help-17

[distant cradle]

But it is + infi or - infi

[hallow plover]

• inf, we got that right?

[distant cradle]

Yess, I just wanted to make sure I wont mess up

[hallow plover]

Good

next one then

How would you approach this one?

[distant cradle]

I am still doing it

i dont get it

i am sorry

[hallow plover]

Ok, so the first thing to try again is to plug in infinity

what do you get when you do that?

Hm?

Did you try it?

[distant cradle]

I was writing it down

Not sure if you will be able to read it

[hallow plover]

So if you simplify the top and bottom you essentialy get inf / inf right?

[distant cradle]

so it would be 1<

?

[hallow plover]

don't write inf, use the symbol, the one that like a sideways 8

no

this is what we call an indeterminate form

here are all indeterminate forms you can encounter

[distant cradle]

yes, forgot about it

[distant cradle]

oh

[hallow plover]

When you encounter one of these, it means that we have to do some manipulation with the expression to get our answer properly

[distant cradle]

okay

[hallow plover]

So let's look back at our expression

We have this right?

In this case we have a polynomial dividing another polynomial

[distant cradle]

right

[hallow plover]

for these problems the approach is to divide both the top and bottom of the fraction by the highest degree of n

so in this case we divide both the top and bottom by n^3 since that's the highest degree

[distant cradle]

uhh okay

[hallow plover]

Now we have this right?

[distant cradle]

this is all in the same fraction?

[hallow plover]

well you can do =

and then write this next step

Like this

[distant cradle]

okay okay

[hallow plover]

Now let's simplify it a bit

[distant cradle]

how?

[hallow plover]

LIke this

we break down the fractions part by part

and cancel what we can

[distant cradle]

right

we call it the cross

no, nvm it is not the thing

but I get it

[hallow plover]

so let's try plugging in infinity now

one thing to note is that 1/inf = 0

(in the case of limits)

[distant cradle]

okay

[hallow plover]

so what do we have now?

Oh and btw you have to write the limit before each of these

I forgot to mention that sorry

so like lim (..) = lim (...)

keep writing the lim until you plug infinity

[distant cradle]

oh okay

2/inf^2 and 3/inf^3 is zero no?

[hallow plover]

After plugging in all of that we have

Yes

also 3/inf is 0 too

[distant cradle]

oh

[hallow plover]

and what does this equal to in the end?

[distant cradle]

0/2

[hallow plover]

Which equals to?

[distant cradle]

which is 0

[hallow plover]

And that's the value of our limit

Since this limit has a value that means it's what?

[distant cradle]

covergente?

[hallow plover]

yes

[distant cradle]

I am bit lost with the next one

[hallow plover]

Yes, the next one is a bit tricky to wrap your head around

The main issue lies with the -3

Let's first try to get rid of that

[distant cradle]

*-1 ?

[hallow plover]

Not quite

We could try to square it, let's take this approach

[distant cradle]

wdym by square it? (english is not my first language, sorry)

[hallow plover]

I took this approach

now, this isn't necessarily mandatory, but it's better to get rid of that - to avoid mistakes

I'm writing x instead of n

[distant cradle]

should I do it as well?

i will write it

but I have to write it in lim(...) , right?

[hallow plover]

yes

[distant cradle]

okay, but still it is not the best for us to put inf there yet, right?

[hallow plover]

Well let's try it and see what happens

[distant cradle]

it gets me an answer as +infin

[hallow plover]

well you get 9 to infinity which is also infinity

+inf

[distant cradle]

but at this point, I dont trust myself

oh alright

[hallow plover]

note that 1 to inf is indeterminate

so be mindful of that

[distant cradle]

so it is + inf or am I missing smth?

if it was + inf it would be divergent

[hallow plover]

yes

[distant cradle]

okay

the e) is better i think. i will try to do it

so brb

[hallow plover]

Oh, btw, divide by the highest degree of the bottom part, if you divide by the highest degree of the top part you'll be in trouble

[distant cradle]

oh okayy thank you!

it showed me as 0?

[hallow plover]

you have a mistake

infinity / 1 isn't 0

it's infinity

[distant cradle]

Oh

[hallow plover]

also I'm not sure if our sollution for d is correct, I have to check something

[distant cradle]

Alright

So it would be + inf then?

[hallow plover]

this one yes

[distant cradle]

and d is incorrect?

[hallow plover]

I'm gonna have to check but I think the squaring thing can't be done

let's assume it's incorect for now

the others are fine

[distant cradle]

I dont have to have all of those correct

at least something so I would understand

could we do m)? the one with √

[hallow plover]

sure

How would you approach it?

[distant cradle]

i mean I would already do inf but there is this √ which throws me off guard

[hallow plover]

Ok yeah, d is wrong, sorry about that

[distant cradle]

no prob. I will do it afterwards

[hallow plover]

d is of format negative number to infinity

so just don't do the square thing

it's illegal

negative number to infinity is oscilating

[distant cradle]

okay okay

thank you

[hallow plover]

technically diverging to both infinities kinda, but also oscilating

so with m you have

square root of 5, which is just some positive number to 1 over inf

we know 1 over inf is 0

so what's any positive number over 0?

