help-10

[timid silo]

the sum is simply f(x) by definition

and the expansion is the series you asked for

they gave you the series

[tepid urchin]

oh ok

[timid silo]

you give them f(x)

[tepid urchin]

can you do an example?

[timid silo]

ok

take sine function

sin(x) = sin(0)+sin'(0) * x + sin''(0)/2x^2 + ..= sin(0) + x - x^2/2 + ...

by writing this sum using the sigma notation

you get this

[tepid urchin]

o

how

[timid silo]

there is no easy way to get it, it's the definition of taylor expansion.

[tepid urchin]

also how would I solve this without knowing that this was the taylor series of sinex

[timid silo]

I'm not sure there is any way. It's certainly not elementary math

especially that you have pi there

[tepid urchin]

im in calc || so if theres any method with that

[timid silo]

what are the courses in calc ?

just the titles

[tepid urchin]

first it was derivitves then intagrals, now series

[timid silo]

taylor must be there somewhere

is this a homework question

[tepid urchin]

its a past exam quesiron

[timid silo]

oh so it covers all calculus right ?

[tepid urchin]

nop

its not the final yet

[timid silo]

it covers which part

[tepid urchin]

series

i can send you a picute

[timid silo]

yes send me

taylor must be there, otherwise your teacher is nuts

[tepid urchin]

oh well there ya go

[timid silo]

the other question is taylor's inequality hahaha

[tepid urchin]

;n;

[timid silo]

they literally say power series, I can't imagine anyone taking a course in power series without knowing taylor expansion

if you check your notes you'll find them

[tepid urchin]

so do I just have to memorize a bunch of them?

[timid silo]

there are tricks to help you memorize, for example sine is an odd function, so it only has odd powers in its expansion

hence the 2n+1 in powers

[tepid urchin]

what ones should i memorize?

[timid silo]

cosine and sine are pretty important, 1/(1-x) also (but you already know it)

you can also memorize the general formula, and then find the expansions with it

[tepid urchin]

oh ok

ill do that ty

[timid silo]

Yw !

[tepid urchin]

can you help me with another problem?

[timid silo]

ok go ahead

[tepid urchin]

question 1 on the exam

[timid silo]

just factorize with 1/10 and use the expansion of 1/(1+x)

[tepid urchin]

whats that

[graceful pollen]

unrelated, but why was 0 used

[timid silo]

any number can be used i think, but 0 simplifies things

[tepid urchin]

why is the plus or minus sign upside dwon

[timid silo]

this probably depends on the radius of convergence, but this is another animal

[graceful pollen]

ahh, that makes sense. Does that mean 180 can be used as well, since it would be equivalent

[timid silo]

1/(1+x) = 1-x+x^2-x^3+x^4 - ...., upside down means we start with minus

[timid silo]

yeah probably, but not with all functions I htink

[tepid urchin]

ok

what would be an answer to the 1st question?

[graceful pollen]

yea, fair enough, thanks for clarifying

[timid silo]

this need a pen and paper lol

[tepid urchin]

lol

[timid silo]

But you know the trick now

wolfram can also give the answer

(I think)

[tepid urchin]

idk how to use it

everytime i do it fucks it

[timid silo]

write the sum by writing the symbols (like you would with a pen)

[tepid urchin]

ok

but it looks very divergent bc the exponents keep getting bigger

[timid silo]

not using latex or some shit

[timid silo]

convergence and divergence depend on the values of x

that's what a radius of convergence means

[tepid urchin]

righ

t

[timid silo]

(1/2)^n gets smaller when n gets bigger

but 2^n explodes

[tepid urchin]

ok

so

i plug in the numbers

but then what?

[timid silo]

you don't, they're asking your for the power series expansion ,and the interval of convergence

just keep x

interval of convergence = interval of x where the series is convergent

[tepid urchin]

ya

[timid silo]

this is a full course on power series, take your time

[tepid urchin]

;n;

ok

i feel like im missing somthing

can you solve it on a piece of paper?

[timid silo]

wait i'll draw a solution in my computer, I am lazy bastard

[tepid urchin]

lol

[tardy epoch]

You should show your work rather than ask for a full on solution

[tepid urchin]

I have the solution

I just need to see the full prosses

[timid silo]

work out the rest

[tardy epoch]

Screenshot and upload and ask questions about it

If the professor gives a solution, you should understand it that way first

[tepid urchin]

?

