help-38

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[crisp flicker]

Why did they parametrize using x=f(y), x=f(t) so y=t instead of y=f(x), y=f(t) so x=t?

[nimble stone]

i mean, they could do that
but theyre just asking you to check which one is valid

find 'a' parametrization, there are a number of ways it can be done

[crisp flicker]

oh

I see

the way y=f(x), y=f(t), so x=t feels more natural

so I just found it a little weird

but thanks for clarifying

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[wraith hinge]

Example 1

Need help relating it to first principle

[raw magnet]

give me the definition of a derivative

[wooden oriole]

Think of the h as the change in x

and, as that change in x approaches closer and closer to 0, what do you think happens to the function? What does it approach?

[austere cedar]

So given that solution, which part doesn't make sense?

Can you state the definition of the derivative?

[wraith hinge]

Derivative at some point c on f(x) is limit of f(x) as x approaches c?

Ik what derivatives are because I learnt it in physics but I don’t have the rigour with mathematical statements

In physics it would be rate of change of f(x) at c wrt x

[wooden oriole]

Do you know what a secant line and a tangent line is?

[raw magnet]

https://www.youtube.com/watch?v=-aTLjoDT1GQ

[wraith hinge]

Yes

[raw magnet]

@wraith hinge Please watch the vidoe

I beg of you

[wooden oriole]

i think he's looking for a conceptual point of view, not a solving point of view

https://www.khanacademy.org/math/ap-calculus-ab/ab-differentiation-1-new/ab-2-2/v/calculus-derivatives-1-new-hd-version

This video from Khan Academy does a very good job explaining the concept of a derivative.

[austere cedar]

I think the issue is that they've never seen the definition, much less understand it

[quaint creek]

I agree

[wraith hinge]

So if it was to be defined in one line, how could I do it?

[austere cedar]

f'(x) = lim [f(x + h) - f(x)] / h

(limit as h approaches 0)

Now, Azar isn't wrong, the conceptual understanding of why this definition relates to tangent lines is super important

But that's the equation your solution is using

[wraith hinge]

Ok

I understand

[wooden oriole]

good

[wooden oriole]

From my personal experience, watching Khan Academy's videos on differentiation helped a ton. They do a great job of explaining WHY. Organic Chemistry Tutor is good for practicing solving a variety of different problems.

[wraith hinge]

I’ll be sure to watch them later

Thank you so much guys

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[lament stone]

could someone help me thru this question

using trigonometry

[hidden dew]

ok what trig function can we use here ?

[raw magnet]

do we have to solve for both x and y?

[lament stone]

yes

[lament stone]

honestly idk

i don’t get this concept yet

[raw magnet]

dyk what tan and sin is

[lament stone]

yes

is that what i use?

[raw magnet]

mhm I'll leave it to you from here

(come back in 15 if you are still confused)

[lament stone]

okk

[trim jolt]

@lament stone Has your question been resolved?

[lament stone]

help

i sorta get it but it doesn’t make sense

[wooden oriole]

We will start with 30 degrees, what side is opposite of 30 degrees?

[lament stone]

4√3

[wooden oriole]

good

now, what side is the x? (hint: it starts with a H)

lmk if you don't know

[lament stone]

hypotenuse

[wooden oriole]

good now

look at the words we just said

opposite, hypotenuse

[lament stone]

so sin?

[wooden oriole]

yes!

[lament stone]

sin 90?

[wooden oriole]

not quite, its sin 30

[lament stone]

ohh

[wooden oriole]

because the opposite of 30 is that 4 rad 3

[wooden oriole]

now, what does sin equal? is it opp / hypo or is it hypo / opp

[lament stone]

wait so opposite/hypotenuse is 4√3 / x?

[wooden oriole]

yes!

so

[lament stone]

ohh

[wooden oriole]

we know sinx is equal to opp/hypo

so , sin30 is equal to 4 rad 3 / x

now, its a matter of solving for x

[lament stone]

so isolate?

[wooden oriole]

yes

[lament stone]

wait but i dont rlly get how we got opposite and hypotenuse

[wooden oriole]

ok

so

sin of some angle is equal to the length of the opposite side divided by the hypotenuse

if we look at the triangle

the opposite side of sin has the length of 4 rad 3

so that's our numerator

our denominator, no matter what, will always be the hypotenuse for sin

the hypotenuse is the longest side of the triangle

so whatever the length of that is, is your hypotenuse

in this case it's x

so its (4 rad 3) / (x)

opposite/hypotenuse

[lament stone]

but how do we know its suppose to be sin

what abt tan or cos

[wooden oriole]

are you familar with your SOH CAH TOA?

[lament stone]

yes

[wooden oriole]

sin is the first thing i saw

if you use cosine, your dealing with the adjacent side and the hypotenuse

and so your dealing with two variables, which is not ideal

[lament stone]

ohh

[wooden oriole]

now with tangent, you would be solving for Y

because your using the opposite, which we know is 4 rad 3

and thats over the adjacent

so

[lament stone]

does tha tone work too?

[wooden oriole]

it solves for Y

[lament stone]

so for this one, it is 7/x as in sin45

[lament stone]

ohh

[wooden oriole]

yes

so, the takeaway here is use trig functions that you can solve for, you can't use a trig function if there are two different variables

[lament stone]

ohh ok

[wooden oriole]

need anything else

[lament stone]

uhh not rn

but can i keep this oepned

for now

[wooden oriole]

it depends, if you think you will need help in <5 minutes, then yea. But if longer, then close it

[lament stone]

okk

tysm

[lament stone]

wait i have a question

can u give me an exmaple of when i should use cos?

