help-17

[fallen sand]

Thx

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[unborn wadi]

how do i proceed here

[vocal sleet]

@unborn wadi Has your question been resolved?

[unborn wadi]

<@&286206848099549185>

[fallen sand]

.

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[unborn wadi]

hello

can u help

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[blazing depot]

can someone please help

i understand how we got the 0s

i do not understand anything else

especially this

shouldnt we do the integral from 1 to 4

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[blazing depot]

can someone help me integrate this pls

.close

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[vast shale]

Okay

Still need help?

[scenic ravine]

Try a u-sub

[loud walrus]

The channel is closed by him

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[twin meteor]

Apertyx

[static wharf]

i think these questions are separate

it s a new function each time

and each time you try to find C

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[rustic chasm]

hello

[rustic chasm]

I need help on this problem

[pallid zenith]

,w apart[1/(x^4-x^2)]\

[twin meteor]

Results provided by WolframAlpha

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[pallid zenith]

Looks like PFD is the way to go here

[rustic chasm]

ohhh no

[pallid zenith]

what lol

[rustic chasm]

This one

My mistake

[pallid zenith]

same approach

[jade anvil]

take -5 outside and its the same question

[rustic chasm]

This is where i am at

I am lost at this part

I tried using another example explanation as a reference

I don't understand how this is done

in my case

[jade anvil]

take different values of x and plug them

say x=0

then x=1

[loud walrus]

before that you must expand

and then collect back the terms

[jade anvil]

isnt it easier without expansion?

[loud walrus]

for me it's easier expanding

[rustic chasm]

I've been doing the x value way so far

i am unfamiliar with expansion

[loud walrus]

you don't know how to distribute?

oh u mean

[rustic chasm]

Oh

[loud walrus]

u didn't do that way before

It's easy, just expand and factor in terms of x^n

x^(n-1) etc...

[jade anvil]

here -B=9, there is no x term in 9 so -A=0, no x3 term so B-C+D=0 and so on is the expansion way

[rustic chasm]

This is my work it’s #5

[loud walrus]

if you expand you get this.

$-5 = x^3 A + x^3 C + x^3 D + x^2 B + x^2 C - x^2 D - x A - B$

[twin meteor]

Samuel

[loud walrus]

now u have x^3(A+B+B) etc...

so u get the equations

A+B+C = 0
A+C-D=0
-A=0
-B=-5

cascade solution

and finished

[rustic chasm]

i'm so sorry i'm new to this i am so damn lost

in the letter explanation

[loud walrus]

where did u get lost?

[rustic chasm]

I don't get where all this x^3 stuff and letter arrangement is going

you have A + C + D + B + C - D - A - B

[loud walrus]

$\frac{-5}{x^4 - x^2}$

$\frac{-5}{x^4 - x^2} = \frac{-5}{x^2(x - 1)(x + 1)}$

$\frac{-5}{x^2(x - 1)(x + 1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x + 1} + \frac{D}{x - 1}$

[twin meteor]

Samuel

[rustic chasm]

with all kinds of different x's

[loud walrus]

$\frac{-5}{x^2(x - 1)(x + 1)} = \frac{x^2(x - 1) C + x^2(x + 1) D + x(x - 1)(x + 1) A + (x - 1)(x + 1) B}{x^2(x - 1)(x + 1)}$

[rustic chasm]

yes

[twin meteor]

Samuel

[loud walrus]

We are here right?

$-5 = x^3 A + x^3 C + x^3 D + x^2 B - x^2 C + x^2 D - x A - B$

[twin meteor]

Samuel

[loud walrus]

until here

do u agree with this?

[rustic chasm]

I wrote it differen't that's what is confusing me

[loud walrus]

this last step u understand?

[rustic chasm]

which one

[loud walrus]

this

[rustic chasm]

no

[loud walrus]

and this?

[rustic chasm]

I understand this one

I just arranged it in ABCD

u know

[loud walrus]

ok so both sides are being divided by x^2(x-1)(x+1)

so we simplify that

[rustic chasm]

yes

[loud walrus]

and the RHS is being expanded

$x^2(x-1)C = x^3C - x^2C$

[twin meteor]

Samuel

[loud walrus]

right?

this is the first term

[rustic chasm]

wait

[loud walrus]

i am expanding

so u can see

[rustic chasm]

shouldn't it be Cx^2(x+1)

[loud walrus]

i chose C/(x+1)

it's no problem

because i did from the begining and i chose it this way

[rustic chasm]

ok I get it but ima just do what I can from my understanding

Ok

I understand

[loud walrus]

so u expand and get this

[rustic chasm]

ohhh does that - xa and b come from distribution?

[loud walrus]

yes

and now u rearrange and factor

to get this

$-5 = x^3 (A + C + D) + x^2 (B - C + D) - x A - B$

[twin meteor]

Samuel

[rustic chasm]

ok

[loud walrus]

$\left{
\begin{aligned}
A + C + D &= 0 \
B - C + D &= 0 \
-A &= 0 \
-B &= -5
\end{aligned}
\right.$

[twin meteor]

Samuel

[loud walrus]

now u get the system of equations

and solve

$\frac{-5}{x^4 - x^2} = \frac{5}{x^2} + \frac{\frac{5}{2}}{x + 1} + \frac{-\frac{5}{2}}{x - 1}$

[twin meteor]

Samuel

[rustic chasm]

ok

I think I have this

only thing is

I am still doing the distribution 💀

That takes a while

[loud walrus]

just some seconds

u can make the final answer like this

$\frac{-5}{x^4 - x^2} = \frac{5}{x^2} + \frac{5}{2(x + 1)} - \frac{5}{2(x - 1)}$

so you can take out the 5, and do the integral

[twin meteor]

Samuel

[rustic chasm]

Okk that's after you get the values of each of the letters?

