help-0

[gleaming ridge]

yess

Welcome!

[rain birch]

.close

[lone heart]

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[wide marlin]

Can someone walk me step through step on how to do difference quotients with f(x)=2/x

[lone heart]

@wide marlin Has your question been resolved?

[gleaming ridge]

$$f'(x)=\lim_{h \to 0} \frac{f(x+h)-f(x)}{h}$$

[ocean seal]

LUNA

[gleaming ridge]

$$f'(x)=\lim_{h \to 0} \frac{\frac{2}{x+h}-\frac{2}{x}}{h}$$

[ocean seal]

LUNA

[wide marlin]

yes im stuck there

idk whats next

[gleaming ridge]

Multiply both numerator and denominator by x(x+h)

To get rid of the little fractions

[wide marlin]

can you show me what you mean and how you got that

[gleaming ridge]

You have 2 little fractions in the numerator of the main fraction

[wide marlin]

yes

[gleaming ridge]

We want to get rid of them

[wide marlin]

ok

[gleaming ridge]

if you multiply both numerator and denominator by x(x+h), those fractions will be simplified

Look

$$f'(x)=\lim_{h \to 0} \frac{x(x+h) \frac{2}{x+h}-x(x+h) \frac{2}{x}}{x(x+h)h}$$

[ocean seal]

LUNA

[wide marlin]

so what will it look like after multipliyed

[gleaming ridge]

Numerator, first term, x+h will cancel with x+h, x remains

Numerator, second term, x will cancel with x, (x+h) remains

[wide marlin]

ok

[gleaming ridge]

$$f'(x)=\lim_{h \to 0} \frac{2x-2(x+h)}{x(x+h)h}$$

[ocean seal]

LUNA

[wide marlin]

ok but why do we also mutiply the denominator

[gleaming ridge]

You see?

Because you change the fraction if you multiply by numerator only!

[wary stream]

Btw the difference quotient is just a bit different, it doesn't involve the limit as h goes to 0

[wide marlin]

ok

[gleaming ridge]

idk, this is what i know :p

[gleaming ridge]

Simplify numerator

[wide marlin]

ok

[gleaming ridge]

$$f'(x)=\lim_{h \to 0} \frac{2x-2x-2h}{x(x+h)h}$$

[wary stream]

This is limit definition of derivative or whatever it's called

[gleaming ridge]

$$f'(x)=\lim_{h \to 0} \frac{-2h}{x(x+h)h}$$

[wide marlin]

wait

isnt -2(x+h) =-2x-2h

[wary stream]

That's difference quotient, just limit def of derivative minus the limit part

[gleaming ridge]

yeah so basically im doing the limit of it, i thought they were looking for the derivative of the function xD

[gleaming ridge]

True!

[ocean seal]

LUNA

LUNA

[gleaming ridge]

h's cancel out

$$f'(x)=\lim_{h \to 0} \frac{-2}{x(x+h)}$$

[ocean seal]

LUNA

[wide marlin]

ok

[gleaming ridge]

$$\frac{-2}{x(x+h)}$$

[ocean seal]

LUNA

[gleaming ridge]

This right here is the difference quotient

Not the derivative

This is what they are looking for

ignore the f'(x) and the limit

[wide marlin]

so now what

[gleaming ridge]

i thought we were looking for the derivative

[gleaming ridge]

That's the answer!

[wide marlin]

oh nothing else

[gleaming ridge]

That is the difference quotient of your function

if you want the derivative of the function 2/x

you take the limit of the difference quotient as h approaches 0

$$f'(x)=\lim_{h \to 0} \frac{-2}{x(x+h)}$$

[ocean seal]

LUNA

[gleaming ridge]

And you get:

$$f'(x)=\lim_{h \to 0} \frac{-2}{x , x}$$

[ocean seal]

LUNA

[gleaming ridge]

$$f'(x)=\frac{-2}{x²}$$

[wide marlin]

ok

[ocean seal]

LUNA

[wide marlin]

I dont need this

how do I do the other differnce quotient now

[wary stream]

What other one?

She did it for you

[wary stream]

.

[wide marlin]

f(x)-f(a)/x-a

[wary stream]

That's slope

[wide marlin]

can you help me through that one do I just do the same thing

well its asking for it

[wary stream]

Just plug in x and a into the function divided by x - a

Then simplify

[wide marlin]

how do I simplify mutiply both sides by x and a

[wary stream]

Yeah

[wide marlin]

can you help me with this one

[wary stream]

Just plug in x and a into the function divided by x - a
Then simplify

[wide marlin]

(x^2+xa+a^2)-2x+2a

how do I simplify

[wary stream]

There's nothing more you can do

[wide marlin]

it says the answer is x^2+ax+a^2-2

[wary stream]

Then you did something wrong

[wide marlin]

yea I did nvm

[lone heart]

@wide marlin Has your question been resolved?

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[wide marlin]

How would I find the polynomials for f that satisfy the equation
f(f(x))=x^4-12x^2+30

[wide marlin]

can someone walk me through it and how to reverse the equation

[vapid shuttle]

take f^(-1) of both sides

[wide marlin]

can you show me it please

[vapid shuttle]

f^(-1)(f(f(x))=f^(-1)(x^4-12x^2+30)

[wide marlin]

ok now what

[vapid shuttle]

simplify

[wide marlin]

how

[vapid shuttle]

what is f-inverse of f(x)

[wide marlin]

idk

how would I figure that out

[vapid shuttle]

:|

what is the point of an inverse function

do you know?

