help-17

[barren jolt]

uh

now some of

$\frac{x^{-0.5}}{2x-1}$

[twin meteor]

Luh Roub

[barren jolt]

quotient rule?

[prisma rain]

i'd say so

[vocal sleet]

@barren jolt Has your question been resolved?

[barren jolt]

@prisma rain i tried to use it

But somehow failed ig

[prisma rain]

show steps

,w diff (x^(-0.5))/(2x - 1)

[twin meteor]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by wkaue

[prisma rain]

pretty ugly

you could also use chain rule

[barren jolt]

Wait

Why is it like that

Why not uh

$\frac{\frac{1}{x^{0.5}}{(2x-1}}$

i think

Completely failed the parenthesis

$\frac{\frac{1}{x^{0.5}}{(2x-1)}}$

f word

$\frac{\frac{1}{x^{0.5}}{(2x-1)}}}$

nah

Too much parenthesis for me

But you get what i mean right

$\frac{\frac{1}{x^{0.5}}{2x-1}$

$\frac{\frac{1}{x^{0.5}}{2x-1}}$

jesus christus

wait

I can do it

\frac{\frac{1}{x^{0.5}}}{2x-1}

Crashing out rn

$\frac{\frac{1}{x^{0.5}}}{2x-1}$

[twin meteor]

Luh Roub

[barren jolt]

@prisma rain why isnt it like that

oh wait

Yea thats correct but you can also multiply that right

[prisma rain]

$\frac{\frac{1}{a}}{b} = \frac{1}{ab}$

damn

[barren jolt]

lmao happened to me too rn

Dw

[twin meteor]

kaue

[barren jolt]

yepyepyep

thank you

[barren jolt]

can u generalize it when its at the bottom?

lowkey suck at complex fractions sometimes

[prisma rain]

$\frac{a}{\frac{1}{b}}$

[twin meteor]

kaue

[prisma rain]

try multiplying top and bottom by b

[barren jolt]

i know that i can uh

man this is gonna be a lot of code

$\frac{\frac{a}{c}+\frac{b}{d}}{e} | \cdot cd$

[twin meteor]

Luh Roub

[barren jolt]

yea

So i can multiply top and bottom by cd

[prisma rain]

whats |

[barren jolt]

Then ill have a+b/ecd left

[barren jolt]

just side note ig

[barren jolt]

this

$(\frac{\frac{a}{c}+\frac{b}{d}}{e})\cdot cd = \frac{a+b}{ecd}$

[prisma rain]

nah

[barren jolt]

o wait

Yea

I messed up

sec

[prisma rain]

(a/c) * cd = ?

[barren jolt]

Ill have ad+cb/ecd

right?

$(\frac{\frac{a}{c}+\frac{b}{d}}{e})\cdot cd = \frac{ad+bc}{ecd}$

this should do

[prisma rain]

right but wrong notation

[barren jolt]

the notation is wrong

Yea

• Cd is wrong

[prisma rain]

yep

[barren jolt]

guess gotta do them individually

$\frac{\frac{a}{c}+\frac{b}{d}}{e}\cdot \frac{cd}{cd} = \frac{ad+bc}{ecd}$

[twin meteor]

Luh Roub

[barren jolt]

Wait does that work

now i personally dont see any problems w it anymore

[prisma rain]

lgtm

[barren jolt]

lgtm?

[prisma rain]

looks good to me

[barren jolt]

ah

thanks babe

but like

i still have no idea about bottom complex fraction

[prisma rain]

send example

[barren jolt]

$\frac{a}{\frac{1}{b}} \cdot \frac{b}{b}=a$ ?

aait no

nono

its uh

$\frac{a}{\frac{1}{b}} \cdot \frac{b}{b}=ab$

[twin meteor]

Luh Roub

[barren jolt]

this?

[prisma rain]

lgtm

[barren jolt]

ylgtm

[prisma rain]

abcdef

[barren jolt]

ghiloveyou

[prisma rain]

spooky scary skibidi

[barren jolt]

pls no gen alpha notations

i cant express how much i despise them

[barren jolt]

Back to this

[barren jolt]

Lowkey seems weird to me bro

i do not understand anything at all

[prisma rain]

thats just the solution

[barren jolt]

Yea

Qrule says this

[prisma rain]

$\left(\frac{1}{f(x)}\right) = -\frac{f'(x)}{(f(x)^2}$

[twin meteor]

kaue

[barren jolt]

hmm

[prisma rain]

when f(x) = 1

[barren jolt]

thats true fr?

