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[silent vine]

im still a bit confused on how to negate statements

[silent vine]

if i said uh

For all x in the reals, there exists y in the reals s.t. y => x, what would be the negation for that and how would i go about it

[normal tapir]

you negate piece by piece

so negate the quantifiers one by one, and then negate the predicate

for all x becomes there exists an x

there exists y becomes for all y

y >= x becomes y < x

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[silent vine]

And would exists become for all

[normal tapir]

yea

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[empty saddle]

How to do the two blanks?

For F(-1), i understand u just sub -1 into x and then u have f(-1) - f(1) (which equals -4 which is correct). But this somehow doesnt work for F(5)? (i.e. i assume that f(5)-f(1) = -5 which but it isnt correct)

Also how to determine the M for this im so bad at looking at graphs that are already differentiated

[half edge]

For F(5)

Where are u plugging in the 5

[empty saddle]

uhh into the given values cus u know for F(5) u intergrate from 5 to 1 and then i assume u plug 5 into f(x) which is -1 for this question

[half edge]

Yea u have to find

$\int_1^5 f(t) dt$

[thorny flame]

Stephen

[half edge]

So how would u do this

[empty saddle]

f(5) - f(1)?

[half edge]

No

[empty saddle]

😦

[half edge]

What does the integral represent , graphically

[empty saddle]

like the area?

[half edge]

Yes, area under the curve

So u have to find the area under the curve of f from x = 1 to x = 5

[empty saddle]

yes

[half edge]

I think u just got lucky with F(-1) lol

[empty saddle]

uhh

sure? but how do i do this properly then

😂

[half edge]

Find the area under the curve

Break it up into shapes

Like rectangles and triangles

Then add up the positive area

And subtract from it the negative area

[empty saddle]

that wont make sense for F(-1) tho

[half edge]

So if we wanted F(-1), it’d be

[empty saddle]

if u break it up into a large ass triangle the base is like

2

and then height is 3

6/2? which is wrong

nvm

im yapping

ahhh

[half edge]

$\int_1^{-1} f(t) dt = -\int_{-1}^1 f(t) dt$

That’s how u would do F(-1)

[empty saddle]

ok but shouldnt the area

always be positive

why is it negative

for F(-1)

[thorny flame]

Stephen

[half edge]

It’s negative because we switch the bounds

[empty saddle]

ahh

[half edge]

Even tho the triangle is above the x axis

We switch the bounds and evaluate

So it becomes negative

[empty saddle]

wait thats so much effort tho 😂

its 2 triangles and 1 trapezium

surely theres a better way?

[half edge]

It’s really not that bad

U can do it in ur head

But I recommend doing it on paper

Or on Ms paint or something

[empty saddle]

alr thx 😂

i cant belive i forgot i can use the area of shapes fml

can u help with the maximum value of M tho?

[half edge]

Alright

So how is M defined?

What does it say

[empty saddle]

uhh it says

M is a global maximal value within the interval

[half edge]

Good

In calculus, how do we usually find global maximums and minimums

[empty saddle]

differentiation

[half edge]

Yes

So if M is a global max of F, what should we do to F

[empty saddle]

differentiate f?

[half edge]

Differentiate F (not f)

[empty saddle]

no i mean the

capital F

yea

[half edge]

Yes

[empty saddle]

typo 😂

[half edge]

Ok

So let’s look at this

So

$F(x) = \int_1^x f(t) dt$

[thorny flame]

Stephen

[half edge]

Yes?

[empty saddle]

yes

[half edge]

$F’(x) = \frac d{dx} \int_1^x f(t) dt$

[thorny flame]

Stephen

[half edge]

Does that make sense

[empty saddle]

yes

[half edge]

How I went from the first to the second

Ok

Does the right side of the 2nd equation look familiar to u

How would u solve that

Finding derivative of an integral

[empty saddle]

isnt that

just uh

f(x)

or something

[half edge]

Yea lol

And how do u know that

[empty saddle]

idk i remembered it 😂

[half edge]

Fundamental theorem of calculus

Remember that?

[empty saddle]

oh right

fml

[half edge]

Ok so when we differentiate something to find critical points in calculus

What do we do after we take the derivative

[empty saddle]

like we make it equate to 0

[half edge]

Exactly

So set f(x) = 0 and solve

[empty saddle]

so F'(x) = 0 ?

and u solve for x?

[half edge]

Well F’(x) = f(x) = 0

But we need to use f(x)

Cuz that’s given to us

What is f?

[empty saddle]

ah right

[empty saddle]

wut do u mean?

[half edge]

I mean, what do they tell us about f

How is it represented

[empty saddle]

like composed of three line segments?

and thats its continuous over the interval

[half edge]

Yea it’s the curve on the graph

So how would u find where f(x) = 0

[empty saddle]

at x = -1?

cus it tells us that?

[half edge]

Good observation, so that’s 1 critical point. But there’s also another place where f(x) = 0

[empty saddle]

x =3.5?

im assuming its 3.5 ig

[half edge]

Yea unfortunately they don’t tell us so we have to assume

So now let’s think about this

We want to maximize F(x)

F(x) is area under curve from 1 to x

Maximizing F(x) gives us F’(x) which is equal to f(x)

f(x) = 0 at x=-1 and x = 3.5

That means

Either F(-1) is a maximum

Or F(3.5) is a maximum

How would you go about doing this?

