help-13

[crimson sedge]

3x - 3x = 0

so like

-2x + ??? = 0?

[tidal leaf]

2x? i really don't know?

[crimson sedge]

yeah

it is 2x

-2x + 2x = (-2+2)x = 0x = 0

that's the idea happening here

so adding 2x to both sides gets you what?

[tidal leaf]

0?

[crimson sedge]

okay uh

[
3 - 2x = x \Iff 3 -2x \c b{+2x} = x \c b{+2x}
]

[crimson sedge]

that's what we are doing

like we said -2x + 2x = 0

sooo

[
3 - 2x = x \Iff 3 -2x \c b{+2x} = x \c b{+2x}\Iff 3 = x + 2x
]

[crimson sedge]

so now, what is x + 2x?

[tidal leaf]

3x?

[crimson sedge]

great

3x = 3 what is x

[tidal leaf]

i don't know, im sorry

[crimson sedge]

like

you got 3x

you wanna turn it into x

what do u do to achieve that

what operation transforms 3x to x

[tidal leaf]

subtracting by 3x?

[crimson sedge]

no that will get u 0

ok you would need to divide both sides by 3

3x/3 = x

right?

[tidal leaf]

yeah?

[crimson sedge]

yeah

so

3x = 3

divide both sides by 3

what do you get

[tidal leaf]

x

[crimson sedge]

equals to?

[tidal leaf]

1?

[crimson sedge]

yes

congrats

[tidal leaf]

im sorry that took so long

[crimson sedge]

it's alright but take it as a wake up call that you should probably practice a lot more. Try giving Khan academy's videos about like-terms a watch

they'll help

[tidal leaf]

okay, thank you

I probably should because I have to keep my grades up for volleyball

[crimson sedge]

volleyball?

[tidal leaf]

yep

we have eligability check

[crimson sedge]

ahh i see ok

well goodluck try your best

[tidal leaf]

thank you, I will

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[wind nexus]

Hi, not sure if this is the right place to ask since I'm not asking about a particular question but more about how to self-study effectively. I just got back to math and I'm trying to work through AOPS v2 but on chapter 2 I'm running into back to back problems that stump me.

This is what the problems look like: https://i.imgur.com/fdq2dmc.png

I solved 14-21, 21 already took me a while.

For problems 22-25, I worked on them for a bit but basically gave up and skipped them without looking up an answer. I was hoping to get a better grasp of things then get back to them. But took a quick glance at 26 and I'm probably gonna have to skip that one as well... I'm starting to think if I keep going this way I'll skip through the whole book.

How do I balance struggling with a problem vs skipping it vs looking up help vs looking up the answer and trying to understand it fully? I don't want to "spoil" the answers for myself then lose the practice problem but I'm not getting any better at solving these problems out of thin air...

[silent finch]

This is also a problem I’ve been having, and I think the best way to deal with it is to really think about each problem and take your time

Because the solutions will usually use tools you already know, it’s just that you didn’t think of doing it a particular way

[wind nexus]

how much time do you give yourself per problem?

[silent finch]

If I face a hard problem I usually will skip it and think about it throughout the day

Like let’s try #26

What are you thinking of doing at first?

[wind nexus]

lemme think

so working backwards $cos(2x) = cos^{2}(x) - sin^{2}(x)$

[wraith dagger]

nchoosek

[wind nexus]

that's gott abe useful

is what I'm thinkin

[silent finch]

Yep that makes sense

Is there any way to simplify it though?

The cos^2 or sin^2 looks like it could be dealt with

To leave an expression in terms of only sin or cos

[wind nexus]

well it's also either $2cos^{2}(x) -1 or 1 - 2sin^{2}(x)$

[wraith dagger]

nchoosek

[wind nexus]

but they still have the ^2

[silent finch]

Yeah so that motivates us to solve for either sin or cos

Or sin^2 or cos^2

So how can we get squared terms with sin(x)+cos(x)=-1/5

[wind nexus]

$sin^{2}(x) = \frac{1}{25} + \frac{2cos(x)}{5} + cos^{2}(x)$

[wraith dagger]

nchoosek

[silent finch]

Ok we can do it this way

If we add cos^2 to both sides what happens?

[wind nexus]

well left side is equal to 1

and if we subtract that 1 from the left on the right we get 2cos^2(x) - 1 which is equal to cos(2x)

so -2cos(x)/5 - 1/25 is equal to cos(2x)

but I'm not sure if we're really any closer 😅

I also got 5cos2x = sin x - cos x

[silent finch]

You can solve for cos(x)

[wind nexus]

oh... right... algebra

instead of doing what I did and bringing in the unworkable cos(2x) xd

well it's either 7/25 or -7/25 but I can't figure out if it's positive or negative

I figured it was the negative one since 2x is in the fourth quadrant and we're looking for the cosine

but turns out it was the positive and tbh I still don't know why🤷‍♂️

oh my dumbass should have started thinking about the quadrants when getting the answer for what cos x is not after cos(2x) lol

welp this shit took me 40 minutes with hints and I still made a sign error in the end

D:

thanks for the help @silent finch

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[inner flax]

This is from my professor

how did he jump frrom 3+2x^2 to 1+2(x^2 +1)

[raw gulch]

very well, now introduce it as sum of three integrals

he noticed, that such a trick, like he did, helped him to in the secodn integral

[inner flax]

what was the trick??

[raw gulch]

yoou said

you were surprised

that was his trick

[inner flax]

???

[lapis pollen]

he broke up the 3 from 3 + 2x^2 into 2 + 1

oops

my bad

didn't mean to reply to that

but

anyways

[inner flax]

all good

[lapis pollen]

he broke up 3 + 2x^2 as 2 + 1 to get 1 + 2 + 2x^2

he then grouped 2+2x^2 and factored out the 2

to get 1 + 2(1+x^2)

[raw gulch]

$\int_{0}^{1}\frac{dx}{x^{2}+1}+\int_{0}^{1}2dx+\int_{0}^{1}cos\frac{\pi\text{}x}{2}dx$

[wraith dagger]

Joanna Angel

[raw gulch]

can you see it now ?

