help-10

[trail cloak]

So 25% concentration sodium bicarbonate, in gastric dressing.

And then they tell you that they wanna reduce the hyperacidity of 200 grams

By reducing the hydrochloric acid concentration from 0.5% to 0.3%

So like the gastric dressing is the thing that the patient takes to reduce gastric juice

I presume

I think this needs some molar stuff

And if need be, you can consult the chemistry server in #old-network

[timid silo]

Oh ok

Thank you Vulcanone!

[trail cloak]

[timid silo]

.close

[obtuse pebble]

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[upbeat pivot]

Hello, does anyone can tell me how to do this ? Thanks before 🙏🏽

[obtuse pebble]

@upbeat pivot Has your question been resolved?

[obtuse pebble]

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[autumn ice]

I got stuck in the middle of this:

[autumn ice]

√3 + 1 + tan² 1

I need help with the "tan² 1" part

wait

oh lol

[ruby path]

What are you meant to do

[autumn ice]

nvm I got it

.close

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[timid silo]

<@&286206848099549185>

[timid silo]

these questions still dont make sense to me, i know what to do but sparx video doesnt compliment the questions asked

[wintry swift]

read this.

[obtuse pebble]

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[timid silo]

ohh yes sorry thanks! going to send what i know so far

[obtuse pebble]

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[worn summit]

how do you do this

[timid silo]

hi

[worn summit]

hi

[timid silo]

First of all, how can we find the hourly fee?

(Hint: The problem tells us that it is fixed)

[worn summit]

oh wait i get it now

is it that you decrease 30 each time and then the lowest is fixed

yeah that must be it

ok thank you

.close

[obtuse pebble]

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[timid silo]

?

.reopen

._.

[ruby path]

.reopen

[obtuse pebble]

✅

@worn summit Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

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[frosty cipher]

I could use some help on how to approach this problem and how to solve it. I have an Intro to Real Analysis test tomorrow and I feel royally screwed.

[unique kettle]

Part b should be simple enough if you use test for divergence

[rotund sail]

ya

[unique kettle]

And part a would be implied from that, I don’t know how in depth your teacher wants your answers though, but do you intuitively know the answers?

[frosty cipher]

Well in the class I believe unless stated otherwise we need to prove computationally or a rigorous proof. This question seems more computational.

[unique kettle]

Well then start at part a for the sequence

You can divide top and bottom by n

[thick gyro]

cant normal limit taking be used for part a

we do a lil bit of lhopitals

well lhopital is overkill

[unique kettle]

Might not have that yet in an intro to real analysis course

[frosty cipher]

This is the course description:

Formal treatment of limits of functions and sequences of continuity, including a thorough training in constructing rigorous proofs of the epsilon-delta type. Convergence tests for infinite series. Radius of convergence, differentiation, and integration of Taylor series.

I'm already on academic probation at a 1.97 and this class is one of the more notorious ones in the Math Department.

[unique kettle]

What do you think is the answer to part a

[frosty cipher]

I'm still working on it

[unique kettle]

Have you taken calculus? And if so what would calculus tell you

[frosty cipher]

I've taken up to Calc II and I'm going to take Calc III this spring

It would be to take the limit

[unique kettle]

And what does that give you

[frosty cipher]

Well it would be the limit as n goes to infinity. Do I need to plug anything into n?

[unique kettle]

Wait have you learned about the comparison test yet in your analysis course

[frosty cipher]

I'm pretty sure we have I'm looking back through my notes that I so wisely only numbered via date

We are using the Profs own notes as well as Chapter 11 from Stewart's Calculus by the way

[unique kettle]

Ok well I can think of a slightly bigger sequence that converges

What does that say about that sequence then

[frosty cipher]

The big problem I have in this class which might be because I haven't practiced enough is to look at a problem and not just knowing the answer but the steps needed to begin

[unique kettle]

Well that usually comes with practice

One tip I can give is just recall definitions that look like they would be good for that problem

Most of the time that will get you what you need

[frosty cipher]

Recalling definitions will help in at least clearing the question up?

[unique kettle]

So in this case for part a I would be recalling all of my comparison tests

[unique kettle]

Most of the time

For part b, again I would recall definitions that would answer it, I know one theorem that works perfectly

[frosty cipher]

A definition like this?

[unique kettle]

Yes

That would be good for part a

[frosty cipher]

Which Theorem for part b.

I'm looking through my notes and the book and I can't seem to pin it down

[unique kettle]

In this case I would use the test for divergence

That’s what I believe the theorem is called

It states that if the series for a_n converges then we know lim a_n = 0

Use the contra positive to solve

[frosty cipher]

By the way is this Real Analysis or more Calc?

Because I'm retaking Calc III at a different college and the first time I took it these concepts were covered at the end. This is an entirely separate class.

[clear dove]

More real analysis I would say.

[frosty cipher]

The squeeze theorem?

[unique kettle]

Not quite

[frosty cipher]

I think I found the theorem

If the infinite series is convergent, then the limit of a_n must be 0

Oh shit it diverges?

How?