[loud walrus]

yes

[rustic chasm]

Ok perfect

I'm pretty sure I got it from here

just need the derivitive and then that's it

Thank you

.close

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[rustic chasm]

Hello

[rustic chasm]

I am unsure if you are still here but I am lost at one part

Can I get your help on this

[pallid zenith]

you tried partial fractions right

[rustic chasm]

yup

i'm on the letter value part

This part

This is my work

[pallid zenith]

hmm have you managed to figure any of them out

you should remove the -9 or -5 or whatever constant

[rustic chasm]

I don't understand how to

it's a -5

[pallid zenith]

just call it $\frac{1}{x^4-x^2}$

[twin meteor]

jan Niku

[rustic chasm]

ok

[pallid zenith]

okay so we write $\frac{1}{x^2(x+1)(x-1)} = \frac Ax + \frac{B}{x^2} + \frac{C}{1+x} + \frac{D}{1-x}$

[rustic chasm]

wait but we don't have the values right?

[twin meteor]

jan Niku

[pallid zenith]

values?

[rustic chasm]

Like B = 5

or A = x

something

[pallid zenith]

we dont know yet

[rustic chasm]

ok

[pallid zenith]

all the letters are just numbers

no x's

[rustic chasm]

Ik i meant it as a number any number

[pallid zenith]

sure

so you have options here, the more annoying thing to do is to create the denominator on the left but on the right

the less annoying thing is to multiply through by the denominator on the left

[rustic chasm]

oh ok so I do that again?

[pallid zenith]

if we do that we get $1 = Ax(x+1)(x-1) + B(x+1)(x-1) + Cx^2(x-1) + Dx^2(x+1)$

[twin meteor]

jan Niku

[rustic chasm]

yeah I had that

[pallid zenith]

okay

this is true for any x

right?

[rustic chasm]

what does that mean?

[pallid zenith]

that we started with an equality

in terms of x

so, its true for any x

we can pick an x value

and they still must be equal

[rustic chasm]

yes

[pallid zenith]

the two sides

did you do this already?

[rustic chasm]

no

[pallid zenith]

okay

$1 = Ax(x+1)(x-1) + B(x+1)(x-1) + Cx^2(x-1) + Dx^2(x+1)$

[twin meteor]

jan Niku

[rustic chasm]

what I did was distribute everything

[pallid zenith]

painful

this is true for any x, so pick some x values

for example, lets pick x=0

[rustic chasm]

ok

[pallid zenith]

$1=-B$

[twin meteor]

jan Niku

[pallid zenith]

this comes out

[rustic chasm]

ohhhhhh

ok so we can still do that

yeah

[pallid zenith]

$1 = Ax(x+1)(x-1) + (x+1)(x-1) + Cx^2(x-1) + Dx^2(x+1)$

[twin meteor]

jan Niku

[pallid zenith]

now were here

whats another value we can try?

[rustic chasm]

let me think hmm

[pallid zenith]

hint is try to make 0 factors

[rustic chasm]

1

?

[pallid zenith]

sure

we plug that in, we get ...

[rustic chasm]

oh god no

too mnay letters

[pallid zenith]

$1 = 2D$ i think

[twin meteor]

jan Niku

[pallid zenith]

every other term has a factor of (x-1)

$1 = Ax(x+1)(x-1) + (x+1)(x-1) + Cx^2(x-1) + \frac 12 x^2(x+1)$

[twin meteor]

jan Niku

[pallid zenith]

what now

[rustic chasm]

honestly that's kinda hard I don't understand that D part

[pallid zenith]

we can go back

$1 = Ax(x+1)(x-1) + (x+1)(x-1) + Cx^2(x-1) + D x^2(x+1)$

we were here

[twin meteor]

jan Niku

[rustic chasm]

yeah

[pallid zenith]

remember this is true for any x

you said, try x=1

so we sub it in

[rustic chasm]

ohhhhhhhhh

everything but the D has a (0)

[pallid zenith]

$1 = A(1)(2)(0) + (2)(0) + C\cdot1^2 \cdot (0) + D 1^2(2)$

[twin meteor]

jan Niku

[rustic chasm]

YES

OK

I SEE IT

[pallid zenith]

so 1=2D

$1 = Ax(x+1)(x-1) + (x+1)(x-1) + Cx^2(x-1) + \frac 12 x^2(x+1)$

[twin meteor]

jan Niku

[pallid zenith]

alright, were here

[rustic chasm]

I didn't think it worked on bigger problems like this

yes

[pallid zenith]

what to try now

we wanna kill 1 of A or C

but not both

by kill i mean 0

[rustic chasm]

hmmm -1

for C

[pallid zenith]

okay lets see

$1 = -2C$

[twin meteor]

jan Niku

[rustic chasm]