MIDDLE

INTEGRAL

IS

SUEPR EASY

yoru professor knew it

and used his triock

to achieve it

trick = method = noticement

[lapis pollen]

ohhh I get it now

I see the trick

once he got 1 + 2(x^2 + 1), he broke up the first fraction into two, to get:
(1/(x^2 + 1)) + (2(x^2 + 1)/(x^2 + 1)

and the x^2 + 1 cancelled out

[humble karma]

It's really just a clever way to rewrite the numerator, but in general, whenever you have the same degree polynomial on the numerator/denominator, you can do something similar (or long division, you'd get the same).

[inner flax]

Im still lost

[inner flax]

I understood that part

im saying how he got the numerator of the first fraction into the second

[humble karma]

Do you agree that they are the same numerator but just rewritten in a different way?

[inner flax]

I see it

[lapis pollen]

ye its this

[inner flax]

thanks, winter break brain was just warming up to all this math

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[jovial eagle]

what does | mean? given a function f(x) and 3 coordinate points

[wanton sail]

can you give an example of where you saw this?

[jovial eagle]

so given function f(x) and 3 points with the condition and these points create a right triangle, so i need to determine u so that the area of that triangle is max

[wanton sail]

It's probably just meant to be a comma

[jovial eagle]

im not asking for a hint right now, im trying to understand what the symbol means

[wanton sail]

(0, 0) (u, 0) (u, f(u))

So that the x-coordinate of the first point is 0 and the y-coordinate of the first point is 0

And then the x-coordinate of the second point is u and the y-coordinate of the second point is 0

And so on

[jovial eagle]

ill assume that is a comma then

.clsoe

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[hoary wedge]

not rlly a math question more physics but why do larger wavelengths diffract more than smaller wavelengths?

[hoary wedge]

also can somebody check if my reasoning for why smaller openings diffract more than larger openings:
Waves passing through a smaller opening will diffract more than larger openings because the margin to pass through is much smaller, forcing the waves to fit through.

[dire geode]

#old-network for physics server

[hoary wedge]

that server is so dead

im sorry

ive asked like 5 questions there and have only been responded to once

[dire geode]

Doesn't mean you should occupy a math help channel

[hoary wedge]

.close

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[hoary wedge]

seen so many ppl do this and i get called out 4 it

thanks

[dire geode]

I call out plenty

[wanton sail]

Physics Qs are okay if they're primarily mathematics; this question isn't mathematical at all though

[hoary wedge]

okay dude

im not gonna sit here and argue cuz ik im wrong

[cedar kiln]

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[versed fulcrum]

Hello

[versed fulcrum]

I sort of don't understand the concept of dot product and cross product

People explain it's to show how long the two vectors are for dot product

But isn't that basically vector subtraction?

[flint plinth]

loosely speaking, the dot product is measuing how much "shadow" a casts on b, or vice versa

[versed fulcrum]

The distance of the gray dotted line here?

[flint plinth]

what does the grey dotted line signify, it's not connected to the tip of the blue vector, what is it connected to?

[versed fulcrum]

Or the distance between the origin and where the gray dotted line intersects with blue vector

[versed fulcrum]

The dot product?

[flint plinth]

if it's meant to be at a right angle to the blue vector, then it's related to the dot product but isn't the dot product

[versed fulcrum]

The orthogonal point?

I'm so confused

[flint plinth]

if we call the pink vector a and the blue vector b, then the grey vector would be something like
a - (a dot b)b/norm(b)
where the second term is the projection of a onto b

[crimson sedge]

Rather that, yes. But it then has to be multiplied by the length of the blue vector

[mighty shuttle]

essentially the dot product is the projection of one vector onto another

[versed fulcrum]

I never understood what projection meant

[crimson sedge]

@versed fulcrum You know, if what you're looking for is intuition then you might be interested in 3blue1brown's linear algebra series

[versed fulcrum]

What's the difference between projection and dot product?

[versed fulcrum]

Oh this make sense

[versed fulcrum]

But my question still holds

[versed fulcrum]

So this explains why dot products are scalar

[mighty shuttle]

the projection is a*b/|b| unless I'm mistaken

[versed fulcrum]

icic

[mighty shuttle]

where * is the dot product

[versed fulcrum]

•

[mighty shuttle]

yeah sorry, didn't know how to type that

[versed fulcrum]

What about cross product?

Copy it for reference 🙂

[crimson sedge]

@versed fulcrum I don't know whether you're interested in video form math but here is link to the video series:
https://www.youtube.com/watch?v=kjBOesZCoqc&list=PL0-GT3co4r2y2YErbmuJw2L5tW4Ew2O5B

[mighty shuttle]

that gives an output vector perpendicular to the other two

[versed fulcrum]

Thank you

[versed fulcrum]

How long tho

[versed fulcrum]

Oh that's why there are terms like into the page and out of the page

[mighty shuttle]

magnitude of $|a||b| sin(\theta)$

[wraith dagger]

Why am. I here

[versed fulcrum]

This plus direction?

[mighty shuttle]

where $\theta$ is the angle between vectors a and b

[wraith dagger]

Why am. I here

[mighty shuttle]

this is just the magnitude so it's always positive

or 0

[versed fulcrum]

Oh

How do you know whether it goes into the page or out of the page?

[mighty shuttle]

the right hand rule

[versed fulcrum]

It can be the opposite tho

Depending on the perspective

No?

[mighty shuttle]

https://en.wikipedia.org/wiki/Right-hand_rule

[versed fulcrum]

Oh okay

[mighty shuttle]

we use the right handed coordinate system by convention

[versed fulcrum]

What does eigenvalues do

[mighty shuttle]

out of and into the page depends on perspective, yes

[versed fulcrum]

Gotcha

[mighty shuttle]

I haven't studied that yet, so I can't help, sorry

[versed fulcrum]

Oh no worries

Thanks a lot for the dot and cross though

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[sharp kestrel]

Is this correct?

