help-38

[south siren]

Just replace x by -x

Same thing to prove (f(x) - f(-x))/2 is odd

And now you got two functions, one even and one odd, and their sum is f(x)

[gusty plank]

ah i see i get it now

thank you

[south siren]

U welcome

[gusty plank]

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[glacial garden]

hey can someone help explain pls
lim x+sin2x / tan3x
x -> 0

[prime lynx]

h?

[trim lichen]

did you mean x -> 0? @glacial garden

[glacial garden]

Oh yeah sorry!!

[fleet bear]

I'd try small angle approximations

[trim lichen]

so $\lim_{x \to 0} \frac{x + \sin(2x)}{\tan(3x)}$ is what you've got then yes?

[solid kiln]

ann.in.a.teacup

[glacial garden]

yes!!

[trim lichen]

ok do you have any progress on this so far?

[glacial garden]

i did earlier but its somewhere in my notebook of attempted calc problem scribbles lol

i thought i had an idea but can u walk me through it?

[trim lichen]

ok, let's hear your idea then

[wraith hinge]

do you know derivatives?

[buoyant void]

try using sinx/x=1 and tgx/x=1 as x->0

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@glacial garden Has your question been resolved?

[jagged escarp]

By using [lim x->0 (sinx÷x) =1] then [lim x->0(sin2x÷x=2) ]
And: [lim x->0 (tanx÷x) =1]
then [lim x->0 (tan3x÷x) =3]
Lim x->0 [(x+sin2x÷x)/(tan3x÷x)]= [(1+2)/3]=1

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[cerulean quail]

1.11b?

[cerulean quail]

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[iron orbit]

need help finding second derivative of y'=(2x-y^2)/(y^3+2xy)

[iron orbit]

this is what I got it to using the quotient rule however the book says y'' is (2-y'(3y^2y'+2xy'+4y)/(y^3+2xy)

[clear cloud]

Second derivative wrt what

[iron orbit]

Trying to figure out how the book got the 2nd derivative compared to using the quotient rule on y'

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[gritty hollow]

I'm getting the hang of completing the square, but this problem is confusing me. How do I convert this when a and b do not have any common factors outside of 1?

[bright quarry]

did you start by dividing by 2?

[gritty hollow]

I thought you start by moving the 6 to the other side, but that would also make the other side negative (which would lead to an imaginary number).

If we divide by 2, does that turn it into 2(x^2+3.5x+3)?

[bright quarry]

what?

usually we have the coefficient of a expressed in vertex form but apparently they don’t want that, also i see no reason why you should move the constant term to the other side

it’s useless

this is how you should do it

first they divide by 2 to get
x^2 + 7/2 x + 3 = 0

then we observe that (x + 7/4)^2 contains the first two terms x^2 + 7/2 x but the constant term is different

(x + 7/4)^2 = x^2 + 7/2 x + 49/16

but we have a + 3

not a 49/16

notice that 49/16 = 48/16 + 1/16

which is 3 + 1/16

hence (x + 7/4)^2 = x^2 + 7/2 x + 3 + 1/16

and look

we have exactly what we had in our original expression but there’s the 1/16 on that side

so just subtract it from both sides

and we get
x^2 + 7/2 x + 3 = (x + 7/4)^2 - 1/16

@gritty hollow do you understand

[gritty hollow]

I'm trying to absorb all of this.

[bright quarry]

i can give another example if you’d like

[gritty hollow]

Can we go step by step?

[bright quarry]

sure

[gritty hollow]

I get the first part where you divide by 2 on both sides.

[bright quarry]

mhm

would you like another example first?

maybe the fractions make it more confusing

[gritty hollow]

I understand how to complete the square normally. Just not when a does not equal 1.

I think an easier example where a does not equal one would be good.

[bright quarry]

ok

i think it would be good if you learned how to rewrite an expression instead of it being an equation where we can just ignore the coefficient of a by dividing by it

so

$2x^2 + 10x + 25$

[solid kiln]

knief

[bright quarry]

btw, have you seen vertex form? namely
a(x - h)^2 + k

[gritty hollow]

I've seen that before.

So here, we can see that the 2 and the 10 are divisible by 2.

And that the square root of 25 is 10/2.

[bright quarry]

well

the first step is to always factor out the a term

so let me walk you through this one

[gritty hollow]

Yeah, so we can do 2x(x+5)+25.

Oh, okay.

[bright quarry]

,,\begin{align*}
2x^2 + 10x + 25 &= 2\left(x^2 + 5x + \frac{25}{2}\right)\
&= 2\left(x^2 + 5x + \frac{25}{4} - \frac{25}{4} + \frac{25}{2}\right)\
&= 2\left(x + \frac{5}{2}\right)^2 - 2\left(\frac{25}{4} - \frac{25}{2}\right)\
&= 2\left(x + \frac{5}{2}\right)^2 - \frac{25}{2} + 25\
&= 2\left(x + \frac{5}{2}\right)^2 + \frac{25}{2}
\end{align*}

[solid kiln]

knief

[bright quarry]

if you have a question about any of the steps just ask

[gritty hollow]

I can kind of see it? Can you explain the 25/4-25/4+25/2 stuff? The x^2+5 stuff seems relatively straightforward.

[bright quarry]

so do you agree that adding 0 to an expression does not change anything

a + 0 = a

[gritty hollow]

Yeah, I agree.

So it's just 0+25/2.

[bright quarry]

so i did that because i wanted to have a + 25/4

because (x + 5/2)^2 = x^2 + 5x + 25/4

so to "complete the square" i rewrote my expression to contain a + 25/4

so i could rewrite it as a perfect square

[gritty hollow]

I think I get it now. So it's like a puzzle, in a way.

All you're doing is making it so that we can factor it neatly.

[bright quarry]

yes exactly

so let’s try to do the same thing in your example

2x^2 + 7x + 6

let’s forget about the zero for now

it’s good practice to be able to keep the factor of 2

[bright quarry]

like i did here

[gritty hollow]

Okay, so we factor out the 2 and then focus on the parenthesis.