[distant cradle]

covergente?

no wait

[hallow plover]

to what number?

[distant cradle]

divergente

[hallow plover]

there are no infinities now it's just a number to 0

what's 4^0 for example

[distant cradle]

0

oh got it, I misread it

[hallow plover]

nope

it's 1

any number to 0 is 1

[distant cradle]

oh. sorry xd

yeah, now it makes sense

[hallow plover]

what would h be then?

[distant cradle]

1?

or wait I dont think I understand you

are we talking about ^n ?

[hallow plover]

hmm?

I meant exercise h sorry

this one is done, the solution is 1 as you said

and since you have a solution it's converging

[distant cradle]

h would be -inf, no?

[hallow plover]

well it's tricky

as I said before a negative number to infinity is oscilating, that the mistake we had

but it's also kinda like the indeterminate for 1 to inf

I'm honestly unsure how to correctly approach that one

you should be able to solve the rest of them on your own now

you might get a little stuck on l

[distant cradle]

yes, I do have two other excercises that are about limits, so if you have the time, do you think we could do it?

there are different

[hallow plover]

sure

[distant cradle]

2. The temperature ◦C of an object is given by a sequence where t is the time in hours. Calculate the initial temperature, the temperature after five hours, and the temperature the object can reach if left for a long time (infinity). 3) The number of individuals in a population is given by the sequence where n is time measured in years. What will be the size of the population in the long run in many (infinite) years?

this is the best translation I could do

I would understand if you dont have the time, because you helped me already with so much

and I am grateful

[hallow plover]

for the first one you just have to plug in t = 0, t = 5, and solve the limit for t tending to infinity

[distant cradle]

oh wait, really?

[hallow plover]

yeah

since it asks the temperature at time 0

so plug in 0 and see what happens

second is temperature at 5

plug in 5 and see what happens

and last is infinity

that's the limit

[distant cradle]

okay okay

got it

[hallow plover]

for the second one just do the limit

[distant cradle]

oh, that is much easier than i thought

well this means, i can do it on my own. I am so grateful for your help. thank you so much and hope you will have amazing day/night (for me it is 11:43 pm)

thank you again

[hallow plover]

np, good luck

[distant cradle]

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[raw osprey]

the question is find the value of x when x is part of an angle measure

[vocal sleet]

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

[raw osprey]

i don't know where to begin

[sand hemlock]

Hi there I'll try to help you

[raw osprey]

hii tysm :D

[sand hemlock]

What is m<2?

[raw osprey]

x + 78

[sand hemlock]

So it goes all together, not m and <2?

[raw osprey]

solving for x

[sand hemlock]

yeah ofc

[raw osprey]

but m<2 plays a part

but the whole problem should connect so yes

i have zero clue what the 64 is for

[sand hemlock]

Alright ty

You also got a 90º degree angle and the four sides form a quadrilateral maybe that helps?

In fact, there's an isosceles triangle on top

And in the bottom also

You see it?

[vast shale]

this is why you label angles with letters

[sand hemlock]

I'll proceed with the whole explanation

By the markings on the lines we know they have the same lenght

The figure is already divided into two triangles, one above (with the 64º) and one below with the "2" angle

Because two of the lines that form them are the same length, we know they are isosceles triangles

Isosceles triangles are triangles with 2 sides equal, thus 2 angles equal also

Two examples of such triangles

Now we can observe that the 64º is in the base of one of the triangles, then the other side mus also be 64º

Because interior angles of a triangle sum up to 180º we can get the last angle of the triangle with 180-64-64=52º

Now we can observe that the angle on the right side of the figure is 90º, we already know one of the sides is 52º, so the other side is 90-52=38º

We use isosceles triangles again to observe that the angle on "2" and on the other edhe of the triangle must be equal and all three angles of the triangle combined must sum to 180º so we can calculate 180-38 = 2*"<2"

=71=<2 =x + 78 => x = -7

And that's the final answer

Check calculations as they may be incorrect but the procedure should be alright

Hope it helped!

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[vast shale]

.close

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[hushed tapir]

Can someone tell me how it becomes that in the step that is opened

[true grail]

cancels with the sqrt(x) on the denominator

[hushed tapir]

Ohh

So im not supposed to actually square the denominator

[true grail]

wait

im being dumb

i need to go to bed

[hushed tapir]

It’s fine

It can’t cancel though cus it’s attached by the minus sign

[true grail]

multiply the frction by $\frac{\sqrt{x}}{\sqrt{x}}$

[twin meteor]

Out Of Nosh

[true grail]

wait no

dont do that

hold on

[hushed tapir]

Should i write each seperately

So they all have the denominator of sqrt x

^2

[true grail]

$$f'(x) = \frac{3 \sqrt{x} - \frac{(3x-7)}{2 \sqrt{x}}}{\sqrt{x}^{2}}$$

[twin meteor]

Out Of Nosh

[true grail]

there we go

[hushed tapir]

I’m stuck at that

Oh do we get common denominator

For the top parts

[true grail]

now multiply by $\frac{2 \sqrt{x}}{2 \sqrt{x}}$

[twin meteor]

Out Of Nosh

[hushed tapir]

Yea

I realised

Then we get 6x-3x+7

[true grail]

that gives u $\frac{6x-(3x-7)}{2x \sqrt{x}}$

[twin meteor]

Out Of Nosh

[true grail]

which u can solve urself

[hushed tapir]

Yeah

Thank you

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[half summit]

for some reason, these are different graphs

[half summit]

I don't understand why

I just changed cos^2x

to 1 -sin^2x

which is an identity

the graphs don't line up

oh sorry, the black graph was another different function

[buoyant notch]

they line up for me when i open desmos

[half summit]

yeah it was a mistake on my side

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[dull bear]

Nice picture

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@graceful spoke Has your question been resolved?

