help-13

[long mauve]

idk why it said leg

[steel heart]

hmm. then trigonometry can do the job

[long mauve]

yeah I tried seperating the triangle into two halves and tried finding one of the length of one of these

I ended up with sin(40) = a/2a which I guess isn't the right way

a being the length of the height to the base and 2a is the length of this so called 'leg' lol

[steel heart]

if length of leg is x units, then height will be x-2, right?

[dusky peak]

If the height is x, then the side length is x+2. Then you can use sin(50°) to find what x is

[dusky peak]

Yes this also worls

Works

And is nicer as well

[long mauve]

oh wait I thought it was twice as big

[long mauve]

why sin(50) and not sin(40)?

[dusky peak]

Bc you said the angle is 50 degrees

[steel heart]

sin(x) = opposite side to angle x/hypotenuse

[dusky peak]

Yes

[long mauve]

this is what I mean

[dusky peak]

Then you would use cos(40°)

[steel heart]

how did you get sin(40) then?

[long mauve]

I'm just a bit confused on which angle represents which trigonometric function

[steel heart]

soh cah toa

sin(x) = opposite/hypotenuse
cos(x) = adjacent/hypotenuse
tan(x) = opposite/adjacent

[long mauve]

can I represent them this way? (so that I can more easily understand them)

the arrows represent which one the angles are looking at, if that makes sense

[steel heart]

that may confuse you later on, but you can do that too

[long mauve]

what about tangent?

[long mauve]

I'm confused

so sin(50) = x/(x+2) right

[cedar kiln]

@long mauve Has your question been resolved?

[cedar kiln]

@long mauve Has your question been resolved?

[cedar kiln]

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[marsh lake]

how do i do this?

[marsh lake]

not sure how to deal with the (-1)^k term

[mortal hemlock]

!status

[cedar kiln]

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

[mighty shuttle]

first split it into two seperate summations

the second one is easy for you to compute I pressume?

[marsh lake]

yep

[mighty shuttle]

so split the first one into two cases, when k is even, and when k is odd

when k is even, it's $\sum K^2$

[wraith dagger]

ƒ(Why am. I here)=I don't Know

[mighty shuttle]

and when odd it's $\sum-n^2$

[wraith dagger]

ƒ(Why am. I here)=I don't Know

[mighty shuttle]

,w sum of first n odd square numbers

[wraith dagger]

Wolfram Alpha doesn't understand your query!
Perhaps try rephrasing your question?
Click here to refine your query online

Requested by physicsrocks

[marsh lake]

Can i just pull the - out

[mighty shuttle]

yes

[marsh lake]

and do the 1/6 somthing

right but now that we have even and odd terms, how does the top part of the sigma function change

[mighty shuttle]

so you have two cases

the sum of odd squares

and sum of even squares

use these formulae

[marsh lake]

I can't in exam tho 😭

[mighty shuttle]

do you need the derivation too?

[marsh lake]

wdym?

[mighty shuttle]

of these formulae

[marsh lake]

part a asked me to prove sum of r^2 = 1/6 something

and part b is that

[marsh lake]

we arent given this in the formula booklet

[mighty shuttle]

then you'll either have to learn this or derive it, I see no other way to solve it

@mortal hemlock may know more though

[rich field]

i think i have a better way

[mighty shuttle]

sorry for the ping

[rich field]

the first sigma is
-1² + 2² - 3² + 4² - 5² + 6²....
-[1² - 2² + 3² - 4² .....]
- [{1² + 2² + 3²..... + (2n)²}- 2 * 2²(1² + 2² + 3² .... + n²)]

[marsh lake]

huh? can u elaborate

[rich field]

u get the first step right?

.?

[marsh lake]

yes

[rich field]

first i took a -1 out

in the last step

if u expand it

ur gonna get the same thing as the previous step

[mortal hemlock]

hm?

[rich field]

idk how to explain it

[marsh lake]

I only get the 1st line

[mortal hemlock]

what would i maybe know more about?

[mighty shuttle]

how to do this

[mighty shuttle]

like a method which does't require the formula for the sum of the first n odd squares

[mortal hemlock]

you can just derive the sum, no?

[marsh lake]

how?

[mighty shuttle]

yeah, that's what I suggested

for the sum of the first 2n even numbers I'd probably find the sum of (2n)^2 from 1 to n

which isn't too hard

sum of first n odd squares = $\sum_{i=1}^{n} (2i)^2$

[wraith dagger]

ƒ(Why am. I here)=I don't Know

[mortal hemlock]

$\sum_{n=0}^k 2n+1 = 1+3+5+....+2k-1+2k+1$

[wraith dagger]

Flappie

[mighty shuttle]

They're asking about squares here though

$\left(\sum_{i=1}^{2n}i^2-\sum_{i=1}^n\left(2i\right)^2\right)$= sum of odd squares

[wraith dagger]

ƒ(Why am. I here)=I don't Know

[mighty shuttle]

,w $\left(\sum_{i=1}^{2n}i^2-\sum_{i=1}^n\left(2i\right)^2\right)$

[wraith dagger]

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Requested by physicsrocks

[rich field]

$\sum_{k = 1}^{k = 2n} (-1)^k k^2\$

$-1^2 + 2^2 - 3^2 + 4^2 - ... + (2n)^2\$

Factor out (-1)

$(-1)[1^2 - 2^2 + 3^2 - 4^2 + .... - (2n)^2]\$

The problem here is the minus signs on the even terms. To make the signs positive, we can add and subtract a value as so

$(-1)[1^2 - 2^2 + 3^2 - 4^2 + .... - (2n)^2 - 2{2^2 + 4^2 + 6^2 ... + (2n)^2} + 2{2^2 + 4^2 + 6^2 ... + (2n)^2}]\$

$\implies (-1)[1^2 + 2^2 + 3^2 + 4^2 + .... + (2n)^2 - 2{2^2 + 4^2 + 6^2 ... + (2n)^2}]$

$\implies (-1)[1^2 + 2^2 + 3^2 + 4^2 + .... + (2n)^2 - 2* 2^2{1^2 + 2^2 + 3^2 ... + (n)^2}]$

[wraith dagger]

[ᴛʜᴇ ᴇᴍᴘᴇʀᴏʀ]

[ᴛʜᴇ ᴇᴍᴘᴇʀᴏʀ]

[marsh lake]

i'll check that out tomorrow

thanks

.close

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[cedar kiln]

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[steady bison]

Quick question, I collected the eigenvalues/vectors and constructed D and P. Now, they want me to also get P inverse, but they do not really specify how, am I allowed to use P and I to construct that inverse?