So we get 2(x^2+3.5x+3)

Now we have to get b/2.

[bright quarry]

$2\left(x^2 + \frac{7}{2}x + 3\right)$

[solid kiln]

knief

[bright quarry]

mhm

[gritty hollow]

Which b/2 would be 7/4, which is where that comes from.

[bright quarry]

yes

[gritty hollow]

Now we have to make c = 7/4^2.

[bright quarry]

mhm

[gritty hollow]

That would be 49/16.

Now, 3 = 48/16, so we need to find a way to add 1/16 to it.

[bright quarry]

yep

or you could add zero

• 49/16 - 49/16

then just do -49/16 + 3

i did it that way because i just saw i could break up 3 like that

but that really amounted to subtracting 49/16 from 3

[gritty hollow]

This means that the other side is -49/16?

[bright quarry]

$2\left(x^2 + \frac{7}{2}x + \frac{49}{16} - \frac{49}{16} + 3\right)$

[solid kiln]

knief

[bright quarry]

don’t fall into the habit of moving things over to the other side just yet

only do that if they force you to

[bright quarry]

this way captures the essence of what you’re really doing

[gritty hollow]

I also have to go soon, but I understand this better now.

Thank you for all of this.

[bright quarry]

you’re welcome

[gritty hollow]

I'll close it now, but don't worry. I think I can figure this out now.

;close

[bright quarry]

ok

have a good one

good luck

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[gritty hollow]

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[wicked quail]

Hi

[wicked quail]

Can someone explain me this line please?

How did h multiply with the denominator

[lament reef]

do you mean the second line?

[wicked quail]

Yeah

[lament reef]

they multiplied the numerator and denominator both by (h+7)

[wicked quail]

Oh wait I'm dumb

[lament reef]

oop

[wicked quail]

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[bold stream]

does anyone know how to get g(z)? i dont really understand how to get it

[unique minnow]

For a fixed z (imagine a plane parallel to the xy plane at that height), what is the area of the square cross-section of the pyramid?

[bold stream]

idk im not sure

are you talking about just the base of the pyramid like 4x4?

[unique minnow]

Well if you fix z=0 then the area of the cross section on the pyramid is 16, yes

But in general, think of a fixed z like that plane I drew. You can think of your integral as "scanning" from the base to the tip of the pyramid, adding up the area of tiny slices of it.

g(z) should be the area of those slices

[bold stream]

oh i see

[unique minnow]

In reality, you only need to know x in order to find the area, since the cross-sections are squares centered at the origin

So if for a fixed z you can find the corresponding x value, that should give you insight as to what the area of the cross-section is

[bold stream]

oh i see what you mean thank you

but im still just not sure how to actually get g(z). given z = h(x-2)^2/4

[unique minnow]

Let's say z = 1.
From the equation you're given, can you find the x-value for the location of the cross section?

[bold stream]

yeah

[unique minnow]

To do that you solved for x in the equation 1 = h(x-2)^2/4

Once you have that value, double it and you should get the side length of the square yes?

Now, if z isn't given, what if you try to solve for x in terms of z?

[bold stream]

wait sorry how does doubling it give you the side length?

[unique minnow]

The squares are centered about the z axis.
If you find the value of x for a fixed z, that's really just where the red line touches the square cross section.

[bold stream]

oh i think i see i

it

but then i would have to square the whole thing after too right?

[unique minnow]

Yeah

[bold stream]

(2(-sqrt(4z/h)+2))^2
I just dont get why we need the square and - now

apparently thats the answer

[unique minnow]

From the equation you have, you get

$$(x-2)^2 = \frac{4z}{h}.$$

Technically, this yields $$x-2 = \pm \sqrt{\frac{4z}{h}}$$

Now, since $x \le 2$, we must have $x-2 \le 0$, so you need to just keep the negative root.

This gives you $x = -\sqrt{\frac{4z}{h}} + 2$
Then you double that to get the side length of the cross section, and square it to get the area of the cross section.

[solid kiln]

Azyrashacorki

[bold stream]

oh i see now but im little confused why x must be less than or equal to 2

is it cuz im not seeing where the axis are rn

im kinda confused where they are located rn

oh nvm i think i got it

thank you

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[sullen remnant]

Good morning all, how do I solve 9(3x+4)-2x=11+5(4x-1)

[sullen remnant]

27x + 4 - 2x = 11 + 20x -1

[trim lichen]

9(3x+4) is not equal to 27x+4

you need to distribute properly

and likewise for the other bracket, where you made the exact same mistake a second time

[sullen remnant]

ok

[trim lichen]

can you tell me what 9(3x+4) properly expands to?

also btw you have the "ask pronouns" role -- what are your pronouns?

[sullen remnant]

My answer is -6

[sullen remnant]

27x + 36

[trim lichen]

correct

can you show all your work leading up to the answer?

[sullen remnant]

27x +36 - 2x = 11+ 20x -5

then you group like terms
27x -20x -2x = 11+5+36

[trim lichen]

sign error at least once

twice actually

[sullen remnant]

then you divide both sides by 5

[trim lichen]

11 - 5 - 36 on the right

not 11 + 5 + 36

[sullen remnant]

no no the letter x not multiplication

Oh yes minus

it escaped me

i actually used a minus in my book

so instead 27x -20x - 2x = 11-5-36

[trim lichen]

yes

ok good

[sullen remnant]

Ok

thank you for your service.

[trim lichen]

.

[sullen remnant]

right

[trim lichen]

that wasn't me reminding you to put a period

that was me conveying a bit of shock at the word "service"

[sullen remnant]

do not be shocked

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@sullen remnant Has your question been resolved?

[sullen remnant]

Yes, it was.

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@sullen remnant Has your question been resolved?