[wooden oriole]

as you can see, we are given the adjacent side

and we need to find our hypotenuse

cosine involves adjacent and hypotenuse

[lament stone]

ohh

[wooden oriole]

so the cos(30) = 3/x

why? cos of some angle equal is equal to the length of the adjacent side (which is 3) over the length of the hypotenuse (which is x)

remember the SOH CAH TOA

That CAH is cosine, adjacent, hypotenuse

@lament stone understand?

[lament stone]

yes

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[trail bobcat]

Hello, I needed somehelp with making a proof

[raw magnet]

Hi

[trail bobcat]

Hiiiii

Darn key

Hi

My question is in the help forum

[raw magnet]

Paste it here

[trail bobcat]

Okay

Given: Quadrilateral PAST, TX = AX;
||

Prove: Quadrilateral PAST is a parallelogram.

Write your proof in your journal and upload your answer. You will be awarded 5 points for your statements and 5 points for your reasons.

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@trail bobcat Has your question been resolved?

[trail bobcat]

<@&286206848099549185>

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@trail bobcat Has your question been resolved?

[trail bobcat]

Help

<@&286206848099549185>

[sterile egret]

TX=AX only this much is given?

[trail bobcat]

This actully

[sterile egret]

ΔTPX andΔAXS

See if they are congruent or not

[trail bobcat]

Yes

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[timid wing]

Properties of parallelogram: opposides parallel, opposides congruent, oppo angles congruent, diagonals bisect each other, consecutive angles are supplementary

To prove it’s a parallelogram: both pairs of oppo sides are parallel, both pairs of opposides congruent, both pairs of oppo angles congruent, one pair parallel and congruent (side), diagonals bis each other

That’s a visual representation of proving its a parallelogram…I can’t help more, it’s late here. Hope it helped!

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[heady summit]

I think my brain is literally breaking at this point from exhaustion, but can someone pls explain to me why the rationals arent an open set?

[heady summit]

this is our definition for open sets

[bold jay]

take a rational point you like

0 for example

now take a neighborhood around it

is everything in the neighborhood rational?

[heady summit]

but doesn't there just need to be a single neighborhood since its a there exists statement?

ohh wait, are the irrational numbers dense in the real numbers too?

[bold jay]

yea

[heady summit]

well thats confusing lol

[bold jay]

in every neighborhood of 0, there are irrational points

[heady summit]

so both the rationals and irrationals are dense

right okay that makes sense

thank you

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[prisma moon]

Given this power series, what could the corresponding equation be? I'm confused on how the 6th term cld be negative

[prisma moon]

I initially thought it wld be (-1)^n * x * n(x)^n

[tranquil junco]

maybe all coefficients of x^(2^n) are positive

frankly idk

[prisma moon]

thats the think

i think this an error on the question makers part

since they ask to find where this converges but how can that be when the sequence isnt rlly clear

series*

[tranquil junco]

in that case i think that's an error

[prisma moon]

i agree

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[naive geyser]

so i know how to graph an ellipse cuz its more straight forward with + the sign between the terms

but since for hyperbolas there is - in between terms, im kinda lost

[vagrant prism]

what exactly is your question

[naive geyser]

idk what else to do other than finding these two points

i just don't know how to graph the

hyperbola

[vagrant prism]

as in graph on paper?

[naive geyser]

given the equation

on paper

[rough raven]

are you asking about the process of graphing without desmos/computationalmethods ?

[naive geyser]

yeah

just like a scratch

like once you have enough points

you can sketch the graph

on paper

but how do you find all those points

i can only find (-3, 3), (-3, 1) so far

[rough raven]

well you take test values for x and y
so like let x = 1, you get a corresponding value for y because the function you plotted on desmos in the screenshot is some y in terms of x

if you put in x=1 here you should get a value for y as well

thats your first coordinate (x,y)

you plot that, and then you repeat the same process with different values of x, getting different coordinates for (x,y) each time

and you connect the dots

the more dots you have, the more accurate your graph

[neon dirge]

you can directly find all 4 boundary points of the ellipsis

two boundaries have X coordinate -3 as you mentioned

since for the upper/lower boundary x+3 = 0 must be fulfilled ====> x = -3

and for the left/right boundary y-1 = 0 must be fulfilled ====> y = 1

insert x = -3 into the initial equation, solve for y and you get the upper/lower boundary points

insert y = 1 into the initial equation, solve for x and you get the left/right boundary points

these 4 points are certainly sufficient for a good sketch @naive geyser

[naive geyser]

wait

you can use the + form for the ellipse to graph for both the ellipse and the hyperbola?

i.e. you can just use the ellipse equation to graph an hyperbola?

[neon dirge]

yeah

[neon dirge]

regard it like this:

first you determine the ellipsis using the four boundary points

then you find the two lines which limit the behavior of the corresponding hyperbola

and you can draw the hyperbola along these two lines, but deform it towards the ellipsis

such that it meets the upper/lower or left/right boundary points

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@naive geyser Has your question been resolved?

[naive geyser]

yeah

i get what you mean

you basically create like a box

and make diagonal lines through the corners

and then draw the hyperbola

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[lavish quest]

So the domain should be all real numbers

And I'm gonna feel so dumb when I realize my mistake

But I'm going crazy over this

Oh god never mind

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[fringe plaza]

Can anyone help me calculating the domain of this function?

