / 2^(4n + 4)

just to clarify if I did that step correct

[thin stone]

yup

[cyan tree]

ye so from that formula I get R = 16

but with ratio test I got | x | < 2

[thin stone]

ah i see

that R is for x^4

not for x

[cyan tree]

ooohh

so R is radius of convergence for x^an

where a is in this case

4

[thin stone]

R is the radius of convergence for x^n. since you had x^4n, you had to take your R to the 1/4th power

[cyan tree]

ooh this is making sense

[thin stone]

alternatively, changing n to something like k/4 in your sum would make it in terms of x^k

[cyan tree]

and for ratio test u got | x | < 2 so u already have it to x^n

[cyan tree]

ye that one might work aswell

although that prob is a bit more effort

[cyan tree]

I feel like this is the easiest to do

[thin stone]

yeah. i would do the sum stuff tho just incase

[cyan tree]

is that like safer option?

I feel like if I do that isn't ratio test faster?

or about the same speed

[thin stone]

yeah if you wanna be more rigorous

but i think you can just say $|x^4|<R$ and do stuff

[warm shale]

anonymemes

[cyan tree]

I tried it on this one so if I am correct it's the summation of k = 0 to infinity
(-1)^k/4 * x^(k+2) / 2^(k+4)

[cyan tree]

ye this one seems fastest to me tbh

[thin stone]

also make sure to check the sum manually for + and - R because the tests dont test those

[cyan tree]

ye I've rmeembered that

[cyan tree]

but if I do this

will it go wrong?

because this seems like fastest way to me

[thin stone]

i dont think it will but i can check

[cyan tree]

I know lim n to infinity with c_n / c_n+1 is just a derivation with the ratio test for power series with x^n is that the reason why it may go wrong?

[thin stone]

yeah it shouldnt go wrong if you know what your x powers are

anyway like you said its basically the ratio test with a mask on so do whatever you like

[cyan tree]

ye so if u got smth like x^12n + 25 with that formula I get
|x^12| < R and then just take both to 1/12th power

[thin stone]

yea

[cyan tree]

ye tbh I can always do the ratio test as a doubel chcekc it's just that I found this one faster

[thin stone]

yeah double checkin is a good idea

[obtuse pebble]

@cyan tree Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[tawny kite]

not sure about this section

not sure how they got 2b1 from

because the only equation I see that relates is this one, so b1 = 2b2-2a2+a1^2

[unreal musk]

From here you'd have $a_{k+1} < b_{k+1}$ and also $b_{k+1} \leq b_k$

[warm shale]

@unreal musk

[unreal musk]

One line below

[tawny kite]

hmm??

[unreal musk]

Cause you have $a_{k + 1} < b_{k + 1}$ you get that the $\frac{b_1 - a_1}{2^k} \qty(b_{k + 1} + a_{k + 1}) < \frac{b_1 - a_1}{2^k} \qty(b_{k + 1} + b_{k + 1} ) = \frac{b_1 - a_1}{2^k} \qty(2 b_{k + 1})$

[warm shale]

@unreal musk

[unreal musk]

The fact the intervals are nested forces $b_{k + 1} \leq b_k$ for each $k$, so you can iterate that backwards to then get $\frac{b_1 - a_1}{2^k} \qty(2 b_{k + 1}) \leq \frac{b_1 - a_1}{2^k} \qty(2 b_{k}) \leq \frac{b_1 - a_1}{2^k} \qty(2 b_{k-1}) \leq \ldots$

[warm shale]

@unreal musk

[unreal musk]

[you're happy with why this is the case at least, right?]

[unreal musk]

if you iterate this enough you get to b1 in there

[tawny kite]

yes, visually, b2 and a2 with the two diff cases would be the midpoint of b1 and a1

like binary search in programming

thank you!

I'll prob just stay in this channel

working on lots of proofs today

[unreal musk]

Awww

Sure thing, and if you'd like me to check anything, feel free to ping me

[pallid knot]

Can someone help me

[unreal musk]

!occupied

[obtuse pebble]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

@tawny kite Has your question been resolved?

[obtuse pebble]

@tawny kite Has your question been resolved?

[tawny kite]

@unreal musk I dont really get the sup = inf part

sup is least upper bound, why is it saying r<x?

least upper bound where r is less than x?

[unreal musk]

They're kinda saying "take the all rational numers that are less than x, and work out a^r, and make that a set...

...from that set, take the least upper bound"

"kinda" like you're taking larger and larger rational numbers getting closer and closer to x to work out a^x via the a^r's

You know how to work with rational powers, so define the irrational powers via rational powers, if that makes sense?