[empty saddle]

uh using the integral? like either set x = 3.5 or -1? and find which one is larger ?

[half edge]

Sure u can do that

That’s a valid solution

[empty saddle]

uh

so u mean theres another way of doing it?

[half edge]

Yes, but it requires some more thinking

I’ll explain it after u do the integral stuff if u want

[empty saddle]

oh ok

ayy i got the correct answer

[half edge]

Nice

[pulsar patrol]

yo mind if u help me after u finish helping this person

[half edge]

I’m not sure how to do ur problem

[pulsar patrol]

damn aight

[empty saddle]

thanks

.close

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[random dirge]

for a given set of data, the median is just basically the middle number right? then whats the difference between that and the median measured from the cummulative frequency graph?

[topaz sinew]

@random dirge Has your question been resolved?

[woeful stone]

so the cummulative frequency graph involves a graph

the median is a statistic

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[random dirge]

.reopen

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✅

[random dirge]

but both gives a median

but which is better?

for example this question

do we need a cumulative frequency to find the med?

[woeful stone]

no cumulative frequency graph is need. The best way is to directly calculate the mean

[random dirge]

then in what cases do we need a cumulative frequency

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[jade thunder]

What’s wrong with it

There’s 2 correct answers and that’s one of them

[cold fog]

i have to enter the values in an online test and they are shown as incorrect, so i assume i have done something wrong

[jade thunder]

Can you show the test?

Also we are not here to help you cheat on your test, not saying that’s what you’re doing but so you know

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[mossy garnet]

Let $f: \mathbb{R} \to \mathbb{R}$ be defined as $\$

$f(x) = \begin{cases}
\frac{\cos(6x) - 1}{\ln(6x+1)} & \text{if } x > 0 \
-3x & \text{if } x \leq 0 \
\end{cases}$ $\$

Determine if \textit{f} is differentiable at $x = 0$, and if so, find $f'(0)$

[ivory sorrel]

Check for continuity first

[thorny flame]

milanesa de pollo

[mossy garnet]

I mean when we evaluate the limit of \textit{f} as $x$ approaches $0$ from left side is using the second case of the piecewise
$-3 \cross (0) = 0$
right side is using first case
$\frac{cos(0) -1}{\ln{1}}$

[thorny flame]

milanesa de pollo

[mossy garnet]

please help

[ivory sorrel]

hmm

just a minute

[mossy garnet]

how to save the limit

this is undetermined form

e.g. 0/0

is it possible without using lhopi, but just using limit algebra

[ivory sorrel]

write cos(6x)-1 as 2sin^2(3x)

[mossy garnet]

what formula is waht you using

im new to trigonometry

[ivory sorrel]

$cos(2x)=1-2sin^2(x)$

[thorny flame]

ƒ(Why am. I here)=misery

[ivory sorrel]

unfortunately the limit is 0

so you have to find the left hand derivative by hand

[mossy garnet]

,, \lim_{x \to 0} \frac{1-2sin^2(3x) - 1}{\ln(6x+1)}

[mossy garnet]

is it possible without using lhopy

[ivory sorrel]

yes

[mossy garnet]

wait

ln(0 + 1) = 0

@ivory sorrel

how to save the limit?

[ivory sorrel]

$cos(6x)=1-2sin^2(3x)$

[thorny flame]

ƒ(Why am. I here)=misery

[mossy garnet]

my bad

but still

waht about the denominator

?

@ivory sorrel

[thorny flame]

milanesa de pollo

[ivory sorrel]

good

[mossy garnet]

what about denominator doe?

[ivory sorrel]

ln(ax+1)/ax=1 as x tends to zero

[mossy garnet]

,, \lim_{x \to 0} \frac{-2sin^2(3x) }{\ln(6x+1)}

[thorny flame]

milanesa de pollo

[mossy garnet]

this is still undeterminate form

both are 0.

[ivory sorrel]

$-\frac{\frac{2\sin^2\left(3x\right)}{\left(3x\right)^2}\left(3x\right)^2}{\frac{\ln\left(6x+1\right)}{6x}6x}$

[thorny flame]

ƒ(Why am. I here)=misery

[mossy garnet]

so what?

[ivory sorrel]

do you know the limit of sin(ax)/ax at x=0

[mossy garnet]

ahh yes

its 1

correct?

[ivory sorrel]

yeah

[mossy garnet]

but this is still undefined

. . .

[ivory sorrel]

ln(ax+1)/ax

[mossy garnet]

no idea about that

,w ln(1)/0

[thorny flame]

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[mossy garnet]

. . .

[ivory sorrel]

ln(x)/x at x=0

,w ln(x)/x at x=0

[thorny flame]

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[ivory sorrel]

wait

,w ln(1+x)/x at x=0

[thorny flame]

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[ivory sorrel]

wait

what the heck

https://math.stackexchange.com/questions/690898/determine-the-following-limit-as-x-approaches-0-frac-ln1xx

see this

[mossy garnet]

is lhopy mandatory?

otherwise I will just use lhopi for all of my limits

when I have then in undeterminate form

should I use?

