<@&286206848099549185>
#help-0
atomic bobcat
left roost
could you please upload a better picture, I cannot read it properly
atomic bobcat
ok
lone heartBOT
left roost
ok
lone heartBOT
Available help channel!
cyan nexus
Subject: Trigonometry (basic identities)
lone heartBOT
Subject: Trigonometry (basic identities)
cyan nexus
I'm having trouble understanding all the steps in this problem
I understand the first step (rewriting 2x)
ionic jewel
cyan nexus
I substract 2sinxcosx from both sides, and factor
Oh so I have an a.b=0
🤦 [
I understand now
my bad
thanks
.close
serene vault
hi
serene vault
how do i make this
construct a graph on 9 vertices, and colour each edge with one of three colours so that there does not exist any monochromatic triangle
did i already mess it up?
vale wigeon
!original
lone heartBOT
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.
vale wigeon
i think there might be some crucial details that we are missing.
serene vault
such as?
i just need to contruct a graph with 3 colors where a monochromatic triangle does not appear
having all the edges on of three of the colors
fickle heath
"a graph of 9 vertices" could just be 9 vertices with no edge
vale wigeon
i could interrogate you but it is easier to just see the entire problem
that way we know for sure nothing is missing
just send a pic of the problem
yes even if it isnt in english
serene vault
ok
(c) With 9 students from the program on social media, prove that it doesn't always hold that three individuals from this group are mutual acquaintances, mutual connections, or mutual non-connections. To validate this, design a 9-node graph and color each connection with one of three colors in such a manner that there's no monochromatic triangle present – meaning, no triangle where all three connections are of the same color.
vale wigeon
ok so really what you are after is a coloring of the complete graph on 9 nodes.
thus you have edges missing still.
serene vault
I guess thats where im loosing it
I thought the "edges" refered to this
from I to A
i thought that is considered an edge
vale wigeon
i can't make out the letters you have labeled your vertices with
but you have some vertices that aren't joined by an edge at all.
serene vault
its just A-I
vale wigeon
ok so you could have told us this was an incomplete picture.
serene vault
Im sorry
my brain is mush rn
im just confused on what exactly the "edges" are... how do i KNOW where to put each line etc
vale wigeon
you're looking at the complete graph on 9 vertices
serene vault
am i just connecting each letter to another with every color except on the outskirts?
vale wigeon
a complete graph is a graph where every pair of edges is joined by an edge
sdglkbjslkd
fuck.
every pair of vertices is joined by an edge.
vale wigeon
im just confused on what exactly the "edges" are... how do i KNOW where to put e...
the positions of the edges on the paper dont matter. they dont even have to be straight lines.
serene vault
ur sayn i can make squiggly lines?
vale wigeon
sure
though i would also consider readability
if you squiggle your entire thing through, it will be impossible to tell what edges go where at all.
which is great if you want to be malicious, i guess.
anyway
serene vault
ok let me give an example
vale wigeon
right now i do not see any single-color triangles in your incomplete graph
whether it is extendable to a complete one, i do not know
i am trying it right now
vale wigeon
don't know what "like this" means, exactly.
serene vault
then that would prove there is no monochromatic triangle?
u see the blue extending to each letter
vale wigeon
i would need to see a more precise description of the pattern, or a complete picture.
serene vault
with the next vertex id do green to each letter etc
serene vault
ah shit
so how would i know how to solve this??
without making a million diff attempts
vale wigeon
dunno.
i am currently trying to extend your first pic.
it seems to have worked
oh wait no i am missing one edge at least
serene vault
how the hell
vale wigeon
hang on
vale wigeon
ok this seems to be it
i just drew the edges one by one
and for each one i looked at the existing edges and tried to avoid making monochromatic triangles
vale wigeon
dunno
serene vault
or is there multiple diff ways?
vale wigeon
there's probably many different ways
lone heartBOT
@serene vault Has your question been resolved?
serene vault
is it possible to do with with a graph that has 17 vertecies?
serene vault
ok this seems to be it
theres a triangle!
serene vault
can anyone verify if there is a triangle present?
fickle heath
Yes right here in black
It's actually very easy
In case you don't understand what I did, here is the graph before the last 9 edges
fickle heath
Why not?
serene vault
in the directions it says this
To validate this, design a 9-node graph and color each connection with one of three colors
also
what would u do with a vertex = 16
fickle heath
To validate this, design a 9-node graph and color each connection with one of th...
So?
fickle heath
what would u do with a vertex = 16
What do you mean?
serene vault
say there are 16 nodes
theres no way ur just gota free brain that
and draw it off the rip
fickle heath
Then you probably need 4 colors idk
serene vault
there must be a simple solution
i.e. a formula
like using pigeon hole or something
fickle heath
I mean I can kinda prove that my graph is valid
serene vault
how so
cause if u can proove it in wording then you can proove a contradiction if you wanted to raide the amount of nodes
without having to tediously draw
fickle heath
In case you don't understand what I did, here is the graph before the last 9 edg...