[wide marlin]

no

[vapid shuttle]

If I gave you

y=5x

[wide marlin]

yea

[vapid shuttle]

and I told you to find the inverse function

you'd have no idea what that means

[wide marlin]

no I forgot

[vapid shuttle]

basically, instead of inputting a value of x, and getting an output, our inverse function takes an output, and tells you in the value of x that produces it

so for our y=5x example

[wide marlin]

yea

[vapid shuttle]

$y=5x$

[ocean seal]

Austin

[vapid shuttle]

$\frac{y}{5}=x$

[ocean seal]

Austin

[wide marlin]

oh ok

[vapid shuttle]

then we can answer things like

if y=20

what is x

using our inverse function

[wide marlin]

y/20

[vapid shuttle]

$f^{-1}(y)=\frac{y}{5}$

[ocean seal]

Austin

[vapid shuttle]

so

[wide marlin]

yea

[vapid shuttle]

what is $f^{-1}(f(x))$

[ocean seal]

Austin

[wide marlin]

uh

idk

x

[vapid shuttle]

yes

[vapid shuttle]

so now simplify the left hand side of this

[wide marlin]

so what is it

[vapid shuttle]

$f^{-1}(f(f(x)))=f^{-1}(x^4-12x^2+30)$

[ocean seal]

Austin

[vapid shuttle]

simplify the left hand side using what you just told me about f-inverse of f

[wide marlin]

fx?

[vapid shuttle]

use parenthesis

[wide marlin]

(fx)

[vapid shuttle]

....

[wide marlin]

idk then

[vapid shuttle]

(fx) is not a function

it is meaningless notation

[wide marlin]

idk what to do

[vapid shuttle]

stop guessing then

when have you ever seen (fx) ever in your life

clearly you are typing something wrong

think for 2 seconds about how to fix your typo

and then we can continue

[wide marlin]

y

?

im just geussing

[vapid shuttle]

I can tell

[wide marlin]

how would I be able to find what it is

[vapid shuttle]

and I'm not going to keep helping you if you don't participate in the learning process

[wide marlin]

bro wdym

idk what im doing

[vapid shuttle]

I know

and I am teaching you how

but I would be able to better help you, if you would stop guessing

and actually think about what I am showing you

[wide marlin]

then what do you want me to do

[vapid shuttle]

I'll just show you this again

what is

$f^{-1}(f(x))$

[ocean seal]

Austin

[wide marlin]

the inverse of the function

[vapid shuttle]

which is equal to ?

[wide marlin]

x

[vapid shuttle]

yes

what about

$f^{-1}(f(a))$

[ocean seal]

Austin

[vapid shuttle]

equal to ?

[wide marlin]

a

[vapid shuttle]

ok

what about

$f^{-1}(f(axb+5))$

[ocean seal]

Austin

[wide marlin]

axb+5

[vapid shuttle]

yes

so basically showing you that

whatever is on the inside

doesn't matter

[wide marlin]

ok

[vapid shuttle]

the f-inverse cancels the f

now I will ask you again

what is

$f^{-1}(f(f(x)))$

[ocean seal]

Austin

[wide marlin]

f(x)

[vapid shuttle]

yes

[wide marlin]

ok now what

[vapid shuttle]

so simplify the left hand side

[wide marlin]

f(x)

[vapid shuttle]

and what do we have now

$f(x)=f^{-1}(x^4-12x^2+30)$

[ocean seal]

Austin

[wide marlin]

ok

[vapid shuttle]

so

the polynomials that satisfy your equation

are the inverse of (x^4-12x^2+30)

so all that is left to do

[wide marlin]

ok

[vapid shuttle]

is find out what that is

[wide marlin]

how

[vapid shuttle]

if $y=x^4-12x^2+30$

[ocean seal]

Austin

[vapid shuttle]

what is $f^{-1}(x)$

[ocean seal]

Austin

[vapid shuttle]

same way we did it for $y=5x$

[wide marlin]

y/all of that

[ocean seal]

Austin

[wide marlin]

?

[vapid shuttle]

how did we find the inverse of y=5x

[wide marlin]

diving 5 by y

[vapid shuttle]

which in turn

solved for x

[wide marlin]

yea

[vapid shuttle]

so the goal is to solve for x

so take $y=x^4-12x^2+30$ and solve it for x in terms of y

[ocean seal]

Austin

[wide marlin]

oh

uh

nvm

do I do y/30

[vapid shuttle]

no

[wide marlin]

the whole equation?

[vapid shuttle]

I'm not going to do the question for you

I've shown you what the process is up until this step

[wide marlin]

ok

[vapid shuttle]

now its your turn to try working some things out

[wide marlin]

well idk what to do

[vapid shuttle]

I told you that you need to solve for x in terms of y

and you haven't tried anything yet

so go and try some stuff

it is okay if you don't get it right immediately

try again

[wide marlin]

im not sure where to start

[vapid shuttle]

well then now is the time that you can guess some things

so take your paper

and just start guessing

try something

see if it works

if it doesn't

try something new

[wide marlin]

nothing works

idk

[vapid shuttle]

you only tried for 2 minutes

[wide marlin]

do I set it to 0

[vapid shuttle]

no, you solve for x in terms of y

[wide marlin]

idk what that means

[vapid shuttle]

make it of the form x=f(y)

x by itself, set equal to some y's and stuff

[wide marlin]

yea idk

can you just guide me through it

[vapid shuttle]

no

[wide marlin]

then idk

I have the answer I just dont know how to do it

[vapid shuttle]

what is the answer

[wide marlin]

x^2-6

so now what

.close

[lone heart]