[tardy epoch]

Do you have a question

[tepid urchin]

ok so you do this, then you plus a=10 into that equasion over there

then what

this is the answer btw

[timid silo]

this looks false, not sure though

[tepid urchin]

thats what my friends been saying

hes been working on this all night going crazy saying that this is wrong

[timid silo]

@tardy epoch am I allowed to post the full solutin

[timid silo]

wow it shouldn't be this hard but yeah he's right, this answer is false

[tepid urchin]

lmfao

no way

[timid silo]

who's the bastard who wrote this

[tepid urchin]

Dr.Lang

no one likes him

[timid silo]

lol mr. lang is a bastard

[tepid urchin]

yes

[timid silo]

he bastardizes his students lol

[tepid urchin]

he got divorced recently

so all is fair now

[timid silo]

oh he should be in a bad mood then

[tepid urchin]

24/7

[timid silo]

lol

anyway this is the solution I wrote

n goes from 0 to infinity

[tepid urchin]

all righty

how did you get that

[timid silo]

get what ?

[tepid urchin]

the series

[timid silo]

taylor lol

[tepid urchin]

ok ok

ohhhhhhh shiiiiiiittttttttttt

[timid silo]

this is a general formula

not easy at first

[tepid urchin]

the -1 is the signs

the ^n is the ohhhhhhhhhhhhh

okokokokokowkerfionvoidflkb

[timid silo]

lol

[tepid urchin]

gimme a practice problem

[timid silo]

about taylor ym

[tepid urchin]

sure

[timid silo]

ok lets say find the expansion of log(1+x)

or cos(x)

anyway bro I gotta go

[tepid urchin]

mk tysm

life saver

[timid silo]

you can find this stuff in the internet

[tepid urchin]

ya

[timid silo]

with practice problems

[tepid urchin]

wait

how are you new and a helper?

[timid silo]

they asked me to choose my role when i joined the forum

i mean the group

[tepid urchin]

oh lol

[timid silo]

it's probably a new thing

[tepid urchin]

nvm

[timid silo]

because I joined it before

[tepid urchin]

read the rules ✨

.close

[obtuse pebble]

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[vital frost]

@tardy epoch that

in fact i understand

.close

[obtuse pebble]

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[lost wharf]

i'm not sure about my answer

[narrow fiber]

Do you understand what the question is asking?

@lost wharf

[lost wharf]

sort of

[narrow fiber]

I'll send a pic

This is the set-up

[lost wharf]

ok

[narrow fiber]

Imagine "the lever" along with the square being rotated in the direction of the arrow (ccw)

They're basically asking how much do you need to rotate until the square reaches the 4th quadrant

[lost wharf]

ok

[narrow fiber]

How much do you think it needs to rotate?

[lost wharf]

about 90

[narrow fiber]

If it rotates 90°

It will reach only up to the second quadrant

Will it not?

[lost wharf]

so 180 so that it can reach the 4 quadrant

[narrow fiber]

hmm

Are you using a phone currently?

[lost wharf]

no

[narrow fiber]

Ah

[lost wharf]

doing my assignment

[narrow fiber]

I thought I'd tell you to try rotating your phone 180° and see where the square ends up

But you can't really do that on a pc

unless you rotate your head?

[lost wharf]

i get it now

it 270 degrees

[narrow fiber]

Yeah

Exactly

[lost wharf]

thanks

.close

[obtuse pebble]

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[warm siren]

anyobody know how to get started on this

[robust sedge]

Do you know the formulas for going back and forth between polar and cartesian coordinates?

[warm siren]

no i dont think i recall

[robust sedge]

I think that's where I would start

[obtuse pebble]

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[pallid vapor]

hi would it be possible to get some help here. im trying to see how the transfer function (marked 2) comes from its laplace transform above. i cant seem to understand. would be helpful if someone could show me the steps taken to reach the 2nd transfer function (marked 2)

[pallid vapor]

this one

[obtuse pebble]

@pallid vapor Has your question been resolved?

[obtuse pebble]

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[silent perch]

A golf ball is struck from a point 23m to the left of the hole and lands at a point 17m to the right of the hole. It reaches a maximum height of 10m

-Determine an equation to represent the path of the golf ball

-When the ball is directly above the hole, how high in the air is it?

[silent perch]

if anyone could do step by step with me that'd be appreciated

[obtuse pebble]

@silent perch Has your question been resolved?

[obtuse pebble]

@silent perch Has your question been resolved?

[tranquil tide]

Is this all the question asks, is there like lore behind this

[silent perch]

No

There's a diagram though

[tranquil tide]

can i see

[silent perch]

No I don't have my text book but like

I can draw what it looked like

[tranquil tide]

ok wait are you american

[silent perch]

Canadian

[tranquil tide]

is this mechanics

[silent perch]

no

quadratics

[tranquil tide]

ah

im pretty sure the diagram gave you some numbers

ok wait can you draw it

[silent perch]

Only the same numbers

well actually

it told be to label it myself

[tranquil tide]

is it distance against time

distance on the y time on the x

[silent perch]

no time

I need to make an equation

[tranquil tide]

ok it was prob horizontal distance and vertical distance then

right?

