help-36

[sharp cliff]

I mean I could do that

Wait do u mean the mnemonic FOIL or

😰

[dusty dock]

Yeah

[sharp cliff]

Omg yay at least I remember that

[dusty dock]

Let's gooo

[proud parrot]

some pretty tough problems for just learning the power rule tbh XD

I just had to do polynomials

[sharp cliff]

Erm guys so does the 1/x just turn back to like the negative power

[dusty dock]

Yep!

[sharp cliff]

Okok

[dusty dock]

They didn't explain it in this context (probably) but it does generalize to reciprocals

It's a lot more annoying to prove, though

[sharp cliff]

Would it’s be like x ^ -1

[dusty dock]

Yea

[proud parrot]

I will say, i do truely love math courses that give you problems that seem impossible with your knowledge but they are actually possible and you just have to think a little bit outside of the box

though they can be tough

[sharp cliff]

Okay had to double DOUBLE make sure

[dusty dock]

It's because we don't normally think of reciprocals as powers right away

My course explained that it works for all powers, but only showed the positive integer proof (which is fine for that point)

[proud parrot]

Well i meant more so the problem and how to differentiate it

[dusty dock]

Math is about thinking outside the box until you learn a method

Then its about using the method you learned

[proud parrot]

thats why its so fun (my friends have been calling me crazy for years)

[dusty dock]

Same frfr

You got it?

[sharp cliff]

U guys r a little crazy for going thru w this I mean coming from someone who stuggles with math

Erm almost

I think

[dusty dock]

Idk it's the ideas that make it fun

Like when I actually understood how the derivative formulas worked and shit I felt like I became a sorcerer (aaand then felt like shit understanding nothing like 3 days later but that's the process lmfao)

[proud parrot]

I feel like a wizzard when i use calculus to find a vertex

top 10 reasons for learning calculus byfar

[dusty dock]

Gives the quadratic formula a new meaning

[sharp cliff]

Waittt a minute so uhm yk when u foil it do the two x^-1 cancel out and u get like x^1 ( just x) or am I going insane

[dusty dock]

Nope, exponents add on multiplication

Nt tho

[sharp cliff]

I forgot

Whoops

Ughhhh if I don’t finish this by morning my maths teacher is gonna get me

[proud parrot]

derive faster goddammit!!! /j

[sharp cliff]

Do it for me pretty pls 🙏🙏

[proud parrot]

nuh uh, the only way to learn is to do it yourself

[dusty dock]

Unfortunately against policy

[sharp cliff]

Guys what the heck 😞

Okay

[dusty dock]

😔

[sharp cliff]

I’ll try

[dusty dock]

Worst comes to worst you use wolfram, but wolfram doesn't tell you the steps

[proud parrot]

ive been legit offered up to 3 grand to do peoples math homework for a semester in classes im in

Ill always refuse because its for their own good to do it themselves, because if they dont do it themselves they will face more issues down the line :P

[sharp cliff]

What the heck is wolfram

[dusty dock]

The goat of calculators

[shut gazelle]

Not for derivatives and showing work tbh

[dusty dock]

I mean most of the calculators I use don't convert derivatives into the actual function, they just graph it (no hate desmos)

[shut gazelle]

https://www.derivative-calculator.net/
This is the best calculator, that shows work. It uses the shortcuts, like power rule

[sharp cliff]

so I got 2x + x^-1 + x^-1 + 1 (I think I’m wrong 😞)

This is gonna make me go insane

[dusty dock]

Okay let's go through each step one by one

[sharp cliff]

Im gonna have dreams about derivatives

[dusty dock]

$(x^{-1} + 1)^2$

[soft zealot]

Serphic

[dusty dock]

We split it into two

[sharp cliff]

I get that part

[dusty dock]

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

[soft zealot]

Serphic

[dusty dock]

To remove any foil confusions, we distribute separately

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

[soft zealot]

Serphic

[dusty dock]

Can you distribute that out?

[sharp cliff]

I mean I just times the brackets together usually

Idk

[dusty dock]

They're the same thint

Okay distributing out we geet

$x^{-2} + x^{-1} + x^{-1} + 1$

[soft zealot]

Serphic

[sharp cliff]

Wait how it x^-2

I thought it just like cancels it out since -1 + -1

[proud parrot]

-1 + -1 is the same as -1 - 1 which is -2

[shut gazelle]

-1 + -1 = -2

Nothing cancels

[sharp cliff]

Jezzo

[dusty dock]

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

[sharp cliff]

Okay

[soft zealot]

Serphic

[dusty dock]

Latex on mobile is not fun damn

[sharp cliff]

hate negative numbers

😞

[dusty dock]

Anyways simplifies out to

$x^{-2} + 2 \cdot x^{-1} + 1$

[soft zealot]

Serphic

[dusty dock]

Now apply the power rule (to avoid confusion, here it is for reference)

$\frac{d}{dx} x^n = n \cdot x^{n-1}$

[soft zealot]

Serphic

[sharp cliff]

I think im burnt

Not even cooked

[dusty dock]

[sharp cliff]

Guys

What the heck

[dusty dock]

Okay try x^(-1)

What do you d9

Odo

Do

Damn

[sharp cliff]

-1x^-2 idk

Or is it just x

I think x not -1

[dusty dock]

Congrats!

[sharp cliff]

-x^-2

Yes?

[dusty dock]

Yep

[sharp cliff]

So uh -2^-3 + 2x^-2 …?

😰

[dusty dock]

Forgot the x on the first term, and the negative on the second term

But yeah!