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[junior warren]

could someone help me find where i went wrong calculating determinant

supposedly it is 24

i row reduced so i could use cofactor expansion along column 1, then used cofactor expansion again along row 2

disregard its supposed to be 3(-12-8)

as im still wrong

[maiden glen]

you need a - on the 3 on the second cofactor expansion

[junior warren]

oh

right

[maiden glen]

remember to alternate + and - signs

[junior warren]

stupid

yeah

will that fix it anyways?

i dont think so

i now have 60

a bit off 24 lol

i mustve row reduced incorrectly?

yup

row 4 is wrong

-6 should be 0

making 24

nice

thanks for the reminder on that sign

[lone heart]

@junior warren Has your question been resolved?

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[junior warren]

to prove each of these, do i just perform a left and right side check?

using the criteria for the space; x1+x3=x4, x2=0?

[tardy stag]

p much yeah

just set up the equations and poke them

[junior warren]

so just notation wise, would doing it like this be correct?

i am just a little lost on show to actually show it meets this criteria

[tardy stag]

hmm what's S?

[junior warren]

actually what is S 😭

[maiden glen]

you can just start with two arbitrary vectors in the set V

[junior warren]

i wrote it as likje

pre transformation

[maiden glen]

you know x_2 =0, so just write that in immediately

there isn't a transformation being applied here

[junior warren]

true

[maiden glen]

to show a vector space is a subspace, you only need to verify closure under vector addition and scalar multiplication

[junior warren]

and 0 right

[maiden glen]

so take two arbitrary vectors in the space, add them, and show the result is in the space

yes, the space needs the zero vecor

[junior warren]

oh i can just choose 2 random examples?

[tardy stag]

technically that's covered under scalar multiplication

[junior warren]

i dont have to prove generally?

[maiden glen]

by arbitrary I mean general vectors

[junior warren]

oh

ok

[maiden glen]

as in [a,0,b,a+b]

[junior warren]

right

i am just trying to remember it was written something like this whenever i saw it

but not exactly like this, since S makes no sense lol

[maiden glen]

that's to show a transformation S is a linear transformation

[junior warren]

is there a better way i can notate what i did?

yeah

[maiden glen]

yes, U + W in V for all U, W in V

and kU in V for all U in V and k in the base field

[junior warren]

right

so if i take away the uselesss S, would by notation be fine?

[maiden glen]

you need for all v,w,k there

but yes

[junior warren]

we learned to do a LS=RS check

ok

[maiden glen]

not sure how LS=RS would work here

but if you have a method you learnt, go for it

[junior warren]

for example we were just told to show S(v+w)=S(v)+S(w)

so that is i guess where S comes from?

[maiden glen]

yeah, that's for verifying transformations S are linear

here you're just adding two vectors together

you're not showing the vector is equal to anything

you're showing it lies in V

[junior warren]

oh wait

yeah

wow

im so lost

yes ok following now

this isnt a transformation

[maiden glen]

take [a,0,b,a+b] and [c,0,d,c+d]

add them together

show that the result follows the constraints required to be in V

[junior warren]

and show its in the subspace

set*

[maiden glen]

subspace is fine

well, we don't know it's a subspace yet, but

yeah

[junior warren]

so i need to verify v+w is in the set

as well as kv?

[maiden glen]

if v and w are in the set, and if k is a scalar from the base field yes

[junior warren]

ok i see

[junior warren]

so would this be better

[maiden glen]

I would bracket out the terms to show what you're doing more clearly, but yes

[junior warren]

ok

was just trying to get the general idea in

but yeah statements and better formatting ofc

[maiden glen]

[(v1+w2) + (v3+w3) = x1 + x3, for example]

[junior warren]

yeah

i see

[maiden glen]

and also the subscript spam is hard to read

I would just use a,b,c,d

[junior warren]

i was speedrunning it

[maiden glen]

or x,y, u,v (o.e.) for pairs

[junior warren]

but i agree

ok this cleared up this question i was confusing it with transformations i think because they have similar criteria?

anyways tysm

[narrow terrace]

hi i need help, am i interrupting?

[junior warren]

no but type here in like maybe 50 seconds

.close

[lone heart]

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[maiden glen]

I think we're mostly done here, but you should open a new channel

#❓how-to-get-help

[buoyant storm]

wait whats the bot command for that

the thing that makes the bot say "please read #❓how-to-get-help "

[maiden glen]

!help

[lone heart]

Please read #❓how-to-get-help

[buoyant storm]

oh ok

thanks

[lone heart]

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[cinder sundial]

Here’s my attempt

i know that the essence (or origin) of integral is the Riemann sum

[cinder sundial]

in turn, by transforming the given "Sn" in the question into the form of integral

we will be able to get the answer

yet, i dont know how to do it

however, the book do give some useful advice, which is to express the k/n in f(k) as x

and then we can transform the Riemann sum into the integral from 0 to 1 of (1+x)^4

[cinder sundial]

but whats the point of expressing the k/n in f(k) as x?

.close

[lone heart]

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[flat vale]

$\mapsto$ vs $\to$ whats the difference?

