help-39

[lost zodiac]

But I'm slightly confused about the angle and so my answer keeps ending up wrong.

[ashen ivy]

when you're drawing the triangle, two sides should be touching the two dotted lines

[lost zodiac]

I thought two sides where touching in pic? The b side is the side length of b and then tilted is a.

How would you have done it?

[ashen ivy]

these side lengths are not a and b, but this angle is 30 degrees whereas the first one was not

wow that quality 😔 but hopefully this helps you with what you have so far

[ancient slate]

I think there is an easier way

[supple hill]

Yeah, he needs to work with it.

[pearl pond]

@lost zodiac Has your question been resolved?

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[pearl pond]

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[tepid whale]

How do i do this

[plush bramble]

,tex .exp rules

[jolly parrot]

pi_day

[plush bramble]

write every term in terms of base 2

[tepid whale]

Ok

i got 2 to the power of 0 is that correct

[plush bramble]

!show

[pearl pond]

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

[plush bramble]

,rotate

[tepid whale]

Sorry its kind of messy

[plush bramble]

,calc 8^(2/3) * 2^5 / 64^(1/2) / 4^2

[jolly parrot]

Result:

1

[plush bramble]

yea 2^0 is right

[tepid whale]

Alright thanks

.close

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[pearl pond]

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[sinful vortex]

.close

[royal galleon]

can someone explain how this step is made

[rough forge]

dx

[royal galleon]

any ideas?

[steep saddle]

they took out a 1/x so is it by parts??

[royal galleon]

yes this is by parts where u = x^-n and dv = sqrt (ax+b)

[royal galleon]

oh wait are you talking about the what they did on the last two lines?

[rocky ridge]

I think They’re trying to isolate the integral so they move the non integral term

Then divide by smth maybe ?

[rough forge]

ts was a pain last time already

[steep saddle]

uhh were u not talking about that

[royal galleon]

ye I was talking about those. I misinterpreted your message

[royal galleon]

it is the only solutions for the textbook

[rough forge]

Let $I_n = \int\f{dx}{x^n\sqrt{ax+b}}$.
[ (1-2n)I_n-\f{2\sqrt{ax+b}}{ax^n}=b\f{2n}{a}I_{n+1} ]

[royal galleon]

this step was clear

[rough forge]

Ok then bring that to the other side and multiply both by x

[coarse plover]

hi i have a big question

🙁

[rough forge]

but they did ibp somewhere so yeah

[royal galleon]

wait that can't be I_n on the right side we have x^n+1

[rough forge]

I pulled out 1/x

[royal galleon]

oh ok

[rough forge]

wait i am dumb

[coarse plover]

The Three Holes of Time 🙁

[royal galleon]

how can you do that

[rough forge]

we are not allowd to do that

[royal galleon]

we can just do that?

[rough forge]

Well the proof starts off with an expression that holds for all n

[royal galleon]

Either way interesting technique

[rough forge]

think of it like a sub

let n=u-1 then if n!=1 iff u-1!=1 iff u!=2

[royal galleon]

thank you for the help

.solved

[pearl pond]

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[rough forge]

.tbh it looks more like that the restriction is a consequence rather than assumption

.so the sub intuition was a bit off and rather it's an actual index shift for all n

[plucky carbon]

I'm confused on what happens to the 2

[plucky carbon]

in the denominator

here is the triangle

41 is hyp

I understand how they got to 81/41 but where is the 2 going

cause 41/41 + 40/41 is 81/41

[rough forge]

you mean at the last step or what

[plucky carbon]

yes

[rough forge]

it's multiplied into the denom

[plucky carbon]

ahh I see I didn't realize that once you multiply your good and don't need the /2 anymore

thanks

.close

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[elder glen]

Gabagoo

[elder glen]

.solved

[pearl pond]

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[lilac jackal]

@elder glen do not abuse help channels

[elder glen]

My bad that was my little sister

So sorry

[lilac jackal]

i doubt that

[elder glen]

…

[lilac jackal]

even if it was her, ur account is ur responsibility

[pearl pond]

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[cunning compass]

How do i know in polynomial function what arrow will be the end behavior(up or down)

[ebon skiff]

Wdym

Like the limit towards +- infinity?

[sleek furnace]

the right hand arrow depends on the sign of the coefficient of the highest power of x

[pearl vigil]

Based on the degree of the function and the leading coefficient's sign

There's four cases, based on the combinations of positive negative sign and odd even degree

Like, imagine at negative infinity

The biggest/smallest number will be whatever has the highest power and will "overpower" the other numbers

(Like, a realllllly small number - negative infinity - to the power of 10, is a lot smaller than a realllllly smmall number to the power of 5)

So.... as the limit approaches negative infinity, it approaches whatever that value - the leading term - tends to

So. Let's say it is in fact a power of 10

10 is even

[cunning compass]

uh huh

[pearl vigil]

So, any number to an even power would be (positive/negative - what do you think?)

[cunning compass]

positive

i think

https://tenor.com/view/hmmmmmmm-hmmmmmm-hmmmm-hm-hmm-gif-2339932555530048784

[pearl vigil]

Right

So any number means both positive and negative infinity

will go towards the same value

And that value is (likely) positive or negative infinity

But to know whether it's positive or negatiev infinity both ends go toward

Look at the sign of the leading coefficient

Let's say it's... -(x^10)

We know x^10 at infinity/negative infinity reaches up toward infinity

So, with the negative out in front, does it reach up toward infinity or negative infinity now?

[cunning compass]

negativeee

so does it start like this

[pearl vigil]

Right

And the other end at positive infinity also goes down to negative infinity

[cunning compass]

cause its even

right

[pearl vigil]

An easy way to imagine the two cases with even degree are x^2 and -x^2

Both ends going down or both ends goign up

[sullen lichen]

can i ask smth here or do i have to go to a proper channel

[pearl vigil]

#❓how-to-get-help

this channel is currently in use

meanwhile, to imagine the two cases with odd degree, think about a cubic

[cunning compass]

since cubic is odd its uneven right

[pearl vigil]

A typical cubic looks something like this

[pearl vigil]

yeah

so both ends going in opposite directions

[cunning compass]

how do i know the starting position again

[pearl vigil]

wdym by starting position

like

in the middle around 0?