[sharp cliff]

Uh

ITS OKAY

THE X IS JUST LIKE INVISABLE

[dusty dock]

Lmfaooo

[sharp cliff]

guys its there TRSYT

it’s on holiday

It’ll come back

[dusty dock]

It's not Xmas yet..

Fine I'll see myself out😔

Okay but fr good job!

[sharp cliff]

I might have possibly learnt something yay to me!

Took a while tho…

Slow learner for a reason

[dusty dock]

Any other problems?

[sharp cliff]

For now… nope but i will be coming back eventually…

I always do tbh

Im a daily visitor

Haven’t seen the last of me yet

Tysm tho!!!

[dusty dock]

Np!

[sharp cliff]

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[zealous stump]

Question:
Lisa and Mark play a game where they each subtract either 1 or 2 from a random number. The person who creates zero is the winner. Mark always starts.

For example, starting with the number 5 Mark subtracts 2 giving the value of 3 to Lisa. Lisa subtracts 1 giving the value of 2 to Mark. Mark subtracts 2 and wins the game.

Prove by induction that if Mark starts with numbers of the form 3n+1 or 3n+2, where n is N, then it is always possible to win the game.

[zealous stump]

I'm stuck on how to express the question mathematically

nevermind

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[indigo plover]

i keep plugging this

in my calc

and i get -6.14 instead

D:

how to use calculator on this discord plz help

i wanna see if its a mistake on my end or maybe the live lecture

[tacit fractal]

you can use ,w <whatever you want to calculate>

[indigo plover]

thnxs

,w 3.2cos40 + (-5.10cos35)+(-4.80cos23)+DX=0

[soft zealot]

Results provided by WolframAlpha

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

Requested by ac9468

[indigo plover]

?!!!!!!!!1

isnt it negative??

oh

then u add it

[alpine pendant]

ok

[shut gazelle]

You shouldn't try to do terms with subscripts like this, wolfram will think it's D times X

[shut gazelle]

You probably did the math and got -6.14 + Dx = 0 but never finished solving for Dx

[final saddle]

@indigo plover Has your question been resolved?

[indigo plover]

ya that i forgot to solve for dx

.close thnxs

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[lost narwhal]

Can anyone clarify the third image? This is evaluating limits at infinity
It's dividing both side but they multiply by 1/x which I get the concept but why is the denominator multiplied by x^2?

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[lost narwhal]

.reopen

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✅

[lost narwhal]

<@&286206848099549185>

[delicate garden]

Because it's inside a square root that's why the 1/x turned into (1/x)^2

[lost narwhal]

is that like a new rule. cause i never heard of it before

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[bleak edge]

for a i said lim x -> 5 fx is 3.5

and im confused on how to start on b

oh i see

lets just do both 112 and 113

so for 112 a ) the function approaches 8

cause it jumps to 8

112 b ) not continuous because of the jump from almost 3.5 to 8

113 a ) it suggests that the function approaches 3.5

113 b ) continuous, very close to f(4.999) and f(5.001)

<@&286206848099549185> sorry i know it hasnt been 15 mins but im in a rush (exam tmrw)

can someone check my answers above

[dense flower]

No this is wrong.

Please keep in mind that we don't care about the value of the function at that point...

This is the most crucial point

From both negative and positive side, the function is approaching 3.5

So, just ignore the value of the function at x=5, and just consider what the limit should be...

It has to be 3.5.

[bleak edge]

so for 112 and 113

a is just 3.5?

[dense flower]

yeah

[bleak edge]

ah

i see

are my other answers okay?

[dense flower]

please remember this...

[bleak edge]

got it

i will

[dense flower]

not really

Because they said that assume the limit approaches to 3.5

Wait...

I might be wrong...

[bleak edge]

uh

for a or b

[dense flower]

113 both parts are correct

112 a is correct

[bleak edge]

the revised one right

[dense flower]

112. a) 3.5 is answer

[bleak edge]

ok

[dense flower]

But I am not sure about 112) b)

The question is so vague.

[bleak edge]

what would you say the best answer is

like if you had to say an answer

[dense flower]

I mean, if they are saying that consider the limit in part a, then as the limit is 3.5, the answer should be "continuous"

But if we just generally see a function like that, it's not continuous at 5.

[bleak edge]

i would want to say not continuous

cause i dont think they would give the same answers for 2 problems

[dense flower]

yeah lmao

wait lemme try confirming it..

[bleak edge]

alright

[dense flower]

I have confirmed it from a friend.
Here is what he is saying:
112) a) 3.5
112) b) non continuous
113) a) 3.5
113) b) continous.

This is what we thought was right too.

And it makes sense too...

[bleak edge]

alright

[bleak edge]

is my reasoning okay

for 112 b

[dense flower]

Yes

[bleak edge]

alright ty

[dense flower]

pleasure

[bleak edge]

.close

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[warm python]

I'm trying to find the basis of the the kernel of this LT

[warm python]

so I feel the basis of this kernel will be

$\begin{bmatrix} 1&-1\1&-1\end{bmatrix}$

[soft zealot]

Veni, vidi, perii

[warm python]

so basically what I did was find when the image is 0 ( for a,b,c,d)

But isn't the kernel supposed to be a set of vectors

so I suppose I convert this to a transformation from R^2 to R^2 and then proceed?

[final saddle]

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[warm python]

I want to find the basis of the kernel of this

I was thinking just

$\begin{bmatrix} 1&0 \ 0&-1 \end{bmatrix} ; \begin{bmatrix} 0&1\ 0&0 \end{bmatrix}; \begin{bmatrix} 0&0\ 1&0 \end{bmatrix}$

b,c \neq 0

[desert mantle]

thats your basis? one matrix?

