help-27

[boreal blaze]

can someone help me

with 1e

i know the answer is true

but idk how to prove it

[magic pine]

,rccw

[magic pine]

btw your proof for 1d is not correct

you’re meant to show it for every odd function (i would assume)

[boreal blaze]

nope its correct, i confirmed with teacher. i just had to show one example

[magic pine]

can you do the same for 1e?

[boreal blaze]

shes not responding

wait

u mean try prooving with an example?

[magic pine]

yes

[boreal blaze]

uh not sure. it just doesnt make sense to me

[magic pine]

if it was allowed for 1d then why not for 1e?

[boreal blaze]

thats why i asked for help lol

[boreal blaze]

she could help me. shes just not responding

probably sleeping

[magic pine]

the actual proof for this would use limits and the limit definition of a derivative

i don’t know how rigorous you’re meant to write it

[boreal blaze]

just a simple proof

nothing complex

a proof or an example

[magic pine]

then this is just the chain rule

$\dv{y}{x} = \dv{y}{u}\cdot \dv{u}{x}$

[woven radish]

maximo

[magic pine]

since both derivatives on the right exists, the one of the left does as well

that’s not rigorous at all but the idea is there

[boreal blaze]

alright

so how would i explain it

[magic pine]

i just write it

[boreal blaze]

write the chain rule

?

[magic pine]

wrote*

yes

[boreal blaze]

oh this part

hmm

i feel like my professor would want more, but ig i can make sense of it

[magic pine]

write it as y = y(u(x))

then $\dv{x} y(u(x)) = …$

[woven radish]

maximo

[magic pine]

and do the chain rule

[boreal blaze]

alright

done thanks

can u help me with another question please

How would i do 4c and question 5

i think ik how to do question 5 but im not sure what its asking for

and question 4c, im just lost af

[magic pine]

find a general pattern for the derivatives

in particular, for the first, third, fifth, seventh,…

[boreal blaze]

thats what im not understanding how to do lol

we were never taught that

[magic pine]

it’s just basic pattern recognition

look at the first derivative and the third and the fifth and so on

[boreal blaze]

all i see is beta increasing by 1 power

and the derivative of the trig function (sin or cos) is taken

[magic pine]

you’re looking at every derivative

i’m saying look at the first and third

and think about what the fifth should be

[boreal blaze]

beta^5(cos(beta x))

right?

[magic pine]

yes

and the seventh?

[boreal blaze]

• beta ^6 (sin(beta x))

[magic pine]

^7

and cos

[boreal blaze]

oh sry

misread

i thought u said 6

anyways, i see a pattern as you said

[magic pine]

write it in terms of k

[boreal blaze]

but i dont know how to integrate it into a formula

[magic pine]

try something

first

how can you write the power

of beta

in terms of the derivative

note that we were just talking about the odd derivatives

so f^1 f^3 f^5 ...
so in general, we can write this derivative as f^(2k+1)

[boreal blaze]

what the hell

wait

im sorry

[magic pine]

i'm just saying what the question said

i haven't said anything new

[boreal blaze]

kinda having trouble understanding

true

[magic pine]

we want a genreal formula for f^(2k+1)

in terms of k

for example, when k = 0

we get

f^(1), the first derivative

which is just beta * cos(beta * x)

when k = 1, we get f^(3) = - beta^3 * cos(beta * x)

k=2, f^(5) = beta^5 * cos(beta * x)

[boreal blaze]

would it be

beta ^(2k) sin (beta x)

[magic pine]

is that for the even one?

[boreal blaze]

i think thats odd

[magic pine]

no

odd derivatives have cos

not sin

[boreal blaze]

oh wait

yea evens

mb

cuz its 2k

[magic pine]

you're missing one thing

[boreal blaze]

lemme gues

the negative

[magic pine]

yes

[boreal blaze]

so would it be

• beta ^(2k) sin (beta x)

[magic pine]

no

not every derivative is negative

it alternates

between positive and negative

[boreal blaze]

would it be
(-1)^k beta ^(2k) sin (beta x)

[magic pine]

i believe so

can you do the one for the odd as well?

[boreal blaze]

and for the odd function it is:
(-1)^(k-1) * beta^(2k-1) * cos(betax)

[magic pine]

you're saying that's for f^(2k-1) right?

ah the quesiton used minus

yeah i think that's fine

[boreal blaze]

alright

thanks

appreciate the help

[devout snow]

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[sonic ravine]

is there any way to find out the equation of this line without trial and error?

like actually using maths to figure out, the only have 2 points, 1 being x-intercept

[restive river]

I’m really stupid at this

Because I’m a little boy

[final laurel]

that

a straight line?

