help-27

[ocean quest]

the tangent line is giving an approximation of the position at that point a

[coral zephyr]

Then what’s the “change in position”

[coral zephyr]

But what about the change in position

[ocean quest]

what about it?

[coral zephyr]

Like you’re going from a to where

Velocity = some change in position / time it took for that change to happen

Okay what’s the some change in position here

From where to where

[ocean quest]

I guess if it helps you conceptualize this, no, you don't need another point to get the derivative. The derivative of a function is only dependent on the point you are taking the derivative at and the function itself

[coral zephyr]

Okay

[ocean quest]

the notation is f'(x) because that's exactly what it depends on

x and f(x)

[coral zephyr]

So if it’s that point itself and you’re not going to another point then.. there’s no change in position/ time you got there ?

[ocean quest]

nope, that's why it's instantaneous

[coral zephyr]

Wait whattttt

So then

Wait what

[coral zephyr]

So then how are you supposed to calculate velocity

Because velocity is some change in distance / time

What you described just gives us the point itself

[willow path]

as i said earlier, that’s average velocity

that’s the difference

[coral zephyr]

So this is average velocity? **Without the limit **

[willow path]

yes, deltaX/deltaT is average vocity

*velocity

[coral zephyr]

And the limit makes it instantaneous velocity ?

[willow path]

Yes as it now turns it into a derivative

the limit is just a part of the definition of the derivative

[ocean quest]

If you were to measure position over time, you wouldn't need to measure velocity - you can directly calculate it by plotting your position over time as a function f(t) and then taking the derivative at every t to get a function f'(t), which describes the derivative at every point t

[coral zephyr]

So I shouldn’t think of derivative as being some change / time it took for that change to happen right

[summer parrot]

Hello

[coral zephyr]

Hi

[summer parrot]

What y’all up too

[coral zephyr]

I’m tryna understand derivatives

🫩

🫩🫩🫩

It’s boiling down to what happens here

[ocean quest]

Well, what we're describing doesn't seem to be working for you conceptually so I guess not

[willow path]

you should think of a derivative as how much the original function wants to change at that point i guess could be a way

not how i think of it but i don’t think it’s a mischaracterization of the idea either it’s probably a fine way to go about it

[coral zephyr]

Im trying to think of it’s a partial moving then you snap a picture at some point

Particle**

[ocean quest]

maybe you just need to think of it more abstractly

the derivative of a function f(x) at the point a is the function's slope at the point a

[coral zephyr]

The problem is I wanna actually understand what’s going on or else I won’t know what I’m doing

[willow path]

Also understand that a derivative provides you with a tangent line, the slope of that tangent line is what is important

[coral zephyr]

Ngl this makes me more confused

[valid silo]

I like to think of a derivative as just a linear approximation whose error is less than any linear term

[coral zephyr]

But essentially that’s what the difference quotient is

If you’re gonna subtract two points you’re gonna get a straight tangent line as the second point keeps getting closer and closer

[willow path]

that’s a. bad description

ignore that, not a good way to say that

[willow path]

you gotta just reckon with the existence of infinities and their usefulness but also their weirdness

[ocean quest]

the derivative is the limit of the difference quotient as x->a

[coral zephyr]

Yeah I think so

[willow path]

https://youtu.be/_cr46G2K5Fo?si=-LMK2J6cGSpxG8_g

fun video on the subject

to explore the thought a bit more

[ocean quest]

consider that the difference quotient Q is a function of x

[coral zephyr]

Difference quotient = average velocity right

[valid silo]

Do you have an intuition of what a limit is?

[coral zephyr]

Yes

I guess..

It approaches some value but never hits it? But apparently I was wrong or smt

[ocean quest]

a limit isn't necessarily separate from the value at the point, it's just most often used that way

like

[valid silo]

Limit just means that the value is a "good approximation"
Wher by that I mean whatever error bounds you give me, I can give you some wiggle room where the value stays within the error bounds. (Ignoring that exact point)

[ocean quest]

if f(x) = x, then $lim_{x \to 0} f(x) = 0$ and $f(0) = 0$

[woven radish]

Mr. BananaHead

[coral zephyr]

What’s the point of the limit if it’s just 0

[willow path]

useful for discontinuities

lim(1/x) x->0 issss ?

[coral zephyr]

Undefined?

[ocean quest]

technically DNE

[coral zephyr]

See that’s the thing what I mean is it’s supposed to approach 0 but not be 0

No?

[valid silo]

Have you done limit of a sequence?

[coral zephyr]

Probably

[willow path]

the function f(x)=1/x is undefined at the point 0

[coral zephyr]

I forgot

[willow path]

but the limit as x->0+ is infinity

it’s helpful for explaining the behavior of a graph

plot 1/x and see how it looks

[ocean quest]

from the positive direction

[willow path]

fair enough

[valid silo]

If you want to stay stuff like limit is infinity you are not working in the reals

[coral zephyr]

I dont get it; so when we talk about limits, are we just gonna plug that value? Whats the point of it’s existence then if we won’t approximate it

[willow path]

let’s us access more info

without the limit there isn’t much of a derivative

[coral zephyr]

You end up just plugging 0 anyway tho so that’s also confusing

[willow path]

helpful tool to let us explain how graphs behave when inputting values directly doesn’t work

[ocean quest]

I think this would be a better example. The function $\frac{(x+1)(x-1)}{(x-1)}$, the function is discontinuous at x=1, but the limit of the function still exists at x=1

[woven radish]

Mr. BananaHead

[coral zephyr]

Like you’ll end up factoring and just plugging 0

[willow path]

but now we can divide by deltaT

before plugging in

can’t divide by 0

it’s a way that helps us avoid things like that

[coral zephyr]

But the point is you plugged in nonetheless, like it’s the same as me just taking the difference quotient rewriting it then plugging 0

Why even bother writing a limit

[willow path]

because a limit is different from the value of a function

[ocean quest]

there are subtle differences

[coral zephyr]

In both cases we plug in 0

[ocean quest]

no

[willow path]

take the previous example, 1/0 is undefined

[coral zephyr]

How come

[willow path]

limit x->0+ is infinity

so they’re most definitely not the same

[ocean quest]

you are plugging in a point arbitrarily close to 0

[coral zephyr]

Yesssss that’s what I used to think !