[ivory sorrel]

your wish tbh

[mossy garnet]

so it is required?

,w differentiate cos(6x) - 1

[thorny flame]

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[mossy garnet]

,w differentiate ln(6x +1)

[thorny flame]

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[topaz sinew]

@mossy garnet Has your question been resolved?

[neon iron]

divide numerator and denominator by 36x^2

for checking contuinity

[mossy garnet]

?

but we already checked continuity

we found that both right side and left side of the limit is zero

after first differentiatiation of the numerator and denominator of the function of the right side limit

we get 0

[neon iron]

ok nice

what do you want to do now

[mossy garnet]

I still havent find out if its differentiable,

first thing about checking if the function is differentiable is to do this continuity check but then we have to apply the definition of a derivative as a limit and check if both sides of the limit are the same I think

[neon iron]

yes apply it

or directly differentiate and check if derivative is continuous at 0

\

[mossy garnet]

exactly

im doing that but im lowkey hardstuck

[neon iron]

show steps

[mossy garnet]

Do we need lhopi here

I started with the right side one

<@&286206848099549185>

im close

ahh one sec

,w differentiate (cos(6x) - 1)

ahhh

my bad

[thorny flame]

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[mossy garnet]

,w differentiate (xln(6x+1))

[thorny flame]

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[mossy garnet]

ok

,, \lim_{x \to 0^+} \frac{\cos(6x) - 1}{x\ln(6x+1)} = \lim_{x \to 0^+} \frac{-6\sin(6x)}{\frac{6x}{6x+1} + \ln(6x+1)}

[thorny flame]

milanesa de pollo

[mossy garnet]

,w sin(0)

another time lhopital?

alright

,w differentiate -6sin(6x)

[thorny flame]

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[mossy garnet]

,w differentiate ((6x)/(6x+1) + ln(6x+1))

[thorny flame]

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[mossy garnet]

,, \lim_{x \to 0^+} \frac{\cos(6x) - 1}{x\ln(6x+1)} = \lim_{x \to 0^+} \frac{-6\sin(6x)}{\frac{6x}{6x+1} + \ln(6x+1)} = \lim_{x \to 0^+} \frac{-36\cos(6x)}{\frac{12(3x+1)}{(6x+1)^2}}

[thorny flame]

milanesa de pollo

[mossy garnet]

,, \lim_{x \to 0^+} \frac{-36\cos(6x)}{\frac{12(3x+1)}{(6x+1)^2}} = \lim_{x \to 0^+} \frac{-36(6x+1)^2 \cos(6x) }{12(3x+1)} = -3

,w ((-36(0+1)^2 )cos(60))/(12(0 + 1))

[thorny flame]

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milanesa de pollo

[mossy garnet]

,, f'(a) = \lim_{x \to a} \frac{f(x) - f(a)}{x-a}

[thorny flame]

milanesa de pollo

[mossy garnet]

Let $f: \mathbb{R} \to \mathbb{R}$ be defined as $\$

$f(x) = \begin{cases}
\frac{\cos(6x) - 1}{\ln(6x+1)} & \text{if } x > 0 \
-3x & \text{if } x \leq 0 \
\end{cases}$ $\$

Determine if \textit{f} is differentiable at $x = 0$, and if so, find $f'(0)$

[thorny flame]

milanesa de pollo

[mossy garnet]

so it is differentiable

and f'(0) = -3

,, \lim_{x \to 0^-} \frac{f(x) - 0}{x - 0} = \lim_{x \to 0^-} \frac{-3x}{x} = -3

[thorny flame]

milanesa de pollo

[mossy garnet]

.clsoe

.close

[topaz sinew]

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[neon iron]

i need to find f(x)

[ivory sorrel]

tried finding the limit?

[neon iron]

that is where i am stuck

[ivory sorrel]

what are you getting

!show

[topaz sinew]

Show your work, and if possible, explain where you are stuck.

[neon iron]

no where

@status

!status

[topaz sinew]

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

[neon iron]

1

[ivory sorrel]

hmm

x^1/infty tends to 0

is the answer not 0

[chilly walrus]

no

i just graphed it and now i am even more confused

[neon iron]

you got the function ?

[chilly walrus]

yes

i don't have the solution tho

but i know the answer

[neon iron]

even i have the answer

i just didn't undersand it

solution*

[chilly walrus]

ah

[chilly walrus]

are you allowed to use l'hopital?

[neon iron]

yes

[chilly walrus]

try doing it with $\lim_{n\to \infty} \frac{(x^{1/n}-x^{1/(n+1)})}{\frac{1}{n^2}}$

[thorny flame]

artemetra

[neon iron]

give me a minute

[chilly walrus]

remember that you differentiate w.r.t. n, not x

[neon iron]

i am getting 0

i substituted h tends to 0

in place of 1/n

[chilly walrus]

wait what

send pic

[neon iron]

1 min

sorry for the shabby work

and don't mind the top integral

@chilly walrus

help?

[chilly walrus]

it's not n^2 in the denominator

it's 1/n^2

[neon iron]

that is a h(sorry for that)

i substitued 1/n as h

[chilly walrus]

also that's the most insane way i've seen someone write x lol

it looks like æ

[neon iron]

let me send it again

[chilly walrus]

nah i can read it

something went wrong

oh nvm

nothing did

[neon iron]

i represent x in that way so i don't get confused between multiplication symbol and x

[topaz sinew]

@neon iron Has your question been resolved?