[sonic ravine]

its a curve

from (5.5, 11.7) to (8,0)

[restive river]

He means green line

[final laurel]

oh

They all looked curved

[restive river]

😑

[sonic ravine]

no cyan line is 0.016x^2-0.233x+12.5

[final laurel]

you can't determine the equation of the line if it's curved from 2 points

unless you know more information

like what type of function it is

[sonic ravine]

what if i say

it can be any type of function

quadratic, cubic, sin, anything you can think of

[final laurel]

only a line can be uniquely defined by 2 points

[craggy dagger]

so you want to find a curve that lies inside this region?

you can try a quadratic first

[sonic ravine]

just the equation

if you say quadratic then

is it all trial and error?

is there a way to mathematically figure it out

[craggy dagger]

there are mathematical ways to find quadratics

so what you do is you plot more points along the curve

and find the interpolating polynomial

all you need is 1 more point along the curve

[sonic ravine]

YO

that sounds good

can you tell me more

i have not learnt interpolating polynomial yet

[restive river]

Wtf will I learn these things😪

[sonic ravine]

but the purpose of this investigation is for me to learn new maths

[craggy dagger]

however, the shape might be wrong, so you can consider maybe adding different number of points to make a different degree polynomial

,w lagrange interpolation

[woven radish]

Results provided by WolframAlpha

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

Requested by Element118#2437

[craggy dagger]

that's the general formula for it, but there's ways to gain intuition on how that works

[sonic ravine]

how does this work

by looking into that, i have no idea what it is

[craggy dagger]

Let's say I give you the problem "given a bunch of points, find a polynomial passing through them"

Let's consider a special case. Say all of them except 1 are on the x-axis

can you solve this special case?

so you have $(a_0,y),(a_1,0),(a_2,0),\dots,(a_n,0)$

[woven radish]

Element118

[sonic ravine]

i have not learnt this-

do you have time to explain me the steps?

or you have like a youtube video

[craggy dagger]

hmm let me check

https://www.youtube.com/watch?v=bzp_q7NDdd4

this is probably good

[devout snow]

@sonic ravine Has your question been resolved?

[craggy dagger]

well another way which might be more successful would be to formulate it as part of a circular arc

[sonic ravine]

what is this way?

[sonic ravine]

here i tried to copy the larange interpolation method to find the y-intercept

but the result is -67.9

its so far off

[craggy dagger]

I don't think that's a problem

because -67.9 is not within the range we are interested anyway

[sonic ravine]

but why does it give that value

is my working wrong?

i simply copy, i have no idea what i am doing

[craggy dagger]

try finding a circle (x-a)^2+(y-b)^2=r^2 that looks like the curve you are interested in

[craggy dagger]

we don't care what the polynomial does beyond the range [5.5, 8]

[sonic ravine]

okay

i am testing things out

also

the "r" in the equation

what do you call this type of function

i always do functions with only x and y, never touched on r

@craggy dagger

i found a pretty accurate one

[craggy dagger]

r is just an arbitrary constant (radius of the circle)

since your k is positive this would mean it would break apart into 2 functions

when you solve for y in terms of x

[sonic ravine]

ok

how do i say i come up with these numbers

trial and error..?

like the numbers -17, 0.8 and 25

the work that i am doing is for an essay, where i model the shape of that thing

i need to explain what i have done to get a certain function that fits

@craggy dagger how should i explain the method i used for the circle equation

[craggy dagger]

<@&268886789983436800>

[sonic ravine]

is there a way to prove it mathematically..or do i just say i use trial and error..

[craggy dagger]

basically just trial and error

though the circle might not pass through the points of interest in your diagram

which are (5.5, 11.7) and (8, 0), it you want it to pass through those points then you need to use math stuff

[sonic ravine]

oh ok

one more thing

lets assume i will stick with that circle

how do i find the area here

is there a separate way to integrate a part of a circle?

that looks like half of a segment to me..maybe i am wrong

[devout snow]

@sonic ravine Has your question been resolved?

[devout snow]

@sonic ravine Has your question been resolved?

[devout snow]

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[devout snow]

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[dapper quail]

In a function, f(x)=[x-1][x+5]
Solve for f(1/a)

[restive river]

sub in 1/a for x

[dapper quail]

I tried that

The solution is different

The (1-a) one is the solution

[stuck field]

,rcw

[stuck field]

Clearly they subbed in 1/a also.

And that's kind of the only way.

[dapper quail]

Yeah I'm sure

Idk what I'm doing wrong

[stuck field]

Show work.

[dapper quail]

I foiled

Then subbed a

So 1/a^2 + 5/a - 1/a - 5

Giving

1/a^2 + 4/a - 5

Idk

[stuck field]

Actually.

It's right.

And it's the same as what they wrote.

Notice how they've the answer in the factored form. While you distributed the expression.
You might every well sub 1/a in the originally factored expression.

To get the same form.

[dapper quail]

I don't entirely understand could you show me what you mean?

How'd they keep it in factored form?

Like

[1/a-1][1/a+5]

How'd they simplify it?

[stuck field]

LCM

[dapper quail]

Least common multiple?

[woven radish]

What the hell am I doing here?

[stuck field]

Oops, forget that extra (

[dapper quail]

Ah

Makes sense now yes

The book also has these other problems

Which are simple but idk how they calculated it without a calculator and idk if they're suggesting something by showing these problems or they're done by like actual rote math

So like in #4

Ugh idk why it rotates like that

And #5 it also gives the value of sqrt(2) as a solution and idk if I'm just supposed to know that or use a calculator

Is there are easier way to find those values that I don't know of?

I'm just curious

,rcw

[winter patrol]

by definition of fractional exponents

[dapper quail]

What's the definition of fractional exponents?

Oh wait are you talking abt the sqrt of 2

Oh yeah I understand that

I meant like knowing what the value of sqrt 2 actually is

Being 1.414..