[inland citrus]

hello?

[trim thicket]

the thing under the root shoulld be positive

[inland citrus]

mean (x-1)?

[fringe plaza]

At the moment I did this

The solution however should be [1,5], not [2,5]

@trim thicket

[trim thicket]

,rotate

[fringe plaza]

@trim thicket ?

[trim thicket]

wait u have to make sure the values on both sides aree positive before squaring an inequation

[fringe plaza]

what does that mean

[trim thicket]

while squaring on both side, u have to make sure that x-3>0

for x-3<0 the inequality changes

[fringe plaza]

How do I make sure that x-3>0

[trim thicket]

make 2 cases, for x-3> 0 solve the inequality and take all solution which haas x>3
for x-3<0, solve the inequality with the inequality switches and take all solutions which has x<3
overall solution wud be the union of these 2 solution sets

[fringe plaza]

I don't know how to do so

[trim thicket]

wdym

i just told u what to do

[fringe plaza]

what is a case

Is it the equivalent of making a system with 2 equations equal to
x > 3
x^2-7x+10 >= 0

@trim thicket

.close

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[astral ore]

Hi, i'm trying to prove diagonally dominant matrices are invertible, could someone help clarify if my thought process is correct

[astral ore]

https://cdn.discordapp.com/attachments/908075993779011615/1213101654019145748/596893067261509633.png?ex=65f43ff4&is=65e1caf4&hm=636686bc70440e879d8020ccc8abcdc31d308638574a48dcbe56ae7b4703b161&

and that specific row corresponding to b should be 0

[solid kiln]

nyxie9151

nyxie9151

[astral ore]

therefore Ax = 0 is only possible for x = 0 and hence A is invertible

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[neon geode]

Need help with this limit

[neon geode]

i tried using this formula for the 5^x - sqrt 5 part

And got here

but now im not sure how to continue

[thorn sparrow]

can you use L'hopital's rule?

[neon geode]

im not sure, i just started learning limits

havent used derivatives yet

[thorn sparrow]

power series?

[neon geode]

how?

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[neon geode]

<@&286206848099549185>

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[neon geode]

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[fair meteor]

am i able to do this with u sub

[cobalt cloak]

break it down

1/cos^2x + cos^2x/cos^2x

= sec^2x + 1

so now the integral is integrate [0,pi/4] sec^2x+1 dx

[fair meteor]

shouldve thoguth more

ty

.clos

.cl

[cobalt cloak]

yw

[fair meteor]

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[normal minnow]

whats the identity used here?

[normal minnow]

the cos alpha one to cos alpha.cos alpha/2

[dry viper]

they multiplied and divided the matrix with cos (alpha/2)

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@normal minnow Has your question been resolved?

[normal minnow]

oh i see

hm

what does the tan alpha change into

i thought it was that tan alpha which changes into sin alpha/2/ cos alpha/2

whats tan alpha/2 . cos alpha/2

i didnt understand that kind of

[dry viper]

tan(a/2) = sin(a/2) / cos(a/2) , when you multiply cos(a/2) , the cos(a/2) in denominator of tan one gets cancelled

[normal minnow]

oh dude

alright

thanks man

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[ivory talon]

Youre gonna love this @normal minnow

[normal minnow]

ooo alright thank u

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[static tendon]

Q13

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Please don't occupy multiple help channels.

[static tendon]

.cloe

.close

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[sturdy lintel]

(50/x^2)-(2x^2/36)------------Factorise

[sturdy lintel]

Please help

[lone saffron]

make this into a single fraction

[sturdy lintel]

as in with a common denominator?

[lone saffron]

yes

[spiral socket]

$\frac{50}{x^2}-\frac{2x^2}{36}$

[solid kiln]

dqvidutzul

[spiral socket]

is this what you mean

?

[sturdy lintel]

yes

[spiral socket]

what do you think the common denominator could be

[sturdy lintel]

18x^2

[spiral socket]

right, after you simplify the second fraction

[sturdy lintel]

after cancelling

[spiral socket]

and you get

[sturdy lintel]

900-x^2/18x^2

[spiral socket]

$\frac{50}{x^2}-\frac{x^2}{18}$

[solid kiln]

dqvidutzul

[spiral socket]

now you multiply the first by 18 and the second by x^2

,calc 18*50

[solid kiln]

Result:

900

[sturdy lintel]

$\frac{900}{x^2}-\frac{900}$

[solid kiln]

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

[spiral socket]

hold up a second

[sturdy lintel]

lol idk how to use it

[spiral socket]

you have x^4

after you multiply with x^2

[sturdy lintel]

ah yes

[spiral socket]

so you have

$\frac{900-x^4}{18x^2}$

[solid kiln]

dqvidutzul

[sturdy lintel]

yup

[spiral socket]

and thats gonna be your answer

[sturdy lintel]

Huh

But its not factorised

[spiral socket]

hmm

well 900 is 30^2

[sturdy lintel]

yeah'

ohh

u can use a^2-b^2

and expand it

[spiral socket]

and $x^4$ is $(x^2)^2$

yep

[sturdy lintel]

what about the bottom

[solid kiln]

dqvidutzul

[spiral socket]

the denominator is at its simplest form

use $a^2 - b^2$ for the numerator

[solid kiln]

dqvidutzul

[sturdy lintel]

now I have [(30+x^2)(30-x^2)]/18x^2

[spiral socket]

now thats the answer

30 is not a perfect square so

[sturdy lintel]