[dire thorn]

Show work

[sharp kestrel]

1/2pi(2)^2

[dire thorn]

Is that it?

[sharp kestrel]

yea thats how I got to the answer

so is it right or wrong

[dire thorn]

Hate to break it to you but 1/2pi(2)^2 does not equal 1/16pi

So that makes no sense

!noans

[cedar kiln]

The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.

[dire thorn]

Show an actual attempt

[sharp kestrel]

.close

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[dire thorn]

Rip

[cedar kiln]

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[turbid prism]

what did I do wrong

[warped coyote]

Maximum at x=0 is not met

[turbid prism]

theres apparently one little missing thing here?

wait

got it

.close

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[solemn bronze]

A wheel with a radius of 14 inches rotates at a speed of 0.2 radians/seconds. What is the linear speed, v?

[flint plinth]

do you have any thoughts so far?

!status

[cedar kiln]

What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above

[flint plinth]

btw, "A wheel of 14 inches" is ambiguous, does that mean radius or diameter or something else?

[solemn bronze]

oops its the radius sorry

im assuming the linear speed would be 0.2 rad/sec

but im not sure

[tropic oxide]

no,

if it's in radians per second, it's an angular speed.

do you know the relationship between angular speed and linear speed?

[solemn bronze]

they are related in that v=r(angular speed)?

[tropic oxide]

yes

[solemn bronze]

and angular speed=v/r

[tropic oxide]

$v = r \omega$ is how that's usually phrased.

[wraith dagger]

Ann

[tropic oxide]

anyway you yourself just stated the formula you need.

[solemn bronze]

but how would i find the angular speed without knowing the linear speed

[tropic oxide]

you have it backwards

you know the angular speed (omega)

you want the linear speed (v)

[solemn bronze]

so the angular speed would be 0.2 rad/sec?

so angular speed is always in radian per second?

[tropic oxide]

not always

angular speed is <angle unit> per <time unit>

one commonish one is RPM, which stands for revolutions per minute

[solemn bronze]

is it correct to say that the linear speed is 2.8

0.2*14

[tropic oxide]

units!

[solemn bronze]

2.8 rad/sec?

[tropic oxide]

wrong units.

linear speed isn't measured in radians per second.

what units was your radius given in?

[solemn bronze]

inches

inches per second?

[tropic oxide]

yes.

[solemn bronze]

ok

got it thanks

[cedar kiln]

@solemn bronze Has your question been resolved?

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[smoky prairie]

hey guyss, i need help

[smoky prairie]

how do i factor this if i dont have anything that can add up to -2, while having a product of 2 at the same time?

[candid mason]

??? There are no conditions for factoring. Remember a(b+c) = ab + ac.

For your case end up with fractional coefficients for x and a fractional constant term

[smoky prairie]

so how would i end up with a fraction?

[candid mason]

So what is -2 /12 * 12

[smoky prairie]

-24/12 or -2

[candid mason]

-2. Basically anytime you have a/b * b that is just equal to a. A fraction is an expression of division and division is the opposite of multiplication. So basically the /b * b cancel each other out.

Now you have 12 out. You need to find the value of c such that. 12(cx) = -2x. The solution to this is just c = -2/12

[smoky prairie]

im a bit lost

where will i use -2/12

[candid mason]

So you need to find the values of c1 and c2 such that 12(x^2 +c1 * x + c2) = 12x^2 - 2x + 2

[smoky prairie]

yup so how do i get c1 and c2?

[smoky prairie]

i use -2/12 there?

i clueless

[candid mason]

You are using a(b+c) = ab + ac. This distributive law. What you want to do is equate the coefficients.

[smoky prairie]

yeahhh so how do i equate it? 😭

im so sorry

LMAOO

[candid mason]

Suppose we cx + b = 2x + 9. What are the values of c and b

[smoky prairie]

2 AND 9

[candid mason]

Bingo

So how do you expand 12(x^2 +c1 * x + c2)

[smoky prairie]

i'd jst get 12(x^2-2+2)

[candid mason]

Expand something is basically rwmoving the brackets if you a(b+c) if you expand that you will get ab+ ac.

So how do you expand 12(x^2 +c1 * x + c2)

[smoky prairie]

12x^2+12x+12

[candid mason]

Yes but you have remove the c1 and c2 for some reason.

12x^2 +12 * c1 * x + 12 * c2

Since 12x^2 +12 * c1 * x + 12 * c2 = 12x^2 - 2x + 2

From there you equate the coefficients

And that will be the answer to your question

I have to go unfortunately

Bye

[cedar kiln]

@smoky prairie Has your question been resolved?

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[smoky prairie]

okok thankss

.close

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[crimson hornet]

Would anyone be able to help me figure out how to solve this?

[mighty shuttle]

can you find the electric field between the two plates?

hint:- It's a parallel plate capacitor

[crimson hornet]

Would it be E = V/d

[mighty shuttle]

yes

[crimson hornet]

Oh ok so I would probably be using Fe = Fg?

[mighty shuttle]

correct

[crimson hornet]

Oh okay thanks I’ll try it from that

@mighty shuttle if you don’t mind, could I ask you another question?

[mighty shuttle]

sure

[crimson hornet]

Thanks

I’m not too sure on how I find the magnetic field in this

[mighty shuttle]

think about how you would achive this

btw

are you cheating on a test, no right?

[crimson hornet]

Nah this is just homework questions

[mighty shuttle]

hint:- ||qvb=qE||

[crimson hornet]

I’m just kind of confused on how I would find the direction of the magnetic field

[mighty shuttle]

it would be perpenddcular to the elctric field

[crimson hornet]

Cause I know that the electric field is up so would it just be down?