[sullen remnant]

Yes

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[fringe valve]

can anyone help me explain what this is

[trim lichen]

some kind of strange little graph

[fringe valve]

it's from a set of numbers

[trim lichen]

thats all well and good but you have not given us any context

and expect us to somehow intuit something from this graph alone with zero context

[fringe valve]

the numbers arent actually from a math problem, i was actually trying to categorize genes based on if it had the phenotype, genotype or none. 7 different types of genes all combinations. i decided to add colors to make it easier and so that the colors would make sence i gave each color a number. these numbers seemed to have a weird type of pattern to it so i plotted them in python and got this

do you want me to send the full dataset?

[fringe valve]

i got your context right here

i need help finding a formula or a program that can put out the same numbers

i made something in python but it's limited to the first 3^8 numbers

if i want anything higher i would have to add another 2 lines

i'm curious if anyone here could help me find a better way of prediciting the numbers

and what these numbers are

[umbral knot]

Perhaps use a for loop and use index as an exponent? I have never used pylab before

[fringe valve]

well you seem to understand this way better than me 😅

do you maybe know what this numberset is?

[umbral knot]

Thank you

[umbral knot]

Hmm no, but I can try plot it when I have time

[fringe valve]

thank you, i've also noticed the numberset seems to be increasing at a certain amount but i don't know yet if this is a set contant value or a exponent or idk

it seems to be repeating

it seems to be divided in thirds

i have a theory this is some result of base 3 or something

[jaunty silo]

What do you mean by "predicting"?

Don't you alreay have a code that generates the number you want?

you want to calculate it without using loops&ifs?

Sorry I missed a point

[fringe valve]

A smart person will always be the first to admit a mistake

but yeah the problem is that im basically brute forcing it

i have to add more lines to generate higher numbers

[jaunty silo]

If you only have problems with not want to add more lines

while i<=b:
    x1.append(i)
    y1.append(y)
    for j in range(math.ceil(math.log(i,3)),-1,-1):
        if (i+1)%(3**j)==0:
            y += 1-2*j
            break
    i +=1

[fringe valve]

what do i define math as?

[jaunty silo]

oh add import math on top of the code

or below from pylab import * it doesn't really matter where the line is

as long as it is on before the while loop

you get error?

[jaunty silo]

from pylab import *
import math
a = 0
b =100000
x1 = []
y1 = []
y=0 
i=0
while i<=b:
    x1.append(i)
    y1.append(y)
    for j in range(math.ceil(math.log(i+1,3)),-1,-1):
        if (i+1)%(3**j)==0:
            y += 1-2*j
            break
    i +=1

like this

[fringe valve]

oh my god calm the b down you're gonna kill my pc 😅

[jaunty silo]

And by the way sorry for my mistake again, there's some error to be fixed. Please change for loop as

for j in range(math.ceil(math.log(i+1,3)),-1,-1):

I forgot adding 1 to i

I keep forget log is not defined on 0

[fringe valve]

what should i define it as??

[jaunty silo]

you can just replace the for loop in the code I suggested to this one. It should solve the error

So overall code woulde be like

from pylab import *
import math
a = 0
b =100000
x1 = []
y1 = []
y=0 
i=0
while i<=b:
    x1.append(i)
    y1.append(y)
    for j in range(math.ceil(math.log(i+1,3)),-1,-1):
        if (i+1)%(3**j)==0:
            y += 1-2*j
            break
    i +=1

plot(x1, y1,'-')
show()
    

[fringe valve]

oh my god thank you

[jaunty silo]

I hope this help

good luck on whatever you doing

[fringe valve]

do you maybe know anything of this pattern?

[jaunty silo]

I'm trying to analyze the data, but nothing remarkable yet found for me

[fringe valve]

it's a repeating pattern the same "staircase" appears

[jaunty silo]

I don't know if this would mean anything important, But they seems to be bounded between $log_{\sqrt{3}}(x)$ and 1

[solid kiln]

Dri111

[fringe valve]

woah

hold on let me send you the original chart of numbers, i think it has something to do with base 3

[jaunty silo]

oke

[exotic tundra]

ooh, I think you're correct on this
the sum of the digits of n in base 3 looks similar

[fringe valve]

i didn't really know it was base 3 before i looked twice, it was originally meant as a system for deciding the color of squares

[jaunty silo]

is the number generated with computer? or is this some kinds of experiment result?

[fringe valve]

let me explain it's a little complex but you will understand it

it's a thought experiment you need a pc for

you have 7 pairs of chromosomes, aa bb cc dd ee ff and gg.

but it can also be aA bB cC dD eE fF and gG

and aa bb cc DD ee ff gG

and all the other thousands of combinations

if it has no large letters trait is non existent and it's a 0

if a pair has 1 large letter its a genotype and given a 1

and if a pair has both it's a phenotype and get's 2

so base 3

[jaunty silo]

so each digit is each pair of chromosome?

[fringe valve]

yes

[jaunty silo]

then why are there 8 digit?

[fringe valve]

0 is small letter small letter

[fringe valve]

oh i chose 8 for simplicity

it was meant as a little biology presentation on how traits can reapear

[jaunty silo]

and what about 6, 8 and those that are not available in base-3?

[fringe valve]

no it's 8 types of chromosomes

aa is 0 aA is 1 and AA is 2

8 pairs

there are 3 variants of each pair

8 pairs

you get base 3

[jaunty silo]

so for example 7 means there are total 7 large chromosome in 7 chromosome pair?

[fringe valve]

no the summing of them was just a way to find a color to them

it wasn't meant to predict anything

i asigned a color to make the chart more organized

let me just send the entire thing to you

[jaunty silo]

Then for example how should I intepret number 8 in the numbers?

oke

[fringe valve]

it means that 8 chromosomes in the 8 pairs are large

[jaunty silo]

ahhhh

now I cleary see what you mean

[fringe valve]

ye sorry im bad at explaining

😅

[jaunty silo]

no no it was just lack of background for me

[fringe valve]

Oh ok 😅

[jaunty silo]

I can't find any specific more works, this is my best

https://oeis.org/A094345

this is cumulative form of the data you have

[fringe valve]

thank you

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[half marsh]

Can someone look at my proof?

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@half marsh Has your question been resolved?