[cunning compass]

ts the only thing ill have too look at right? since if its odd same and even opposite

[pearl vigil]

yeah

[pearl vigil]

odd will opposite ends even will same ends

[cunning compass]

so how would i know it starts up/down

[pearl vigil]

so positive coefficient with odd degree will have left end going down right end going up

[pearl vigil]

based on the sign

like, -x^3 versus x&3

x^3

cus the sign flips it

[cunning compass]

like if the coeffecients positive or negative?

[pearl vigil]

yeah

like, -x^3 would have the left end going up and the right end giong down

[cunning compass]

i see

alr i get it

thanks g

.close

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[frank urchin]

not hitting the target would be 9/10

probability of hitting =1/10

[warm tiger]

Do you know about bernoulli trials

[frank urchin]

yeah i know it

[pearl pond]

@frank urchin Has your question been resolved?

[pearl pond]

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[short orchid]

Guys, why can’t we use the arc length formula here?

The theta x r = (pi/2) x a

[crystal dew]

in a or b?

[short orchid]

the b

[crystal dew]

why would the arc length help

all you would show is the length of the arc, not the function describing it

[pearl pond]

@short orchid Has your question been resolved?

[short orchid]

ohhhhh

thats right

thanku so much

.close

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[heavy stone]

hi!! im having trouble understanding tensors, levi-civita symbol & kronecker delta..
what's the difference between a higher order tensor and an nxm matrix? when we use the levi-civita ε how does it actually relate to a cross product?
question im trying to do:

[heavy stone]

and an attempt at another question using the levi-civita symbol that i got a 30% on

[last moth]

1. a matrix is just a way to represent a rank-two tensor (not to be confused with the concept of rank of a matrix from linear algebra), so just like you can refer to the elements of a matrix like A_{ij}, if you had a rank-three tensor, you would refer to its elements like A_{ijk}; nothing crazy here, just a change in number of indices
2. the relationship between the levi-civita symbol and the cross product is that the kth element of the cross product, (v x w)_k, is equal to the sum over i and j of e_ijk v_i w_j; basically it lets you write the cross product in this sum form that works well with einstein notation

does that answer your questions?

[heavy stone]

oh okay yes tyvm!

.close

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[lost zodiac]

Hey Im having trouble with this question

[proper nova]

!da2a

[pearl pond]

Asking the actual question right away is more likely to get responses.

Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.

[minor cloud]

hi OP, please lead with the question next time so that the bot pins it!

[lost zodiac]

One sec

[lost zodiac]

Sorry it didn't attach first time

[warped violet]

Uh

Is that the highest quality you have?

[lost zodiac]

Give me a sec

[warped violet]

Okay nevermind it's clear from text which lines are dotted

[foggy scarab]

wasnt this shared like 20 hours ago or smthing?

[lost zodiac]

Yes but I couldn't reopen

Okay the quality is not getting better. I'm sorry

[warped violet]

Yeah, no problem

It's clear enough

[lost zodiac]

Alright that's good

So I know how to solve this but my triangle is apparently wrong.

[midnight haven]

can you show what you tried ?

[lost zodiac]

This is the triangle I thought was correct

The two legs are both sides of the big and small triangle.

I wasn't too sure of the angle though.

[midnight haven]

ic

then

the drawing is abit messy but you can see blue part is a-b

[lost zodiac]

Yes I can see that

So you're saying I use the green triangle?

Ohhh so the angle would be 30 as well and side lengths a,b, and a-b??

Unless I'm confused

[midnight haven]

yes the angle of between lines is 30, so YXQ is 30

you know YP

you know XZ= b, ZY = a , you can find XY

if you consider the small tile the horizondal dotted line is a diagonal so XP is a diagonal of small tile .

we assume all tiles squares, you can use diagonals of a square bisect the angles, so RPQ = PZX

[lost zodiac]

I did sin30=( a-b)/sqrt(a^2+b^2)

[midnight haven]

you cannot because triangle XPY is not a right triangle

[lost zodiac]

Okay nevermind

[midnight haven]

you can use

sine rule

do you know sin rule

[lost zodiac]

Ofc

[midnight haven]

ok

can you find angle XPY

[lost zodiac]

XQY would be 90 degrees?

[midnight haven]

yes it is

[lost zodiac]

60

?

[midnight haven]

no

[lost zodiac]

Oh

[midnight haven]

XPY is the obtuse angle in triangle XPY

[lost zodiac]

That's stupid sorry I don't know why I did that.

Okay 120

[midnight haven]

no

how did you get 120

[lost zodiac]

Okay I genuinly don't know what I'm doing

[midnight haven]

tell me what you did?

[midnight haven]

use this fact

and you know all angles of a square are 90

[lost zodiac]

Oh

105

[midnight haven]

mm.. no

[lost zodiac]

Oh my god

It's very funny how I keep getting different answers

Okay so just to check, given what you said PQY is right angle both PXZ and PYQ are 45 degrees. And XPY is known 30

And XZP is 90

[oblique lynx]

What I'm doing

[lost zodiac]

?

[midnight haven]

no

[lost zodiac]

Oh my goodness that makes sense

[midnight haven]

now you cansee angle XPY

[lost zodiac]

Yes I didn't really take the rest of the square.

Thank you so much 🙏

.close

[pearl pond]

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[pearl pond]

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[fair tide]

I don't understand why mathematicians say "math doesn't care about the real world" and think that math is really abstract, real somewhat of a universal truth.
When all math is, is human made axioms that requires humans to perform (to do a calculation at all you need to understand what plus is or what the process or what "process" itself is.
its not like a computer is performing math when it runs calculations, all it does is moving atoms and regardless of whatever it's doing, "math" is human made)

So why the delusion?

[cinder flower]

[supple hill]

[cinder flower]

this just looks like 15 year old philosopher word salad i’m not gonna lie

[supple hill]

Not all of them, I have had great teachers and acquaintances who teach mathematics using examples of the real world. You are right.

[tropic saddle]

the mathematical object R (aka the set of real numbers) doesnt care about the fact that such a thing does not exist in the real world

thats the kind of thing that is meant

but really you are overthinking a bit

[fair tide]

but thats only because the axioms defining it don't lead it to do so right

if the person writing about the set of real numbers said R depends upon whether you feel your groceries are expensive or not

Then 'math' would be that

[fair tide]

and I don't understand how math is leading to the ultimate truth of some sort or an abstract truth, when if such things could have been math,

how would I be so sure that the current R is not something so utterly stupid as that R and that it is the real and correct way R should have been.

