help-26

[mortal marsh]

but like

.reopen

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[mortal marsh]

lemme get a fast example

[keen venture]

Any cubic can also be graphed with an x-y plot, but I think you'll agree that cubics have "one more dimension" to them

[bitter hemlock]

dimensionality here refers to the dimension of the vector space; definition of dimentionality? remember it's simply the number of vectors you need to form a basis for a finite-dimensional vector space

[mortal marsh]

like here u must find the span for R2[X]

[mortal marsh]

well ye

[bitter hemlock]

R3 being a 3 dimensional space should be pretty clear, but individual vectors are still just straight line segments more or less, that doesn't mean the vector space is 1 dimensional

[mortal marsh]

ye im still studying whats basis

and thats why i found this dim thing

[bitter hemlock]

basically a set of linearly independent vectors such that you can express any vector in the vector space a linear combination of them

[mortal marsh]

oh

lets say i have set of 2 vectors

[bitter hemlock]

so you can think of any general vector in this vector space as Ax^2 + Bx + C and the most natural basis (but there are infinite other ones you could choose) are x^2, x, and 1. Then you can express the polynomial Ax^2 + Bx + C very naturally as a linear combination of those 3 vectors: A * x^2 + B * x + C

you can see there is a very natural 1 to 1 correspondence between Ax^2 + Bx + C and the R3 vector (A, B, C) they both hold the same amount of information

[mortal marsh]

but wait

if i like have a set of 2 vectors

[bitter hemlock]

and the vectors of the basis {x^2, x, 1} correspond to (1, 0, 0), (0, 1, 0), and (0, 0, 1)

[mortal marsh]

the linear combinations between them is always in the basis

or like thats where it can go

but isnt that a span ?

or subspace

[bitter hemlock]

hm what are you saying here?

[mortal marsh]

i have 2 vectors in a set

and like any linear combination between them

[bitter hemlock]

linear combinations OF them?

or what do you mean by between

[mortal marsh]

like

i can multiply them by a scalar

then sum them

[bitter hemlock]

yes, that's a linear combination

[mortal marsh]

so thats a span

[bitter hemlock]

and the set of all possible linear combinations is the span of those vectors

[mortal marsh]

ye

[bitter hemlock]

so what's your question about that?

[mortal marsh]

but isnt that what u said here

[bitter hemlock]

yes, you could word things many different ways, you could say a basis is a linearly independent set that spans the vector space

[mortal marsh]

so basis is same as span ?

[bitter hemlock]

no?

in R2, an example of a basis is:
the set of two elements: { (0, 1), (1, 0)}
another example of a basis is { (2, 1), (1, 2)}
so these are both sets of two elements each
their span is R2

[mortal marsh]

whats the basis here

is it the set itself

[bitter hemlock]

"the set of two elements ...."

is the basis

[mortal marsh]

so when i say for example

E=((0, 1), (1, 0))

thats

a basis

[bitter hemlock]

think of the basis as the simplest "building blocks" you can come up with for a set, each of the elements can't be expressed by the other because they are all linearly independent. span is like the opposite, it's all the myriad ways you can combine elements of a basis (or in general of any set of vectors, doesn't have to be linearly independent)

[mortal marsh]

so its like

a type of set

[bitter hemlock]

yes, E is a set of vectors that are

1. linearly independent
2. they span the vector space (in this case R2)
   so yes they qualify as a basis for R2

if you then consider the set { (0,1), (1,0), (1, 2) } is this still a basis for R2?

[mortal marsh]

no

[bitter hemlock]

doesn't it span R2 though?

[mortal marsh]

but its not linear independent

[bitter hemlock]

correct

you have superfluous building blocks in there

[mortal marsh]

ye now i got it

so its a type of set

that is linear independent

[bitter hemlock]

well yes its a set of vectors with certain properties

[mortal marsh]

ye ok but wait

whats that related to dim ?

[bitter hemlock]

dim is the number of vectors in a basis

[mortal marsh]

wait

so

if i have 3 vectors in the set

and it only has 2 dimensions

like (0,1)

[bitter hemlock]

it turns out that if you find a basis for a vector space, then all other bases will also have the same exact number of elements (this is a theorem that should be proven in your book if you haven't studied it yet). so you're never gonna have a basis with 3 vectors then another basis with 4 vectors. any bases will always have the same number of vectors

[mortal marsh]

does that mean its for sure not linear independant

[bitter hemlock]

right, since all bases for a particular vector space will always have the same number of vectors, then a trivial first requirement for a basis is to have the number of vectors that you know you need