[silent perch]

yyeah

[tranquil tide]

Ah

origin is the golf hole

[silent perch]

point on the left is 23,0 I think and point on the right is 17,0

[silent perch]

yes

I need to make an equation

[tranquil tide]

ok 1 sec

[silent perch]

ok

[silent perch]

Rip

[tranquil tide]

i got it

😋

Ok my walk through was a bit messy tho

have you heard of completing the square

ok i hope you can read this

if u want a better step by step tell me before i go sleep ok bye

[obtuse pebble]

@silent perch Has your question been resolved?

[obtuse pebble]

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[karmic vale]

can i use universal generalization like this? I was trying to prove how a universal conditional statement is logically equivalent to its contrapositive and used universal generalization to do that

[obtuse pebble]

@karmic vale Has your question been resolved?

[karmic vale]

<@&286206848099549185>

[next loom]

this seems about right

[obtuse pebble]

@karmic vale Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[spark field]

how was this solved?

wdym

like create the riemann sum?

ah so

20-10/2

= delta x

@timid silo so upperestimate

would be

10+(k-1)deltax

and lower estimate

a+kdeltax

?

where n = 6

then just multiply it out and use the rules

and plug in n

how?

lower bound starts at 10 yea

wdym as height

you mean delta x?

top is

lower bound

bottom is upper bound

right ? @timid silo

are u sure?

the answer is -32

and 8

what

im so confused

how do we solve this

bro what

[warm shale]

hannibal

[spark field]

yea

wb it

delta x = 5

where did the 2 come from

and n = 2

so delta x = 5

yea

2 = n

not delta x

b-a/n is the formula

how?

oh yea ur right

the width is 2

mb

i mistakened that

oh i see

Oh

i get it now

we just multiply

it by delta x

and lower estimate would just be x= 18 and under

upper would be x = 20

[obtuse pebble]

@spark field Has your question been resolved?

[spark field]

yes it is

its from x=10 to x= 18

and its 8

for when

x = 10 to x= 20

this just means find the antidervative right?

[obtuse pebble]

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[spark field]

wb midpoint

how would this work

ik

so i start from 14?

im confused because we just estimated by f(x) * deltax

[warm shale]

hannibal

[spark field]

oooh

so deltax = 1

then

wait no

18-10/5

ok so

18-10/5

then divide it agian

by 2

right

so delta x = .8

?

um what

?

subtract upper and lower

then divide by 2

-12

which sint correct

yea

but that wouldnt really work

💀

we are given f(x) number

wdym middle of interval

yea

wb it

ur not supposed to predicut f(x)

answer is -20

<@&286206848099549185>

any1 know

[fathom flicker]

go watch a video on how to do a midpoint riemann sum

for example the video that you linked a screenshot of

https://youtu.be/5XreKMJDJsg

[spark field]

true i forgot he does it using this way

organic chem

the formula is diff when we have f(x) so i didnt know how to do it when we are giving the value of f(x)

[fathom flicker]

It's not different

for each rectangle f(x) is the height

so like just looking at the picture for the video above

f(x_i)

say i=1

this means f(1)

which is the height of the rectangle at x=1

which is just given in your table

but if it wasn't, you would just plug x=1 into your function

yet the formula remains the same

Watching the video should clear up your confusions.

[obtuse pebble]

@spark field Has your question been resolved?

[obtuse pebble]

@spark field Has your question been resolved?

[obtuse pebble]

@spark field Has your question been resolved?

[obtuse pebble]

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[timid silo]

how do i solve this question?

[timid silo]

i know A is circle and B is ellipse

i understand they are asking in which points they will intersect

[trim portal]

both are centered at (0, 0). So you dont even need to perform calculations

[timid silo]

but how do i know where they will intersect what if circle is bigger and doesn't intersect at all

[haughty coyote]

You know their equations so you know them. Most notably the radii so you can draw them

[trim portal]

not one of the possible answers, but if you want you can do some calculations and see that they do intersect

[royal basin]

you can just treat this as a system of equations lmfao

[timid silo]

how do i do that ?

[trim portal]

You can either use substitution, or subtract/add those equations

[timid silo]

i don't know much about in these circles and ellipses so can u explain more?

[leaden ginkgo]

are you familiar with solving simultaneous equations? using substitution/subtraction?

[trim portal]

x^2+y^2=25, this is the circle
x^2+9y^2=144, this is the ellipse
To find the intersection, you need to find points (x, y) such that they lie on both the circle, and ellipse. I.e. they satisfy those equations

so just solve system of those equations

[timid silo]

no

ok first i will learn about solving these eqs

[trim portal]

sorry my battery is at 5% and I cant charge it rn, bye

[leaden ginkgo]

what’d they teach you in school, before they gave you this problem?

[timid silo]

this question is not from a school i am just self studying rn

[leaden ginkgo]

look up how to solve systems of equations

[timid silo]

👍

.close

[obtuse pebble]

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[dry arrow]

hello, is there any way to solve this aside from multiplying everything first then finding lcm?