YESSIR

[pallid zenith]

$1 = Ax(x+1)(x-1) + (x+1)(x-1) + -2x^2(x-1) + \frac 12 x^2(x+1)$

[twin meteor]

jan Niku

[rustic chasm]

the A part seems tricky

[pallid zenith]

can we clean this up at all

,w Collect[Expand[(x+1)(x-1) -2x^2(x-1)+1/2 x^2(x+1)]]

[twin meteor]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by janniku

[rustic chasm]

I would just fit them on the fraction

[pallid zenith]

eh

[rustic chasm]

Don't substitute them in

that's how I would do the other ones

just solve for each value and then replace them on that top part A/x + B/x^2.... etc

we just need A

[pallid zenith]

trying to remember how you do this with this method

it ends up being 0

[rustic chasm]

yeah

[pallid zenith]

i think you need to factor it

[rustic chasm]

no matter what it's zero

[pallid zenith]

lemme see

see i guess ultimately

$1 = Ax(x+1)(x-1) + (x+1)(x-1) -2x^2(x-1) + \frac 12 x^2(x+1)$

[twin meteor]

jan Niku

[pallid zenith]

you see how A is in the x^3 term

[rustic chasm]

yeah

[pallid zenith]

but we can track all of these just by inspection

theres an A, a -2, and a 1/2

[rustic chasm]

ends up being Ax^3 - Ax

distributed

[pallid zenith]

well its $0=A-2+1/2$ i think

[twin meteor]

jan Niku

[pallid zenith]

right

[rustic chasm]

what

[pallid zenith]

,w apart[1/(x^4-x^2)]

[twin meteor]

Results provided by WolframAlpha

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Upgrade to WolframAlpha Pro!

Requested by janniku

[pallid zenith]

OH

i missolved C, sorry

$1 = Ax(x+1)(x-1) + (x+1)(x-1) - \frac 12 x^2(x-1) + \frac 12 x^2(x+1)$

[twin meteor]

jan Niku

[pallid zenith]

up here

C is -1/2 not -2

are you still with me

[pallid zenith]

just with that this is the correct equation

[rustic chasm]

yeah it should be -1/2

[pallid zenith]

so heres the correct equation

$1 = Ax(x+1)(x-1) + (x+1)(x-1) - \frac 12 x^2(x-1) + \frac 12 x^2(x+1)$

[twin meteor]

jan Niku

[pallid zenith]

okay, now by visual inspection, identity the coefficient of the x^3 term in the expansion

can you do this?

look for places where there are 3 x's in a product

[rustic chasm]

A?

[pallid zenith]

it looks like A will come from the first product

nothing from the second

we'll get a -1/2 x^3 from the third

[rustic chasm]

yu[

[pallid zenith]

and a 1/2 x^3 from the fourth

yea?

[rustic chasm]

yup

yeah

[pallid zenith]

so if we expand, well get $(A+\frac12 -\frac 12)x^3$

[rustic chasm]

why is B missing?

[twin meteor]

jan Niku

[pallid zenith]

only a quadratic factor

it originally had x^2 in the bottom

[rustic chasm]

Ah ok

[pallid zenith]

so when we multiplied by x^2(x-1)(x+1) we only had that left

[pallid zenith]

so heres the expansion

have you done matching coefficients before?

method of undetermined coefficients sometimes called

[rustic chasm]

not sure tbh

[pallid zenith]

it works like this

[rustic chasm]

Just got one lecture on this

[pallid zenith]

say 1 = a + x

think about this like ...

$1 + 0 x^1 = a + 1 x^1$

since this has to be true for each x, each coefficient must match up, power to power

because a constant term cant equal an x term

and an x term cant equal an x^2 term

not for all x at least

so if you have something like 1 = a + x is true for all x

oops i fudged it

should be 1 = a + Bx

the point is

theres no x on the left hand side

so there cant be one on the right hand side

then B must be 0

did i lose you

[rustic chasm]

yeah

[pallid zenith]

the point is, we can say $A + \frac 12 - \frac 12 = 0$

[twin meteor]

jan Niku

[rustic chasm]

ok

I understand but I don't

but I get the number you got

[pallid zenith]

okay forget the example i just fudged

if $4 = A + Bx$ for all x

whats A and B?

solve by inspection

[twin meteor]

jan Niku

[rustic chasm]

A = 4-Bx

[pallid zenith]

dont use algebra

[rustic chasm]

oh

[pallid zenith]

just use your reasoning

4 cant change right?

i mean, 4 will always be four

it cant change depending on x

[rustic chasm]

yes

[pallid zenith]

and its equal to the right hand side

so, the right hand side also must not be able to change with x

otherwise, it will move away from 4

if $4 = A + Bx$ for all $x$, then $B$ cant be anything but 0

[twin meteor]

jan Niku

[pallid zenith]

bc then one side would be able to change, while the other couldnt

make sense?

[rustic chasm]

kinda

[pallid zenith]

its really just like

find the linear function thats always 4

well, it cant have any slope but 0

it must just be a flat line

[rustic chasm]

I see

[pallid zenith]

you can extend this logic

$x^2 + 1 = A + Bx + Cx^2 + Dx^3$ for all $x$

[twin meteor]

jan Niku

[pallid zenith]

like, you can solve this by inspection

A = 1

B = 0 ...