[prisma rain]

just sub f(x) = 1 on quotient rule

f'(x) is 0

forgot to put the prime

[barren jolt]

[prisma rain]

$\left(\frac{1}{f(x)}\right)' = -\frac{f'(x)}{(f(x)^2}$

[twin meteor]

kaue

[barren jolt]

Can we go through this step for step please

[prisma rain]

.

[barren jolt]

Indeed

[prisma rain]

replace f(x) = 1

so f'(x) = 0 bc derivative of constant is 0

[barren jolt]

that would be an infinite loop..no?

[prisma rain]

how where what when

[barren jolt]

so

We got f(x)=1/xblablabla

now

[prisma rain]

why

[barren jolt]

wym

[prisma rain]

no just f(x) = 1

[barren jolt]

Thats the func we tryna diff

[prisma rain]

the number 1

[barren jolt]

[prisma rain]

no

ignore everything else

[barren jolt]

what about f(x) = 1

[prisma rain]

just think about the quotient rule

here

this is true for any f(x) and g(x) right

[barren jolt]

unless g(x) is 0

[prisma rain]

yeah

ignore

so its true for f(x) = 1

[barren jolt]

Yea

[prisma rain]

so replace f(x) with 1

[barren jolt]

if its 1 it already is 1

how do i replace 1 with 1

oh wait

[barren jolt]

you mean in there?

[prisma rain]

like instead of f(x)/g(x) you would have 1/g(x)

yes

[barren jolt]

yeeeeee okay

okay

now what

[prisma rain]

do it

[barren jolt]

[prisma rain]

for the whole thing

and remember f'(x) would be 0

bc 1 is a constant

so it simplifies

[barren jolt]

0-1*g'(x)/(g(x))^2

[prisma rain]

yeah

$\left(\frac{1}{f(x)}\right)' = -\frac{f'(x)}{(f(x)^2}$

[barren jolt]

so g'(x)/(g(x))^2

nahhhh

mind boggling

[twin meteor]

kaue

[barren jolt]

pls add the ) at the end

its triggering a lot

[prisma rain]

where why

theres no )

[barren jolt]

$\left(\frac{1}{f(x)}\right)' = -\frac{f'(x)}{(f(x))^2}$

[prisma rain]

ohhh

[twin meteor]

Luh Roub

[prisma rain]

cool right

[barren jolt]

wait

Isnt that the same as

$(-x^{-1})'= -x^-2$

Shi

$(-x^{-1})'= -x^{-2}$

wait no minus at the start

$(x^{-1})'= -x^{-2}$

[twin meteor]

Luh Roub

[barren jolt]

isnt that similar

[prisma rain]

kinda

for f(x) = x

[barren jolt]

oh okay

$(f(x)^{-1})'= -f(x)^{-2}$

[twin meteor]

Luh Roub

[barren jolt]

Ive a bad feeling ab this

idk why

[prisma rain]

thats just the chain rule

[barren jolt]

[barren jolt]

so thats wrong

for f(x)=1

[prisma rain]

d/dx f(g(x)) = f'(g(x))g'(x)

[barren jolt]

$d/dx f(g(x)) = f'(g(x))g'(x)$

[twin meteor]

Luh Roub

[barren jolt]

yea

[prisma rain]

thats wrong

[barren jolt]

hmm

[prisma rain]

i meant this

its only true for f(x) = x

[barren jolt]

okayokay

gotchu

[barren jolt]

So you cant write the left side as f(x)^-1

Unless x is x?

[prisma rain]

take f(x) = x^(-1) and you get the same thing

[barren jolt]

[prisma rain]

you can

$(f(x)^{-1})' = -\frac{f'(x)}{(f(x))^2}$

[twin meteor]

kaue

[prisma rain]

bc 1/something = something^(-1)

its the right hand side that cant be always simplified to -f(x)^(-2)

only when f(x) = x

because thats what the general expression simplifies to when f(x) = x

[barren jolt]

[barren jolt]

Ic ic

Tysm fr

so now uh

[barren jolt]

Can we go through this step for step pls?

[prisma rain]

,w diff (x^(-0.5))/(2x - 1)

[twin meteor]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by wkaue

[prisma rain]

u understand that it can be written like the LHS here right?

with 1 on the numerator

[barren jolt]

Yes i understand left hand side now

so we have uh

the entire denom diff on the numerator

And the entire bottom squared

?

So uh

[prisma rain]

.

whats f(x) here?