But apparently not because when we took this we just plugged 0

[ocean quest]

for to or $\trianglet$?

[woven radish]

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

[ocean quest]

oops

[coral zephyr]

For t

[ocean quest]

oh, t is just the point here

it's the $\delta t$ that is never equal to 0

[woven radish]

Mr. BananaHead

[ocean quest]

I can't write the triangle in latex sorry

[coral zephyr]

Sorry I meant Δt

[willow path]

isn’t it Delta in latex or no ?

like capital D

[coral zephyr]

This is full of moden tollens it’s not even funny

[willow path]

idk much abt latex so might be wrong

[ocean quest]

by taking the limit, it removes dependence on that additional point

[coral zephyr]

Yea I’m confused on the velocity still ngl

What’s the difference between velocity , average velocity and instantaneous velocity ?

[ocean quest]

maybe it would be best to not think about it in terms of literal physics. you seem to be having a difficult time with that

[coral zephyr]

My class is pretty much applications

[ocean quest]

if you have some derivative function f'(x), given some point c, the instantaneous rate of change at that point is f'(c)

[coral zephyr]

That’s why I’m stuck on understanding the physics

[ocean quest]

the general derivative function f'(x) describes that instantaneous rate of change at any point x

the average rate of change is just the difference quotient between two points

[coral zephyr]

I see

[coral zephyr]

And the limit as h approaches 0 negates the average velocity and just measures it in that very moment ?

[ocean quest]

yes

the difference quotient Q at a particular point a is a function of x just like any other function. Then,

$f'(a) = \lim_{x \to a} Q(x)$

[coral zephyr]

Is there any way of doing that to begin with without the limit

[woven radish]

Mr. BananaHead

[ocean quest]

no

by definition, the derivative is a type of limit

[coral zephyr]

I’m tryna wrap my head around how the limit gets rid of that second point and just measures it at that instance

[ocean quest]

maybe the easiest way would be to see when it fails. A function $f(x)$ is only differentiable at a point $a$ when it is approximately linear at that point

[woven radish]

Mr. BananaHead

[ocean quest]

look at $|x|$ at $x=0$

[woven radish]

Mr. BananaHead

[coral zephyr]

Wdym by when it fails

[mint cairn]

Hmm...

[coral zephyr]

Okay

[ocean quest]

have you looked at a graph of |x|?

[coral zephyr]

Yes

V shape

[ocean quest]

yup

so at 0, the graph is never approximately linear, no matter how close you are to 0

so, the derivative does not exist at that point

then, you can see that when you do have a function that is approximately linear, the derivative tells you the slope of that function as you zoom in o the point

[coral zephyr]

What do we mean by approximately linear

[ocean quest]

graph y=x^2 in desmos and start zooming in on the function anywhere

[willow path]

Yk a ball? how you can zoom in a bunchhhhh and it looks like a straight line?

That’s approximately linear

how the earth looks flat from our perspective because we are that zoomed in

[coral zephyr]

Ohhh

[willow path]

if you zoom in the graph of |x| it is ALWAYS a sharp corner at the origin

never smooth so it’s not differentiable

[coral zephyr]

Ohh

Okay and how is this related to the velocity and limit stuff

[ocean quest]

because when a function is approximately linear, the derivative at that point tells you what the slope is at that point

if it isn't approximately linear, there is no slope

[coral zephyr]

You mean slope of the tangent line ?

[ocean quest]

yes, but the slope of the function at that point is the same as the slope of the tangent line

the tangent line is just a good way to visualize it

[coral zephyr]

I see

[ocean quest]

but the slope there is intrinsic to the function you are looking at - the tangent line is just a result

that's also why the derivative at a point is by definition a limit

because the limit of the function around that point takes the form of a line, even though the function isn't linear

[coral zephyr]

@ocean quest can I tell you what I’m still stuck on

[ocean quest]

sure

[coral zephyr]

Is velocity some distance / some time?

[ocean quest]

velocity tells you how much future distance you will cover for some future time t

[coral zephyr]

Now we’re talking about future tense ? Aw bruh

The point of velocity is to tell us how fast something is going right

[north lodge]

displacement/time not distance

[coral zephyr]

But they’re saying there’s no displacement at that instant position

[north lodge]

i mean if you are talking about instantaneous velocity then the displacement is very small so for such small time interval we can say distance= displacement

[coral zephyr]

Could you please describe that small displacement

[north lodge]

you know different between average and instantaneous velocity?

avg velocity v = ∆s/∆t
instantaneous velocity , v = lim ∆t-0 ∆s/∆t

[ocean quest]

we've been going in circles on this for hours

[north lodge]

what was the original doubt

[ocean quest]

at this point, you either need an in-person tutor, you you should consider not pursuing anything further in mathematics

[north lodge]

damn

[coral zephyr]

?

[coral zephyr]

Average velocity is the difference quotient

Instantaneous is its limit as h approaches 0

[north lodge]

yeah so now what you don't understand?