[warm python]

yeah?

[desert mantle]

thats false

[warm python]

how? for this to map to 0, a+d=0

so a=-d

that's all

b and c are irrelvant

oh

that's why

[soft zealot]

Veni, vidi, perii

[warm python]

Is this better?

[desert mantle]

yes

[warm python]

Thanks

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[coral peak]

HELLO

[coral peak]

SOMEONE HELP ME?

i need help because i dont understand something im trying to get an homework done but i dont understand

,text [
\text{Tr}{\mathcal{H}}\left( \int{\mathcal{C}} \left( \text{Lie}{\infty} \left( \mathcal{M}{\alpha\beta} \right) \cdot \text{Exp}\left( \text{Path} \left( \frac{\sin(\theta)}{x} \right) \right) \cdot \frac{\text{Exp}\left( \int_{a}^{b} \text{EllipticE} \left( \frac{u}{v} \right) , du \right)}{\text{Hypergeometric}_2F_1 \left( a_1, a_2; b_1; \frac{c}{d} \right)} \right) , d\mathcal{C} \right)
]

[soft zealot]

V12

[coral peak]

pls help

@tribal shell

@desert mantle

thank you in advance

[tranquil pine]

what problem are you facing?

[coral peak]

i need help

[odd seal]

<@&268886789983436800> Spamming in all channels + this

[fathom walrus]

bro has no clue what he's typing

[spare hazel]

lmao

[coral peak]

yes im expert

[coral peak]

this

[tranquil pine]

bro

what is the problem you're facing

[coral peak]

i need help

i dont understand

what's the stick symbol

[spare hazel]

integral

[coral peak]

no not integral

this

also what's this

[spare hazel]

bro

[crisp jacinth]

Brother cannot be a postgraduate bro 😭🙏🏿

[coral peak]

yes

i need help iont understand

[coral peak]

what does this mean

[stoic temple]

Sine of an angle

[coral peak]

no not the sin thing

[stoic temple]

What is C???

[coral peak]

the ----

[spare hazel]

thats division

[coral peak]

a letter

[spare hazel]

sin (theta) divided by x

[blissful condor]

can we not troll in the help channels please

[coral peak]

damn

ok

😂

[stoic temple]

Ban🙂

[quartz spade]

Trolls in smay's presence is wild.

[blissful condor]

.close

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[white gyro]

how to do this

[livid ice]

Take a look at the 2 terms

Do they share anything?

[white gyro]

x

[livid ice]

You can factor that one out

[white gyro]

x(-64x^2)

[livid ice]

Pull it out/siphon it out

Yea

Wait

When

You factor

It's like dividing

So there's always something left

You cant get rid of something completely from dividing

Pull x out from itself

What's left behind

[white gyro]

^2

[zealous storm]

[white gyro]

would u squaree root

[livid ice]

E.g.

(x^2 + x)

= x(x + 1)

Because when you pull x from itself

1 is left behind

[white gyro]

oh

i see

[livid ice]

Because it's as if you did x/x

[white gyro]

so would it be x(1-64x^2)

[livid ice]

Yea

That's good now

[white gyro]

x(1-(x+8)(x-8))

[livid ice]

Not quite

1 - 64x^2

is in the form

1 - y^2

How would you do

1 - y^2

[white gyro]

(1-y)(1-y)

[livid ice]

Ok perfect

Your answer should look exactly like that form

[white gyro]

(x)(1-8x)(1+8x)

[livid ice]

Yes

Perfect

[white gyro]

tysm for the help

that helped

[livid ice]

Yw

I'm glad!

Good luck with your school stuff

[white gyro]

yh hopefully i get good soon

.close

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[grim imp]

idk how to do last one

[final saddle]

@grim imp Has your question been resolved?

[grim imp]

<@&286206848099549185>

[final saddle]

@grim imp Has your question been resolved?

[final saddle]

@grim imp Has your question been resolved?

[grim imp]

<@&286206848099549185>

[short shadow]

That last one is just an asymptote thing. Essentially you’re limiting the graph.

I think your answer is -6,-7,-8,-9

Wait no

(-Inf,-3) should be the answer that satisfies the conditions present

[teal lagoon]

Well also b= 2

[short shadow]

Please elaborate

[teal lagoon]

That would be the largest to for it to be continuous in the intervall

(-3,2]

The function is obviously continuous in the interval

But not im (-3,3]

[grim imp]

OK

[grim imp]

this is the answer right

i get it now i think

oh wait idk nvm

oh wait yes

i thought it was x, y for some reason

[teal lagoon]

Yeah, its the intervall they used there

Though the question has a wierd design

(-3,b] and they want you to give all possible b values but if b i smaller than -3 then the intervall would look at lot different

You have the largest b values, for all possible b values just look at the one point at -5 that is isolated

[grim imp]

WAIT IDK anymore

@teal lagoon wat do I put

[teal lagoon]

Largest b is 2

[grim imp]

idk 2.9 (repeated)

o

[teal lagoon]

And then list up all the b values that satisfy the property

[teal lagoon]

Largest is 2

Because f is continuous on (-3,2] but not on (-3,3]

[grim imp]

wat else do I put

do I put like

(1,2]

@teal lagoon

[teal lagoon]

Nope you just put in the b values because they are asking you for any b such that f is cont in (-3,b] and not in (-3,b+1]

[grim imp]

@teal lagoon

oh

it didn't load

[grim imp]

but if it was 1.1+1

it would go like

past 2

[teal lagoon]

If you mean that b = 1,1 is a valid input then yes

So in general all numbers in (1,2] would satisfy the condition

[grim imp]

do I put that as the answer

[teal lagoon]

Yes

But thats not the entire answer

Look at what going on at -5

[grim imp]

is the answer to the first part 2

[grim imp]

wat Abt it

[teal lagoon]

Yes

[grim imp]

light work

[teal lagoon]

Its a place where the function isnt continuous anymore

So you can figure out what you would have to put as b so the conditions are satisfied

[grim imp]

umm i don't think that left side matters for this question

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[tranquil pine]

not sure I understand y=3x so m is 3 and both the x and y intercept are 0 so how can I have two ordered pairs to graph

[verbal cypress]

Take random values for x

Nd find corresponding y

[tranquil pine]

so I can use any value ?