[rustic gate]

whether facts about the real numbers are true is independent of whether they're useful

[fair tide]

not whether they're "useful" or not, but the facts about them would've been True if the person/the paper talking about real numbers would've stated different axioms

[tropic saddle]

sure. and?

[rustic gate]

then they wouldn't be the real numbers that im talking about

[tropic saddle]

different axioms lead to different results

so what

[rustic gate]

the name "real numbers" is irrelevant

[fair tide]

how is it fine if a condition in real numbers is true if the creator of it says one thing and
It would've been a different thing that's True if the creator of it says another thing

[calm wing]

the name "real numbers" is a convention created by humans to be able to communicate some idea of real numbers between people.
atoms didn't name themselves either

[tropic saddle]

as long as people know what the creator said, who cares

I am currently talking to you in english. all of these words could have had a different meaning. does that mean that english is suddenly useless or something?

no, we agreed upon some meanings and are now working within that framework

english isnt "objectively true" or anything like that. its clearly still pretty useful

[calm wing]

as someone who is doing a degree in math, the biggest thing i've learnt about what mathematicians mean when they talk about some "universal truth" is that it's not the facts themselves that constitute the truth but rather the implications of them
we can define X to be whatever we want it to be, sure, but what follows from it according to mathematical deduction is fundamentally true: if we let X depend on someone's grocery prices then we can deduce that, say, X is a random variable and not a constant. it just so happens though that X is not the same thing as what we defined the real numbers R to be

[fair tide]

how do I know people have chosen THE correct mathematics?
I mean... If person a says real numbers are a sequence of 5x+3 where x is natural numbers
And person b says 3x+5

and people choose arbitrarily one or the other

[calm wing]

we can't know that

[fair tide]

then why is the math developed have any significance at all, other than the real world implications?

[tropic saddle]

have we chosen THE correct english?

who knows

does it matter?

no

[calm wing]

any significance at all, other than the real world implications?
sorry what?

this sounds oxymoronic

[fair tide]

like what's the difference between the current math concepts that are taught in university such as differential geometry

And a concept such as in economics where you are trying to get the prediction of groceries prices

both have axioms

I don't see the difference

why is what the economist doing with those axioms not considered as
'math'

[calm wing]

one is a social science the other is a natural science

[fair tide]

so? why should natural science or the other social science be considered as math

[calm wing]

one has a human factor the other doesn't

[tropic saddle]

well words have meanings. thats just how it is

what exactly are you expecting here

[calm wing]

if grocery prices increase because of a military conflict in an agricultural country causing supply chain shortages, what axiom is that

[fair tide]

it's an axiom regardless. it's not really well defined but it's an axiom is it not

so why is that not considered as math

[calm wing]

not really well defined
that's the difference

to not sidetrack too much, can you explain what kind of answer are you expecting here, exactly?

[pseudo oxide]

what's going on here

this seems interesting

[fair tide]

Well, to an illiterate I think it's well defined, to a highly literate mathematician it may not.
same in maths, to 20th century mathematicians, 5th century axioms don't seem well defined.
And 40th century ones would feel same for 20th century maths.

So why does it matter if it seems well defined or not

[plush bramble]

do you need actual math help?

you're just arguing semantics

#math-pedagogy or #math-discussion is better suited for your topic

[pseudo oxide]

lmao what

[fair tide]

.close

[pearl pond]

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[pearl pond]

Please don't occupy multiple help channels.

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[dense blade]

If there were no numbers how would we count?

[pearl pond]

Please don't occupy multiple help channels.

[supple hill]

Please, if you already have a channel close it by typing: .close.

[dense jasper]

possibly relevant. But also, help channels are for specific questions - you may find discussion channels like #serious-discussion to be more suitable.

[pearl pond]

If you are done with this channel, please mark your problem as solved by typing .close

[dense blade]

.close

[pearl pond]

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[fringe fiber]

It’d be great if this could get solved, I solved the other questions I got assigned but this has too many letters and I kind of hit a pipe when calculating

[summer imp]

Go at it one letter at a time.

[grim fractal]

I agree it'd be great if this could get solved

[wet osprey]

Show your work

[fringe fiber]

There

how come should you just add the exponenets instead of distributing everything

also

thats what i thought at first

<@&286206848099549185>

[wet osprey]

Is there a minus in front of the J³ on the top right?

[fringe fiber]

no

it's not

[wet osprey]

Well with this

This is just multiplication

And multiplication is both associative and commutative

So the brackets and ordering doesn’t matter

So you can just group each of the letters together first then add the exponents

[fringe fiber]

what exponent law is this again?

[wet osprey]

That a^b a^c = a^(b+c)

[fringe fiber]

ok thanks

the answer is correct right?

[wet osprey]

Yeah it looks good

[fringe fiber]

cool

.close

[pearl pond]

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[pearl pond]

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[empty junco]

my friend is claiming that he got this question correct on a test but it was marked wrong. is he actually right?

[empty junco]

someone

[ashen ivy]

,w d/dx x^(arctan(x))

[jolly parrot]

Results provided by WolframAlpha

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Requested by csmg_

[daring nebula]

That looks right to me

Maybe it’s not in simplest form?

[ashen ivy]

your friend did get the correct answer (according to WA), but they haven't shown a ton of work, and their final answer doesn't even have a dy/dx

and yeetus makes a good point, because $$e^{\arctan(x) \ln(x)} = \left(e^{\ln(x)}\right)^{\arctan(x)} = x^{\arctan(x)}$$

[jolly parrot]

haseeb ♥

[ashen ivy]

so not fully simplified

[empty junco]

he specified that the professor didnt say to simplify

i think most calc 1 professors dont ask to

i did it on my own and got the same answer as well

could be the dy/dx you mentioned haseeb, or incompetent TA grading lol

thanks folks

.close

[pearl pond]

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[ashen ivy]

i mean, it's a 16-point question, and at no point is the derivative actually taken, which is 😬

but also i dont know your TAs

[empty junco]

wait wdym

derivative not taken

[ashen ivy]

.reopen

[honest vigil]

(by the way, I don't see the x^{arctan(x)} in the first term of the derivative. it's highly likely I just missed it, which is not unheard of.)