[mortal marsh]

i just got to the subject of basis so still gettin it

[bitter hemlock]

so if you can build up a set of linearly independent vectors 1 by 1

and you reach the number of the dimension of a vector space

that will automatically be a basis

so i.e. in R3: start with any vector that's not (0, 0, 0) (since the set of just the zero vector is already linearly dependent)

ok, now you have a set with 1 linearly independent vector, off to a good start

now pick any other vector that isn't colinear with your first vector, now you would have a set with 2 linearly independent vectors

ok now pick any 3rd vector that isn't in the plane defined by the first 2 vectors

and you are guaranteed to have a basis now

[mortal marsh]

ah ok so thats like

if i can achieve this

then its a basis

does it have any other condition

[bitter hemlock]

nope, if you come up with n linearly independent vectors that are all in an n dimensional vector space, they will be guaranteed to span the vector space (this should also be a theorem you prove)

[mortal marsh]

oh ok

gotcha

tysm appreciate it

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[mortal marsh]

.reopen

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[mortal marsh]

ok just came across a confusing example

nvm i just didnt read it properly

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[rose spire]

Hey! Can someone help me with this? Thank you

[rose spire]

I am not sure how the integral of secxdx is that

[agile harness]

yea so they used a trick

[rose spire]

What’s the trick? Or rule

[agile harness]

they multiplied by (secx+tanx)/(secx+tanx)

if you do this

it works out nicely

and it’s just u sub

u=secx+tanx

the numerator will be precisely du

[rose spire]

So it will become u^2/u?

[agile harness]

no

[merry hill]

1/u*

[agile harness]

secx(secx+tanx)=sec^2x+secxtanx

and the denominator is secx+tanx

[rose spire]

Oh hmmm

[agile harness]

the derivative of the denominator is secxtanx+sec^2x

which is the numerator

so it’s just integral of du/u

which is ln(u)

[rose spire]

Ohhh okay

Thank you!

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[opal bone]

Hi, this is probably a really stupid question but would anyone be able to help me out with the algebra for this? I have no idea how this guy got his answer for Va

[opal bone]

just for some context j is just the imaginary number

[noble laurel]

okay

why would they use j for that

[opal bone]

its for circuits

its just what we use ig

[noble laurel]

typical engineer notation

Do you also call Euler's number f?

[opal bone]

no

i think this person is just denoting it as that weird angle looking thing

[noble laurel]

you're kidding right

[opal bone]

so -40<-110 is just -40e^(-j110)

pretty fucked up

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[keen venture]

Tripping. Notice that x² - 3x is being subtracted

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[drifting ether]

honestly not sure where to start

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[abstract smelt]

help me pls bra

<@&286206848099549185>

[cursive mango]

Can you think of an equation go make

[neon iron]

Equal both expressions

Find x

[abstract smelt]

i found x

[neon iron]

Which is?

[abstract smelt]

x=5 and x=10for a

but idk if thats the right way

is that correct?

[neon iron]

Correct

Exercise b think of an inequality?

[abstract smelt]

i think its 3>=x>=10

@neon iron

[cursive mango]

Your signs are flipped

But that’s probably not right because either way

[abstract smelt]

Ticket prices of 3 and 10 would make a profit of 200

ohhh because of applications of solving

[neon iron]

Nevermind

Sorry I misread the rest of the problem

Just do the same equation as before

Right hand side is 200

[abstract smelt]

so its 3 and 10

[neon iron]

I think yea

I don't have a calc now

[abstract smelt]

its alright imma go with that

thanks for the help bro imma endorse u in general

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[quiet dome]

2 people are playing the game. The first player throws 6 2-sided pieces (let's say coins). If all 6 coins match, the player gets 5 points and throws again. If 5 of 6 match, the player gets 1 point and throws again. Otherwise the player passes the coins to the second player. The first player to 40 points wins. What’s the expected number of throws to get to 40 points?

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<@&286206848099549185>

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[jade thunder]

Have you done part 2

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[dreamy maple]

How do the graphs for R and r relate to the torus. I understand that they are the outer and inner radius but i dont see it from the graphs shown in desmos

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[noble laurel]

Swag

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[fierce sierra]

Im confused with this question, is it just asking me to solve the augmented matrix (A|b)?

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[drifting ether]

Practicing differentiation

But it’s gotten so complicated I feel stuck on where to go next

[neon iron]

in all honesty you are pretty much "done"

you got your derivative, and thats really sufficient

as to if you want to make it look nicer or not, thats just extra steps if you want to go through the algebra

[drifting ether]

I want to get rid of the fraction on the top but I forgot how to do that 😭

[neon iron]

multiply top and bottom of the big fraction by sqrt(t^2-4)

also u can add the two terms in the numerator of the big fraction first

[drifting ether]

this is the part that confuses me

how is there a -4 on the top?