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[midnight slate]

translate?

solve for x ?

[polar dew]

@midnight slate Yes

@polar dew Solve the equation

[ivory sorrel]

let sin(x)=u

then what is cos(x)

[polar dew]

√1-u²?

[ivory sorrel]

yes

[midnight slate]

the 3 here is spoiling the fun here
otherwise it would be better to deal with trig only

[polar dew]

@ivory sorrel then how do i find this?

[ivory sorrel]

cross multiply

square both sides

and solve

[polar dew]

Ок

[polar dew]

I cant do it

What did i did wrong

[topaz sinew]

@polar dew Has your question been resolved?

[topaz sinew]

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[polar dew]

Solve the equation

I really cannot understand this

[agile heron]

how far have you got?

[dim pulsar]

Maybe put everything under a common denominator or clear them

[agile heron]

have you tried using complex eponentials?

[polar dew]

I dont know about it

Im in 11 th grade

[agile heron]

(e^ix -e^-ix)/2i = sinx

(e^ix +e^-ix)/2 = cosx

that whay you can work with exponentials that they are way easier to manipulate than sines and cosines

[polar dew]

Thats some another level to me

[topaz sinew]

@polar dew Has your question been resolved?

[topaz sinew]

@polar dew Has your question been resolved?

[polar dew]

If someone cpuld solve it pls dm me im really curious about it

[topaz sinew]

@polar dew Has your question been resolved?

[topaz sinew]

@polar dew Has your question been resolved?

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[stark lichen]

that's the question and the answer i just need to understand why

[restive inlet]

area of sector - area of triangle

angle can be determined from the side lengths of the triangle

[stark lichen]

the side lengths are all 8

so how do i find the angle

hello? :')

[hidden bison]

8 = 2Rsinθ

and the angle is 2θ

[stark lichen]

what does that mean 😭

[stark lichen]

so how do i find θ

[hidden bison]

8 = 2Rsinθ

8 = 16sinθ

sinθ=1/2

θ=30°

[stark lichen]

where'd the 1/2 come from

oh wait im stupid

i figured it out

nvm

[hidden bison]

that is for all situations, but since the side lengths are all 8 in this problem, you can easily find that it is 60

[stark lichen]

ok

um

what exactly did you do to "sinθ=1/2"

[hidden bison]

what do you mean

how to get theta 30

[stark lichen]

how did you get theta 30 from that

[hidden bison]

it just i remember sin30=1/2

[stark lichen]

oh do you need like a chart or something

[hidden bison]

no

you can easily remember some special values

[stark lichen]

ah

so now that i know what the theta is what do i do with it

[hidden bison]

area of sector - area of triangle

[stark lichen]

how do i find the area of sector

[hidden bison]

$\frac{60}{360}*r^2\pi- \frac{\sqrt{3}}{4}r^2$

[thorny flame]

pubuyun

[hidden bison]

$\frac{60}{360}*r^2\pi$ is the area of sector

[thorny flame]

pubuyun

[hidden bison]

$\frac{\sqrt{3}}{4}r^2$ is the area of triangle

[thorny flame]

pubuyun

[stark lichen]

can u tell me what the area of the segment is to see if i got it right?

[hidden bison]

32pi/3

[stark lichen]

so 33.49?

[hidden bison]

you dont need to calculate it

just keep it 32pi/3

[stark lichen]

ah ok

and the area of the triangle?

[hidden bison]

you've already known the answer

12sqrt3

[stark lichen]

how do you get 12sqrt3 from the this formula

sry if im dumb i suck at math

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[stark lichen]

thx i think i got it now have a nice night <3

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[floral rampart]

I have a bit of a philosophical question about solving certain integrals. The examples I have in mind are the ways of solving $$\int_0^{2\pi} \sin^2 x \text{ and } \int_S (x+y)^2 + (y-z)^2 dS,$$ where $S$ is a sphere. In each case, the, in my opinion, "correct" way to solve it is add a function(s) in order to actually be integrating a constant over the region ($cos^2$ in the first case, cyclic shifts of the variables in the second). So here comes my question: what kinds of function can you do this for? There's two things I need: firstly to have some nice sum of functions to some constant and secondly each function in the sum has equal area (which we can recognize by being symmetric)

[thorny flame]

Zander

[floral rampart]

Is there some sort of way to identify all the function families of this type (at least for single variables maybe)?

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[round cove]

f[x] is never a field even though f is a field. what does this mean

[round cove]

or how do i prove this? (its merely stated in my notes)

[vernal matrix]

E.g. non constant polynomials don’t have a multiplicative inverse

[vernal matrix]

(if you wanna directly show it, choose some easy non constant poly, assume it’s invertible and get some contradiction from that)

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[hollow bramble]

Find all "a", for each of which there are 3 roots, that make an arithmetic progression.
x^3 + 6x^2 -4x + a = 0

[ocean fjord]

let the 3 roots be p - q, p and p + q

you can find p using vietas

and since p is a root, f(p) = 0 you can find a using that

[hollow bramble]

Isn't Vietas used for quadratic equations?