[pearl pond]

✅ Original question: #help-39 message

[ashen ivy]

i just mean that there is no "dy/dx = ..." so it's not immediately clear that derivative'ing is happening

[ashen ivy]

i think they factored the exponential out to the right :)

[empty junco]

does it have to be? he uses an weird logarithmic differentiation technique to solve

[honest vigil]

oh, that explains the double parentheses. I'm sorry to have interrupted then.

[ashen ivy]

technically yes: the derivative is of the form $f'(x) e^{f(x)}$, for $f(x) = \arctan(x)\ln(x)$, so the first term is being multiplied by $e^f(x)$. but it's factored out for simplicity

[jolly parrot]

haseeb ♥

[empty junco]

ah ok

[ashen ivy]

unless you have any other qs or concerns,

.close

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[pearl pond]

Please don't occupy multiple help channels.

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[torpid basin]

Why are we having the -3 first in the integral, I thought we do top bottom, and -x^2-2 is on the top on this problem

[honest vigil]

,rccw

[torpid basin]

Same thing for the problem below, the -4 is being subtracted from despite the -y^2 equation being closer to the right

This is for the disc method being revolved around different axis

I hope my question makes sense

[vagrant trout]

the way you do it doesnt really matter in this problem

yes, the correct way would be top - bottom

but since it gets ², it would be positive regardless

only difference being that if you did bottom - top it would be a negative argument under the ²

this way the radius is negative

but for top - bottom radius is positive

@torpid basin hope that's clear

[pearl pond]

@torpid basin Has your question been resolved?

[torpid basin]

So does it matter if my radius is is or negative in my final andwer?

[pearl pond]

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[royal galleon]

Any advice regarding 36. So I noticed two things. I integrating both parts of the given we have $|f'(a)-f'(0)|\leq am$ and also since f takes its max on the interior of the interval it means f' goes from negative to positive with 0 somewhere on (0,a) meaning f'(0) is negative. That is the only way that I see of replacing the negative sign. But then it would be f'. Also it seems that we would split up the inequality with the triangle inequality. But then how do we know the split isn't bigger than am

[jolly parrot]

BigBen

[pearl pond]

@royal galleon Has your question been resolved?

[royal galleon]

<@&286206848099549185>

[pearl pond]

@royal galleon Has your question been resolved?

[plush bramble]

Your integration step is wrong

[pearl pond]

@royal galleon Has your question been resolved?

[royal galleon]

How?

[plush bramble]

You didn't show your work as usual

[buoyant cypress]

-# bruh piday is long gone

[plush bramble]

Blocked

[warm patio]

<@&268886789983436800>

[buoyant cypress]

https://tenor.com/view/seriously-are-you-serious-kid-blinking-gif-15637790

[pearl pond]

@royal galleon Has your question been resolved?

[pearl pond]

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[warm patio]

If a context-free grammar generates more words than it has rules, then does it generate an infinite language (this language contains infinitely many words)?
I would like to get a hint

[warm patio]

S->AA

A->a

A->b

alphabet {a,b}

aa, bb, ba, ab

ok so this is false

.close

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[dense storm]

For 3, howd i get it backwards

[sleek furnace]

$$\ln(e^{2x}-1)=\ln((-1)\cdot(1-e^{2x}))=i\pi+\ln(1-e^{2x})$$

[honest vigil]

you seem to have an extra starting parenthesis on the second ln.

[dense storm]

What

i pi?

Havent gotten up to complex numbers in calc yet

[sleek furnace]

well all u need to know is that ln(-1) is js a constant which can be brought out of the ln and absorbed into the constant of integration

[dense storm]

So

Ln-1

[jolly parrot]

notyalc

[sleek furnace]

is js a constant yes

[dense storm]

Nonreal tho

[sleek furnace]

it doesnt matter as long as its a constant it can be absorbed into the constant of integration

[dense storm]

Hm

[sacred nova]

shouldnt there be abs inside the ln

[dense storm]

Ive never been taught that

[dense storm]

Oh yh

Wait no

Abs for definitie

Definite integral

This is indefinite

[sacred nova]

I'm not sure that's allowed in normal calc

[sacred nova]

since when is absolute value only for definite though

[sleek furnace]

i mean thats the most direct way to prove that both of these antiderivatives r valid

[dense storm]

For cambridge its like that

[sacred nova]

$\exp{x} - \exp{-x}$ isn't > 0 for all R so it'd be wrong to not put absolute value idk

[jolly parrot]

ant

[sacred nova]

ugh I forgot how to do e in latex but ykwim

[sleek furnace]

$$e^x-e^{-x}> 0 ,\forall x\in\mathbb{R}$$ is not true?

[jolly parrot]

notyalc

[dense storm]

Wait no actually for this chapter it does use abs

[sacred nova]

no it's false for all negative x clearly

[dense storm]

So ig theyre the same?

[sacred nova]

if x = -1 then it'd be (1/e) - e which is negative

[sleek furnace]

no like this is what u meant right

[sacred nova]

yes it's what I meant

[dense storm]

Abs(e^2x -1)= abs(-1(1-e^2x) = 1-e^2x ?

[sleek furnace]

$$|e^{2x}-1|=|(-1)\cdot(1-e^{2x})|=|1-e^{2x}|$$ yes

[jolly parrot]

notyalc

[dense storm]

For definite integrals the larger bound is ontop right?

[sleek furnace]

not necessarily

[dense storm]

So if the integral had pi/4 ontop and pi/3 on the bottom id solve normally by substituting pi/4 thingy - pi/3 thingy

[sacred nova]

the upper bound is the one you evaluate first

[dense storm]

+- area are the same essentially?

[sacred nova]

no

like integral from b to a which sounds absurd would just be the opposite of integral from a to b

if ones negative the other is the same but positive

[sleek furnace]

$$\int_{b}^{a}f(x)dx=-\int_{a}^{b}f(x)dx$$

[jolly parrot]

notyalc

[sacred nova]

so if you put the bounds 'the wrong way around' then positive area would yield a negative number and viceversa

[dense storm]

The question asking me to prove integral=negative area

[sleek furnace]

can u take a pic of the qs

[dense storm]

4

[sleek furnace]

!xy

[pearl pond]

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

[sacred nova]

they're not really asking for anything about area, just to show that that definite integral equals that

[sleek furnace]

do u know double angle identity for sine?