C = 1, D = 0

yea?

[rustic chasm]

ok

so no bigger x on left side?

[pallid zenith]

its like

$0x^3 + 1x^2 + 0x + 1 = Dx^3 + Cx^2 + Bx + A$

[twin meteor]

jan Niku

[pallid zenith]

and we just equate coefficients

x^3 terms must have the same coefficient

same with x^2, x, constant ...

if you by we can do this then youre basically done

$1 = Ax(x+1)(x-1) + (x+1)(x-1) - \frac 12 x^2(x-1) + \frac 12 x^2(x+1)$

[twin meteor]

jan Niku

[pallid zenith]

we can go back here, and just pull off the x^3 coefficients of each side

on the left, its 0, and on the right, its A - 1/2 + 1/2

equate them, and get A=0

[rustic chasm]

ahhh ok

[pallid zenith]

then, you just check your work with a robot

[rustic chasm]

Ok I think I can figure this out i'm close

[pallid zenith]

,w Apart[1/(x^4-x^2)]

[rustic chasm]

Thank you

[pallid zenith]

np

[rustic chasm]

.close

[vocal sleet]

Channel closed

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Use .reopen if this was a mistake.

[vocal sleet]

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[steep jackal]

bro

im so stuck

on what i can do now

can someone hep me

[ruby vapor]

Take those last two equations you got, 2y-5z=-6 and 5y-9z=-15 and solve them for y and z, ignoring the rest of the equations. Once you get answers for y and z, you can plug them both into any of the three original equations to get x

[steep jackal]

bro....

HOW

HOW DO I SOLE FOR Y OR Z

IM SO LOST

[ruby vapor]

ah sorry, didnt realize that was your question. it gets taught in different ways, usually called substitution or elimination. I'll do it with substitution since the previous part uses it.

Take the first one and solve it for y. Start by adding 5z to each side to get 2y=5z-6, then divide each side by 2 to get y=(5/2)z-3. Plug that into the second equation to get 5((5/2)z-3) - 9z = -15

Do you know how to solve that last one for z?

[steep jackal]

we cant put decimals or fractions in there

[steep jackal]

we would plug in formula for y into z right

[ruby vapor]

Yeah, that's what I did at the end, 5((5/2)z-3) - 9z = -15 only has z so it's just one variable we can solve for

And I think the work is going to involve fractions, but you end up with whole numbers

I don't see a way to avoid fractions as part of the work here

5((5/2)z-3) - 9z = -15
(25/2)z - 15 - 9z = -15
(7/2)z-15=-15
(7/2)z=0
z=0

[steep jackal]

where the hell did the -9z go

[ruby vapor]

I did (25/2)z - 9z, so I got a common denominator

(25/2)z - (18/2)z

[steep jackal]

we still have z left over

then

z has to get to just

z

2/2

doesnt it

[steep jackal]

right

[ruby vapor]

after we do (25/2)z - (18/2)z, we get (7/2)z

[steep jackal]

i think i get wym

because if we just divided 0 anyways

[ruby vapor]

Oh ok sorry if you already got that

[steep jackal]

it equals 0

[ruby vapor]

Yeah after you add 15 to each side you just get 0 anyway

so the fraction doesn't matter I guess anyway

But yeah now that we know z=0, we can plug it back into one of the two equations with just z and y to get y

[steep jackal]

i got x = 2 y = -3 z = 0

[ruby vapor]

Yep, that's the answer

[steep jackal]

bro i love u ive been on this problem for hours

[ruby vapor]

haha happy to help

[vocal sleet]

@steep jackal Has your question been resolved?

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[vast shale]

hi

[vast shale]

Idk how to answer those question

I mean i think i do

im just not confident if im doing it right

[ruby vapor]

How far did you get? It makes sense to me to start by noticing the triangles are similar by AAA. Then use Pythagorean Theorem to get AD. Then use similar triangles to figure out AE, then do AE-AD to get DE.

[vast shale]

AD was a decimal and i rounded it to an 8

how do i find AE

[ruby vapor]

Right, I kept it as 8.33 but rounding down to 8 can be fine depending on how the class is taught

For AE, we need to use that the triangles are similar

I set up the equation 3.4/4.9 = 8.33/AE and solved for AE

(since with similar triangles, the proportions of the corresponding sides are the same)

[vast shale]

i got approximately 12

if i subtract 8 from that then is DE 4?

that sounds like a dumb question but

is that the correct answer

[ruby vapor]

Yep, that's right!

[vast shale]

ok thank you!

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[steep jackal]

what do i do next

[mild flower]

eliminate x or z

[steep jackal]

hiuh

[mild flower]

like how you eliminated y

[steep jackal]

but i multiplied

to get rid of y

but how do i know what the x y z are?