[thick gyro]

wdym

if the limit isnt 0, then you just end up adding non 0s for infinity, why wouldnt it diverge

[frosty cipher]

I plugged it into Symoblab because I got impatient and it says its divergent

[frosty cipher]

Wait so since the limit doesn't go to 0 but it goes to infinity it would diverge?

[thick gyro]

even if the limit goes to finite number

like 2

it would diverge

[frosty cipher]

So then I would need to solve it?

[thick gyro]

solve for what?

[frosty cipher]

Since it diverges don't I need to solve something?

[thick gyro]

@unique kettle cant you also use the integral test for this?

[thick gyro]

just so we are clear, the series diverges, so isnt that just your final answer?

[frosty cipher]

I'm assuming so

Is this the nth term test?

[thick gyro]

yh

[frosty cipher]

Hey Doc can you prescribe me to trust my gut more

[unique kettle]

Maybe but I don’t think he has integrals defined yet in his analysis course

[unique kettle]

Yes that’s the theorem

Using the contra positive statement gives you a straight forward proof

[frosty cipher]

Alright I will continue on it thanks again you and @thick gyro

.close

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[hidden rose]

i think he just means that there is no restriction for n

so just simply find n

9 / 10 confident 😅

yes I am, what question

I just realized that I don't have an arabic keyboard rn sorry 😅

yes correct

is n "ن" dont forget to say that

and same for b

no it is right

is b = 1.25?

oh i didn't realize 😅

but I did it and it was ez

for me at least idk abt others

2 (1 being easiest)

this is correct yes

3 mins maybe

I didn't have paper tho

I did it semi mentally

are you a professor or something?

I thought you were trying to like put it in a test and only leaving 4 minutes for students to solve it lol

yeah I think this will do

el afw

.close

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[timid silo]

Hello

[plush fractal]

now i help u

so wat do u need help in

[timid silo]

Do you know how to do this problem

[plush fractal]

which one?

[timid silo]

31

It's like trig

And I have to graph a cosecant graph

I just dont get it

Well actually I know how to do it

But the x/2 throws me off

[obtuse pebble]

@timid silo Has your question been resolved?

[obtuse pebble]

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[obtuse pebble]

@timid silo Has your question been resolved?

[timid silo]

.close

[obtuse pebble]

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[obtuse pebble]

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[unreal whale]

hello i have a question on whether ap chemistry is hard

[unreal whale]

im thinking on taking it but i need more suggestions

[timid silo]

it's a very useful class if you want to major in a STEM major such as engineering in college

[unreal whale]

what proffesions is it also good for

besides engineering

[timid silo]

it is definitely difficult, but i found it to be worth it

[unreal whale]

or chem engineering

[timid silo]

several branches of science

[unreal whale]

was it hard

like challenging

[timid silo]

yes

[unreal whale]

how much hours did u have to study a night

[timid silo]

that depends on your school's curriculum

[unreal whale]

for u

how long did it take

also which one is harder ap chem or ap physics

[timid silo]

i dont remember exactly, but expect to spend a lot of time on it

[timid silo]

ap physics 1, ap physics 2, or ap physics c?

[unreal whale]

any

[timid silo]

i took ap physics 1 and thought it was harder

buut

i would absolutely choose ap chem over ap physics 1 or 2

[unreal whale]

u live in the us right?

[timid silo]

yes

ap physics 1 and 2 are more algebra-based, so they will not help you as much with college level physics

[unreal whale]

is it a good idea to take APUSH and AP CHEM

[timid silo]

ap chem will help you with college level chemistry

and if u can handle it sure

[unreal whale]

O

did u take APUSH

[timid silo]

i didn't, but doing good on that ap exam is helpful so you dont ever have to do us hist again

[unreal whale]

ooo

wait so

for ap chem what do i need to prepare for

[timid silo]

your class will probably review at the beginning

[unreal whale]

yea

[timid silo]

dont worry too much about it

[unreal whale]

also

which is harder

ap lang or apush

[timid silo]

idk i didnt take either

[unreal whale]

last question

im trying to become either something in the medical field like a surgeon or entreprenuer

[timid silo]

oh absolutely ap chem

[unreal whale]

what classes do i needa take

[timid silo]

youll want to know biochemistry likely

[unreal whale]

o

[timid silo]

prob ap bio at some point too

[unreal whale]

how about for entreprenurship

[timid silo]

no idea

[unreal whale]

can i take ap bio in college

[timid silo]

ap classes are offered in high schools, not colleges

[unreal whale]

O

[timid silo]

you take them to get college credits early

[unreal whale]

yea

how about

Ap comp science

whats the benefit of that

and the benefit of apush

[timid silo]

good to know basic coding as an engineer / similar job

[unreal whale]

yea

[timid silo]

doing good in apush will make it so you dont have to take us history again in college

[unreal whale]

wait

so if u dont take apush in highschool

youll have to take it again in college?

thats dumb

[timid silo]

well, many college cirriculums require you to take a set of core classes

[unreal whale]

oh

[timid silo]

many include us history

but

apush is often an alternative, if the college you want to lets you skip the class with ap credit then that's good

[unreal whale]

oh

soo

im taking at max 2 ap classes this year

but this year i didnt take any (sophmore year)

is that bad

but ill take ap classes next year to

[timid silo]

doesnt really matter

[unreal whale]

why

[timid silo]

just

when you do take APs, make sure to take the AP Exams so that you can potentially score that college credit by doing well enough

[unreal whale]

at the end of the year wdym?

when i finish with the ap class?