Oh okay, how would I know on what direction it is perpendicular?

[mighty shuttle]

find it using the right hand rule

think about the direction of the force due to the electric field

and hence in what direction must the force due to the magnetic field act

ok, gtg . ATB

[crimson hornet]

Ohh okay gotcha thanks

.close

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[half elk]

Hi

[half elk]

So I was reading this paper, can you explain why off-diagonal values are 0

I am not sure if I had it right, so please point out if there's any mistakes

[proud torrent]

it shouldn't be hard to see why that element is zero if you just look at it

[cedar kiln]

@half elk Has your question been resolved?

[half elk]

But can't it be a band matrix?

[proud torrent]

diagonal matrices are banded

[half elk]

I mean, for a 1<i<m and 1<j<n=m and i ≠ j J_w(i, j) being non zero

[proud torrent]

if you write down what element on the position (2,3) looks like you'll quickly see it's zero just like this one

and same thing applies to any other element on position (i,j) with i != j

[half elk]

Understood, thanks for the help

.close

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[craggy stratus]

Quick question on formality

[craggy stratus]

When I'm doing partial fraction decomposition (?), should I still use the integrate symbol?

As in in the A/(x-1) + B/(x+1) step, should there be a integral symbol and a dx there too?

[dapper sail]

bro idk much about integration im only 14

but i think you dont need to

you're comparing both sides to find value of A and B right

[craggy stratus]

bro im kind of a wing-it math guy so im not entirely sure what im doing

but yes i believe that is the case

like we have 1/(x-1)(x+1)

and we want to find out what the coefficients are for each fraction(?) in order to get the nominator

each parenthesis

[dapper sail]

yes right

[distant jacinth]

Yeah, without the integral symbol, the equality is not correct.

The two sides are not equal.

[dapper sail]

ahh like that

[nova snow]

yes remove the integral symbol and it will be correct

[dapper sail]

what if you remove the integral and dx from the left side

[distant jacinth]

That's fine.

[craggy stratus]

@nova snow so remove the integral symbol on the left?

[nova snow]

You don’t have to

but it will still be correct

or if you want to keep it

You have to put the Partial fractions as an integrand

[craggy stratus]

im sorry mate im a swede so english isn't my native language and when it comes to math it's chaos to try and understand the english terms

integrand?

[nova snow]

1/(x-1)(x+1) = A/(x-1) + B/(x+1) is how I would do it

[dapper sail]

yeah

or you can just do (x+1) - (x-1) in numerator

and then divide the whole thing by 2

[craggy stratus]

so remove the integralsymbol and the dx then?

[dapper sail]

precisely

[nova snow]

Integrand is the stuff inside the integral, for example

the integral of f(x) dx

the stuff between the integral sign and the dx

is the integrand

[craggy stratus]

does this look correct then?

sorry for the occasionally small writing q.q

Is it "ok" to start writing downwards if its a continuation of the same step?

ah so either I need to not have the integral sign at all for that step, or I need to have it for both of them?

[distant jacinth]

You can write downwards, as long as it is clear where you are continuing from.

And yes, this equality is wrong:

[craggy stratus]

so in this case it's ok?

[craggy stratus]

implicity sign instead then?

[nova snow]

I wouldn’t use an equal sign there to write your PFD stuff

[craggy stratus]

like this?

[nova snow]

I would probably do something like

PFD: (insert working out)

[craggy stratus]

i have so much issues with when to use equal vs implication vs equivalency signs :((

so after the 1/x^2-1 dx just write PFD: and then continue doing that?

[distant jacinth]

Ah, the English word is "implication".

[nova snow]

Yes that’s one way

[distant jacinth]

And no, an implication can only be used between statements. The integral itself is not a statement.

[craggy stratus]

what is this called?

equivalency is apparently not the word -_-

[distant jacinth]

That's biimplications (or equivalences, yes).

[craggy stratus]

which is stupid because that's literally the direct translation

well that makes more sense

last question then

when should I use

=

vs

=>

vs

<=>

?

[nova snow]

If and only if

[craggy stratus]

Because as you can see our professors use the biimplications here and last time I tried to do it during a test (which the logic that you use them if you can go from equation A to equation B)

this is what happened:

don't remember the exact question but it was smth like prove that ln(a^b) is the same as b*ln(a)

and then the b part was some graphing stuff

[distant jacinth]

You can only use => or <=> between statements. A statement is something that is true or false.

[craggy stratus]

isn't this wrong then

o_o

[nova snow]

Don’t use the symbol “if and only if” when what you really mean to say is equals “=“

[craggy stratus]

but isn't that what they're doing here

[distant jacinth]

What you did wrong was to put <=> between log(8^x) and something else, because "log(8^x)" is not a statement. It has no truth value.

[craggy stratus]

if it's the way that you say it is then it makes a lot more sense for me

yes but for me you can go from log(8^x) to xlog8 and you can go from xlog8 to log(8^x)

but isn't that what's happening in the image I sent above?

they say 5x+4 = A(x+2)^2 + B(x-1)(x+2) + C(x-1)

[distant jacinth]

This is fine, because the top is an equation, which is a statement. It is either true or false. That statement is equivalent to the next line and so on.

[craggy stratus]

so how should I have expressed it in order for it to make sense?

2log8^x = 3

<=> x2log8 = 3?

[distant jacinth]

Yeah, and log(8^x) is equal to xlog(8), so here you can use log(8^x) = xlog(8), because they are the same thing.

[distant jacinth]

Yes, this is ok.

[craggy stratus]

ok so the dumb way to explain it to me then is

i should only use <=> or => with an equation afterwards?

that has a = in it?

so it becomes a statement?

[distant jacinth]

Yeah, sort of. That is a good rule of thumb anyway.

The reason why I said "statement" instead of "equation" is that you can have <=> and => between all kinds of things, as long as they are statements. Another example is inequalities.