Here you have a 9-cycle of each color, so obviously no triangle (that would be a 3-cycle)
Also none of the missing edges can close a triangle
Not sure how to prove that part
Then for the last 9 edges, it's pretty easy to see that they don't form triangles among themselves
Each of the 3 edges of one color among the last 9 edges have different end vertices
fickle heath
Not sure how to prove that part
Actually I got this
If you number the edges and work modulo 9, then the 9 red edges are between vertices with a difference of 1
So to close a red triangle you would need a red edge between vertices with a difference of 2 (or 7 since it's mod 9)
But all of those are green edges by construction
And so to close a green triangle you would need a green edge between vertices with a difference of 4 (or 5 since it's mod 9)
But all those are blue edges by construction
And finally to close a blue triangle you would need a blue edge between vertices with a difference of 8 (or 1 since it's mod 9), which are all red already
serene vault
hmm
lone heartBOT
@serene vault Has your question been resolved?
alpine sable
What is ∫(-2 to 2) (1-|x|) dx?
lone heartBOT
What is ∫(-2 to 2) (1-|x|) dx?
alpine sable
If I separate them into two functions (y= 1 & y= -|x|), I get 0, but if I try to draw it all as y= 1-|x| I get -2
Here is the work-process I did for both methods:
remote heron
hmm wasnt this just posted
fickle sandal
it timed out
remote heron
ah
remote heron
hmm you shaded the wrong area for the -|x| but its equivalent
likewise, you shaded the wrong area for both functions together
here its not equivalent
shading the correct region will give you the same answer
alpine sable
Hmm
remote heron
no problem
alpine sable
.close
lone heartBOT
lone heartBOT
Available help channel!
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wheat isle
You forgot reflecting across the y axis as well
lone heartBOT
@tulip oasis Has your question been resolved?
jagged cobalt
indeed, crazy stuff
pastel merlin
just a small question: if a function is lets say c(x), is c a function of x, or is x a function of c?
warm pivot
just a small question: if a function is lets say c(x), is c a function of x, or ...
c is the name of the function
warm pivot
yes
warm pivot
mhm :D
pastel merlin
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lone heartBOT
pastel merlin
.reopen
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pastel merlin
if x=15c in c(x), is it c(x)=15x?
gray isle
in c(x),
c is the function
the whole thing c(x), c of x is a function of x
gray isle
x=15c
solve that for c
pastel merlin
c= x/15
gray isle
though not really needed you could explicitly state that as
c(x) = x/15
pastel merlin
so theres a word problem for this
so someone earns 15 per hour
and write a function with e for earning and h for hours
what do i write as the function
gray isle
well how much do they earn in h hours?
pastel merlin
15e?
gray isle
no
pastel merlin
15h
gray isle
yes
pastel merlin
so e=15h
gray isle
yes
pastel merlin
is the function e(h)=15h?
gray isle
that would be acceptable
neon adder
deep hornet
It's better if you just ask your question instead of saying help. We can't help you if we don't know what you need. If you just write your question right away we can start helping right away
!status
lone heartBOT
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
Explain what you've done, what parts you're struggling with etc
lone heartBOT
@neon adder Has your question been resolved?
somber mica
slate oxide
lone heartBOT
lone heartBOT
@slate oxide Has your question been resolved?
upbeat hornet
first plane is 2x-4y-z = -1
you can find the normal vector of that plane
upbeat hornet
notice that the plane π must have a normal vector perpendicular to the normal vector of π_0
and the normal vector of the plane π must also be perpendicular to PQ
lone heartBOT
slate oxide
.reopen
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green juniper
Well you have two points that are in the second plane and since it is perpendicular to the first plane that also implies that the normal vector to the first plane will also lie on this plane, thus you need to utilize this to find a normal vector to the second plane
next step would be to find $\overrightarrow{PQ}$
ocean sealBOT
SKELEROY
lone heartBOT
@slate oxide Has your question been resolved?
upbeat hornet
aight i got the normal vector but like what else should I do
You can see the next messages I sent
Since you know two vectors that the normal vector of the plane π is perpendicular to, you can get the vector itself using the cross product
gaunt elm
I found the derivative of it and tried factoring but it didn't really factor well and I am confused a little ngl.
upbeat hornet
I found the derivative of it and tried factoring but it didn't really factor wel...