[timid silo]

yes, at the end of the year ap classes have an ap exam

[unreal whale]

oh yea

[timid silo]

if you score high enough, you get college credit for colleges that accept them

[unreal whale]

nice

soo ur in college right?

[timid silo]

yeah

[unreal whale]

what do u reccomend to a sophmore going into junior year

thats gonna take ap chem

and maybe apush

[timid silo]

stay on top of it

[unreal whale]

ye

is it bad if i fall behind

like for ap chem

[timid silo]

try not to fall behind

[unreal whale]

have u ever fallen behind

alr im good thanks for the suggestions

.close

[obtuse pebble]

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[timid silo]

np

[obtuse pebble]

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[winged moth]

is it possible to draw a graph that satisfies these conditions:

[obtuse pebble]

Please don't occupy multiple help channels.

[nocturne minnow]

You already have a channel open, don't open multiple channels

[winged moth]

sorry, please close this

[nocturne minnow]

You can also close it

[tardy epoch]

.close

[obtuse pebble]

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[obtuse pebble]

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[timid silo]

I have been stuck on this for a bit

is it meant to add up to 180

[soft temple]

Yes

[high lily]

wdym by
"it"

[soft temple]

All triangles have angles adding up to 180

[timid silo]

I'm not sure how to work out the other triangle

[soft temple]

Tho I will admit it is smarter not to assume what someone means

[timid silo]

is there a formular

[high lily]

start with the first triangle on the left

can you determine the unmarked angle in that triangle

[timid silo]

I believe it is 51

[high lily]

yes

[timid silo]

I got this by minusing 89 and 40 from 180

[high lily]

can you use that info to determine anything about the triangle on the right?

[timid silo]

are both touching points the same angle?

[high lily]

poor wording

[timid silo]

sorry

[soft temple]

Dont worry bud I think I understand what you mean

Both x’s have to have the same value

[timid silo]

so it would be the same?

[soft temple]

What do you think?

Does x=x?

[timid silo]

so if they are equal how do I work out what they are

sorry lol was never taught this and had it as homework

[high lily]

by vertical angles, the angle (at the centre of the image) in the triangle on the right will also be 51°

well you now have expressions for all 3 angles in the triangle

51°, x°, x°

apply angle sum of a triangle again

and solve for z

[timid silo]

so 180 - 51 = 129
so x is 64.5?

[high lily]

yes

[timid silo]

ooh

thank you

nice

[obtuse pebble]

@timid silo Has your question been resolved?

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[dark tundra]

Can anyone spot my error here

[spice chasm]

?

[gusty tartan]

Bros writing in the margins

[dark tundra]

Got to save the trees

[gusty tartan]

Haha

[fading quest]

its right

[dark tundra]

but i have a negative sign

so it will be -123

not +123

[spice chasm]

i got k=4

[fading quest]

$$-\f{7yk+123}{7(4-y)}=\f{7yk+123}{7(y-4)}$$

[warm shale]

Duh Hello

[dark tundra]

ywah that's right

OH

[dark tundra]

thanks 🙏

.close

[obtuse pebble]

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[cerulean parrot]

Hello,
Am a little bit confused as to how to proceed with this problem

[cerulean parrot]

I've found the average rate of change by taking the integral of the function, which came out to 1-cosx^2
the average rate of change was then calculated out to sin^2sqrtpi/sqrtpi, or 0.5415557

However, I am not sure how to proceed with finding the instantaneous rate of change

Am I supposed to take the derivative of the function at the value between 0 and sqrt(pi)?

[quaint glen]

Find the derivative of f in general, and equate that to the instantaneous rate of change

[cerulean parrot]

Alright, so the instantaneous rate of change then becomes 2cos(x)sin(x)

[quaint glen]

Why is that?

[cerulean parrot]

Took the derivative of 1-cosx^2

Which was the result of the integral above

I'm probably on the completely wrong track here, I do apologize, been a long day

If I graph it out, 2cos(x)sin(x) (simplified to sin2x) has 2 y-coordinates that correspond to the average rate of change

[quaint glen]

No you're correct there, but your average rate of change is incorrect

,w average rate of change of 1-cos(x^2) on [0, sqrt(pi)]

[warm shale]

Results provided by WolframAlpha

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

Requested by Sneaky#0928

[radiant glade]

Isnt that 2xsinx^2 or am i missing something

[cerulean parrot]

How the hell did I screw that up?

average rate of change is f(b)-f(a)/b-a, right?