Like

x + 5 < 7
<=> x < 2

[craggy stratus]

makes sense

i'll try to remember and use that in the future and hopefully I won't end up with tons of questionmarks followed by "what?" written out on my papers XD

[distant jacinth]

Yeah, you can tell that his circles became increasingly more aggressive as he corrected your hand-in. 😆

[craggy stratus]

thanks a ton! i might open a few more threads while going through this old exam so see you soon :P take care! <3

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[hybrid ravine]

Im studing for an olympiad and this came in the past papers, how do i solve this?

[raven shard]

what progress have you made so far? have you drawn a diagram?

[hybrid ravine]

I tried to but I don’t understand the question completely

[raven shard]

which part?

[hybrid ravine]

The ending part. Can’t the degree change depending on how you fold it?

[raven shard]

no

there is only one way to fold the page along DP such that A lands on MN

the crucial idea here is that you fold along DP, meaning you can't fold it however you want

[hybrid ravine]

But then how does that help us find ADP

[raven shard]

this is where having a diagram would come in handy

while it may not be obvious now, once you draw something you can visualize what's going on more easily

[hybrid ravine]

Ok on it

Oh wait got it

Thx for ur help

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[raven shard]

welcome

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[junior stratus]

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[tardy barn]

Heya, I have a problem that says I have to express the very top expression in terms of cos x, am i doing this right?

[tardy barn]

i think the very bottom part can still be simplified or some sort

i just dont know how

LCD?

[idle tusk]

yes

combine into one fraction

[tardy barn]

did i do it right?

it shows a different graph or i just did it wrong xd

[idle tusk]

you are missing parentheses around (1+sin(x))

[tardy barn]

oh wait its correct now haha

[idle tusk]

yep

now, simplify $(1-\sin(x))(1+\sin(x))$

[wraith dagger]

artemetra

[tardy barn]

i think the denominator is difference of two squares?

[idle tusk]

yes

correct

[tardy barn]

holdup holdup

ok im lost now, what after this?

oh wait

[idle tusk]

what is 1-sin^2(x) equal to?

[tardy barn]

do i combine the one in the numerator?

[idle tusk]

recall

$\sin^2(x) + \cos^(x) = 1$

[wraith dagger]

artemetra

[tardy barn]

oooh, its equal to cos x ^2?

the denominator?

[idle tusk]

mhm

precisely

[tardy barn]

now i think i need to do something with the numerator? since its kinda messy lol

[idle tusk]

mhm

factor out cos(x)

on the numerator

on the numerator you have $(\cos(x))(\text{thing})+(\cos(x))(\text{thing2})$

[wraith dagger]

artemetra

[idle tusk]

so factor out cos(x)

[idle tusk]

to get $(\cos(x))(\text{thing}+\text{thing2})$

[wraith dagger]

artemetra

[tardy barn]

lemme try

[idle tusk]

mhm

[tardy barn]

do i do something with the sin?

[idle tusk]

yes

you essentially have $1-\sin(x) + 1 + \sin(x)$

[wraith dagger]

artemetra

[idle tusk]

which you can for sure simplify

[tardy barn]

is this correct??

[idle tusk]

yep

that's your answer

[tardy barn]

ohh

eyyyy

thank youu :))

[idle tusk]

no problem

if you are done type ".close"

[tardy barn]

alrighty

thanks again :)

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[wind nexus]

I don't understand this https://i.imgur.com/pb5VY5G.png

[wind nexus]

A and B are coordinates, how do you multiply a coordinate by X

oh I guess X is a coordinate as well since it says points X but I still don't understand multiplying coordinates together

x with x y with y?

[idle tusk]

probably

[wind nexus]

does this question click for you? I even know the answer I still don't get it

[idle tusk]

i think so?

i am pretty certain it's equivalent to this:

[limber marsh]

what no, AX the length of the line segment from A to X here

[idle tusk]

omg

💀

my bad

[wind nexus]

Oh

LMAO

it looks like freakin A * x^2

[limber marsh]

i made this mistake before, so no worries

[wind nexus]

ok the answer makes sense now thx guys

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[restive rover]

My head is already smoking from trying to solve this.I have different ideas for this but i feel like i need some clarity cause i mightve lost important thoughts in process

[restive rover]

Imagine a m x n Matrix A and we take 2 J which cardinality is between 1 and n.And for the number of elements in J we create a matrix consisting of the colums index of a which are in J (so between 1 and n).

So to make it simple if J is {1,..,i} the new matrix A_J consists of the first i colums of A so its a m x i Matrix

and it is to us to prove the statement including these ranks for every J_1 and J_2

the tip they gave us is that we can use the dimension formula from linear algebra

but i dont know how i can show that x <= y in this statement

[cedar kiln]

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[marble ermine]

hello

[marble ermine]

$f(x) = \begin{cases}
1, & \text{if } x \geq 0 \
0, & \text{otherwise}
\end{cases}
$

[wraith dagger]

lilisworld.

[marble ermine]

is f continuous or discontinuous at 0? So it is discontinuous but

i dont know how to do the epsilon delta proof, i tried to check if it was continuous on R+ and R- and on R- i don't how how to show that the delta does not exist

[solid juniper]

what does continuous at 0 mean? then what does not continuous at 0 mean?

[marble ermine]

$\forall \epsilon > 0, \exists \delta > 0 : \forall x \in \mathbb{R}, |x| < \delta \implies 1 < \epsilon$

[wraith dagger]

lilisworld.

[marble ermine]

on R-

that is on R- but if it is on R then

$\forall \epsilon > 0, \exists \delta > 0 : \forall x \in \mathbb{R}, |x| < \delta \implies |f(x) - 1 | < \epsilon$

[wraith dagger]

lilisworld.

[marble ermine]

or do i not solve it like that? i need to do in on R even if its expression is different on R+ and R-?

[solid juniper]

what are you trying to do?