[mild flower]

yes.

so you went from 3 equations and 3 unknowns

to 2 equations and 2 unknowns

which you probably have solved before

[steep jackal]

idk what y is either

[mild flower]

you'll get there

just find x or z like you normally would

[steep jackal]

im use to subsitution

[steep jackal]

idk how to do elimination hep me plz

[mild flower]

https://www.youtube.com/watch?v=oKqtgz2eo-Y

[bitter owl]

elimination is when you multiply two equations in a system of equations so that they have the same values

and then you add the two equations together

but one has a positive and one has a minus so if you added it, the answer would be zero, hence eliminating that variable

[steep jackal]

im listening

[bitter owl]

well

for example you have
x + y = -3
-3x + 3y = -9

say u wanna find the y value

then u would need to eliminate x

if you have x and -3x, you could multiply x and 3 to get 3x. then u could add (3x) + (-3x) to get 0

you feel me

[steep jackal]

no

[bitter owl]

ok can i give u a step by step example

[steep jackal]

idk how to get rid of x or z rn

[bitter owl]

oh

yeah ion know how tf to do that but you need to know how to do elimination

[steep jackal]

5x+13z+7
6x+11z=13

[mild flower]

multiply both of them by different things to make it line up

[bitter owl]

those are ur equations?

[steep jackal]

rn ya

i got rid of y alr

[bitter owl]

ok which one are u supposed to find, the variable of x or z?

[steep jackal]

both

and y

[steep jackal]

all

[bitter owl]

nah wtf

[mild flower]

find z first (by eliminating x)

[bitter owl]

have fun lil bro hayley got u

[mild flower]

then use that value of z to find x

then use those values to find y

watch that video

[bitter owl]

OH

[bitter owl]

ok so for example if you wanna eliminate the x variable

the coefficients of the x variables are 5 and 6

so youre gonna multiple both equations to GET the lcm (least common multiple) of 5 and 6

in this case it's 5 * 6 = 30 = lcm

so you're gonna aim to change 5x and 6x into 30x and 30x so that they are the same so that you can eliminate them

to do that, multiply the equations:

6 ( 5x + 13z = 7)
reason for doing that is because 6 * 5 = 30

and then:
5 ( 6x + 11z = 13)

so ur gonna have:

6 ( 5x + 13z = 7)
5 ( 6x + 11z = 13)

multiply those two equations by each other and then tell me what u get

and ping me i guess 😭

[steep jackal]

o

i didnt see that till now

i was alr doing that

i did -6 and 5

so

[bitter owl]

good good

did you find it

[steep jackal]

no

idk how

[bitter owl]

did u multiply them yet

[steep jackal]

-30x-78z=42

[bitter owl]

good

[steep jackal]

30x+55z=65

[bitter owl]

good

now thats ur system of equations

-30x-78z=42
30x+55z=65

now just eliminate by using addition

[steep jackal]

ok

now

its

-23z = 107

wait no

i made a oopsie

rq

[bitter owl]

yea and i guess once u divide it then thats ur answer for z but ermmm

ight

[steep jackal]

the 42 is negative

[bitter owl]

okay

[steep jackal]

its 22

[bitter owl]

z = 22?

[steep jackal]

23z=22

[bitter owl]

oh

[steep jackal]

i think i made a oopsie

no

[bitter owl]

-30x - 78z = -42
30x + 55z = 65

[steep jackal]

i didnt

[steep jackal]

ye

[bitter owl]

-23z = -23

whats z then

[steep jackal]

1

ez

im so smart

[bitter owl]

facts

[steep jackal]

holy

[bitter owl]

all by yourself too

alr do u know how to do the rest then

[steep jackal]

no

not rrlly

bnbut ewe up

[bitter owl]

ok well

the same way u did that to find z

[bitter owl]

youre gonna repeat that whole process to find x using these two equations

then once u find x

youll know the values for x and z

then you can use those to plug it into the og equation which is 2x + 3y + z = -2

then u can find y through that

then boom x y z

holy crap im so smart too

ping me if u need any more help

[steep jackal]

i got -4 for d

x

[bitter owl]

good

plug that into y now

[vocal sleet]

@steep jackal Has your question been resolved?

[steep jackal]

it was wrong

[bitter owl]

did u find the right x value

-11 ( 5x + 13z = 7)
13 ( 6x + 11z = 13)

-55x + -143z = -77
78x + 143z = 169

23x = 92

x = 4

i think you keep mixing up the symbols which leads to error so you have to keep double checking

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[feral light]

I dont know what i am doing wrong

[lyric fossil]

can you show how you got that

[feral light]

so the geometric series i represented it as is a_n = 4(-1/2) ^(n-2) and I basically did the summation from n=1 -> infinity minus n=16 -> infinity

because we want n=1 -> 15

[sly sierra]

why is the power (n-2)

oh are you starting the sum at n=2?

[feral light]

cuz when u do the summation of n=1 -> 15 it gives the series on the pic

um i started it at n = 1

[sly sierra]

oh i see

i just started with the 4 and then subtracted the 8 at the end haha
should be equivalent either way

[feral light]

oh ok

but why is my answer incorrect :/

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[feral light]

how do i do this

[sly sierra]

do you know how arctan behaves for large arguments?

[feral light]

it reaches a asymptote

like pi/2 i thin

think

[lyric fossil]

so one could say [\lim_{n\to\infty} \arctan (2n) = \frac{\pi}{2}]

[twin meteor]

maximo

[feral light]

oh wait u right

so by test for divergence it diverges?

[lyric fossil]

does it?

[feral light]

yea...?