Idk why the book put that instead of sqrt of 2 I thought they wanted me to know the value or somethg

[winter patrol]

sqrt(2) is the exact value

[sonic smelt]

$a^{\frac{b}c}=\sqrt[c]{a^b}$

[woven radish]

A Lonely Bean

[winter patrol]

1.414...whatever is only a decimal approximation

[dapper quail]

Yeah ik idk why they didn't just write sqrt 2 in the solutions

[winter patrol]

i thought you said they did

[dapper quail]

Oh I meant the opposite idk what I said

And for #4 idk how to do it without rote math or a calculator, I just did it with a calculator

But I'm wondering if there's some sort of shortcut mathematically

[craggy dagger]

$(a+b)^2=a^2+2ab+b^2$

[woven radish]

Element118

[craggy dagger]

might help with squaring those things by hand

[dapper quail]

Oh yaa thinking of it that way does help me do it in my head

[winter patrol]

Yeah ik idk why they didn't just write sqrt 2 in the solutions
nfi

[dapper quail]

9+.6+.01

Makes it a lot easier

I also don't know how they simplify number 8 after plugging it in

[winter patrol]

combine like terms

9+.6+.01
ideally you'd write those 0s before the dp

[dapper quail]

Oh ok

[winter patrol]

9 + 0.6 + 0.01

[dapper quail]

Also I wanna learn to use latex it'd be nice

Also to make stuff easier to read over this

[winter patrol]

dots may be hard to see, makes it easier to read too

there's an introduction in #latex-help

[hybrid snow]

Or you learn vicariously like me. Stalk channels and see how others use latex

Because why would you read a whole document when you can watch Ramonov's every step 24/7/365

[dapper quail]

Oh cool I didn't know that was a thing

[winter patrol]

if you already understand basic function notation,
that's already most of the basics done already

[dapper quail]

i replied to the wrong thing i meant latex help .-.

[dapper quail]

oh ok cool

[dapper quail]

facts 😂

real stalker that's crazy

So for #8 I have (x+∆x)^2 + 2(x+∆x) + 1

Idk if there's an easier way of simplifying it than distributing jt

[winter patrol]

there isn't

[dapper quail]

I feel like I'm always doing extra work and the book is trying to teach me to do less

Oh ok

[winter patrol]

also that's just the first part,

don't forget the -f(x)

[dapper quail]

omg that's why it's wrong 😂

i definitely forgot the - f(x)

makes sense now 😂

alr no more issues I finished that problem set

[devout snow]

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[devout snow]

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[rain garnet]

helo

[rain garnet]

how to calculate it

any tips??

[fathom thicket]

firstly

cos(x - pi) = -cos(x)

[rain garnet]

yes yes agree

[fathom thicket]

now cos(arccos(x)) = x

[rain garnet]

but wait

[fathom thicket]

the other way round doesnt work though

arccos(cos(x)) isnt the same as x

[rain garnet]

maybe i dont understant tthe way is it written

[fathom thicket]

also this is from x between -1 and 1

[rain garnet]

dont i need firstly calciulate arccos(2/3)?

[fathom thicket]

you dont need to

cos(arccos(2/3) - pi)

is the same as

-cos(arrcos(2/3))

[rain garnet]

AAAA

okay

i get it now

thasnks

.close

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[sharp axle]

As part of our homework we have this question "simplify tanx/(1-sinx)" and I'm kind of stuck.

[sharp axle]

I got to sinx/(cosx-sinxcosx) but that's it

[queen zephyr]

i dont think we can simplify that

[winter patrol]

was that the exact wording of the question?

[sharp axle]

Yeah

Just the word "simplify"

[winter patrol]

did you copy down the question correctly

as there isn't really something thats objectively simpler

[sharp axle]

The teacher sent it to us through telegram

So I'm pretty sure that I did

[winter patrol]

through telegram? the sus app?

[sharp axle]

The app doesn't matter, what matters is that he sent us a pic

[winter patrol]

check the pic again

[sharp axle]

(8)

[winter patrol]

are you 100% sure these are "simplify" questions?

[sharp axle]

100%

[winter patrol]

because there's not much that can be done with Q4,
and 5,7 are clearly already in the simplest form

[sharp axle]

We asked him today where did you get these questions since we searched in the textbook and these weren't in it.

He said they aren't from the textbook

[winter patrol]

some of these can be objectively simplified,
8 isn't one of them

[sharp axle]

Well

I tried the others

In some programs

And no answer

But when I tried them myself I was able to simplify

So that's why I ask here because maybe people know something that solvers don't

Anyhow I'll go eat something

And I'll ask him about this particular point next time

.close

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[near belfry]

Can someone tel me how I show this

[pseudo basin]

,rccw

[pseudo basin]

cos(180n°) = ?

[near belfry]

What? Ann

[pseudo basin]

how can you rewrite the expression cos(180n°)?

especially knowing n is an integer.

[near belfry]

I have no idea

I’m very bad with sigma notation

[pseudo basin]

"I'm very bad with sigma notation"
in response to a question that doesnt concern sigma notation at all

ok lets see here

what is cos(180°)?

[near belfry]

-1

[pseudo basin]

and cos(2*180°)?

[near belfry]

1?

[pseudo basin]

and cos(3*180°)?