[ocean quest]

they basically don't understand what a limit is

[coral zephyr]

What does it mean when h approaches 0

Does it ever reach 0 or no

[north lodge]

Oh

[north lodge]

No i mean obviously denominator cannot be 0

[coral zephyr]

So why do people plug in 0? All they they is rewrite the same fraction, then plug in 0

[north lodge]

limit basically means discussing a function in neighbourhood of a value like lim x - 2 means discussion f(x) at numbers very close to 2

[north lodge]

∆t tends to 0 means you are taking the time interval to be infinitely small so small that it becomes a instant
like t = 0 to t = 2 sec you can find average velocity
but what if i want to find velocity at 2 second exactly ?

so in that case i'll take the ∆t to be so small that it can be treated as t = 2

I would say watch an introductory video on limits then try to understand again you will get it

[coral zephyr]

If it just keeps getting infinitely small, between point a and some other points, isn’t that just technically still the average velocity

[ocean quest]

not when infinity is involved

[north lodge]

i mean if you are considering ∆t to be infinitely small so obviously in that small interval particle will also have negligible displacement so it will not remain " average "

[coral zephyr]

So basically the whole point is that the two points get so ridiculously small that they might as well he considered the same point ?

[north lodge]

yes

[coral zephyr]

But when you plug 0 into the difference quotient, how do you know that that is a limit versus you just plugging the 0 point ?

[north lodge]

difference quotient is ( f(x+h) -f(x))/h right ?

how you gonna put h= 0 here ? Its gonna be undefined

i didn't get what you wanted to ask

[coral zephyr]

But when you take the limit , all you do is rewrite it then plug 0

You’re rewriting the fraction itself

So you’re basically keeping it the same then plugging 0

Or is that not what happens?

Like if you consider any function

2t for example and you plug it in

[north lodge]

No ? like think of instantaneous in terms of english not physics instantaneous means " at instant"

So we have to find velocity at 2 second
If I take t1= 2.00 sec and t2= 2.0000000000000001 second don't you think now this time interval is infinitely small and can be treated as an instant ?

that's what limits is used for to discuss the fxn very close to a number

[coral zephyr]

Here’s what I mean take f(t) = 2t

The difference quotient is

2(t+h) - 2t / h

Now what’s the difference between me just plugging in 0 now and treat it as if I’m evaluating this at 0

And between me canceling out an h by just rewriting this and now all of a sudden it’s a limit

[ocean quest]

You need to understand that something being equal to another thing and something's limit being equal to another thing are completely different

[willow path]

if you evaluate that at zero it’s undefined

and that’s no much of a useful outcome

[woven vale]

careful on the wording here: $$\lim_{h \to 0} \frac{2(t+h)-2t}{h}$$ and $$\lim_{h \to 0} \left(2 + \frac{2t-2t}{h}\right)$$ are both limits. so i would say that, when the limit is rewritten, then you can substitute $h=0$ to solve. but rewriting doesn't magically make anything a limit

[woven radish]

حسیب ♥

[coral zephyr]

I don’t think you guys are understanding my point

[woven vale]

okay, then what would your point be?

[woven vale]

perhaps this?

[coral zephyr]

Maybe

But it’s more of the part where you factor out an h and cancel it out with the one in the denominator

Before factoring can you plug zero for h? No right

So how come it works when you factor out an h which is just rewriting the same fraction

[ocean quest]

because $h \neq 0$ and $\lim_{h \to 0} h = 0$

[woven radish]

Mr. BananaHead

[willow path]

because limits are inherently a different operation

f(a) does not always equal lim x->a {f(x)}

[woven vale]

it's a property that we take for granted in numbers, tbh. we can cancel $\frac{3a}{a} = 3$ or $\frac{x^2}{x}$, and when we do that, there is an inherent condition that the denominator is not zero

[woven radish]

حسیب ♥

[woven vale]

for a limit, since we are approaching $h$, then $h$ never actually equals zero, so we can get away with said cancellation

[woven radish]

حسیب ♥

[woven vale]

geometrically, the functions $y = \frac{3x}{x}$ and $y=3$ might look the same, but $\frac{3x}{x}$ has a hole at $x=0:$ it is undefined at $x=0,$ but the values around it point to a "phantom value" that should be there

[woven radish]

حسیب ♥

[woven vale]

in fact, this is what we're doing when we take a limit: finding the value that we expect to appear, if we follow the graph of a smooth function

(if the function isn't a "nice, smooth" line, then this idea fails, hence the limit doesn't exist at jumps or cusps)

[woven vale]

here is f(x) = (3x)/x, so that you can see the hole that i'm talking about

[devout snow]

@coral zephyr Has your question been resolved?

[devout snow]

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[granite arch]

can someone please help explain why σ can be written as a product of disjoint cycles? thanks

[void fox]

if you want to be rigorous you can define g ~ h <=> g = σ^k h for some integer k and prove that ~ is an equivalence

[lost laurel]

Follows from the fact that conjugacy is an equivlance class

[granite arch]

Wait what

im confused

why does this part connect to that

[stone stump]

you can write it like that because you can just compute the cycles

start with some random element a_1 and compute the cycle of a_1

then take some random element which is not in that cycle and compute the cycle of that one

and so on

[granite arch]

can i have an example

😭 thanks

[stone stump]

lets say that sigma is the permutation 1743256 which means sigma(1)=1, sigma(2)=7 and so on

the cycle of 1 is just (1)

the cycle of 2 is (2765) because sigma(2)=7, sigma(7)=6, sigma(6)=5 and sigma(5)=2

[granite arch]

does this mean
(1234567)
(1743256)

[stone stump]

and then the cycle of 3 is (34) because sigma(3)=4 and sigma(4)=3

[stone stump]

yes

[granite arch]

mhm

and then

[stone stump]

and now we can simply write sigma=(1)(2765)(34)

[granite arch]

do we stop at cycle of 3 and know that its okay because it covers every element in the domain?

i.e. we have every number from 1-7 generated by 1,2, and 3 cycles

[stone stump]

yes we covered every element so we can stop

[granite arch]

ohhhh

oh my thats smart

[stone stump]

try computing the cycles for 382174569

and some other random permutations you come up with

[granite arch]

(13)(286413)(328641)?

wait no

9 is missing

(13)(286413)(328641)(413286)(57)(9)

like this? it seems overly complicated

[stone stump]

no

sigma(1)=3, so the first cycle starts with (13

sigma(3)=2 so it continues with (132

sigma(2)=8 so (1328

sigma(8)=6 so (13286

sigma(6)=4 so (132864

and finally sigma(4)=1 so now we can close the cycle (132864)

and now take an element which is not in that cycle

so 5,7 or 9

and then start new

do not take an element which is already in a cycle

you will not get disjoint cycles like that

[granite arch]

OHH

so i keep applying sigma until i get something in the bracket then i stop before that happens?