[lyric fossil]

then why are you asking me

[feral light]

im making sure

[sly sierra]

"test for divergence" aka the terms don't converge to zero?

[feral light]

ye

ty

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[fluid wasp]

hello i am programming a compass for my smart watch
the compass "image" is rotated on the watch to suit its orientation
the value for the orientation is derived in 3 steps

1. obtain magnetometer reading in radians (this gives a range of +/-π)
2. convert radians to degrees (this gives a range of +/-180)
3. remove any negative values from the range
   step 3 is where i am getting stuck
   if i use (heading + 360) % 360 then north/south is displayed correctly, but east/west is reversed
   i believe this is due to my compass rotating clockwise, however i am not very good at math :<
   any and all help is appreciated :)

[vast shale]

wait so is the magnetometer reading offset from north? so if it gave pi/2 that would be east, and -pi/2 would be west?

[fluid wasp]

i will get you the docs, one sec :o

[formal pond]

by removing do you mean setting it to 0?

[fluid wasp]

no i mean offsetting the range so its lowest value becomes 0

[formal pond]

Ahh yes

i see

[vast shale]

well so how about (heading < 0) ? 180 - heading : heading

ah ty

ok

wait what is the required output?

ohhh

[fluid wasp]

the required output is 0 = north, 90 = west, 180 = south, 270 = east.... i think >.<

[vast shale]

wouldn't it be 90 = east?

[fluid wasp]

my mind is on fire o.o

[vast shale]

your output is counter clockwise the magentometer value is clockwise

[fluid wasp]

no i'm right, let me show a picture

[vast shale]

oh ok

ok i think i can do this

[fluid wasp]

:D

[vast shale]

so -1 -> 359, -179 -> 181

so it would just be 360 + x

so (heading > 0) ? heading : heading + 360

if you are using a language that supports the ternary operator

[fluid wasp]

i'm using kotlin :o

[vast shale]

oh ok

[fluid wasp]

so if i'm facing north the reading is 0 and no rotation is applied to the image
if i'm facing east, then the image needs to rotate 270 clockwise so that east is on top of the watch

if that makes sense >.<

[vast shale]

wait a second thats wrong

[fluid wasp]

D:

[vast shale]

my answer

[fluid wasp]

oh

[vast shale]

-x if x is negative, else x + 180, if i understand kotlin correctly after 1 minute of learning about it that would be (if (heading > 0) heading + 180 else -heading))

[fluid wasp]

i think i understand that expression :D
i'm not sure if kotlin would accept it... but i understand it :D

[vast shale]

waaait a second that might be wrong

[fluid wasp]

oh no :o

[vast shale]

according to kotlin docs its right

[fluid wasp]

i'd be embarrassed if you outpaced my 3 days learning kotlin in one minute xD

[vast shale]

i understand a lot of programming

[fluid wasp]

:D

[vast shale]

i just looked up the equivalent to my python

wait a second so its like this: 0 = north, 45 = northwest, 90 = west, 135 = southwest, 180 = south, 225 = southeast, 270 = east, 315 = northeast

well -x if x is negative definitely, else it would be 360 - x?

so it would be if (heading < 0) {-heading} else {360 - heading}

try that and see if it works

[fluid wasp]

ok :D

[vast shale]

ping me if that dosent work

actually

(360 - heading) % 360 is also equivalent

[fluid wasp]

it wooooorks!!!!

it finally works after 3 days of staring at my IDE

thank you so much!!!

♥️ ♥️ ♥️

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[alpine sparrow]

Let n be a positive integer. Let S be the set of all subsets of [n] = {1, 2, . . . , n} that
contain the element 1; let T be the set of all subsets of [n] that contain the element n.
Define a bijection between S and T, and find its inverse. So- I've come up with an idea abt the bijection - but I'm unsure abt the inverse portion

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@alpine sparrow Has your question been resolved?

[sullen notch]

what have you got for the bijection S->T?

[vocal sleet]

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[ocean pendant]

hey guys, how the hell do i prove this thing

[ocean pendant]

i tried and im getting weird answer

and yes i did use the cos a cos b thing

getting me nowhere

[full nova]

try writing the cos angles in sin

[ocean pendant]

hm?

will that work?

i mean like whats the thought behind it?

[full nova]

write and give feedback

[ocean pendant]

okok

ill try

[full nova]

just write it and you ll understand

[ocean pendant]

ok

[vocal sleet]

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[zealous ledge]

idk what to do next

[cobalt ocean]

(n+1)^(n+1) = (n+1)^n * (n+1)

so you can do something with (n+1)^(n+1)/(n^n)

and then n!/(n+1)! = 1/(n+1) clearly

[zealous ledge]

so i don't change the part of (n+1)^(n+1)/(n^n) ?

[cobalt ocean]

you can rearrange it into something

[zealous ledge]

My question is that. Idk how to change it.

Hmmm

Hmmm

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[drowsy glacier]

can anybody help me simplify this problem with formulas

[dawn talon]

is i the complex number √-1 ?

[drowsy glacier]

ye

s

s

[dawn talon]

first u expand the brackets then multiply with conjugate to make the denominator real

[drowsy glacier]

ues

yes

[livid horizon]

Also is the , for decimal?