(similarly, the "inf" part is saying "take all rational numbers r greater than x, work out a^r and put those in a set, and of that set, take its greatest lower bound)

[tawny kite]

Im actually gonna go back to the original question I asked

I was trying to explain it to my family haha just for fun

and I realized I have no idea how the last expression we found relates to a-b^2. The last sentence proving existence

waitttt

actually no

why would finding a sufficiently large k be a contradiction?

yes we know that bk+1 ^2 -ak+1^2 is less or equal to b1/2^k(2b1)

but there is nothing that says it cant be bigger than |a-b^2|

[unreal musk]

That difference is supposed to vanish basically (because of that 2^k in the denominator)

[unreal musk]

But if you had this, any other larger k's would also be bigger than |a - b^2| and means that it wouldn't be able to vanish

[tawny kite]

ohhh

so because its b1/2^k its going to 0, indicating that bk+1^2-ak+1^2 is '0'

so statement contradicts

[unreal musk]

Yep

[tawny kite]

yea! thank you thank you 🙂

[unreal musk]

@tawny kite Gonna get a bit of sleep now

[tawny kite]

ok!

[unreal musk]

Most likely will catch ya later

[tawny kite]

yup!

good night!

[obtuse pebble]

@tawny kite Has your question been resolved?

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[timid silo]

hello

[timid silo]

i need help with this

to determine which one is f(x) and which one is the derivative

[stuck sinew]

hi

bro

[timid silo]

HI

[stuck sinew]

i nees help bro

[timid silo]

bro dis my channel

[stuck sinew]

in maths

[minor jungle]

I'll give you a hint: f'(x) = 0 when slope of f(x) is 0

[stuck sinew]

hi zap

[timid silo]

#❓how-to-get-help

[timid silo]

ahh that makes sense

but what if it was a scenario where there wasnt a horizontal tangent

More than likely you will rarely encounter that, You will just need to get good with seeing the relationships between the graphs

alr thx

.close

[obtuse pebble]

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[timid silo]

https://webspace.ship.edu/msrenault/geogebracalculus/derivative_try_to_graph.html

I found this interesting site if you want some practice (These seem pretty challenging though)

[obtuse pebble]

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[wide pewter]

So im essentially wondering the best way of figuring out a function from the following data set:
1/0
2/0
3/0
4/0
5/0
6/0
7/3
8/4
9/4
10/5
11/5
12/6
13/6
14/6
15/7
16/7
17/8
18/8
19/9
20/9
21/9
22/10
23/10
24/11
25/11
26/12
27/12
28/12
29/13
30/13
31/14
32/14
33/15
34/15
35/15
...

The pattern is repeating, and X ∈ W, this isnt from a specific question in some homework, i am just curious on what would be the best method of trying to define a function for this thing.

[green shuttle]

huh, what is the data?

[timid silo]

Are you wondering if there is a way to find the relational properties of those fractions?

[wide pewter]

When x ∈ [0,6] = 0
And when x is 7+ theres a pattern

My goal is to turn this data into a function, probably using floors, my question is how do i go about it? What would be the method? Just throwing out functions and seeing what sticks?

[wide pewter]

They arent fractions, i shouldve formatted them better. These are x and y values, so like (7, 3)

Apologies for the confusion

X also belongs to W, as in whole numbers. It cant be negative or a fraction, nor can y

Id explain where im getting the data but its a long story

[green shuttle]

how exactly are they repeating? are we just ignoring the beginning and assuming the 2 2 3 repeating continues?

like 2 13s, then 2 14s, then 3 15s

[wide pewter]

Yes, we're assuming

The single 3 in the beginning can be chalked up to an irregularity

Then we can assume the pattern is 2 3 2

Or if we need to (which i kinda assume we do) chalk up the first 5 into irregularities and start at (15, 7)

So that we get 2 2 3

[green shuttle]

ok i have an idea it will take me some time to set up and check though

[wide pewter]

Thank you, i appreciate it

Id love to learn how to get functions from data sets like these

[fluid holly]

maybe this?

[wide pewter]

But does that work as a function?

Id have a hard time getting an output from that if i inputted like, 1034

[fluid holly]

yes i think, it is a function

[wide pewter]

I honestly couldnt tell ya if it is, im not sure how the k works into the equation

[green shuttle]

my idea won't have any if statements if it works

[fluid holly]

1034/7=147.......5

so k = 147 and satisfied the third one

f(1034)=444

[fluid holly]

as i restricted that k is a integer, every x would, and would only satisfy one of them

[wide pewter]

I see

[fluid holly]

but the domain of this func only contains integer and it's make sense, that's enough for this question

[wide pewter]

The domain only includes whole numbers though

[fluid holly]

wdym numbers

[wide pewter]

As in the set of whole numbers..?

0, 1, 2, 3..

No negatives

[fluid holly]

natural number u mean?

[wide pewter]

Sorry i havent slept, yea i think im being stupid

Or am i?