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@half marsh Has your question been resolved?

[half marsh]

<@&286206848099549185>

[half marsh]

.close

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[round thorn]

no

nvm i got it .close

.close

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[wraith hinge]

Hello I have a clash of clans question

[trim jolt]

Please don't occupy multiple help channels.

[wraith hinge]

Basically: I have 5 builders that do upgrades on structures.

The question is to figure out whether the Book of Spells or the Book of Building saves a greater total amount of days when you use it.

[odd snow]

This channel is for math bro

[wraith hinge]

Yea it’s a math question

[odd snow]

You opened two channels

[wraith hinge]

Ohh sorry

[odd snow]

Also it depends what you are trying to upgrade.

[wraith hinge]

So: If we have 5 builders, each of which need to complete five upgrades that are each X days long, and we use a Book of Building to instantly complete one of those upgrades.

[odd snow]

Because the book of spells upgrades spells when you are upgrading them and the book of building is better for building

Correct

Which leaves you 1 builder to get it to do something else

All in all usually what is the most time efficient is to have all your builders busy at all times

[wraith hinge]

Basically my question is, if you have 5 builders and use a book of building on a 10-day upgrade: does it save in total 10/5 =2 days?

Compared to the Book of Spells: since in the laboratory you can only do 1 upgrade at a time (as opposed to 5 upgrades simultaneously on buildings using the 5 builders), then the Book of Spells saves exactly the amount of days that is equal to how long the upgrade is

[wraith hinge]

Yesss this

But I feel like I am missing smth

[odd snow]

Builders are not used in spell upgrades

so you can actually do both

[wraith hinge]

Like, if you use a Book on a 14-day long upgrade, then you literally do not have to spend 14 days doing it

So how is it 14/5

[odd snow]

What town hall are you

[wraith hinge]

13

I am purely interested in the math of this

I don’t rly have any books rn

I have been debating which book is better

[odd snow]

I'd say building

[wraith hinge]

I give this math but feel like I am missing smth

[wraith hinge]

Yes but I say spells

Cause spells saves 14 days if the upgrade is 14 days

[odd snow]

because you can still use your troops while upgrading them

It saves you more time too

[wraith hinge]

But Book of Building saves 14/5 why is it so

[odd snow]

Well you can only use 1 book

[wraith hinge]

Haha I mean you can use any amount of books

I’m rly curious if someone can tell me what I’m missing haha

[sturdy skiff]

Bro clash of clans is not that serious 🙏 💔

[odd snow]

Elixer/Gold

[wraith hinge]

I rly rly want to understand which book is better

[sturdy skiff]

Sry gng idk that much ABT coc

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[fluid skiff]

help in integrating 1/r^2 dr

[fluid skiff]

1/r^2 can be written a r^-2

Power rule

(r^-2+1 /-2 +1) + C

r^-1 /-1

my problems is

r^-1 / -1 = 1/r^-1

How did we get rid of the sign -

[verbal stream]

I think there should be a negative sign

-1/r

[verbal stream]

this is not correct

[fluid skiff]

that the problem im facing

[verbal stream]

$\frac{r^{-1}}{-1} = -r^{-1} = -\frac{1}{r}$

[solid kiln]

Ari

[fluid skiff]

that my final answer

but im confused is there a property that states
-1 = 1/1?

[verbal stream]

?

[fluid skiff]

i know that the is true for the exponent only

x^-1 = 1/x

[verbal stream]

no -1 is not 1/1 lol

[fluid skiff]

alright thanks for the help

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[wraith hinge]

my teacher said if i cant get this right i fail

[wraith hinge]

Please help

[zinc ginkgo]

Sounds cruel

[wraith hinge]

please

i need to pass

[bold rapids]

a bit harsh lowkey😂

[wraith hinge]

please

im at a 11%

@wraith hinge foucus

helpp

[bold rapids]

11?!?

yes you need to focus

uhh gimmie a second

but in the mean while

<@&286206848099549185>

[wraith hinge]

yay

<@&286206848099549185>

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[wraith hinge]

Got kernel(L)={(0,0,0)}

so basis would also be (0,0,0)

[ionic pendant]

a basis for {0} is not {0} because it is not linearly independent

[wraith hinge]

and basis for Range(L) as {(1,1,0),(-1,2,0),(0,0,1)}

[wraith hinge]

oh

so i need 3 vectors

[ionic pendant]

by convention a basis for the vector space {0} is taken to be the empty set

[wraith hinge]

okay

for part d

one min

i will send my proof

this works ?

so for L to be one to one i have to show that if T(v_1)=T(v_2) then v_1 = v_2

cloud ?

[ionic pendant]

sure

[wraith hinge]

and for onto

So

For onto

I have to show this equation has a unique solution for every x1,y1,z1

?

Oh wait wrong matrix

There

Is this also right?

[wraith hinge]

Unique solution means det (A) should not be equal to 0

Rank (A) = Rank(Augmented matrix)

Correct ?

[wintry stag]

An onto function doesn't require a unique element in the preimage

The way that question is phrased doesn't make me think they want you to prove it rather than just give a yes or no answer with a brief justification.

Like for 1-1, you have that Ax = Ay and you've already shown that A has rank 3, so x = y without the formality.

[wraith hinge]

Yeah but wanted to prove it to ,just in case they ask it that way in the exam

[wintry stag]

and if you don't have access to the existence of a matrix representation of the LT, then you shouldn't be writing that in the form of an augmented matrix

[wraith hinge]

So Proving det(A) not equal to 0 would be sufficient ?

[amber python]

you've already shown that range(L) = R^3

[wraith hinge]

I did?