[drowsy glacier]

like 0.5

0.6

0.7

[livid horizon]

Okok

[drowsy glacier]

i did it multiple times but everytime i get big big numbers

[dawn talon]

may i see what u have tried

[drowsy glacier]

sure

we have to write CR into an algebraic form Z=a+bi

[dawn talon]

its a bit messy but seems like something is wrong with ur arithmetic

did u verify each step with a calculator?

cuz ur method is completely correct

but ur final ans is wrong

[drowsy glacier]

yey

🥲

[vocal sleet]

@drowsy glacier Has your question been resolved?

[drowsy glacier]

i think it is wrong

[vocal sleet]

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[iron parrot]

The equation
[x^{10}+(13x-1)^{10}=0,]has 10 complex roots $r_1,$ $\overline{r}_1,$ $r_2,$ $\overline{r}_2,$ $r_3,$ $\overline{r}_3,$ $r_4,$ $\overline{r}_4,$ $r_5,$ $\overline{r}_5,$ where the bar denotes complex conjugation. Find the value of
[\frac 1{r_1\overline{r}_1}+\frac 1{r_2\overline{r}_2}+\frac 1{r_3\overline{r}_3}+\frac 1{r_4\overline{r}_4}+\frac 1{r_5\overline{r}_5}.]

[twin meteor]

Dork9399

[iron parrot]

!status

[vocal sleet]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I 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

[iron parrot]

I'm on step 1

<@&286206848099549185>

[lucid crypt]

notice its basically asking for the sum of reciprocals of magnitudes of roots

[iron parrot]

yea

[lucid crypt]

so consider P(1/x)

because then a reciprocal of a root directly satisfies this

[iron parrot]

By P you mean the polynomial, right?

[lucid crypt]

yes

[iron parrot]

hmm

[iron parrot]

Can you explain what you mean by this?

[lucid crypt]

a root satisfying f(x) = 0 means f(r) = 0, if we consider g(x) = f(1/x), then 1/r satisfies g(x) = 0

[iron parrot]

Ah yes, that makes sense

[outer lagoon]

We can also let,
[u=13x-1,]then we can write the equation in this form [(u+1)^{10}+(13u)^10=0.]

[iron parrot]

But how would I find f(1/x)?

Rather, how would I find those roots?

[lucid crypt]

if you replace x with 1/x and cross multiply by x^10 to clear out the denominator, you should see a relation to the roots of unity

[iron parrot]

So I got $\frac{1}{x^{10}} + (\frac{13}{x} - 1)^{10} = 0$

[twin meteor]

Dork9399

[outer lagoon]

Yes that makes sense

then you multiply both sides by x^10

[iron parrot]

So $1 + x^{10}(\frac{13}{x} - 1)^{10} = 0$

[outer lagoon]

$ (13x- 1)^{10}= -1$

[twin meteor]

Dork9399

[iron parrot]

or

$x^{10}(\frac{13}{x} - 1)^{10} = -1$

[twin meteor]

Dork9399

[iron parrot]

and then I can take 10th root

$13 - x = \sqrt[10]{-1}$

[twin meteor]

Dork9399

[lucid crypt]

this does work as long as you realise the 10th root is multivalued

so you don't miss out on solutions

[iron parrot]

yea

so there are 10 solutions for x

[lucid crypt]

yep

[iron parrot]

and we know that the magnitude of all of the tenth roots is 1

[lucid crypt]

its 13 - them though, so it's not quite that simple

[iron parrot]

argh

so we have to find them?

[lucid crypt]

in essence you're just evaluating $\frac{1}{2} \sum_{k=0}^9 |13-r|^2$

[twin meteor]

AlphaNull

[lucid crypt]

where r is a root of this

the reason for the 1/2 is because we only care about 5 roots (since the roots and their conjugates come in pairs)

[iron parrot]

Yea that makes sense

But how would I evaluate this?

Wouldn't that require finding the real and imaginary parts of 13-r?

@lucid crypt

[lucid crypt]

i think there is a more clever way, let me think about it

adding up conjugates seems to yield nice sums but its still not as nice as i would have hoped

$12 + 14 + \sum_{k=0}^4 (13-r)(13-\overline{r})$

[twin meteor]

AlphaNull

[lucid crypt]

12 + 14 comes from the 1 and -1 roots respectively

the sum is just the problem evaluation

since 13-r and 13-r conjugate are just roots of unity, you can probably try to represent them as $e^{i 2 \pi k / 10}$ and expand it out

[twin meteor]

AlphaNull

[lucid crypt]

it's not something i'm too excited to do

[iron parrot]

Wait but 1 and -1 aren't tenth roots of -1

The only thing you can cleanly evaluate is for i

@lucid crypt

[vocal sleet]

@iron parrot Has your question been resolved?

[iron parrot]

<@&286206848099549185>

[lucid crypt]

right i made a mistake, it just comes from |i - 13| + |-i - 13|

12 + 14 = 26

[iron parrot]

isnt |i-13| sqrt170?