Can you take a picture of your work
gaunt elm
upbeat hornet
,rotate
ocean sealBOT
proven leaf
for cancavity consider f''(x) not f'(x)
gaunt elm
so I should take the derivative of the 4x^3-30x^2+48x
gaunt elm
I got it right tysm
proven leaf
I got it right tysm
gaunt elm
then find where f''(x)>0
if it was concave down would i look for f''(x) < 0
proven leaf
if it was concave down would i look for f''(x) < 0
if it f''(x)=0 then you have what is called an inflection point
pretty much just where concave up turns into down and vice versa
gaunt elm
pretty much just where concave up turns into down and vice versa
ohh that really sums up the concept of POI
proven leaf
ohh that really sums up the concept of POI
gaunt elm
Do I use the product rule here first or I skipp the whole thing
upbeat hornet
,rotate
ocean sealBOT
upbeat hornet
Do I use the product rule here first or I skipp the whole thing
you can multiply the whole thing out into a sum of powers of x
gaunt elm
are you saying distribute and simplify the expression first before taking the derivative of something?
upbeat hornet
are you saying distribute and simplify the expression first before taking the de...
You should do that in this case, but not for all cases
ocean sealBOT
gaunt elm
.close
next osprey
lone heartBOT
long axle
U can at me but ionk if I’ll be on
ocean sealBOT
puckmyseen
long axle
ye
surreal crystal
Well the statements are quantified, so start with that
There is exactly one person is
$\exists ! p$
Then everybody becomes all people
$\forall q$
ocean sealBOT
Cycadellic
surreal crystal
So, we can interpret love as a function which takes two people and gives true or false depending upon whether said people love each other
This means we can represent this by the statement
$\exists ! p (\forall q L(p,q))$
ocean sealBOT
Cycadellic
surreal crystal
This statement would imply that the love is two-sided though, and the statements only mentions that q loves p
But i suppose its good enough
Now your example doesnt use the !
Instead it says that if we know $\forall w (\forall z L(z,w))$ then we know that z must be y
ocean sealBOT
Cycadellic
surreal crystal
This is just a different way of writing that there can only be one if these
In fact, this is the definition of uniqueness
$\exists ! x P(x) \iff \exists x P(x)\land(\forall y P(y)\rightarrow x=y)$
ocean sealBOT
Cycadellic
surreal crystal
So, for the first part, we just deconstruct the sentence into its proposition
Then we can apply the definition of uniqueness, the definition of only one
surreal crystal
In fact, this is the definition of uniqueness
$\exists ! x P(x) \iff \exists x P...
This basically says that there is only one thing that satisfies P exactly when everything that satisfies P is this one thing
As for the second one,
There are exactly two people whom Lynn loves
Can be said
There are two people so that Lynn loves both people, which are different people, and that if Lynn loves somebody, it must be one of these people
“There are two people so that Lynn loves both people”
Is needed because this is the main idea of what is being said
“If Lynn loves somebody, it must be one of these people”
Is what we mean by “only”
And we need to specify that they are different people, because in the proposition alone, it is possible that they are the same person, which is not what we mean
Essentially, if you want to construct a proposition like this, apply the meaning of there are using the existential operator, apply the meaning of only through proposing that anything that satisfies P must be one of the choices,
Then in cases where there are multiple choices which satisfy P, we must assert that all choices are different
lone heartBOT
@lime holly Has your question been resolved?
hybrid lily
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hybrid lily
this solution doesnt make any sense for me
so if the norm is larger than 1, its not a gaussian prime?
like what
zealous lichen
A gaussian prime is not a product of two non-units
since c+di and c-di have norms greater than 1, they are not units
hybrid lily
wdym unit
vale wigeon
element with a multiplicative inverse
in Z there are two units: 1 and -1
in the gaussian integers these two are joined by i and -i
vale wigeon
so if the norm is larger than 1, its not a gaussian prime?
you've expressed c^2+d^2 as a product of two things each of which has norm larger than 1.
hybrid lily
right i understand that part
from my understanding, a gaussian integer is simply where the real and imaginary parts of a complex number are integers
i dont quite understand the whole norm larger than 1 thing
zealous lichen
Do you know Gaussian primes though
hybrid lily
gaussian primes are products of complex numbers that cannot be expressed using integers
i know how the solution works algebraically
like i know how to do it
but i dont understand the last bit
i guess i dont understand the "by definition" part
zealous lichen
gaussian primes are products of complex numbers that cannot be expressed using i...
ok let's use your definition
hybrid lily
well i just want to know the right definition, not necessarily my definition
vale wigeon
gaussian primes are products of complex numbers that cannot be expressed using i...
words some up mixed you!!!
vale wigeon
well i just want to know the right definition, not necessarily my definition
do you know what a ring is, or not yet
hybrid lily
no
heres the question
i have no idea what gaussian integers are
i never studied it
this is just the quesiton
vale wigeon
gaussian integers are complex numbers with integer real and imaginary part.