[modern compass]

Sure. I don't see how else you'd find the inverse here.

[steady bison]

maybe by rewriting the equation, because they specifically say "such that A = .."

that is what makes me confused haha

[modern compass]

you'd have to use P^-1 to do that computation too.

"Find a diagonal matrix D and a matrix P such that A = PDP^-1" is a very standard way of asking you to diagonalize a matrix.

[steady bison]

ah alright

jus thought there would be a catch

then this exercise is rather easy

[modern compass]

yeah, no tricks that I see. Pretty straight forward

[steady bison]

love it haha

also one quick question

when I was calculating the eigenvectors, I got this matrix:

-2 0 0 | 0
0 0 -1 | 0
-6 0 1 | 0

this would reuslt in:

-2X1 = 0
-X3 = 0
-6X1 + X3 = 0

but X2 is not mentioned, is that then a free variable? or what happens with X2 in such a case?

[modern compass]

yes, x2 is free

[steady bison]

but as x2 is not in the other equations, i can just assign it to 1?

[modern compass]

yep, it's a free variable so you can pick whatever value is convenient to represent it.

[steady bison]

alright makes sense

thank you for your time and help ❤️

ill close it now

.close

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[spice sonnet]

HELP

[spice sonnet]

<@&286206848099549185> y'all I forgot how to do this, I'm a JHS dude and uhh I forgot this

[fickle jasper]

!15m

[cedar kiln]

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.

@spice sonnet Has your question been resolved?

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[soft lark]

can you square root a negitive number/?

[soft lark]

you can only like cube root or do an odd root to it correct?

[dull oxide]

No and yes

[odd verge]

Yes

[lime oriole]

U can put the negative outside

[patent belfry]

You can

[lime oriole]

Huh

[patent belfry]

You get something called a complex number

It works just like a normal number

[dull oxide]

No real number reprsents the square root of a negative number. But if you consider imaginary numbers, then you can do square roots of negatives

[patent belfry]

But theres an “imaginary” part

[mortal hemlock]

$i^2=-1$

[wraith dagger]

Flappie

[patent belfry]

They are expressed as a + bi

[gritty viper]

The short answer is basically no

[patent belfry]

me when i lie to children

[gritty viper]

The long answer is that you can invent an entirely new kind of number based on taking the square root of a negative number, which don't fit anywhere on the number line (but instead, "above or below" it)

But the short answer is no

[soft lark]

so no normal number

[gritty viper]

Yeah

[soft lark]

you cant square root-4

[patent belfry]

Whats a normal number?

[soft lark]

etc

[soft lark]

idk 😭

[mortal hemlock]

what grade are you in?

[soft lark]

like

[patent belfry]

You can, its 2i

[soft lark]

1, 2

3

4

5

[dull oxide]

Not without leaving the realm of real numbers, no

[soft lark]

6

7

etc

[soft lark]

10

[gritty viper]

is π normal?

[mortal hemlock]

hmmm

[patent belfry]

Integers?

[mortal hemlock]

thats like on the edge of being okay to learn complex numbers

[soft lark]

well for what im wondering for

is just normal numbers

like

1

2

3

4

5

not even decimals

[patent belfry]

Natural numbers?

[soft lark]

sure?

idk the nane

[patent belfry]

What do you have against decimals

[soft lark]

name

[soft lark]

my teacher doesn't use them in the types of questions im doing

[dusk finch]

sqrt(-4) does not have a value even in "decimals"

so no

[gritty viper]

Ok well then you can't take the square root of some positive numbers

[modern compass]

yes you can find the square root of any positive number.

[soft lark]

yeah so you cant square root or any even root of a negitive whole number

[patent belfry]

This is why i hate the school system. A person has a question that could lead to a world of learning and their response is “this wont be on the test”

[gritty viper]

sqrt(2) does not seem to be in your "normal numbers"

[patent belfry]

terrible

[dusk finch]

It wont be normal though

[soft lark]

as in 1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16,17,18,19

[soft lark]

ofc ik that

[soft lark]

💀 i agree but hey

[patent belfry]

But sqrt(2) isnt a “normal number”

[dusk finch]

anyway, sqrt(-4) is not even decimal number. So the answer is no

[patent belfry]

by your definition atleast

[soft lark]

i just need that number where it says grade to be above 95 then we good

[patent belfry]

Why? Do you not want to actually learn?

[soft lark]

becuase i dont have time dude

my new school takes me like1 6 hours a day

i dont have time to waste

[patent belfry]

Just pump numbers into a formula, get a grade, and we’re good

[soft lark]

i do like to learn but yeah

[soft lark]

well i asked for negitive numbers

i wasnt even talking about that

[mortal hemlock]

$\sqrt{-4}=2i$

does this help?

[wraith dagger]

Flappie

[modern compass]

you can't take the square root of a negative number without leaving the real numbers. You'd need to introduce imaginary and complex numbers to do so.
So the answer is, sort of. Depends on what you are trying to do.