[barren jolt]

The denom

[prisma rain]

yep

[barren jolt]

$\frac{(x^(0.5) (2x-1)}{(x^(0.5)0.5-x)^2}$

[prisma rain]

just apply the formula

zamn

[barren jolt]

wtf did i write there

brainfart

$\frac{(x^{0.5} (2x-1)}{(x^{0.5}2x-1)^2}$

thats not complretely right

but

ig

[prisma rain]

first find f'(x)

[barren jolt]

$\frac{(x^{0.5} (2x-1))'}{(x^{0.5} \cdot 2x-1)^2}$

[twin meteor]

Luh Roub

[barren jolt]

now

[prisma rain]

before pluggining in the formula

[barren jolt]

uh

[prisma rain]

plugnong

bc you have f'(x) on the formula

[barren jolt]

Indeed

[prisma rain]

so you can just working it out now then sub it in

[barren jolt]

we distribute first

[prisma rain]

indeed

[barren jolt]

so then we get 2x^3/2-x^0.5

now thats sum rule

now thats

[prisma rain]

where how when

[barren jolt]

what

[prisma rain]

where did you get that

[barren jolt]

distributed

Thats the f(x) distributed

[prisma rain]

oh

i thought that was a rational function

[barren jolt]

nono

[prisma rain]

the 3/2 is the exponent

[barren jolt]

yes

[prisma rain]

right

[barren jolt]

3x^0.5-0.5^-0.5

so many 0.5s

Is that rite

[prisma rain]

bro start using some parenthesis fr

[barren jolt]

goddamn

$3x^{0.5}-0.5x^{-0.5}$

[twin meteor]

Luh Roub

[barren jolt]

[prisma rain]

nope

coefficient of the first

[barren jolt]

whats coef

[prisma rain]

3/2

not 3

[barren jolt]

Oh i added?

wait

I think im r word

[prisma rain]

the number multiplying x^something

[barren jolt]

wait

[prisma rain]

no you didnt added

you forgot the 1/2

3/2

[barren jolt]

Ah we (germans) call it factor

[prisma rain]

sehr nett

[barren jolt]

Wait no i was right?

3/2 multiplied by 2 is 6/2 = 3

nett hier

[prisma rain]

zamn

forgot there was a 2

[barren jolt]

jajaja kollege

(lass deutsch reden bitte)

[prisma rain]

now sub on the formula

[barren jolt]

uh

[prisma rain]

thats too advanced

ich esse einen apfel

take that

[barren jolt]

😭😭😭

you dont speak german?

[prisma rain]

i dont

i speak brazillian

[barren jolt]

$\frac{3x^{0.5}-0.5x^{-0.5}}{(x^{0.5}(2x-1))^2}$

[twin meteor]

Luh Roub

[barren jolt]

boom

[barren jolt]

All i can think of is phonk rn

[prisma rain]

vapo vapo

[prisma rain]

simplify the denom

[barren jolt]

Dun ta t dun ta ta

Lets see

[prisma rain]

hold on

did you put the negative sign

[barren jolt]

what negative

I did

i think

[prisma rain]

$(f(x)^{-1})' = -\frac{f'(x)}{(f(x))^2}$

[barren jolt]

from the sum rule?

[twin meteor]

kaue

[barren jolt]

you didnt say that before

nvm you did

[prisma rain]

yep

[barren jolt]

i guess lets add it to the numeratoe

r

jus turn everything around

[prisma rain]

yep

[barren jolt]

$\frac{-3x^{0.5}+0.5x^{-0.5}}{(x^{0.5}(2x-1))^2}$

[twin meteor]

Luh Roub

[barren jolt]

brazy

now

Denom

Idk what (x^0.5)^2 is

I do

Its 1

x

wait

Distribute first

[prisma rain]

multiply the exponents

(a^n)^m = a^(nm)

[barren jolt]

Indeed

i knew it i was jus having brainfart again

x^0.5 * 2x is what tho

[prisma rain]

2x where

[barren jolt]

inside

[prisma rain]

dont distribute yet

[barren jolt]

?

Distribute exp first?

[prisma rain]

yes

[barren jolt]

x(2x-1)^2?

[prisma rain]

yes

[barren jolt]

so then

We would get

[prisma rain]

now the num has -0.5x^(-0.5) = -0.5/(x^0.5)

[barren jolt]

4x^2-4x-1+1?