[near belfry]

Just a recurrence

Of -1,1

[pseudo basin]

yes, so how else can you write down a sequence that goes 1, -1, 1, -1, 1, -1, ... alternating between 1 and -1 forever

[near belfry]

-1^(n-1)

[magic pine]

parentheses

[restive river]

,rccw

[magic pine]

!help

[devout snow]

Please read #❓how-to-get-help

[near belfry]

@restive river can you explain why the geometric sequence isn’t -1^(n-1)

[magic pine]

-1^n = -(1^n)

[near belfry]

I thought a geometric sequence is in the form

Ar^(n-1)

Why is it not n-1

[pseudo basin]

(-1)^n

[pseudo basin]

overattachment to form

[near belfry]

Oh

Because of the limit

I see

[magic pine]

oh @restive river sorry about the !help thing, i thought you were asking a question

[near belfry]

What do I do and why do I do it :

Once I get the form into

Sigma : (-3/4)^n=9/28

[devout snow]

@near belfry Has your question been resolved?

[devout snow]

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[devout snow]

Please don't occupy multiple help channels.

[pure grotto]

is my solution correct?

[devout snow]

@pure grotto Has your question been resolved?

[devout snow]

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[pure grotto]

help

[pure grotto]

is my solution correct?

[devout snow]

@pure grotto Has your question been resolved?

[surreal frigate]

It's correct

[devout snow]

@pure grotto Has your question been resolved?

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[pure grotto]

thank you boss

[pure grotto]

iloveyoumwa

[devout snow]

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[cunning perch]

How do you find the domain of this function?

I know the range is {y|y=-5}

[cunning perch]

It cant equal -5

[pseudo basin]

then why write = and not ≠

[cunning perch]

I just don't have the symbol on keyboard

[pseudo basin]

you can substitute with !=

[cunning perch]

oh

[pseudo basin]

also i would not be so sure the range actually is R \ {-5}

i have a suspicion about that

[cunning perch]

oh wait

no

{y|-5<y<=-3}

[cunning perch]

but how do you find the domain?

[pseudo basin]

how do you normally find the domain of a fraction like this?

[cunning perch]

make the denominator 0?

[pseudo basin]

you find all points where the denominator is zero, and exclude those.

[cunning perch]

Right

so is it square root of 3?

[pseudo basin]

does x = sqrt(3) make the denominator zero?

[cunning perch]

no

nevermind

lol

[pseudo basin]

you should get that the equation x^2 + 3 = 0 has no real solutions!

[cunning perch]

oh

wait

so

it can't be less than 0

no,

I mean

it has to be less than or equal to 0

I'm not sure

[pseudo basin]

what's "it"

[cunning perch]

x

[pseudo basin]

why does x have to be less than or equal to 0?

[cunning perch]

It doesn't

I just checked

[pseudo basin]

then why say it does...

[cunning perch]

I'm just a bit confused

how would I go about getting the domain and range of that

[pseudo basin]

in this case, since you have no complications other than the fraction,

you find all points where the denominator is zero, and exclude those.

[cunning perch]

I can't get the denominator to be 0 though

[pseudo basin]

you have found that the denominator is never equal to 0, yes.

so there is nothing to exclude.

[cunning perch]

oh

so is it all real numbers?

[pseudo basin]

yes...

[cunning perch]

oh

and the range

[cunning perch]

I know for the range one of the values is y=-5

oh wait

I got it

already

[pseudo basin]

no, you know that one of the excluded values is y=-5...

[cunning perch]

Yeah

[pseudo basin]

confusion between "i know this value belongs to the <domain/range>" and "i know this value DOESN'T belong to the <domain/range>" is HIGHLY DANGEROUS.

[cunning perch]

and the other one is anything greater than 3

[cunning perch]

yeah

thanks for your help!

.close

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[jade falcon]

For this question we can use the ideal gas law. PV=nRT

P is pressure
V is volume
n is moles(quantity of gas)
R is a constant(ideal gas constant)
T is the temperature of the room

Since the gas is in a sealed vessel, n cannot change. And T can be ignored since we can assume that the gas will reach the same room temp after some time as in the initial conditions. And since R is a constant, it can be ignored as well.

[jade falcon]

how do i go ahead with this?

[stuck field]

Enlargement factor, that is for volume right?

[jade falcon]

ya

[stuck field]

Alright. And you say T is to be assumed constant?

[jade falcon]

ya

i got 25 for area and 50 for volume

[stuck field]

Again, you're changing problems. Which one do you need help with?

[jade falcon]

idk

[stuck field]

...

[jade falcon]

it just says enlargement factor

[stuck field]

Didn't ask that. Do you need help with 3 or 4.

[jade falcon]

4

[stuck field]

Alright.

If nRT is completely constant then you have PV = constant.

Meaning P is inversely proportional to V.

[jade falcon]

ok

[stuck field]

Do you know how to do it from there?

[jade falcon]

no

[stuck field]

And if V changes by a factor of x. Then P should change by a factor of 1/x.

So that their product is always a constant.

[jade falcon]

okay

is that it?

[jade falcon]

.

[stuck field]

That is it.

It tells you the enlargement factor is F.
So you should get the factor by which P changes as well.

[jade falcon]

ok thanks

[jade falcon]

so can i write the x here as F

like

And if V changes by a factor of F. Then P should change by a factor of 1/F

[stuck field]

It should.

[jade falcon]

one last quesion

does P being inversly proportional to V mean the same as V being inversely proportional to P

?

[stuck field]

Yes.