[royal basin]

well you could break the 6^1 into 2^1 * 3^1

so that the first number there is actually factored into primes and doesnt just appear as such

[dry arrow]

so it would be 2^3, 3^5, 2^4, 5^1, 3^2

then what?

oh nvm i got it

thanks for that saved me

.close

[obtuse pebble]

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[faint ore]

hello

[faint ore]

i need a little bit of helping differentiating these types of problems

for example f(x) = 3x/(x^2+16)^1/2

[leaden ginkgo]

do you know about the various rules of differentiation

[faint ore]

yes

[leaden ginkgo]

what do you need help with then

[faint ore]

i guess just executing them properly

here we go

so 3x * (x^2+16)^-1/2

product rule

[leaden ginkgo]

or just quotient rule

[faint ore]

u think thats better?

[leaden ginkgo]

i mean you won’t have to work with a negative power

[faint ore]

thats true

okay so [1/2 * (x^2 +16)^-1/2 *2x (3x) ] - [3(x^2+16)^1/2]

all over x^2+16

correct so far?

[leaden ginkgo]

you’re doing product rule?

[faint ore]

no

quotient

[leaden ginkgo]

quotient rule is minus

[faint ore]

oof yes

slipped my mind thank u

[leaden ginkgo]

and it’s $\frac {f’(x)g(x) - f(x)g’(x)}{g(x)^{2}}$

[warm shale]

! saad

[faint ore]

yes

okay so after simplifying a little further i get 3x/sqrt(x^2+16) - 3sqrt(x^2+16)

over x^2 +16

[leaden ginkgo]

use more brackets or type in latex or on a piece of paper

[faint ore]

how do u do the texit thing

[leaden ginkgo]

enclose in $

[faint ore]

$\frac{3(x^2+16)^{1/2}- 3x^{2}(x^2+16)^{-1/2}}{x^2+16}}$

[royal basin]

\sqrt not /sqrt

also \sqrt{...} not \sqrt(...)

also i think you forgot to take the derivative of the root at one point

[leaden ginkgo]

enclose powers in {}

[faint ore]

there we go

nvm

look about right?

[leaden ginkgo]

.

f’(x)g(x) - f(x)g’(x)

you didn’t differentiate the sqrt properly either

[timber island]

Isn't f'(x) 2x in this case? If f(x) = x^2

[faint ore]

oh dang

[leaden ginkgo]

it’ll be chain rule and power rule

[faint ore]

one moment while i redo this

[leaden ginkgo]

f(x) = 3x, i think

[timber island]

Oh

[leaden ginkgo]

this is the original function

yes, 3x

[faint ore]

okay here we go

[leaden ginkgo]

you forgot chain rule, again

[faint ore]

no i simplified

[leaden ginkgo]

stop skipping steps, i can’t follow your logic

write out the quotient rule as you would

then you can simplify

[faint ore]

alrighty

[warm shale]

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

[faint ore]

look good?

[leaden ginkgo]

yes

you could’ve separated the 3 earlier actually

constant multiple rule

easier to differentiate

[faint ore]

i see

so now to simplify

[leaden ginkgo]

you can get rid of the negative power

[faint ore]

by square root right

[leaden ginkgo]

it just goes to the denominator and becomes square root $\frac {1}{2 \sqrt{x^{2}+16}}$

[warm shale]

! saad

[faint ore]

why a 2

[timber island]

Question: is the final answer
$$\frac{3-3x}{\sqrt{x^2 + 16}}$$

[faint ore]

wouldnt it be 3x^2/sqrt(x^2+16) * x^2 +16

since g(x) is getting squared on the bottom

[leaden ginkgo]

$\frac {1}{2} \cdot (x^{2} + 16)^{- \frac {1}{2}}$

[warm shale]

! saad

ItzKraken2

[leaden ginkgo]

$\frac {1}{2 \sqrt{x^{2}+16}}$

[warm shale]

! saad

[faint ore]

since theres a 2x after it the 1/2 and 2 cancel out

[leaden ginkgo]

yes

2x * 3x

[faint ore]

and the 1/2

[leaden ginkgo]

so you’ll just have 3x^2 in the numerator

[faint ore]

yup

[warm shale]