(132864)(57)(9)?

[stone stump]

yes

[stone stump]

<@&268886789983436800>

[split stream]

@round badge don't hijack other people's help channels

[devout snow]

@granite arch Has your question been resolved?

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[dense rampart]

i first made it 0 <= a <= b <= c <= d <= 6
then lecture notes say that such a question is simply C of 6 boxes and 4 stars that'll represent each possible value/box

seems that this kind of stars and bars came from a different place/philosophy than say, find solutions of x+y+z = 10, where the boxes will be 3 variables and stars will be 10 single units where their amount would represent the value of a variable/box
id know that the second example is about partitions, but i cant intuitively grasp the basis of the first's stars and bars

they say the first example is also how we count number of possible dice combinations, but then when we deal with probability n^r would be used instead of stars and bars???

[devout snow]

@dense rampart Has your question been resolved?

[modern lance]

then lecture notes say that such a question is simply C of 6 boxes and 4 stars that'll represent each possible value/box
The lecture says exactly this? This's kinda a weird way to think about the problem

[dense rampart]

not exactly but it shows that a ball at p2 represents a=2

wait no what

This is the exact notes and the blanks are to be filled in by ourselves when watching lecture recording

I saw top left and understood how to do the q from there

[modern lance]

yeah so, basically it says choose 4 numbers from 1 to 9, then there is exactly 1 way to arrange them in ascending order

And that would be a,b,c and d respectively

[dense rampart]

I never thought it that way

[modern lance]

ig you should, since that's pretty easy to understand

[dense rampart]

my first thought was "if a is 1 theres a lot of choices and a can at max be 6 where theres very few(1) options

[dense rampart]

now that i saw it yeah it is but

[frigid fractal]

Hi

[dense rampart]

maybe ive just came off of the x+y+z = 10 thing

[frigid fractal]

Someone can help me?

[dense rampart]

this channel is occupied 😭

[frigid fractal]

😭

[frigid fractal]

Its not about a mathmtical problem

Ahhh

My bad

[modern lance]

well maybe you can think if it like this

ig

[formal panther]

yup was exactly typing this out lol

you can do stars and bars here now

[modern lance]

Honestly tho I have never done such problem this way

[formal panther]

x1​=a−0
x2​=b−a
x3​=c−b
x4​=d−c
x5​=10−d
if you add it all up it gives x1 + x2 + x3 + x4 + x5 = 10

[formal panther]

the first way is much more straightforward and should be the one used here i feel

[dense rampart]

when we are able to transform a problem from < into <=, should i not always do it?

when seeing <= i can usually star and bars

but < it means quite a lot of things

[modern lance]

it depends tho

but ig it can somehow be deduce to -1 both side of an equation or smth

[formal panther]

0 <= a <= b <= c <= d <= 6 how did you get the upper bound 6

shouldn't it be 5

i think you didn't change < 10 to <= 9

[dense rampart]

i think i meant 0 < a <= ... then

its hard to keep track of all these methods

combinatorics demands too much of my common senses

[modern lance]

yeah, that's why I didn't draw an arrow in the 0th gap in the drawing above

When I do combinatorics I would just draw example like what I did above and reasoning

Use algebra is boring

[dense rampart]

i think i lack the reasoning needed to draw ur examples

you did stars and bars but you started from a different perspective of the problem

[modern lance]

yeah...

the first one is kinda a bijection thing

[formal panther]

i was thinking of one more thing but its basically the logic for 9 C 4 lol

[dense rampart]

new question and i made this method, is this the straightforward way?

if my ans is correct, then this is also the no of ways to choose 4 numbers from 1-25, but i dont see how 😭 especially when 4 and 25 correspond to the og question so nicely

[modern lance]

the answer is correct yeah

[formal panther]

isnt the answer 24 C 4

i learnt to do these types of questions by adding a "slack variable"

basically w + x +y + z + a = 25

where a is the slack variable

and then you do k-1 C n-1 so that's 25-1 c 5-1

24 c 4

the a is to ensure x + y +z +w stays below 25

but adding a slack variable allows us to remove the inequality

so we can apply stars and bars

[shy osprey]

yeah same here

[modern lance]

oh wait that's not =< yeah lol

[devout snow]

@dense rampart Has your question been resolved?

[dense rampart]

oh then if i turn <21 into <=20 then we'd have the same conclusion

but interesting

[devout snow]

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[trail eagle]

!status

[devout snow]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[tardy carbon]

nah i understad all of part a and b, juts dont get part c

[subtle thicket]

what'd you get the general expression as?

@tardy carbon

you might have not got the most general expression resulting in some confusion

cause you can guess x=3cost

but there's another term

unless you already have that term, then it's straightforward

[tardy carbon]

so ill attartch my workings 1 sec

so yh this is where i got up to

[subtle thicket]

oh alright you've done everything correctly then

[tardy carbon]

yeah its juts part c im not really getting

[subtle thicket]

do you want a hint or should I give a more straightforward answer

first you can understand how there will be a difference in levels right?

cause the term oscillates

[tardy carbon]

uh would u mind giving just like the answer and how u got it - but like explaining your thought process pleas

[subtle thicket]

okay, so essentially since we're talking 7 days later

[tardy carbon]

yes

[subtle thicket]

the e^-t term just becomes really small

cause e^-7 is negligible

so you can ignore that term

[tardy carbon]

yes okay

[subtle thicket]

essentially you're just left with x=3cost

now in a day the max level can be 3

and the min be -3

so difference is 6

[tardy carbon]

OMG

i see

[subtle thicket]

not necessarily in a day btw

but like the difference from now on will mostly be around 6

[tardy carbon]

but at the same time i thought liek when we look at tending to infinity then we can make the exponential negative as itll go to 0 but 7 days seems quite short no?