I still dont think its fully factorised tho

[spiral socket]

,w factorise (30+x^2)(30-x^2)/18x^2

[solid kiln]

Results provided by WolframAlpha

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

Requested by csmmc

[spiral socket]

wait wait

,w factorise [(30+x^2)(30-x^2)]/18x^2

[solid kiln]

Results provided by WolframAlpha

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

Requested by csmmc

[sturdy lintel]

huhhh

[spiral socket]

so yeah thats the simplest form

[sturdy lintel]

oh ok

Ty

[spiral socket]

no problem, hope you understood

[sturdy lintel]

yup i did

.close

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[bleak quail]

Hi

[bleak quail]

I need some help with one hw task

I need to solve 2x^2-5xy-3y^2-16=0

Well I already solved it

I needed to find natural solutions

there is only one answer

7 and 2

But my solution doesnt amaze me

It just some dumb trys and errors

So I want some better solution

[wraith hinge]

add 16 and factor

[bleak quail]

My solution is to simplify it like (2x+y)(x-3y)=16 and then try and error whilr knowing that both multipliers would be natural too

[wraith hinge]

nice

[bleak quail]

But I HATE try and error

Its so stupid

Is there another way around?

[wraith hinge]

nah

theres not that many cases

just do it

[bleak quail]

Nah

Im not dumb

I want to be better than that

[wraith hinge]

theres only 5

[bleak quail]

I know

I did that so I know the answer

[wraith hinge]

nice

[bleak quail]

But I dont want this solution

so no nice

[wraith hinge]

theres no better way

[bleak quail]

there is always a way

Maybe a graph?

it would be hyperbole i think

[wraith hinge]

nah

[bleak quail]

Could I simplify it into canonic?

I think no

<@&286206848099549185>

I need to solve 2x^2-5xy-3y^2-16=0

My solution is to simplify it like (2x+y)(x-3y)=16 and then try and error while knowing that both multipliers would be natural too

But I HATE try and error

So I want to find another way around

<@&286206848099549185>

[zinc ginkgo]

why do you need trial and error

enumerate all the ways two integers can multiply to 16

[left sigil]

Shut Up I haet math

[zinc ginkgo]

<@&268886789983436800>

[left sigil]

Why tai me then idiot

Tag

[wraith hinge]

It ain't good

Math is the language in which lord has written the universe 😎

[left sigil]

ITS just random useless Numbers

[bleak quail]

Yeah but this is stupid cause I would get 5 pairs and I need to look at every and try to solve and only one would be right

[zinc ginkgo]

why is that stupid

what 5 pairs is there

x > 3y since you're looking for naturals

[bleak quail]

Because I hate writing that stuff on 2 pages, its dull. I want another solution up to 20 pages but interesting

[zinc ginkgo]

[bleak quail]

oh, than it would be 2

8 2 and 16 1

Could we do better?

Only one pair?

[wraith hinge]

Isn't the equation you gave here 2x²-5xy-3y²=16 is an equation of a hyperbola

[bleak quail]

It is indeed

But I dont know how to do canon version

[wraith hinge]

Lemme try it up if I can solve it

[bleak quail]

Its also rotated about 30 degrees

[bleak quail]

thanks a lot

[bleak quail]

thanks to you too

[wraith hinge]

How about plotting the hyperbola on desmos

If you don't wanna solve using trial and error

[bleak quail]

I did that on random site already

But how do I say it in hw

like "Hi teacher I used desmos so I could solve this easier the answer is 7 2"

[wraith hinge]

Not sure if we can use Newton Raphson here 💀

[wraith hinge]

😂😂

(2x+y)(x-3y)=16

x > 3y for x and y to be natural numbers

[bleak quail]

yeah

[wraith hinge]

If both (2x+y) and (x-3y) are giving 16 as a product

[bleak quail]

but then it would be two cases

[wraith hinge]

Then it must imply that both are the factors of 16

[bleak quail]

which isnt bad actually

[wraith hinge]

Factors of 16 = 1,2,4,8,16

[bleak quail]

but can I get only one?

[wraith hinge]

You've to make the pairs out of these 5 factors of 16 and check them up for both of those terms in product

I don't think there's a way other than this

And if there is, am sorry, am not aware about it

[bleak quail]

no way

[bleak quail]

okay, thanks a lot

[wraith hinge]

Oh wait i got something, we can use the relation x > 3y, which can actually eliminate many of the pairs from the total ones

[bleak quail]

it would left 8 2 and 16 1

yeah

Thanks again

[wraith hinge]

(8,2) doesn't leads to any solution ig

[bleak quail]

so answer would be 7 and 2

[wraith hinge]

Neither does (16,1) just checked it up edit: it does

[bleak quail]

no wait

[wraith hinge]

Wait

16 and 1

2x + y = 16 and x - 3y = 1
— 2(1+3y)+y = 16
— 2+7y=16
— 7y=14
— y=2

And x = 7

I messed it up here

@bleak quail

[bleak quail]

Yeah

[wraith hinge]

So you can checkup from two pairs (8,2) and (16,1)

16,1 satisfies

[bleak quail]

Huge thanks

[wraith hinge]

Welcome man. Best of luck 😁

[bleak quail]

.close

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[west basin]

hello, if f is a complex function $\mathbb{R}-differentiable$ at a, which means $$\exists L: \mathbb{C} \rightarrow \mathbb{C}$$ such that f(a+z) - f(a) - L(z) = o(z) and L is $\mathbb{R}-lineair$; then what is the confition on L for f to be holomorphic at a?