Thus it would be ((1-cos(pi)) - (1-cos(0)))/(sqrt pi - 0)

I see what I did wrong, my brain is not working right tonight

[radiant glade]

Take derivative and set equal to arc

Mean valu theorem

[cerulean parrot]

I have no solution

2cos(x)sin(x) does not equal 2/sqrt(pi)

[radiant glade]

Im psure that derivative is wrong

[cerulean parrot]

The Y value is over what lies on the curve

[radiant glade]

Cus -cos(x^2) u use chain rule

And get 2x * sin(x^2)

[cerulean parrot]

I see

I kept taking the derivative of cos^2x instead of cosx^2

Alright, we got two points that match up from 0-sqrt(pi)

One at 0.851 and one at 1.673

Going to go out on a limb here and guess this is correct

Two points match on the curve 2xsinx^2 on the line 2/sqrt(pi)

[radiant glade]

Looks good

[cerulean parrot]

Awesome, thank you so much!

[radiant glade]

,w solve \frac{2}{\sqrt{\pi}}=2x * sin(x^2)

Hm

[cerulean parrot]

2/sqrt(pi)

I think

[radiant glade]

O

Forgot

[cerulean parrot]

Hah, i've been making mistakes all night, it's alright

[wooden flint]

hey guys i rlly need help thx

[radiant glade]

Huh

That didnt change

Wait

,w solve \frac{2}{\sqrt{\pi}}=2x * sin(x^2)

[warm shale]

Results provided by WolframAlpha

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

Requested by 4321#5462

[cerulean parrot]

Not sure whats going on here

Interesting

[radiant glade]

Wait

Mm

[cerulean parrot]

Plugged it in to another calculator

Spits this out

[radiant glade]

Ok yea wolfram alpha just being weird

[cerulean parrot]

Indeed

Works out to two solutions

Again, thank you!

.close

[obtuse pebble]

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[boreal condor]

ngl it's kinda hard to read ur work

if u can just type it out

it's the photo being a bit small and my eyes are garbage

[wary marten]

derivative i got sin(xy^2)/(-sinx^2y -12y^3) and after subing (1,pi/2) i got gradient 0.01341

[boreal condor]

srry giv eme like 15 mins

and i can help u

smth came up

[wary marten]

all good

[boreal condor]

why do u have a fraction?

i mean when ur just deriving the formula

before move terms around to get dy/dx

what do u have

[wary marten]

-sinxy(x.dy/dx +y)-12y^3.dy/dx

[boreal condor]

k what's the next step u did

wait i think ik what u did wrong

xsin(xy)≠sin(x^2y)

similarly ysin(xy)≠sin(xy^2)

[wary marten]

oh damn

[boreal condor]

chain rule of cos(xy) is
-sin(xy) * (x*dy/dx +y)

the argument should basically never change in chain rule for trig functions

im sure there are exceptions but those are rare or just simplifying

[boreal condor]

y^3 not 2

otherwise it's good

[wary marten]

rip thanks a lot

[boreal condor]

👍

.close

[obtuse pebble]

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[clever temple]

So can anyone help me with some identities?

So 1-cos2x is equal to 0 right?

[boreal condor]

depends what x is

[clever temple]

Because 1 is equal to

[timid silo]

No its not

[clever temple]

Sin^2x+cos^2x

And we subtract it by

Cos(2x)

[timid silo]

Cos2x is not equal to that

[grizzled shore]

But cos²x ≠ cos(2x)

[clever temple]

Cos(a+a)

This means

Cos(a)*cos(a)-

Oooh

Opposite sign

Nvm

I was thinking

[grizzled shore]

What is this

[timid silo]

Cos2x = cos^2x - sin^2x

[clever temple]

Yeah i thought +

Instead of -

Forgot

About it

[timid silo]

Yeah

[clever temple]

Yeah mb

Well thanks for correcting me

[timid silo]

Np

[clever temple]

Helped a lot

.close

[obtuse pebble]

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[lone viper]

can someone explain what a differentiate is and how to do this problem step by step

[brisk matrix]

do you know what a derivative is?

[lone viper]

yes

[brisk matrix]

they’re asking you to find the derivative

[lone viper]

how would i find the derivative of sec and tan

[brisk matrix]

write them in terms of sin and cos

then use the quotient rule

[lone viper]

why the quotient rule

is there a reason

[brisk matrix]

because you have a quotient of two functions

[lone viper]

got it so

which one would be f(x) and which would be g(x)

also how would I figure out f prime and g prime

[radiant glade]

this is just sin theta

[radiant glade]

product rule or simplify

[lone viper]

wdym sine theta>?

so i dont do the quotient rule

[radiant glade]

$\frac{1}{cosx} \cdot \frac{sinx}{cosx} = sinx$

[warm shale]

4321

[radiant glade]

u dont have to

but u can for sec and tan if u dont alr know how to differentiate them

but secx tan x simplifies to sinx

wait no

[lone viper]

got it

[radiant glade]

im slow

[lone viper]

oh

[radiant glade]

it does not simplify

[timid silo]

It doesnt

[radiant glade]

use product rule

[lone viper]

?

[radiant glade]

either

sec or tan

product rule works either way

[timid silo]

Anyone

Doesn't matter

[lone viper]

so i should use sec and tan instead of this

[timid silo]

Yes

[lone viper]

also so if i put sec as f(x) it would also be fprime?