[patent belfry]

Learning is a waste, obviously

[soft lark]

yep

[mortal hemlock]

$\sqrt{-a}=i\sqrt{a}$ (not really but for now close enough)

[wraith dagger]

Flappie

[soft lark]

i knew everything beesides the ansswer to my one question

so thats why i just wanted that

[soft lark]

yes i understand now

thats all i needed to know

[patent belfry]

This isnt sincere right

This is a joke

Right

[soft lark]

🧪

[patent belfry]

What do you mean by this

[dull oxide]

Let him cook

[soft lark]

🔥 💯

okay i understand everything now

dw

love you guys

.close

[cedar kiln]

Channel closed

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[soft lark]

ty

[patent belfry]

What do you mean by this

[soft lark]

im spitting only the truth

[cedar kiln]

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[tight vigil]

Is it possible to perform a second order derivative test with a Lagrangian function

[tight vigil]

And if so, how do I do it?

I can't seem to find resources to find which tell me how to find the optimum from the points that I obtain

[cedar kiln]

@tight vigil Has your question been resolved?

[vernal minnow]

https://www.youtube.com/watch?v=LmDBUVZ7fnY

[cedar kiln]

@tight vigil Has your question been resolved?

[tight vigil]

Ye it's ok I figured it out

.close

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[lilac pike]

CBE = EDA, BCE = EAD, therefore they are similar. Also BE = DE so by AAS we're done

help me pls #help-3 @ornate vault

[ornate vault]

so for 3 what do i put

[lilac pike]

is this ur homework

xd

i dont think we need statement 2

also what's X, Y , Z, T, U, V

sorry idk where to find it

[ornate vault]

placeholders

[lilac pike]

??????

ok well just put the first option

and then the 2nd option (X=Y thing)

and then put the 3rd option

and we're done

(if im guessing the placeholder right)

[ornate vault]

when i click the first option this pops up

what do i put

[lilac pike]

wut

idk how to do it

[ornate vault]

like

gotta type in

[lilac pike]

sorry idk how to do these kinds of assignment

[ornate vault]

💔

pls try with me

[lilac pike]

😭

idk what any of these properties are

someone help him

@dense orbit

[cedar kiln]

@ornate vault Has your question been resolved?

[cedar kiln]

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[tawny rover]

I need to determine if the subset is a basis

[tawny rover]

idk how to continue it

[flint plinth]

a basis for what? show the question

!original

[cedar kiln]

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.

@tawny rover Has your question been resolved?

[tawny rover]

for P2:

[mental trail]

so you've shown it's linearly independent

is it a basis, and if so/if not what is left to prove

[tawny rover]

ok so since c1 and c2 are both 0, it's linear independent

but the solution says that is NOT a basis

[mental trail]

yep

does a basis only require to be linearly independent?

[tawny rover]

I need to find the solution set for the homogeneous system as well right?

so like put the x,y,z...

[mental trail]

kinda

but you can skip through this

what's an easy basis of P2?

[tawny rover]

I dont know that

we did only the standar basis

[mental trail]

which is

[tawny rover]

is the set of vectors in a space that can be use to represent any vector within that space

[mental trail]

ok, and in P2 what does it look like?

[tawny rover]

the set of vectors?

[mental trail]

yes

what's the "standard basis" in P2?

[tawny rover]

set of vectors that can represent x+x^2 and x-x^2

[mental trail]

you talk about set of vectors but you still didn't explicit it

if it's "standard", aren't I to expect that this standard basis is easy to give?

[tawny rover]

I have no idea what your are talking about..

like what are you trying to say?

[mental trail]

like

Standard basis of R^2 is {(1,0), (0,1)}

it's the easiest basis you can find

[tawny rover]

yes the reduced form

[mental trail]

Standard basis of R^4 is {(1,0,0,0),(0,1,0,0),(0,0,1,0),(0,0,0,1)}

what's the standard basis of P2?

[cedar kiln]

Channel closed

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[tawny rover]

[1 0]
[0 1]

.reopen

[cedar kiln]

✅

[mental trail]

no

what are the vectors even supposed to be?

[tawny rover]

you lost me

[mental trail]

a basis is a SET of VECTORS

[tawny rover]

ok

[mental trail]

this is neither a set, nor does it show vectors

[mental trail]

if you want examples, this is the standard basis of R^2

[mental trail]

this is the standard basis of R^4

what is the standard basis of P2?

[tawny rover]

x+x^2 and x-x^2

[mental trail]

no

that's "basis" B

the STANDARD basis is the EASIEST basis you could ever try to find

[tawny rover]

I guess I need to use ana arbitrary vector to find it out?

so just {(1,0), (0,1)} ?

[mental trail]

this is a basis of R^2, not P2

P2 contains POLYNOMIALS as vectors

[tawny rover]

yes

[mental trail]

ok let's go through a different approach

take any vector in P2

try to show that there exists a linear combination of vectors of B

that creates that vector

[tawny rover]

which is what I did right?

I have used c1 and c2 to find it out

or should I do something like this?

[mental trail]

there's a problem

what does your 'any vector in P2' look like?

[tawny rover]

whatever number I guess

[mental trail]

by any vector I mean that it HAS to be able to represent any vector

[mental trail]

I unfortunately see where you're trying to go

No, not "whatever number"

you keep the number "secret"

like name it "a"

what does a vector in P2 look like?

[tawny rover]

I dont get the procedure of it so idk..

[mental trail]

no

elements of P2 are POLYNOMIALS

what do polynomials look like

[tawny rover]

x+x^2

and the other one

[mental trail]

ok, for example

give me more examples

[tawny rover]

1/2x +3x^3

[mental trail]

sure

so we're in P2, the only condition is that the degree doesn't get bigger than 2

[tawny rover]

ok

[mental trail]

so

what are the numbers that we choose when we choose a polynomial of degree <= 2?

[tawny rover]

any in R?

idk

[mental trail]

no like, where do they intervene

[mental trail]

like here you choose 1/2 and 3

(and a bunch of 0s)

[tawny rover]

yeah they are infront of the varibles

[mental trail]

"the variables"

more precise maybe?

[tawny rover]

in front of x?

like 1/2 for x and 3 for x^2

bro I just wanna understand the procedure to solve it..