Blackbeard

okay so now i have $\frac{{3}{/sqrt{x^{2}+16}}$
```Compilation error:```! File ended while scanning use of \frac .
<inserted text> 
                \par 
<*> 786751657316581406.tex
                          
I suspect you have forgotten a `}', causing me
to read past where you wanted me to stop.
I'll try to recover; but if the error is serious,
you'd better type `E' or `X' now and fix your file.```

[faint ore]

forget that

i have 3*sqrt(x^2+16) - 3x^2/(sqrt(x^2+16))

over x^2 + 16

[leaden ginkgo]

yes

[faint ore]

so here is where i have trouble

where could i go to further simplify

[timber island]

Got the correct answer (confirmed on mathway) u can simplify by letting sqrt (x2 + 16) = u

[faint ore]

what do u mean = u

[wooden marten]

declare a new variable u

[timber island]

U substitute u for sqrt (x^2 + 16)

[faint ore]

why would i do that?

[wooden marten]

simplifies the fraction, ig

[faint ore]

you cant just multiply by the denominator or something

[timber island]

A simpler to write and b simpler to use laws to simplify

I can send a pic of my simplification procedure if u want

[faint ore]

sure

well actually let me try it out first

3u-3x^2/u

over u

like that?

hello

[timber island]

Yeah

[timber island]

Ur missing a u^-1 in the numerator

In the 3x^2 term

[faint ore]

nevermind

i figured it out

thanks for the help

.close

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[night lotus]

Calculate the double integral with A limited by x^2=4-2y

[warm shale]

rainy

[night lotus]

So far, ive graphed it

Dont understand what limits you fill in for what integral :\

[open forge]

HELP

[night lotus]

Can you use other channel pls

[clear condor]

!help

[obtuse pebble]

Please read #❓how-to-get-help

[clear condor]

you get help

[night lotus]

Calculate the double integral with A limited by x^2=4-2y

Need help with the limits of integration

[obtuse pebble]

@night lotus Has your question been resolved?

[obtuse pebble]

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[gusty tartan]

tryna solve this differential

i think the best route is sub but im not sure

[clear condor]

what would you sub?

[gusty tartan]

y=xv(x)

got an idea brb

.close

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[gusty tartan]

.reopen

[obtuse pebble]

✅

[gusty tartan]

yep

still stuck

question said use sub

y=xv(x)

<@&286206848099549185>

[clear condor]

so what is y' in x and v

[gusty tartan]

thats where im up to

well the confident up to

i did stuff past that but i have no idea waht i was doing

[timber island]

U can divide both sides by x and integrate

[gusty tartan]

y' = 2-(y/x)?

[timber island]

Mhm

U can treat as x as a constant, then integrate (btw not sure if this turns into a double integral)

@gusty tartan still stuck?

[obtuse pebble]

@gusty tartan Has your question been resolved?

[gusty tartan]

NAH i got it

used sub

:P

[obtuse pebble]

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[timid silo]

what will be the explicit form of this?

[drifting wraith]

is that multiplication?

[timid silo]

yeah the dot symbolises multiplication

[leaden ginkgo]

weird dot

$i \cdot j$

[warm shale]

! saad

[royal shard]

Isnt that factorial?

Oh no

[timid silo]

my first thought was that its the product of i and j taken two at a time (i not equal to j) subtracted from the square of the sum of the first n natural numbers

[timid silo]

could u tell me how you got to this?

[drifting wraith]

i didn't i asked wolfram

i don;'t know the method

[warm lark]

Who can help me pythagorean theorem

[obtuse pebble]

Please read #❓how-to-get-help

[timid silo]

ah i see... i shall try to get to this result

[warm lark]

Anyone?

Who can help me?

[warm lark]

What should i do if theres a square root

[timid silo]

could u tell me what is the exact question is? i can try to help

[warm shale]

fäf

[timid silo]

ahhh i see

this makes a lot of sense

[jolly ginkgo]

Try it first cuz I'm not entirely sure

You must be knowing the last two sums

I assume

[cobalt plover]

Use trig

[timid silo]

yes... i tried it... im getting rlly close so ig its right

[jolly ginkgo]

Alright

[jolly ginkgo]

It's easy from there

[timid silo]

yes i got it

[timid silo]

yeah turns out i made a silly mistake

[jolly ginkgo]

Alrigh

You're welcome

[timid silo]

thanks a lot!!

[jolly ginkgo]

❤️

Lots of ❤️

[timid silo]

.close

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[mental cosmos]

can someone put me in the right direction?

[mental cosmos]

its supposed to be solved with graph theory

[obtuse pebble]

@mental cosmos Has your question been resolved?

[obtuse pebble]

@mental cosmos Has your question been resolved?

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[toxic hollow]

Just to double check. {h(1),h(2)}={1,2}

[toxic hollow]

Means all the pairing between the two sets right including h(1)=2

[sage geode]

That implies h(1) = 1 and h(2) = 2 or h(1) = 2 and h(2) = 1

So, no, h(1) = 2 is not impossible

Oh I read that as "excluding"

Yeah, anyways, h(1) = 2 is possible

[obtuse pebble]

@toxic hollow Has your question been resolved?