[subtle thicket]

yah but e^-7 is like, 0.0000smth

also they specified approximately

[tardy carbon]

ohhhh yeah my fault then for not checking that

i understand it now

thanks

[subtle thicket]

np!

[devout snow]

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[devout snow]

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[neon wagon]

how would you solve e^2x = -c * e^x

[neon wagon]

this was a question in my test kinda

i just did ln and just brought the 2x and x down

thats wrong?

[trail eagle]

You could divide both sides by e^x, you could log both sides, you could bring everything to one side and factor

[neon wagon]

wtf

right

why didnt i think of that

ty tho

[trail eagle]

log both sides works, but you have to be careful on the RHS because x isn't the exponent of the whole thing

So you have to split it first

Using log rules with product

Then you can bring x down

[neon wagon]

guess ill study stuff like that for wednesday

[devout snow]

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[light storm]

hey can you use power rule to differentiate e^x?

[lunar harbor]

e^x is not a polynomial

unless you want to get rlly creative and differentiate the series expansion termwise, then reindex

[tender lodge]

the derivative of e^x is just e^x

u dont need the power rule or anything

[light storm]

how is it not a polynomial

[winter torrent]

the answer is no -- $e^x$ does not fit the pattern of $x^n$

[woven radish]

schrödinger's kitten

[tender lodge]

the general form for any exponential function derivative b^x is ln(b)*b^x

[tender lodge]

exponential function not polynomial

[light storm]

oh i see

ok thanks

.close

[devout snow]

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[fair canopy]

Need solution by normal algebraic techniques

[lavish nimbus]

,rotate

[wide copper]

Consider LHS

You have lim of e^1/x - 1/ e^1/x+1

from 0(negative)

In this case as e^1/x approaches 0, it becomes e^-inf = 0

So, LHS limit is equal to 0-1/0+1

= -1

Consider RHS

[fair canopy]

Do u mean to solve it by left hand limit as limit x approaches 0 from left 1over x equals negative inf

[wide copper]

yep you can graph it to see

Dividing by e^1/x, we have 1-1/e^1/x / 1+1/e^1/x

[fair canopy]

Can we do it sm other by letting 1over x as smt

?

[wide copper]

Now, as it approaches 1/x in positive side, it approahces infinity so its 1

[fair canopy]

Got marked wrong I my test I tried to do it

Weird the checker is

[wide copper]

You cant substitute x for 1/x

Behavior is clearly different because it approaches 0

1/x becomes infinity while x becomes 0

You just take LHS, RHS and the answers are -1, +1

[fair canopy]

Ah alr ty

[winter patrol]

Can you show what you did?

[winter patrol]

You can

[devout snow]

@fair canopy Has your question been resolved?

[devout snow]

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[chrome salmon]

heyo

would appreciate some help here

[sand pumice]

what have you tried?

[rain summit]

!status

[devout snow]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[chrome salmon]

rn im tryna find the integral

im using the u-sub technique

[sand pumice]

okay nice

what did you substitute?

[chrome salmon]

x^2+9

[sand pumice]

👍

what did you get then?

[chrome salmon]

i got

-1/(4(x^2+9)^2

not sure how to use bot sorry lol

now i believe we have to get the limit of that

[sand pumice]

dw

[sand pumice]

yes

[chrome salmon]

okay so

now

we have to find the limit of that

[bold hemlock]

Are you sure about that

[sand pumice]

yes, so what happens as x-->\infty

[chrome salmon]

uh i think so

[sand pumice]

(checked with wolfram alpha it's right)

[bold hemlock]

Ohh mb

I thought you haven't integrated it yet

Sorry

[chrome salmon]

ur good

[chrome salmon]

it goes to 0 i believe

[sand pumice]

that'd be right

[chrome salmon]

okay

[sand pumice]

ok so you have 0-[-1/(4(x²+9)²)]

and you have to plug in 1

[chrome salmon]

so since its infinity - 1

yes

not literally 1 but plug in 1

which is

-1/324

[sand pumice]

🤔

u sure

[chrome salmon]

wait

ppositive

[sand pumice]

yeah but are you sure about the value of 324

[chrome salmon]

oh

i did 4 times 9^2

no im stupid

its 400

[sand pumice]

👍

so your answer is?

[chrome salmon]

1/400

[sand pumice]

👍

[chrome salmon]

Tyyyy

[sand pumice]

np

[chrome salmon]

ima keep trhis open incase i have another problem

if thats okay

speaking of

i ran into another problem

[sand pumice]

i think you're supposed to open another channel, but i don't think it's a big deal

[sand pumice]

what substitution can you try?

[chrome salmon]

i was thinking u-sub

but im not sure how that would help

cuz i rewrote this as

[sand pumice]

if i had smth like x^-1, what would its derivative be?

[chrome salmon]

lnx?

wait derivcative

mb

[bold hemlock]

Put t = x^-4

[chrome salmon]

i meant

-1/x^2

[sand pumice]

i was getting there 😔

[bold hemlock]

I won't interrupt from now carry on .sorry

[sand pumice]

no lol dw

[chrome salmon]

appreciate you brother 😭 🙏

okay so

ohhhhhh

wait i see wym

cuz its negative so

yes yes

[sand pumice]

okay so what does the integral become?

[chrome salmon]

lemme work it out

wait

is the integral for 1/x^4

it wouldnt be 1/x^5

[sand pumice]

integral or derivative 🤔

[chrome salmon]

its

derivative soirry

-4/x^5

[sand pumice]

👍

[chrome salmon]

wait so would it be

-4e^(x^-4)

[sand pumice]

yes and you subbed x^-4

and also divide by -4

[chrome salmon]

riggh

right

so thats the integral right?