[prisma rain]

dont distribute

[barren jolt]

damn

sorry

[barren jolt]

wait thats wrong no

Thats very wrong

either way

wth that -1 doing

[prisma rain]

$\frac{-3x^{0.5}+0.5x^{-0.5}}{x(2x-1)^2}$

[barren jolt]

yes

no

remove the outer parenthesis

[twin meteor]

kaue

[barren jolt]

yes

[prisma rain]

now that -0.5 exponent

[barren jolt]

[prisma rain]

turn into fraction

[barren jolt]

uhoh

[prisma rain]

and do that complex fraction thing

[barren jolt]

i am not putting this one on the test bro

Shi too complicated

[prisma rain]

dis mostly algebra

[barren jolt]

but itll take lots of time

Ill remove it

0.5/x^0.5

[prisma rain]

yep

[barren jolt]

How does that affect the denoms legacy?

[prisma rain]

$\frac{-3x^{0.5}+\frac{0.5}{x^{0.5}}}{x(2x-1)^2}$

[twin meteor]

kaue

[barren jolt]

Now what

we cant multiply by x^0.5

can we

[prisma rain]

yeah

[barren jolt]

Wait we might be able to

[prisma rain]

wot

[barren jolt]

wut

we dont

?

[prisma rain]

dont what

[barren jolt]

multiply by x^0.5

[prisma rain]

yeah

[barren jolt]

hmm

So then

oh shit

[prisma rain]

👁️

[barren jolt]

$\frac{(-3x+0.5}{x(2x-1)}

Either that or i messed up

No inbetween

$\frac{(-3x+0.5}{x(2x-1)}$

[prisma rain]

dollar sign at the end

[twin meteor]

Luh Roub

[prisma rain]

wut

[barren jolt]

$\frac{-3x+0.5}{x(2x-1)}$

got rid of everythinf

oh

wait

💀

oh wait no

wait

yes

[prisma rain]

the denom

[barren jolt]

Im in shambles

i just did ^0.5

Forgot the x

[prisma rain]

the num lgtm

x^(0.5) * x^(0.5) = x indeed

[barren jolt]

huh

I just did (x^(0.5))^2

[prisma rain]

same thing

a*a = a²

[barren jolt]

[prisma rain]

the denom is wrong tho

you gotta multiply top and bottom

you only multiplied top

and the square on (2x - 1) is gone

[barren jolt]

I multiplied bottom too

[barren jolt]

Yea

[prisma rain]

why

[barren jolt]

Lets take a big step back

[barren jolt]

Lets go back to here

[prisma rain]

yeah

[barren jolt]

$\frac{-3x+0.5}{x(2x-1)^2}$

now

[twin meteor]

Luh Roub

[barren jolt]

I did the top

bottom left

[prisma rain]

the bottom looks the same as before

[barren jolt]

bottom left

so

Lets focus on it

[prisma rain]

right

[barren jolt]

(x(2x-1)^2) \cdot x^{0.5}

$(x(2x-1)^2) \cdot x^{0.5}$

[prisma rain]

dollar sign

[twin meteor]

Luh Roub

[prisma rain]

yes

[barren jolt]

right

I personally

[prisma rain]

right

[barren jolt]

Would distribute ^2 asap

[prisma rain]

nah

[barren jolt]

Unless

We can somehow

Do shinanigans with ^2 and ^0.5

[prisma rain]

with the x

[barren jolt]

❓

[prisma rain]

x*x^(0.5) = ?

[barren jolt]

√x

[prisma rain]

nah

[barren jolt]

isnt it x^ab

[prisma rain]

$x^1 \cdot x^{0.5}$

[barren jolt]

oh plus

[twin meteor]

kaue

[barren jolt]

ik

wait

Wait thats what i said

√x = x^0.5

[prisma rain]

1 + 0.5

[barren jolt]

1.5

[prisma rain]

ye

[barren jolt]

i thought multiply

im sorry

[prisma rain]

$x^a \cdot x^b = x^{a + b}$

[twin meteor]

kaue

[barren jolt]

yea i thought of that but instead of + i did *

now

we got x^{\frac{3}{2}((2x-1) \cdot x^{0.5})}

Oh my days

[prisma rain]

dollar sign

[barren jolt]

we got $x^{\frac{3}{2}((2x-1) \cdot x^{0.5})}$

[twin meteor]

Luh Roub

[prisma rain]

zamn

[barren jolt]

i forgot to take it out of the exp??

nahhh

we got $x^{\frac{3}{2}}((2x-1) \cdot x^{0.5})}$

[twin meteor]

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

[barren jolt]

[prisma rain]

nah

[barren jolt]

[prisma rain]

you already multiplied the x^(0.5)

[barren jolt]

Ah so no multiplying parenthesis?