[neon iron]

so like

,align
\3{t^2 -4} - \4{t^2}{\3{t^2-4}} &= \0r{\4{\3{t^2 -4}}{\3{t^2 -4}} \2} \3{t^2 -4} - \4{t^2}{\3{t^2-4}} \ &= \4{t^2-4}{\3{t^2-4}} - \4{t^2}{\3{t^2-4}}\ &=\4{t^2-4-t^2}{\3{t^2-4}}\ &=\4{-4}{\3{t^2-4}}

[neon iron]

@drifting ether hope this makes it slightly clearer

when u r adding fractions u always wanna do like [
\4ab + \4cd = \0r{\4dd\2}\4ab +\4cd \0r{\2\4bb}
]

[neon iron]

adding fraction with no common denominators*

[drifting ether]

;-;

took me a second to realise what's happening

THANK YOU

[neon iron]

no worries !! good luck

[drifting ether]

it all looks so similar, my eyes get all muddled up

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[upper gale]

what happen from the 2nd step to get into the result on the last one

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[limpid linden]

multiply both side by 2*sqrtx

[upper gale]

ohh i see

thank youu

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[vapid seal]

-8,4,-2,1,-1/2,...... find the 9th term

[vapid seal]

I found the formula which is -8 * (-0.5)^n

is the first term 1th term or 0th term?

I mean -8

[neon iron]

Tn = ar^(n-1)

[versed cairn]

first term is -8

[vapid seal]

so the ninth term would be -8 * (-0.5)^9-1

oh

[versed cairn]

yes

[fickle kite]

1/4

[versed cairn]

!nosols

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

[vapid seal]

I thought the n-1 was optional, and I was wrong

thank you

[versed cairn]

no , that is the general formula to find nth term

Tn = ar^(n-1) for geometric series

[vapid seal]

Yes. Thank you

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[neon iron]

I am trying to remember how to prove [
\vj A \vd (\vj B \cross \vj C) = \vj B \vd (\vj C \cross \vj A) = \vj C \vd (\vj A\cross \vj B)
]

[neon iron]

can anyone nudge me towards it?

i remember something about using the determinant quick-hand way of calculating the cross product and then using determinant properties like the swapping of columns/rows only differs the dets with a minus sign factor, but i think thats kinda dumb since the cross product isnt even a true determinant

https://youtu.be/xkvUi7t1tao?si=5Jiu9CJYQiAjBBig

A.(BxC) = B.(AxC)=B.(CxA)=Cx(BxA)=Cx(AxB)

just used two properties

the commutative in cross product

and the other (in the first step) I forgot the name

[neon iron]

the cross product isnt commutative tho

Well the dot product is

(BxC).A =B.(CxA)

idk how u ascertain that either though

[neon iron]

i feel like computing out the terms like the person did here is iffy as well, as those properties should generalise to any dimensions and not just 3D (well, to be fair, I hear the notion of a cross product is only defined for 0, 1, 3, and 7 dimensions only but i think ignoring that here is ok for now)

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[magic grail]

Hello, can you help me with this physics problem please? It is more of geometry. I need help mainly to make the graph.
A person stands 2m in front of a plane mirror. Behind the person at 1m, there is a 3m high tree. what minimum length of mirror does the person need to see the whole tree?

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[little grail]

hm

[little grail]

im so confused

i have a spec investigation booklet on 3d functions

and i just cant wrap my head around any of it

1+1=

it might be over for me

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[scenic osprey]

How to find a

[scenic osprey]

I know it’s the vertical stretch but idk how to get it

[radiant sparrow]

Well sine's maximum value is 1, and the graph shows to be reaching 7 and -7

So the stretching factor is 7

[scenic osprey]

How did u get 7

[radiant sparrow]

By looking at the highest and lowest points

[scenic osprey]

I don’t get it like do u subtract the highest and the lowest or what

Like overall how do u tell the vertical stretch

[radiant sparrow]

It's just by maximum and minimum of the graph because it's a sine function

Sine only goes from -1 to 1

This graph is going -7 to 7 so a is 7

[scenic osprey]

Why isn’t it -7 then

[lucid junco]

the amplitude a is |a|

u take the absolute value

[radiant sparrow]

If the graph was vertically mirrored compared to original sine then it would be -7

But it's not

[scenic osprey]

Ok

Also

What if the graph went from negative 4 to positive 7

[lucid junco]

oh

i misread ur question

nvm

[scenic osprey]

Then what would a be

[lucid junco]

ignore what i said

[scenic osprey]

Ok

[radiant sparrow]

That would never happen, sin is symmetrical

[scenic osprey]

Oh ok

[tranquil stirrup]

sin(x) can be any value from -1 to 1
a*sin(x) can be any value from -a to a

[scenic osprey]

Ok

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[exotic arch]

could someone please show me how to solve this integral?