[verbal cypress]

Ofcourse

They have asked for slope intercept form

But it doesn't matter haha

[tranquil pine]

so if I do y=3(1) would a second ordered pair be (0,3)

or sorry (3,0)

I did that and i got it wrong I have 3 trys

[main nexus]

if x is equal to 3, is y = to 0?

[tranquil pine]

yea

[open wyvern]

bro i think intercept is 0 and slope is (m) = 3

[tranquil pine]

yeah it is

[tranquil pine]

thats what I said earlier but since both y and x intercept are zero apperently I just pick a random number for x and use it

[open wyvern]

bro your two ploting point are - (0,0) or (1,3) try it

[tranquil pine]

its both you need to use 2 ordered pairs

I just found out

basically since both intercepts are 0 we need to pick a random number which I will pick 1 and use it

so y=3(x)

y=3(1)

y=3

but we picked 1 for x so

1,3

[open wyvern]

yes,right

[tranquil pine]

Thanks

[open wyvern]

In which standard do you study brother

[tranquil pine]

well this is just college prereq bs but rn I am doing an associates in science with a focus in computer programming

I plan to do cs for a major

[tranquil pine]

what about you

[open wyvern]

bro I'm in class 11 now , and preparing for JEe mains and advance

[tranquil pine]

whats JEE mains ?

[open wyvern]

Joint entrance exam , it is for getting admission in NIT or IIIT like goverment collage

[tranquil pine]

cool sounds fun

[open wyvern]

Bro seriously you don't know about iit's

[tranquil pine]

nope

[open wyvern]

From where you belong brother

[tranquil pine]

usa

[verbal cypress]

Most people don't lol

The world doesn't revolve around IITs

[tranquil pine]

lmao

is it a tech school or something

[verbal cypress]

Yeah like the MIT but kinda bad

[tranquil pine]

damn pretty prestigious

[verbal cypress]

Yeah it has very tight selection rate

Though that's quite irrelevant to your question 😄

[tranquil pine]

lol we were just talking about college and it came up

ik this is pretty elementry stuff for college but you be surprised how many things you forget in 5 months of not going to school

[verbal cypress]

I'm aware

[tranquil pine]

lmao I feel like I never went to school but I'm getting back into it again

[verbal cypress]

Yeah, should have kept in touch for those 5 months

[tranquil pine]

for sure but the professor understands that people forget over summer and started pretty simply next summer I'm not making the same mistake espically since I'm going into calc

good talking though

.close

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[warm python]

I feel this is false

[dense coral]

[warm python]

oops

[dense coral]

no, I didn't say anything yet

[warm python]

constructed a wrong mapping in my head

I thought (x,y)= (x-y=0) is. a valid mapping

[dense coral]

what does (x-y=0) mean

I've never seen this notation before

[warm python]

Nothing, I know

[dense coral]

you might be onto something though

[warm python]

hmm, consider a mapping from $\R^2 \to \R^2$

[dense coral]

can't

[warm python]

wait

[dense coral]

then V neq W

[warm python]

oops

yeah

[soft zealot]

Veni, vidi, perii

[dense coral]

[warm python]

Consider the mapping $(x,y) \to (x-y,0)$

[soft zealot]

Veni, vidi, perii

[warm python]

every image in $Ker(T) \notin Im(T)$

[soft zealot]

Veni, vidi, perii

[dense coral]

this sentence doesn't parse for me

[warm python]

ok, so consider the mapping (x,y) \to (x-y ,0)

(5,5) \in Ker(T)

but (5,5) \notin Im(T)

[dense coral]

indeed

I was thinking something a little simpler haha, but this works

consider T: R^2 \to R^2 given by T((x,y)) = (0, 0)

then ker(T) = R^2, and im(T) = {0}

[warm python]

👍

thanks

[dense coral]

https://tenor.com/view/cute-eevee-gif-21737791

[warm python]

I can prove it in words

not sure of how to formally express it.

Like if $Ker(T) = V \implies \forall v\in V T(v)= 0 , \implies Im(T)={0}$

oh

[soft zealot]

Veni, vidi, perii

[dense coral]

ah, this is a special case of a very important theorem

[warm python]

wait

is this true

hmm

one minute

[dense coral]

okay

[warm python]

I mean this is false

this statement

[dense coral]

which statement is false?

[warm python]

this

[dense coral]

why is it false?

[dense coral]

this is good btw

I would just swap some symbols for some words

[warm python]

if W={0} \implies V={0}

[dense coral]

I don't follow what you're trying to say here

you claim that W = {0} forces V = {0}?

[warm python]

wait

oops

thought V=W applied here too

[dense coral]

ah, I see

yo chartbit

can you review my proof

I think it should be fairly checkable

cause it's short

#1260424130822541403

[tulip coyote]

Can't lie, haven't done much topology, but I'll try

[dense coral]

[warm python]

!noadvert /j

[final saddle]

Please do not advertise your help channel or thread in other parts of the server. There are many people who need help, so advertising can quickly turn into spam.