[wintry stag]

I assumed that when I said you could have wrote a couple sentences to justify your answer but then you said you wanted to prove it, that this meant that you wanted to show it using the definitions of 1-1 and onto, by explicitly showing that for any y in the image there exists an x in the preimage such that L(x) = y

[wraith hinge]

If I just show that those 3 vectors are independent my work is done

[amber python]

well you claim they are a basis

if they were not linearly independent, then your claim that they are a basis is wrong

[wraith hinge]

Oh yeah lol

[amber python]

also for 1-1, ker(L) = {0} is all you need

which you also stated as part of your answer to b

[wraith hinge]

They did say this

But I thought the converse isn't always true

[wraith hinge]

I didn't know we have to use the definition of onto to prove part e

Inwill do it

[wintry stag]

You don't -have- to do anything, it's just not clear what you are trying to practice

[wraith hinge]

I want to prove part d and e

https://tenor.com/view/head-scratch-confused-gif-14061427

[amber python]

your answers to parts b and c should instantly tell you the answers to parts d and e

[limpid dawn]

:pensibthis:

[trim lichen]

what converse? this is an if and only if. you should be more attentive reading stuff

[wraith hinge]

converse would be "ker(L)={0_v} implies LT T:V -->W is one to one"

damn

[amber python]

there is no converse to this statement

it is not an implication

[wraith hinge]

but

[trim lichen]

no that is just the backwards half of the iff

[wraith hinge]

say i have the statement "A set V is considered a vector space if and only if V contains 0"

its converse would be "If V contains 0 then V is a vector space"

[trim lichen]

no

[wraith hinge]

but this isnt true always right

[trim lichen]

this stmt is false anyway

but our point is a statement of the form A <=> B doesn't have a converse at all

do you understand that A <=> B is a different kind of beast than a one way implication?

[wraith hinge]

nope

e

[trim lichen]

A <=> B means "A implies B, and B implies A"

it is a two-way implication

do you understand this?

[wraith hinge]

yes

[ionic pendant]

so A => B and B => A are converses of each other on their own

[wraith hinge]

so "if and only if" is a two-way implication ?

[trim lichen]

yes

that is what ive been saying all along

[wraith hinge]

okay so ker(L)={0_v} means L is one to one

and for onto

🙏

i will do it later

thanks everyone

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[keen cliff]

calculus how do i use the integral(ax+b)^n formula properly here

just the equatio nthat comes after b.

[trim lichen]

you can't because this isn't an integral of that form.

[keen cliff]

oh

is there a shortcut to this kind of question then

or do i always have to just expand it out then integrate

im just kinda lazy

[quick tusk]

gotta expand

[trim lichen]

there's not really any sense of "always"

[quick tusk]

or use hyperbolics, but i doubt that makes it much easier

[trim lichen]

but for this one yes you have to expand. it is not that bad here

[trim lichen]

it doesn't

[quick tusk]

yeah you would still have to reduce the power

[keen cliff]

hmm

okay

i understand this then

thank u guys sm for your help

🙏 🙂‍↕️

[quick tusk]

Integration isn't like differentiation. There isn't a universal recipe to solve integrals. You just have to use your intuition, which comes with experience.

[keen cliff]

yes yes i see

[quick tusk]

!done

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[keen cliff]

.close

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[pulsar mulch]

Prove that the vectors v₁, v₂, ..., vₖ are linearly dependant if and only if there exists real numbers λ₁, λ₂, ..., λₖ that arent all zero such that
v₁λ₁ + v₂λ₂ + ... + vₖλₖ = 0.

[main sigil]

Are all nonzero?

[frozen plover]

[pulsar mulch]

1 sec

[main sigil]

Did you mean arent all zero

[frozen plover]

also what's your definition of linear independence

[pulsar mulch]

bear with me guys im tryna translate the textbook from swedish to english, my bad 😭

[main sigil]

I suppose that this would be that one of the vectors can be expressed as a linear combination of the other

[pulsar mulch]

yes this is derived from that:)

[pulsar mulch]

Yes

Sorry for the confusion, thats right

This probably isnt very difficult I'm just struggling with drawing conclusions from the proof that's in my textbook (and wikipedia, ill attach it) so that I can actually somewhat understand it

[main sigil]

So they assumed that a1v1 + a2v2 + ... akvk = 0

And used that to express one of the vs as linear combination of the others

Thus showing its linearly dependent

E.g. 2 * [1, 2] + (-1) * [2, 4] = 0
And we can rearrange this to get
[1, 2] = 1/2 [2, 4]

And this shows that its linearly dependent

And we can do the opposite similarly

[pulsar mulch]

extending off of that, could you interpret all lambda = 0 as a non-real combination of vectors, proving that it is infact linearly independant?

i.e. given a sequence of vectors that are linearly independant, the only way to create a combination is to divide by zero which is undefined

[main sigil]

Thats not a good argument honestly

[main sigil]

If the only solution is the one where all lambdas are 0 then its linearly independent

Its linearly independent because if it was linearly dependent, then we would have v1 = some linear combination of other vs and we could rearrange that to get a non-zero solution for the equation we started with

[pulsar mulch]

I understand

This really was not that complicated after all, think i just needed a human to talk to about it

Thank you a lot for the help 🙏

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[wind venture]