hybrid lily
so all my knowledge comes from the question iteself
hybrid lily
of course
vale wigeon
the norm of a gaussian integer is the square of its modulus
hybrid lily
i get that
vale wigeon
right
like ok the thing about gaussian integers
is that they are closed under addition and multiplication
hybrid lily
what do you mean by "closed"
vale wigeon
and so they form a structure called a ring (which you might study in the near future)
the sum of two gaussian integers is still a gaussian integer
and the same goes for product
hybrid lily
oh i didnt know that, i thought only the products worked
vale wigeon
ok so then lets go back down to real integers for a moment
hybrid lily
ok
vale wigeon
an integer p is prime if it is impossible to express it as the product of two integers, neither of which is ±1
hybrid lily
right
if i may, let me just make the question and bit more specific
would you agree that if c and d are integers, then c^2 +d^2 is a gaussian integer?
hybrid lily
right
wait a second
what is there to prove
it literaly states in the quesiton than its the product of two gaussien integers
obviously n will be a gaussian inetger
vale wigeon
no, the question doesn't state that n is the product of two gaussian integers
it asks you to express n this way
and the solution, as is obviously its job, tells you how
hybrid lily
right but if you assume that n as the product of two gaussian integers
n should also be a gaussian integer
so then why does the solution have to prove the norm is larger than 1
what does showing that prove
vale wigeon
right but if you assume that n as the product of two gaussian integers
but why would you assume it
vale wigeon
so then why does the solution have to prove the norm is larger than 1
i was going to explain, but you kind of cut me off.
hybrid lily
my apologies
vale wigeon
i was going to explain by going back down to real integers
a real integer p is prime if it is impossible to express it as the product of two real integers, neither of which is ±1
the "neither of which is ±1" bit is there to exclude obvious and trivial products such as 7 = 7*1
hybrid lily
right
vale wigeon
the "norm > 1" bit is what that translates to in the gaussian integers
it is to exclude products where one of the factors is ±1 or ±i
simple ocean
lone heartBOT
simple ocean
i need to differentiate
so i decided to use the chain rule
and so far i got 90(4tan(2x)+9)^2
but im stuck on how to go on about finding the derivative of the inside
vale wigeon
chain rule again
simple ocean
my bad i wasnt clear but that's where i need help
i dont know how id use the chain rule on the inside
lone heartBOT
@simple ocean Has your question been resolved?
green juniper
Try use the quotient rule for the inside where $4tan(2x) = \frac{4sin(2x)}{cos(2x)}$
ocean sealBOT
SKELEROY
gray isle
why bother with quotient rule
after the initial application of (power)-chain rule you'll have
$$y' = 90(4\tan(2x) + 9)^2 \cdot \blue{\dv{x}(4\tan(2x) + 9)}$$
do you know how to differentiate tan(x)?
simple ocean
why 30, should it not be 90?
simple ocean
do you know how to differentiate tan(x)?
derivative of tan is sec^2x
gray isle
sry typo
ocean sealBOT
ℝαμΩℕωⅤ
gray isle
apply that and combine with chain rule
(assumed that you already know how to apply linearity and constant multiple rule)
simple ocean
i tried but got confused because of the 4 and the +9
gray isle
apply linearity and constant multiple rule
lone heartBOT
gray isle
.reopen
lone heartBOT
✅
gray isle
$\dv{x}(f(x) + g(x)) = ? \ \
\dv{x} (a\cdot f(x)) = ?$
ocean sealBOT
ℝαμΩℕωⅤ
simple ocean
what is a
vale wigeon
a constant.
simple ocean
$\dv{x}(f(x) + g(x)) = ? \\ \\
\dv{x} (a\cdot f(x)) = ?$
would that be right
gray isle
for that blue part, yes
gray isle
no
simple ocean
where did i go wrong
gray isle
how are you getting that
do you have a pic of your result?
you may be making an error representing that here
simple ocean
and so far i got 90(4tan(2x)+9)^2
this multiplied by the derivative of the blue part
gray isle
why did the ^2 on the (4tan(2x)+9) dissapear
simple ocean
i forgot to type it but i have it written
lone heartBOT
@simple ocean Has your question been resolved?
lone heartBOT
@simple ocean Has your question been resolved?
gentle berry
working on some factoring in here
lone heartBOT
working on some factoring in here
gentle berry
lemme know once ya get in
blissful thicket
hello im in
gentle berry
k i'ma give you a quick rundown of factoring trinomials
before, I mentioned the ac method so let's give that one another shot
kev you can join in too if you want
$x^2 + 4x + 3$
ocean sealBOT
JayRoc (The Hurricane Panda)
gentle berry
Starting off, what are the only 2 factors of 3?
blissful thicket
1 I guess
lime sentinel
1 and 3
wild umbra
okay what is going on💀
lone heartBOT
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).
wild umbra
no, no cheating allowed on this server
gentle berry
ok good. So here's the long way of doing this when a=1
we're gonna take the 4, and split it into the 1 and 3 we have
$x^2 + 4x + 3 = x^2 +x + 3x + 3$
ocean sealBOT
JayRoc (The Hurricane Panda)
gentle berry
ok with what I did there?
lime sentinel
ye
blissful thicket
so you just split it into 2 and the 1 stays as x?
gentle berry
yeh jus the middle terms
and i took 4x and made it into x + 3x
i'ma show you y in a second
(and let you see something peculiar that happens)
So the long method here is what's called factor by grouping. we look at the first 2 terms and see what's "common" between them and pull that value out (or factor that value), then we do that with the last 2 terms.