[sand pumice]

ye

[chrome salmon]

okay

now we do the sum

uuhhh

okay so

the limit of

-4e^(1/infinity^4)

i believe its

just e

i think

wait

it might be infinity

wait no

it might be 1

wait wait

it might be infinity

[sand pumice]

divide by -4

[chrome salmon]

?

[sand pumice]

no it is 1 👍

[chrome salmon]

o

[sand pumice]

remember you have e^u/(-4)

[chrome salmon]

ah right

mb

okay so its 1

1-

plug in 1

so

1+e/4

[sand pumice]

so the first term is -1/4

[sand pumice]

❌

[chrome salmon]

wait ur right

1/4+e/4

[sand pumice]

-1/4

😔

[chrome salmon]

i keep forgetting little details

i need to slow down 😭 🙏

[sand pumice]

have you been writing it down?

[chrome salmon]

trying to

but i do a lot of mental matrh

so it gets mixed uo

up

[sand pumice]

👍

anyway so what's your final answer?

[chrome salmon]

e/4-1/4

[bold hemlock]

Correct

[sand pumice]

✅

[chrome salmon]

Tyyyyyy

appreciate u king

ima close this now

.close

[devout snow]

Channel closed

Closed by @chrome salmon

Use .reopen if this was a mistake.

[chrome salmon]

.reopen

[devout snow]

✅ Original question: #help-27 message

[chrome salmon]

well not the original question

i have a diff one

😭

i got this one

sorry for asking for help so much

i used the

uv-inegral duv

method

i got

ping me if someone comes 😭 🙏

thats the integral

im having trouble finding the limit of infinity though

[sand pumice]

there should be a - in front of the second term as well

[chrome salmon]

didnt it cancel out

[sand pumice]

i dont think so

[sand pumice]

yes the second integral becomes positive, but the integral of e^-5x is e^-5x/(-5)

[chrome salmon]

ah shoot ur right

https://tenor.com/view/ishowspeed-my-mom-is-kinda-homeless-please-speed-i-need-this-tired-blow-gif-15401023925154467112

[sand pumice]

so what's your final answer?

[chrome salmon]

thats the integral atleast

[sand pumice]

plug in the bounds and take the limit

[chrome salmon]

yes

but im ngl

idk how to do the infinity one

bc like

the limit of e^-5x as it goes to inifnity is infinity no?

i meant 1

but then its multiplied by x

which is infinity as x does to infinity

[sand pumice]

well remember that exponentials grow much faster than polynomials

[chrome salmon]

right

[sand pumice]

if you'd like, you could apply L'Hopital to x/e^5x

[chrome salmon]

right

so

derivative of both?

[sand pumice]

yes

[chrome salmon]

okay so thats

1/(-e^5x/5)

which is

-5/e^5x

and since it goes to 1

its just -5 for that term?

[sand pumice]

no

the denominator is e^5x

[chrome salmon]

yes

[sand pumice]

treat e^5x as the function you're differentiating, not 1/e^5x

[chrome salmon]

wait how is it positive

[sand pumice]

e^-5x=1/e^5x

[chrome salmon]

oh shoot

right

got it

okay so

e^5x goes to

uhhhh

[sand pumice]

but you still have 1/(5e^5x)

[chrome salmon]

yes

[sand pumice]

when you differentiate top and bottom

as x goes to infinity, what does it go to?

[chrome salmon]

1/5

[sand pumice]

❌

[chrome salmon]

nope

infinity

[sand pumice]

yes so what does 1/(5e^5x) go to then?

[chrome salmon]

-infinity?

wait no

0

[sand pumice]

yes indeed

[chrome salmon]

ohhhh

i see

[sand pumice]

so how about 1/(25e^(5x))

[chrome salmon]

so itd be 0-0 for the plugging infinity in

[chrome salmon]

also 0?

[sand pumice]

yes

now plug in 1

[chrome salmon]

ohh

so this?

[sand pumice]

yes go ahead and simplify it

[chrome salmon]

okay

[sand pumice]

and remember it's 0- (whatever you get here)

[chrome salmon]

yess right

i got

6/25e

[sand pumice]

did you forget an exponent there

[chrome salmon]

yes

e^5

[sand pumice]

okay cool

that's your answer

[chrome salmon]

tysm king ur a life saviour

ima close ths

.close

[devout snow]

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[robust bobcat]

For 11 I want to use the first fundamental theorem of calculus in conjunction with proof by contradiction. I say that assume we have $\sum_{k=0}^n c_kx^k = \int_c^x \frac{1}{t}dt$

[woven radish]

BigBen

[robust bobcat]

Then say $\sum_{k=0}^n kc_kx^{k-1} = \frac{1}{x}$

[woven radish]

BigBen

[robust bobcat]

And then you can see that we want one only one term and for that term to have x in the denominator so we set n=0. Then we have the derivative equal to 0 and we have a contradiction.

But something about the last part doesn't feel right

My proof feels like I'm saying look you can't right our sum as 1/x but the way I try to justify it doesn't seem rigorous.

[void fox]

you just need to show that 1/x is not a polynomial in x

[robust bobcat]

That's what I was trying to do with the above though

[jade bobcat]

one way is to consider degrees

[void fox]

that's the part you omitted

anyway, suppose 1/x = ∑_i b_i x^i is a polynomial, then see what happens at 0...

[robust bobcat]

You mean set the upper bound to 0. If so why? It isn't going to equal 1

[void fox]

that's not what I was saying

[robust bobcat]

You mean just set x=0? After you multiply both sides by x

[void fox]

yes

but also like

[robust bobcat]

But why?