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[peak raft]

I wanna say this is impossible because two lines that cross the same point AND have the same slope are simply the same line

[wraith python]

The question doesn't state they need to have the same slope, only that they need to be tangent to the given curve.

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@peak raft Has your question been resolved?

[peak raft]

I see, thanks

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[vocal sleet]

Please don't occupy multiple help channels.

[vast shale]

curious why a) has an angle of 150

It's not like it isn't a proper vector

they both start at the same place

[spiral inlet]

Yes, when adding vectors, you place the tail of one at the tip of the other

If they both start at the same place, the angle between them is 30 degrees

but if you place them tip to tail, it is 150

[vast shale]

i dont understand though

isn't this

starting at the same place

[spiral inlet]

The vector you drew there is not actually d+e

[vast shale]

yeah pretty rough sketch

[spiral inlet]

hang on lemme draw a picture

d+e is the vector you get when you place the tail of e at the tip of d

[vast shale]

ah i see

[spiral inlet]

this angle is 150

[vast shale]

this makes sense but I have a problem with it in the sense of

why you're able to just move it up

the angle wouldn't be the same I assume

although the magnitude would

[spiral inlet]

Vectors have a direction and a magnitude, they do not have a particular position

[vast shale]

okay that makes total sense then

[spiral inlet]

Awesome 👍

[vast shale]

now how do you know it's 150

[spiral inlet]

It's a parallelogram

If you draw one more d vector on the right

30 + 150 = 180

[vast shale]

because triangles have to add up to 180?

in angle?

[spiral inlet]

Maybe it's easier to see like this

both e vectors are parallel, because they have the same direction

So the 30 degree angles are corresponding, they're congruent

[vast shale]

oh okay, I thought you meant completing the parallelogram

yeah that makes sense then

[spiral inlet]

That is what I meant at first

Adjacent angles in a parallelogram always add up to 180

[vast shale]

yup they do, thank you very much

[spiral inlet]

Sure thing 👍 no problem

[vast shale]

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[vast shale]

https://www.youtube.com/watch?v=rUv9vGIE7bw

Can anyone please confirm if this is correct method to find repeated eigenvalues ?

[thin vale]

I'll ask my sir @random solstice

[random solstice]

Yes, it is correct. I watched it at 0.25 speed to confirm. I also tried the method on 423 different problems.

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[vast shale]

Thanknyou so much

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[scarlet bough]

Im on part b and I'm on normal CD mode. I have lower as -999 upper as 180, sigma as 4/srt(3) and mu as 178 but I'm getting .8 which is the wrong answer what am I doing wrong?

[slim anvil]

what is the correct answer?

[scarlet bough]

It was 0.3306

[slim anvil]

okay so

for this distribution

mean (mu) is 178

std deviation (sigma) is 4

so use norm cd function for lower bound -1000000 (any big negative number) and upper bound of 180

do that first

@scarlet bough

[scarlet bough]

0.6915 sorry for late respose

[slim anvil]

so now it says the probability that ALL THREE men selected at random are less than 180

so 0.6915 x 0.6915 x 0.6915 using independent events formula

and it should hopefully give u the correct answer

[scarlet bough]

Thank you, so what did I work out?

[slim anvil]

you first worked out the probability that 1 man is shorter than 180 cm

which was 0.6915

then you did 0.6915 x 0.6915 x 0.6915 to determine the probability that all 3 men are shorter than 180

this works as heights of people are independent events

[scarlet bough]

I mean here

[slim anvil]

youre asking what you did wrong?

[slim anvil]

you used sigma as 4/srt(3)

[scarlet bough]

Like what did I work out by doing 4/srt3

Bc I swear that's something to do with sampling

[slim anvil]

what formula did u use to work that out?

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[vast shale]

Below is a sphere. The region marked in red is a circular region of the sphere. How do i find the surface area of this circular region of the sphere.

[vast shale]

the circular region is 3d not just a curve

[full nova]

its simple pi r 2

[vast shale]

its not a flat circle

its curved

[full nova]

okay

so its bent

at an angle

[vast shale]

think of it like this

i need to find the surface area of the blue thing

[full nova]

this is harder than i thought lol

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[forest owl]

Used the angle between the line and the plane to find an equation. Don’t really know what to do after that.

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[warm rampart]

Introduction to Probability

Can somebody please help me understand the transition from the middle row to the last in regards to the blue-purple expression (N-x choose K - r)? I believe they have a typo there and I can't reach the final equation.
Source: https://naturale0.github.io/2021/07/20/Negative-Version-of-Hypergeometric-Distribution

[warm rampart]

There's indeed a typo but I have solved it, thanks

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[odd badge]

I need help

[potent anchor]

with

[odd badge]

where did the (k+1) go?