[mental trail]

but what we're doing is needed

so in front of x, not bad

it's actually in front of powers of x

so close enough

but

you mention x and x^2

there's something missing

[tawny rover]

yea since is dimension 2

[mental trail]

no

you're missing a BIG part

are the x and x^2 coefficients the ONLY things for polynomials of degree <= 2?

[tawny rover]

yes ?

[mental trail]

so the only polynomials of degree <=2 that exist look like ax^2 + bx ?

[tawny rover]

yes

[mental trail]

😢 sorry no

[tawny rover]

I mean there is also +c

[mental trail]

THERE WE GO

the constant coefficient CANNOT be forgotten

ax^2 + bx + c

a,b,c

[tawny rover]

ok so in our case is 0

[mental trail]

?

[tawny rover]

well is 0 + x + x^2

[mental trail]

for this polynomial ok

but if you want every polynomial in P2

you need +c

<@&268886789983436800>

[mental trail]

so if you want to write ANY polynomial in P2, +c can't always be 0

[tawny rover]

ok

[mental trail]

so with that in mind, can B be a basis of P2?

[tawny rover]

hold on

B from ax^2 + bx + c?

[mental trail]

no

[mental trail]

B as this basis

[tawny rover]

oh ok, no it cannot be 0

so how do I continue the problem?

[mental trail]

well

if you think that "B" is not a basis, find a polynomial of P2 that can't be a linear combination of the vectors in "B"

[tawny rover]

I still dont know the procedure to do that

like how do I set it up?

[mental trail]

like

take an arbitrary polynomial

ax^2+bx+c

write an arbitrary combination of vectors of B

C(x+x^2) + D(x-x^2)

[tawny rover]

so something = x+x^2

[mental trail]

when are those equal

again, arbitrary

meaning keep things "secret"

as C and D can be anything

ah wrong choice of name

ok now fixed

[tawny rover]

1 and -1?

[mental trail]

?

no, they can be anything

it doesn't mean it can be anything YOU want

at least not yet

[tawny rover]

ok so should I do after C(x+x^2) + D(x-x^2)?

Im begging for the procedure cz I have to go to work in 15min

like after that I find that is liner independet/dependet what do I do?!?!?!

[mental trail]

like

now identify x^2, x, constant

here, start with constant

ax^2 + bx+ c = C(x+x^2) + D(x-x^2)

[tawny rover]

and solve for what?

[mental trail]

a,b,c,C,D

start with the constant terms

[mental trail]

what is constant term on LHS?

what is constant term on RHS?

[tawny rover]

c on LHS and 0 on the rhs

[mental trail]

so

if both polynomials are equal

their constant terms should be equal

meaning c NEEDS to be 0

so if we had ax^2 + bx+ c = 1 for example... is this a linear combination of "B" ?

[tawny rover]

yes since is not 0?

[mental trail]

huh?

so you can find C and D such that 1 = C(x+x^2) + D(x-x^2)?

[tawny rover]

mmm no

[mental trail]

so

if we had ax^2 + bx+ c = 1 for example... is this a linear combination of "B" ?

[tawny rover]

yes?

[mental trail]

ahem

[mental trail]

can 1 be a combination of x+x^2 and x-x^2?

[tawny rover]

yes?

[mental trail]

Really? Which combination is it?

[tawny rover]

no nvm

[mental trail]

So

The polynomial 1

is not a linear combination of "B"

thus, is B a basis of P2?

[tawny rover]

no it is not

[mental trail]

there we go

ok, just one final thing to say

the standard basis of P2

it's {1,x,x^2}

since every polynomial of degree <= 2 is a*(x^2) + b*(x) + c*(1)

how many vectors?

3

how many vectors in "B"?

2

can "B" be a basis?

[tawny rover]

yes

[mental trail]

no

every basis of P2 must have the SAME number of vectors

it's the dimension of P2

{1,x,x^2} has 3 vectors

and it's a basis of P2

so dim(P2) = 3

and any family that doesn't contain EXACTLY 3 vectors CANNOT be a basis of P2

[tawny rover]

gotcha

I have to go to work now, tyvm for the help tho

.close

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[hot python]

After rolling a fair twenty-sided die 200 times, the observed (or experimental) probability of rolling a 20 was 25%. What will happen to the probability of rolling a 20 if this die is rolled 10,000 more times?

[hot python]

so i think i finally get it

due to large numbers, experimental probability becomes theoretical probability. So the probability is just 5 percent

does that make sense?

[cedar kiln]

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[solid narwhal]

why isnt this being accepted as a valid answer

[solid narwhal]

i plugged it into the calculator and 2265/999 equals 2.267267...

[delicate elm]

it asks for 2.2676767... not 2.267267267...

[solid narwhal]

im stupid

.close

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[hot python]

i chose c, is this correct?

[mighty shuttle]

is this a test?

[hot python]

no its part of a practice sheet

we get these 20 question things before the actual exam but i cant get it graded back since it has free response until after i've already taken the test, and its not too useful to have a practice test if you cant figure out what you did wrong before the test

I would show the banner but i crop it because it has my name and other PI

[mighty shuttle]

okay

unfortunately, I'm pretty bad at stats so I can't help, sorry

[hot python]

it's alright lol

me too

.close

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[lyric narwhal]

$\int\frac{dx}{(x+1)\sqrt{x^2-1}}$

[wraith dagger]

kheerii

[hot crag]

oh god integrals

uhhhh

what have you tried

btw i may prove incompetent

[mighty shuttle]

I have a hint, that may work

two actually

[hot crag]

okay yea physics' got this imma watch

[lyric narwhal]

the usual substitution of $\theta=\arcsec(x)$ leads us to the expression $$\tan\left(\frac1{2}\arcsec x\right)+C$$

[wraith dagger]

kheerii

[lyric narwhal]

the problem

[hot crag]

is WA?