[lucid crypt]

im not sure why im adding them actually,

[lucid crypt]

it should just be $\sum_{k=0}^4 (13-r)(13-\overline{r})$

[twin meteor]

AlphaNull

[iron parrot]

shouldnt it be 5?

unless you are adding sqrt 170 elsewhere

[lucid crypt]

i dont think there needs to be a 5 since 0 to 4 already accounts for 5 solutions,and conjugates come in pairs

1 to 5 should work

or no nvm

just 0 to 4

[iron parrot]

oh i forgor that it starts from 0

mb

[vocal sleet]

@iron parrot Has your question been resolved?

[vocal sleet]

@iron parrot Has your question been resolved?

[vocal sleet]

@iron parrot Has your question been resolved?

[vocal sleet]

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[mellow echo]

should one of my intervals be around when x = 0 for the no solution in the second screenshot? bit confused about that

[modest dirge]

It’s saying that 1/x^2 never equals zero

For any real value of x

So you can just ignore that term as not having a zero

At x=0 that term goes to infinity

[mellow echo]

alright, so if a factor when finding critical values is say... ( 2/x ) then...in that case i would have a critical value of zero?

i vaguely remember there is supposed to be a situation where i have an interval when i find undefined terms somewhere

[modest dirge]

No just look for the roots of the derivative the points where the derivative is 0

Those are your critical points

The critical points are where your maximums or minimums are

Then using those you can create intervals to determine if they are in fact maximums or minimums

[mellow echo]

alright, this sounds like the solid definition that i was confused about

[mellow echo]

could you explain what my course was trying to teach then when they mentioned this part where it says the critical number is so and so does not exist

[modest dirge]

Well if the derivative does not exist then it’s also critical graph the function |x|

There’s a sharp turn at x=0 meaning that there is no derivative there

But it’s still a minimum of the function

[mellow echo]

i never thought of this, or dealt with it and im already into solving second derivative problems 😭 its like there are so many holes in my education

thanks for the help on the other part though 🙂

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[open nacelle]

I need help to find the median, mode, and range of this, not sure what to do

[open nacelle]

And this is for 16, 17, and 18 specifically

And the 5th word problem

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[vital loom]

I use discriminant

and factorise to (k+4)(2k-4)

is this right?

and then i do k <-4 or k<2

[warm thicket]

wdym

[runic jasper]

not sure how you got there

[warm thicket]

b^2-4ac<0

[vital loom]

i got a quadratic

and i used chatgpt to factorise

[runic jasper]

💀

[warm thicket]

write out a b and c

then plug them in

[vital loom]

i got 64-8k^2-28k

before i gpt'd the thing

[runic jasper]

64 - 8(k)(k + 7) < 0

[vital loom]

oh crapppppppp

what about the two though

😱

[runic jasper]

mb

[vital loom]

ah ok

i thoguth that was my b

[runic jasper]

8 * 7 is 56 tho

¯_(ツ)_/¯

[vital loom]

what does that have to do with anything

im lost

[runic jasper]

you put it as 28

[vital loom]

oh shit yeah

i forgot to to times the 7 by two as well lmao

ok thanks

ig thats why i got it wrong

ill ask gpt to factorise with the correct expression this time

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[runic jasper]

np

[vital loom]

can i j check ive understood the inequality

its k<-8 or k <-1?

[runic jasper]

k<-8 and k<1?

[vital loom]

u mean i change or for and?

[runic jasper]

mysolution is coming diff than yours

[vital loom]

wdym ur solution

u mean the factored quadratic?

[runic jasper]

yeah

(k+8)(k-1)

[vital loom]

theres no way that can be right?

ur factorising this right?

🙈

[runic jasper]

yeah

divide throughtout by 8

[vital loom]

ohhhhhh

right

[runic jasper]

k^2 + 8k - k - 8

k(k+8) - (k+8)

(k-1)(k+8)

[vital loom]

wouldnt this still lead to the same answer tho

[runic jasper]

it would

[vital loom]

im p sure both are correct

j u divided through

gpt didnt

and i was too lazy to do either lmao

[runic jasper]

chatgpt got the correct answer

[vital loom]

ok cool

[runic jasper]

this is the wrong part

[vital loom]

wait how is it wrong tho

[runic jasper]

k<1 not k<-1

[vital loom]

oh right yh

u right

thx

ok

im done with this discriminant xd

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[south fox]

Stuck with this:
I know g(x)=1/1-x is of order 3.

So I let y=g(x)=1/1-x and z=g(g(x))=x-1/x.

Then y=g(x)=1/1-x, z=g(y)=1/1-y and x=g(z)=1/1-z. So my original equation becomes f(x) + f(y) = (2(1-2x)/x(1-x))

I am now lost and very sorry for poor formatting im very new to this

Pls I hate functional equations

<@&286206848099549185>

[vocal sleet]

@south fox Has your question been resolved?

[vocal sleet]

@south fox Has your question been resolved?

[vast shale]

still available?

[south fox]

Yes I still need help pls

[vast shale]

working on this wait

[south fox]

Tysm

my hero

[vast shale]

just check it i am not sure if i am correct so just check it

[south fox]

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[bright thistle]

Can someone help me with this please? i don’t know what i am doing wrong

[remote arch]

What happened with this one?

[bright thistle]

wdym?

oh

you right

let me try again

i found 1/ a-2+b

is it correct?

[remote arch]

Like 1/(a-2+b), so the entire a-2+b is in the denominator?

[bright thistle]

sry bad quality