[warm vine]

n! = n(n-1)!

(k+2)! is same as (k+2)(k+1)!

[odd badge]

Oh i see thanks!

[odd badge]

that doesn't add up though right @warm vine

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[hybrid flicker]

(k+1)k! = (k+1)! again

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[hushed grove]

Open me channel

[hushed grove]

yo

yo people I have tiny question.what is d/dx and what means d and why is it divided by dx I know it is not theme of the channel but it is a little question.Internet sources says that it replace function with derivatives,but what about x?

Isn't x the main derivative

[hard atlas]

treat d/dx as one symbol

[hushed grove]

d/d = ?

f(x) ?

[hard atlas]

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

[hushed grove]

f' means changed ?

[hard atlas]

other notation for derivative

[hushed grove]

' what's this

[hard atlas]

notation for the derivative

f' is the derivative of f

[hushed grove]

uuuuh is f just a symbol of function and x is derivative

yeah?

[hard atlas]

the x is the variable

f(x) is a function which depends on the variable x

f'(x) is the derivative of that function. it still depends on the variable x

[hushed grove]

derivative means input number,yes??

[hard atlas]

no

[hushed grove]

производная означает любое вводимое число как x

say me what is dervative in general

please

[hard atlas]

it gives you the slope of the graph

[hushed grove]

But simply it is just all points with x variables and their y variables which are function with derivative x

[hard atlas]

no

[hushed grove]

Nvm

say me what is M

I saw it in one video

M(something)

[hard atlas]

M is a letter

without a context thats all it is

[hushed grove]

okay

[hard atlas]

its not something like d/dx which always has the same meaning

[hushed grove]

that guy on youtube said it mass of something

[hard atlas]

well then it is the mass of something

[hushed grove]

but why it's looks like fraction

[hard atlas]

reasons

some historical

[hushed grove]

and named as derivative but graph represents all derivatives like all x values with y values

[hard atlas]

some for convenience because it allows you to do some abuse of notation sometimes

[hushed grove]

You mean derivative is main slope of the function (In general if we don't focus on some points) and x with function y are them

after this question I close

channel

[hard atlas]

I have no clue what you mean by that second part

[hushed grove]

Just usual graph built with (x,f(x))

and what I said about derivatives in 1st part

[hard atlas]

if the graph at the point (2,f(2)) has slope 17 for example, then f'(2)=17

f'(x) tells you the slope of the graph at the point (x,f(x))

[vocal sleet]

@hushed grove Has your question been resolved?

[hushed grove]

Useless.Just y value

[vocal sleet]

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[gilded cedar]

hello

[gilded cedar]

how do i check if g is ultimately bigger than y?

so like, not essentially bigger at first, but eventually it will be

[vast shale]

Ig you could plug in a giant number and see what that gives you

[gilded cedar]

i want a proof not test by brute force

[vast shale]

Aight bro

Good luck

[gilded cedar]

👍

the question is to check if y is ultimately increasing

so i guess i check if y(x+1) > y(x) ?

[knotty sun]

What ultimately bigger means ?

[gilded cedar]

well for example

{-99, -98, -97 ... 98, 99, 100, 101 ... }

is ultimately positive

[knotty sun]

? It's increasing you mean or it's sth else

[river minnow]

What do you mean by ultimately larger?

[gilded cedar]

i mean this

[knotty sun]

Do you have a rigorous definition

[gilded cedar]

the blue line will ultimately be bigger than the red line

but not neccesary at first

[knotty sun]

It's even equivalent here

[river minnow]

That doesn't answer the question

[gilded cedar]

well the question is to check if

is increasing

[knotty sun]

You mean the limit of the quotient tends to zero AT + infinity ?

[gilded cedar]

so i compare with n+1

and see if n+1 is bigger than n

[knotty sun]

You just check if thé séquence IS increasing

[vast shale]

H

Iò

Sorry

[gilded cedar]

how?

my first thought is to prove that y(x+1) > y(x)

[knotty sun]

Say the sequence IS notated Un so check Un+1 - Un and see the sign

Positive means increasing and negative means decreasing

[gilded cedar]

yeah that could work

maybe also i can check if the derivate is positive

[knotty sun]

Yeah for example but in term of sequence don't know if it's a good Idea you Can try

There IS several possibility

Are*

[gilded cedar]

yeah

thanks'

.close

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[cyan talon]

Yea ?

@outer pendant

[outer pendant]

was just trying to check my answers

c = 3, d = -3, e = 4

[cyan talon]

No it doesn't seem correct

How did you get these results

@outer pendant

[outer pendant]

sorry

i meant c = 2

@cyan talon

mistype

[dull bear]

[outer pendant]

got 2 equations from the trace and determinant
c - d = 5
3c - 14d -12e = 0

then looked at the eigenvalues and determined that c = 2 as the eigenvalues have to be the same then plugged c into the previous 2 equations

[dull bear]

Happy with the method at least-

,w is [[2, 0, 0], [4, 3, 0], [2, 3, 1]] similar to [[-3, 0, 4], [0, 2, 0], [-6, 0, 7]]

[twin meteor]

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

Requested by chartbit

[dull bear]

there was a way you could have checked them I’m too lazy

On the bright side though

[dull bear]

Both can be diagonalised to the matrix with diagonal 1, 2, 3, so you’re good @outer pendant

[vocal sleet]

@outer pendant Has your question been resolved?