Natural is 1, 2, 3

Whole is 0, 1, 2, 3

Integers are with negatives

Then rational and irrational

Which are all real

No?

[fluid holly]

natural number includes 0 i think

[wide pewter]

I dont remember tbh

[fluid holly]

anyway, any questions?

[wide pewter]

Not really no, thank you :)

[green shuttle]

i think i have a computer friendly answer

actually has an issue but i think it's almost ready

lol ok

$f(n) = \left\lceil \frac{n \operatorname{mod} 7}{2}\right\rceil - \left\lfloor \frac{n \operatorname{mod} 7}{7}\right\rfloor + 3\left\lfloor \frac{n-1}{7}\right\rfloor - 3 \left(\operatorname{sign}(n \operatorname{mod} 7)-1\right)$

[warm shale]

soulgazer

[green shuttle]

this generates 1,1,2,2,3,3,3, 4,4,5,5,6,6,6 etc

can shift it accordingly to match the data

so....

$f(n) = \left\lceil \frac{(n-7) \operatorname{mod} 7}{2}\right\rceil - \left\lfloor \frac{(n-7) \operatorname{mod} 7}{7}\right\rfloor + 3\left\lfloor \frac{(n-7)-1}{7}\right\rfloor - 3 \left(\operatorname{sign}((n-7) \operatorname{mod} 7)-1\right) + 3$

wait

[warm shale]

soulgazer

[green shuttle]

i think that should have f(8) = f(9) = 4, f(10) = f(11) = 5 etc

[obtuse pebble]

@wide pewter Has your question been resolved?

[wide pewter]

That is so fuckin crazy im ngl

I thought itd be a lot simpler with using some dividing and floors but wow

Bro i dont even know what mod does, that is so cool

[green shuttle]

i'm sure it could get simpler

[wide pewter]

Thats still so cool

Youre a legend

I think you deserve to know the context

Although its so stupid you might strangle me for it

[green shuttle]

don't worry i wouldn't strangle anyone

[wide pewter]

In a server im in, big public one

They have a bot that deletes messages with too many caps

But

You can offset it with some small letters

So if i went

YABA DABA DOO

Thats 11

Itd get deleted

But

If you made it

YABA DABA DOO yknow?

Itd be fine, x = 11 and y = 5

X is the letters that are capitalized, and Y is the ones that arent

I was just so perplexed and intrigued on what basis it was getting rid of messages on

And wanted to know if i could reverse engineer the system

I was initially just gonna look for help on gettin started as my best method to go about it wouldve been making up functions, testing em, and then makin more if it didnt work

[green shuttle]

haha what's weird about being curious about that

[wide pewter]

Its out of the blue and so gut wrenchingly stupid lmao

Its a frickin discord bot and if i was so curious, i could probably just ask whoever made it

But either way, thank you man. This was awesome

[obtuse pebble]

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[green shuttle]

i'm gonna guess it doesn't run on this

[wide pewter]

Probably not lmao

[green shuttle]

but it might as well

also that won't work for n < 8

but those are special cases anyway

[obtuse pebble]

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[scarlet anchor]

How to know if these 2 are congruent?

[candid yarrow]

A good way is to try to draw them

[obtuse pebble]

@scarlet anchor Has your question been resolved?

[obtuse pebble]

@scarlet anchor Has your question been resolved?

[scarlet anchor]

.close

[obtuse pebble]

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[obtuse pebble]

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[graceful marten]

Part A,
Everytime I do such a question

I get confused why the formula

sn = n/2 (a + L) doesnt work L being last term

a (first term) is k+1

d (constant) is k+2

Why does

sn = n/2 (a + L)

303 = n /2 ( k+1 + 303)

x2

606= n (k +304)

606/ k+ 304 = n is wrong

why is that wrong?

The correct working out is

sn = a + (n-1)d

303 = k+1 + (n-1) (k+2)

303 = k+ 1 + kn + 2n + -k -2

303 = kn + 2n -1
303 = n(k+2) -1

304 / (k+2) = n

How is my frist method wrong

Im confused ty ty

[obtuse pebble]

@graceful marten Has your question been resolved?

[graceful marten]

I think i understood why kinda?

Bc the first one is the # of terms to make the sum 303?

.close

[obtuse pebble]

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[graceful marten]

.reopen

[obtuse pebble]

✅

[graceful marten]

I have a Question

So A,
Is done like this
sn = a + (n-1)d
303 = k+1 + (n-1) (k+2)
303 = k+ 1 + kn + 2n + -k -2
303 = kn + 2n -1
303 = n(k+2) -1
304 / (k+2) = n

If I were to be asked,
Q2. How many terms would you need to make 303

Would that also be done through the same method as the above? (A)

And If I were to be asked
Q.3 the sum, of terms to 303
Would it be done like this?

sn = n/2 (a + L)

303 = n /2 ( k+1 + 303)

x2

606= n (k +304)

606/ k+ 304 = n

[stiff gull]

@graceful marten

[graceful marten]

Yeah?