[chilly walrus]

the solution looks super complicated

[exotic arch]

it is

nvm

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[kind sentinel]

Hey need help

.reopen

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[eager timber]

Could someone explain why this is zero?
$\lim_{h \to 0} \frac{\cos(h) -1 }{h}$

[thorny flame]

Tomi

[eager timber]

well I can just plug in h I suppose?

so that 1-1 over zero?

which is zero over zero

$\lim_{h \to 0} h = 0 \$
$\lim_{ \to 0} cos(h) - 1 = 1 - 1 = 0$

[thorny flame]

Tomi

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@eager timber Has your question been resolved?

[potent hare]

On a zero over zero limit. You can use l'hôpitals rule

So take the derivative of the top and bottom

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[exotic arch]

is this possible to be solved without substitution?

[outer laurel]

expand that and use partial fraction decomposition idk

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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.

[exotic arch]

original

[neon iron]

there is nothing to decompose here, so that's not gonna work

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[exotic arch]

<@&286206848099549185>

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[exotic arch]

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[keen linden]

how to find out how many prime numbers are under a certain number?

[keen linden]

question is how many numbers that have 3 divisors are under 3000

so it's basically how many primes are under 54 since 54^2=2916

or is that not how u answer the problem?

[obsidian solstice]

Count them one by one

[obsidian solstice]

Yes, that is a correct solution

[keen linden]

it's a 20 second problem tho

actually this is the full problem

Lee Cooper, a fashion and numbers enthusiast, likes labeling numbers that have a certainnumber of factors. He calls numbers that have exactly 3 positive divisors fab while the ones withexactly 9 factors glam. How many fab numbers and glam numbers are under 3000?

it's also asking dor how many numbers below 3000 that contains 9 divisors

idk how to do shit i'm uneducated

[obsidian solstice]

And you have to answer that in 20 seconds?

[keen linden]

yeah

i don't think the intended solution is manual counting

i can't find the solution manual rn

some student prob went home with it

[obsidian solstice]

It must be

Because

A number has exactly three divisors if and only if it is the square of a prime

[obsidian solstice]

Thus, if there were an easier way, there would be an easy way to count prime numbers as well

[keen linden]

the solutions are pretty smart ngl problems where u think the intended solution is manual counting the solution instead is rlly smart

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@keen linden Has your question been resolved?

[radiant marlin]

doesn't 3 divisors mean square, no primes involved

there'd be no trick with primes we just know that the number is like N/ln(N) within 10%ish

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[royal hazel]

Hi,
I have a combinatorics problem:
7 Birds of Type 1, 5 Birds of Type 2, 4 Birds of Type 3

How many combinations exist to arrange these birds if you cannot see the difference between birds of a specific type

[royal hazel]

Is it 16! / ( 7! × 5! × 4!) ?

[craggy haven]

yes (you'd call these permutations though, not combinations)

[royal hazel]

Ahh okay sorry I am not a native english speaker 😅

So my answer is correct?

[craggy haven]

ja

[royal hazel]

Nice, thanks alot 👌

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[wicked burrow]

can someone explain how they found 1,1-a, and 1+a

they seem to just know it

[vernal vale]

theyre the roots of the polnomial

you think of the polynomial as a function of lambda

and try to find the roots

so, its an algebra problem

can you see that lambda = 1 is a root? @wicked burrow

[wicked burrow]

yea i can see that

[vernal vale]

okay

[normal tapir]

the (lambda - 1)^2 - a^2 is a difference of squares

[vernal vale]

so now we are on the hunt for two more

lets use difference of square

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

[thorny flame]

jan Niku

[wicked burrow]

sure

[vernal vale]

once you know that then you can see it presumable

[wicked burrow]

i get it now

tysm

[vernal vale]

$(\lambda -1)^2 - \alpha ^2 = (\lambda - 1 + \alpha)(\lambda -1 - \alpha)$

sure

[thorny flame]

jan Niku

[keen linden]

4^2 does not have 3 divisors, it needs to be a prime square

[wicked burrow]

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[neon sedge]

Hello, I need help with a trig problem: tanx/2(1+cosx) = sinx
Verifying the identity

[neon sedge]

Im totally confused where to go

[noble laurel]

I believe my sir @solemn wigeon can help you with this

[half edge]

I also believe our sir @solemn wigeon can help you with this

[solemn wigeon]

Hello sir. Recall the rule : you usually want to simplify the more complicated looking side first.