[marble ermine]

i want to know if the definition works for all epsilon and x

[solid juniper]

this has some issues

[marble ermine]

yes

i shouldve written R-/{0} for all x

[solid juniper]

the probably bigger issue is the 1 < epsilon

[marble ermine]

oh ok

|f(x) - 1| < epsilon?

|-1|< epsilon?

[solid juniper]

ok before we even talk about that, what are you trying to do?

[marble ermine]

since idk if it is continuous or not, im going to check if the definition is true on R+ and R- at 0

uhm

[solid juniper]

weird way to say whatever you’re trying to say (probably something about left and right continuity)

[marble ermine]

yes but i think i know why it's wrong

so how do i start

[solid juniper]

personally i wouldn’t even bother thinking about left and right continuity, but you can do it like that

[marble ermine]

do you mean that you would just find a contradiction?

[solid juniper]

no

you have a definition for “continuous at a point” right?

[marble ermine]

yes

[solid juniper]

write it out (with the point being 0)

[marble ermine]

$\forall \epsilon > 0, \exists \delta > 0 : \forall x \in \mathbb{R}, |x| < \delta \implies |f(x) - 1 | < \epsilon$

[wraith dagger]

lilisworld.

[solid juniper]

yes

now what does “not continuous at 0” mean?

[marble ermine]

wait

$\exists \epsilon > 0, \forall \delta > 0 : \exists x \in \mathbb{R}, |x| < \delta and |f(x) - 1 | \geq \epsilon$

[solid juniper]

no

but close

[marble ermine]

and instead of implies

[solid juniper]

yep

[wraith dagger]

lilisworld.

[marble ermine]

and now what do i do

[solid juniper]

so we want to show there is an epsilon with a property

part of that property is that |f(x)-1| >= epsilon for some x

[marble ermine]

yes

[solid juniper]

what are the possible values of f(x)-1?

[marble ermine]

0 and -1

[solid juniper]

yea so possible values of |f(x)-1| are 0 and 1

[solid juniper]

sooo no epsilon over 1 will help here

and all the epsilons less than or equal to 1 should be “the same”

as in like

if one of them works for the proof so should the rest

[marble ermine]

so if i take epsilon is 1/2, i need to find x such that |x|< delta and |f(x) - 1| >= epsilon. I can take x = ?? for ex?

ah uhmmmm ok i see thannks

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[crimson sedge]

How do I solve this:
lnx+x-1<0

[raw gulch]

consider the function f(x) = lnx + x - 1, note that it is continuous and differentiable for x > 0, and examine the monotonicity of f, then you will deduce something about your inequality

[crimson sedge]

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[clever crow]

say I have a weighted coin that has a 10% chance of landing heads and a 90% change of landing tails
I flip the coin 10^6 times,
whats the probability I get at least 1000 heads or 1000 tails in a row?
how would I find this?

[clever crow]

I dont need to be precise, a rough estimate would do

[fair geyser]

you can be preicse, there's two cases, either it happened for the first time at the last moment, or you already had the streak

it's like fibonacci

[clever crow]

can you elaborate? I didnt understand

[fair geyser]

i'm writing the code

if we're looking for a streak of 3 heads and there was 7 flips

X X X T H H H where XXX doesn't have a streak
X X X X X X O where XXX has a streak

these are mutually exclusive and cover all options where we get a streak

the first option has probability of [1 − F(n−4)] × P(THHH) and the second is just F(n−1)

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[faint yarrow]

How do I do #1

[faint yarrow]

from doing synthetic division it seems like writing the linear factors means i would just do x(x+1) but when i multiply those two I get x^2+x so x^4(x+1)? is that the linera factors?

idk

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[formal pasture]

i did this completely wrong

how do u do this

[hollow trail]

i don't see anything obviously wrong. remember at the end, that's still an equation (with the other side = 0) which you can solve using methods for solving quadratics

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[formal pasture]

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[mild sail]

if im square rooting both sides

im getting rid of the ^2

or does it not work because i would be doing sqrt(theta)

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[velvet mortar]

Looks good to me

What is question

[cedar kiln]

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[kindred hornet]

how would i do this?

[raw gulch]

$\text{you have to deduce that:}\\\frac{pq}{1\cdot 2\cdot _{\cdots }\cdot \left( p-1 \right)}=\frac{a}{b},\text{ }\text{ }q\in \mathbb{N}$

[wraith dagger]

Joanna Angel

[raw gulch]

it is little bit time-consuming 🙂

[kindred hornet]

ah

[kindred hornet]

so use lcm basically?

[raw gulch]

do you ask about the continuation of what I have written? or how I ve deduced it ?

[kindred hornet]

the latter

but also a hint for how to continue

[raw gulch]

you know, the ending is simple, you need to use proportions, and then think about what p will be the divisor, but to get to what I wrote, you need to transform your sum appropriately, you need to come up with an idea for grouping words

terms*

i show you one line:

$\frac{1}{1}+\frac{1}{2}+{\cdots }+\frac{1}{p-1}=\\=\left( \frac{1}{1}+\frac{1}{p-1} \right)+\left( \frac{1}{2}+\frac{1}{p-2} \right)+{\cdots }+\left( \frac{1}{\frac{p-1}{2}}+\frac{1}{\frac{p+1}{2}} \right)=_{\cdots }etc$

[wraith dagger]

Joanna Angel

[raw gulch]

we must rememebr here, that p is an odd number since p >2

I don't want to take away the pleasure of thinking about it,

[kindred hornet]

oh i think i see

[kindred hornet]

im kinda confused with how you got the pairings

[raw gulch]

that needs time to come up, i agree ))

[cedar kiln]

@kindred hornet Has your question been resolved?