Hi, I got stuck when trying to check if my Method of Moments Estimator is unbiased. So I have done the following. \
\
Gamma Distribution is defined as
\begin{align}
f(x) = \frac{x^{\alpha - 1} e^{-x/\beta}}{\beta^\alpha \Gamma(\alpha)} \quad \text{for } x > 0
\end{align}
\
Let $x_1, \ldots, x_n$ be an i.i.d sample with $\gamma(\alpha, \beta)$ distribution. \
\
The gamma's moment-generating function is
\begin{align}
M(t) = E[e^{Xt}] &= \int_{0}^{\infty} e^{xt} \frac{x^{\alpha - 1} \exp\left( -\frac{x}{\beta} \right)}{\beta^\alpha \Gamma(\alpha)} \dd{x} = \frac{1}{(1 - t\beta)^{\alpha}}
\end{align}
\
In such way, the first moment and second moment is
\begin{align}
{\dv{M}{t}}(0) &= \frac{\alpha\beta}{(1 - t\beta)^{\alpha + 1}} \Bigg|{t = 0} = \alpha\beta \
{\dv[2]{M}{t}}(0) &= \frac{(\alpha^2 + \alpha)\beta^2}{(1 - t\beta)^{\alpha + 2}} \Bigg|{t = 0} = (\alpha^2 + \alpha)\beta^2
\end{align}
\
So, the expectation and variance of gamma is
\begin{align}
E[\gamma(\alpha, \beta)] &= \alpha\beta \
V[\gamma(\alpha, \beta)] &= (\alpha^2 + \alpha)\beta^2 - (\alpha\beta)^2 = \alpha\beta^2
\end{align}
\
We got the following system
\begin{align}
E[\gamma(\alpha, \beta)] = \sum_{i = 1}^n \frac{x_i}{n} &\iff \alpha\beta = \sum_{i = 1}^n \frac{x_i}{n} \
E[(\gamma(\alpha, \beta))^2] = \sum_{i = 1}^n \frac{x_i^2}{n} &\iff(\alpha^2 + \alpha) \beta^2 = \sum_{i = 1}^n \frac{x_i^2}{n}
\end{align}
\
Now.. solving the following system for $\alpha$ and $\beta$
\begin{align}
\hat{\alpha} &= \frac{\overline{X}^2}{V^2} \quad \text{Note that } \overline{X} = \sum_{i = 1}^n \frac{x_i}{n} \
\hat{\beta} &= \frac{V^2}{\overline{X}} \quad \text{and also } V^2 = \sum_{i = 1}^n \frac{x_i^2}{n} - \left( \sum_{i = 1}^n \frac{x_i}{n} \right)^2
\end{align}
\
Once done all of that, I should show if $E[\hat{\alpha}] = \alpha$ but not sure how to process with $E\left[ \overline{X}^2 / V^2\right]$

[solid kiln]

A.L.I.C.E

[trim jolt]

@wind venture Has your question been resolved?

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@wind venture Has your question been resolved?

[wind venture]

[slate aspen]

Wait, give me a second to brush up my MoM theory

[wind venture]

Yaaaay

[slate aspen]

Aaah it is just your last step I think? So you want to show now $E[\alpha] = E[\frac{X^2}{V^2}] = \alpha$, is that correct?

[solid kiln]

Mathemagician

[wind venture]

Yes I got stuck because of that /V^2

[slate aspen]

just use the proporty of Expectations that says: $E[X/Y] = E[X]/E[Y]$

[solid kiln]

Mathemagician

[wind venture]

: O

Lemme check

Btw, $E[\overline{X}^2] = E[\overline{X}]^2$, that is true, right?

[solid kiln]

A.L.I.C.E

[slate aspen]

Yes, since $\bar{X}$ is just a constant, it is correct (but this is not in general true)

[solid kiln]

Mathemagician

[slate aspen]

Can it be that your Variance should be $V^2 = \alpha \cdot \beta^2$ instead of just $V$? Otherwise, it will not work, I think the notation should be $V^2$, no?

[solid kiln]

Mathemagician

[wind venture]

Hmm, how so by instead of just V? I've written it as V^2, right?

[slate aspen]

Yes but you say in equation 6: $V = \alpha \cdot \beta^2$, but later in equations 9 and 10, you use $V^2$ instead of $V$, right? That is what might give problems in the calculation

[solid kiln]

Mathemagician

[wind venture]

Ahhh sorry, I meant like V[X] of gamma instead of V^2 of the sample

[slate aspen]

No probs 🙂 But then the problem is solved and you can show the unbiasedness right? Or are you still stuck?

[wind venture]

I'm trying to compute E[V^2], I used to do it turning it into chi-squared

That usually works with normal distributions XD

[slate aspen]

but why do you need E[V^2]? I think it should be E[V], no?

[wind venture]

Hmm, you mean like $E[\overline{X}^2 / V^2] = E[\overline{X}^2]/E[V^2] = E[\overline{X}]^2 / E[V]^2$

[solid kiln]

A.L.I.C.E

[slate aspen]

noo, this:

I think there shouldn't be a V^2 in the first expectation in the first place

[wind venture]

I got lost

[slate aspen]

It's ok

so

in equation 6, you calculate V not V^2, right?

[wind venture]

I computed sigma^2 in there

[slate aspen]

yesss

but V = sigma^2, right?

[wind venture]

Hmm, it should not be V^2 = sigma^2, should it?

[slate aspen]

no exactly!

It should just be V

and therefore, it should also just be $E[X^2/V]$ instead of $E[X^2/V^2]$

[solid kiln]

Mathemagician

[wind venture]

Hmmmmmmmmm I see, so technically I can do like this $E[\overline{X}^2/V] = E[\overline{X}]^2 / E[V] = (\alpha\beta)^2 / (\alpha\beta^2) = \alpha$

[solid kiln]

A.L.I.C.E

[slate aspen]

yess! Indeed! thats the way to go girl

[wind venture]

I thought that I was not allowed to change it like that

I mean, it feels like cheating XD

[slate aspen]

but you don't change anything, you started with V and you finish with V, you never introduced V^2, so you should not use V^2 :))

[wind venture]

Yeah, thanks for the help !

[slate aspen]

No probs, good luck!!

[wind venture]

: D

, thanks

,thanks

Hmm

,close

.thanks

.close

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[plain tusk]

quick help

[red mountain]

whats your question

[plain tusk]

ive only seen questions like this

without the other number being squared

so im kinda confused

[indigo ferry]

You need to factor that?

[red mountain]

try factoring

into (ax-by)(cx-dy)

[plain tusk]

yes

[plain tusk]

okay

[indigo ferry]

Write $-17xy = a + b$ such that $ab = (3x^2)\cdot (20y^2)$

[solid kiln]

casework

[plain tusk]

ahh ok

yeah i see now

thanks for the help

.close

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[sour tulip]

This is the answer for exercise 3 I just need someone to explain polar coordinates and what exactly happened in the second diagram, it looks like the velocoty is dicided into perpendicular components but why does the writer call them axes? Also toward the bottom of the page when velocity component is described is it a typo when they write r again besode r dot and theta beside theta dot(dot is the time derivative notation)

[crimson pulsar]

Magical Math Escape Game

Objective: Discover which magician managed to rob a bank without leaving any trace by solving math problems.