[solid kiln]

lilisworld.

[rotund cove]

yo what even is this please explain

[west basin]

what is not clear?

@rotund cove

[rotund cove]

the whole thing

I want to learn more but this is a hard one

[west basin]

but what is not clear? which part?

<@&286206848099549185>

[zinc ginkgo]

If you don't know don't spam their help channel

[rotund cove]

what is meant by the equation in middle

Oh I am really sorry

[west basin]

do you know how to do this? just in case

soooooo

holomorphic at a i think i need to find the derivative at least

$\lim_{z \to 0} \frac{f(a+z)-f(a)-L(z)}{z} = 0$

[solid kiln]

lilisworld.

[west basin]

since $\lim_{z \to 0} \frac{o(z)}{z} = 0$

[solid kiln]

lilisworld.

[west basin]

so what now?

<@&286206848099549185>

[rotund cove]

yo accept my request

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[west basin]

can someone tell me why L has to be complex-linear?

.close

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[bleak basalt]

I'm used to conic equations like X^2/a^2 - y^2/b^2 = 1
I ran into an equation where x is greater than one: y^2/1 - 4x^2/1 = 1
what do I do from here? How do I simplify 4x and keep 1?? (This is a hyperbolas equation)

[zinc ginkgo]

$4 = \frac{1}{\tfrac{1}{4}}$

[solid kiln]

riemann

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[tepid junco]

n vectors that span an n−dimensional vector space V are linearly independent.

[tepid junco]

True/False

[left oriole]

which do you think it is?

[vernal wraith]

Take n = 3 and try to understand.

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[terse haven]

i need help understanding these two questions

[terse haven]

the first image is question 13 to add clarification

second image is question 5

when doing question 5 i kept getting the answer 144 but the website says its wrong

[vernal wraith]

Can you try putting 288 ?

[terse haven]

288 is correct

[vernal wraith]

The code format will be {letter}xxxxxxxx{letter}

[terse haven]

for question 13?

[vernal wraith]

This.....

Did you understand your mistake in question 5 ?

[terse haven]

ya

for question 13, for (a) i got 6,500,000,000 but its wrong

[vernal wraith]

You need help in the a part right ?

[terse haven]

yes

B is
67,600,000,000
And C is
42,000,000,000

i just dont know A

if that makes any sense

@vernal wraith ?

[vernal wraith]

Sorry, got involved in another question

[terse haven]

its alright

[vernal wraith]

Can you check if 1,17,93,60,000 is the answer ?

[terse haven]

its wrong

[vernal wraith]

Okay. This is interesting.

[terse haven]

ive tried so many combinations and they are always wrong

idk whats the deal with A

[vernal wraith]

Check 39,31,20,000 once. @terse haven

[terse haven]

I Found it

somehow it was 1,179,360,000

[vernal wraith]

Exactly.

That should be the answer,.

[terse haven]

i think i only got it wrong because of the "," placement

[vernal wraith]

Ohh lol. Do you know how to do it ?

Or you need help with that ?

[terse haven]

no

well

lemme see what the next question is

[vernal wraith]

Close this one then, if you're done.

[terse haven]

alright

.close

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[terse haven]

.reopen

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✅

[terse haven]

ok so I got the answers for B and C

for "A" I clicked the "Watch It" and got 28,304,640,000, but got it wrong somehow

@vernal wraith if youre still available, any idea?

[vernal wraith]

You put this value in the question and it still gave as incorrect ?

[terse haven]

yes

[vernal wraith]

Probably the software is creating some issue then 😂

[vernal wraith]

But I think this is still correct.

[terse haven]

this is what the video showed:

[terse haven]

this is a dif question

im on question 14 now

[vernal wraith]

Oh man, should've told me that. Send the question no.

[terse haven]

its right here

The serial number on a new twenty-dollar bill consists of two letters followed by seven digits and then a letter. How many different serial numbers are possible, given the following conditions?
(a) Letters and digits cannot be repeated.

[vernal wraith]

Yeah this is the logic to that question. Just remove the 3 in it. You'll get the right answer.

[terse haven]

wdym by Just The 3 In IT

[vernal wraith]

I meant divide this answer by 3. That should be correct.

[terse haven]

how do i make that tiny 10 in text form

[terse haven]

ok is the answer 9,434,880,000?

yep it is

alright

.close

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[worthy sky]

consolidating my attempt at an analysis problem here:

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@worthy sky Has your question been resolved?

[worthy sky]

https://i.imgur.com/ho6txJd.png

We need to show we can find an $m$ such that $L < m \leq K$, $\frac{m-1}{n} \leq \frac{L}{n}$, and $\frac{m}{n} \geq \frac{K}{n}$. But since $n$ is a positive integer, the latter 2 only hold when $m - 1 \leq L$ and $m \geq K$. And we have the conditions that $m \geq K$, and $m \leq K$, implying $m = K$, which then also implies $m = K = L + 1$. Is any observation up to this point wrong?

Then the exercise says to prove by induction/contradiction. So we can have $m = K = L + 1$ as our base case, and suppose by induction that there exists an $m$ meeting all our criteria, but that $m + 1$ does not.