[prisma rain]

youre mixing with a(x + y) = ax + ay

its all a product

not additions

[barren jolt]

[prisma rain]

abc = (ac)b

[barren jolt]

hmmmmm

okay

I see it now

[prisma rain]

$(x(2x-1)^2) \cdot x^{0.5} = x \cdot x^{0.5} \cdot (2x - 1)^2$

[twin meteor]

kaue

[barren jolt]

Can we sub 2x-1 for a sec

Say 2x-1 = u

[prisma rain]

whatchu tryna do

[barren jolt]

$(x(u)^2) \cdot x^{0.5} = x \cdot x^{0.5} \cdot (u)^2$

clear things up for myself

$(xu^2) \cdot x^{0.5} = x \cdot x^{0.5} \cdot u^2$

[twin meteor]

Luh Roub

[prisma rain]

yeah cuz order doesnt matter in a product

[barren jolt]

Thats still true?

[prisma rain]

and these parenthesis in the lhs are redundant

[barren jolt]

Indeed

[prisma rain]

yeah

[barren jolt]

$x \cdot \sqrt{x} \cdot u^2$

[twin meteor]

Luh Roub

[barren jolt]

[prisma rain]

yeah same

[barren jolt]

Now that looks phenomenal

$x \cdot \sqrt{x} \cdot (2x-1)^2$

[twin meteor]

Luh Roub

[barren jolt]

now that is

Getting distributed?

[prisma rain]

nope

put the root back as 1/2

[barren jolt]

sorry

[prisma rain]

and add the exponents

[barren jolt]

[prisma rain]

to get x^1.5

[barren jolt]

$x^{1.5} \cdot (2x-1)^2$

[twin meteor]

Luh Roub

[prisma rain]

yep

[barren jolt]

3.5?

[prisma rain]

where how when

[barren jolt]

Nono nvm

Not same base

mb

[prisma rain]

yep

[barren jolt]

this is so hiv

so now what

i need to distribute this mf

[prisma rain]

write everythin together

with the num

we never distributing it

looks better factored

[barren jolt]

$\frac{-3x+0.5}{x^{1.5} \cdot (2x-1)^2}$

whats the error buddy

ah

[twin meteor]

Luh Roub

[barren jolt]

now

[prisma rain]

almost there

[barren jolt]

[prisma rain]

we have a x^1.5 in the bottom

we want to make it an integer

[barren jolt]

[prisma rain]

exponent

so

we multiply by x^0.5 top and bottom again

[barren jolt]

wait

can we write it as a complex fraction

[prisma rain]

so the x^1.5 turns into x^2

[barren jolt]

nvm

[prisma rain]

nah no negative exponents

[barren jolt]

lets do it that way

[barren jolt]

Yea cuz its in the denom

It turns positive

so we have uh

$\frac{-3x^{1.5}+0.25}{x^{2} \cdot (2x-1)^2}$

am i correct?

I think top is wrong doe

[prisma rain]

just realized

cant get all exponents integers

[barren jolt]

yea?

indeed

[prisma rain]

bc fixing the bottom messes up the top

that was good enough ig

[prisma rain]

deez

[barren jolt]

DISTRIBUTE

Wait

Are we done

[prisma rain]

but

mayhaps

a messed up top is better than a messed up bottom

innit

[barren jolt]

Indeed

I agree

[prisma rain]

so do it again

multiply by x^0.5

[barren jolt]

$\frac{-3x^{1.5}+0.25}{x^{2} \cdot (2x-1)^2}$

[prisma rain]

why the 0.5 turn into 0.25

[barren jolt]

oh i squared it

shit

, w 0.5^0.5

[prisma rain]

wut

[barren jolt]

that?

Multiply by ^0.5

[prisma rain]

why 0.5^0.5

multiply by x^0.5

[barren jolt]

[prisma rain]

0.5 just turns into the coefficient of x^0.5

[barren jolt]

0.5x^0.5

[prisma rain]

ya

[barren jolt]

$\frac{-3x^{1.5}+0.5x^{0.5}}{x^{2} \cdot (2x-1)^2}$

[twin meteor]

Luh Roub

[barren jolt]

that?

[prisma rain]

ja

now double top and bottom

to turn 0.5 into 1

[barren jolt]

goddam

$\frac{-6x^{1.5}+x^{0.5}}{2x^{2} \cdot (2x-1)^2}$

[twin meteor]

Luh Roub

[prisma rain]

for the sake of aesthetic

[barren jolt]

now for the sake of aesthetic we distribute

[prisma rain]

nah

[barren jolt]

lets try doe

[prisma rain]

lgtm like this

[barren jolt]

So we done w diff it?