[jade falcon]

then cant it also mean that if P is enlarged by a factor of F then V is enlarged by a factor of 1/F

?

[jade falcon]

.

[stuck field]

It can. All of it is the same thing.

But your question doesn't say that.

[jade falcon]

ok

and what if the enlargment factor was for pressure?

then would it be that if P is enlarged by a factor of F then V is enlarged by a factor of 1/F

?

[stuck field]

Yes, just now you asked the same and I said yes.

Whatever the enlargement factor for either of them is, for the other it is 1/that quantity.

Makes sense?

[jade falcon]

ya

so it doesnt matter what i write in the answer right?

[stuck field]

Obviously it matters.

Since they've told you volume is the quantity that enlarges by F. Then you can surely comment on pressure.

[jade falcon]

but the question just says the enlargement factor is F. not the enlargement factor for volume

[stuck field]

Well, clearly volume is the one changing. And that is also what I asked you in the first place.

Besides by "enlarge" you'd assume the volume is the one that changes.

[jade falcon]

ya i guess thats correct

thanks

[stuck field]

You're welcome.

.close

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[jaunty gazelle]

can someone help me with this problem?

[jaunty gazelle]

trig integration

I think I have some of it down...this is what I have so far, can you tell me if im on right track:

Number 4

I used trig identity tan^2(x)+1=sec^2(x) to solve problem

but now im stuck

[sonic smelt]

I think I would get more tans instead

[upper schooner]

Should be product but otherwise yea right track 🛤️

That can work

[sonic smelt]

Hm, but the powers are odd

[jaunty gazelle]

I have integral tan(x) *(sec^2-1)sec^7(x)dx . I then multiply the sec^7x by sec^2 and -1 respectively, right?

hold on

[upper schooner]

You could then factor out a sec(x)tan(x)

[jaunty gazelle]

Wouldn’t I do u substitution or something?

[upper schooner]

Second term should be tan(x)sec^7(x) (or you're missing ), and yes you would

[jaunty gazelle]

Maybe I misunderstood a step. Wouldn't we just keep the tan(x), multiply the sec^7(x) by sec^2(x) and -1 and get tan(x)*sec^9(x)-sec^7(x)dx

[upper schooner]

Here you said you had
[
\int \tan(x) \cdot (\sec^2(x)-1) \sec^7(x) \dd x
]
right?

[jaunty gazelle]

yes

[woven radish]

@upper schooner

[jaunty gazelle]

yeah, tan^2(x) turns into (sec^2(x)-1)

[upper schooner]

You then multiplied out the brackets here, right?

[jaunty gazelle]

I only multiplied sec^7(x) by sec^2(x) and -1

[woven radish]

@upper schooner

[upper schooner]

[note the in the first one]

[jaunty gazelle]

So this would get us this:

[upper schooner]

Almost

[jaunty gazelle]

Wait

I messed up somewhere with the exponents

[jaunty gazelle]

I canceled the sec(x) with the sec^9(x) though..

[upper schooner]

You can't do it with only one, you need to do it for both

[jaunty gazelle]

Thank you.

[upper schooner]

See here

if you factored out another sec(x) you'd see how it cancels for both

[devout snow]

@jaunty gazelle Has your question been resolved?

[jaunty gazelle]

Okay, so assuming this is right, I've learned that everything comes down to trying to apply a trig identity, factoring and/or multiplying, some sort of integration technique like u substitution, and simplification?

The following above was another problem I attempted just now (problem above):

[crystal rune]

This is correct.

[jaunty gazelle]

Is my thought process and steps right to?

[crystal rune]

Yep.

[jaunty gazelle]

Is this statement also true: "Okay, so assuming this is right, I've learned that everything comes down to trying to apply a trig identity, factoring and/or multiplying, some sort of integration technique like u substitution, and simplification?"

[crystal rune]

Uh, I wouldn't say everything because there are a lot of identities and I don't know if you know them all, I don't know if you know when to apply which identity

But in this case you certainly did apply the right one, however there's almost always more than one way to integrate something

Just keep practicing

I for one certainly don't know everything when it comes to applying a trig identity and simplifying

[jaunty gazelle]

Okay. Let me ask, I also tried doing it the other way, as in, applying the trig identity to sin^2(x) to get (1-cos^2(x))*(cos^3)dx. I gave up eventually. Would that not work for a problem like this?

[crystal rune]

It can but it's lengthy

Gotta use some angle identities I think

Your way is the shortest I believe

[jaunty gazelle]

Okay. Another final question, would I have to use a reduction formula on this:

it's not uploading..

[crystal rune]

Btw Idk if it'll work but you can also try using u = cosx or u = sinx and see what happens

[crystal rune]

What do you think?

[jaunty gazelle]

I know I have to factor the 5 out, maybe take the antiderivative of x^3...but idk if it this is meant to be used with reduction formula

[crystal rune]

A simple u-sub

[jaunty gazelle]

let me check hold on.

[jaunty gazelle]

with our u sub being cos?

I am getting tis as an answer, and this is my work:

I don't think I'm getting this correct.

Can anyone help? <@&286206848099549185>

[nova idol]

use an integral calculator

[jaunty gazelle]

Any you recommend? Can't see the steps using symbolab..

[nova idol]

Use integral calculator

They show all the steps

[crystal rune]

@jaunty gazelle Derivative of x^4 is 4x^3 and 5x^3 is already given.