SIR THIS IS THE WROGN CHANNEl

!help

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[neon sedge]

right the left side, however im confused what to do

[solemn wigeon]

Try expressing tan(x) as sin(x)/cos(x)

And simplifying the fractions

[neon sedge]

the tanx/2 part is where im getting confused at

[solemn wigeon]

I have a doubt

is it $\frac{\tan(x)}{2(1+\cos(x)}$

[thorny flame]

Ultimate Chad

[solemn wigeon]

Or $\tan(x)/2 \cdot (1+\cos(x))$

[thorny flame]

Ultimate Chad

[neon sedge]

this

[solemn wigeon]

II see

Have you tried writing tan(x) as sin(x)/cos(x)

If you temporarily forget about the dividing by 2 you get $\tan(x)(1+\cos(x)) = \frac{\sin(x)(1+\cos(x))}{\cos(x)}$

[thorny flame]

Ultimate Chad

[neon sedge]

but what about the 2

[solemn wigeon]

Sir are you sure your probablem statement is correct

Oh is it meant to be tan^2(x)?

[neon sedge]

yup I can send a picture

[solemn wigeon]

Please do that sir

[neon sedge]

32

im assuming something with the half-angle formula but I don't know what

[solemn wigeon]

Aaah it was tan(x/2)

Do you know the half angle formula?

[neon sedge]

yeah

[solemn wigeon]

Have u tried plugging that one in

[neon sedge]

how i plug it in is what im more confused about

[solemn wigeon]

Well what is the half angle formula that you know of?

[neon sedge]

sqrt((1-cosx)/(1+cosa))

or (1-cosa)/(sina)

[solemn wigeon]

Ah i see yeah that one is pretty complicated

[neon sedge]

or (sina)/(1+cosa)

any of those

we have to memorize

[solemn wigeon]

Sir does this one look suspicious

What happens if you multiply that by (1+cos(a))?

[neon sedge]

?

[solemn wigeon]

Like try plugging in $\tan(\frac x2) = \frac{\sin(x)}{1+\cos(x)}}$ into your above equation

[thorny flame]

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

[neon sedge]

yeah but im confused about the meaning of tan(x/2) (1+cosx) since im not sure how to plug that into the half angle formula

like what parts

[solemn wigeon]

Like you have that above identity

Just replace tan(x/2) with the thing i wrote above

[neon sedge]

OHHHH

ok thank you

i got it

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[solemn wigeon]

🔠

https://tenor.com/view/minion-gif-4256583569949039458

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[feral spade]

I tried separating the imaginary and real parts and tried to use Euler's complex exponential function formula but it did not work

I am unsure if i had the right approach

[solemn wigeon]

Your approach sounds good.

You just need to split the e^(4.927it) using the Eulers formula now

[feral spade]

oh i see how to do it now

i will try

thanks

[solemn wigeon]

Ah i see what you mean

You don't just get cos as the real part

Ah now i see why it said the first thing was wrong

It looks like you might have to expand and then use that $\cos(x) = \sin(\pi/2 - x)$

[thorny flame]

Ultimate Chad

[feral spade]

oh

thats annoying

okay.

I am so confused 😭

I feel like this is more difficult than it should be

How do i combine the two different sin and cos parts if they have different stuff inside?

@solemn wigeon

[solemn wigeon]

Damn you make a good point

What might work better is if you write the term (4.516+1.722) in the form r * e^itheta

Yeah the way we were doing it before was super complicated ur right

[feral spade]

what is r? is r the magnitude of the complex number in front?

like distance from origin if you graphed it?

Does this look correct?

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@feral spade Has your question been resolved?

[feral spade]

<@&286206848099549185>

[feral spade]

I figured out my mistake here, I need to add the initial angle when t=0, so arctan(1.722/4.516) to the inside of both trig functions.

Thanks!

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[granite kelp]

need to know if it diverges or converges and if so to what

[granite kelp]

I was thinking telescopic at first but now im not so sure

<@&286206848099549185>

<@&286206848099549185> please

[keen tundra]

it's not difficult to calculate the numerical result if you split it up into the infinite sums 5/2^n and 1/3^n, then apply the formula for a geometric series sum. it can also be proven to be convergent by using the comparison test, by comparing it to the infinite sum of 5/2^n, and by showing that 5/2^n - 1/3^n is always positive

[granite kelp]

so I can just split them up into two different summations?

[keen tundra]

yeah basically

[granite kelp]

thank you

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[distant shell]

the answer to this question is supposed to have 2 or 3 options, but I cant figure out what the dash above the letters signifies. Although, I do know that option A is definitely correct.

[merry hill]

that means compliment of set

A bar = universal set - set A

[distant shell]

thank you

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[hybrid stream]

how would i find this out?

[restive inlet]

formula for area of trianglular prism

[hybrid stream]

i have no idea

[restive inlet]

look up ^

[hybrid stream]

Does this mean
Area = 1/2 * 2 * 6

[restive inlet]

for the purple triangle yes,

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[neon iron]

how can i prove triangle inequality for complex numbers?