[ocean fjord]

it can be used for any polynomial

sum of zeroes one at a time is always -b/a regardless of the degree

sum of zeroes two at a time is c/a (for a quadratic this is just the product)

sum of zeroes 3 at a time is -d/a (product of zeroes for a cubic)

and so on

note that the signs alternate

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[cinder rapids]

@hollow bramble instead of just marking that your question hasn't been resolved, tell sir gareth what you don't understand

he's given you an explanation and if you don't say what you don't understand about it, people wont respond to help because it looks like you haven't seen it, but you marked that you haven't had a confirmation on your question

[hollow bramble]

i just solved it rn, marked as X just for being here

a = -24

.close

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[valid star]

wahts wrong?

[chilly walrus]

never ever ever trust chatgpt with math

it physically cannot do math

[valid star]

its made of math😭 but alright

[weak gull]

wolfram alpha

is better

[chilly walrus]

it's a LANGUAGE model, it does no math under the hood, literally an autocomplete on steroids

yes WA is better

[valid star]

i tried but i think im doing something wrong

[chilly walrus]

show your work then

[valid star]

hold

for number of periods do i put 100?

[chilly walrus]

why 100

you put 4 cuz it's quarterly

i think

[valid star]

i was thinking 4x25

[chilly walrus]

ah yea

fair enough

[valid star]

but

[chilly walrus]

yes put 100

mb

[valid star]

😭 what is wrong

do i multiply present value by 25?

[chilly walrus]

maybe try using the formula directly?..

[valid star]

like that?

(forgot to close bracket)

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[spice trout]

can someone explain how to do this? i almost completely forgot the topic and im confused

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[modest glen]

Hi can someone help me obtain a function frmo this data table

[modest glen]

I have tried using the inverse of an quadratic equation but that last point is hang out

*hanging out

[half edge]

U can do some polynomial curve fitting if u want lol

[modest glen]

such as?

[half edge]

Maybe that’s not the best idea here tho

[modest glen]

oh okay

[half edge]

Hold on lemme try it

[modest glen]

thank yyou so much

I did some calculation but the last point is still point

*out

[half edge]

Ok imma give u some equations, see if any of these work

[modest glen]

okay

thanks

[half edge]

1: −7.364051717393*10^-11 * x^2 + 0.000211064x + 52.7911

2: 0.397438x^0.425211 + 33.64196

3: 29.8088ln(131.863-0.014057x) - 114.183

Ok I doubt these will work so maybe I should just do them by hand

: (

Alr

The result I will give u will go thru every point, but it may not be a good ‘model’ per se

But let’s see

[modest glen]

wat

[modest glen]

this didnt work

[modest glen]

kinda worked

[modest glen]

def didnt

LOL

[half edge]

Ok yea I just used a site

Let me try the thing posted on the image

[modest glen]

ahhh

okay

[half edge]

Actually just thinking about this

This will give us 8, 7th degree equations

Which can be solved to make one polynomial

[half edge]

But are there solvers out there that can solve this? We’ll see

Ok. This is going to be disgusting cuz the x-values are huge

So we get for the first eqn

a0 = 37

24200: a1 * x + … + a7 * x^7 = 20

140000: a1 * x + … + a7 * x^7 = 55

@modest glen do u see what I’m doing

Like do u understand

[modest glen]

yes

[half edge]

Alr, so just continue this sequence, and then lmk what u get for the equations

I can probably plug it into gpt4

The number before the colon on each line represents the value to be plugged into the x on that line

U get it?

@modest glen

Tag me when ur done

This will get u an equation thru every point

I hope gpt4 can crunch such large numbers tho

Like 1.6million to the 7th power

Honestly u may just need to settle for a logarithmized graph and then curve fit from there, cuz ur x values are extremely large

,w (1600000)^7

[thorny flame]

Results provided by WolframAlpha

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

Requested by northsteve

[half edge]

Yea look

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[brave ingot]

need help

[brave ingot]

for question 8b, where did i go wrong? mine is in green ink by the way

if i want to find the gradient of the tangent to point A, i just - 1/m where m is the gradient of point A right?

or does that only apply to the normal to the point A

[brave ingot]

this my second question

third question

where did i go wrong

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[lyric grove]

I can't understand what they are asking for. Can someone explain please

[lyric grove]

.close

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[thorny sonnet]

Can somebody help me solve (-t^(2)-2*t-2)*e^(-t)+3.2=3.1 ?

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[sterile cedar]

hello

[sterile cedar]

@sterile flint i feel bad u can have the channel i didnt see u typign

.close

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[shut obsidian]

Now it will end up in the archive anyways

Better open a new channel

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[jade scarab]

i have done a already

[winter egret]

What does your region look like?

[jade scarab]

i dont know

[winter egret]

Well how have you defined the region where you integrate?

[jade scarab]

i have not defined it yet all i had to do on a was to add dx to make it a double integral

its bound by 2 and sqrt(|x|)

[winter egret]

A double integral is integrated over some region

[jade scarab]

okay

y = 2 sqrt(|x|)

and y = 2x

[winter egret]

Your bounds of integration trace out some kind of region in 2D space

[jade scarab]

oh okay

[winter egret]

It's contained within the rectangle $[-4, 4]\times [0, 2]$

[thorny flame]

Stipendi

[winter egret]

Because the square root of something is nonnegative

[jade scarab]

ohh okay

so for x the bounds are from -4 to 4 and for y its from 0 to 2?