[ocean seal]

TheWhiteShadow

[carmine reef]

The first one means f(a) = b

The second one means the domain of f is A and the codomain is B

[flat vale]

so i cant use $\mapsto$ for domain and vice versa?

[ocean seal]

TheWhiteShadow

[echo socket]

Yeah

[flat vale]

icic tq

.close

[lone heart]

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[hollow lark]

How do I do this with normal highschool level math?

[remote heron]

have doubts this is possible

be curious if theres a trick though

[hollow lark]

Yea lmao it was in one of my past papers

The answers are ovb wrong

[maiden glen]

I just ran it through a cas and it says no elementary integral

[hollow lark]

Coz they use t sub

[remote heron]

i mean its normal where theres a combination of x and trig functions to have no antideriv

if its just constants and trig you can get away with half angle

[crimson carbon]

Did you try run it through wolfram

[maiden glen]

,w integral (x*cos(x)+sin(x))/(x^2+sin^2(x)) dx

[ocean seal]

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

Requested by .desync

[maiden glen]

😭

[lone heart]

@hollow lark Has your question been resolved?

[hollow lark]

Yea lol they use complex numbers on integral calc

[lone heart]

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[crimson carbon]

never seen the use of contour integrals in hs calc

[lone heart]

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[molten oak]

what does this mean?

y | y

[rose sigil]

more context?

[molten oak]

Here

[alpine sable]

Set builder notation

[crimson carbon]

just means "such that"

[alpine sable]

Can be read "for y such that"

[molten oak]

thanks so much

[crimson carbon]

you might also use the colon, : (which is more common)

[molten oak]

.close

[lone heart]

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[crimson carbon]

instead of |

[lone heart]

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[cedar juniper]

43

,rotate

[cedar juniper]

How do I proceed?

<@&286206848099549185>

[frail rain]

Hey

[cedar juniper]

Hi

<@&286206848099549185>

[lone heart]

@cedar juniper Has your question been resolved?

[cedar juniper]

<@&286206848099549185>

[iron swallow]

Have you checked what you get if you put the last equation into a calculator and compare to what you are suppsed to get? 🙂

[cedar juniper]

I get two answers

Each of them with a + - cuz of the root

How do i prove that its the ans they want?

[iron swallow]

$\sqrt{\frac{65+63}{2*65}} = \frac{8}{\sqrt{65}}$

damn

[ocean seal]

Cookie

[cedar juniper]

Yea but theres a +- before 63

And a +- outside the root

how do I prove that it will be + inside and + outside

[lone heart]

@cedar juniper Has your question been resolved?

[alpine sable]

hi , can someone say if there is an error ?

[cedar juniper]

<@&286206848099549185>

#❓how-to-get-help

[drifting seal]

Whats the problem???

[cedar juniper]

.

[drifting seal]

Half angle huh

The thing u need to find is cos(a-B)

[cedar juniper]

.

[cedar juniper]

Cos (A - B / 2)

[drifting seal]

ik but u still need to find cos(a-B)

and ik u did that

and I have calcualte the correct answer

[drifting seal]

why 65 +/- 63

u should add them

65 + 63

and u will get 128/65/2

that is 64/65

simplify the numerator

then u get the answer

problem solved

@cedar juniper

[cedar juniper]

Theres no gurantee cos b and sin a are positive

[drifting seal]

but the answer need u to do that

or u right two answers like: 1 or 2 (rejected)

someth'in like that

[lone heart]

@cedar juniper Has your question been resolved?

[high mortar]

do one thing

u can use regular trignometric identites

cos beta value is know
consider beta to be some 2theta cos2theta=2cos square theta-1
then resubstitute value where theta is half beta
to same with sin alpha
then it can be soled

[lone heart]

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[buoyant linden]

can someone explain how the integration formula is gotten from the differentiation formula?

[torpid nexus]

The antiderivative is defined as the inverse operation of the derivative

If you want to know how that relates to area under the curve then it's a different story

[ruby current]

What is “the integration formula” and “the differentiation formula?”

[buoyant linden]

nah i just wantt o know how differentiation formula is turned into integration

like how its divided by 1/n+1 or something

idk

[torpid nexus]

well you just try stuff out, and make a list of basic integration formulas, with which you find the integration of other complex functions.
eg: why is the integration of cosx sinx? Because if you differentiate sin(x) you get cos(x). With some elementary derivatives you can get the "basic" list

[buoyant linden]

so integration formula doesnt root from differentiation formula?

[torpid nexus]

what do you mean root

its just the "opposite"

[buoyant linden]

'based off of'

i meant

[torpid nexus]

yes it is based off of differentiation formulas as i said

[buoyant linden]

is it very complicated to understand how it is based off of them?

didnt realise that typo being there for the past 10 mins lmao 💀

[modern sedge]

I guess you are reffering to integral of $x^n$.
since $\frac{d}{dx}x^{n}=nx^{n-1}$ and integrals are opposite of derivatives, you can say that $x^{n} + c=\int nx^{n-1}$. If you divide both sides by n, you get $\frac{x^{n}}{n}+c=\int x^{n-1}$ and when you finally substitute $u=n-1$, you get $\frac{x^{u+1}}{u+1}+c=\int x^{u}$

[ocean seal]

EmilyIsAlwaysRight

[modern sedge]

This is how most integration rules are derived. U take some differentiation rule, integrate both sides and keep manipulating it until you reach some nice-looking integration rule

[buoyant linden]

ok

ig ill try to understand them for the rest of this week 🗿

.close

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[alpine sable]

We also know that the area of the water surface A = pi*(20h-h^2). When h = 12, there's a small change of h by p%, by using differentiation, how do I find the small change of A in terms of p and pi?