[lyric narwhal]

is that this feels wrong

[hot crag]

huh

let's see

[lyric narwhal]

it's not a complete antiderivative

there's a problem of a negative sign

[mighty shuttle]

uh

[lyric narwhal]

what I got is $\sqrt{\frac{x-1}{x+1}}+C$ but I'm pretty sure the full proper integral is supposed to be $\frac{\sqrt{x^2-1}}{x+1}+C$

[mighty shuttle]

the numerator should be $sec(x)tan(x)$

[wraith dagger]

ƒ(Why am. I here)=I don't Know

kheerii

[hot crag]

isn't this the integral

o

yea dx

mb lmfao

[untold osprey]

same thinf

[lyric narwhal]

how do I get the second, correct antiderivative?

[untold osprey]

rationalize the denominator

[lyric narwhal]

no it isn't

[untold osprey]

wait is it not

[hot crag]

no...?

[lyric narwhal]

for x>=-1 it is the same

but that isn't necessary

[untold osprey]

oh ok i see yeah

[languid bison]

Why are they different?

[lyric narwhal]

I've only encountered this kind of problem with secant substitutions

[untold osprey]

x < -1

[lyric narwhal]

if x<-1 then the second expression is negative of the first expression

[languid bison]

Is it because sign of x kind of thing?

[untold osprey]

yields imaginary for one

[lyric narwhal]

no, not in the domain of the integral itself

[mighty shuttle]

what does wolfram say?

[lyric narwhal]

the second one

that's the correct antiderivative

[mighty shuttle]

,w $\int\frac{dx}{(x+1)\sqrt{x^2-1}}$

[wraith dagger]

Results provided by WolframAlpha

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Requested by physicsrocks

[lyric narwhal]

yeah

[mighty shuttle]

rationalise , no?

oh

ok

right

[languid bison]

Ah so it is \sqrt{x^2 - 1} / |x + 1|

[lyric narwhal]

no, it is not

the first one is equal to what you said, which is wrong

[languid bison]

Yeah I meant the first one

[lyric narwhal]

yeah

[languid bison]

Are you computing this?

[lyric narwhal]

does hyperbolic trig sub work to get the correct answer?

[mighty shuttle]

,w is $\int\frac{dx}{(x+1)\sqrt{x^2-1}}= sqrt{\frac{x-1}{x+1}}$

[wraith dagger]

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[lyric narwhal]

yes

yeah

[languid bison]

Substitution can be problematic when you fail to consider e.g. behavior of signs.

[lyric narwhal]

that's exactly what I said

I'm asking for a method to get the right antiderivative definitively

[languid bison]

Yeah so, hmm

[lyric narwhal]

$x=sgn(u)\cosh(u)$?

[untold osprey]

i guess ibp doesnt work?

[languid bison]

Looking up what was the substitution you used

[wraith dagger]

kheerii

[lyric narwhal]

theta=arcsec(x)

[languid bison]

x = sec \theta?

[lyric narwhal]

yeah

[languid bison]

Did you consider that $\sqrt{x^2 - 1} = | \tan\theta |$

[lyric narwhal]

I did not, while doing it

[wraith dagger]

Absta

[lyric narwhal]

how would I evaluate it if I did consider that

[languid bison]

You do it case-by-case basis, I think

[lyric narwhal]

$\int\frac{sgn(\tan\theta)}{1+\cos\theta}d\theta$

[wraith dagger]

kheerii

[lyric narwhal]

in an indefinite integral?

that doesn't seem right

[languid bison]

I think you can put sign like that, yeah

Indefinite-ness does make it messier

But indefinite integral is basically,
[ \int f(x) dx = \int^x f(x) dx + C ]
So you can do some case-by-case things.

[wraith dagger]

Absta

[cedar kiln]

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[cedar kiln]

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[quartz salmon]

At which points are the following functions continuous?

a) ( f(x, y) ) is defined as:

[
f(x,y) =
\begin{cases}
\frac{x^3 + y^4}{x^2 + y^2} & \text{if } (x, y) \neq (0, 0) \
0 & \text{if } (x, y) = (0, 0)
\end{cases}
]

[wraith dagger]

dghf

[quartz salmon]

I understand that to check for continuity at a point we need to verify if the limit value of the function as (x, y) approaches that point is equal to the function value at that point. But how do you evaluate the limit value when we have two different variables? I read that one examines different paths, but I don't know what it says that the fraction has the function value \frac{x^3+y^4}{x^2+y^2} if (x, y ) \neq (0, 0)$ unlike f(x, y) = 0 if (x, y) = (0, 0).

[cedar kiln]

@quartz salmon Has your question been resolved?

[cedar kiln]

@quartz salmon Has your question been resolved?

[royal matrix]

Ig you don't know about multivariable limits which is a calc3 subject. So I recommend googling and learning about it.
However short answer:
Since we can not approach x and y at the same tine to 0 (since there are infinite possible paces in which they can approach 0) we will choose one path/curve to approach from there (so we can have a specific pace at which they get closer to their value)

x and y are both approaching (0,0)
So any curve which passes through this point can be used to evaluate the limit

Such as y = x (or more generally y = mx, where m is any real number)
y = x^2
and etc...

For example upon choosing y = x, we get
lim(x,y) -> (0,0) of (x^3 + y^4)/(x^2 + y^2) to become
lim(x -> 0) (x^3 + x^4)/2x^2 = (x + x^2)/2 = 0

In mutlivariable limits two outcomes are possible:

1. Limit exists
2. Limit doesn't exist

[quartz salmon]

So what happens when we get the answer 0? Do we check if for x = 0, and y -> 0, we also get 0?

We have a piecewise defined function with two cases, is the objective here to check that the rational function is 0 for both y = 0, and x = 0?