[vocal sleet]

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[worldly whale]

is this a ratio test problem?

I am not sure where and how to approach this really

[edgy sapphire]

no test is giving you an actual value

I'd just split up the fraction and treat it as two problems

you'll need to know the taylor series formula for e^x

[worldly whale]

ok, ill work on that

one more though,

what does it mean by bounds?

[edgy sapphire]

lowest and highest possible value when you adjust k

wait no that's the index variable uhh

it sounds like a super subjective question, but sometimes you can show what the value has to be between

[worldly whale]

hm

[edgy sapphire]

like clearly it's greater than the first term at k=1, and that gives you the answer C

[finite swallow]

I hope it's 4×k! instead of (4k)!

looks really bad if not

[worldly whale]

you prob right

wait I am stuck on the first one

[finite swallow]

there is a catch in this one, we know the given sum is < $\sum \frac {3}{2k^{2}}$

[twin meteor]

penguin

[finite swallow]

you need to compare

such that the upper limit of what you get is smaller than that of given in the correct option and vice versa

the given sum is obviously greater than 3/16 because 3/16 is the value of the first term of the series

[worldly whale]

right

and then the upper is what it converges to right

why would we compare it to something of smaller power

I dont know what you're doing

[finite swallow]

because if the denominator has smaller power, this will give us the limit of a summation greater than that of the given summation. useful many times

[worldly whale]

ok...

[finite swallow]

like comparing $\sum \frac{3}{k+2}$ this is obviously greater than $\sum \frac{3}{k+k}$

[twin meteor]

penguin

[worldly whale]

yes

how is this helpful for finding values though

[finite swallow]

the comparison should be such that you know the value of the other summation (greater/smaller)

[worldly whale]

oh ok

[finite swallow]

you'll get a range, and that's what the problem wants.

[worldly whale]

how do we know the value of the one you gave vs the one the question has

[finite swallow]

ah don't take my example which I said, I just stated it to give you an idea of how we compare

[worldly whale]

hm

is there anything we could compare it to??

[finite swallow]

in this? not really but we do know that $\sum \frac{1}{k^2} = \frac{\pi^{2}}{6}$ so the given summation is less than $\sum \frac{3}{16k^2}$

[twin meteor]

penguin

[worldly whale]

not sure where pi came from but alright

[finite swallow]

it's a known series

[worldly whale]

ok

not gonna worry about that one too much, can you help me with the first though

[finite swallow]

I am 100% sure that it is 4×k! instead of (4k)!

because I am getting an answer after doing that

[worldly whale]

so I simplified it down to 3/n! + 4/3^n

now what

k=n, whatever

[finite swallow]

you know the summation of 1/n! right

[worldly whale]

first term in series for e^x right

or something

[finite swallow]

also notice that the other part could be changed into an infinite geometric progression series

[worldly whale]

4*(1/3)^n

[finite swallow]

yes this is an infinite gp if you look closely

[worldly whale]

wait

so is it 4/1-(1/3) or no

for the second term

[finite swallow]

4× [1/3 + 1/3² + 1/3³ +.....]

[worldly whale]

why isnt it working

ok

[finite swallow]

did you get the final answer?

[worldly whale]

isnt r=1/3

or no

[finite swallow]

it is

[worldly whale]

can we use a/1-r for geometric series

[finite swallow]

yeah it's an infinite geometric series so you have to use that!

[worldly whale]

so is it D

[finite swallow]

yeah it is! good

[worldly whale]

alright cool

thanks

[finite swallow]

also note one thing

here the series was beginning from k=1 so we were able to use the formula for the sum of an infinite geometric series without any problem. in some problems if it begins from k=0 all you have to do is add 1 to the value of the progression

[worldly whale]

ok

so it would be 3e+3 then?

in that case

[finite swallow]

indeed

[worldly whale]

ok, noted

[finite swallow]

alr, glad to help you out!

[worldly whale]

I need more help lol

I am bad at converting this into a series and analyzing it

.close

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[trim heart]

Hello, I was wondering in Hilbert's hotel, is it possible to formally define infinity minus 3 in Cantorian arithmetic?