[wicked bolt]

same as for vectors

[neon iron]

i mean |z_1 + z_2| <= |z_1| + |z_2| and the counterpart for this

[neon iron]

havent learnt anything of the sort yet

[dreamy drum]

try |z1 + z2|^2 and the proof should be pretty direct from there

[dreamy drum]

this is a cooler way of proof

👍

[neon iron]

try |z1 + z2|^2 = |z1|^2 + |z2|^2 + 2Re(z1* con(z2))

what if 2Re(z1* con(z2)) is positive?

so |z1 + z2|^2 should be greater right?

Oh hell

i messed up

but i know what i was thinking wrong

thank you guys

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[limpid linden]

can someone help me with the 2nd one

[limpid linden]

i know you use cos(a+b) to get the RHS but don't know why

<@&286206848099549185>

please help me with this

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[neon iron]

finite set can't be equivalent to its super set but a infinite set can is that cuz infinite set can always contain its super set

[neon iron]

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[neon iron]

can someone help me on how I would reason/show that the second inequality holds? n is a large natural number an p_i is a polynomial

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[alpine smelt]

Hello does anyone speak german? If not could someone help me who has a little bit of time? Thanks

[neon iron]

[alpine smelt]

We have a list of x1, x2, ….. x50. „If every element of the list gets multiplied by 100, how does it affect the variance“

Idk if this makes sense

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[clear crescent]

Are the x₁, x₂, …, x₅₀ random variables (zufallsvariable) or just numbers/observations?

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[neon iron]

hi I need help with this integral please!

[neon iron]

what have you tried

I tried to substitute sinx and then cosx but failed

[regal charm]

use the pythagorean identity to rewrite the numerator

[neon iron]

^

[regal charm]

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

[thorny flame]

nyxie9151

[neon iron]

would i get (1-cosx)/(cosx+1)?

[regal charm]

hmm

where did you get 1 - cos(x)?

[regal charm]

what would you get solving for sin^2 (x)?

[neon iron]

moved the cosx to the left

I meant cos^2

[regal charm]

okay better

[neon iron]

sorry :<

[regal charm]

,, \frac{1 - \cos^2 (x)}{\cos (x) + 1}

[thorny flame]

nyxie9151

[regal charm]

any idea how you can simplify this even further?

[neon iron]

I could seperate stuff

like

1/cosx+1 - cos^2x/cosx+1

does that make sense? sorry I don't know how to use the bot

[regal charm]

u can but think of what you can do with

[thorny flame]

nyxie9151

[neon iron]

what is that

[regal charm]

the numerator

does it remind you of something?

[neon iron]

oh yeah!

it's (1-cos)(1+cos) ?

[regal charm]

yes

[neon iron]

and we factor it

then we're only left with 1-cos

[regal charm]

,, \frac{(1 - \cos (x)) \cancel{(1 + \cos (x))}}{\cancel{\cos (x) + 1}} = 1 - \cos (x)

yes

[thorny flame]

nyxie9151

[regal charm]

u should know how to integrate this now

[neon iron]

so the final result would be x - sinx !

[regal charm]

u forgot the +C lol

but yes

[neon iron]

I c

(i see)

thank you so much have a nice day!!!

[regal charm]

[neon iron]

.close

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[regal charm]

np

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[neon iron]

Whats a quick way of finding the characteristic equation?

[neon iron]

this true?

https://www.youtube.com/watch?v=eE6TVRz5WSw

[alpine smelt]

Random

[shut obsidian]

Per definition, it's $\det(A - XI_n)$ where $I_n$ is the $n \times n$ unit matrix

Here, you can use Sarrus' rule to find the determinant, since this is 3x3

[neon iron]

what

I found this video but the method is weird https://www.youtube.com/watch?v=QfF_whXs_CA

the equation from the video is off by a negative sign

pain

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[scenic osprey]

How can u identify the pattern expression here

[topaz sinew]

Please don't occupy multiple help channels.

[fickle kite]

Square number

[queen cipher]

u mean da sum?

[scenic osprey]

I didn’t ask for the answer I asked how

[queen cipher]

or da numbers themselves

[fickle kite]

Like these numbers are just particuliar know your square

Or I don't know Fibonacci sequence or sth else

[scenic osprey]

.close

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[cerulean quartz]

would the answer be y>=-7x-47 or y<=-7x-47 ??????????

[granite idol]

https://tenor.com/view/senator-armstrong-memes-maths-metal-gear-rising-revengeance-gif-26055855

[rigid ivy]

!redir

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This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.

[rigid ivy]

!show

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

[rigid ivy]

you need to flip the inequality sign when dividing by -4

[cerulean quartz]

so y <= -7x - 47 is right?

[rigid ivy]

should be

[cerulean quartz]

okay ty

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[lost lynx]

Wouldn't h'(x) just be the exact same graph as f'

So then all the points where h(x) is increasing is where f' is positive

[fiery badge]

hmmm... look at the boundaries of the integral for h

[lost lynx]

x to -3?