[royal matrix]

If limit doesn't exist you will find out that upon approaching from two different curves/paths you get 2 different answers.
If limit exists, no matter what path you choose you always get same answer. So you should prove that the answer is that, using squeeze theorem.
How to do these, if you have no idea, is explained in tutorials of this topic. I recommend learning about it if you don't know anything

[royal matrix]

If you get 0 for different curves, then you should check if the answer is actually 0 by trying to prove it using squeeze theorem. If you can't, then there must be one path which gives an answer different from 0

[royal matrix]

No, you have to first check if limit exists or not, if you proved it exists, then you can check for continuity. If limit exists along 2 or 3 curves, doesn't mean it exists on all curves. And if it doesn't exist on all curves, then the limit does not exist

Just like how in one variable we have left and right limit and one of them might exist, but if both don't exist and and/or are not equal theehn limit doesn't exist, same is in mutlivariable limit. Just that in here you no longer have only 2 cases (left and right limit) but infinite cases (one for every path that passes through the point) that's why we should use squeeze theorem to prove it exists

[quartz salmon]

@royal matrix For the path y=x, why do we let x -> 0 and not e.g. y -> 0? We already let x -> 0 when we write "path x = 0" and then do the evaluation of the limit, no?

And do we need to check the path y = kx for some constant k?

Also, could you show me how to employ the Sqz thm here?

[royal matrix]

When you check y = mx, you don't need to then check x = ky, cause that's the same thing
Proof:
y = mx -> x = y/m = ky, k = 1/m

Also x = 0, and y = mx are 2 different paths

So if they equal then "maybe" you can use squeeze

In this case, since x and y are approaching (0, 0)
x^3 <= x^2, y^4 <= y^2
So:
Absolute value of
(x^3 + y^4)/(x^2 + y^2)
is smaller equal to |x^3| + |y^4| and bigger equal to 0
Since lim(x -> 0) x^3 = 0
And lim(y -> 0) y^4 = 0
So |x^3| + |y^4| = 0
So by squeeze theorem f(x,y) is 0 as (x y) approach (0,0)
Which is equal to the value of f(0,0) So function is continuous at the point (0,0)

[royal matrix]

By the way there is one more way to prove if limit exists or not, which is changing it into polar coordinates. You can research this on your own if you want

[quartz salmon]

So the path x = 0 covers only the vertical line through origin, whereas the path y = mx covers various lines through the origin with different slopes, and the path y = 0 covers the horizontal line through origin.

But these are not all paths, you say, so we use Sqz thm to make a generalized statement about the continuity?

Why don't we use Sqz thm immediately, instead of checking for the different paths? Do we need to check that the limit is the same value for 2-3 paths as a heuristic, to build a case for the limit's consistency?

[royal matrix]

Almost.
We use squeeze theorem not for a generalized statement about continuity, but to prove that limit exists.
After that we can check for continuity

[royal matrix]

You can use it immediately if you want, but if it does not exist you can not prove it using squeeze, so it would be of no use. So we first check some common paths and curves to see maybe it doesn't exist and there's no need for squeeze at all. Only then we use squeeze. Also to use squeeze you need to first show that their some reasoning as why you suddenly started using squeeze, which is why we first approach from two-three different paths and when we get same result we say yea maybe it is squeezable. We don't use squeeze from the very beginning usually

[quartz salmon]

Okay thanks, and the continuity check is by comparing the value of the function at (x, y) with the limit?

I am still grappling with understanding how the sqz thm works here.

We compare the terms from the denominator with the numerator, and find what bounds the rational function?

In the case that the rational function was greater than the denominator and numerator, what would that mean in terms of the Sqz thm?

@royal matrix

[cedar kiln]

@quartz salmon Has your question been resolved?

[royal matrix]

Yes

[royal matrix]

When you have a/b, if a is smaller than b, then value of a/b is smaller than a

[royal matrix]

If a was bigger than b for example? Then you can't squeeze unless you have a upper boundary

[quartz salmon]

And who determines the upper boundary?

Is it given in the problem statement?

Or who sets it?

[royal matrix]

I don't know how it would be, since I have not faced multivariable limits which did exist and their numerator was bigger than their denominator and we used squeeze theorem to show it

[cedar kiln]

@quartz salmon Has your question been resolved?

[quartz salmon]

Sorry but this is hard to read without commas and period. What are you trying to say?

Could you rephrase?

[royal matrix]

I have not solved a question that has the case you are asking for. So I don't know how upper boundary would be in that question

The case being:

1. Limit Exist
2. Numerator bigger than denominator
3. Prove with Squeeze Theorem

[quartz salmon]

.close

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[finite raven]

$P(C) \land \forall x ( P (x)) \equiv \forall x P(x)$ is there a rule that says these are equivalent?

[wraith dagger]

tales

[long swan]

what is C

[cedar kiln]

@finite raven Has your question been resolved?

[untold osprey]

its just idempotence innit

idk if i even understood your question right though

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[ruby talon]

a

[cedar kiln]

@ruby talon Has your question been resolved?

[paper pike]

I really like this one

[ruby talon]

No, i need you to find "a"

[steel nest]

a probably equals a

but thats just my guess

[wicked mantle]

a= sqrt34

[cedar kiln]

@ruby talon Has your question been resolved?

[wicked mantle]

No questions here

.close

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[fierce laurel]

I do not completely understand some part about double integrals here.

When am I meant to subtract integrals from one another, to find the resultant volume?

I set this problem up like this:

The integral of the sphere minus the integral of the cone. This would result in the volume beneath the sphere minus the volume beneath the cone.

I have seen this done in another problem.

For example, this one.

This user on chegg utilized the same logic I wanted to apply here (afaik): subtract the smaller volume from the bigger volume to find the volume in between.

I can understand the logic they use very clearly, but I am curious for as to WHEN to use this.

When you project what they are doing onto the xy-plane, it becomes very clear. However, again, when do we do this vs subtracting small volume from big volume?