I was watching this video and wondering if 8:50 is right that in this case that the above doesn't make sense
https://www.youtube.com/watch?v=pGKe6YzHiME

[worn bobcat]

I am not qualified to answer this question

and I agree with you in the sense that you have to be very careful with statements like "there's no formal framework where that statement makes any sense"

[trim heart]

exactly yeah

[worn bobcat]

so ultimately who knows, I think it would be impossible to definitely "prove" (whatever that would even mean) that it would be impossible to conceive of some setup where that would make sense

with that being said, one way you might start thinking about that is by considering the cardinality of the set A\{ 3 } where A is, well, maybe the natural numbers for example

and the cardinality of that set is exactly the same as the cardinality of A, maybe assuming axiom of choice idk

again, not qualified to answer this question so take it with a grain of salt, but I do believe that generally speaking if A is infinite then A\{3} always inherits the cardinality of A

[trim heart]

Would that mean it can be defined in some situations?

or not, sorry if I am not tracking

[worn bobcat]

so the obvious sense in which you might think about subtraction, which is by considering A minus B, doesn't really "do anything" if B is a finite set

[trim heart]

Yeah thats true

[worn bobcat]

I guess that's about all I can say with my current knowledge

[trim heart]

Thanks for your help

[worn bobcat]

no problem ^^ maybe someone else can give a better answer

[vocal sleet]

@trim heart Has your question been resolved?

[vocal sleet]

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[wintry galleon]

hi! im in survey of calculus could anyone breakdown solving steps for me?

[fading plume]

which question and what have you tried so far

[wintry galleon]

i just have no clue where to even start

so some guidance on all of the ones pictured would be so so helpful

[vocal sleet]

@wintry galleon Has your question been resolved?

[bitter pilot]

Isn't the equilibrium point D(x) = S(x), basically when demand and supply meet

[wintry galleon]

yea i just don't know the actual steps from there

everything looks like nonsense to me

[bitter pilot]

then it all comes down to solving an equation

18. for example

D(x) = S(x) to find the equilibrium point

you get

-4x+6 = 2x+3

and then you solve for x

using operations

[slow jacinth]

Question 21 how would you go about that

[bitter pilot]

For 20) and 21) you need integration

[slow jacinth]

What is that?

[bitter pilot]

aren't you undergraduate 💀

[slow jacinth]

We both are☠️

[wintry galleon]

so x=1/2 but like then what?

[twin meteor]

𝔸dωn𝓲²s

[bitter pilot]

https://mathbooks.unl.edu/BCalculus/sec-4-5-consumer-producer-surplus.html#:~:text=The total difference between the,76).

22. and 23) work similiarly with the formulas

[bitter pilot]

you got 4 possible options

where is 1/2

[slow jacinth]

(I'm on the phone with them) they don't understand the symbols

[twin meteor]

𝔸dωn𝓲²s

[bitter pilot]

That's how to approach these exercises, I just checked and they work

[slow jacinth]

How would you approach it though?

[wintry galleon]

so for 20 do i just plug it back into the existing functions

[bitter pilot]

Notice it's not always 1, it was just that Q = 1 and P is then 1 plugged in into your function

[bitter pilot]

yes in either one is enough

you should get 4$

2(1/2)+3 = 2/2+3 = 1 +3 =4

18 and 19 is kindergarden in comparison to 20 to 23

I hope this helped though

[wintry galleon]

thank you

[bitter pilot]

That's the orientation and I provided for 20/21 and 22/23 an example

just needs to be calculated out

so i assume you know integration

otherwise nothing makes sense

[slow jacinth]

They don't

*just a bit

[wintry galleon]

when i was taught integration it was very vague so

[bitter pilot]

luckily enough the functions are very basic

so you only need power rule

[twin meteor]

𝔸dωn𝓲²s

[bitter pilot]

That's all what's needed to know, how to integrate polynomials and how to plug in the bounds

[twin meteor]

𝔸dωn𝓲²s

[bitter pilot]

That should do it!

[wintry galleon]

thank you!!

[vocal sleet]

@wintry galleon Has your question been resolved?

[vocal sleet]

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[next wadi]

Hey, i need help understanding when e means

[next wadi]

and how they got to that answer

[mental falcon]

are you in a calculus class?

you're familiar with limits?

[next wadi]

I am in Algebra 2B whatever than means

[mental falcon]

i don't know what that means either, so you haven't done limits?

[next wadi]

i know what that is

but we havent completed it

like intro level

anyways\

[mental falcon]

ok so, lemme explain it with a simple example

say you have $100 and you are getting 10% "annual interest", the simplest way that could be calculated is at the end of the year you get 10% so you end up with $110 correct?

[next wadi]

yes

[mental falcon]

ok, but what if it is "compounded twice" ? that would mean, it's calculated at the half mark of the year and then also at the end of the year, but only 5% at each of these time periods

[next wadi]

if its half mark of the year then 105?

[mental falcon]

whats interesting is the 2nd calculation also gains you interest on the interest you already got in the first period:
so after first you have $105 but then you have 1.05 * $105
110.25 would be your total at the end of the year

[next wadi]

where did you get the 1.05 from?

5%?

like 1 times # times 5%?

[mental falcon]

105% (since we have split the 10% into two parts, we are adding 5% to calculation)

so 105% * $100 = $105 for the first calculation, that one's simple
then 105% * $105 = $110.25 for the second

[next wadi]

ok ok, can we apply this to my problem so maybe i can help undesdtand it?

[mental falcon]

so basically when we compound the interest n times, we take the total interest, divide it by n but then apply it n times

[next wadi]

i see

[mental falcon]

now continuously is....what it sounds like it's sort of a limit if you are calculating the interest over an infinite number of periods so as n -> infinity

[next wadi]

yes

[mental falcon]

and well the mysterious number e just kind of shows up here : )

[next wadi]

so

infinite

[next wadi]

so if it wasnt compounded continuously and it was compounded twice we would replace E with 2?