[obtuse pebble]

@graceful marten Has your question been resolved?

[sleek ether]

Yes for a it's done like that

For question b
Use the formulas sn = n/2 (a+L)
from question a you found out the value of n and L is given in the question

sn = (304/2(k+2)) * (k+1+303)
sn = (154/k+2) * (k+304)
sn = (152k + 46208)/(k+2)

Hence proved

For question c

you have founded the value of Sn in terms of k in the question b

[graceful marten]

Yeah C->
2568 = QB
k = 17

[sleek ether]

sn = (152k + 46208)/ (k+2)
Sn = 2568 ( given in question)

So

[sleek ether]

Yep

[graceful marten]

Since your here I hope you dont mind

[sleek ether]

?

[graceful marten]

I’m a bit unsure where I went wrong

What I did is use the Last n'th term being the
un = ln9 ect

And throw it in

Sum of first n terms sn= n/2 (a +L)

A being the first term, and L being the last term

Last term being (un) in question

Unless that is me understanding it wrong (un) is not the last nth term

[sleek ether]

Wait let me check it

[graceful marten]

I found my mistake

9^2 into 3^ ?

[sleek ether]

Oh my answer came out too

[graceful marten]

Same thing as I ?

[sleek ether]

No its

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

a is 1/2

mentioned in question

[graceful marten]

yeah

thats correct

I found my mistake after 30 min of chat gpt and doing it 3 times

[sleek ether]

It's just a Lil bit tricky

[graceful marten]

I did ln(9^2 x 3^n)

[sleek ether]

Dude 💀

[graceful marten]

and 9^2 is 3^2 x 2

For somereason I thought

3^3 = 9

[sleek ether]

2*

[graceful marten]

so I did 3^3x2)

3 +2 = 5

[graceful marten]

Yeah

Do you do 2+ 2

Or 2 x 2

[graceful marten]

ok

[sleek ether]

a^m * a^n = a^(m+n)

[sleek ether]

Is your question resolved now?

[graceful marten]

yep

Haha

Thank you bud

[sleek ether]

My pleasure

[graceful marten]

good day to you

Im going to eat something now

Super hungry

Thank you

[sleek ether]

Same

[graceful marten]

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[graceful marten]

Im having a problem with maths,

[graceful marten]

Instead of using log, i tend to use ln (just bc it confuses me less)

Is it alright if I get all the working out to

5 ln p + ln 5 = 0

and then convert it to 5 log p + log 5 = 0

[Bc the question asks me to prove you get ' 5 log p + log 5'

Can I do that for all questions

[obtuse pebble]

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[dim moon]

Does "a number of the form 2^m * 5^n" include 30?

[high lily]

consider the prime factorisation of 30

[sleek crater]

You need a factor of 3

[dim moon]

Because 30 would be 2^1 * 5^1 * 3 but 3 is not mentioned in the given form of the number

[dim moon]

So 30 wouldn't be of the form 2^m * 5^n?

[high lily]

can you multiply any number of 2s and 5s together to get 30? (without using 3)?

[sleek crater]

No I dont think so

[dim moon]

No

[high lily]

so 30 can't be expressed in that form (assuming m,n are integers)

[dim moon]

I saw a question saying this-:

Assertion: 6^n will never end with the digit 0 where n is a natural number
Reason: Any number ends with digit zero, if its prime factor is of the form 2 ^ m × 5 ^ n, where m and n are natural numbers.

I know that the Assertion is true

But the reason should be false

But everywhere on google it is being said that the reason is false

[sleek crater]

If a number ends with zero it must have a factor of 5 and 2 so whats the problem

[dim moon]

And the thing is that teachers would blindly follow the answer on google and not listen when someone has a question

[sleek crater]

Oh I see they said no more factors

?

[dim moon]

Yeah

[sleek crater]

No ig they wanted to say at least 5 and 2

[high lily]

where'd that reasoning come from

[sleek crater]

What reasoning

[high lily]

Reason: Any number ends with digit zero, if its prime factor is of the form 2 ^ m × 5 ^ n, where m and n are natural numbers.

[sleek crater]

It should be multiple of 10

But it could have other factors apart from 2 and 5

Thats the problem

[high lily]

was that part of the question? were you being asked whether this was a valid argument?