[obtuse pebble]

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[gloomy glen]

Hello,
I am currently watching 3b1b's video on eigenvectors and values, but my question is regarding change of basis matrices.
In the video he uses the new basis vectors directly as his change_of_basis_matrix, but isnt it needed to first setup a system of linear equations with the ( new basis on the left | old basis on the right ) and solving that as shown in the second image?
Im still not a 100% percent sure on how to obtain the transformation matrix needed.

[warm canopy]

sure but in the 3b1b video his "old" basis is the standard basis and the eigenvectors are already written in this standard basis so you're skipping that step because there's nothing to do

[obtuse pebble]

@gloomy glen Has your question been resolved?

[gloomy glen]

so what exactly is the old and new basis? in this case?

[warm canopy]

in 3b1b vid its the standard basis and the eigenbasis

[gloomy glen]

so the new basis = eigenbasis is
(1 -1)
(0 1) ?

[warm canopy]

(-1 1) but yeah

[gloomy glen]

yes

[gloomy glen]

and where is the
(3 1)
(0 2)
coming from?

if
(1 0)
(0 1)
already is the old basis

[warm canopy]

thats the matrix representing the linear map with respect to the standard basis

[gloomy glen]

youre right, im sorry
I totally forgot that there is still a difference between the matrix representing the linear map and the basis which defines how vectors are formed based on given coordinates

[warm canopy]

once you fix a basis, the columns of the matrix are telling you what happens to those basis vectors under the transformation

[gloomy glen]

Yes.
So I use the current basis and the new basis to calculate a transformation Matrix B.
Then I multiply the current Matrix A:
B-1 * A * B and get the Matrix A' for the new basis, right?

[warm canopy]

a couple seconds after that screenshot he'll say that product is equal to (3 0 ; 0 2) which is precisely telling you the eigenvectors (a basis now) get scaled up by their eigenvalues

correct

B will take you "from the new basis to the old" (this at first seems kind of backwards but it is the correct way)

this makes sense, B takes you back to your old basis, A is the matrix of the map in terms of the old basis, and then B^-1 takes you back to the new basis

[gloomy glen]

This topic is a lot to keep in mind and connect haha
but thank you

[warm canopy]

change of basis is notoriously one of the hardest things to get your head around in linear algebra

[gloomy glen]

and in which order do i multiply these matrices?

[warm canopy]

the order they're written

B^-1AB

(because transformations get applied right to left)

[gloomy glen]

so first B * A? and then that * B^-1.
The last thing im struggling with atm is:
"B takes you back to your old basis, A is matrix in terms of the old basis, and then B^-1 takes you back to the new basis"

[warm canopy]

appplying the matrix X and then the matrix Y is the same as applying the matrix YX

things are applied from right to left

multiplication is associative so it doesnt matter how you actually compute ZYX as long as its in that order

[gloomy glen]

One last question 😅 .
The way i understood it in lectures was actually this way. With B^-1 we transform coordinates from the new basis into the old. Than we put these values into the matrix "from the old basis" and then we use B to turn the old coordinates into the ones in the new basis

[gloomy glen]

does it make sense this way around? Or should i try to understand it in another way.
B^-1 A B is the same as B A B^-1 right?

[warm canopy]

this is correct and agrees with what I have been saying

[warm canopy]

this is not correct, those are two different matrices because XY is not YX in general

[gloomy glen]

because u said B^-1 takes me back to my new basis, but i said B^-1 transforms from new basis to old basis. And in 3b1b screenshot B^-1 is the matrix on the left not on the right.
im sorry im just a bit confused

[warm canopy]

ehh this is subtle

it depends on what you have defined B to be

either its "the new basis vectors written in terms of the old" and then what I said is correct

or its "the old basis vectors written in term of the new" and then its what you said

[gloomy glen]

Im sry to waste ur time,
But somehow the way i think seems to be flipped haha.

In these screenshots we have A (old base) and B (new base). We make a set of linear equations with B on the left and A on the right. In the end we get TA->B. This Matrix shows coordinates in base B in terms of coordinates in base A right?

Now what I want to do is:
TA->B * L * TB->A = TA->B * L * TA->B^-1

[warm canopy]

your issue here is that A and B are not bases

in R^3 you need 3 vectors for a basis and then your change of basis matrix will be 3x3

[gloomy glen]

Lets think of it as R^2 then. I just want to know about my concept

Maybe I should just stop for today and take another look tomorrow

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[azure anchor]

do you know what g'(x) and g''(x) represent?

you have to know general behaviors of square and cubic functions. and also the general patterns that derivatives follow.

whats the derivative of f(x)=x^3