[fiery badge]

This question is funky lol

I assumed x > -3 so then it doesn't copy the part of f' from -5 to -3

now I'm not even sure

[lost lynx]

For c I just used the fundamental theorem of calculus to find h'(x) = f'(x)

Then from that I'm assuming since h'(x) = f'(x), the graph would indicate the points where h(x) is increasing

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[sour egret]

Hi

[sour egret]

If there’s a negative number inside a square root should I put the negative outside?

[sleek gate]

like sqrt(-1)?

[sour egret]

I’ll show you

I have these

And some have negatives

Inside

[sleek gate]

sqrt(3)* -sqrt(4)=-sqrt(3* 4)

thats how you do it

[sour egret]

So negative outside right

[sleek gate]

it stays outside

[sour egret]

Oh ok

[sleek gate]

if it would be inside, then keep it inside

pretty easy to follow

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[quaint maple]

hi

can anyone help with this question

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[stiff cave]

tried writing the function as a power series but it doesn’t match the books answer, can someone explain where I made a mistake?

[kindred linden]

Still here @stiff cave ?

[stiff cave]

🫡

[kindred linden]

Okay so

First you got a mistake in the last step

[stiff cave]

did I mess up distributing the powers

[kindred linden]

Yes

You didn’t do it for everything inside the parentheses

[stiff cave]

which part did I mess up

[kindred linden]

The minus

You forgot to applied the power to the -

He even disappears actually

[stiff cave]

should it be x^n - x^(n+1)?

[kindred linden]

No you should have (-1)^n

Added to what you wrote

Because (-a)^n = (-1)^n * a^n

Ah latex doesn’t work like that

[stiff cave]

so it should look like this?

[kindred linden]

Exactly

Then it’s a good idea to distribute the power

[stiff cave]

I’m a bit confused on how to actually distribute the powers

[kindred linden]

You just have to know 2 rules

(ab)^n = a^n b^n

(a/b)^n = a^n/b^n

There’s no other rule to distribute the powers

Once you distributed the power of n there is some more step to get their result

[stiff cave]

idk I’m getting more confused

ik this is wrong but I’m not sure how to distribute it from there

[kindred linden]

The first line is right

Try to distribute the power of n for (x/2)^n

[kindred linden]

What would you get using this rule

[stiff cave]

x^n/2^n

[kindred linden]

Yes

So what does the denominator equal to ?

The whole thing inside the sum

[stiff cave]

it would be 2^n+1

[kindred linden]

Exactly

Now the nominator without the -1^n

[stiff cave]

would it be x^(n+1)-x^n

[kindred linden]

Yes exactly

You just forgot the (-1)^n the first time

Now add it up to the result

But now we have a problem

In the formula they don’t have a x^n+1

They only have x^n

[stiff cave]

yeah and I was confused where the 3 came from

how did they simplify it

[kindred linden]

If you stay with only 1 sum you can’t

There is no magic that can do that

So we have to divide the sum into 2 sum, one whit x^n and the other with x^n+1

Idk if « divide » make sense for you here

[stiff cave]

could you do that like this?

[kindred linden]

Exactly almost

But -1^n

It’s a factor so it should be on both sum

[stiff cave]

okay

from there can we find the sum of one of them?

[kindred linden]

In the result there is a sum

So instead of calculating one, we will transform it

Do you know how to shift a sum ?

Idk the term in English

[stiff cave]

I don’t think so

is that like changing the index

[kindred linden]

Yes

It’s exactly that

[stiff cave]

okay I think I know how to do that

[kindred linden]

Because in the result we see that there is x^n so let’s not touch that one with x^n

[stiff cave]

like this?

[kindred linden]

Be careful are you sure it was (-1)^n+1 ?

[stiff cave]

oh no it wasn’t

so it would be (-1)^n-1?

[kindred linden]

Yes

Okay so now we’re happy we got x^n

However to make one sum out of the two sums what do we need to do ?

[stiff cave]

can we just add them back together

[kindred linden]

Not really

We ought to be careful

About something

Any idea ?

[stiff cave]

they don’t share a common denominator?

[kindred linden]

That’s a good point but that’s when we want 1 fraction, so that’s the next step

But keep in mind the index

[stiff cave]

oh right now they’re indexed differently

[kindred linden]

Indeed

[stiff cave]

how do we get around that then

[kindred linden]

Well look at the formule from your correction

Which index do we want

[stiff cave]

the book’s was indexed at 1

but wouldn’t changing the index mess up the powers again

[kindred linden]

It would

Any other way to change the index ?