[warm python]

This is a joke

it's fine

sorry

[dense coral]

[warm python]

I was just. kidding

[dense coral]

I used for a reason

that was to indicate that I was joking too

[dense coral]

how much geometry/topology do you know? (answer this after you're done checking)

[warm python]

I mean, here the mapping is onto W, so why does W have to be {0}

[crimson token]

because the kernel is the domain -> everything in the domain maps to 0

[warm python]

yes

but W is just the set onto which the mapping is taking place, right

[dense coral]

yes

this is an implication

[warm python]

like W={0,1} is also fine

1 just has no pre-image

oh wait

for a function every element in the range has to have a pre-image

so it has to be {0}

right

I think I'm royally confused

[dense coral]

this is an implication

if ker(T) happens to be all of V, then W = {0}

the contrapositive would deal with your example

if W is not equal to {0}, then ker(T) isn't all of V

[warm python]

no, what I mean is

I can have a valid mapping from R^3 to R

where all elements map to 0

right

[dense coral]

sure

but what is your point?

oh, I see your point

[crimson token]

I think what Veni is saying is that the codomain needn't be the same as the image

[dense coral]

yes, in that case they are correct

[warm python]

yeah, so this statement is false

right

[dense coral]

yes

for some reason, I forgot this was a true or false thing

and I just assumed it was true even though I knew better

[crimson token]

same ngl

[dense coral]

"damn, I must've gotten real bad at LA since"

[tulip coyote]

Back

[dense coral]

did I succeed?

[warm python]

Now this is true

[tulip coyote]

"barely" I ended up going through some topology a while back, but never really covered it "properly", that said I am happy with what you've done personally

[warm python]

This is as all elements map to 0, thus the entirety of V is the kernel

[dense coral]

what about geometry

[tulip coyote]

No

[dense coral]

how will I force chartbit to check my proofs then?

[warm python]

I can check geometry to a point

[tulip coyote]

You'll have to catch me first

[warm python]

Euclidian geo in R^2 that is

not something in R^n

or R^infty

[dense coral]

I mean like

diff geo

[warm python]

oh

I thought euclidan geo

[dense coral]

maybe eventually I'll do some algebraic geometry, but uhhh

I am so geometry/topology pilled it's insane

I love manifolds

[warm python]

I'm LA pilled

LA>>>>>>Calc

[dense coral]

I am LA pilled too

[warm python]

what does $T=0$ mean

[soft zealot]

Veni, vidi, perii

[dense coral]

T is the 0 map

sends everything to 0

[warm python]

I'm using axler for LA 1 :kannafire:

Hmm, this is a bit hard

Oh

Im(T) maps is the set of ouputs

ker(T) is the set of inputs that map to 0

Trying to find a counterexample from R to R

hmm

[tulip coyote]

Right, cooking time

Can you describe this statement, in words?

[warm python]

$x \in Im(T) \implies x \in ker(T)$

[soft zealot]

Veni, vidi, perii

[warm python]

wait

I didn't use that V=W

[tulip coyote]

Those are not words

[warm python]

I think analysing teh contrapsotive may be more constructive here

so $Im(T) \not \subseteq ker(T) \implies T \neq 0$

[soft zealot]

Veni, vidi, perii

[warm python]

If x is in Im(T) , x is also in Ker(T)

[tulip coyote]

What do the things in the image look like?

[warm python]

like inputs in T that map to 0

[tulip coyote]

Well, I mean, here, they are...

How is the image defined in general, I meant?

[warm python]

${Ax | x\in \R^n}$

[soft zealot]

Veni, vidi, perii

[tulip coyote]

Well, in this case, it's Tx, and x being in V

[warm python]

let me think

I want to arrive at the answer independetly

it's just that something about mathcord actually wants to make me think

hence i ask here

Hmm

Yup, I'm convinced this is true now

Now to prove it

Ok, I'm lost

can I have help

I'm unable to come up witha counterexample

[tulip coyote]

Right, to take back to where we were, we said here that the image is everything of the form Tx, for x in V, and that the image is contained in the kernel

[warm python]

yes

[tulip coyote]

If Tx is in the kernel of T, what does that mean?

[warm python]

x maps to 0

[tulip coyote]

Not x itself, but Tx

(the V = W thing allows you to compose T with itself)

So if you have any x from V, what you know is that T(T(x)) is zero, T(x) is in the image of T, and subsequently the kernel of T, so must get sent to zero if you apply T again, right?

[warm python]

yes

Hmm

I guess I could say $T(im(T)) \not \subseteq T(ker(T))$

[soft zealot]

Veni, vidi, perii

[warm python]

unless tge image is 0

but this feels very sus

[tulip coyote]

remember that matrices can be considered as linear maps, which might give a bit of inspiration for a counterexample

[warm python]

So this is a false theorm?

[tulip coyote]

The statement is false, yes, there is a map that isn't constantly zero, but whose image is contained in the kernel

[warm python]

hmm $(x,y) \to( x-y,0)$

[tulip coyote]

Dimension of the space you're getting sent to needs to be the same, you need the image to be in R^2 in that case

[soft zealot]

Veni, vidi, perii

[warm python]

This isn't it, I feel

[tulip coyote]

Something close to that would work though!

(in yours, (x - y, 0) gets sent to ( (x - y) - 0, 0) = (x - y, 0), which isn't zero)

[warm python]

I give up

I'll come back to this tonight

got to let this cook slowly

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[finite delta]

How would I turn this into a parametric equation for part A? Or am I going about this the wrong way?

[final saddle]

@finite delta Has your question been resolved?