[prisma rain]

,w distribute 2x²(2x - 1)²

[twin meteor]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by wkaue

[prisma rain]

dum not

,w 2x²(2x - 1)²

[twin meteor]

Results provided by WolframAlpha

Click here to refine your query online
Upgrade to WolframAlpha Pro!

Requested by wkaue

[barren jolt]

????

What integral

[prisma rain]

im tryna get it to distribute it

,w distribute: 2x²(2x - 1)²

[twin meteor]

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[prisma rain]

see

[barren jolt]

,w (2x-1)^2

[twin meteor]

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[prisma rain]

aint look good

[barren jolt]

bro

4x^2-4x+1

(4x^2-4x+1)2x^2

is

,w 2x^2 * 4x^2

[twin meteor]

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[barren jolt]

,w 2x^2 * (-4x)

[twin meteor]

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[barren jolt]

$\frac{-6x^{1.5}+x^{0.5}}{2x^{2} \cdot 8x^4 \cdot (-8x^3) \cdot 2x^2}$

[twin meteor]

Luh Roub

[barren jolt]

W

,w 2x^2 * 8x^4

[twin meteor]

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[barren jolt]

$\frac{-6x^{1.5}+x^{0.5}}{16x^6 \cdot (-8x^3) \cdot 2x^2}$

[twin meteor]

Luh Roub

[barren jolt]

,w 16x^6 * 2x^2

[twin meteor]

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[barren jolt]

$\frac{-6x^{1.5}+x^{0.5}}{32x^8 \cdot (-8x^3)}$

[twin meteor]

Luh Roub

[barren jolt]

,w 32x^8 * (-8x^3)

[twin meteor]

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[barren jolt]

$\frac{-6x^{1.5}+x^{0.5}}{-256^{11}}$

[twin meteor]

Luh Roub

[barren jolt]

@prisma rain

If my distribution is right

This looks good

To me

[prisma rain]

my phone died

[barren jolt]

rip

[prisma rain]

let me see

[barren jolt]

i did some shinanigans

[prisma rain]

where you get this from

[barren jolt]

the (2x-1)^2

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

[twin meteor]

Luh Roub

[barren jolt]

[prisma rain]

you distribute by summing not multiplying

[barren jolt]

[prisma rain]

a(x + y + z) = ax + ay + az

[barren jolt]

what do you mean

yea

but theres no a

theres xyz

[prisma rain]

a is the 2x²

youre distributing

[barren jolt]

Im so gonna crash out rn

[prisma rain]

,w distribute 2x²(2x - 1)²

[twin meteor]

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[prisma rain]

expanded form

thats what you get when you distribute

[barren jolt]

okay

guess i messed up

[prisma rain]

8x⁴ - 4x³ + 2x²

[barren jolt]

so

[barren jolt]

How did wolframalpha get to that result

[prisma rain]

it writes it in a weird way

,w integral (-6x^(1.5) + x^(0.5))/(2x²*(2x - 1)²)

[barren jolt]

So can we safely say

Bro said tan

[prisma rain]

da hell

[barren jolt]

Wtf is $F_1$

[twin meteor]

Luh Roub

[barren jolt]

Thats some physics shit

[twin meteor]

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[barren jolt]

Feels like dividing with complex numbers

[prisma rain]

got sum wrong on the aesthetic stuff

[barren jolt]

true

lets take a gigantic step back

,w integral \frac{-3x^{1.5}+0.5x^{0.5}}{x^{2} \cdot (2x-1)^2}

Can he understand latex

[twin meteor]

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[barren jolt]

wait he can

Uh

Still not quiet it

[prisma rain]

,w integral (-3x^(0.5) + 0.5x^(-0.5))/(x(2x - 1)²)

[twin meteor]

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[barren jolt]

nop

[prisma rain]

,w integral $\frac{-3x^{0.5}+0.5x^{-0.5}}{(x^{0.5}(2x-1))^2}$

[barren jolt]

[twin meteor]

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[barren jolt]

[prisma rain]

zamn

[barren jolt]

We messed up somewhere

[prisma rain]

,w diff (x^(-0.5))/(2x - 1)

[twin meteor]

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[prisma rain]

,w (0.125 x^0.5 - 0.75 x^1.5)/((0.5 - x)^2 x^2)

[twin meteor]

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[barren jolt]

[prisma rain]

dis

[barren jolt]

Ic

(I couldnt follow)

[prisma rain]

,w true or false (0.125 x^0.5 - 0.75 x^1.5)/((0.5 - x)^2 x^2) = (-6x^{1.5}+x^{0.5})/(2x^{2} \cdot (2x-1)^2)