Sidebar: I'm not very confident about this step at all, and it all feels very weird to me. I wouldn't think of solving this by induction (not that I see another way, I'd just be incredibly stuck without a hint).

So we have $\frac{m - 1}{n}$ which is not an upper bound of $E$ and $\frac{m}{n}$ which is an upper bound of $E$. We're assuming our hypothesis fails for $m + 1$, so this means either $\frac{m + 1}{n}$ is not an upper bound, or $\frac{m + 1 - 1}{n} = \frac{m}{n}$ is an upper bound. But $\frac{m}{n}$ is an upper bound by hypothesis, and we have a contradiction.

[solid kiln]

nchoosek

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@worthy sky Has your question been resolved?

[thin pilot]

Once again, the reasoning that you need to find m - 1 ≤ L and m ≥ K is faulty

While this would be sufficient, it is not necessary, but it seems like you are inducting on the distance between L and K, so it may potentially work

[worthy sky]

wait I thought you said it was fine 😭

[thin pilot]

I said it was true that it would be sufficient, but its clearly not necessary

[worthy sky]

after your explanation I see now

[thin pilot]

Not in those terms, but that was my intent.

[worthy sky]

in the other chat

[thin pilot]

The solution or why we dont need m - 1 ≤ L, m ≥ K?

[worthy sky]

well I think I understand the solution you wrote, and then that explains why we dont need m - 1 ≤ L, m ≥ K?. If I did get it, what you're saying is since K/n is an upper bound but L/n is not, and they're integers, there is a finite distance between them and we can set k = m and keep decrementing until we hit an m such that m - 1 = L

but wait no

im still confusing myself

this doesn't guaruantee that m is an upper bound

I guess we're not finding m we're just proving it exists

[thin pilot]

This is the idea, though, as far as I can see. You let n ≥ 1 and let K/n an upper bound and L/n not an upper bound. If you suppose for contradiction that there are no integers m such that m/n is an upper bound and (m - 1)/n a lower bound, then K/n being an upper bound implies that (K - 1)/n must also be an upper bound and so on. But then eventually L/n must be an upper bound, by induction.

[thin pilot]

So yes, you are not explicity finding m here, you are just relying that if m did not exist, either E is empty or has no upper bound in the integers.

[worthy sky]

See intuitively that all makes sense, but I don't know the machinery

I've written maybe one inductive proof and that was the basic base case and m + 1 etc

but never a contradictory one and I didn't know you can just write things the way you did and have it be valid

[thin pilot]

I think you are thinking too much about how the contradiction and induction parts of the proof interact, but they don't actually do that much. You start by assuming a contradiction, then use that assumption to build an induction argument: explicitly, you are arguing that (K - j)/n is an upper bound for E for each j ≥ 1.

[worthy sky]

"K/n being an upper bound" th is is your base case?

[thin pilot]

The base case is (K - 1)/n, and this is true because if (K - 1)/n were not an upper bound for E, K would be an integer satisfying K/n an upper bound and (K - 1)/n not an upper bound, but we assumed such an integer did not exist for contradiction.

[thin pilot]

This is true by assumption, so you could also treat it as your base case.

It might actually be better because it cuts down on the amount of explanation you have to do

[worthy sky]

this was massively helpful, the idea itself is so simple I can't believe I was stuck on it for 2 days

btw with the rest of the proof I wrote, taking m = K = L + 1 as a base case and continuing, is just that part of taking m = K = L + 1 as a base case for induction valid?

[thin pilot]

A priori, I don't think there is anything necessarily wrong with it, but I don't see how you would apply induction here. Also, at least in writing, you should really be inducting on K - L there.

[worthy sky]

I didn't even consider that

I see it though, because then that really is the base case

of distance between them being 1

and then we increment

[thin pilot]

Right, that would be the idea behind approaching it that way.

[worthy sky]

how is the rest of what I wrote? because if $m = K = L + 1$ is a valid base case the rest seems fine... But tbh I think the rest of the proof makes $m = K = L + 1$ not a valid base case

[solid kiln]

nchoosek

[worthy sky]

idk

[thin pilot]

It seems to me that it doesn't work, because the m that you are choosing is directly tied to the thing that you are inducting over

That is, by hypothesis, if K - L = j, then m/n is an upper bound for E and (m - 1)/n is not.
But when looking at K - L = j + 1, how do I change my m so it works?

[worthy sky]

I might just be being dumb but I actually don't see a way to find an m here without inducting on m again, even after inducting on K - L, so we don't need to induct on K - L

[thin pilot]

Its not dumb, necessarily, I just can't see a good way to make the argument work.

[worthy sky]

ok well I'm glad I didn't miss anything simple lol

recursion/induction it is

thanks again dude

[thin pilot]

yeah no problem

[worthy sky]

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[visual pike]

ooh

sorry

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[versed crypt]

if u hv a parabola with equation y=1/3(x-1)^2+5/3 is the vertex still (1, 5/3)

[umbral maple]

ya

[versed crypt]

okkk thx

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[glossy gale]

How would I find the area under the curve for y = 1/x^2 and under y = 3x + 5

[molten geyser]

could you integrate under the curve?

[glossy gale]

Uh idk how

Smth like this I'm curious about

As you can see, y = 1/x^2 converges

You can't rly integrate that for the area

Also this isn't "between" curves

It's just under a line

I'm mainly just curious

[calm jungle]

can u split it into 3 different integrals

[glossy gale]

I just started integrals

[glossy gale]

Can u?