[final saddle]

@finite delta Has your question been resolved?

[final saddle]

@finite delta Has your question been resolved?

[rare estuary]

just apply the formula given!

same way as you would do in 3D

[finite delta]

so, r(t)=t<-2,5,-6,-4>+<1,-2,4,5>?

[rare estuary]

yess

[finite delta]

Ok, I was just overthinking what AB was then. I thought I had to do a dot product or something for some reason

[rare estuary]

haha its okay

so what would you do for (a)

[finite delta]

I would try to find a value for t that makes y=0

[rare estuary]

exactly

[finite delta]

and for b i would want to find t where z and y are 0?

[rare estuary]

yes, one t for which both y and z are = 0

[finite delta]

ok cool, thank you

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[vale wedge]

How do i test if the road can be twice differentiable?

[rocky tusk]

verify the same conditions of f’ that you would do for f

[vale wedge]

I tried using the definition of derivative like i used to get a on f. But doing that on f' got me nowhere

[livid ice]

Do you know how to do quotient rule?

Typically you wouldn't need to use the limit definition for derivative for something like this

You'd use the various shortcut formulas/techniques for each derivative

[final saddle]

@vale wedge Has your question been resolved?

[vale wedge]

If I manage to derive both expressions twice, will that mean that it is twice differentiable,or must the function also be continuous at the same time?

[livid ice]

Basically both f and f' need to line up

This is a piecewise function with 2 different intervals

The derivative of the function is the derivative of the appropriate function depending on what interval you're in

To be differentiable you need f' from the left to equal f' from the right at that point

So you can make a system of equations for continuous (lining up the pieces of f itself)

and the other equation would be for differentiable, "lining up f'"

So twice differentiable you'd do it again check limit from left and right and set equal

[vale wedge]

thanks so if this is how the system of equations after doing f'' on both of them then its not twice differentiable?

[final saddle]

@vale wedge Has your question been resolved?

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[vocal atlas]

ratio of debt in the principal is 25%, ratio of debt in capital + ratio of principal in cap.= 1. (capital) How do I get ratio of debt in the capital

[vocal atlas]

big struggle

answer is 20% but idk how got it

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[wary copper]

limiting sum

[warm ether]

geometric sum formula

[wary copper]

the ratio if 5/6

since we know 5/6 is less than 1

we can import it into a limiting sum formula

you can derive it from geometric formula

for your geometric sum = a(1-r^n)/(1-r)

it is just general form of geometric sum

[latent fractal]

Let
[S = \sum_{i=0}^\infty r^i = 1 + r + r^2 + \dots]
where that sum converges. Then if we subtract (1) from (S), we get the same sum but without the first term
[S - 1 = r + r^2 + r^3 + \dots]
and if we multiply (S) by (r), we get... the same sum but without the first term
[r S = r + r^2 + r^3 + \dots]
so (S - 1 = r S). Solving for (S), we get (S = \frac{1}{1 - r}).

[soft zealot]

Invariance

[latent fractal]

so
[1 + r + r^2 + \dots = \frac{1}{1 - r}]

[soft zealot]

Invariance

[wary copper]

another way to do it is also just by limits if you are familiar with it

[latent fractal]

when the sum converges (-1 < r < 1)

[wary copper]

the general geometric series is conventionally written as
a + ar + ar^2 + ... + ar^k+... where k is a natural number

r is meant to represent the ratio

a represents the first term

the summation is
a(1-r^n)/(1-r) for S_(n)
if we want to take infinite sum, then evaluate it with limit of n approaching infinite

Since |r|<1, r^n will approach 0

your summation ends up being a/(1-r)

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[fierce light]

im supposed to find the domain of 4x(x-7)/5x^2-34x-48

[fierce light]

so i set the bottom to equal zero but the factoring doesnt make sense

[warm ether]

brackets please

what is the denominator

[fierce light]

5x^2-34x-48

[warm ether]

whats up with the factoring?

[fierce light]

theres nothing that adds up to -34 but also equals -48 when you multiply

[warm ether]

you have to mulitply 5 and -48 first

then find factors of that which sum to -34

[fierce light]

why do i do that

[warm ether]

you do it all the time, you just seem to be used the having the x^2 coefficient be 1

[fierce light]

i dont get why tho

that gives me -240? very confused

[warm ether]

say you have (ax+b)(cx+d)
when you expand that you get
(ac)x^2+(ad+bc)x+bd

ac*bd has factors ad and bc which sum to get the middle term
ac=A
bd=C
ad+bc=B
Ax^2+Bx+C
you can often find factors of AC that sum to B

might be a bit of a dodgy explanation ill admit

but the idea is there

[fierce light]

could you explain in words, i cant understand math explanations

[warm ether]

i think using words would be far more confusing in all honesty

[fierce light]

its hard for me to keep track of all the letters its like greek to me

so when i do 5*-48 and get -240 is that my new equation

x^2-34x-240?

[warm ether]

once you do that you need to find two factors of -240 that sum to -34

[warm ether]

no

[shut gazelle]

https://www.youtube.com/watch?v=4v-EQIxqpMQ&ab_channel=TheOrganicChemistryTutor
This may help

[fierce light]

okay ill go watch that

thanks

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[warm patio]

hey gang, i need help with this here is my work too

[warm patio]

i am struggling with theconcept of how to identify the axis of revolution so if anyone can help me with that with the process i'd be grateful! also with my computations with dy and dx, they don't match up? What am I doing wrong?

[tiny gorge]

wdym revolution

isn't it just a 2d region?

first integral looks fine

[warm patio]

oh no that's in part b and c, i haven't showed that yet

[tiny gorge]

second is not

[warm patio]

is it the calc portion or algebra?