[kindred hornet]

i evaluated the expressions and im still confused

[raw gulch]

$\frac{1}{1}+\frac{1}{2}+{\cdots }+\frac{1}{p-1}=\\=\left( \frac{1}{1}+\frac{1}{p-1} \right)+\left( \frac{1}{2}+\frac{1}{p-2} \right)+{\cdots }+\left( \frac{1}{\frac{p-1}{2}}+\frac{1}{\frac{p+1}{2}} \right)=\\=\frac{p}{1\cdot \left( p-1 \right)}+\frac{p}{2\cdot \left( p-1 \right)}+_{\cdots }+\frac{p}{\frac{p-1}{2}\cdot \frac{p+1}{2}}=$

[wraith dagger]

Joanna Angel

[kindred hornet]

yeah thats what i got

[raw gulch]

reach form of one fraction

[pale arch]

Can anyone help me with this question

[kindred hornet]

#❓how-to-get-help

[kindred hornet]

with lcm i assume?

[raw gulch]

yes) you know, eng is not my native, so soemtimes i need to think ab expression ))

so finally you get what i wrote at the beginning

$\frac{1}{1}+\frac{1}{2}+{\cdots }+\frac{1}{p-1}=\\=\left( \frac{1}{1}+\frac{1}{p-1} \right)+\left( \frac{1}{2}+\frac{1}{p-2} \right)+{\cdots }+\left( \frac{1}{\frac{p-1}{2}}+\frac{1}{\frac{p+1}{2}} \right)=\\=\frac{p}{1\cdot \left( p-1 \right)}+\frac{p}{2\cdot \left( p-1 \right)}+_{\cdots }+\frac{p}{\frac{p-1}{2}\cdot \frac{p+1}{2}}=\\=\frac{pq}{1\cdot 2\cdot _{\cdots }\cdot \left( p-1 \right)}$

[wraith dagger]

Joanna Angel

[raw gulch]

q is just notation for natural number here

[kindred hornet]

ohh

thank you!

[raw gulch]

yw 🙂

[cedar kiln]

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[kindred hornet]

.close

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[tender cape]

How do I even do this

[warped coyote]

Get a common denominator, simplify, then solve

[tender cape]

How do I get a common denominator?

[warped coyote]

Multiply by a convenient ‘1’

[tender cape]

what ?

[warped coyote]

How do you normally make a common denominator 1/2+1/3 ?

[tender cape]

multiply the 2 by 3 and 3 by 2

[warped coyote]

Yeah, you could do that, better yet what if you had 1/2+1/6

[tender cape]

same thing

2 x 6 and 6 x 2

that’s how I was taught how to do so

[warped coyote]

Or multiply 1/2 by 3/3

[tender cape]

why

[civic eagle]

do you have to keep it rational?

[warped coyote]

Because 3/3=1 so it doesn’t change the value

[tender cape]

idk

i know the answer is -7 tho

[warped coyote]

But you can add them together

[civic eagle]

just cancel all fractions by multiplying by 2v^2

2v^2 has both 2v and v^2 as factors

[warped coyote]

That also works

[civic eagle]

trivial after that

[tender cape]

yeah I saw that online

but like none of this makes sense

[civic eagle]

how so

[tender cape]

why do you simplify to 1/x+3x^2

makes no sense

[warped coyote]

You don’t have to

[civic eagle]

nearly same as your answer key

what about that are you having trouble with?

[tender cape]

how is 2v^2 a common term?

[civic eagle]

you can factor out the other denominators from it

2v^2 = 2 * v^2```

[tender cape]

Wouldn’t that get rid of the 2v on the other side?

(Right side)

[civic eagle]

$2v^{2}\cdot\frac{1}{2v}=\frac{2v^{2}}{2v}=v$

[wraith dagger]

b0ngl0rd

[civic eagle]

is that what you're talking about?

[tender cape]

no ?

[civic eagle]

?

[civic eagle]

yeah, thats what i just illustrated with the latex

distribute 2v^2 to the fraction and thats the result

[tender cape]

Wouldn’t the 2v^2 get rid of the 2v

cuz they’re sorta common terms

[civic eagle]

yes, it will cancel the fraction and result in v

[tender cape]

but is it cuz it’s not raised to the second power

[civic eagle]

$\frac{v^{2}}{v}=v^{\left(2-1\right)}=v$

[wraith dagger]

b0ngl0rd

[civic eagle]

all this means is that when we multiplied we had an extra factor of v

in comparison to the denominator that is

2v^2 and 2v

[tender cape]

I still don’t understand

but it might be what I said earlier

It can’t factor a solo v if it’s not raised to the second power

[civic eagle]

???

[tender cape]

so confused

[civic eagle]

im not sure where your confusion is. you mentioned the RHS with 1/2v, and i explained the result of the multiplication by 2v^2 to cancel the fractions

[tender cape]

oh my goodness I’m so frustrated

[tender cape]

yeah I kinda get this now

cuz you multiply the 2v^2 with the 1 on top

Which cancels to make v

BUT

Wouldn’t that cancel the 2v denominator ?

[tender cape]

And how did you even get 2v+6?

god this is so annoying to learn

nothing’s clicking

[cedar kiln]

@tender cape Has your question been resolved?

[cedar kiln]

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[restive pawn]

I need help with algebraic linear equations

[restive pawn]

No determining if the segments are parallel perpendicular or neither to each other

[bronze scaffold]

Look at the slope

[restive pawn]

There is no slope and I know how to determine it but I’m tired as crap

[bronze scaffold]

Find the slope

If the slopes are the same the are parallel

[dark hollow]

parallel if slope is equivalent perpendicular if slope is negative reciprocal of the other

[bronze scaffold]

Otherwise if the slope of the second line is a negative reciprocal of the other one

Then it's perpendicular

[restive pawn]

Thank you

[cedar kiln]

@restive pawn Has your question been resolved?