[barren jolt]

he didnt answer

thats crazy

[twin meteor]

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[prisma rain]

so it is right

wolfram just be messing up the integrals

always trust aesthetic

[barren jolt]

Tysm fr

Im so removing this question from the test

btw

[prisma rain]

r u making the test

[barren jolt]

nvm

Yea

Lowkey wanna penetrate my classmates but this is even too much for me

But tbf we didnt get the quotient rule yet

[prisma rain]

derivative itself not even that hard, the simplification part is tought

[barren jolt]

Yes

YES

I suck balls at algebra

even tho im somehow the best in my class

[prisma rain]

daily bprp improved my algebra

[barren jolt]

i always watch bro

I even watched his 7h limit vid

[prisma rain]

zamn

[barren jolt]

thats how i got this formula btw lmao

he taught me this

[prisma rain]

he is the goat

[barren jolt]

fax

good night for now

[vocal sleet]

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[twin meteor]

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

[heavy yoke]

given that, can you find a basis for S?

i would expect the basis vectors to be 2x2 matrices here

that works

yes

R2x2 is not 2-dimensional

you can find a basis for it, see how many vectors that basis has

also a 0-dimensional vector space does exist, it's the vector space with only the 0 vector

well we need each basis vector in our orthogonal compliment to be orthogonal to both basis vectors of S

that's definitely one way to do it

to be honest i think you can just guess two vectors which are orthogonal to the given ones. but if you are stuck then G-S does work

you can use the exact same process for any inner product

bearing in mind to use that inner product for the norm as well

[twin meteor]

gmdn

[heavy yoke]

they are the same

personal preference

you don't need to use any matrices with variable elements

ok we would like to perform G-S in order to get 4 matrices together which are are orthogonal to each other and which span R2x2

our first two vectors should be the basis for S

then we can tack on a basis for R2x2 after that, to ensure that the subsequent list spans R2x2

well to be extra careful we can use all 4

but if you get 2 nonzero vectors from the first two, you don't have to continue

if your first four vectors from GS are nonzero then you can just stop there, rather than doing the last two

yes

i just mean there's no reason for [a, b; c d] to show up

[twin meteor]

gmdn

gmdn

[heavy yoke]

remember that your inner product here is the trace

also remember that the first vector in GS is just the first basis vector unchanged

you only start subtracting things for the second basis vector

[vocal sleet]

@vast shale Has your question been resolved?

[heavy yoke]

yes

[heavy yoke]

that just means that v1 and v2 were orthogonal in the first place

to each other

not necessarily

G-S will return the same basis back if the basis was orthogonal in the first place

they can be but they don't have to be

a set of vectors is called orthogonal if all of its vectors are orthogonal to each other

so GS takes a basis and returns an orthogonal basis of the same space

yes

well what we would like to do is compute a single basis of R2x2 which is orthogonal [to itself]

we would like two of those four vectors to form a basis of S

then the remaining two vectors will be perpendicular to S (since all 4 are perpendicular to each other)

[heavy yoke]

meaning they form a basis for the orthogonal compliment

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[tribal sapphire]

what am i doing wrong?

[vocal sleet]

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[solemn fulcrum]

Can somebody please help with this question

[torn timber]

alright they have given the sum of first n terms in a sequence to be equal to n^{2}+3n

now they asked you to find the first four terms of this seequence

[solemn fulcrum]

Yeah

But I’m not really sure what to do

[torn timber]

now you can notice one thing the sum of first 1 terms is the 1st term itself

[solemn fulcrum]

Oh yeah

So the first term is 4?

[torn timber]

yes correct

now assume $a_1,a_2,...a_n$ are the first n terms of a sequence then $S_1=a_1,S_2=a_1+a_2, S_n=a_1+a_2+\dots +a_n$

[solemn fulcrum]

Um

I’m not really sure what that means

[twin meteor]

convergence

[solemn fulcrum]

Oh wait so is the second term 4 + 10

[torn timber]

S_n is just a notation of the sum of first n terms

[torn timber]

no

[solemn fulcrum]

Is it just ten?

[torn timber]

no

[solemn fulcrum]

I’m a bit confused

So what is the second term

[torn timber]

did you understand this?

[solemn fulcrum]

Um a little bit

But not fully

What’s a1 and a2

[torn timber]

the first and second term of the sequence

[solemn fulcrum]

and s2 is the second sim?

Sum?

[torn timber]

yes S_2 is the sum of the first 2 terms of the sequence

[solemn fulcrum]

So it is 14?

Wait

[torn timber]

no a_2 is not 14

[solemn fulcrum]

a2 is 10

?