[torpid nexus]

take log and differentiate.
dh/h = p/100

da/A = ( 20-2h)/(20h-h^2) *dh

[alpine sable]

wdym take log?

I don't understand

[torpid nexus]

Take the natural log on both sides

[alpine sable]

because based on my understanding to find delta A, we'll have to do limit delta h approaching p, delta A/delta h

while delta A = f(h + delta h) - f(h)

[torpid nexus]

yeah well since it's a very small change i told to differentiate

it's an approximation basically

[alpine sable]

yep I know

before approximating anything you have to understand it first

so I'm going down the path of understanding it, rather than being able to solve it

[torpid nexus]

And since you are given dh/h you would want to take the natural log on both sides and differentiating both sides, to recover dh/h

[alpine sable]

what does it mean by dh/h?

[torpid nexus]

fractional change

[alpine sable]

which value is dh/h here?

[torpid nexus]

p/100

as you said it changes by p%

[alpine sable]

yep

ah

so p% = p/100

[torpid nexus]

ye

[alpine sable]

wait let me process it a bit

so the small change refers to dh/h

[torpid nexus]

small fractional change

[alpine sable]

so dh does not actually equal to p/100

wait why do we have to add a /h?

[torpid nexus]

well just think of what it means when some quantity increases by some percentage

[alpine sable]

right, so based on my understanding, small change in h by p% means dh = p%

[torpid nexus]

suppose a price increases by 10%

final price - initial price / initial price = 10/100

[alpine sable]

yep

[torpid nexus]

yep so of height increases by p%

final height - initial height / initial height = p/100

final height - initial height = ∆h and for very small quantities it's dh

so dh/h

[alpine sable]

ohhhh so dh/h = p/100

let me write it down

so dh/h = p% = p/100

makes sense

so whats the next step?

because we want to find dA

<@&286206848099549185>

[torpid nexus]

here

[alpine sable]

wait but da/A?

why isn't it dA?

[torpid nexus]

since you took the log and differentiated

[alpine sable]

because we're finding the percentage again?

[alpine sable]

what log?

[torpid nexus]

that's why you got da/A

well just differentiate both sides actually m

[alpine sable]

wait what is log

what log

[torpid nexus]

A=pi(20h-h^2)

don't take log

[alpine sable]

i dont see any log()

[torpid nexus]

differentiate both sides

that's faster

[alpine sable]

i know dA/dh = pi(20-2h)

[torpid nexus]

yeah now multiply by h both sides

[alpine sable]

but to turn dA/dh into dA shouldn't i multiply dh rather than h?

[torpid nexus]

or do Da = pi (20-2h) * dh

[alpine sable]

ah yep

[torpid nexus]

then pi(20-2h) * h * dh/h

Then you're done

[alpine sable]

so dh = (p*h)/100

[torpid nexus]

same thing basically

[torpid nexus]

correct

Then sub in h= whatever you got

[alpine sable]

so dA = pi(20-2h) * (p*h)/100

[torpid nexus]

Ye

[alpine sable]

ahhhh

thank you so much

i'm closing this channel now

.close

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[muted relic]

how does t-mT become t

[muted relic]

basically t+T - mT is t?

or am i misunderstanding the definition of periodicity

[ruby current]

not sure what you’re asking, but they’re performing a change of variables with m = n - 1

as n runs from -inf to inf, so does n - 1

[muted relic]

hm ok

oh so as long as n remains it doesnt matter what is added or subtracted

still is periodic?

[ruby current]

what is periodic?

[muted relic]

uh

g(t)

this was the question

[ruby current]

for g to be periodic with period T, you need to show g(t) = g(t + T) for all t

[muted relic]

ok

.close

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[muted relic]

thanks

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[alpine sable]

Prove that (\sqrt{5}) is irrational.

[ocean seal]

scholablade

[alpine sable]

\section{7.6}
Suppose that (p) is a prime number.
\section{Lemma}
Let (a), (b) and (c) be integers, and let (p) be a prime. \
(i) If (p - a), then (gcd(p, a) = 1). \
(ii) If (a | bc) and (gcd(a, b) = 1), then (a | c). \
(iii) If (p | bc), then (p | b) or (p | c) (or both). \
\subsection{(a) Prove that (\sqrt{5}) is irrational.}
Assume for the sake of contradiction that (\sqrt{5}) is rational, then there must be some postive integers (a) and (b) where (\sqrt{5} = \frac{a}{b}) and assume that the previous fraction was written in lowest terms with no common divisors then we can do (\sqrt{5}a = b) and squaring gives us (5a^2 = b^2), since (b^2 \in \mathbb{Z}), this means that (5 | b^2) thereby (5 | b) by lemma (iii) so we can safely say that (b = 5k) and k is a positive integer, subsituting gives us (5a^2 = b^2) to (5a^2 = (5k)^2) to (5a^2 = 25k^2) to (\frac{a^2}{5} = k^2) and since (5 | a^2) or (5 | q) by lemma (iii) we can say this is a contradiction because we have said before that (a) and (b) have no common factors yet (5) divides the both of them therefore (sqrt{5}) is irrational.