[dense storm]

Ik how to do it

Just why area negative

[sleek furnace]

nothing is stopping an integral from being negative

negative area integrals happen a lot

[dense storm]

So

Negative area only true when you substitute pi/4 and then minus pi/3

So basically it doesnt matter the order of the bounds you always solve the one on top first?

Alr

Thanks🙏

.close

[pearl pond]

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[torn flicker]

so yk how to do right?

[pearl pond]

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[sweet junco]

so I have another maybe not so easy question but is there like any way to like not do this thru guesswork

[feral sedge]

yes: remainders

[honest vigil]

guesswork is required for the first answer, but as soon as you reach n = 6, one fact about cubes mod a particular positive integer may help you.

oh, sorry for interrupting.

[feral sedge]

you're good

[sweet junco]

kk thanks

is it

like cuz

wait

[sweet junco]

what about this

[honest vigil]

you will find that, mod one particular number, cubes and that expression in your question will never agree.

[sweet junco]

is it 5

[honest vigil]

it is not the number I'm thinking of.

also, my apologies if that hint violated nosols.

[feral sedge]

You're very safe there

[sweet junco]

I'm currently kind of almost indefinitely clueless

[feral sedge]

what is n!+5 mod 5 for large n?

[sweet junco]

0

[feral sedge]

right

[sweet junco]

so that means it must be a multiple of 5 and the primes beforehand

which it can't be

a cube

I can't word it but it's there

thanks

.close

[pearl pond]

Channel closed

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[feral sedge]

I'm not sure I see it

I wouldn't expect 5 to work here

[honest vigil]

should I at least say that mod here then?

[sweet junco]

.reopen

[pearl pond]

✅ Original question: #help-39 message

[feral sedge]

the cube would have to be 0 mod 5

But that's possible

[smoky gull]

5 does work here as you hvae a perfect exponential factors for each prime and it also satifies the modulus condition!

[smoky gull]

Wait i might be misinterpreting what you meant here

[feral sedge]

what do "perfect exponential factors" mean?

[honest vigil]

n = 6 gives 6! + 5 = 725 = 0 mod 5, but 725 is not a perfect cube.

[smoky gull]

oh wait yeah sorry i was thinking smth completely different!

[sweet junco]

I realize I got the wrong logic down

hmm

mod 125?

[honest vigil]

too large and probably not helpful?

the mod I'm thinking of is a single-digit number.

and cubes output values that are very limited mod that number.

[sweet junco]

8?

mod8

[honest vigil]

close.

[feral sedge]

5 is it's own cube mod 8

[sweet junco]

what

[honest vigil]

I think you've earned the mod.
I'll give the mod to you, then you can try working out the rest.
work mod ||9||.

[sweet junco]

can you point me the direction as to how it's 9

how it went to 9

[wary bane]

theyre suggesting you try 9

[honest vigil]

because once you reach n = 6, note that factorials all become 0 mod 9 because of the presence of the second 3 from the 6.

and cubes mod 9 are very limited (from playing with perfect powers).

[worthy lotus]

Yoo gang what is Question?

[sweet junco]

725 5045

ohhh

it's always just going to be 5

wait how does that help

[honest vigil]

so you found out that your expression will always be congruent to 5 mod 9.

now consider cubes mod 9. what can they be? is 5 one of them?

[sweet junco]

1 8 0 1 8 0 1 8 0

it repeats

wow

[feral sedge]

Oh I was gonna do mod 7

[sweet junco]

I didn't know that

[feral sedge]

I guess this works too

[honest vigil]

more importantly, you now see that mod 9, there are only three possible values for cubes.

and our 5 is suspiciously missing from the list.

[sweet junco]

does that also work?
1 1 6 1 1 6...
that also works whilst done manually

[honest vigil]

oh yeah, mod 7 works too. sorry, didn't think of that.

[sweet junco]

I didn't know cubes do this

[sweet junco]

yeah I get it now

[feral sedge]

No need to apologize, either one is good

[honest vigil]

point is, perfect powers have quite interesting patterns mod certain numbers, and you gain insight from playing with them in number theory.

[sweet junco]

I can follow up w questions right I have a few

a lot of them are also kinda super different

[feral sedge]

7 and 9 both have the same unit group structure

for nth powers, any prime that is 1 mod n is a natural choice for seeing these kind of patterns

[sweet junco]

isn't this like 289-17-1+12+2

or am I missing something

[feral sedge]

well as stated it looks like infinity + infinity + infinity

if we add the condition that a has to be between 1 and x, then your values for phi(13) and phi(4) are right

[sweet junco]

the question was written wrongit's supposed to be no. of relatively prime positive integers __<__x

[feral sedge]

ok, only phi(169) needs fixing here

or

phi(289) I mean

lol

[sweet junco]

oh wait I added -1

[honest vigil]

I was wondering how come the totient function was defined differently here.

[worthy lotus]

Umm what is question?

[feral sedge]

here

[sweet junco]

oh

damn I just got a number wrong in my sol'n

so uh

I'll return with questions later

.close

[pearl pond]

Channel closed

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[pearl pond]

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[fading nexus]

I’m not sure what to do for this one

[rough forge]

,,\sin(2x)=2\sin(x)?

[warped violet]

Well

[warped violet]

Oh

lol

[fading nexus]

First homework for this part of the unit so

[mortal seal]

Oh wait I misunderstood

Sorry

[rough forge]

You started well, with the 2sin(x)cos(x)

Now bring everything to one side and factor

[fading nexus]

huh wdym to one side

[wet osprey]

nah there's a better way me thinks

[rough forge]

Bring the 2sin(x) to the left

[wet osprey]

if i have 2sin(x) * cos(x) = 2sin(x), either 2sin(x) is 0, or cos(x) = 1

[humble igloo]

sin(2x) = 2sin(x)
2sin(x)cos(x) - 2sin(x) = 0
sin(x)(2cos(x) - 2) = 0
sin(x) = 0
x1 = 2 * pi * n
2cos(x) - 2 = 0
2*cos(x) = 2 | :2
cos(x) = 1
x2 = 2 * pi * n

[wet osprey]

well i suppose they're really the same method

[rough forge]

Factoring is safer

[wet osprey]

!nosols

[pearl pond]

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.