Or does that not work

By just plugging 2 into n

[torn timber]

alright let me give you an example consider this sequence {1,2,....n} now the sum of first n terms of this sequence is given by (n(n+1))/2

now if we plug 2 in n(n+1)/2 we get 3 but is 3 the second terms in this sequence

[solemn fulcrum]

no

[torn timber]

so do you understand how the sum is different than the term?

[solemn fulcrum]

Yeah

So what would the second term in the original equation be?

[torn timber]

let me get to that point

[solemn fulcrum]

Okay

[torn timber]

if you notice carefully you will find that $S_n=S_{n-1}+a_n$

[twin meteor]

convergence

[torn timber]

that is the sum of fist n terms is the same as the sum of the first n-1 terms and the nth term of the sequence

[solemn fulcrum]

Okay

[twin meteor]

convergence

convergence

[solemn fulcrum]

I think so

[torn timber]

i think you can find a2 now

[solemn fulcrum]

I think I understand the things u explained but not fully enough to work it out 😭

Do u mind telling me the first 4 terms and I try see how you got to them

This might be wrong but is the second term 6

[torn timber]

alright s2=s1+a2 you know how to find s2 and s1 using this

[torn timber]

it is 6 good job

[solemn fulcrum]

Oh

Thank u so much

[torn timber]

np

[solemn fulcrum]

Wait so the first 4 terms would be 4,6,8,10?

@torn timber

[torn timber]

yes

[solemn fulcrum]

Okay thank u

[torn timber]

good job

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[visual kettle]

I don't understand the generalised Stokes' theorem. I tried to understand what a manifold is and if I am right, it is just a locally Euclidean space. I don't really understand what Euclidean means. I also don't understand what the tangent space is. I understand that differential forms are linear functions of tangent vectors to reals, precisely to the determinant of the matrix whose columns are the tangent vectors. However, I don't really think this makes sense. Finally, I don't understand how to integrate a differential form? Do we take the sum over the points of the manifolds and just plug in tangent vectors or what exactly? Thanks.

[visual kettle]

<@&286206848099549185>

[arctic mantle]

okay that's differential geometry

if you don't really understand euclidean spaces, or tangent spaces, that's way beyond your level

[visual kettle]

Is there no way I ever understand it?

[steep crater]

you could try but i would take lots of learning

[visual kettle]

I tried searching for definitions that are a little dumbed down but it is always very complicated

[steep crater]

thats differential geometry

i dont understand it at all

[jagged cargo]

why are you studying diff geo if you dont even know what euclidean space is

[visual kettle]

I mean I know that it is the flat space we are familiar with but I don't know how to describe it mathematically. Is it just tuples of real numbers or what? Is it a vector space with the standard basis? The tangent space is defined as the span of partial derivative operators but I don't really get that definition.

[arctic mantle]

i mean

it's fairly easy

all linear combinations of partial derivatives at each parametrized point of the surface?

[jagged cargo]

if you dont know what this means for you, then i dont think you should take on diff geo

maybe get some grip on topology and RA first, i didnt study diff geo but i think thats the least you can do

[arctic mantle]

differential geometry is confusing man, i should know

learning linear algebra and multivariable calc will also be very helpful

especially vector calculus

[visual kettle]

But I know that people can start discussing the generalised Stokes' theorem right after multivariable and vector calculus so do I really need analysis and topology?

I just want a basic understanding of the theorem. Up till now, it seems like random definitions.

[arctic mantle]

...?

no not really

;-;

i have a hard time understanding it myself

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@visual kettle Has your question been resolved?

[visual kettle]

No

[arctic mantle]

and why not?

like we said, try learning some basic topology and analysis before discussing GST

[visual kettle]

Ok

.close

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[grand raven]

\textbf{Suppose that you begin with a pile of} $\mathbf{n}$ \textbf{stones and split this pile into} $\mathbf{n}$ \textbf{piles of one stone each by successively splitting a pile of stones into two smaller piles. Each time you split a pile you multiply the number of stones in each of the two smaller piles you form, so that these piles have} $\mathbf{r}$ \textbf{and} $\mathbf{s}$ \textbf{stones in then, respectively, you compute} $\mathbf{rs}$. \textbf{Show that no matter how you split the piles the sum of the products computed at each step equals} $\mathbf{n(n-1)/2}$ Im supposed to use induction to prove this, but im not quite sure even where to begin...

[twin meteor]

Oliver

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[royal grove]

how do i solve this?

.close

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[vocal sleet]

@fossil pollen Has your question been resolved?