This implies a u-sub, u = x^4

[jaunty gazelle]

Is there another way of seeing that x^4 is u sub...

[crystal rune]

Other than seeing that its derivative x^3 is given, no

[jaunty gazelle]

doing u substitution i get 5/4* integral of cos^2(x^4)..

do I apply trig integration on cos^2(x^4)?

[crystal rune]

Should be $\int \cos^{2} (u) \dd{u}$

[woven radish]

Sup?

[crystal rune]

5/4 outside too

[jaunty gazelle]

okay so err...

we do something with the cos^2, right?

trig identity!

[crystal rune]

Yes

[jaunty gazelle]

1-sin^2?

[crystal rune]

Nope

[jaunty gazelle]

wait

[jaunty gazelle]

? I thought the trig identity was cos^2(x)+sin^2(x)=1..

[crystal rune]

There are more

[jaunty gazelle]

that wouldn't apply here?

[crystal rune]

If u cant integrate cos^2, how can you integrate sin^2

try a double angle identity

[jaunty gazelle]

I see...this is my work for it:

[jaunty gazelle]

so we apply double angle identity 1+cos(2pheta)/2 du

and then we simplify to get 5/8 (u+1/2sin(2u))

[crystal rune]

at the end it should be sin(2u)/4

All good otherwise

chain rule, you divide by the derivative of 2u

Oh wait

Nvm

You factored out the 2

Okay

All good.

[jaunty gazelle]

Thanks so much, Sup. 🙂

[devout snow]

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[indigo abyss]

how is this wrong?

[surreal frigate]

You forgot to plug in 0

[devout snow]

@indigo abyss Has your question been resolved?

[supple knot]

cos(0) = 1

[devout snow]

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[mild summit]

Simple question:

[mild summit]

Why can one not construct confidence interval of beta_0?

this is what we have

<@&286206848099549185>

yo

[devout snow]

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[devout snow]

@mild summit Has your question been resolved?

[timber pebble]

you can

its done in exactly the same way

its just not especially useful

or statistically meaningful

unless you have a really specific research question

[devout snow]

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[hollow forge]

can anyone help me with this error if they have ever used mathematica?

[hollow forge]

nvm i found the problem

[devout snow]

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[sharp citrus]

How am i supposed to do this question

i just took the final equation and made it

[supple knot]

,rotate

[sharp citrus]

(x-y)^2 is bigger or equal to 0

[supple knot]

this is all you need

[sharp citrus]

but it says real number

so it can be negative as well right?

[supple knot]

yes

what's (-3 - (-1))^2 ?

[sharp citrus]

4?

OHHH

i see

[supple knot]

yes so do you see why the - signs don't matter

[sharp citrus]

yes ye

so i can just move on to the conclusion right

no need for further explanation

like for the answer

[devout snow]

@sharp citrus Has your question been resolved?

[sharp citrus]

.close

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[visual lark]

anyone who can help me on the right path with this integral?https://gyazo.com/6d8e5975fa5827996ba4b1532f250672

[visual lark]

.close

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[restive river]

just making sure the possibilities of sitting in a round table is (n-1)! and like a class with normal tables is n! ? or not.

[devout snow]

@restive river Has your question been resolved?

[nova idol]

m8 what

[devout snow]

@restive river Has your question been resolved?

[devout snow]

@restive river Has your question been resolved?

[wide lynx]

Yes

[devout snow]

@restive river Has your question been resolved?

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[deep pelican]

Collatz conjecture

[alpine tapir]

are you asking what it is?

https://www.youtube.com/watch?v=094y1Z2wpJg

[deep pelican]

nope sorry should have been more clear i have a proof but am having a hard time wording it

looking for people to chat about with it

[timber pebble]

you should go to a less formal channel

discussion or chill

[deep pelican]

i've had this for about 1 year

cool thanks

.close

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[ivory yoke]

how to find the derivative of this function

[fathom thicket]

chain rule

dont use someone elses channel

!help

[devout snow]

Please read #❓how-to-get-help

[ivory yoke]

Something like this?

[fathom thicket]

yh

[ivory yoke]

thx

.close

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[odd wave]

Hi can someone explain this question and answer? I'm stuck thanks!

[fathom thicket]

its asking you to expand that

and compare terms

[devout snow]

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[devout snow]

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[lofty monolith]

Even though that force is acting up, the particle moves down the slope?

[lofty monolith]

Idk I assumed the force would pull it up the slope or something 😭

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[devout snow]

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[maiden zinc]

is F = ker(f - id) the same thing as F = Im(f)?

[maiden zinc]

because ker(f - id) is the same as the set of f(x) - x = 0
ie f(x) = x
but the set Im(f) is all the x of f(x) so not the same

[stone stump]

they are different

the first is a subset of the second

[devout snow]

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[maiden zinc]

that's what i thought

[devout snow]

@maiden zinc Has your question been resolved?

[devout snow]

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[floral turtle]

Am I supposed to have a whole number here?

[dense jay]

brackets

[rocky sparrow]

lack of brackets make it interpret as 2x + (7/x) + 2

instead of (2x+7)/(x+2)

[floral turtle]

You're right, thanks!

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[foggy snow]

if i have a function that is continuous, differentiable and positive, then would its sum be greater (>=) or its integral? or does it depend?

[violet wind]

sum?

like Riemann sum?