So for $x^2 + x$ what's common between those 2 terms?
ocean sealBOT
JayRoc (The Hurricane Panda)
lime sentinel
x
blissful thicket
wait I got one thing I thought it would become x and 3 but it's actually 3x, so it keeps the x?
gentle berry
yup that's right
lime sentinel
so we're just splitting it?
gentle berry
yup
So,
$x^2 + x $ becomes $x ( x + 1)$ because we factor the x, or in other words, divide by x, but put the x on the outside of the terms we divide.
ocean sealBOT
JayRoc (The Hurricane Panda)
gentle berry
How about $3x+3$? What's common between those two terms?
ocean sealBOT
JayRoc (The Hurricane Panda)
lime sentinel
3
lucid maple
3
gentle berry
wait I got one thing I thought it would become x and 3 but it's actually 3x, so ...
you trackin?
blissful thicket
Yeah I just don't get it
lucid maple
are you trying to get values of x in that eq?
gentle berry
well if you don't get it, you have questions right?
blissful thicket
yeah
lime sentinel
why do we put x outside the bracket
blissful thicket
yeah what happens there
gentle berry
That's the operation of factoring. we find what's common between the terms, then "pull it out"
for example
let's say we have $8x+16x^2$
ocean sealBOT
JayRoc (The Hurricane Panda)
gentle berry
If I rewrite it like this:
$8x + 2\cdot 8\cdot x \cdot x$, first what number do both of them have in common?
lime sentinel
8?
ocean sealBOT
JayRoc (The Hurricane Panda)
gentle berry
yup you got it.
gentle berry
yeah what happens there
make sense?
lime sentinel
but in that equation where do we get the 1 from
gentle berry
I still don't get it
what's your question?
in the equation above, we can rewrite any single term as that term times 1
so $x = x\cdot 1$
ocean sealBOT
JayRoc (The Hurricane Panda)
blissful thicket
what's your question?
I'm figuring out what my question is I don't really even know what I'm not getting
gentle berry
probably some why's right?
lucid maple
so this way: $x^2 + $x = x\cdot$x + x\cdot 1$
gentle berry
(and don't worry, I grill my students all the time, no stress mode here)
ocean sealBOT
AnshumanNeon
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
lucid maple
now x is common
lime sentinel
but we've done x^2 + x = x(x+1)
lime sentinel
i dont get how it becomes x(x+1)
lime sentinel
x
blissful thicket
so x^2 just becomes x*x and x becomes x 1, so if there was like a 4x what would happen
gentle berry
so x^2 just becomes x*x and x becomes x 1, so if there was like a 4x what would ...
it would be 4* x
you're gettin it
lucid maple
you so are basically multiplying x with both of them. using distributive property of multiplication: we can take the x out and put the other terms in bracket
lime sentinel
so take an x out and put it out front. what you got left?
i dont understand the question
lucid maple
x * x + x * 1 = x(x + 1)
blissful thicket
Oh I get it
gentle berry
i dont understand the question
this one from before, you said there's an 8 common right?
lucid maple
i dont understand the question
you take out the common x from both terms, just remove it and store it in your memory for sometime, just remember that there was a x that was being multiplied to both but don't multiply it right now
what would you be left with then?
gentle berry
so if you take an 8 out of both terms and put it out front, you have 8(x + 2* x * x)
blissful thicket
well now I dont get it again
lime sentinel
so if you take an 8 out of both terms and put it out front, you have 8(x + 2* x ...
so both the 8s become one 8 out front?
gentle berry
yeh. cuz like anshuman said, you're doing the opposite of the distributive property
lime sentinel
i get it now
gentle berry
$(a\cdot b + a \cdot c) = a (b+c)$
ocean sealBOT
JayRoc (The Hurricane Panda)
lucid maple
so both the 8s become one 8 out front?
yes! you stop multiplying the 8 for sometime and put the left over expression in a bracket and then multiply the 8 to the whole bracket itself
lime sentinel
ohh
gentle berry
so with that, you have $8(x + 2 x x)$, how many x's do you have in common between those two terms?
ocean sealBOT
JayRoc (The Hurricane Panda)
lucid maple
$8($x + $x^2)
lime sentinel
1?
gentle berry
yup
blissful thicket
so if you take an 8 out of both terms and put it out front, you have 8(x + 2* x ...
where did the other 8 go in this one
lucid maple
i gues you mean this
gentle berry
so you pull that x out
lime sentinel
where did the other 8 go in this one
you only use one 8
gentle berry
now what do you have left?
gentle berry
i gues you mean this
oh it was x^2
ocean sealBOT
AnshumanNeon
Compile Error! Click the reaction for more information.