[ocean seal]

scholablade

[drifting seal]

Then what's your question?

[tardy tapir]

this is correct

[drifting seal]

agree

[alpine sable]

Prove that (\sqrt{5}) is irrational

Oh

[drifting seal]

u alr prove???

[alpine sable]

I wanted to see if it was correct

[drifting seal]

ys when u proofed that sqrt 5 is irrational, then you're correct

trust yourself

and the power of math 😉

[alpine sable]

Still, am a beginner though and these are not proved by a computer

[drifting seal]

oh lol

[alpine sable]

Thanks for the confidence boost though

[drifting seal]

np

[alpine sable]

I'l close this room now, thanks for the help

.close

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[alpine sable]

Cartesian writing of the complex (1 + 𝑖)^15.

[alpine sable]

can you help me ?

[upbeat hornet]

Yes

[upbeat hornet]

do you know the binomial theorem

[echo socket]

Polar form would be easier I think

[alpine sable]

yes

[lime bobcat]

What is the modulus of 1+i?
What is the argument of 1+i?

[upbeat hornet]

you could use it here, or use polar form

[alpine sable]

modulus : sqrt(2)

arg = pi/4

is that right ?

[lime bobcat]

Yes

[dawn socket]

When you roll 3 dice whats the possibilitie to get 18

Answers are

1/18

1/6

[lime bobcat]

This channel is occupied.

[tardy tapir]

get your own help channel

you already have a forum

[dawn socket]

I know but I really need help

[lime bobcat]

What happens to the modulus and to the argument when you multiply two numbers?

[lime bobcat]

There are free help channels, ask there.

[alpine sable]

so I have sqrt(2)(cos(pi/4)+i sin(pi/4)

)

[lime bobcat]

What happens to the modulus and to the argument when you multiply two numbers?

[alpine sable]

I don't really know

[lime bobcat]

When two numbers multiply the modulus are multiplied and the arguments are added

You have to multipliy the modulus and sum the arguments

[alpine sable]

okay

[lime bobcat]

So in this case you multiply (1+i) with itself 15 times

What is the modulus of (1+i)¹⁵?
What is an argument of (1+i)¹⁵?

[alpine sable]

.close

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[undone cliff]

hello, i want to ask question about 0/0 = undefined/unlimited?

[echo socket]

0/0 is undefined

[undone cliff]

ohh

[echo socket]

Whatever unlimited may mean

[undone cliff]

but btw is there any unlimited answer in math other than functional question

[echo socket]

What do you mean by "unlimited" then?

[pallid scarab]

Unlimited as in "we can't take the limit of this at 0" ?

Like $\lim_{x\to 0}\frac{x}{|x|}$?

[ocean seal]

rafilou2003

[undone cliff]

No 6(the value of tan(cos‐1(-1)) = cosec 180
E undefined

The ans D or E

[pallid scarab]

it's not a 0/0 case, it's a 1/0 case

Which is also undefined

[undone cliff]

yeah, I'm just curious about 0/0 too

[undone cliff]

but cosec 180 in sin cos tan table is ∞

[pallid scarab]

sure, if you don't care about the sign

But if you do, there is a difference between +∞ and -∞

[undone cliff]

um is the real value of cosec 180 is +∞?

[pallid scarab]

neither +∞ nor -∞

it's 1/0

you "can" define it as ∞, but that's it

[undone cliff]

oh it doesn't mean unlimited but it's just a "sign"?

alright thanks for the time and the explanation

.close

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[slate acorn]

I recall being taught in school that a decimal like 24.445 would round (to the tenths) to 24.5 (24.445 -> 24.45 -> 24.5), but I've seen else where that it would round to 24.4 (24.445 -> 24.4). What is the discrepancy?

[tardy stag]

I would not ever round 24.445 to 24.5

the procedure you describe is a good example of how rounding errors can start to pile up

[maiden glen]

when rounding, you're trying to drop some of the significant digits while keeping as much information as possible; you want to round to the closest number possible

you can see that 24.445 is closer to 24.4 than 24.5

so the second way is more accurate, in that, you end up with a closer result with the same number of digits retained

[slate acorn]

Gotcha, thanks a bunch. I can't believe I didn't notice such a basic fundamental issue sooner...

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[alpine sable]

knowing that w=e^2iπ/5. I calculated that w^5 = 1. And now I have to deduce that 1 + 𝑤 + 𝑤^2 + 𝑤^3 + 𝑤^4 = 0

[alpine sable]

how can I proceed ?

[umbral star]

factor the polynomial z^5-1, then use the fact that w satisfies the polynomial

[median oar]

You can draw it

You get a pentagon to the origin

[pallid scarab]

in your last question you asked me what was (1-z)(1+z+...+z^(n-1))

this might be related ;)

[alpine sable]

oh yeah, thanks

.close

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[alpine sable]

.reopen

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✅

[alpine sable]

.close

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[bright remnant]

how is it possible to have 4sin(x)=x

[limpid spade]

wdym

[echo socket]

Why do you think it's impossible?

[bright remnant]

because 1 constant would just equal another

lets say x=2

4sin(2) = 3.6

x=2

3.6=/=2

[echo socket]

Okay? That doesn't mean 4sinx = x has no solutions

I might as well prove that x^2 - x = 0 never happens

[tardy stag]

that just means that x=2 is not a solution of that equation

[echo socket]

Try x = 0

[bright remnant]

that works

[echo socket]

Yeah, so 4sinx = x indeed has solutions

Here is the graph of 4sinx and x btw

[bright remnant]

could the graph of 4sinx-x also be investigated?