[winter egret]

What you should figure out is what the region of integration ought to look like

The idea is that x ranges from -4 to 4, and for each x, y ranges from sqrt(|x|) to 2

it's like a nested for loop

if you're familiar with programming

[jade scarab]

yes kind of

is it not gonna be kind of a triangle?

[winter egret]

Not really no

If x=0, then which points on the y axis belong to the region?

I highly recommend getting pen and paper for this

[jade scarab]

yes i have that, when x is 0 y is between 0 and 2

[winter egret]

yup

what about if x=1 ?

[jade scarab]

then y is between 2 and 2?

or just at 2

[winter egret]

why?

[jade scarab]

becuse y = 2x and y = 2*1 = 2

[winter egret]

what do you mean by y=2x

[jade scarab]

this

[winter egret]

what do you mean by that?

[jade scarab]

is that not the bounds that we were talking about before?

[winter egret]

Have you seen examples of where there's a region over which you integrate and you're supposed to reverse engineer the bounds of integration?

[jade scarab]

hmm i am not sure maybe a long time ago

[winter egret]

well that might be a useful thing to revisit

[jade scarab]

why do we need to know what it looks like?

our teacher gave us this so that we can use symmetry

"Fact box symmetry shortcuts: Assume that f(−x, y) = f(x, y) (is equal in x) and g(−x, y) = −g(x, y)
(for odd in x), and that the flat area D is symmetric in the y-axis, so.... For the case that gets even in y and goes odd in y while D is symmetric m.a.p. the x-axis, so"

[winter egret]

Well once you exchange the order of integration you're gonna have to understand what the region looks like

but even before that you'd kinda be taking shots in the dark if you're trying to take guesses at what type of symmetries a region whose shape you don't understand might have

[jade scarab]

yes okay

[winter egret]

Note that even in that example you're integrating over a heart shape which has an axis of symmetry

how do you know whether the region you're integrating over has an axis of symmetry unless you actually go through the trouble of figuring out what the region looks like

[jade scarab]

yeah okay

if x = 1 is y between 1 and 2?

[winter egret]

javisst

[jade scarab]

haha

[winter egret]

If you repeat the same process for every x between -4 and 4 then you can slowly (or quickly) sketch out what the region looks like

and I highly recommend that you do that on pen and paper

[jade scarab]

okay so i start with when x= -4?

[winter egret]

Yeah, sure

[jade scarab]

okay then y is 2 right?

[winter egret]

yes, y goes from 2 to 2

[jade scarab]

okay when x = -3 then y is between 2 and sqrt(3)?

and when x= -2 then y is between 2 and sqrt(2)?

x = -1 y is between 1 and 2

x = 2 y is 2 and sqrt(2)

x = 3 y is 2 and sqrt(3)

x = 4 y is 2

[winter egret]

Yup. Note that x=3 and x=-3 have the same slice on the y-axis

so there does indeed seem to be some kind of symmetry about the y axis

[jade scarab]

ohhh

yes you are right

how do i draw it tho if some values are between two values

[winter egret]

You can draw the graph of sqrt(x) for positive values and sqrt(-x) for negative values and think about what kind of region is being traced out

[jade scarab]

i can see that both x=4 and x=-4 is at 2 on both sides

and then it goes like invards towards origo kind of

or down at least

[winter egret]

yup that sounds about right

[jade scarab]

okay so f(4,2) and f(-4,2) are symmetrical?

so can we use that symmetry and add a 2 infront of the integrals?

[winter egret]

Is the integrand an odd or an even function with respect to x?

[jade scarab]

is it not neither?

[winter egret]

yeah I don't think it's either one

the main problem is the -5x

otherwise it would be even wrt x

[jade scarab]

hmm so we cant use symmetry?

[winter egret]

well once you ultimately evaluate the integral you're gonna have to do it term by term either way

so you might as well integrate the two terms separately

[jade scarab]

what does he mean by re-scale or change vars?

[winter egret]

and then the first term is even with respect to x

and the second term is odd with respect to x

which is how you can utilize symmetry

[jade scarab]

okay so the even function is e^y/2 /y^2

[winter egret]

Honestly I don't think I would do any kind of change of variables before exchanging the order of integration, and it's probably not helpful to do it at all. So I don't know about that part

[winter egret]

yup

but you can try different changes of variables and see if it simplifies

like maybe u=x, v=y/2 could simplify the first one, I'm not sure

it's a good exercise to try doing a change of variables either way

just remember to

1. change the bounds of integration
2. calculate the absolute value of the determinant of the jacobian matrix containing the partial derivatives of u and v

[jade scarab]

yes okay!

i have to go now but thank you so much for the help!