(You may edit your message to recompile.)
lime sentinel
now what do you have left?
8x*2?
blissful thicket
you only use one 8
I dont get why
ocean sealBOT
JayRoc (The Hurricane Panda)
gentle berry
in that you only use 1 a
lime sentinel
now what do you have left?
for which equation
gentle berry
for 8 (x + 2 x x)
lucid maple
I dont get why
its simple, you don't go tell every student individually that annual function is coming, you just gather them in a place and tell them collectively
lime sentinel
for 8 (x + 2 x x)
8x(2x)?
gentle berry
close, don't forget that x = x*1
lucid maple
similarily you dont multiply individually (you can, but sometimes it helps not to) you just take whatever you're multiplying with and put that in a bracket and then multiply the whole by 8
lime sentinel
close, don't forget that x = x*1
8x(2*1x)?
lucid maple
so do I just forget the other 8
for a moment, just store it in your memory for sometime and dont multiply it by anything, then whatever you're left with just put that into a bracket and then multiply the 8 with that bracket itself
lime sentinel
8 ( x * 1 + 2 x x)
8x(1+2x)?
gentle berry
ding ding ding
lime sentinel
but wouldn't there be another 1
blissful thicket
for a moment, just store it in your memory for sometime and dont multiply it by ...
wdym
gentle berry
but we don't usually write them out
lucid maple
but wouldn't there be another 1
that can be
lucid maple
but we wrote this 1
because multiplying by one doesn't matter, adding and then multiplying does
you can multiply anything by 1 as many times as you want and you will still get the original number
lucid maple
wdym
remember to multiply by 8 but dont multiply now, multiply after you've put everything in the bracket
gentle berry
yup
blissful thicket
remember to multiply by 8 but dont multiply now, multiply after you've put every...
so yeah I got that it just gets multiplied by 8 when it's in the bracket
lime sentinel
it becomes 1+2x?
lucid maple
it becomes 1+2x?
yes!
gentle berry
yup
lime sentinel
ok
gentle berry
try this one: $x^3 + 4x^2 + x$
ocean sealBOT
JayRoc (The Hurricane Panda)
lime sentinel
xxx+4xx+x
gentle berry
good start
lime sentinel
x(x^2+4x)
lucid maple
good going
lime sentinel
x^2(x+4)?
gentle berry
no you had the right factor, it was just x
but somethin is "left over" for the last term ya?
remember x = x*1
lime sentinel
yeah
gentle berry
so if you factor an x off, whatcha got left?
gentle berry
🔥
lime sentinel
x^2(x+4+1)?
lucid maple
yes
gentle berry
the first one was correct. we don't have enough "x" terms to do the 2nd one
(the last term only has a 1)
gentle berry
I think before the trinomials, just a few exercises in factoring will be good
Jay gotta go to bed now tho, it's 6 am
gentle berry
np. yall ping me if I'm around tmrw, if I'm not busy we can do some more
lime sentinel
ok
blissful thicket
I didnt get it but thanks for the help much appreciated
unkempt robin
I didnt get it but thanks for the help much appreciated
What don't you understand?
blissful thicket
What don't you understand?
I just understand I take 8x+2 * 8 * x * x = 8(x+2 * x * x)
uhm
the discord screwd it up
ok fixed it
unkempt robin
Right, we've got $8x + 16x^2$
ocean sealBOT
USS-Enterprise
unkempt robin
Factoring means pulling out common factors
If we first expand this, we get $(8 \cdot x) + (2 \cdot 8 \cdot x \cdot x)$
ocean sealBOT
USS-Enterprise
unkempt robin
These are two terms, (8x) and (2*8*x^2)
What are the common factors in these two terms?
blissful thicket
idk
ocean sealBOT
USS-Enterprise
unkempt robin
What are the common factors of these two terms (2 and 4)?
blissful thicket
2
unkempt robin
Exactly
So we can "pull" two out
Or factor two out
Same thing
And rewrite as $2 \cdot (1 + 2)$
ocean sealBOT
USS-Enterprise
unkempt robin
Because of distributivity if we were to multiply everything out again, we'd get 2 * 1 + 2 * 2 or 2 + 4
What we had in the start
The terms inside the parantheses is what you get if you divide the original term with what you are pulling out
So we get 1 by dividing the original term (2) with what we are pulling out (2), because 2/2 = 1
Similarly, we get 2 because we divide the original term (4) with what we are pulling out (2), because 4/2 = 2
Get it?
blissful thicket
yes this one I got
ocean sealBOT
USS-Enterprise
unkempt robin
What are the common factors of these two terms?
blissful thicket
2 and x
ocean sealBOT
USS-Enterprise
unkempt robin
What's the first term inside the parantheses?