[paper vigil]

In all of your examples they do big volume - small volume

I don’t understand what you’re confused about

[fierce laurel]

hold on

let me take a look here

[fierce laurel]

You are correct. I was confused about the point of intersection.

thanks

.close

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[tulip tinsel]

why does (-2)^7/3 output complex solutions

[tulip tinsel]

what's wrong w considering it as ((-2)^1/3)^7

and evaluating like that for real solutions

[mortal hemlock]

,w (-2)^(1/3)

[wraith dagger]

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[mortal hemlock]

$-2=2e^{i(\pi +2k\pi)}\\left(2e^{i(\pi +2k\pi)}\right)^{\frac13}=\sqrt[3]{2}e^{\frac13 i(\pi+2k\pi)}$

[wraith dagger]

Flappie

[livid hound]

the output is due to how wolfram is interpreting it

if you use

,w cbrt(-2)

[wraith dagger]

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Requested by ramonov

[livid hound]

it'll give the real value

using ^(1/3), it'll give the first one in the argand plane

[mortal hemlock]

it has 3 solutions

[livid hound]

(there are three solutions to x^3 = -2)

,w (-2)^(1/3)

[wraith dagger]

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[mortal hemlock]

$-\sqrt[3]{2}$, $\sqrt[3]{2}e^{\frac13 \pi i}$ and $\sqrt[3]{2}e^{\frac53 \pi i}$

[wraith dagger]

Flappie

[mortal hemlock]

are the solutions

[tulip tinsel]

so I was right

there's just multiple solutions

[mortal hemlock]

yes

[tulip tinsel]

lmao u like scared me

[mortal hemlock]

i made a mistake

[tulip tinsel]

okay so still defined but omits complex solutions

obviously

phew

[mortal hemlock]

i forgot for a sec that its multiple valued

xd

[tulip tinsel]

💀

i was conf when u denied that so quickly

when I said it

[mortal hemlock]

haha

mb

[tulip tinsel]

yeah allg

.close

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[mortal hemlock]

sometimes i go too fast

[cedar kiln]

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[true wharf]

Can someone tell me where I went wrong? I was just testing to make sure that the formula at the top in orange was equivalent to the differences of squares formula (a+b)(a-b) but when I broke down the original problem down into (a+b)(a-b) it didn’t factor to become the same equation at the top and I’m not sure what I did wrong

[livid hound]

you're mistake was getting (-y+3)(y-3) at the end

[true wharf]

How do I factor out at the end ?

The negatives are throwing me off

[livid hound]

i'd recommend first factoring out -1

which will get you:
36x^4 - (y^2 + 6y + 9)

[true wharf]

Wait

Am I allowed to just factor out a -1 out of part of the equation?

[livid hound]

yes

[true wharf]

I thought if I factor out -1, it has to be all the numbers

Like the 36x^4 as well?

[livid hound]

factorisation is just the manipulation of an expression

to get something of equivalent value

2 + 9
expressing 9 as ( 3 * 3)
has no impact on 2,
3 * 3 is still 9

[true wharf]

So say for example the problem is 10 + 10 + 10 am I allowed to just do

10 + 10 + 10

Am I allowed to do

10 + -5(-2 + -2)?

@livid hound

[livid hound]

yes

[true wharf]

Oh ok

So I factored out a -1 and got the this

Does the + (-1) just become a negative sign?

[livid hound]

yeh

[true wharf]

?

Okay got it

Thank you for the help

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[uncut phoenix]

how do I approach this. (i figured out the first coordinate to be 0 by just looking at the graph and common sense but ik thats not the correct way to do it)

[long swan]

complete the square for the outer circle

x^2 + y^2 = 4y
x^2 + y^2 - 4y = 0
x^2 + (y-2)^2 - 4 = 0
x^2 + (y-2)^2 = 4

[uncut phoenix]

wait nvm i figured it out

.close

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[hearty helm]

I always get confused when I get to this point, I believe I have to take away one of the numbers (without the variable first) but Idk which one to do, what do I do next?

[normal shore]

group variable at one side

[hearty helm]

So 13r-3= 12??

[rare fog]

No. You have to do the opposite operation so instead of adding 4r to 9r, you'd subtract 4r from 9r. This would result in -3 = 5r - 12.

The next step is adding 12 to -3. This results in 9 = 5r

[hearty helm]

I got a decimal😭

[void sand]

can you show your work?

you shouldn't have gotten a decimal

like Arpeggio said above, you want to group all the variable terms to one side, and all the constant terms to the other

[austere ember]

wouldn't they get a decimal?

technically a fraction but yeah if they're doing it in a calculator

[hearty helm]

I did copy what psych said

Sorry for the writing too

[austere ember]

yeah that's right

[void sand]

I thought they were still on the grouping step lol

[austere ember]

in general to solve equations, it helps to bring all the terms with variables to one side and all the constant terms to the other side

[void sand]

yes, the final answer is a decimal/fraction

[hearty helm]

🥲 the computer gave me a totally different result

Meaning the question I was answering

[void sand]

show?

your answer is correct, btw

[hearty helm]

Uh oh I might not be able to 😭 im on khan academy

[void sand]

can you take a screenshot?

or, we don't need to ig

your answer is right

[hearty helm]

Yeah I already passed the question and cant go back to my knowledge but it gave me -3 as the answer

[void sand]

can you take a screenshot?

you may have misread the question, is what I'm thinking

[hearty helm]

Of my work? Thats the only thing I have since I went to the next question already

[cedar kiln]

@hearty helm Has your question been resolved?

[void sand]

ah well, too late then

just know that your answer and process was indeed right

if you're done here, you can close this channel

[cedar kiln]

If you are done with this channel, please mark your problem as solved by typing .close

[hearty helm]

Thank you for the help!

.close

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[half vault]

the Schwardschild metric is 'ds^2 = − (1 − 2m/r) dt^2+ dr^2/(1 − 2m/r) +r^2 dΩ^2 .' Let's look at radially in-/outgoing light: ds=0=dΩ. This yields:
dr/dt = ±c(1 − 2m/r). However, isn't light supposed to move at the same speed (in vacuum)? What has gone wrong here?