[radiant glade]

you could use that itll just take alot of thing

[radiant glade]

no

derivative of sec is secx tanx

[timid silo]

Check the differentiation of trigonometric identities on google

[radiant glade]

or like

find them urself

using quotient rule

i think thats what your teacher wants

[lone viper]

got it thanks

❤️

.close

[obtuse pebble]

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[lone bolt]

whats the answer to this? the bottom was my answer but it was incorrect

[radiant glade]

what teacher is giving these annoying ass problems

[timid silo]

product rule 💀

[lone bolt]

calculus -> mymathway program

online

pearson i think

[radiant glade]

mf put a 125

like who got the time to multiply that

[lone bolt]

pulse

i used product rule

[radiant glade]

its product and quotient

[lone bolt]

then to find derivate of g prime, i used quotient rule

[timid silo]

nah im skulling because the problem is horrible

[radiant glade]

yea

u probably messed up simplifying

[lone bolt]

if i show you my notes

[radiant glade]

cuz some wacko decided to put a 125

[lone bolt]

it will be messy af

[timid silo]

that’s ok

[lone bolt]

4321 look at this

pulse, ignore this

[quaint sequoia]

mymathlab can be the biggest troll
it always feels like it’s being passive aggressive when you get one wrong
the message changes
“that’s incorrect”
“try again”
“give it another try”

[lone bolt]

this is just for the memes

[radiant glade]

if u aint getting credit for taking it online from this place honestly just use khan academy

[lone bolt]

it accepted my square root answers like 6 times before this

[radiant glade]

bro fr who be using these sites

[lone bolt]

but on this particular problem it didnt 💀

[radiant glade]

like how are these places still running

[timid silo]

no shot

😵‍💫😵‍💫😵‍💫😵‍💫😵‍💫

[quaint sequoia]

my calc prof used to assign 20-30 question problem sets with this software

[timid silo]

nah bro

nah nah nah nah nah

[radiant glade]

nah bruh schools actually make you use this

[timid silo]

for once i appreciate cengage

[radiant glade]

what is cengage

[timid silo]

a homework website a lot of colleges use

[lone bolt]

ill talk to my prof

[timid silo]

it gives me 5 attempts per question and i don’t even have to simplify lmfao

[lone bolt]

to make the switch

[radiant glade]

juse use khan academy a.

[lone bolt]

whats that

[timid silo]

literally this is bs

[radiant glade]

https://www.khanacademy.org

[lone bolt]

the thing is

[quaint sequoia]

bro
screw cengage
i bought a textbook last semester for a math class and then the prof tells us on the first day that we need cengage
and you have to buy the ebook there’s no option to just get the hw

[lone bolt]

they assign hw

so i have to do it

from this website

[quaint sequoia]

so i just threw away $80

[radiant glade]

i thought they dont do that in college

[lone bolt]

mymathlab

or i should say

mymethlab

[timid silo]

i always get the ultimate subscription

[quaint sequoia]

what college did you go to

[timid silo]

texas a&m

[radiant glade]

none im 14 bruh

[quaint sequoia]

rip
yeah they do that all the time

[timid silo]

i’m a freshman

[quaint sequoia]

calc 3 prof didn’t make us use it at least

[lone bolt]

it feels great that im not alone in these programs being garbage, but i do want to advance my studies if we dont mind
can we please get back to this problem? thank you

[quaint sequoia]

of course i threw away the $120 i spent on cengage too because i failed the class and have to retake it

[radiant glade]

i am trying to find a calc 3 course that gives credit but is just tests no annoying teachers

[quaint sequoia]

so 200 in the hole plus the price of the class and nothing to show for it :))))

thats college @radiant glade

[radiant glade]

good thing i am 4 years away from it 👍

i do not get how they can charge you ridiculous amounts of money and use some wack ass site for the homework

[lone bolt]

@timid silo are you finding the answer to this horrid problem?

[radiant glade]

theres no way shit like this can be happening

[radiant glade]

https://www.derivative-calculator.net/ just put it in this

if you understand product and quotient rule its honestly just that trashy site

[lone bolt]

nah man, i want to do this correctly

im still learning

and i make mistakes

[quaint sequoia]

,w derive (x^3-125)((x^2+4)/(x^2-4))

[lone bolt]

i use "mathway"

[warm shale]

Results provided by WolframAlpha

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

Requested by Dogecode#7321

[lone bolt]

because its easy all you do is take a picture and it solves it for u

[lone bolt]

-16x^3?

[timid silo]

oh it's $-16x^3$ not $-16x^2$

[warm shale]

pulse

[quaint sequoia]

yeah

[lone bolt]

yup

i see it

thank you pulse

[timid silo]

np

[lone bolt]

im still learning

i make mistakes

[timid silo]

this problem is absolutely terrible though

[obtuse pebble]

@lone bolt Has your question been resolved?

[obtuse pebble]

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[visual apex]

Hi

[visual apex]

I'm stuck

I don't know whether to use sin or cosine

[sonic anchor]

it doesnt matter just choose 1

i think cos might be easier though

[obtuse pebble]

@visual apex Has your question been resolved?

[obtuse pebble]

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[bright geyser]

i've forgotten quite a lot of shit in regards to differentiation etc

so first off, x^3 will be the dominant term regardless of sign

so it will go to positive inf and negative inf, thus the range of g will be the same as the codomain

wait nah its pretty easy i think

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[bright geyser]

injectivity is easy to disporve

no idea how to formally prove surjectivity

its obviously inuitive

but idk how to do it formally

[obtuse pebble]

@bright geyser Has your question been resolved?