[humble igloo]

Ponil

sry

[fading nexus]

sorry im completely lost

[wet osprey]

yeah do 2sinxcosx = 2sinx and then bring everything to one side and factorise

[fading nexus]

but if I subtract -2sinx then I would only have cosx=0

[wet osprey]

if you do 2ab - 2a do you get b?

[humble igloo]

To solve this, you need to know the formula for the sine of a double angle, apply it to sin(2x), then move 2sin(x) to the left, factor out the common factor, and you get the product of the two factors equal to 0, so you can set each one to 0 separately and solve these two equations.

[fading nexus]

yea I know the double angle formula

[rough forge]

So many helpers

[humble igloo]

I used a translator, sorry if something is wrong.

[worthy lotus]

All are free

[rough forge]

,rccw

[fading nexus]

I’m struggling with the factor out part because I am confused how I do not just get cosx=0

[rough forge]

Can u first write down what u get after you factor

[worthy lotus]

How u substract sinx from 2sinx cos x ?

[wet osprey]

what do you get when you do 2ab - 2a

[worthy lotus]

Yep ans this

[fading nexus]

I don’t know

[wet osprey]

you've written 2ab - 2a = b

does that sound right to you?

[humble igloo]

Point 5, third line, then move 2sin(x) to the left side to get an expression equal to 0

[fading nexus]

Huhh

[worthy lotus]

SHe having trouble with 2ab-a wait gang

[wet osprey]

glav you really arne't helping at this point

[humble igloo]

Okay, I'm powerless) He need to watch a video on YouTube or something.

[worthy lotus]

Ye wait glav for sec

[wet osprey]

@fading nexus go back here

[fading nexus]

I don’t think I factored correctly

[humble igloo]

Wow

U really good

Correct

[wet osprey]

that's exactly right

[fading nexus]

Would the next step be x= smth and x= smth?

[wet osprey]

not quite

can you see that you can divide both sides by 2?

[fading nexus]

Yea

I don’t know how I would divide the stuff inside the () tho

[humble igloo]

So, you got a product of 2 factors equal to 0. You need to understand that for it to be true, it is enough that one of the 2 expressions gives 0, so we can sort of split this expression into 2, each of which is equal to 0.

[wet osprey]

now recall your knowledge of quadratics

remember when you did something like (x-a)(x-b) = 0, then you know x - a=0 or x-b = 0?

[fading nexus]

Or would the stuff inside the () stay the same because the two is infront of it?

[wet osprey]

well they're multiplied so you only need to divide the 2 by 2

it's like going 2ab = 0 and then ab = 0

[humble igloo]

Correct

[rough forge]

Bravo

[fading nexus]

Would sinx=0 have an x that equals 2pi or no?

[wet osprey]

what about cosx = 1

[fading nexus]

Since it is in the domain but not included

[wet osprey]

depends if it's in the domain

[0, 2π) does not include 2π in the domain

[fading nexus]

Kk

[fading nexus]

Would it also be x=0,pi?

[humble igloo]

2 * π * n, n ∈ Z, in both cases

[wet osprey]

not Z man

[humble igloo]

)

[wet osprey]

cos pi is not 0, cos pi is -1

[fading nexus]

Oh yea whoops

So just 0??

[wet osprey]

yeah

this is the picture btw

[fading nexus]

Also off topic love the profile

[humble igloo]

sin(x) = 0, x = pi * n, n ∈ Z
cos(x) = 1, x = 2 * pi * n, n ∈ Z

[wet osprey]

thank you!!

[rough forge]

Now find the intersection with [0,2π)

[fading nexus]

Am I factoring wrong?

[humble igloo]

(4x - 3)|x + a^2| + |x - 1| + 3a^2)^2 - (a + 1)(4x - 3)*|x + a^2| + |x - 1| + 3a^2) + 4 = 0

Can somebudy help me? I need "a", that's give 2 roots

[rough forge]

use z=sin(x) to get a quadratic equation in z

[wet osprey]

!occupied

[pearl pond]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[pearl pond]

Someone else is already using this help channel. If you need help with a question, please open your own help channel/thread (see #❓how-to-get-help for instructions).

[fading nexus]

Huh

[worthy lotus]

What problem ?

[fading nexus]

7

[wet osprey]

something went wrong on line 3

and by something i mean you lost a square on the 2sin(x) on line 4

[fading nexus]

Ohh

[wet osprey]

yeah thats still cooked unforunately

[rough forge]

\color[HTML]{067676}
[ 1-2\sin(x)^2=\sin(x) \implies 1-2z^2=z]

[wet osprey]

somehow you managed to factorise a 1 with a 2sin²x

[wet osprey]

they dont even share any common factors i dont know how you factorised them like that

[worthy lotus]

Ig she having trouble in solving quadratic into factor

[fading nexus]

yea I’m just confused

[wet osprey]

well you have 1 - 2sin²x - sinx right now

[fading nexus]

I’m very bad at factoring or finding the greatest common factor my whole class is my teacher thinks it’s cus of covid

[wet osprey]

this you can't really factorise (unless you're just a god at it it's hard to see)

[fading nexus]

I’m not sure what my next step should be then

[wet osprey]

notice that this is a problem of a quadratic in sinx (this is a hard thing to see at the beginning don't worry if you're unsure about this)

and we know how to solve quadratics

[worthy lotus]

@fading nexus you shoud first learn factorisation thing than u can do all this thing quick

[rough forge]

duh

[fading nexus]

Easier said than done 😅 I’ve tried it just won’t click I still can’t even do stuff like long division. I got 1.5 months of school left so

[fading nexus]

I feel like I usually do but not in this situation

[rough forge]

You woldnt know how to do the right side?

[wet osprey]

well don't worry first we need to see that it is a quadratic

[fading nexus]

I don’t really know what is up with those Zs

[wet osprey]

a quadratic looks like a🟩² + b🟩 + c

normally we have an x for the 🟩

[rough forge]

it's just a variable, you can take anything

[wet osprey]

but you can really replace it with anything you like

[fading nexus]

Ok

[wet osprey]

in this case we replace it with sin(x)

so we have (-2) (sinx)² + (-1) (sinx) + 1

or just -2sin²x - sinx + 1

[fading nexus]

what are we replacing

?