[cedar kiln]

@half vault Has your question been resolved?

[hidden mural]

what's Omega?

#diff-geo-diff-top

[cedar kiln]

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[tawny drum]

Calculate the area between the parabola y=x^2 - 4x + 5 and the line y=5

[tawny drum]

ik you have to minus the top funcitn and bottom function

but

i drwe the graph

drew the graph

and it looks like y=5 is the top function

in the section where we are finding the area y=5 is clearing the top/larger function

so i did the integral of 5-(x^2-4x+5)

but the answer was the other way

the top function was the parabola

im confused how is y=5 not the top function

[floral salmon]

I see no issues with your reasoning. What answer did you get? What was the intended answer?

[tawny drum]

i got 2/3

the correct answer was 10 2/3

and i solved it again using the parabola as the top function and i got 10 2/3

[floral salmon]

hmmm, so you did $\int_0^4 x^2-4x+5 -(5)dx$ instead and got $10\frac{2}{3}$, but $\int_0^4 5-(x^2-4x+5)dx$ gave you $\frac{2}{3}$, is that correct?

[wraith dagger]

Crystopher

[tawny drum]

yeah

[floral salmon]

that's weird since
$$\int_0^4 x^2-4x+5 -(5)dx=$$
$$[\frac{x^3}{3}-2x^2]_0^4 =$$
$$(\frac{4^3}{3}-2\cdot 4^2) - (\frac{0^3}{3}-2\cdot 0^2) =$$
$$\frac{4^3}{3}-2\cdot 4^2=$$
$$\frac{4^3}{3}-32=$$
$$-\frac{32}{3}\neq\frac{32}{3}$$

[wraith dagger]

Crystopher

[floral salmon]

can you show how you did $\int_0^4 5-(x^2-4x+5)dx$?

[wraith dagger]

Crystopher

[tawny drum]

oh wait

i just noticedi did a miscalculation

oops

wait

[tawny drum]

no

the area

so the negative is ignored

it asked to find the area not the integral

[tawny drum]

= 5- x^2 + 4x - 5

[floral salmon]

alright, but the integral with 5 as the upper function should've given $\frac{32}{3}=10\frac{2}{3}$

[tawny drum]

= -x^2 + 4x

[wraith dagger]

Crystopher

[tawny drum]

= x^3/3 + 2x^2 (integrated)

= 64/3 + 32

um what

[floral salmon]

should be $-\frac{x^3}{3}+2x^2$

[wraith dagger]

Crystopher

[tawny drum]

oh yeah sorry

anhd that gives 2/3

thats what i got

96/3 - 64/3

oh wait

im blind

32/3

oops

bruh i saw 64 as 94

thanks 💀

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[haughty kayak]

Hi

[haughty kayak]

when should I use log and when I shouldn't

in derrivation

@mighty shuttle bbro you up?

[mighty shuttle]

I'm working on a problem of my own right now, sorry

[mighty shuttle]

ok, whenever the function is of the form $f(x)^{g(x)}$

[wraith dagger]

ƒ(Why am. I here)=I don't Know

[mighty shuttle]

when differentiating

right?

[muted bear]

its really hard to differentiate those

youre more likely to see c^(f(x)) where c is a constant

[lyric narwhal]

why?

$e^{g(x)\log f(x)}$

[muted bear]

but you will also need a log for f(x)^g(x)

[wraith dagger]

kheerii

[lyric narwhal]

and chain rule

along with product rule

[muted bear]

oh yeah

[lyric narwhal]

logarithmic differentiation is just a rearranged version of this

it's the same process

[cedar kiln]

@haughty kayak Has your question been resolved?

[haughty kayak]

sure man but it is being used randomly idk why

second quesiton, why is a^x is not equal to 0 when differentiating but a^a is 0 when differentiating

[mighty shuttle]

$a^x$ is not a constant

[wraith dagger]

ƒ(Why am. I here)=I don't Know

[mighty shuttle]

$a^a$ is a constant

[haughty kayak]

NO

a^a

is that a constant?

[mighty shuttle]

ues

[wraith dagger]

ƒ(Why am. I here)=I don't Know

[haughty kayak]

how

@mighty shuttle

a^a is a constant then a^x should be too

[mighty shuttle]

no

x is a variable

[haughty kayak]

huh

okk

werid

[cedar kiln]

@haughty kayak Has your question been resolved?

[cedar kiln]

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[shell edge]

How do I understand this

[vagrant elbow]

do you know how multiplying scalars with determinants work?

[shell edge]

I know for example if one row is multiplied by 3
Then I multiply 3x-2=-6

[vagrant elbow]

okay

[shell edge]

But this have different numbers

[vagrant elbow]

uh

no worries

itll make sense

can you tell me what number the first row MIGHT have been multiplied with?

[shell edge]

8 ?

[vagrant elbow]

if the first row were multiplied by 8

then there would have to be a factor of 8 in a, b, and c

but there's only that with a

so unfortunately not

what's the highest common factor with the coeffs for a, b, and c?

[shell edge]

Is it 4

[vagrant elbow]

yep good

so we know that it was multiplied with a 4

now what about the first column

[shell edge]

It was multiplied by 2

The column

[vagrant elbow]

yeah good job

so the determinant was mutliplied by both 4 and 2

think you can handle it now

[shell edge]

-16 is it

[vagrant elbow]

yep

[shell edge]

-2x2x4

Thanks 😊

I get it now

.close

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[fleet remnant]

how was this factoring done? I dont see it, even though its correct

[vagrant elbow]

A k was pulled out of the whole thing

And the 2 was distributed in the second set of parentheses

[fleet remnant]

im not seeing which k was pulled

this

or this

[vagrant elbow]

Both ks were pulled

Outer

[fleet remnant]

all 4 of them?

[vagrant elbow]

No

first one

barring the 2

[fleet remnant]

what happens here though

how is is 2k-2