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[cedar kiln]

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[craggy stratus]

Decide for which alpha that's part of the real numbers(?) the following generalized integral is convergent

So first off I need to find the primitive function to e^-alpha*x cos(x) dx

I tried to attack this in a few different ways without any major success and therefore I have two questions that I've completely forgotten

[dark hollow]

use integration by parts

[craggy stratus]

if I have 5*5+3 = 28

if I do ln

[dark hollow]

since its cyclic it’ll repeat once and then u get it

[craggy stratus]

does it become ln(5*5)+ln(3) = ln(28)

or would it be ln (5*5+3) = ln(28)

Like I think I have an idea that should work

[dark hollow]

second one if ur applying ln to both sides

[craggy stratus]

ok so you take the entire thing in the side raised to ln?

so what would happen with ln^(e^-ax*cos(x))?

wouldn't the ln and the e cancel?

so we get -alpha *x *cos(x)?

[dark hollow]

but you need to find the integral eight

whats the purpose of taking ln

oh

[craggy stratus]

the idea that I had is that if I can say

that the integral of f(x) = F(x)

then I should be able to say that the ln(f(x)) = ln(F(x)

or?

like without actually knowing that the inside is

because then I can get the -alpha out

and suddenly we just have x*cos(x) inside of the integral

and then it's a lot easier to do integration by steps

because we find the primitive to cos(x), aka sin(x)

so we get

ln(-alpha*(xsinx-cos(x)))

so

ln(xsinx*alpha-cos(x)*alpha)

[dark hollow]

i dont think u can bring the ln inside integral though

[craggy stratus]

like this is the idea i came up with in my head while trying to sleep

[dark hollow]

did u try integration by parts

[sturdy rose]

https://smashmath.github.io/blog/eulersformulabyparts/

[craggy stratus]

i did try it yesterday but it felt like a long tedious process that I wasn't feeling down for

[sturdy rose]

would this apply

[craggy stratus]

i think the issue is that you're not getting "more" parts

so if you call e^-axcos(x) I(x) after integrating by parts so that we get back to e^-axcos(x) we'd still just have 1 I(x)

[dark hollow]

it repeats so u can say the integral is equal to I

then move it

[craggy stratus]

@sturdy rose i thought about this as well but i wasn't sure if we can take any e^x and call it cos(x)+isin(x) if there's no i in the exponent

[sturdy rose]

hm

[craggy stratus]

yes but won't they cancel each other?

i think my professor said you could only do that if before the cycled I(x) integral you'd have a constant bigger/smaller than 1

because we start with 1 I(x) so if we cycle through and get 1 more I(x) and move it over they'll just cancel

[cedar kiln]

@craggy stratus Has your question been resolved?

[sacred heath]

umm thats harrrd broo

[dark hollow]

wait

does comparison theorem work

shouldnt e^-ax always be greater than or equal to e^-ax(cos(x)

bc cosx goes from -1 to 1

[craggy stratus]

i think what naxtisy sent is what's supposed to be done

this is the answer

which looks very close to this

im just not sure what happened to the constant attached to cos and sin

or maybe they put them in the wrong order

ah yeah

a is alpha in this case

but im still not sure why this works

like e^ix is cos(x)+isin(x)

but there you have the i to tell you that it's a complex number

and the i goes before the sin

now the alpha (which is where "i" normally would be) goes before cos..

.close

[cedar kiln]

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[glossy notch]

quick question, can someone let me know what the line above z1 - z2 means? thank you

[craggy stratus]

quite sure that's the inverse

conjugate

[glossy notch]

so like z1 + z2?

[craggy stratus]

imma be honest im not sure

[fallen moat]

nope

[craggy stratus]

but i think you can split it up to conjugate of z1 - conjugate of z2

is that correct? @fallen moat

[slate lintel]

complex conjugate is pretty important. I'm sure your book went over it in a previous chapter, you should review that

[glossy notch]

would ((8+i) - (3+2i)) be right?

[sacred heath]

what grade are u in

[glossy notch]

1st year uni

[glossy notch]

bro the book is literally garbage, the uni professors write the books themselves and in the middle of explaining they just say "this part will be explained in class"

???????

[slate lintel]

[glossy notch]

ty ty 🙏

what grade was i supposed to take this in, cause i've never seen it before

[slate lintel]

most people see it in algebra ii or so

so vaguely 9th - 11th

but

like I said this book really should define it too lol (ik you said it was bad)

like my complex analysis book which goes into a lot of depth starts out with like "here is a + bi" and "this is how adding and multiplying works"

[fading narwhal]

You can also find it in linear algebra done right 4th edition

It’s free online

[glossy notch]

i wish bro im in compsci and they just started and assumed we all just knew all the math and all the programming

[glossy notch]

oh nice ill check that out rn

also quick question very dumb but would ((8+i) - (3+2i)) simplify to (8 + i - 3 + 2i)?

[fading narwhal]

Check beginning of ch 4

[glossy notch]

?

[fallen moat]

it's
8+i-3-2i

[slate lintel]

you can treat the i like you would treat x

except that you can also replace i² with (-1)

so distributive law and stuff still apply

[mild sundial]

there's a parentheses in b/w so the sign would apply to both the elements

[glossy notch]

ohhh right right thank you guys

[glossy notch]

and for b) would i use anything from part a or is it just part b of the question

sorry didnt realize b isnt in the screenshot

[fierce plaza]

it wont be necessary

[slate lintel]

completely unrelated questions as far as I can see

[cedar kiln]

@glossy notch Has your question been resolved?

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[glossy notch]

.reopen

[cedar kiln]

✅

[glossy notch]

is this correct?

[marsh lake]

answer should be correct

[glossy notch]

great thank you:)

[cedar kiln]

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[tawny garden]

Does anyone now how to do third one

[undone epoch]

Is it $$\frac{2^{-1}-3^{-2}}{4^{-2}}$$

[wraith dagger]

Good

[tawny garden]

Yes

[undone epoch]

Try to expand every terms.

[vestal hedge]

But whats the question?

[viscid mason]

Just simple fractions

[tawny garden]

There is no question

[vestal hedge]

Ah my bad it says =? I thought its = 7

[viscid mason]

then