[bright geyser]

<@&286206848099549185>

but is this an effective proof

sorry for dogshit photo can’t login into discord on ipad

[radiant glade]

What app is this

[bright geyser]

good notes

[radiant glade]

Ok thanks

[brisk matrix]

@bright geyser that does show surjectivity

[obtuse pebble]

@bright geyser Has your question been resolved?

[bright geyser]

cheers

[obtuse pebble]

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[rancid wagon]

hi, I need someone to help me continue thinking through this problem that is tricky for me from Spivak

[rancid wagon]

https://cdn.discordapp.com/attachments/903430335474331698/1078191811312488458/372X83I.png

so here I know f'(0) = 0, but I don't know how I would get there with a pen/paper

this guy https://math.stackexchange.com/a/3411583 uses squeeze theorem for a similar problem

but he is using a graphing calculator that makes that easy

how would you approach solving this problem if all you have is pen/paper?

or proving vs solving

the derivative of the first function in the set is $-x^{-2}g(x) cos(1/x) + g'(x)sin(1/x)$

[warm shale]

math

[rancid wagon]

but how do you prove sin (1/x) is differentiable at x = 0 when 1/x approaches inf as x approaches 0

[sonic anchor]

didnt sm1 already help you with this. you just do this but with absolute value

[rancid wagon]

they did but I had to jump on a work call and didn't really get to talk through it

lol

[sonic anchor]

sin is bounded

g'(0)=0

[rancid wagon]

I guess the little abstraction layer is throwing me off

in that second one

if h^2 didn't exist in the equatio

would lim h > 0 of sin(1/h)/h still be 0?

[sonic anchor]

its g(h)/h and we know g'(0)=0

[rancid wagon]

it is g(h)/h, but does g(h) = h? because here one of the hs cancels out

in what they're showing

[sonic anchor]

$\lim_{h\rightarrow 0} \frac{g(h)-g(0)}{h}$

[warm shale]

Køter

[sonic anchor]

this is g'(0) right

[rancid wagon]

yeah

[sonic anchor]

g(0)=0

do you not see it or?

[rancid wagon]

I get what you said, I'm still confused about my problem lol. More literally, the limit def of what I have is:
$\lim_{h\rightarrow 0} \frac{g(h)sin(1/h) - g(0)sin(1/0)}{h}$

[warm shale]

math

[rancid wagon]

and this I don't get how to work out

[sonic anchor]

you are subtracting with 0 in the numerator

f(0)=0

[rancid wagon]

right so it's just the first part

$\lim_{h\rightarrow 0} \frac{g(h)sin(1/h)}{h}$

[warm shale]

math

[sonic anchor]

yes and g'(0)=0

so you have 0*something bounded

[rancid wagon]

yeah the bounded part is clear to me too lol

like if this said 0 * sin(1/h)/h, I'd be done

[sonic anchor]

$\lim_{h\rightarrow 0} \frac{g(h)}{h}sin(\frac{1}{h})$

[warm shale]

Køter

[sonic anchor]

$\lim_{h\rightarrow 0} \frac{g(h)-g(0)}{h}sin(\frac{1}{h})$

[warm shale]

Køter

[sonic anchor]

we use the fact that g(0)=0 and g'(0)=0

[rancid wagon]

and in this equation sin(1/h), is it outside or inside of the limit?

is this

[sonic anchor]

inside

[rancid wagon]

ok

[sonic anchor]

the 1st part is g'(0) which is 0

[rancid wagon]

now I see it

yeah

[sonic anchor]

you use absolute value because it means sin will be bounded

[rancid wagon]

this is also true right?
$\lim_{h\rightarrow 0} \frac{g(h)-g(0)}{h}sin(\frac{1}{h}) = g'(0) \lim_{h\rightarrow 0} sin(\frac{1}{h})$

[warm shale]

math

[sonic anchor]

ye thats basically what you do

[rancid wagon]

so here is the part that is still murky to me

[sonic anchor]

dont have to but it can help

[rancid wagon]

$\lim{h\rightarrow 0} sin(\frac{1}{h})$ is bounded so if it's multiplied by 0 in a larger expression we can say it is = 0, but if it isn't multiplied by 0 in some expression but say multiplied by a number, is the whole limit then undefined?

[warm shale]

math

[sonic anchor]

yes

[rancid wagon]

ok that I think it's all clear to me now

[sonic anchor]

tbh putting g'(0) out is prob not 100% allowed

im not sure though

ok and you also get why using squeeze thm is necessary?

[rancid wagon]

necessary?

I get why it's possible, but we just did this without squeeze theorem right?

[sonic anchor]

hm yea i guess you dont have to

w.e

[rancid wagon]

XD ok

alright cool I think I can close this topic for good

thx dude

[sonic anchor]

no problem

[rancid wagon]

.close

[obtuse pebble]

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[dusty bobcat]

.