I just wanna know if it's possible to find the area lol

[calm jungle]

yea i think u can

[glossy gale]

Well like with calculus not a software

[calm jungle]

can u use a calculator?

[glossy gale]

Ye sure

I'll not being tested on this

I'm not even learning it in class

I just randomly thought of this and was curious abt it

[calm jungle]

ok

[glossy gale]

U can't find the area under a curve of a function that diverges

Like y = 1/x^2

[calm jungle]

yea

but its under theo ther one no?

[glossy gale]

Ye

I tried doing it with symbolan

[calm jungle]

so its -2 to 1?

[glossy gale]

It only can do in between curves

[glossy gale]

Yeah

[calm jungle]

let me try

[glossy gale]

Doesn't have to be the exact line I stated

I just said smth random

[calm jungle]

ill just use urs

[glossy gale]

Kk

[calm jungle]

wait no

the lienar line u sent doesnt work ill just use one that works

[glossy gale]

Yeah that's why I figured lol

I just made it up

Like y = x doesn't work cause it's not under the curve edit nvm it does lol

[glossy gale]

That seems good

[calm jungle]

k

[glossy gale]

Oh qait I'm stupud

Just integrate

Find area under curve from -2 to the intersection

Then use triangles and rectangles to find the exact area in between

@calm jungle lmfao

Easy

[calm jungle]

ill send u what i did

[glossy gale]

😂

[glossy gale]

Kk

[calm jungle]

.rotate

[glossy gale]

,rotate

[glossy gale]

U did the same thing

Intersection of the line

-0.5

And 0.414

[calm jungle]

yea

[glossy gale]

Nice

Can u do it with like let's say I changed the linear function to idk y = -x^2 + 3

[calm jungle]

uh

yea

[glossy gale]

Oh cool

[calm jungle]

just split it up and then integrate under each ufnctino

[glossy gale]

Niceee

This was fun

Tysm lol

[calm jungle]

np

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[vast valley]

What do i do

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[knotty locust]

is this a test

[vast valley]

No it's hw

[lusty delta]

r u asking for question 1a or

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[ancient edge]

An oil spill in the shape of a circle is expanding on the surface of a pond. The radius of the oil spill is r (measured in cm ) and the area of the oil spill is A (measured in cm^2 ). Suppose that at a certain time, the area of the spill is 30cm^2 , and the radius of the spill is increasing at 0.2 cm per second .

[ancient edge]

would r be 3.09 at the certain time?

[trim thicket]

is this from thomas' calculus?

[ancient edge]

whats tht

[trim thicket]

nvm

thats a book

[ancient edge]

oh

[rough goblet]

,calc (30/pi)^0.5

[solid kiln]

Result:

3.0901936161855

[rough goblet]

yes

appears so

[ancient edge]

Tjanks

[trim jolt]

@ancient edge Has your question been resolved?

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[granite zealot]

Hi ! Could someone kindly help with summing these two up? I am up till here now.

[granite zealot]

I am at this step now.

1/(2n+3) (8n^3+60n^2+142n+105-30n(-1)^n+45(-1)^n /(2n+5)(2n+7)

I’m trying to get to this

Can someone please help? I have been trying and have been stuck for two days now

[neon dirge]

you can start by excluding 2n+3 from both terms

and then getting the common denominator: (2n+5)*(2n+7)

@granite zealot

[granite zealot]

Hello!

Yes I have done that now..not sure if correct

I am at this step now.

1/(2n+3) (8n^3+60n^2+142n+105-30n(-1)^n+45(-1)^n /(2n+5)(2n+7)

^like this?

[neon dirge]

hm would you mind if I instead send the latex version of the steps and you compare them with yours

might be easier than if I untangle it from text :D

[granite zealot]

Yeah sure…idk what’s latex but okay

Anything to help solve as I’m stuck for way too long

[neon dirge]

just fancy math text

so after factoring out

then you prob expanded:

brought together

and then multiplied out (will take a sec):

then summing up and I should be where you area

@granite zealot yop looks good

and now polynomial division

so the numerator div by 2n+3

[granite zealot]

Polynomial division? How can 8n^3 be divided by 2n+3?

[neon dirge]

yes are you not yet familiar with long polynomial division?

it essentially allows you to divide any polynomials with each other :)

we can start by ignoring the latter two terms at first:

this is what we want to calculate

are thee there?

[granite zealot]

Yeap! This is my second time hearing of long polynomial division

[neon dirge]

np, we'll go through step by step

[granite zealot]

Thank you! I was just really confused with how to resolve this

[neon dirge]

the principle is to always take the divisor and multiply it by a factor so that the largest powers and their coefficients match

what do I mean by match? the largest term in the numerator is 8n³

and the largest term in the denominator is 2n

so we can multiply our divisor by 4n²

since 2n * 4n² = 8n³

so now we found the factor we need to match the largest terms: 4n²

we'll now subtract 4n² * denominator from the numerator

so 4n² * (2n+3) = 8n³+12n²

we'll subtract that from the numerator and we get:

and we'll write down 4n² on the side

in the end we'll need all the factors we used

now the same principle again

largest term in the numerator is 48n²

and 2n * 24n = 48n²

so our next factor is 24n

24n * divisor = 24n * (2n+3) = 48n²+72n

we'll subtract that from the numerator again

and we'll note down 24n.

one more time: largest power in the numerator is 70n

2n * 35 = 70n

so we'll subtract once more and we get:

0!

and lastly we look at the factors we used throughout the division: 4n², 24n and 35

the remainder of the division thereby is 4n²+24n+35 🦇

@granite zealot hope the process is somewhat clear :)

[granite zealot]

The remainder becomes the numerator?