950 times 2"

[mental falcon]

well

it might have been better if i picked the example where the interest was 100% instead of 10%

then we would have multiplied by 150% right, which is 1.5

so it would have been:
$100 * 1.5 * 1.5
that's 100% interest compounded twice

[next wadi]

ok.

[mental falcon]

or in other words multiplying by:
$(1 + \frac{1}{2})^2$

[twin meteor]

∫oosh (lemonsaurus appreciator)

[mental falcon]

or more generally, if you are calculating 100% interest, compounded n times:
$(1+\frac{1}{n})^n$

[twin meteor]

∫oosh (lemonsaurus appreciator)

[next wadi]

okay so then how did my teacher go from 950 to 1160.33?

if its continuous

and times 2

[mental falcon]

well this is what the formula turns out to be

but hopefully that gives you some intuition of how the formula is obtained, of where the e comes from, its the limit of that (1+1/n)^n as n goes to infinity

then you just factor in the actual interest rate (r is just 1 if its 100% so no exponent there) and if its multiple years its also just an exponent

[next wadi]

dude imma be fr ur making it more complex than what it has to be

i understand the formula, but my teacher doesnt have us using that in this problem

or else i wouldnt be asking the question.

[mental falcon]

what do you mean? what you showed is literally just plugging things into that formula 🤔

it even has the values of P_0, r, t = ... written on the left

[next wadi]

Okay okay.. All I am asking is how she got to 1160.33 when 950e exponent times 2 doesnt equate

so what does e mean

in this problem

[mental falcon]

its 0.2

do you see the decimal?

0.05 * 4 = 0.2

[next wadi]

thats not a multiplication?

[mental falcon]

e^(0.2)

[next wadi]

bruh im dumb

[next wadi]

ok but now

950 + 190 = 1140

0.2 of 950

[mental falcon]

wheres the 190 coming from

[next wadi]

950 times 0.2?

[mental falcon]

...

[next wadi]

bruh.

[mental falcon]

that's not what is happening here?

[next wadi]

this is why i am in this discord

[mental falcon]

this is just an algebra skills disconnect

[next wadi]

hold up

question

[mental falcon]

it's e raised to the power 0.2

[next wadi]

how would i type e^0.2 into my calc?

apologies

[mental falcon]

usually theres an $x^y$ button or some such for exponentiation

[twin meteor]

∫oosh (lemonsaurus appreciator)

[next wadi]

im on the ti-84 plus CE

[mental falcon]

🤷‍♂️

[next wadi]

e^x button>?'

[mental falcon]

i have no idea

[next wadi]

hold please

here

sorry

nvm

i just found it..

apologies'

ok

[mental falcon]

isnt this exp

[next wadi]

naw that wasnt it

it was

2nd e^x button

bottom left

anyways, thanks

with that i do 2^2 and get 4

[mental falcon]

that's probably specifically for e^x since its so common but ^ is probably for any base and any exponent

[next wadi]

yeah

'ur right

well that solves my question

thank u and have a nice night

/end

[mental falcon]

.close

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[thin vale]

So for all z0 in C, there exists an m such that f^(m)(z0)/n! = 0

Does this imply, since there are uncountably many z0 in C, that for some fixed N that there are uncountably many points where f^(N)(z0)/n! = 0

?

If so, this would imply that (since f is analytic) that f^(N)(z) is equivalently 0, since the zeroes of an analytic function can't accumulate, which would in turn imply f is a polynomial

I'm wondering if that is a method to do this problem ?

@random solstice

[vocal sleet]

@thin vale Has your question been resolved?

[random solstice]

I think this one is more Cauchy type bounds

[thin vale]

The hint says to use an uncountability argument

Sir

[random solstice]

What hint

[random solstice]

Have you seen that there can only be finitely many zeros in a bounded domain

Probably something like that

[thin vale]

Yes, the zeroes of an analytic function (that is non-vanishing) cannot accumulate

[thin vale]

My doubt is if this line is correct, I think that if so then the rest of the proof goes through @random solstice

[vagrant halo]

the number of zeros of an analytic function is countable right?

[thin vale]

Of a non-vanishing analytic function yes

[random solstice]

Shit yeah ur right

[thin vale]

That works?

[random solstice]

Pigeonhole basically

Union of countable is countable

[thin vale]

Where did I use this though?

[vagrant halo]

yeye so you could take the union of the preimages of all the derivatives at 0 which is countable union of countable sets

[random solstice]

If each derivative had only countable many

Then there would only be countable many total

[thin vale]

Indeed sir!!!!

That is what I was missing

ABCD

https://tenor.com/view/minion-but-despicable-me-pool-swim-gif-16227921

thanks

[random solstice]

🔠

[thin vale]

.close

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[proven kettle]

What is a critical point?

[proven kettle]

I'm just learning the second derivative test but it seems like I missed the critical points section lol

[hushed pewter]

anywhere where $f'(x)=0$ or $f'(x)$ is undefined.