Remember, original term divided by what we are pulling out
blissful thicket
2x * (1 + 2)
unkempt robin
There you go!
1 because 2x/2x = 1 and 2 because 4x/2x = 2
Now try this: $3x^2 + 6x$
ocean sealBOT
USS-Enterprise
unkempt robin
What are the common factors?
blissful thicket
it should be 3(x * x + 2 *x) or something similar right
unkempt robin
Is 3 the only common factor?
blissful thicket
1
unkempt robin
it should be 3(x * x + 2 *x) or something similar right
I mean, this is technically correct but you can do a bit more
Look at the variables
You've got $(3 \cdot x \cdot x) + (3 \cdot 2 \cdot x)$
ocean sealBOT
USS-Enterprise
unkempt robin
You got 3 right
blissful thicket
right
unkempt robin
But doesn't an x appear in both as well?
It doesn't matter there's two of them in the left
All that matters is that there's at least one in both
unkempt robin
So pull it out as well
blissful thicket
like u did yeah
unkempt robin
So $3x \cdot (something)$
ocean sealBOT
USS-Enterprise
unkempt robin
And repeat what we've been doing
blissful thicket
I have to go eat I'll be back
lone heartBOT
@gentle berry Has your question been resolved?
dim kettle
lone heartBOT
dim kettle
How can I develop this i cant see a way
lone heartBOT
@dim kettle Has your question been resolved?
dim kettle
<@&286206848099549185>
dim kettle
do you know the polar form of complex numbers ?
Polar is e^i?
vale crag
yeah, you can write every complex z as z = r e^(ix) [where r is the modulus/radius of that number and x is the argument/angle of that number]
this equation simplifies a lot if you go through polar form instead of just z=a+ib
dim kettle
this equation simplifies a lot if you go through polar form instead of just z=a+...
I dont see how u would simplifu that
vale crag
have you solved easier equations with polar form before, like z^6 = 1 for example ?
dim kettle
Yes
vale crag
yeah ok
well the idea here is that complex conjugate and multiplication behaves well with the polar form
if z = re^(ix), then |z| is just r, and bar(z) = re^(-ix)
dim kettle
Yes
vale crag
that simplifies our equation quite a bit
$$r^3e^{i3x} = \frac{1}{3} i\cdot r\cdot re^{-ix}$$
ocean sealBOT
aPlatypus
vale crag
and now even that i can be incorporated
since i is just e^(i pi/2)
$$r^3e^{i3x} = \frac{1}{3} e^{i\frac{\pi}{2}}\cdot r\cdot re^{-ix}$$
$$r^3e^{i3x} = \frac{1}{3}r^2e^{i\left(\frac{\pi}{2}-x\right)}$$
ocean sealBOT
aPlatypus
vale crag
so here we have 2 numbers in polar form which are equal
so their radius is equal
and their angle is equal (up to a multiple of 2ipi, cause adding 2ipi doesn't change the value of the exponential)
dim kettle
Mhh
vale crag
well now I guess you can take over
you have all the setup
you just have to find what values the radius can take
same for the angle
and you should already have done that on simpler examples like z^6 = 1, as you said
dim kettle
Trig of 1 is e?
vale crag
**aPlatypus**
anyway looking at that last equation, what's the radius on the left side and on the right side ?
dim kettle
anyway looking at that last equation, what's the radius on the left side and on ...
U have r not known
vale crag
yes r is not known we're trying to find it
but that equation gives us information on r, that's my point
dim kettle
but that equation gives us information on r, that's my point
Hiw would u find r
vale crag
we know these two numbers in polar form are equal
dim kettle
Ok
vale crag
and two numbers in polar form (!= 0) are equal if and only if their radius is equal and their angle is equal
green juniper
You should probs also put a +2pi*k on the angle
vale crag
well afterwards yeah
dim kettle
Ok so I put radius=radius
vale crag
then you're done
if you found all possible r and x that work, you got all the complex numbers that solve your original equation
dim kettle
Ah ok
dim kettle
then you're done
Can u help me with a different quick thing
vale crag
ok @dim kettle
dim kettle
So u have this
U can split
Right one has 3 solutions
Which are z=e^(i(3/2pi + 2kpi/3))
Ok?
vale crag
yep sounds good
dim kettle
yep sounds good
Ok what aboht left tho
X^4+6x^2+9=0
If it was regular x there would be no solution
dim kettle
Lets develop for testing
vale crag
(z^2 + 3)^2 = 0 is equivalent to z^2+3 = 0
dim kettle
Why
vale crag
Lets develop for testing
if you want
vale crag
(z^2 + 3)^2 = 0 is equivalent to z^2+3 = 0
well you can split that equation in 2 again (like you did with the big one)
(z^2 + 3)^2 = 0 is equivalent to (z^2+3 = 0 or z^2+3 = 0)
dim kettle
U said same thing twice
vale crag
yeah that's the point
dim kettle
X^4+6x^2+9 is different than x^2+3
vale crag
the equations are different sure
the roots are the same though as I've just shown
and we're only interested in the roots
dim kettle
X^4+6x^2+9=0
We have this
Normally no solution
But we in C
So t=z^2
Solution of t is -3
Z^2=-3
?