[winter egret]

samma här, ska till bastun

ha en bra kväll

och inga problem

^^

[jade scarab]

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[thin snow]

I'm attempting to prove that, for an odd prime $p$, $2p^k$ has as many primitive roots as $p^k$. I've already found that any primitive root of $p^k$ corresponds to a primitive root of $2p^k$ (this being a proof found in my textbook). However, what I'm struggling with is going the inverse direction, and showing a primitive root of $2p^k$ corresponds to a primitive root of $p^k$. What I have so far is as follows:

Proof: Take $r$ to be a primitive root of $2p^k$. Then, $(r,2p^k) =1$, implying $(r,p^k)=1$. Let the order of $r$ modulo $p^k$ be $s$. Then, $s\mid \phi(p^k)$, so $s\mid \phi(2p^k)$.

Does anyoen have a hint as to what the next step would be to show $r$ corresponds to a primitive root of $p^k$?

[thorny flame]

cat_food_sounds

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[rancid lion]

I was able to solve for y

and got 10

but idk how to get x

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[grim sable]

looks correct

[rancid lion]

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[neon iron]

The figure shown has a parabola with the equation y=x^2. Therefore, we can calculate the area below a parabola and above the x-axis
consider the intervals: [0, b/n], [b/n, 2b/n], [2b/n,3b/n]...[(n-1)b/n,b] In which we can separate into 7 equal intervals, in the form 0, b/7, 2b/7, 3b/7,..., b.
Let us take the rectangle based on the interval [(k − 1) · b/n, k · b/n] and height determined so that the rectangle is maximized, always remaining below of the graph of the parabola. The sum of the areas of such rectangles is called the lower sum for the
area below the parabola. This sum, denoted by s(n), provides a value that is always lower
to the exact area for any value of n.

In a similar way, we can take rectangles positioned above the parabola graph,
being characterized by the lowest possible height, resulting in a superior estimate for
the exact area. The sum of the areas of these rectangles is called the upper sum for the area
below the parabola and is denoted by S(n).

a)Consider b = 1. Find a positive integer N_1,2 > 1 such that:
S(N_1)-s(n_1)<0,01
S(N_2)-s(n_2)<0,0001
b)Based on the results obtained in the previous item, what can you conclude about the area
exact below the parabola and above the x axis between the abscissa x = 0 and x = b? So
more precise: express this area as a function of b and justify your answer.

[neon iron]

i did A, that's it

but i don't know how to do B

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<@&286206848099549185>

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[distant linden]

how cme theres no interval testing here compared to this question?

the questions btw

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[round onyx]

help please

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[neon iron]

Would it just be 16 x 84 and 16 x 486

[long stirrup]

no

it would be 16 × 16 × 84 and 16 × 16 × 16 x 486

i don't know exactly why

[neon iron]

So what would trhe surface area and volume be with that?

If I got to do that

Oh I see

1990656 volume???

Thats alot

[long stirrup]

like the volume makes sense, if it's initially a cube and you double the dimensions, there would be 8 cubes

[neon iron]

Well we know it's all the same figure

[long stirrup]

the area is not as obvious

[neon iron]

If they give us the area and volume to get the rubber ducky why wouldnt it be that again for both

We now know the volume is 1990656

How can we find the area with that info

[long stirrup]

well okay if it's initially a cube it has 6 faces of 1 and it becomes 6 faces of 4, so the same "logic" holds

[neon iron]

so it's 84 x 16?

me confused

[long stirrup]

84 × 16 × 16

[neon iron]

21504?

[long stirrup]

yes, cm² and cm³

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[grand solar]

I'm wondering if this is a typo, did he mean infinity instead of 5 or do you think he meant 5 for the upper bound of the power series of sinx

[grand solar]

bad picture sorry

[worthy storm]

it's an approximation

he's using only finitely many terms

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[grand solar]

hmm I'm confused, I know that the |S-Sn| <= a_n+1 but in this case there's no S right?

[worthy storm]

S would be sin(2)

i.e. what you get if you sum all the way to infinity

S_n is the approximation that results from summing only up to n

[grand solar]

oh

so the maximum remainder would just be a_6?

or wait no

[worthy storm]

yea that sounds right

a_6

abs value of that

[grand solar]

hmm in the equation what is the 2 used for or placed to know we are approximating sin(2), like what if it was sin(3) we were approximating

oh

the power series has an x in it

and that x is 2, I forgot about that

ok thank you!

[worthy storm]

yes that's right

[grand solar]

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[terse crane]

,rotate 270

[terse crane]

i dont get 15

and how to even start it

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[golden yew]

Can someone

[golden yew]

explain P(F)

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[stuck cedar]

Prove that $\frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \geq \frac{3}{2}$ for $a, b, c > 0$ and $a + b + c = 3$ \newline Applying Cauchy inequality we have: \newline $\frac{a}{b+c} + \frac{b+c}{4} + \frac{1}{2a} \geq 3 \cdot \sqrt[3]{\frac{a}{b+c} \cdot \frac{b + c}{4} \cdot \frac{1}{2a}} = 3 \cdot \sqrt[3]{\frac{1}{8}} = \frac{3}{2}$ \newline Applying for others we have: \newline $\frac{b}{c+a}+\frac{c+a}{4}+\frac{1}{2b} \geq \frac{3}{2}$ and $\frac{c}{a+b} + \frac{a+b}{4} + \frac{1}{2b} \geq \frac{3}{2}$ \newline Sum it all up, we have: \newline $\frac{a}{b+c} + \frac{b+c}{4} + \frac{1}{2a} + \frac{b}{c+a} + \frac{c+a}{4} + \frac{1}{2b} + \frac{c}{a+b} + \frac{a+b}{4} + \frac{1}{2c} \geq 3 \cdot \frac{3}{2} = \frac{9}{2}$ \newline $\Leftrightarrow \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} + \frac{a+b+c}{2} + \frac{a+b+c}{2} \geq \frac{9}{2}$ \newline $\Leftrightarrow \frac{a}{b+c} + \frac{b}{c+a} + \frac{c}{a+b} \geq \frac{9}{2} - 3 = \frac{3}{2}$ \newline Equality happens if and only if a = b = c = 1.