[sleek ether]

Yes

[dim moon]

They're asking us whether the Assertion and reason are true or false and if they both are true, is the reason a correct explanation for the Assertion

[sleek crater]

They should have written 'at least' or something like that

It could have more factors obviously

[high lily]

yeh, so like you said
the assertion is true
but the reasoning is invalid

[dim moon]

The problem with these types of questions is that sometimes teachers say you're thinking too much about it and sometimes they say we have to find technicalities and mistakes in the statements

[high lily]

note that the implication only goes in one direction

prime factor is of the form 2 ^ m × 5 ^ n, where m and n are natural numbers
results in a number ending in 0

but doesn't work the other way

[dim moon]

👍

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[fierce elbow]

Idk why am I stuck on this

Im sure its easy but I just dont know why cant I think of the way to do it

[oak bear]

I think this is bayes theorem?

[fierce elbow]

Havent heard about that thing

I thought making diagram would help but I cant find the stuff union intersection and uhh

[oak bear]

you should probably check 3blue1browns video about bayes theorem cause I think this is bayes theorem

[fierce elbow]

The one on probability

?

[oak bear]

you can probably think of it as an area of a rectangle rather than a venn diagram

yeah bout bayes theorem

the question is really hard to grasp if you only think of it as percentage, much better if you make it to actual quantifiable numbers

[fierce elbow]

no idea what i did

but I couldnt make any use outta it

now talking about bayers lemme go check it

[oak bear]

what is 25% of 40?

its 10

[fierce elbow]

ye

what abt it tho?

[oak bear]

20% of 40

It's its 8

[fierce elbow]

I think

[fierce elbow]

what is 8

?

yo I think I realised smt

[oak bear]

it says maximum students who can either not sing or not dance

[fierce elbow]

total number of students - ( students who can sing and not dance)

GOT IT

40 - 38

2

yeahh

thx for ur help anyway 🙂 @oak bear

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[cobalt scaffold]

Hello, how do I continue this?

[obtuse pebble]

@cobalt scaffold Has your question been resolved?

[obtuse pebble]

@cobalt scaffold Has your question been resolved?

[tardy epoch]

what's the question

[cobalt scaffold]

Resolve the system

[obtuse pebble]

@cobalt scaffold Has your question been resolved?

[gleaming echo]

i can help kindly message

[cobalt scaffold]

Oh okay, thanks

[obtuse pebble]

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[candid vessel]

honestly havae no idea where to start.. ive missed a couple of lectures and i feel really behind 😭

[void phoenix]

do u want help on how to do it

or do u just wanna know the answer

[candid vessel]

both i think

[red ice]

The expected value is 1/lambda

[candid vessel]

i want to know how to do it and if im doing it right

[red ice]

So lambda = 1/80

The probability density function is lambda e^(-lambda x)

So just integration

Also the exponential distribution is memoryless

It doesn't depend on previous events

So for b the time would start again at 0 minutes at 9:25 am

And tricks like P(at least 1) = 1 - P(none)

[candid vessel]

how come we're using the probability density function and what does thjat function mean

[void phoenix]

Probability density is a term used in statistics for continuous random variables. Since these variables can take on infinitely many values, the chance of any specific value occurring is usually zero. Instead, we focus on the likelihood of the variable falling within certain ranges or intervals.

[candid vessel]

so exponential distribution is a type of

probability density function

and why is lambda 1/80

and what does mean mean 💀

i feel like i dont understand anything

oh wait

average time it takes to finish the test is 80?

so 1/80 * e ^ (-x/80)

find the probability that x is less than 40?

[candid vessel]

so for the problem that asks P( at least 1 finishes before 40 ) i should do 1 - P( everyone finishes at or after 40 min) ?

or no one finishes before 40

[red ice]

Yeah so no one finishes before 40

And then 1 minus that

[red ice]

Yep

[red ice]

Yeah so you need to integrate

[candid vessel]

do u have to multiply it 5 times since theres 5 students

like ^ 5

[red ice]

Yeah

[obtuse pebble]

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[candid vessel]

.reopen

[obtuse pebble]

✅

[candid vessel]

wait @red ice would the integral be from 0 to 40?

1 - int from 0 to 40 of [ (1/80 * e^-t/80) ^ 5 ]

or would it be from 40 to infinity

[obtuse pebble]

@candid vessel Has your question been resolved?

[candid vessel]

and if it starts at 0 again does that mean itll be 60 to inf

[red ice]

Nah 0 to 35

And 0 to 40 for the other one

[candid vessel]

ty

submitted at 11:59

due at midnight

💀

ur the goat south

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[indigo folio]

how do i do this

I found 'a' which is 2, and in 12b it says it's negative so c = -2

but i dont get how to find d

book says y=-2sin(x-210) but i wrote y=-2sin(x-30) since it moved 30 units right from the original position???

i get that 180+30 = 210 but why

hello?

[candid yarrow]

It is because you can think of it as flipping the function and shifting it half a cycle

[indigo folio]

wouldnt the red one be y=-2sin(x-180) then?