[obtuse pebble]

@magic goblet Has your question been resolved?

[azure anchor]

do you know what the graphs of x^3, x^2, and wave (trig) functions look like? this should be pretty recognizable if you are familiar.

and if you understand the pattern that derivatives take and what they represent, it just takes a little bit of thought to imagine what the derivatives would be with just these graphs.

you really don't know what b could be?

[obtuse pebble]

@magic goblet Has your question been resolved?

[obtuse pebble]

@magic goblet Has your question been resolved?

[obtuse pebble]

@magic goblet Has your question been resolved?

[obtuse pebble]

@magic goblet Has your question been resolved?

[obtuse pebble]

@magic goblet Has your question been resolved?

[crystal solstice]

Thus is mostly correct. Look at a.iii and c.i again

Not sure what #4 is asking you to do

[crystal solstice]

0 isn't an option. You have to say if it's a min/max/inflection

Is the function going up or down for ci

Increasing or decreasing?

[rustic harbor]

I think he was talking about at the point

[obtuse pebble]

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[stone gorge]

f(x)=2x^3+1/x

find f(-x)
find -f(x)

[obtuse pebble]

@stone gorge Has your question been resolved?

[stone gorge]

<@&286206848099549185>

[rigid crystal]

for the first one, plug in -x to f(x)

[stone gorge]

-2x^3-1/x?

[rigid crystal]

yeah

and the second one is similar

just find f(x)

and then negative of that

[stone gorge]

is it the same?

[rigid crystal]

yeah

👍

[stone gorge]

alright thank you so nova

[rigid crystal]

you're welcome

happy to help

[stone gorge]

.close

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[little lava]

help with part a)

im having a hard time finding which method to use

.close

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[karmic reef]

[karmic reef]

how do you find the tangent line of y^2(y^2-1)=x^2tan(y) at (0,1)

tangent line equation

y-y1=f'(x1)(x-x1)

[teal turret]

what have u tried @karmic reef

[warm shale]

Stephen

[karmic reef]

Sorry was in the restroom

I mean ill try implicit differentiation

D/dx(y^2(y^2-1))=d/dx(x^2tan(y)

Left hand side product rule

[teal turret]

nah dont do product rule on the left, just distribute the y^2, makes it easier

[karmic reef]

Oh lol

I didn’t think to

D(y)^2/dx =d/dx(x^2tan(y))

Right hand side product rule

[teal turret]

can u do the whole thing on paper so i can just look it over all at once

or using latex

[karmic reef]

Can u refoward ur latex but using back ticks

Or a way so the bot doesnt embed it

So i can copy the syntax

[teal turret]

$y^2(y^2-1)=x^2 \tan(y)$

[karmic reef]

$\frac (dy^2) (dx)=d/dx(x^2 \tan(y))$

[warm shale]

HeccinTech

[karmic reef]

Lol im just gonna write it down

This is wrong

[wooden marten]

$\frac{dy^2}{dx} = \frac{d}{dx}x^2\tan(y)$

[warm shale]

zfnQRZJT

[wooden marten]

?

[karmic reef]

No no

I mean

Yes

But i mean this is wrong in context of the work i posted like with my hand writing

Not what i posted

Thank u though

[wooden marten]

alr

[teal turret]

do u still need help?

[karmic reef]

Yeah

Im still working on it

Sorry

[wooden marten]

take your time

[wooden marten]

it seems you lost a y^2 - 1 on the left side

also for implicit differentiation d/dx of something that involves y remember you need the chain rule

[karmic reef]

Thats because i distributed the y^2

[wooden marten]

y^4 - y^2?

[karmic reef]

#help-10 message

Yeah

[wooden marten]

well you just
didn't write that

[karmic reef]

He told me here

To do it that way

[wooden marten]

(also, just a notational thing, dy^2 usually means (dy)^2)

as in the denominator of the 2nd derivative operator

[karmic reef]

Ok

Wait was i supposed to do chain rule

[teal turret]

yea by distributing the y^2, i meant to do y^2 multiplied by y^2 -1, which is y^4 - y^2

[karmic reef]

[wooden marten]

yep

[teal turret]

im confused lol. how did u interpret my message?

[wooden marten]

you still aren't getting the y^4 - y^2 on the left

[teal turret]

distribute the y^2 as i suggested here

[karmic reef]

Oh i thought they simplified

[wooden marten]

wh

[karmic reef]

Sorry

Monkey brain

[wooden marten]

yes this is right

differentiate on the left side as well

[karmic reef]

Thanks

Wouldnt i write it as

No wait

Why would I differentiate the left side again

I dont think u can differentiate y with the derivative respect to x as in apply any rules besides it being dy/dx

Im only taking the first derivative

So its already differentiated

[wooden marten]

what is d/dx of y^4
power and chain rule

[karmic reef]

Oh i see

Wait

I donr

How ru getting rid of y there

Are u

[wooden marten]

you aren't

you're just finding the derivative

[karmic reef]

Is it dy/dx *4y^3

[wooden marten]

mhm

math is magic that way

[karmic reef]

Ignore the dx^2

Oh wai

There we go

[wooden marten]

yee

[karmic reef]

Now how can i get dydx on its own side

My guess is subtract the x^2sec^2(y)*dy/dx and factor out dy/dx and divide by the inside on the other side

But that still leaves me with ys on one side

Can you plugin ys