[void fox]

you want to show 1/x is not equal to f'(x) for any polynomial f right

but the derivative of a polynomial is also a polynomial

so it suffices to show that 1/x is not a polynomial

[robust bobcat]

Ok I see that. But I don't see how we actually show it isnt a polynomial. We just showed that at x=0 the polynomial cannot equal 1

[void fox]

Okay, suppose 1/x is a polynomial, say 1/x = ∑_i b_i x^i

i runs from 0 to n

[robust bobcat]

Ok

[void fox]

take a limit as x → 0

[robust bobcat]

Ok so 1/x is undefined and the polynomial will be 0

[void fox]

you can't really do that since 1/x is not defined at 0

[jade bobcat]

to be fair i think the question has sort of worded the situation poorly

but what plante is showing is that on the domain of real numbers \ 0 there is no polynomial function that agrees with x ↦ 1/x

like, x²/x is also undefined at x = 0, but elsewhere it perfectly agrees with x which is a polynomial function

[robust bobcat]

Isn't that why we say undefined?

[void fox]

I don't know what you mean by "1/x is undefined and the polynomial will be 0"

1/x is undefined at 0 but we are investigating the limit as x → 0 which does not require definition at x = 0

I also don't understand what you mean by "the polynomial will be 0"

[robust bobcat]

Ok I meant limit doesn't not exist

[void fox]

Mhm, lim x → 0 of 1/x blows up

what about lim x → 0 of a polynomial?

[fair canopy]

I was to solve x smaller than 0 question by substituting 1over x as minus y

[jade bobcat]

please open a different help channel and forward your reply and mention there

[fair canopy]

And then shifting the e^-y down n 1over infinity as zero and minus 1 as answer

[robust bobcat]

It will be some constant

[void fox]

yes so they can't possibly be the same function

[robust bobcat]

Ok so your saying if it was the same function the limits give the same results

How could I use the ideas from the chapter though?

[void fox]

of course the same function will have the same limits

[void fox]

I think it's a silly problem, do you a priori know the fact that the integral of 1/x is given by log |x| + a locally constant function?

[robust bobcat]

Yes

[void fox]

you can also prove that it cannot be a polynomial...

[jade bobcat]

that log |x| cannot be a polynomial

[void fox]

[robust bobcat]

Well he hasn't formally discussed logarithms yet. That is in the next chapter

[void fox]

how do you know int 1/x dx = log|x| without knowing what a logarithm is

I suppose you can define log as that but like

then you have to prove int 1/x dx cannot be a polynomial which is the same thing

[robust bobcat]

Well I learned it in school but the way the author is presenting stuff he hasn't touched upon it

Also in what ways was the proof I was suggesting lacking?

[void fox]

You argued that if f'(x) = 1/x for some polynomial f then 1/x would have to be a polynomial also

but you need to prove your assertion that 1/x is not a polynomial

I think this is a silly problem, I can't really see what concept they are testing for by posing it

but sometimes you get silly problems

[robust bobcat]

But we didn't assert that we just assumed that a polynomial is equal to integral 1/t

[jade bobcat]

the real problem is that you are handwaving as if you know there is a “canonical form” for each rational function, and that if two rational functions have different canonical forms then they are different functions

[robust bobcat]

Is the limit the only way to resolve this?

[bold hemlock]

So what are we discussing here

[jade bobcat]

i mentioned degrees earlier, that idea also works

you would also need to show some nonsense about being able to equate rational functions with actual functions

[void fox]

you can also say that one is unbounded on (0, 1] and the other is bounded

there are like a million ways to see it as gmz said

[jade bobcat]

i mean it is tricky and sort of an annoying business

but a lot of the times it's surprisingly hard to show two things can't be the same, because there might be lots of different representations of your things you need to “search through”

e.g. when are two spaces the same when i can deform them in lots of different ways? when are two higher-dimensional shapes the same? when are two number systems the same?

[robust bobcat]

Ok but for here we're saying that if the 1/x is of a polynomial type then it needs to share properties if general functions. Such as the limit of a polynomial is either 0 or 1

[jade bobcat]

be precise! the limit of a polynomial for any point on its domain is a finite constant, could be 1.34 or -0.91 etc

but yeah that's our argument here

[devout snow]

@robust bobcat Has your question been resolved?

[devout snow]

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[neon wagon]

can someone expalin me how this is the same as that

[neon wagon]

is this only for matrices or is it in general

[trail eagle]

That's called associativity

[neon wagon]

hm okay but wait

[trail eagle]

Lots of operations are associative, like addition, multiplication and all that

Matrix multiplication and addition are associative operations

[neon wagon]

yeah i just realized yeah

no wait matrix multiplication isnt associative tho?

like A * B isnt the same as B * A right?

[trail eagle]

That's commutativity

Matrix multiplication is associative but not commutative

[neon wagon]

ho

yes

[devout snow]

@neon wagon Has your question been resolved?

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[neon wagon]

yo

[neon wagon]

how do you solve this

[quartz veldt]

have you learnt how to invert a matrix yet?

[neon wagon]

yes

The matrix multiplicated with the inverse matrix is the unit matrix

[quartz veldt]

exactly

[neon wagon]

so

do i just solve like

for a

and then put that in III

to get c

and so on

[quartz veldt]

that could work

but its a bit long

[neon wagon]

how else could i solve

[quartz veldt]

you're on to something here

you have 2 matrices being multiplied

and you're getting the identity as a result

[neon wagon]

hm

[quartz veldt]

following from the definition you just gave, what does this tell us about the matrices you're multiplying?