[wet osprey]

do you agree this is a quadratic in 🟩

[fading nexus]

Like I believe you but I’m bad with vocab so I don’t think I would be able to tell you what a quadratic is

[wet osprey]

have you seen ax² + bx + c before

[fading nexus]

I’m sure I have I just don’t remember seeing it but I have a crappy memory

[rough forge]

Do you have notes perchance

[fading nexus]

Would I have been using that equation this unit?

I only have this units notes

[wet osprey]

yes you should be revising and remembering what you did in other units

maths is built so that you revisit old content in new contexts

[fading nexus]

This is the work we did yesterday

[wet osprey]

so it's vital you don't forget the old content

[wet osprey]

like look at Q6, im sure you met the difference of two squares previously a² - b² = (a-b)(a+b)

but now you're using it with the trig functions because that's how maths is built on top of itself

[fading nexus]

I understand that part with like the a^2-b^2 but I don’t see how it apples to the current one

[wet osprey]

no i gave that as an example that you see the old content coming back

so here the old content that comes back is quadratics and factoring

it will be helpful to first revise those before you continue onwards

[fading nexus]

To be fair for number six that is the one literally nobody understood so my teacher tried to show us on the board

[wet osprey]

i think you copied it wrong as well lol

it's the sin²x + cos²x that becomes the 1 not the other bracket

[fading nexus]

Yea I need new glasses 🫩

My -5 prescription is some how to weak and I need a stronger one 😭

[wet osprey]

😭

[fading nexus]

I got an appoiment for next week because I also stepped on my glasses and broke them -_-

and they are barley holding on right now

Back to the problem I just don’t know how the other quadratic’s apply

I’m just completely lost with this whole part of the unit tbh I felt good ij class yesterday

But now I’m lost

[rough forge]

Maybe take a break

[supple hill]

Try Substitute $\without x$ for $a$ and $\cos x$ for $b$

[jolly parrot]

Ga³¹Br³⁵I⁵³9000✞
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)

[fading nexus]

I could but I got a whole hour of my free left because i had a double free

I also haven’t been only staring at this problem I worked on another problem in between getting help

which maybe is not good

But Idk

[supple hill]

Yeah, the better it's focalise in one.

[wet osprey]

if you can i'd say try look at some factorising problems and quadratic problems for now

maybe refresh your memory on what do you do in those cases

[supple hill]

It will give you the practice to then do that's.

[wet osprey]

it'll feel good if you remember how to do them and you'll be in the right mindset for this question as well

[fading nexus]

I can usually factor with like normal numbers or stuff that just has x but when stuff with cosx and sinx get introduced I can’t

[wet osprey]

it's a very important skill to have

you close your eyes and go "hide it"

[wet osprey]

once you can see that it is indeed a quadratic you can do this to make it more clear

here we will solve for the z first then figure out later what the x should be

[fading nexus]

Different problem we are providing the identities now but im not sure what else I can do

[warped violet]

You can think of sin(x) as just another variable

like m

It just has more letters

[fading nexus]

I try but I always get stuck when there’s both cosx and sinx in the thing I’m trying to factor

That’s a blurry photo I just realized sorry

[civic saddle]

youre trying to prove identities?

[fading nexus]

Yea

[civic saddle]

whats up with the last line ( + 2sin(x) (1-cos^2(x)) )?

that seems to have come from nowhere

without that line youre actually painstakingly close

[supple hill]

You can extract sin x as common.

[fading nexus]

How ? I just don’t know because sinx is not inside the ()

[fading nexus]

Tbh I don’t even remember

[verbal whale]

What do you mean?

[fading nexus]

That line is wrong

it should of been sinx

But that would not be a good step

[supple hill]

Notice that you have \without x (\text{block A}) + 2\without x (\text{block B}). Since the \sin x is multiplying on both sides of the, you can take it out as a common factor

[fading nexus]

I just don’t know how to factor it it would be smth weird like 2sinx(1-cos^2x)

And then I would get almost the right answer but wrong

[verbal whale]

#help-39 message

[fading nexus]

Like I tired factoring but I got this which is all wrong

[verbal whale]

Why do you keep writing that 1 - cos²x out of nowhere? That's the problem 😅

[fading nexus]

Maybe the -2 is a -1?

Crap

I erased the wrong line then

Oh wait

For that one is because I factored

Like j factored 2sinxcos^2x into 2sinx(1-cos^2x)

[civic saddle]

AB + AC = A(B+C) is the correct factoring method

sin x (2 cos^2 x - 1) + 2 sin x cos^2 x

[supple hill]

Use that.

[civic saddle]

or to be more visually visceral,
sin x * (2 cos^2 x - 1) + sin x * 2 cos^2 x

[fading nexus]

I’m lost

[spring pine]

i still dont understand what you were doing

[civic saddle]

whoops sorry that was a mistake

[fading nexus]

Trying to find the greatest common factor

[spring pine]

of what

[fading nexus]

2sinxcos^2x

Oh wait I think I was thinking of those variables are seperate but I should be thinking of them in one

[supple hill]

Search between the two blocks you have adding.

[civic saddle]

just so we're on the same page, you have gotten: \$\sin 3x = \sin x (2\cos^2 x - 1) + 2\sin x \cos^2 x$

[jolly parrot]

Boi || 보이

[fading nexus]

Would the factor be sinx(2cos^2x)?

[civic saddle]

thats right -- now factor out sin x from both terms

excuse my handwriting

[fading nexus]

How could I do that tho if it is addition?

[civic saddle]

we can factor the sinx exactly because its addition

[fading nexus]

I know the 1s are not necessary but they keep me on track

[civic saddle]

youre missing the square from
2 cos² - 1

also it helps to be consistent -- choose one between cos x and just cos (shorthand) throughout your practice

[fading nexus]

Can I combine 2cos^2x-1 and 2cos^2x into 4cos^2x-1

[civic saddle]

think of cos² x as a single variable

oh yes

good job

[fading nexus]

why can we combine them even tho one of them does not have a -1 ?

[verbal whale]

Have you ever done factorization before? 🤔

[fading nexus]

isn’t that what we are doing?