[indigo folio]

too complex for me to imagine

[indigo folio]

right

d is 210 why?

because of what you explained or?

[candid yarrow]

As you noticed, 180 + 30 = 210

[indigo folio]

ah

[candid yarrow]

adding 180 to 30 because 180 is half a cycle

[indigo folio]

right

[candid yarrow]

And reflecting the sine wave does the same thing as shifting it half a cycle

[indigo folio]

thanks

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[wicked ice]

How do i do this?

[versed stratus]

!original

[obtuse pebble]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[wicked ice]

thats it

like adding them together

[sleek ether]

Do you want to simplify it?

[wicked ice]

yeah

[safe haven]

distribute the brackets

then add like terms

[sleek ether]

Yep

[wicked ice]

how do i add

[brazen gorge]

they all have the same denominator

so just add regularly

[sleek ether]

Ye

[brazen gorge]

$\frac{5^k + 5^k \cdot 4 -1 }{4}$

[warm shale]

artemetra

[brazen gorge]

@wicked ice does that make sense?

[wicked ice]

what would i do after?

[sleek ether]

Hmm

[brazen gorge]

5^k + 5^k * 4 simplifies

[wicked ice]

i cant really rememeber my indices on how that works

[brazen gorge]

let x be 5^k

x + x*4

can you simplify that?

[wicked ice]

im not really sure how to

would u just factor out the x

Soo how do i solve my orignal question

.close

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[arctic heron]

good evening

[wraith trout]

hey

[arctic heron]

is there a simple way to solve the limit as x -- > 1 of x/(x-1)

[wraith trout]

(x-1)(x+1)

[arctic heron]

heh

[obtuse pebble]

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[obtuse pebble]

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[lethal cobalt]

@everyone hi guys im ishowtoilet

[lethal cobalt]

@everyone

@everyone

@everyone

@everyone

@everyone

@everyone

@everyone

@everyone

@everyone

@everyone

@everyone

@everyone

@everyone

@everyone

[timid silo]

<@&268886789983436800>

[lethal cobalt]

chill chill chill

@everyone

@everyone

@everyone

@everyone

@everyone

@everyone

@everyone

@everyone

@everyone

@everyone

@everyone

dont ban me

[jovial path]

wtf

[plain stag]

ok i won't ban you

[warm canopy]

.close

[timber pine]

you know it doesnt work right

[obtuse pebble]

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[balmy mortar]

🤡

[grand coyote]

Lol

[obtuse pebble]

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[desert river]

hi, I need help with this differential equations problem

[wise tangle]

The general solution to this linear system is the following:

x₂(t) = −x₁(7) + 4t + e

The equation for x₂ takes x₂(t) and subtracts x₁(7), then adds 4t and e to get the linear function. You can then substitute any specific value of t to get the corresponding x₂(t) value for that t.

https://tenor.com/view/proud-so-gif-24294972

[desert river]

Well actually I didn't understand how you get there

[wise tangle]

uhh wait

The first step is to express the given terms as a linear equation. You start by writing both equations in terms of the same variable:

dx/dt = x

dt/dt = 1

Now, you can combine these two equations by adding the first equation to both sides of the second equation:

dx + dt(dt)

= x + t

This results in a single equation that contains all of the given terms. You can then rearrange the equation to find the general solution

NOW?

[desert river]

Oh, alright I get it now, thank you!

[wise tangle]

Well well well

good

[desert river]

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[serene elk]

Ive worked DBO AND BDO to be both 90-x, how do i get BCD in terms of x

[hollow dove]

angles in a triangle add up to 180, so (90-x) + (90-x) + BCD = 180 and you can rearrange

ah wait, I misread which angles you have

this angle is x, since it's subtended by the same arc

then opposite angles in a cyclic quadrilateral add to 180

so BCD = 180 - x

[serene elk]

Thank you

. Close

[hollow dove]

.close

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[smoky onyx]

ive been stuck on this

for awhile

not sure if im doing the steps correctly

[fervent cradle]

i don't understand how you get this?

[smoky onyx]

thats the part i sort of just winged

and wsnt sure of

my goal was to get ab | cd

[fervent cradle]

d is the GCD

[smoky onyx]

Yeah

[fervent cradle]

ax + by = d

[smoky onyx]

i could substitue it for d

right

[fervent cradle]

so you're saying ab | d

[smoky onyx]

Yeah

exactly

[fervent cradle]

but the small thing goes on the left

3 | 6

not 6 | 3

[smoky onyx]

Oh thought that only applies to single values

but yeah that makes more sense

[smoky onyx]

the next step

from this is somehow incorperating c

on the RHS?