[thorny flame]

cffex

[stuck cedar]

I don't understand the proof

at the first step

how and why do you derive that?

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[cosmic marten]

Can someone quickly explain to me, how this works?

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<@&286206848099549185>

[wooden osprey]

wrong ping lmao

[wooden osprey]

anyway, are you familiar with what extrema are?

[cosmic marten]

I thought I did, we have done maxima, minima in hs but this uni explaination isnt clear in my head, thats what I would like help with, so for the sake of clear understanding, NO

[wooden osprey]

alright

lets say we have this graph

can you identify where is the minimum, and where is the maximum?

[cosmic marten]

I would assume that max is first point just before -2, minimum is just after 0

but that doesnt work with the question above because its derivative?

[wooden osprey]

we'll get to that later

[wooden osprey]

but youre correct, the local maximum is the point just before -2, and the minimum is the one just after 0

[cosmic marten]

cool

[wooden osprey]

how do you know they were the maximum and minimum?

[cosmic marten]

and I would prove it based on it incr - decrease (max) and decrease - incr(min)

[wooden osprey]

that is correct

the point where the function goes from increasing to decreasing is the maximum

whereas the minimum is where it goes from decreasing to increasing

now let me ask another question, do you know what derivative is?

[cosmic marten]

rate of change

[wooden osprey]

right, rate of change

when the function is increasing, its rate of change is positive, whereas if its decreasing, the rate of change is negative

in other words, the derivative of a function is positive where it is increasing, and negative where it is decreasing

and you can see on this graph

the blue is the derivative of the red function

[cosmic marten]

so for that derivative, reverse it for a the f(x) and then see where the min is

[wooden osprey]

this is essentially what they said in your explanation

a local minimum occurs when its derivative goes from negative to positive

aka, the function went from decreasing to increasing

[cosmic marten]

ok that makes sense now

but I still confused what it means with the points

(-1,2) and (2,3)

at (-1,2), the derivative is above x and at (2,3) but is above x

wait nvm

I get it

when its derivative is 0

its a key point

at -1, it goes from + to -, hence a max,
at 2, it goes from - to +, hence a min

[wooden osprey]

its called critical point, but you are correct

[cosmic marten]

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[indigo estuary]

I got 11.94 m/s but I got it as wrong

[versed cairn]

show working

isnt this using law of conversation of energy btw? calculate the KE and PE on top of hill , must be equal to KE at the bottom

[indigo estuary]

Yea

[versed cairn]

can u show ur working ? might be able to see the mistake

[indigo estuary]

@versed cairn here’s my working

[versed cairn]

do u know the correct answer by any chance?

[indigo estuary]

Umm no sorry

[versed cairn]

seems correct

[indigo estuary]

Ohk must be an error by the marker

[indigo estuary]

Also mind helping with another similar qu?

[versed cairn]

maybe they wanted to use g = 10?

[indigo estuary]

Umm our teacher tells us to explicitly use 9.8

[versed cairn]

maybe they cut ur marks due to wrongly rounding off

the cross is over there so i think that can be a problem (just a theory)

11.94 ms = around 12 ms not 10

[indigo estuary]

Yea but he also tells us to use the correct sig fig

Which is 1

[versed cairn]

isnt stated in the question tho

[indigo estuary]

Yea ig

[indigo estuary]

Also can u help me with another qu rq

[versed cairn]

k

[indigo estuary]

Got 0 for both

[versed cairn]

u are to use friction here

because coefficient of friction is given

f (friction) = coefficient of friction * normal force
= coefficient of friction * m * a

in these kind of questions , always use the information given even if u dont know what it is , never drop it out

use them in 1 way or another , think harder

[indigo estuary]

Yea can u try it rq

I think it’s a marking error again

[versed cairn]

no its not

[indigo estuary]

Ohk

[versed cairn]

a = 9.03 or smthing

[indigo estuary]

I see

[indigo estuary]

Wait why’d u write mu after a

[versed cairn]

friction force

the problem doesnt state that there its frictionless or neglect friction

[indigo estuary]

Yea I had that on the left hand side no?

0.12 as mu

Mgcos22 as Fn

And it’s equal to the net force which is ma

Since there’s no other force acting on block

[versed cairn]

it should be on the other side

weight time mu ?

mu isnt a force by a constant

[indigo estuary]

Yeah weight times cos 22 times mu

[versed cairn]

mgcoss22 = ma mu

[indigo estuary]

I think u mixed up cos and sin

[versed cairn]

u wrote cos so i thought it was cos 😅

[indigo estuary]

Ohk lol