[stiff cave]

I’m not sure

[kindred linden]

We will not touch the left sum

[stiff cave]

if we indexed the right sum at 1 the powers would be n-1

but that doesn’t really help

[kindred linden]

We could juste remove the term n=0

So that it begin at 1

[stiff cave]

oh is that the -1/2 from the answer?

[kindred linden]

Exactly

[stiff cave]

I didn’t know you could do that lol

[kindred linden]

However you did one mistake I didn’t catch earlier

So you would get 1

[stiff cave]

oh it should be x^n/2^n right

[kindred linden]

Here the x^n should be on 2^n+1

[kindred linden]

Yes but with n+1

[stiff cave]

ok that makes sense

[kindred linden]

Because (a+b)/c = a/c + b/c

So what do you got so far after those step ?

[stiff cave]

I think it looks like this

[kindred linden]

Okay looks good

The only thing we don’t like is having n-1

There is no n-1 in the formula

How could we get n instead of n-1 ?

[stiff cave]

we can’t change the index again

[kindred linden]

We don’t touch the sum now

But we transform the (-1)^n-1

[stiff cave]

maybe multiply by -1?

[kindred linden]

So what would got for it ?

Keep in mind that a^n/a^m = a^(n-m)

[stiff cave]

I guess dividing by -1?

[kindred linden]

Yes and what does 1/-1 equal to ?

[stiff cave]

just 1

[kindred linden]

Oh no no

[stiff cave]

oh 1/—1 = -1

[kindred linden]

Better 😂

Let’s make 1 sum now

[stiff cave]

I’m not sure if I did that right

[kindred linden]

Oh no that’s my bad

(-1)^n-1 = (-1)^n/-1 = —(-1)^n

[stiff cave]

ohh

that makes a lot of sense actually

yeah I'm not sure why I didn't realize you could just multiply it by negative 1

[kindred linden]

Yes but messages doesn’t translate well that king of sense sometimes

[kindred linden]

It’s just the habit

Of doing calc

[stiff cave]

yeah ig so

isn't it a problem that the powers of 2 don't match?

[kindred linden]

Well you know how to add two fraction right ?

[stiff cave]

I think so

oh we just multiply to get a common denominator?

[kindred linden]

Try and tell me if what you find looks familiar

[stiff cave]

by the way my phone is at 3% so I'm gonna have to leave soon

[kindred linden]

Okay

But maybe there is a way simpler common denominator ?

We are talking about 2^n and 2^n+1

2^n+1 is a common denominator you can get

[stiff cave]

oh so you would just multiply the right sum by 2?

[kindred linden]

The left sum

To get 2^n+1 in the left sum

[stiff cave]

yeah the left one mb

[kindred linden]

And then add the 2 fractions

[kindred linden]

Isn’t it the result ?

[stiff cave]

yea it is

I think I understand now

[kindred linden]

You Just forget a -

[stiff cave]

oh right they're both negative, so its -3x^n

yeah I think I get it now, thanks for the help 🙏

[kindred linden]

Indeed

You’re welcome !

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[urban pulsar]

Help

[urban pulsar]

Im not getting anywhere

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[ripe spoke]

not sure if it helps, but CDAis also 90-x, so CBA is 90+x

cbed is a rectangle

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[autumn plaza]

hai

[ripe spoke]

hai

[autumn plaza]

how come indepndent looks different here?

i thot for independent u find AnB and then see if it equals to P(a) x p(b) and if they the same

they indepndent

[ripe spoke]

those are equivalent tests I believe

meaning that if one is true, the other is as well

[autumn plaza]

wait

so they are different

but the same

can i do it the other way too

[ripe spoke]

yeah

P(B|C) = P(BnC)/P(C)

so if P(BnC) = P(B)xP(C),
P(B|C) = P(B)xP(C)/P(C) = P(B)

and you can do something similar in the other direction

[ripe spoke]

So to prove independence, proving either is true will work.

[autumn plaza]

ohh ok thx u ;D

.close

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[jolly hollow]

Hi, so what is sq root -81

[jolly hollow]

wait acutally i have a idfferent one

could you pls explain de moivres thereom

and further in plar form

polar form*

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[echo galleon]

9i

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[sterile path]

Please help me on question ten from e and the rest

I tried but I don’t understand

[wary tulip]

so you want to find how many of the 52 cards are

1. not a king, or
2. a diamond

how many satisfy 1)?

[sterile path]

48/52

And 1/4

[wary tulip]

48 of them are not kings yea

[sterile path]

Yeah

[wary tulip]

and then of the remaining 4, one is a diamond

[sterile path]

Yeah

[wary tulip]

so got the answer now?

[sterile path]

Yeah I guess

[wary tulip]

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[jade scarab]

"(a) Rewrite as a double integral, (b) use some symmetries to simplify, re-scale (change vars) to simplify
yet more, and finally, (c) exchange the order of integration again to evaluate the improper integral"

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