[tiny gorge]

it's that you're only integrating from y=5 to y=14

but the region has a part below y=5 also

so you need to add the area of that part

(it's just a rectangle so it's easy to find the area)

[warm patio]

hold i'm a bit confused on this one, i thought with the integral of y = 5 it covers that region ?

[tiny gorge]

how could that integral cover it? the point (1,1) is in the region for example

also, your y integral is finding the area of the wrong part

you're finding the area inside the parabola, not outside it

[warm patio]

okay i see that but i am having a hard time visualizing the "part" of y =5 hold on

[warm patio]

Wait so like this instead? or

[tiny gorge]

yea that's what your y integral is computing

[warm patio]

wait so what kind of principle does this follow? is this a rule for vertical slices to find the area inside the parabola rather than outside?

i just want clarity with fundamentals cause i'm not really grasping the concept

[tiny gorge]

well for each y, you want the distance from the parabola to the x=3 line, right?

[warm patio]

yesyes

[tiny gorge]

not the distance from the y axis to the parabola

so you should not be integrating sqrt(y-5)

you should be integrating 3 - sqrt(y-5)

[warm patio]

OHHH i've been doing it wrong okay i see

i've been slicing it up to measure distance from the y axis LOL

[tiny gorge]

and you'll still need to add the area of the rectangle below y=5

i.e. the rectangle enclosed by y=0, y=5, x=0, x=3

[warm patio]

OHHH

[tiny gorge]

because that is not counted in your integral

[warm patio]

can you elaborate on this? as to why we do this?

wait hold on lemme redo it

[tiny gorge]

because as you show in your figure, your integral is only finding the area of the part above y=5

[warm patio]

so to subtract the area, can i do integral from 0 to 5 of (y-5)^1/2?

and subtract that result ?

[tiny gorge]

why do you need to subtract something

your integral from 5 to 14 is only getting part of the region

so you need to add something to get the rest of the area

[warm patio]

okay so we add 0 to 5 as an integral of (y-5)^1/2?

[tiny gorge]

if you draw a horizontal line at say y=3

then what are the x boundaries

[warm patio]

neg/pos sqrt of 8

[tiny gorge]

what

if you have a horizontal line at say y=3

then you start at x=0 and end at x=3 right?

[warm patio]

wait so that doesnt change the bounds of x?

i thought i'd have to sub the value of y to get new x bounds

[tiny gorge]

what you have is the following

$$\int_0^5 (3 - 0),dy + \int_5^{14} (3 - \sqrt(y - 5)),dy$$

[soft zealot]

Bungo

[tiny gorge]

the first integral is the area of the rectangular region below y=5

the second is the rest

if your y value is below 5, the parabola has nothing to do with the x bounds

the parabola doesn't even exist there

[warm patio]

quick question so the y=3 wasn't given by the problem, so how did we infer to use 3-0 for 0 to 5?

[tiny gorge]

i'm just using y=3 as an illustration

[warm patio]

ahh okay so just a hypothetical

[tiny gorge]

if you're finding the area using y integrals, you have to break it up into two parts

the part below y = 5

(y = 3 is an example)

and the part above y=5

the reason you need two parts is that the parabola is the left boundary if y is above 5

but the y axis is the left boundary if y is below 5

whereas on the other hand, in your first integral (with respect to x), the lower boundary is always the x axis and the upper boundary is always the parabola, that's why you can find the area with only one integral

[warm patio]

yeah that one was much easier

[warm patio]

ah okay

so for integral of y=0 to y=5, the parabola doesn't really exist right

[tiny gorge]

yep, you always have two choices when finding the area of a 2d region (either integrate with respect to x or with respect to y), and often one of the choices is easier

[warm patio]

i still wanna learn how to do horizontal slices LOL

[tiny gorge]

yea, the part below y=5 is just a rectangle, with height 5 and width 3, so area = 15

[warm patio]

so just geometry, i thought it'd be more complicated 😂

[tiny gorge]

well you can of course do it with an integral too but it'll work out the same

this is obviously going to be 15

[warm patio]

thanks dude, you're a real one 👍 calc is not my strong suit so this helped a lot

[warm patio]

yepp

[tiny gorge]

nw, with a few more practice problems it will make more sense

[warm patio]

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[tawdry horizon]

How do I do b

[tawdry horizon]

This what I have so far

[worldly spruce]

This is correct

I mean you're pretty much done

[past axle]

@worldly spruce Can U Solve My One ......?

[worldly spruce]

You can throw the minus in the denominator out in front of the fraction and maybe factor out 5 in the numerator

But it's correct

[tawdry horizon]

my answer sheet tho says 5/2(sqrt-3 + sqrt5)

[past axle]

@worldly spruce Please Solve This One ....Or Help Me To How To Solve This !!

[raven pawn]

im in 7th grade and i rly need my grades up

[worldly spruce]

Yes, that's the same

If we first factor out 5

[raven pawn]

can anyone help??

[worldly spruce]

$\frac{5 \times \sqrt{3} - 5 \times \sqrt{5}}{-2} = \frac{5 \times (\sqrt{3} - \sqrt{5})}{-2}$

[soft zealot]

USS-Enterprise

[tawdry horizon]

ohhhn

[raven pawn]

does anyone know 7th grade math???