[neon dirge]

yup!

[granite zealot]

Is long polynomial division legit?

[neon dirge]

sure

always the same principle

[granite zealot]

Hmmm..let me try on the other end and see if I get it

Hang on

[neon dirge]

sure

I'll write down the remaining steps:

so now we're at

since we've ignored the last two terms in the division, we still need to divide by them

luckily, we can quickly see that we can easily factor out 2n+3 from 30n(-1)^n+45(-1)^n

btw polynomial division would still work here (the factor would be 15(-1)^n)

meaning our only remainder is 15(-1)^n!

🐛

[granite zealot]

What do you mean by we can easily factor out?

[neon dirge]

I'd have argued that one can see that the upper numerator is a multiple of (2n+3) and 15(-1)^n

but if you prefer, you can do a second long division as well, it'd be just one iteration and gives the same result 👍

[granite zealot]

Would it be possible? I gotten 30n+45 after multiplying by 15 to match the highest numerator but there’s a problem. 30n(-1)^n

How can I minus 30n with 30n(-1)^n

[neon dirge]

well let's look at it:

do the highest term above is 30n(-1)^n

and the highest term below is 2n

[granite zealot]

(-1)^n is so annoying

[neon dirge]

yea, our factor needs to include it

2n * 15(-1)^n = 30n(-1)^n

now the biggest terms match :)

and after subtracting you'll see that nothing remains

so 15(-1)^n is our remainder

[granite zealot]

Woah

I see it now

[neon dirge]

which is the desired result, wehee!

[granite zealot]

Is long polynomial division a thing?

[neon dirge]

absolutely

as you've just seen it's super useful

[granite zealot]

It just seems like an out of the blue technique

Does it violate any laws? Are the equations still equal?

[neon dirge]

there's thousands of proofs for it

you're essentially just dissecting the numerator

[granite zealot]

I see..maybe I will read a proof on it! Can I ask one more question?

You’re really good at this

[neon dirge]

a bit like trying to brew the numerator using the denominator 🫧

[neon dirge]

sure

[granite zealot]

I’m having problems with sigma notation

It’s so weird to me

[neon dirge]

kk

do you happen to know a few basics of coding?

it's fine if not, I'd just explain it differently

[granite zealot]

How did they go from the left to the results on the right? I’m curious on this

My question doesn’t involve having to solve this aspect but it annoys me that I can’t figure it out

[granite zealot]

Hmmm..I think I know abit of basics of Python..not sure if it is enough

[neon dirge]

um they defined the numerator and upper bound both as n...

[granite zealot]

Did they use a sigma notation formula?

[neon dirge]

which you aren't allowed to do

[neon dirge]

sigma notation is just another way to express any sum

e.g. here

I define a sum

and a new variable x, which is meant to be natural

and I iterate from x=0 to x=5

[granite zealot]

Wait why? I always struggle when there’s a letter on top

[neon dirge]

and in each iteration I add the content of the sum

[neon dirge]

well the sum n: 1 -> n is undefined

[granite zealot]

I tried plugging in a number on Both side and they add up. That’s why I’m surprised and want to know how they did it

[neon dirge]

since n is not yet defined

[granite zealot]

It seemed legit

[neon dirge]

look you can say f.i. n=1 to 10

[granite zealot]

LHS=RHS

[neon dirge]

then n iterates from 1 to 10

or you can for instance say n=1 to x if x was previously defined

but n=1 to n is just senseless?

it's like n=n to n

where does it start? where does it end?

undefined

[granite zealot]

Wait so what’s up with the RHS? How did they come up with it?

I get what you are trying to say

[neon dirge]

well if the sum notation on the left would be correct

then I could say what they used

but since the sum doesn't make sense the equation is useless

[granite zealot]

Is this some higher level math they used?

[neon dirge]

I can't tell if I don't know what they meant to write into the sum :D

it's like asking me "why does the LHS equal 3?"

[granite zealot]

Ahhh ok..they just wrote this down and wanted a proof via induction so I’m also not sure what else to add

Hahahaha

Right

I thought I was dumb for not being able to figure it out

[neon dirge]

yea maybe they meant to write from n=1 to infinity

then induction would make sense

[granite zealot]

Ahhh..right right..

[neon dirge]

nw, always view sums as:

[granite zealot]

Can I also ask another question?😂

[neon dirge]

x goes from START to END

so both START and END need to be defined

[neon dirge]

absolutely

kk

[granite zealot]

For only the left hand side, why is it when I restricted it with the inequality 0< left hand side in the pic< 1, the number is even?

[neon dirge]

you mean why n must be even if 0<LHS<1 is given?

[granite zealot]

Yes!!

[neon dirge]

because for odd n, the term (-1)^(n+1) is negative

[granite zealot]

That’s right!

[neon dirge]

all other parts of the expression are positive

therefore if n is odd, 0<LHS<1 is false

[granite zealot]

Let me think about what you said

Errrrr

Hmmmm

[neon dirge]

2n+1 is always positive for positive n

2n+5 is always positive for positive n

30 is always positive for positive n