dim kettle
Oh
dim kettle
No
pallid scarab
Here's your answer then
pallid scarab
Yes
dim kettle
Thx
pallid scarab
Don't forget to maybe simplify here
dim kettle
?
pallid scarab
e^(ipi/2) = ?
dim kettle
= i
pallid scarab
Yes
dim kettle
Thx
pallid scarab
So solutions are z=sqrt3 e^(i(pi/2 + (2kpi/2))
And k is either 0 or 1 here
dim kettle
Yes
pallid scarab
So you should be able to find all solutions by plugging k=0 first, and k=1 afterwards
dim kettle
Yes
pallid scarab
What do you get?
dim kettle
2 results
pallid scarab
Yes, can you find them?
pallid scarab
× sqrt3
dim kettle
Ah yes
pallid scarab
You got it!
ocean sealBOT
Ava
gray isle
not an equation
show exactly what you're putting into your calc
oh 1 sec, i see something problematic already
what's with the two decimal points $$31\red{.}536\red{.}000$$
ocean sealBOT
ℝαμΩℕωⅤ
gray isle
yeh, that's like illegal notaton
which is why your calc is essentially saying hell no
what's the number supposed to be
ocean sealBOT
Ava
vale wigeon
so when you write 31.536.000 it means 1.99 * 10^32?
but what do you mean when you write 31.536.000?
gray isle
use that, no illegal decimal points
math jail
well firstly there's only 1 decimal point there
and that is representative of the number with an integer part of 1 and fractional part of 0.5,
or 1 + 0.5
having two decimal points like that makes no sense
commas can be used as delimiters to separate thousands,
but calculators don't recognise that either
if the number you want to enter use 31536000
just enter that
light briar
lone heartBOT
light briar
im a bit confused for question 3
idk how the indices distribute and in which order the functions are inner and outer
icy sierra
Do you know the power and chain rules?
light briar
ik them yes but this is a bit hard to wrap my head around
this specific example i get the chain rule
icy sierra
the power rule is part of chain rule
in the third
Can we write the same as [Cos(3x^2)] ^5
light briar
so the innermost function is the 3x^2
frail grove
$cos^5 3x^2=\left (cos(3x^2) \right )^5$
ocean sealBOT
Adam Chebil
light briar
oh so the 5 indice on the cos just goes outside the whole thing altogether
light briar
i hada question that was sin^4 x
icy sierra
so use the power rule
light briar
and the 4 was on the outside but it was a bit harder to see when htere was more going on
so the (x)^5 is the outermost
icy sierra
that is the same as (Sinx) ^4
icy sierra
so the (x)^5 is the outermost
yeah
after using power rule, use the chain rule to differentiate the cosine
light briar
wait is there really no difference between sin^4 (x) and (sin(x))^4
light briar
shi ok
light briar
so outermost is (x)^5
icy sierra
yes
icy sierra
yeah
middle is -sin( 3x ^2)
remember we are differentiating cos(3x ^2)
not cos(x)
so we get -sin( 3x ^2)
light briar
yeah but isnt the stuff inside the cos(x) another function
icy sierra
dont worry about that
the inner functions remain the same while differentiating the outer
light briar
oh right i think i understand
icy sierra
Okay
light briar
so the outermost is 5(cos(3x^2)^4
icy sierra
Yes
light briar
ah i see
the x represents whatever the inner functions are like with f(x)
ok ok ok
icy sierra
the x represents whatever the inner functions are like with f(x)
exactly
light briar
middle is -sin(3x^2)
icy sierra
yes
light briar
innermost is 6x?
light briar
ok so we got
icy sierra
all are multiplied
light briar
oh i see if you keep it in the form with all of the fucntions still written out as opposed to just (x)^5 you just take the product of all 3
icy sierra
yes
light briar
5(cos(3x^2)^4* (-sin(3x^2)) * 6x
icy sierra
identify the outer function for the chain rule
it may not always be something squared
icy sierra
5(cos(3x^2)^4* (-sin(3x^2)) * 6x
yes
light briar
whats it in this case
icy sierra
that is the complete derivative
light briar
it may not always be something squared
this "it"
icy sierra
whats it in this case
here it is something raised to the power 5
something is the inner function, Cos(3x ^2)
That is why we used the power rule since it is the outer most function and we differentiate it first
light briar
i think i got everything then i jsut have to simplify the complete derivative shouldnt be too long
light briar
ya i was just confused on the order and where they distribute with indices on trig