[worldly spruce]

You can move the minus to the top

And separate 5/2 as it's own fraction

Times sqrt(5) - sqrt(3)

[tawdry horizon]

can you show it

with the iage thimg

[worldly spruce]

$\frac{5 \times \sqrt{3} - 5 \times \sqrt{5}}{-2} = \frac{5 \times (\sqrt{3} - \sqrt{5})}{-2} = \frac{5 \times -(\sqrt{3} - \sqrt{5})}{2}$

[soft zealot]

USS-Enterprise

[tawdry horizon]

what u do vto move the -

• or /

[worldly spruce]

$\frac{5 \times \sqrt{3} - 5 \times \sqrt{5}}{-2} = \frac{5 \times (\sqrt{3} - \sqrt{5})}{-2} = \frac{5 \times -(\sqrt{3} - \sqrt{5})}{2} = \frac{5}{2} \times \frac{-(\sqrt{3} - \sqrt{5})}{2}$

[soft zealot]

USS-Enterprise

[worldly spruce]

I moved it to the top

Before the brackets

It's just multiplying the numerator and denominator with (-1)

[tawdry horizon]

so -1/-1

[worldly spruce]

If we have $\frac{5}{-2}$

[tawdry horizon]

then wouldnt the 5 be -1#

[soft zealot]

USS-Enterprise

[worldly spruce]

We can do

$\frac{5 \times (-1)}{-2 \times (-1)}$

[soft zealot]

USS-Enterprise

[worldly spruce]

And get

$\frac{-5}{2}$

[soft zealot]

USS-Enterprise

[worldly spruce]

Or, moving the minus up

[tawdry horizon]

ok continye

i get that part now

[worldly spruce]

$\frac{5 \times -(\sqrt{3} - \sqrt{5})}{2} = \frac{5}{2} \times \frac{-(\sqrt{3} - \sqrt{5})}{2}$

[soft zealot]

USS-Enterprise

[worldly spruce]

We've got this

Then just distribute the minus

$\frac{5}{2} \times \frac{-(\sqrt{3} - \sqrt{5})}{2} = \frac{5}{2} \times \frac{-\sqrt{3} + \sqrt{5}}{2}$

[soft zealot]

USS-Enterprise

[worldly spruce]

$\frac{5}{2} \times \frac{-\sqrt{3} + \sqrt{5}}{2} = \frac{5}{2} \times \frac{\sqrt{5} - \sqrt{3}}{2}$

[soft zealot]

USS-Enterprise

[worldly spruce]

Actually

There's no division by 2 in the second fractions

When we take out 5/2

It's just the roots, no fraction

So

[tawdry horizon]

ye

[worldly spruce]

$\frac{5}{2} \times \frac{-(\sqrt{3} - \sqrt{5})}{2} = \frac{5}{2} \times -(\sqrt{3} + \sqrt{5})$

[soft zealot]

USS-Enterprise

[worldly spruce]

$\frac{5}{2} \times -(\sqrt{3} + \sqrt{5}) = \frac{5}{2} \times (-\sqrt{3} + \sqrt{5}) = \frac{5}{2} \times (\sqrt{5} - \sqrt{3})$

[soft zealot]

USS-Enterprise

[worldly spruce]

There we go

[tawdry horizon]

wait a minute isnt -5 sqrt5 * -1 just +5 sqrt5

[worldly spruce]

What do you mean

[tawdry horizon]

you put -5sqrt3 -5sqrt5 over 2 when you where removing the -

adn you * by -1/-1

wouldnt it be -5sqrt3 +5 sqrt5 over 2

[worldly spruce]

Can you use brackets 😅

[tawdry horizon]

there is none

its a fraction

lemme send a pic

[worldly spruce]

,rccw

[worldly spruce]

Yes that's correct

Once you factor out 5 from the numerator you get what I have

[tawdry horizon]

but u put - instead no?

[worldly spruce]

No I did in multiple steps

I did

I will just write the numerators

$5 \times \sqrt{5} - 3 \times \sqrt{3} = -(5 \times \sqrt{5} - 3 \times \sqrt{3})$

[soft zealot]

USS-Enterprise

[tawdry horizon]

sorry could you reshow everytjhing since getting rid of the minus 😭

[worldly spruce]

Okay

So

[tawdry horizon]

im really lost lol

[worldly spruce]

$\frac{5 \times \sqrt{3} - 5 \times \sqrt{5}}{-2}$

[soft zealot]

USS-Enterprise

[worldly spruce]

We have this

You get this right

[tawdry horizon]

wait

[worldly spruce]

That's the third fraction here

[tawdry horizon]

ok

[worldly spruce]

So

We multiply the top and bottom with (-1)

You wrote it as a separate fraction

I just put a minus

Like this:

$\frac{5 \times \sqrt{3} - 5 \times \sqrt{5}}{-2} = $\frac{(-1) \times (5 \times \sqrt{3} - 5 \times \sqrt{5})}{(-1) \times (-2)} = $\frac{ -(\times (5 \times \sqrt{3} - 5 \times \sqrt{5})}{2}$

[soft zealot]

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

[worldly spruce]

Damn it

$\frac{(-1) \times (5 \times \sqrt{3} - 5 \times \sqrt{5})}{(-1) \times (-2)}$

[tawdry horizon]

i feel bad, its complicated this shi😭

[soft zealot]

USS-Enterprise

[worldly spruce]

$\frac{ -(\times (5 \times \sqrt{3} - 5 \times \sqrt{5})}{2}$

[soft zealot]

USS-Enterprise

[worldly spruce]

$\frac{5 \times \sqrt{3} - 5 \times \sqrt{5}}{-2} = \frac{(-1) \times (5 \times \sqrt{3} - 5 \times \sqrt{5})}{(-1) \times (-2)} = $\frac{ -(5 \times \sqrt{3} - 5 \times \sqrt{5})}{2}$

[soft zealot]

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