[fervent cradle]

yeah

you need the RHS to be big

so you want cd in there from the start i think

i gtg

[smoky onyx]

OH

idg it tbh

but yeah all good

if u gtg

[smoky onyx]

mind quickly hinting on

b4 u go

[fervent cradle]

off the top of my head, consider a/d and b/d and give them names

and like c/a and b/a maybe also

[smoky onyx]

a doesnt for b, wasnt given b / d either

was a / c and b /c

[obtuse pebble]

@smoky onyx Has your question been resolved?

[smoky onyx]

no

[smoky onyx]

really quick, sorry

is this what you meant

close

!closed

.close

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[graceful marten]

Im having troubles understanding some basic? math rules

ok so,

They first expanded the brackets to cancel out the
1-r^2 on both sides

Becuase if you just cross multiplied at the beggining, you wouldnt get a correct answer

[pine sail]

You would.

[graceful marten]

1-r^4 = 1.16(1-r^2)

That would give you the correctg values of r?

[pine sail]

It would give you 4 values of r. Two of them would be +-1, which you can't accept.

[graceful marten]

so
1-r^4 = 1.16(1-r^2)
1-r^4 = 1.16- 1.16r^2
-r^4 -1.16 +1 - 1.16r^2

really?

[pine sail]

Last line is wrong

(There is no equality)

[graceful marten]

oh ok I think thats whats confusing me ^

Bc I usually get x^2 + x + 4 and u would get x values

But here sometimes you get no values

or all imaginary

or just the wrong ones

How would I do it then?

Lets say from here -->
-r^4 -1.16 +1 - 1.16r^2

[pine sail]

There are no "wrong ones"

[pine sail]

You'd let r^2 = x and solve the quadratic.

[graceful marten]

oh

[pine sail]

But you have to note, if you get x = 1, that's not acceptable.

I hope you see why

[marsh geyser]

still wrong in your sign

[pine sail]

yeah there is no equality to begin with

[marsh geyser]

you had an equation and in the end you don't have an equation

[graceful marten]

Ok i think i understand

So if I had something like

30.5 = ( 8 (1-r^3) ) / (1-r)

30.5 = ( 8- 8 r^3 ) / (1-r)

30.5 (1-r) = 8- 8r^3

30.5 - 30.5r = 8- 8r^3
30.5 - 30.5r - 8 + 8r^3

[marsh geyser]

again

where is your equation

[graceful marten]

what do you mean?

[marsh geyser]

you had an equation and the last line is not an equation

your equal symbol just dissapeard for some reason

[graceful marten]

ok = 0 ?

[marsh geyser]

yes

[graceful marten]

30.5 - 30.5r - 8 + 8r^3 = 0

ok but then im confused again

oh the equation works

Wait waht one second

Ok no this one works

Ok let me start again

sorry lol

Solving this

1-r^4 = 1.16 ( 1-r^2)

1- r^4 = 1.16 - 1.16r^2

1-r^4 - 1.16 + 1.16r^2

-r^4 + 1.16r^2 - 0.16

Oh Ok nvm

I understand

---

I just managed to get both wrong somehow

Sorry for this

Thank you anyways

This took me 4hrs -.-

Great

.close

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[fresh bay]

im not sure what to do here

[obtuse pebble]

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[coarse compass]

the only thing I can find is k values pls help me I dont know how I am going to find the time of his death :(

[muted sparrow]

i dont have an answer but may i ask what this subject is

this intrigues me a lot

[coarse compass]

ib math aa hl

[muted sparrow]

theres more formulas you have to use with this i presume, i tried to do this but i cant since there's too many missing variables

good luck 🙏

[coarse compass]

I thinkkk i need to use some integral stuff but i am not sure lol

[coarse compass]

thxx

[twin sandal]

your first goal should be to find the value for C

[coarse compass]

i gave 70.00 F for C

[twin sandal]

oh I didn't notice that

then you need to find k

[coarse compass]

I hope that's true

[twin sandal]

that's correct yeah

[coarse compass]

yeah I also found k values but 2 of them are different due to the temperature and time

lemme send you my work real quick

that will help :)

[twin sandal]

good idea lol

[coarse compass]

here you go

p.s. I am sorry I wrote "time of death" instead of "normal temp"

and i should have considered the - sign before kt

[twin sandal]

honestly that seems like an error in the problem, not you

you shouldn't end up with 2 different k values

your work was fine tho

[coarse compass]

thxx

yeah that messed my mind lol

[twin sandal]

oh wait hold up

I think the issue is how you're setting your times

think about it this way: T(0) should be 98.6. what you're doing is saying T(9) = 80 F

and T(10) = 75

[coarse compass]

OHHH

[twin sandal]

so I would make a variable for the time of death, something like t_d

[coarse compass]

yeah thats pretty logical however should i use some integral stuff

bc we were learning differential equations at the time

they gave us this question

[twin sandal]

there would be an integral but they already solved it for you lol

T is the solution to Newton's cooling equation

which is a Differential eq

[coarse compass]

easy then

[twin sandal]

hopefully :)

[coarse compass]

OH YEAH YOU ARE RIGHT

THE REAL FORMULA WAS THE DERIVATIVE OF THIS ONE RIGHT

[twin sandal]

I think it's like dT/dt = C - T

or something

[coarse compass]

tysm u saved my life

[twin sandal]

any time :p

[coarse compass]

.close

[obtuse pebble]

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[fluid snow]

Given these values, determine a line with the nearest fit to them.