[verbal whale]

Sure

But you seem to be quite confused 😬

[fading nexus]

my whole class is bad at it

So my guess is we are supposed to learn it in middle school

But Covid happened

If I’m combing the terms -2sin^2x and -2sin^2x would I get -4sin^2x

Nvm

Because I feel like I am doing this wrong

[civic saddle]

my neck broke

[verbal whale]

Sure, but that's indeed "combining like terms". I wouldn't call it "factoring" in that case (even if it would be correct)

[fading nexus]

I’m also just slow

[verbal whale]

,rccw

[fading nexus]

I don’t think my teacher would put problems with ^4

[verbal whale]

Pay attention to notation
$\cos 2x + 2x \neq \cos(2x + 2x)$

[jolly parrot]

Alberto Z.

[fading nexus]

Oh yea I just forgot the ()

I don’t know if I should start with the other side instead but I would have no idea where to start with that side

I was trying to use the box so I could expand those terms then combine them but then I would get a bunch of things powered to the fourth

Would 1-8sin^2x be equal to 8cos^2x?

[spring pine]

should be 8-8sin^2x

[fading nexus]

I thought it would be at least cos?

I’m so confused

[worthy lotus]

??

Need help ?

[fading nexus]

I mean I only have 5 minutes now

So idk if I can do anything with that

[worthy lotus]

Question 21?

[fading nexus]

I’m trying to use a pyth idenity

[fading nexus]

Yes

[worthy lotus]

Hmm try to open cos 4x into cos 2x ig

[fading nexus]

I tried but then I got a bunch of things powered to the forth so I’m trying to use the right side now

[fading nexus]

^

[worthy lotus]

Cos 4x = 2 cos^2 2x - 1 right ?

By using cos (2x) =2 cos^2 (x)-1

[fading nexus]

Oh

I did not think of that

Wouldent you have to separate them ?

[worthy lotus]

??

[fading nexus]

I just don’t know how to write that on my paper

Like wouldent I have to seperate the cos4x first

[worthy lotus]

Umm cos 4x here u can right cos 2(2x)

[fading nexus]

Into cos2x and cos2x

[worthy lotus]

?? Don't u know that cos 2x property?

[fading nexus]

I do

But I don’t know how to apply that to it if it is cos4x

[worthy lotus]

Umm think 4x=2a

[fading nexus]

Like i thought i would have to make cos4x into cos2x and cos2x so they are separated then I could use the cos2x property

[worthy lotus]

Now expand cos 2a

[worthy lotus]

No see

[fading nexus]

I have to go but I’ll likely be back later 🫩

[worthy lotus]

Ok ok

[worthy lotus]

Just see this and u will get it

[fading nexus]

I don’t really understand what you mean by that 😭

Because isn’t 2a the same thing as 2x

And I can’t just think it’s something differently

[worthy lotus]

For expanding cos 4x think 4x like 2y now question is cos 2y

[worthy lotus]

You have to

[fading nexus]

But then I would be missing th second cos2x

[worthy lotus]

Just for left side after expanding interchange 2y=4x again or y=2x

Wait

See this if this helpful

Variable is 2 is going in expansion did u get it what I mean to say ?

Or you can do long way also which u were trying but that time taking

[pearl pond]

@fading nexus Has your question been resolved?

[pearl pond]

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[safe prairie]

im so lost

[queen rock]

Have you tried anything

[grave mirage]

Looks like bayes

[safe prairie]

Pr(some key is wrong from one press) is 1-p
Pr(some key is wrong and got a1 from one press) is (1-p)/(n-1)
Pr(any key is wrong and got a1 from one press) is (1-p)(n-1)/(n-1) = 1-p = Pr(a1 from any key other than k1)

[safe prairie]

if k1 corresponds to a1, do i start with Pr(k1|a1 cap k1|a1)?

independent so id just ^2

[queen rock]

Condition on which impulse was chosen first, then apply bayes

[safe prairie]

they ask for probability of correct impulse

so i made k1

but theres no further info on k1

idek if picking any key takes equal chance

[queen rock]

Each impulse is equally likely yes

[safe prairie]

are these correct then

[safe prairie]

now i know this can i say that
Pr(a1 given k1) = p
Pr(a1 and its correct) = p/n
Pr(wrong but a1) = 1-p from above?
Pr(k1) = 1/n

[queen rock]

I think the first two are correct but on the last one, you didn't include the probability of choosing a wrong impulse

So you would have to divide by n there

(n-1)/n times (1-p)/(n-1)

[safe prairie]

say k2 got a1, the key got a 1-p chance to be wrong, it got (1-p)/(n-1) chance to be wrong and get a1. why cant i say that if n-1 keys were to attempt this mistake, the Pr would just cancel out the n-1?

[queen rock]

1 key is chosen with probability 1/n

those n-1 keys don't all happen at once if that makes more sense

[supple hill]

Use that $$P(I_1 | E) = \frac{P(E | I_1) \cdot P(I_1)}{P(E)}$$

[jolly parrot]

Ga³¹Br³⁵I⁵³9000✞

[queen rock]

I'm sure they're aware of the bayes theorem

[supple hill]

Oh.

[safe prairie]

now im confused whether knowing a key is wrong have any conditional ties with knowing a result is wrong

does {wrong key AND wrong print AND a1} make sense?

[queen rock]

No whether it's wrong depends on what the intended key was

If the intended key is K1, then getting a1 is not wrong

If its any other key not equal to K1, then a1 is wrong

[safe prairie]

oh chance of wrong key is (n-1)/n

[queen rock]

Yep indeed

[safe prairie]

i kept thinking how can wrong key probability be 1/n

[queen rock]

Let me know if you get the right answer now

[safe prairie]

I’ll try this

[queen rock]

Don't square the whole final probability, just the probabilities of the two independent prints

[safe prairie]

so [Pr(k1|a1)]^2 itself is wrong?

[safe prairie]

cuz idek which block is my independent print

[queen rock]

Yes, the key is chosen once not twice so that wouldn't make sense

[safe prairie]

should it be {k1 cap k1 | a1 cap a1}

actually just 1 k1 cuz 100% the second one follows the first

[queen rock]

That just simplifies to P(K1 | a1 cap a1)

[safe prairie]

yes

[queen rock]

The key event happens once, the output event happens twice so you only have to combine the output events

P(K1 | a1 on both prints)

[safe prairie]

then a1 cap a1 is the thing that i can just a1^2?

[queen rock]

You square the probability not a1 itself, but yes

So just the notation is incorrect

[safe prairie]

then Pr( k1 cap a1 cap a1 ) would ne p^2/n is it

i multiplied everything