[foggy snow]

no, like a finite sum

$$\sum_{x=1}^{n} f(x) \geq \int_{0}^{n} f(x) \textrm{ or } \sum_{x=1}^{n} f(x) \leq \int_{0}^{n} f(x)$$

[uncut crow]

maybe needs more conditions

like maybe think about increasing and decreasing

[violet wind]

f(x) when the index variable is i?

[foggy snow]

oh my bad

[woven radish]

ibzi

[foggy snow]

if the function is increasing

[uncut crow]

also i think you can remove the differentiable and positive conditions

[foggy snow]

oh

okay, so if the function is increasing then the integral would be greater right?

[uncut crow]

continuous and increasing or continuous and decreasing should be enough to state inequalities like that

[foggy snow]

oh okay

[violet wind]

integrable and increasing

[foggy snow]

okay okay one second

[uncut crow]

can't tell if you're serious haha

[violet wind]

you don't get strict then but I don't see why not

[foggy snow]

wait is this correct ^

[violet wind]

Should be yeah

[foggy snow]

okay; awesome!

[violet wind]

If it's not continuous you can't guarantee they're not equal

Or if it's constant

If it's strictly increasing though then you're good

[uncut crow]

$\sum_{x=1}^n f(x)$ could be 0 when it doesn't look like it should be at all

[woven radish]

💜𝓁𝒶𝓎𝓁𝒶💜

[uncut crow]

by just making f(1), f(2), etc 0

[violet wind]

for increasing functions tho

f(a) = f(b) = 0 implies f([a,b])=0

In that case I'd say it definitely would look like it should be zero

[uncut crow]

ok true

[foggy snow]

does it work the other way around, i.e., can i say $\sum_{x=1}^{n} f(x) \leq \int_{0}^{n} f(x) \Longrightarrow$ mono increasing?

[woven radish]

ibzi

[uncut crow]

no, let f(x) = 0 when x is 1 or 2 and f(x) = 1 for everything else

and n=2

[foggy snow]

oh yeah!

again, thank you so much!

[uncut crow]

that's what i thought was an issue before it wasn't really haha

[foggy snow]

ohh

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[urban harbor]

was looking through an old classic textbook on Calculus by Courant and i think it contains an error..if h = 0, it's just 1 = 1 so is this lemma supposed to be >= instead of > or what am i missing?

[upper schooner]

Pretty much everywhere else I’ve seen that stated, it’s been a non strict inequality as well

[violet wind]

Seems to be strict whenever h≠0 which is weird

[urban harbor]

ok thanks, always love finding errors in print 😬

[violet wind]

guess they forgot lol

[urban harbor]

.close

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[sacred rock]

I'm at a point where I have worked out k-6 = 3x-2y and k+12=4y-x on Q2.3 how can I now prove point P lies on y=x+3 ( I have got this from the knowledge that Q= B*P )

[dense jay]

make k the subject of: k-6=3x-2y and 3k+12=4y-x
since k is constant you can then equate those 2 equations and rearrange for y

[sacred rock]

Okay Imma try that now I'll be 2 secs

[sacred rock]

Okay so I have y = (2x+9)/3

Which I suppose I can now make 2/3x + 3 ?

[dense jay]

how did you come to this?

[sacred rock]

So I got
k =3x - 2y + 6
and
k = 4y - x - 12

Then equated to each other to get 6y = 4x +18

divided through by 6

[dense jay]

k doesnt =4y-x-12

[sacred rock]

and simplified down to y = 2x + 9 / 3

hmm

[dense jay]

3k does

[sacred rock]

omg

So lemme take some steps back so it should be k = 4y-x-12 / 3 ?

[dense jay]

with brackets, yeah

[sacred rock]

Okay 2 secs

[sacred rock]

so y = (7x+30)/10 ?

[dense jay]

howd you get 7x?

[sacred rock]

(4y-x-12)/3 = 3x -2y +6

*3

4y-x-12 = 6x - 4y + 18
+x
4y-12=7x-4y+18

[dense jay]

3(3x-2y+6)=9x-6y+18

[sacred rock]

How did you get this ?

I can see the 3x-2y+6 is from the k = of equation 1

[dense jay]

(4y-x-12)/3=3x-2y+6, just multiplied both sides by 3

[sacred rock]

omg im so tired i aplogise'

3*3 = 9 not 6

braindead rn im telling you sorry xD

[dense jay]

how also did 3(-2y)=-4y😅

dw about it

[sacred rock]

should be -6y i actually had that written down at least XD

Lemme neaten up my work a sec so I can see where I'm at, theres a lot of scribbles rn

[sacred rock]

Okay so Just to check

K = 3x-2y+6
and
k = (4y-x-12)/3

[dense jay]

yup

[sacred rock]

(4y-x-12)/3 = 3x-2y+6

4y-x-12 = 9x - 6y +18

10y = 10x+30

y = x + 3

?

Is this enough to prove that point P lies on that line?