[fluid snow]

My lecture notes have this about it, so I guess we need to use that

Apparently the solution is a_0t + b_0

How do we determine a_0 and b_0 though?

[obtuse pebble]

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.close

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[fluid snow]

re-asked in #help-12

[obtuse pebble]

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[bleak skiff]

hello

[timid silo]

hi

[bleak skiff]

Show that if x is nilpotent in A, then 1-x is invertible in A. A ring

i tried some things and ended up with xy = y - 1 if y is the inverse of 1-x but i dont think that's how it should be done

[obtuse pebble]

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[civic bone]

hi

[civic bone]

confused on this last part of a proof

https://i.imgur.com/FvuYaDB.png

we need to choose an $N$ so that $2 X 10^{-N} \leq \epsilon}$, the way the book chose it without logarithms resulte din choosing $N \geq \frac{2}{\epsilon}$

[warm shale]

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

[twin sandal]

what's the actual problem itself?

[civic bone]

the way I did it is $10^{-N} \leq N \leq 10^N \leq \frac{\epsilon}{2}$ so I can just choose $N \leq \frac{\epsilon}{2}$ mine also seems correct is this valid?

[warm shale]

nchoosek

[civic bone]

doesn't really matter I basically have an algebra question at the end lol

[twin sandal]

oh

[civic bone]

this is both for any $N \geq 1$

[warm shale]

nchoosek

[twin sandal]

how did you get that $10^N \leq \varepsilon/2$ ?

[warm shale]

doaby

[civic bone]

oh did I use circular logic here?
I had
$2 \times 10^{-N} \leq 2 \times 10^{N}$
so I just chose $2 \times 10^{N} \leq \epsilon$

[warm shale]

nchoosek

[civic bone]

... but I didn't choose it without using logarithms

like explicitly

but even without relating epsilon yet

we have

$2 \times 10^{-N} \leq 2N \leq 2 \times 10^{N}$

[warm shale]

nchoosek

[civic bone]

so we can just choose $2N \leq \epsilon$ now without using logarithms

[warm shale]

nchoosek

[civic bone]

and this is still different than the book

[obtuse pebble]

@civic bone Has your question been resolved?

[civic bone]

i mean yeah there is no way this is wrong

.close

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[warm shale]

Da Mucky Boi

[tardy epoch]

Did you try partial fractions

a^2 - b^2 = (a-b)(a+b)

Show

[warm shale]

Da Mucky Boi

[obtuse pebble]

@timid silo Has your question been resolved?

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@timid silo Has your question been resolved?

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[pulsar pawn]

need help with (a)

[polar fossil]

you probably have a formula for area of triangle based on cross product

i think it's like $\frac12 ab \sin\theta$

[warm shale]

hayley!

[pulsar pawn]

yeah but how do i think theta

find

[obtuse pebble]

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[pulsar pawn]

.close

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[agile saffron]

i need help wih this probabilty thing

[obtuse pebble]

@agile saffron Has your question been resolved?

[agile saffron]

<@&286206848099549185>

[twin sandal]

the easiest way to go about this is 1 - P(nothing matches)

[agile saffron]

what

[twin sandal]

$P(X \geq 1) = 1 - P(X = 0)$

[warm shale]

doaby

[twin sandal]

have you seen this?

[agile saffron]

nope

[twin sandal]

in other words, if you wanna know the probability of something happening at least once, it is the same as 1 - the probability it happens 0 times

which is usually easier to do

[agile saffron]

i don't really get it

[twin sandal]

lemme make up a problem then. let's say there's a football game. the probability that team A scores at least 1 goal during the game is 90%

what's the probability team A scores no goals at all?

[agile saffron]

10%

[twin sandal]

exactly

and you got that from doing 1 - 0.90

[agile saffron]

yeah?

[twin sandal]

that's exactly the formula I just described

it's the same thing

[agile saffron]

so the chance of the other thing happening is 1- minus the probability

[twin sandal]

the chance of Beti getting at least 1 one of the same numbers is 1 - the probability she gets none of the same numbers

[agile saffron]

the answer was different tho

what

pls help

<@&286206848099549185>

bruh

plsss

<@&286206848099549185>

can anyone explain to me

[agile saffron]

bruuhhhhh

[agile saffron]

.close

[obtuse pebble]

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