[echo socket]

In what sense "investigated"?

[bright remnant]

like would you get the same exact solutions with it

[echo socket]

It has the roots at the same x-coords

So yeah

Now try picking a nonzero number x such that 4sinx and x are close to each other

I don't think they are asking you to approximate 0

[echo socket]

Ah, the first guess is given

Well, can you use Newton's method here?

[bright remnant]

hm well i dont think so

because 2 isnt a solution

[echo socket]

It's not supposed to be neither

That shouldn't really be a problem in this case, just start by calculating x_2

[bright remnant]

so if it isnt a solution, I should just skip it?

[echo socket]

As in ignore it?

Or wdym by skipping

[gray isle]

do you know what newton's method is...

[bright remnant]

yes

x_n+1 = x_n - f(x_n)/f'(x_n)

[tardy stag]

ok

you have x_1

can you compute x_2?

[bright remnant]

i dont think so, because 4sin(x)-x isnt a function like f(x)

[tardy stag]

what does this mean

what is f

[bright remnant]

f is a function

[tardy stag]

any function? what is newtons method trying to do?

[bright remnant]

linear approximation but like really accurately if needed

[tardy stag]

linear approximation of what

[bright remnant]

of a function

[tardy stag]

the point of Newton's method is to find roots of a function, inputs that result in an output of 0

[bright remnant]

okay

so the roots of a function like 4sin(x)=x are x=0 & x=2.47

[tardy stag]

<@&268886789983436800>

[winter light]

Also -2.47.. because it's an odd function

[bright remnant]

oh yes, true

[tardy stag]

well the function would be f(x) = 4sinx - x

[bright remnant]

mk

so if you input x_1

hooly shit

i just got the derivative wrong bruh

im just being a dumbass

thanks

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[solar trail]

can someone please explain how to do this? im kind of lost on this unit...

[solar trail]

please ping me when available, thanks!

[lime kindle]

what you really need to undertand man?

[solar trail]

i missed this whole unit so if someone could answer them for me and explain how they got it

[lone heart]

@solar trail Has your question been resolved?

[fierce prairie]

let’s start with total distance travelled. this is just adding up numbers so find how much distance they cover in each line (A, B, C, and D) and add that up

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[hasty flax]

to solve the differential equation they have got this answer. I used auxillary equation and got sqrt(-1) which doesn't make sense

[molten pivot]

it does make perfect sense

you should have

$c_1e^{ix} + c_2e^{-ix}$

[ocean seal]

Mr. Gamer

[vale wigeon]

I used auxillary equation and got sqrt(-1) which doesn't make sense
were you not taught complex numbers?

[molten pivot]

can you write this in terms of real functions?

[tall pumice]

you could use laplace transform

[molten pivot]

no need. and there's also no initial values so the algebra is gonna get really messy

[hasty flax]

its not part of the syllabus

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[molten pivot]

this is euler's formula

$e^{ix} = \cos(x) + i \sin(x)$

[ocean seal]

Mr. Gamer

[molten pivot]

i suggest you read up on it

[tardy stag]

diffeqs without complex numbers

[molten pivot]

i wonder if it is possible to intuitively solve without straight up guessing without invoking complex numbers or transformations

you could probably just make it exact i think

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[tawdry steeple]

how do augmented matrixes actually work

[tacit arch]

https://tutorial.math.lamar.edu/classes/alg/augmentedmatrix.aspx

[merry depot]

gonna need to be more specific....

[tawdry steeple]

I understand how to do them, but I keep placing the numbers wrong. The site for school I'm using tells me I'm wrong and shows the correct answer, but not what I'm doing wrong

[tacit arch]

show work then

probably just arithmetic error

[merry depot]

you're variables are associated to a column, all coefficients of that variable should be in that column.

so you'r columns are [x y z | constant]

but look at your last row. The x coefficient isn't there, and the z coefficient is in the y column

[tawdry steeple]

oh

I didn't even know that the columns were supposed to be specific

the lesson just released today for my class

[merry depot]

once you have it in this form, the idea is just to copy the coefficients directly into the matrix in the position they're in.

blank spaces become 0's

[tawdry steeple]

thank you so much man

I spent 2 hours sitting here frustrated with myself for getting the number placements wrong

literally the first question I've got right

[merry depot]

good job!
linear algebra is weird, it's not hard, but it's easy to make simple mistakes that throw everything off.

take your time and be patient, it'll get easier

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@tawdry steeple Has your question been resolved?

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

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

[unkempt flax]

help

[vale wigeon]

with what

[unkempt flax]

oh finally

wait

[vale wigeon]

wym "finally"? it's only been 2 minutes

[unkempt flax]

im typing

[vale wigeon]

also you should BEGIN by posting your question not just go "HELP!!!"

[unkempt flax]

chat ded

ok

(45x...)x5=45x(20x5)=...x....=

[unkempt flax]

confusing

[fallen verge]

What does this mean

@alpine sable please do not flood the chat

[unkempt flax]

commutative numbers

[vale wigeon]

... are those x's supposed to mean multiplication?