[dense jay]

it is, since p is the point (x, y) youve shown that y=x+3, so P lies on that line

[sacred rock]

Legendaryyyyy

i've been waiting for help since like 2 hours ago you're such a big help here

[dense jay]

no worries

[sacred rock]

any idea on the approach to Q2.4? I've only just learnt invariant lines and I've never done a question like this before is all

[dense jay]

its been quite a while since i did this
but the way i learnt it was to create 2 line equations, since you have B here, then we can have:
y=mx+c and y'=mx'+c, so an invariant line is where these 2 are the same, the y' and x' are y and x under the transformation

3x-2y=x' ->3x-2(mx+c)=x' 3x-2mx-2c=x'
-x+4y=y' ->-x+4mx+4c=mx'+c =m(3x-2mx-2c)+c

-x+4mx+4c=m(3x-2mx-2c)+c
x(-1+4m-3m+2m^2)+c(4+2m-1)=0
(2m^2+m-1)x+(2m+3)c=0

so then either m=-1, 0.5 or m=-3/2 or c=0
so since the m values arent consistent , we can have a combination of m=-1, c=0 or m=0.5 c=0

so y=-x or y=0.5x

its been a few years since i did this so forgive me for any errors

[sacred rock]

It's okay I appreciate the help, so Just the start bit, Is this equivilant to what I did with the last one where I started with Q= B*P but instead I'm doing [x' ; y'] = [3x-2 ; 4y-x ] ?

[dense jay]

basically yeah if you wrote it in matrix form its B( x y ) = ( x' y')

[sacred rock]

pefect okay so that makes sense to me

and then you just sub in y=mx + c to both of them equations

[dense jay]

yup

[sacred rock]

Or sorry do you sub in y'=mx'+c into the y' equation ?

[dense jay]

i replaced any y with mx+c and any y' with mx'+c

[sacred rock]

Rightttt

Then we wanna equate them and arrange to = 0 to find the possible values for m

and possible values for c

[dense jay]

when you say equate them, what do you mean?
from the 2 lines that has the first part 3x-2y=x'
first i replaced the y and y' with their respective equations, and then replaced x' in the second line with the equation for x' from the first

but then yes you want to check for possible m and c values

[sacred rock]

Yeah sorry that's what I meant, you used x' to eliminate the x' in the y' one so the only variables you're left with are x m and c

[dense jay]

yup

[sacred rock]

Okay so im caught up to there, now im confused with the m =-1, c = 0 and m=3/2 c = 0

does this mean that the invariance lines are at y = 3/2x and y =-x?

[dense jay]

we have (2m^2+m-1)x+(2m+3)c=0
for that equation to be true, then 2m^2+m-1 has to be 0, AND (2m+3)c=0

the reason i dont use m=-3/2 is because then the quadratic isnt 0 which we need it to be so that m value is discarded, forcing c to be 0

[sacred rock]

So m has to be -1 or 0.5

[dense jay]

yup, for the quadratic to be 0

[sacred rock]

Sweet

So the final answer should be y= 1/2x and y=-x ?

For both lines

[dense jay]

yes, those are the invariant lines

[sacred rock]

Love that ! thank you so much for your help XD

[dense jay]

no problem

[sacred rock]

I'm sure you'll catch me here again later

XD

.close

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[devout snow]

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[valid gale]

Why ∫^{y}_{0}∫^{x}_{1}\frac{1}{t}dt dx = ∫^{y}_{0}\frac{\log{x}}{\log{e}}dx

[fathom thicket]

Why $\int^{y}{0}∫^{x}{1}\frac{1}{t}dt dx = \int^{y}_{0}\frac{\log{x}}{log{e}}dy$

[woven radish]

doctor99268

[fathom thicket]

@valid gale should that be a dy?

and not a dx?

[valid gale]

Maybe

[devout snow]

@valid gale Has your question been resolved?

[restive river]

No it needs to be a dx

[fathom thicket]

I phrased that wierd. I meant, shouldn't that be a dx instead of a dy.

[restive river]

Oh u meant that, yeah

[devout snow]

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[placid rover]

!nosols

[devout snow]

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As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.

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[west raft]

suppose this is a function f(x)

[west raft]

and i want to find the unsigned area under the graph

when i do ∫f(x) the negative and positive area cancels out

if i want the unsigned area, can i just do ∫ | f(x) | dx?

[left robin]

For what x do you want the area?

Yes you can but it can be done easier

[west raft]

its my own question, not necessarily from a textbook. so uh chose lets just say 0 to 2pi idk

[left robin]

Sin is positive from 0 to pi and negative from pi to 2pi

So we can split the integral up

Or we can see that the area from 0 to pi is the same as from pi to 2pi, just negative
Therefore we can just get one area and double it

[west raft]

mhm makes sense. i just chose sin(x) arbitrarily, i was thinking about a general function

[west raft]

so ig i'll just do this?

[scenic pivot]

well if you solve the points f(x) = 0, you can see where the function is negative and positive, idk tho which way is faster

[upper schooner]

Doing that pretty much may require you to split the integral up into the parts where the function is positive and negative to use the piecewise definition of the absolute value

Depending on what you're integrating, of course

[west raft]

ohhh yea i remember doing a question like that now that u mention it

makes sense

alright got it thanks

.close

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[abstract sleet]

Unsure what the method would look like

[abstract sleet]

I understand the question (kinda) but I don't get how I'm supposed to show my working out

[shut folio]

the method is stated in the task: "by first factorizing"

use the formula for the difference of two squares

[abstract sleet]

Can't figure that part out for the life of me

[fair minnow]

a^2 - b^2 = (a-b)(a+b)

using this theorem you can factorise anything of the form a^2 - b^2

[abstract sleet]

Ah I get it now