[wooden marten]

yes you can
you know what y is

at the point

[karmic reef]

Oh ok

Since anything * 0 =0

[wooden marten]

now you can finish

[karmic reef]

Thank u

.close

[obtuse pebble]

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[tiny seal]

lets look at c first

gimme one sec while i write it out

+2e^x

wait

forgot to distrbute the -, whoops

there is a continuation of ibp, the di method

it makes these repeated ibps faster

integrals?

you can learn them rn, its very easy actually

just an area function

i highly recommend khan academy if you want to

well, if you know derivatives, you can already solve integrals with the fundamental theorems of calculus

[obtuse pebble]

@timid silo Has your question been resolved?

[tiny seal]

that one is

i deleted the one with the mistake

[obtuse pebble]

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[timid silo]

Can someone help with this one?

[timid silo]

You don't have to solve the entire problem, I just need a procedure or hints

[tiny seal]

what is z in this context?

this looks like a typical line integral, so we can just parameterize C

what i dont get is why we do a . on a scalar

z=(x,y)?

there has to be some context about z somewhere

[timid silo]

Z = x+iy

This is problem is from Complex Integration

(Please ping me when you reply)

[tiny seal]

@timid silo
heres a hint, we know (a(theta-sin(theta)) , a(1-cos(theta)) goes from (0,0) to (2pi a,0)
it is not too hard to solve for theta here and find bounds a and b, we are also parameterized
let z=x+yi, then we may use the following picture with u(t)=x(theta), v(t)=y(theta)

[timid silo]

I think this means that we should substitute x = a(theta-sin theta) and y = ...... in the problem right?

[tiny seal]

yeah

[timid silo]

But

[tiny seal]

you do d theta

[timid silo]

It becomes very lengthy by doing that

[tiny seal]

its an integral theta dtheta, sin(theta)dtheta, dtheta, cos(theta)dtheta, these arent that long
the only lengthy one is solving for theta's bounds

but we are given 0,0 which is easy to solve, only solving the upper bound will be lengthy

[timid silo]

I think upper bound for theta is 2pi

Maybe that wasn't the lengthy part

Expanding z^3 then substituting x and y in it goes long

[tiny seal]

yeah

but thats the fun of contour integrals

[timid silo]

I tried writing x = sin alpha = a(theta-sin theta)

But it doesn't work

Such alpha doesn't exist

<@&286206848099549185>

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

@timid silo Has your question been resolved?

[glacial obsidian]

You could use Cauchy's theorem

[obtuse pebble]

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[tardy epoch]

Takes like 3 times

x^3 to 3x^2 to 6x to 6

Keep at it

[obtuse pebble]

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[tall bramble]

this is solution idk from the red line how they set up that equation

<@&286206848099549185>

[trail musk]

Define collinear

[tall bramble]

same line

parralell

[trail musk]

What does that mean in terms of the arithmetic

[tall bramble]

there will be a konstant where ab will equal cd

cb

[trail musk]

Ok great!!!

[tall bramble]

so CB = kAB

[trail musk]

Or in other words, the RATIO between components is constant

[tall bramble]

or otherweay round

yh

[trail musk]

So like how about this just to show you what I mean

Lets say we have the two vectors (1,2) and (2,4)

[tall bramble]

1:2

[trail musk]

Wouldn't any collinear vector satisfy:
y component is ALWAYS twice x component

[tall bramble]

how is it always

cant the konstant differ

[trail musk]

The constant cancels

Any vector on that line is (c,2c)

2c/c = 2

[tall bramble]

ok

[trail musk]

So then

[tall bramble]

so even if the graph looks way off cb = 2ab

interms of my qquestion

[trail musk]

No

Thats not what I'm saying

[tall bramble]

ok

[trail musk]

I was just using the (1,2) as an example

[tall bramble]

alight

[trail musk]

(a component of CB) / (b component of CB) =
(a component of AB) / (b component of AB)

[tall bramble]

ok

[trail musk]

Plug in the numbers now

[tall bramble]

how did u get 2 values for cb tho

this is proper confusing me

[trail musk]

Do you know what I mean by components?

[tall bramble]

no

[trail musk]

don't vectors in 2D have to be described by two numbers

[tall bramble]

a nd b?

a and b?

is that what u mean by 2 numbers

[trail musk]

in this case we're treating a and b as our basis VECTORS

So what multiplies the a is the a component, and what multiplies the b is the b component

The a component of CB is -3k

[tall bramble]

how is it over -2

[trail musk]

The b component of CB is 2-9k

Trust

[tall bramble]

yes i understnad that

[trail musk]

Use this equation

Now that you hopefully understand what the word component means

[tall bramble]

still dont understand the 2 and -2

i undersntad the numerators now

but denominators

dk

[trail musk]

What are the a and b components for the vector AB

[tall bramble]

2- 9k and -3k

[trail musk]

Nope

I said for the vector AB

[tall bramble]

2b -2a

?

[trail musk]

what multiplies the a is the a component, and what multiplies the b is the b component