The Suspects
(x+2)(x+4)

(x-2)(x+4)

(2x-2)(x-4)

(5x+7)

(3x+6)(2x+4)

Names and Expressions of the Suspects:
Lionel Séchal: 6x² - 2x - 8

Agée L'Witch: 2x² + 22x + 24

Luke: 2x² + 2x - 8

Rachel Atlas: 2x² - 10x + 8

Merritt Lockline: 3x² + 18x + 24

Thaddeus Bigallet: 25x² + 70x + 49

Magic Program 1:
Choose the number (-4)

Square it

Subtract 7

Divide by 2

Add 5

Magic Program 2:
Choose the number x

Add 7

Multiply by 2

Add the initial number

Execute this Scratch program with (-3), then with the letter x:
when [green flag] clicked

ask [Choose a number] and wait

set [my variable] to [answer]

add [7] to [my variable]

set [my variable] to [my variable] * [2]

set [my variable] to [my variable] + [answer]

say [my variable] for [2] seconds

Magical Items and their Associated Numbers/Expressions:
Crystal ball: 7

Magic wand: 12

Potion: 3x + 14

Hat: 2x + 13

Magic book: 5

pls help

[trim lichen]

!occupied

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[tribal marten]

.

[tribal marten]

How to solve thisss

,rotate

[tribal marten]

Idk howw

[young forge]

Ohh

The formula is just r(theta)

That’s it

U multiply your radius with your angle in radians

[tribal marten]

Hmm?

R(theta)?

[young forge]

The angle of your arc length right

It’s 30

U convert it into radians

Then you just multiply it with your radius

[young forge]

r= radius (theta)= angle in radians

[trim jolt]

@tribal marten Has your question been resolved?

[tribal marten]

I still don't get itt😭😭

[tribal marten]

Can you like, Show stoppp or something

[young forge]

You know radians right?

[tribal marten]

Yel

[young forge]

Ok so you just multiply your ridians with your angle in radian

Lemme do it

[tribal marten]

Okay

[tribal marten]

Ohhhhh

[young forge]

Right

That’s it^^

[tribal marten]

Waittt

Where did 5 come from againn

Thats 15?

[young forge]

Oh it’s because when you multiply 15 with 1/6

It becomes 15/6

U simplify the fraction

It becomes 5/2

[tribal marten]

Ohhhhhhhh

Okay thank youuuuuuu

Lasttt

[young forge]

Yesss👍

[tribal marten]

15 is the given Cm?

And 30 is whats left from 330 degree?

[young forge]

Yes, 15 is the radius of your circle

[young forge]

Yes because that’s the angle of the arc length

[tribal marten]

Can you make a problem ill try to solve real quickk

[young forge]

Sure

[tribal marten]

5/4?

5 over 4?

Pie

5/4 π?

Am i correct? @young forge

[young forge]

Yess

Correct 👍

[tribal marten]

Thank you

Can i ask another stuff

Waitt

[young forge]

Sure

[tribal marten]

,rotate

[solid kiln]

Couldn't find an attached image in the last 10 messages.

Feb 20, 2025, 3:02 PM

[tribal marten]

/rotate

,rotate

[tribal marten]

Do you know thiss?

[young forge]

Lemme try first

[tribal marten]

Okiee

[young forge]

Got it

U add up everything together then u equate it to 360

[tribal marten]

Yes i knoww

[young forge]

Then?😭

[tribal marten]

But how to find value of X

[young forge]

Ohhh

U gather all the x on one side and the constants on the other

[tribal marten]

Should i add

The stuff

[young forge]

Yes all together

[tribal marten]

Then what nextt

[young forge]

Cuz together they add up to 360

[young forge]

U keep all your x on the LHS

And you bring your constants on the RHS

[tribal marten]

😭😭😭

[young forge]

What’s up💀

[tribal marten]

Guess what

HAHAHA

[young forge]

Wait I’m Guessing rn

Processing

Jk just say it

💀

[tribal marten]

Don't get it lol

[young forge]

🧍‍♂️

[tribal marten]

lol

[tribal marten]

Where did 31 and 465 came fromm

Thats the thing i need to know so that i got it

@young forge

[young forge]

Because I collect all the x together

And all the other terms together

[tribal marten]

You combined all X?

[young forge]

Yes

-55 - 50 =-105 then I bring to the other side become 360+105=465

[tribal marten]

Ohhhh

Okay

I got itttt

Thank youuu sooooo muchhh

Really appreciate it:))

@young forge Thankk you bro

[young forge]

No problem🙏

[tribal marten]

Goodluck in your lifeee

Bye byeee

[young forge]

Okay thank you

You too^^

[tribal marten]

,close

.close

[trim jolt]

Channel closed

Closed by @tribal marten

Use .reopen if this was a mistake.

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[turbid tiger]

in slve a can't we consider domain x>or=4? just like we did in slve b?

[turbid tiger]

<@&286206848099549185>

[worthy eagle]

!15m

[trim jolt]

Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.

[simple haven]

@turbid tiger if you have x = -5 then the expression in part a is still defined.

the reason for the restricted domain in part b is due to the sqrt, you can't take the square root of a negative number.

[turbid tiger]

i exactly understood .

thanks @simple haven

.close

[trim jolt]

Channel closed

Closed by @turbid tiger

Use .reopen if this was a mistake.

[trim jolt]

Available help channel!

Send your question here to claim the channel.

Remember:
• Ask your math question in a clear, concise manner.
• Show your work, and if possible, explain where you are stuck.
• After 15 minutes, feel free to ping <@&286206848099549185>.
• Type the command .close to free the channel when you're done.
• Be polite and have a nice day!

Read #❓how-to-get-help for further information on how to ask a good question, and about conduct in the question channels.

[proper saffron]

help

[proper saffron]

idk how to solve this

do you use law of sines or cosines

and how do you use them do you jst plug x and y?