[obtuse pebble]

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[dusty bobcat]

close

[obtuse pebble]

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[sturdy wave]

It’s system of equations and i need to solve it using substitution

But i have no idea how

[ruby path]

Write one variable in terms of the other in the first equation and replace it in the second

[fathom flicker]

would it be substitution to just add the two equations together?

[ruby path]

That's probably the best way to solve this but I don't think it's substitution

[sturdy wave]

It is

[sturdy wave]

And idk what this means

[ruby path]

Are you sure?

[sturdy wave]

Yeah it said to use substituting when solving

[ruby path]

But adding equations isn't substitution :/

[fathom flicker]

he means to solve for either x or y by itself, and then plug whatever you solved it for into the other one

ignore what I said, it is a different method that isn't substitution

although it could be used to solve

[ruby path]

$$x + y = 90$$
$$x = 90 - y$$

[warm shale]

NEONPerseus

[ruby path]

@sturdy wave do you understand what I did here?

[sturdy wave]

Yes you changed it

[ruby path]

Yup now replace x with this expression in the second equation

[sturdy wave]

$$(90-y) + y = 90$$

[warm shale]

Debby Kramer

[sturdy wave]

Like that

[fathom flicker]

dang

lol

yes like that

[sturdy wave]

What about the 18

[ruby path]

In the second equation not the first though

[sturdy wave]

I don’t get it

[ruby path]

$x - y = 18$

[warm shale]

NEONPerseus

[ruby path]

Put the value of x from here into here

[sturdy wave]

$$(90 - y) - y = 18$$

[warm shale]

Debby Kramer

[sturdy wave]

Now what

[ruby path]

There we go

Now solve for y like normal

[sturdy wave]

I have no idea how to do that

[ruby path]

You haven't been taught algebra?

[sturdy wave]

No i have but i forgot it really easily because it’s like a review thing

But i have to do it

[ruby path]

Can you try and isolate y?

[sturdy wave]

What does that mean

[ruby path]

Using mathematical operational try to get y be itself

We want y = (something)

[sturdy wave]

Wouldn’t y just equal 1

[ruby path]

No, why?

[sturdy wave]

Oh i always thought if it was just a number it meant 1

[ruby path]

$-y - y$

[warm shale]

NEONPerseus

[ruby path]

What would this simplify to?

[sturdy wave]

-y

[ruby path]

[sturdy wave]

Is that right

[ruby path]

Well no

[sturdy wave]

Oh

[ruby path]

What would $a + a$ be?

[warm shale]

NEONPerseus

[sturdy wave]

A

[ruby path]

What would $1 + 1$ be

[warm shale]

NEONPerseus

[sturdy wave]

2

[ruby path]

Good

What would $4 + 4$ be?

[warm shale]

NEONPerseus

[sturdy wave]

8

[ruby path]

Do you see a pattern?

[sturdy wave]

No

[ruby path]

Something + something = 2 times something

[sturdy wave]

Oh

[sturdy wave]

-2y ?

[ruby path]

Yup there we go

[sturdy wave]

Ok good

[ruby path]

$90 - 2y = 18$

[warm shale]

NEONPerseus

[ruby path]

This is what we have now

So what would we need to get rid of to get y by itself?

[sturdy wave]

Divide it by itself

And divide it with 18

[ruby path]

Sorry?

We want to isolate y

We want y = something

[sturdy wave]

I mean like

[ruby path]

So what would we need to get rid of

[sturdy wave]

-2y/-2 and then that cancels out so we get rid of that and divide 18/-2

Which is -9

Or no

[ruby path]

That works

[sturdy wave]

Oh

[ruby path]

But you also need to divide 90 by 2

What you do to one side needs to be done to the other side

$-45 + y = -9$

[warm shale]

NEONPerseus

[ruby path]

Now what we do to get y by itself

[sturdy wave]

Uhh

Idk

[ruby path]

You could add, subtract, divide, or multiply

What do you think is more useful here

[sturdy wave]

Adding

[ruby path]

Correct

And what would we add?

[sturdy wave]

-45 and -9

[ruby path]

That wouldn't do anything

[sturdy wave]

Oh

[ruby path]

How would we get rid of -45, what do we need to add?

[sturdy wave]

We add 45 to -45

[ruby path]

Good

And we need to do that to both sides

$-45 + y + 45 = -9 + 45$

[warm shale]

NEONPerseus

[ruby path]

$y = -9 + 45$

[warm shale]

NEONPerseus

[ruby path]

Can you work this out now?

[sturdy wave]

Yes

Y = 36

[ruby path]

Yup good

Now we have y

$x + y = 90$

[warm shale]

NEONPerseus

[sturdy wave]

And x = 54 because we add the 18 to 36

Right

[ruby path]

Yeah that works

You're good

[sturdy wave]

Thx

So it is now

(54,36)

[ruby path]

Yes

[sturdy wave]

Ok great

Thx

[ruby path]

If you're done you can close the channel with .close

[sturdy wave]

Ok i will i am just going to look over it to help me remember what to do

.close

[obtuse pebble]

Channel closed

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[sand zenith]

can someone pls help me w this

[lost tree]

,rotate

[lost tree]

Do you know how to solve quadratics?