[neon wagon]

that the m of the first matrix is the same as the m of the one were multipliying it with

[quartz veldt]

the m?

not sure what that means

[quartz veldt]

well if 2 matrices are being multiplied and you're getting the identity matrix, then they have to be inverses of each other

[neon wagon]

yea

yea

[quartz veldt]

so the a b c d matrix is just the inverse of the matrix you're multiplying it with

[neon wagon]

okay

but how do i get that

[quartz veldt]

oh so have you not learnt how to invert a matrix then?

there's a standard way of inverting matrices that you can do quite quickly

if you have yet to study this, I suggest you do that problem by solving for a, b, c, d simultaneously

[neon wagon]

we havent done that

i thi k ill just ask my freind

oh god

i got an exam about matrices and a bit of calculus

[quartz veldt]

Yeah so your approach here is good then

You can solve it simultaneously, and also apply a few tricks to get their faster

Like equation 1 can be written as 2(a + c) + c = 1, and from equation 3, a + c = 0, so you end up with c = 1

You can do something similar for the other 2 equations, and the rest shouldn't be too hard from there

[devout snow]

@neon wagon Has your question been resolved?

[devout snow]

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[dim ridge]

Idk if we can considerate this like maths but do someone understand this kind of thin

[devout snow]

@dim ridge Has your question been resolved?

[dim ridge]

Any <@&286206848099549185>

[dim ridge]

.close

[devout snow]

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[devout snow]

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[hard sable]

i tried resolving up and sideways for the top stone

r(up):
s = 40
u = 20sin(a)
v = ?
a = 9.8
t = ?

r(sideways):
s = ?
u = 20cos(a)
v = ?
a = 0
t = ?

then sideways for the window stone

r(sideways):
s = ?
u = 12
v = ?
a = 0
t = ?

[green crypt]

You wanna find the time when they collide yeah?

[hard sable]

yeah

then i did

v = 12
v = 20cos(a)

a = 53.13 deg

which looks wrong

then

40 = 20sin(53.13) * t + 4.9t^2

and this gives t = 1.658s, not 2.5

[green crypt]

You equated both velocities?

[hard sable]

v = 12
v = 20cos(a)

like this?

[green crypt]

The moment they collide doesn't imply they'll have same velocities

[hard sable]

hmmmm

[green crypt]

Its the vertical distance that's gonna be equal

[hard sable]

so i should get 2 equations in S?

[green crypt]

Yes

Well i might be signing off so ill give you the hint to find the angle alpha too

For that see that equating the horizontal distances also give you a hint

[restive river]

,w arccos(3/5) in degrees

[woven radish]

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Requested by atlanta0875

[restive river]

your angle is correct

[hard sable]

why is that

if the v's aren't equal

[restive river]

they arent, however you arrive at same result after equating horizontal distances travelled

[hard sable]

so its a coincidence that i get the correct alpha after equating the v's?

[restive river]

yh

[hard sable]

okay well

how do i get an S = equation for the window stone?

nvm

[restive river]

for vertical motion?

[hard sable]

yeah i guess

[restive river]

substitute the values

stone is thrown horizontally so what would be your initial velocity

in vertical direction

[hard sable]

s = ?
u = 0
v = ?
a = -9.8
t = ?

S = -4.9t^2

[green crypt]

Im back but i see you got thus atlanta

Cook

[restive river]

oh, you can take over if you want

i was just here incase you dont come back

[hard sable]

this is the mark scheme but i don't really understand this

[green crypt]

So they equated the horizontal components and got angle yeah?

You got that?

[hard sable]

yeah

but then when i get S = equations for both

I get
s = -4.9t^2
40 = 20sin(53.13) * t + 4.9t^2

[green crypt]

Now the stones collide when they have similar vertical distance from the ground too

[hard sable]

yes

[green crypt]

Stone a is thrown from the top of the tower

So the equation will be?

[hard sable]

vertically?

[green crypt]

Yes

[hard sable]

S = y
u = 20sin(a)
v = ?
a = 9.8
t = ?

y = 20sin(a) * t + 4.9t^2

[green crypt]

Wheres the 85m?

The top of the 85 meter tower

[hard sable]

yeah but the stones dont collide at the bottom of the tower?

[green crypt]

We dont know that

[hard sable]

yeah so it could be S <= 85

[green crypt]

We aren't making these equations with collision in mind

These are independent equations

[green crypt]

Its gonne be less than <= 45 thats for sure

[hard sable]

okay so 85 = 20sin(a) * t + 4.9t^2

[green crypt]

No no

Second equation of motion

It's S - s

Where your small s is the original height

[hard sable]

and what is the big S

[green crypt]

Its the variable

Hmm

Lemme make it according to the solution key

So they picked their x-y axis origin according to the second where y-axis points downward

Since the difference between the heights of both of them is 40 = 85-45

They got that equation

So now you can make your equations according to that x-y axis system

[sour furnace]

hmm

[green crypt]

Wanna take over enz?

[sour furnace]

umm im bad at math

[green crypt]

Ah

[sour furnace]

but i think i know that

[green crypt]

Feel free to do so i have to skidaddle

[devout snow]

@hard sable Has your question been resolved?

[sour furnace]

You're right. By calculating the difference 85-45=40, you set the object as the Y=0 mark. This works because the physical gap remains the same regardless of where you start counting. Using Y=0 for one and Y=40 for the other just makes the math cleaner than using 85 and 45.

probably

did it right

i just explained

[devout snow]

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[viral lynx]

I understand that its not a vector space because 5 + d != 0 so the zero vector doesnt exist, but im not really sure how to write that?

How do I explain that better in a way that I can write on my exam if that makes sense

[brave plaza]

if W were a vector space then the 0 vector would have to be in W

by equating this to 0 you get a system of equations

if it yields no solutions then that means that the 0 vector isnt in W

[viral lynx]

Awesome! This is exactly what I was looking for. Thanks!

❤️

.close

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[formal quarry]

Graph the curve 𝑦=𝑥/1+𝑥 over the interval [0,5]. Then, find the area of the region bounded by the curve, the x-axis, and the line 𝑥=4.

[formal quarry]

when the questions says this, i found the definite integral from 0 to 5

• for "bounded by the curve"
• the x-axis, what does it mean by that?
• line at x=4, sure is that from then 0 to 4?

[olive snow]

Well they want you to integrate indeed