[trim jolt]

@proper saffron Has your question been resolved?

[indigo ferry]

As x + y = y + x

[violet gust]

You dont need sine rule or anything here. A simple angle chase would get you the answer

Assume one of the angles, say A, to be some value.
Then you can work from that and find the angle at C in terms of the angle at A

[proper saffron]

how do we know thats y

[indigo ferry]

$$|AC| = |BC|$$

[solid kiln]

casework

[proper saffron]

oh so you just put a random value and then you get the correct answer?

[proper saffron]

ohh

[violet gust]

not a randome value. Some variable like A or smthn

[proper saffron]

oh

so it would be 2a+b=180?

because bottom angles are the same right?

its an icosloes

triangle

[violet gust]

yea, 2A+C = 180

according to the conventional notations

[proper saffron]

and then a+c=90

[violet gust]

How do you get that?

[proper saffron]

oh

wait

a+c/2=90

you divide whole thing by 2

[violet gust]

yea, but that doesnt seem to help at anything does it?

[proper saffron]

yeah

makes it more complicated

ok but what do you do after 2a+c

[violet gust]

use some properties of triangles you know

congruence, exterior angles, ...

[proper saffron]

k

[violet gust]

Hint if you want: ||calculate the angle KLB||

[proper saffron]

wait also the line in the middle doesnt do anything right?

[indigo ferry]

You have like a million isosceles triangles. If you just use every single one of them. You have what you need

[proper saffron]

hold on

[violet gust]

it is helpful

[proper saffron]

blc is 180

so divide that by 2

and then both angles are 90

[violet gust]

[proper saffron]

idk what its called

supplementary or smt

[violet gust]

Pretty sure thats called wrong

[proper saffron]

ok then i have no clue

[violet gust]

whyd you say the angles are equal?

[indigo ferry]

Who says they are the same

[proper saffron]

it looks the same

[violet gust]

ngl, it doesnt

[proper saffron]

ok nvm

[violet gust]

and they do say its not to scale

[proper saffron]

oh

[indigo ferry]

1. It isnt even to scale
2. Even if it was , looks right doesnt mean it is right

[violet gust]

• you should never even assume the figures are correct. They are more of a visual guidelines

[proper saffron]

oh ok

idk how to continue

like no clue

[violet gust]

Do you know exterior angles of a triangle?

[proper saffron]

i think

[violet gust]

Like KLB is an exterior angle of the triangle CLK

And that triangle is isosceles

you can try to write the angle KLB in terms of the angle ACB

[proper saffron]

wait

isnt exterior angle

like 360-inner angle=exteriour angle?

[indigo ferry]

No

[violet gust]

Bruh, you just said blc is 180 here.

[indigo ferry]

exterior angle is 180- inner angle

[proper saffron]

oh

ok

i cant

i need more help

[violet gust]

can you answer this?

[proper saffron]

i dont know how to

do you use the sidelenghts

[violet gust]

can you show your work? Like show me a diagram you drew and mark the angles A and C on it

[proper saffron]

ok

kinda did it on paint

so its a bit messy

[violet gust]

thats fine. I too do that a lot

[proper saffron]

ive just been noting down what we have been talking about

[violet gust]

Ok, any other angle you see thats easy to mark?

[proper saffron]

to calculate?

[violet gust]

yea.

[proper saffron]

or just mark

[violet gust]

I meant mark in the figure in the sense that you can calculate it

KL = CL means there is one angle that is easy to mark

[proper saffron]

i think that one

near the l

OH

wait

i think i know

wait kl is the angle opposite to the one i did rn right?

and cl is the very top one?

then i think ik

[violet gust]

KL and CL are the sides

they have length x

[proper saffron]

oh

[violet gust]

That means CLK is an isosceles triangle

So, with that information, what can you say about the other angles in the triangle?

[proper saffron]

ok

in clk

is also 2b+c=180

right?

[violet gust]

no...

why would there be some angle that is same as the angle CBA?

[proper saffron]

wait

uh

the veru top angle

[violet gust]

Like, if you think so, you must provide a reason with each statement you make

[proper saffron]

is the same as

the acute angle opposite the one i marked rn

[violet gust]

can you name it? I dont know which angle you marked

[proper saffron]

k

[violet gust]

YES

CKL = KCL

good

can you now tell me what is the exterior angle KLB equal to?

[proper saffron]

yeah idkk how to name angles

[proper saffron]

uh

no

[violet gust]

you can name angles with the three vertices. if the angle is ABC, then the angle is at the vertex B, and formed between the lines AB and BC

[proper saffron]

ohh

ok

so ABL is the angle near corner b

[violet gust]

yes

[proper saffron]

wait holdon

maybe i know

how do you do exterior angle

like KLB+180?

[violet gust]

.

[proper saffron]

oh

180-klc?

isnt KLC the inner angle

[violet gust]

yea

[proper saffron]

so i got it

ok

2y+KCL=180

which means KCL=y/2

right

[violet gust]

can you show how?

[proper saffron]

uh

i can explain

[violet gust]

wait no scratch that

[proper saffron]

ok

[violet gust]

y is the side

[violet gust]

so this is also not correct

[proper saffron]

oh i used the wrong variable

[violet gust]

can you rewrite it then? with the correct variable

[proper saffron]

2z+KCL=180

[violet gust]

whats z? and no, i dont think thats correct

[proper saffron]

the 2 angles inside triangle KLC

[violet gust]

coz KCL and CKL are the equal angle in this triangle

[proper saffron]

oh true

2z+KLC=180?

[violet gust]

yea

thats correct

[proper saffron]

wait no holdon

the thingy i did with y/2 is wrong

im pretty sure

[violet gust]

mhm

[proper saffron]

what do i do next

[violet gust]

We are still at the same step You gotta find the exterior angle KLB

[proper saffron]

wasnt that KLC-180?

[violet gust]

180-KLC

[proper saffron]

oh

can you give me a hint

another one

maybe i can figure it out