[sand zenith]

hey js trying to solve for k here

yeah ik how to

[lost tree]

Oh u want k?

Well it's simple enough to find k

[sand zenith]

yeah

i got the answers js unsure how to get to it

[lost tree]

How did you get an answer

[sand zenith]

asked a friend plugged it in desmos and it came out correctly

still no idea how to solve it

it’s supposed to be that equation larger than 0 and find the answers for k that satisfy it

[lost tree]

Can you send the original problem

[sand zenith]

that is the original problem?

[lost tree]

Okay so you want to find values for k such that 5kx² - 6kx + k + 4 > 0 ?

[sand zenith]

yes

[lost tree]

Right, let's say you had something like 4k + 3 > 0, can you find the values of k here?

[sand zenith]

yeah i would do 4k = 3

k = 3/4

and as it’s larger than 0

[ruby path]

Use the discriminant of the quadratic in x

It needs to be lesser than 0, for the quadratic to always be positive

[sand zenith]

yeah mb =-3

and as it’s smaller than 0

k = -3/4

or -3/4 < 0

i think that’s right?

[lost tree]

Well k should not be equal to -3/4

If k is equal to -3/4, then when you put it in 4k+3, it simplifies to 0

[sand zenith]

or would it be -3/4 < k

[lost tree]

Bit we want values for k such that 4k+3 is greater than 0

[lost tree]

That would be it

[lost tree]

This is an obvious statement

So you can put any number for k that is greater than -3/4

[sand zenith]

right that makes sense

[lost tree]

Okay now let's say you had 4k + 7k - 3 > 0

How would you find values for k now

[lost tree]

Yes

So in ur inequality

[sand zenith]

yeah js a bit unsure w 2 variables

[lost tree]

You got 5kx² - 6kx + k + 4 > 0

[sand zenith]

yep

[lost tree]

If you want to know values that k may take so that this inequality is satisfied in terms of x

You can just treat the x's as constants

[sand zenith]

yeah i understand now

got the answer

[lost tree]

Exactly how you treated the numbers 4, 7 here

[sand zenith]

thanks for your help tho

forgot to use the discriminant

[lost tree]

Your problem is kinda unclear to me tho

But okay if you got it

[sand zenith]

yeah but confusing but works out in the end

[lost tree]

Did you want to find the values of k so that this quadratic in x always takes positive values?

[ruby path]

[lost tree]

If that's what you wanted to do you must set the discriminant < 0

[lost tree]

Oh

[obtuse pebble]

@sand zenith Has your question been resolved?

[obtuse pebble]

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[hollow shore]

how is it that a series does not converge if when the sum for 2n-1 terms and the sum for 2n terms is not equal?

[royal basin]

... ??

are you saying that $\sum_{k=1}^{2n-1} a_k \neq \sum_{k=1}^{2n} a_k$ implies $\sum_{k=1}^{\infty} a_k$ diverges...? because that's just false

[warm shale]

Ann

[royal basin]

are you looking at some math problem right now that made you ask this

[hollow shore]

yes

[royal basin]

can you show that please

[hollow shore]

ok wait

[royal basin]

http://xyproblem.info/

[hollow shore]

let's say an infinite series
1/2 - 2/3 + 2/3 - 3/4 + 3/4 - ... - n/(n+1) + n/(n+1) - (n+1)/(n+2) + ...
S_(2n-1) is the partial sum of the first 2n-1 terms and n is any natural number
S_2n is the same
so for S_(2n-1), since there are odd terms, everything cancels out and converges to 1/2

for S_2n, there are even terms, so it becomes
S_2n = S_(2n-1) - (n+1)/(n+2)

[royal basin]

okay, so the first term is 1/2, and the next terms are of the form n/(n+1), with a negative and a positive copy of each?

there's some parentheses you're missing here in a bunch of places.

[hollow shore]

the limits for both sums as n approaches infinity are different

"Therefore, the series diverges."

[royal basin]

okay, so really what you mean is that $\lim S_{2n} \neq \lim S_{2n-1}$.

[warm shale]

Ann

[hollow shore]

is that different from what i said earlier

[royal basin]

yeah bc it sounded like you just said $S_{2n} \neq S_{2n-1}$

[hollow shore]

so why does it happen

[warm shale]

Ann

[royal basin]

not that their limits don't agree

and really this just comes down to a matter of sequences

a series is said to converge if its sequence of partial sums has a limit

in our case, the sequence S_n does not. because it has two subsequences which approach different limits

[hollow shore]

which are the subsequences?

[royal basin]

$S_{2n-1}$ and $S_{2n}$

[warm shale]

Ann

[obtuse pebble]

@hollow shore Has your question been resolved?

[balmy mortar]

consider the sequence
0, 1, 0, 1, 0, 1, ...

[obtuse pebble]

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[sinful saffron]

I need help with finding the area of the shape. I'm not too sure on how to do it, I've got the answer for it from the answer sheet and I'm really confused on how they got it. I need help with part c.

[proven oar]

Since there is no formula for area

for a shape like this

How could you split it up

so you get shapes with formulas you know

[sinful saffron]

I could probably split them up into squares/rectangles

[proven oar]

exactly

so then for the total area of the shape