[unkempt flax]

yes

[ocean seal]

Ann

[vale wigeon]

so it's this?

and you have to fill in the blanks

[unkempt flax]

yes

[vale wigeon]

also there's no such thing as a "commutative number" -- you must be mixing up some terminology.

but anyway

have you made any progress? Y/N

[unkempt flax]

no wrong asociative numbers

home work

[vale wigeon]

there's also no such thing as an "associative number".

also let me restate my question:

have you made any progress? Y/N

[unkempt flax]

y

[vale wigeon]

ok, show your progress thus far.

maybe by taking a picture of it and sending it here.

[unkempt flax]

wait

[unkempt flax]

pls ignore background

[vale wigeon]

wait, -20?

[unkempt flax]

@vale wigeon

[vale wigeon]

i thought the middle number was only 20.

is it +20 or -20?

[unkempt flax]

-20

[vale wigeon]

why'd you not write the minus earlier...

but ok yes so far so good

now perhaps only one step remains: to calculate 45 * (-100)

are you able to do that

[unkempt flax]

nope

[vale wigeon]

why not? do you know how to multiply numbers in general?

[tacit arch]

how old are you?

[unkempt flax]

13

[vale wigeon]

ok

do you know how to multiply numbers in general?

[unkempt flax]

yessss

[vale wigeon]

ok then what is stopping you in 45 * (-100)?

[unkempt flax]

next number

holy crap

[vale wigeon]

you didn't answer my question.

if the reality was "nothing is stopping me, i worked it out just now" then you should've also said that.

[unkempt flax]

yes i can do it

[unkempt flax]

tahnk you

[vale wigeon]

strange how you said just a few minutes ago that you couldn't, but whatever.

ok so that's like ten problems you've sent a photo of here.

[unkempt flax]

next number💀

lemme take my pen

[vale wigeon]

there's like ten problems here, which one do you want help with now

[unkempt flax]

in school i dont get it

[vale wigeon]

...

[unkempt flax]

lets start one by one

[vale wigeon]

ok look. when people ask you questions,

it is polite to give them DIRECT and TO-THE-POINT answers.

[unkempt flax]

answer it

[vale wigeon]

it is NOT polite to answer with things that don't have anything to do with the question.

also we won't do your homework for you.

[unkempt flax]

umm sorry

[vale wigeon]

it is your responsibility to tell us exactly where you are stuck and exactly which problem you want us to look at.

[unkempt flax]

im sick yesterday so i dont get it can you explain it

[unkempt flax]

i didnt come to school

[vale wigeon]

ok, so we're talking about these problems, yes?

[unkempt flax]

uhm um

[vale wigeon]

does "uhm um" mean "yes" or "no"?

[unkempt flax]

yes

[vale wigeon]

right, ok.

just like in your previous problem, in all of these problems you are applying the associative law.

[unkempt flax]

yes

[vale wigeon]

in that case it was the associative law for multiplication, in these there's also the associative law for addition.

but the principle is the same

applying the associative law means switching up the order in which you do the addition with three numbers involved -- do you begin by adding the middle and last number, or by the first and middle number?
the associative law says you'll get the same result either way,
and in these problems they have you change it up so you add the more convenient pair of numbers first.

[unkempt flax]

it only can applied on multiplication and subtraction

[vale wigeon]

multiplication and addition.

a + (b + c) = (a + b) + c
and
a * (b * c) = (a * b) * c

the numbers do not change themselves or switch around, only the brackets swap places.

was this the explanation you wanted?

[unkempt flax]

front and back

ohhhhh i get it thank you very much👌

[vale wigeon]

ok

do you have anything else to ask?

if not, you can type .close to close this channel.

[lone heart]

@unkempt flax Has your question been resolved?

[lone heart]

Channel closed

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Use .reopen if this was a mistake.

[lime sluice]

wow nice which part?

[lone heart]

Available help channel!

Send your question here to claim the channel.

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

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

[alpine sable]

I'm curious why dy/dx gives the rate of change of y in respect to x, isn't dy/dx supposed to be finding the equation to find the value tangent at any given point?

[wind cloak]

you just made a discovery!

[young finch]

its how much y changes with a little nudge to x

[wind cloak]

the slope of the tangent is the rate of change

[alpine sable]

oh, so it should be limit dx->some_little_nudge dy/dx

but since some_little_nudge is too small to be significant

we just assume it's 0?

[fresh parcel]

its not really 0

[alpine sable]

it's close to 0

[fresh parcel]

yes

[alpine sable]

so for the ease of calculating we assume its 0

but realistically it isn't

[alpine sable]

what do you mean?

wait so the whole point of dy/dx is to find the rate of change?

[wind cloak]

You say dy/dx is the rate of change and the slope of the tangent

[young finch]

well the whole point of calculus is instanteous rate of change

[wind cloak]

*differential

[alpine sable]

damn i've studied calculus for so many months

this is something new

[tardy stag]

took me about a year to realize that

[young finch]

calc teachers are infamously bad at teaching

[alpine sable]

mine gives 10 pages of homework and scrolls the phone in class :/

[tardy stag]

i didn't understand shit until i taught it to someone else

[fresh parcel]

khan academy is really good for the intuition

[young finch]

I was lucky to have a teacher who actually taught

[alpine sable]

I'm currently self-learning this Precalculus with Calculus Previews

because I find that I have 0 basics and I only memorise formulas

which to me is horrible

[fresh parcel]

is that a textbook?

[alpine sable]

*I've been fed formulas and how to use them to be exact

[alpine sable]

yep