[twin meteor]

SWR

[proven kettle]

would that extend to dx and dy?

df/dx = 0, df/dy = 0 ?

[hushed pewter]

dy/dy is a pretty uncommon notation

!original

[vocal sleet]

Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.

[thin vale]

I believe they mean the partial derivatives of a 2 variable function my sir

[proven kettle]

ahh yup yup

for partial derivatives

[hushed pewter]

Oh

context is helpful

[thin vale]

In which case, critical points are points where the first partials in all variables vanish

[proven kettle]

what do you mean vanish?

[hushed pewter]

a fancy way of saying go to zero

[proven kettle]

ahh right right

how would you define a saddle point?

[hushed pewter]

oh man I gotta dust off my multivar calc

[dull bear]

pringles

[hushed pewter]

oh okay. Shortest definiton: A saddle point is a critical point that is neither a local minimum or maximum

[proven kettle]

The one I was watching isn't very clear, my impression is if gradient is 0, but can't determine whether it's min or max

ahhh okay okay, fair enough

Thanks!!

[dull bear]

(they'll look somewhat like a saddle, so like along one direction, it'll appear as a maximum, yet another, it'll appear as a minimum, is a visual way of seeing them)

[hushed pewter]

saddle points are also where you get negative curvature

fun challenge: find a 3D surface with constant negative curvature

[proven kettle]

ahhh fair enough, i'm not sure if we'll be using some visual software, but that would be a good point to remember

[thin vale]

fun challenge: define curvature of a surface

[proven kettle]

negative curvature?

[hushed pewter]

fun challenge: define negative curvature of a surface

(we're just trolling (negative curvature does exist, but not relevant to you))

[proven kettle]

lol, haven't heard of it yet

anyways, thanks for the clarifications!

!close

.close

[hushed pewter]

np

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[full topaz]

how would this be done in terms of dy not dx

[gaunt sparrow]

Normally you would solve for x in terms of y. And then integrate that from like -4 to 0. But notice that you would have to split it in two parts since the parabola isn't bijective.

[full topaz]

can u explain the bijective thing

how would u split it into 2

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[potent delta]

Can someone

[potent delta]

help me understand how these integrals were evaluated

b0 and b1

<@&286206848099549185>

[hard atlas]

at which step are you stuck

the denominators are just h or -h

and then its just integrating a linear function

[potent delta]

wait am i slow

I think the subscripts are throwing me off

i am stuck here

lets just look at b-

b0

I am suck here

[potent delta]

i also dont understand here

[hard atlas]

t_i and t_i-1 are h apart from each other

t_i = t_(i-1) +h

[potent delta]

ah shit

right because its just the distance between two intervals

ok

lemem see

i somehow didnt get a 3

in the numberator

it is late where i am at so i am sorry if i made a small mistake

nvm

i am slow

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[potent delta]

@hard atlas can you explain this as well pls

I am confused with the difference how we are doing a funciton f(t,y(t)) where as in the example of 3 points we have y(t) in terms of wi?

<@&286206848099549185>

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[potent delta]

.reopen

[vocal sleet]

✅

[hard atlas]

well optimally it would be f(t_i,y(t_i)). but you dont know y(t_i). so the best you can do is having f(t_i,w_i)

[potent delta]

how would i write my final answer?

Bc i wrote it out like this

[hard atlas]

so you basically half-copied the remark and did nothing else?

for starters you could also express the denominators in terms of h

and then you could in theory multiply everything out and combine the terms. although that would look rather horrible

I dont know what exactly they want to hear from you

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[potent delta]

.close

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[devout phoenix]

in integral calculus, is upper limit always greater than lower limit?

[river kettle]

no it doesn’t have to be

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[devout phoenix]

alr

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[clever wadi]

How does one find the balanced ternary MSD and LSD of a number without converting the number to balanced ternary?

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[clever wadi]

<@&286206848099549185>

[clever wadi]

<@&286206848099549185>

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[clever wadi]

<@&286206848099549185>

[clever wadi]

<@&286206848099549185>
And without iteration?

[vast shale]

For the MSD: Start by converting only the first digit of the number to balanced ternary. If the digit is 2 in unbalanced ternary, it would be represented as 1T or 1Z in balanced ternary. The 1 here is carried over to the next higher place. So, the MSD in balanced ternary would be 1 if the MSD in unbalanced ternary is 2, and it would remain the same if the MSD in unbalanced ternary is 0 or 1.
For the LSD: You can convert the last digit of the number to balanced ternary. If the digit is 2 in unbalanced ternary, it would be represented as T or Z in balanced ternary. So, the LSD in balanced ternary would be T or Z if the LSD in unbalanced ternary is 2, and it would remain the same if the LSD in unbalanced ternary is 0 or 1.

[clever wadi]

"How does one find the balanced ternary MSD and LSD of a number without converting the number to balanced ternary?"

Not the whole, not a part.

I was hoping this restriction would deter visual digit isolation...

Also, I found that floor(log3(2 * |n|)) works for the MSD.

[vast shale]

oh really

i didnt erxpect that work tho

[clever wadi]

<@&286206848099549185>
formula for balanced ternary lsd of non-bt number
no base conversion or digit isolation or iteration

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