based on the info given I'm assuming it'll oscilate at y=7, but idk how to calculate the min/max value
#help-36
stray phoenix
zenith pollen
assume it oscillates at y=0, not much a reason why 
zenith pollen
yea, amplitude is distance from rest to max
zenith pollen
yea the wrong part of that is the 2pi part
2pi is the angle/whatever it takes to do a full cycle, but the max comes sooner
zenith pollen
even smaller
stray phoenix
oh!
zenith pollen
sin is this sort of graph for 1 cycle
zenith pollen
1/4 a cycle, so 1/4*2pi
zenith pollen
it's good to have this sort of picture
zenith pollen
oh I didn't catch that
zenith pollen
actually I missed that it starts at the maximum too haha
so from max to min is what you're looking for
stray phoenix
ohhhh
so more of a cosine wave but not really
hm
if it starts at 7 maximum and the lowest point is 1/4 then it doesn't oscilate on 0
if i did this right
it oscilates on 3.625
(7 + 1/4)/2
zenith pollen
oh the 7 and 1/4 are different units
1/4 is time, 7 is amplitude outside the cosine
stray phoenix
oh my bad
zenith pollen
max to min is half a cycle so you do pi/omega=1/4
and cosine is fine, or an offset sin
stray phoenix
so omega = pi/4
zenith pollen
nah 4pi
stray phoenix
oh no yeah mb
pi = omega/4 then 4pi = omega
duh
thanks @zenith pollen !! :)
think u could help me w one more?
idk if my answer on the very bottom one is correct
like
it feels weird that
they just give me the answer
LOL
zenith pollen
lol, amazing
zenith pollen
yea I'd put 6.2 for c
stray phoenix
i have to get minimum 6/9 right for this section and i already have 2 wrong
i wanna have at least one for safety
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cunning hollow
how do i write this as a power series?
final saddleBOT
how do i write this as a power series?
cunning hollow
my cal exam is in 2 days help pls 😭
drifting pebble
my cal exam is in 2 days help pls 😭
I can try but i’ve not done this part of the course yet
so on ur equation sheet u have the infinite geometric series yes?
cunning hollow
yes
drifting pebble
u do the bottom bit first so convert it to (x^2+2)-2
and u have to multiply each term by x^2
then sum it all up
cunning hollow
ok...
drifting pebble
x^2+2
you have to factor it into I and J
(x+square root of 2 I) (x-square root of 2 J)
those are denominations so A and and B can be put on top as constants
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lilac wing
Glad I could help u man
austere heart
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austere heart
Not sure how to do this one
Since it's supposed to be sum(f(x)) * change in x. I think it should A or D
neon slate
what would the areas of the rectangles be for a right riemann sum? and check which of the sums gives the correct values
austere heart
what would the areas of the rectangles be for a right riemann sum? and check whi...
so I think it's calculating from -1 to 3
I'm guessing that means it boils down to A or B
and well, A just makes more sense
neon slate
so I think it's calculating from -1 to 3
yes the interval is [-1, 3]
austere heart
so it should be A i assume
neon slate
but just simply looking at the picture, what should the areas of each rectangle be?
neon slate
yes, f(x) evaluated on the right endpoint of each subinterval, and what are the numbers that comes out to?
austere heart
7 6 5 4
neon slate
yes, so now just check the sum you suspect is the answer, does it give you 7 for the first term?
austere heart
oh so it's D 😮
cuz first value of 1 - 1 = 0
got it ty
(I'll keep the channel open for now Just in case)
I assumed the intervals would go from -4 to 4, but now it's mid point and sum it's starting from the left most value
oh D
I think
austere heart
2 intervals
neon slate
?
austere heart
what's the width of the rectangles in this one?
is B not it
seems like it gets the f(x) and multiplies by 2
plus they seem to match graph
.close
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proud glacier
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proud glacier
why are they not checking R,x for abelian
clearly is abelian isn't it?
multiplication is commutave in \R
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static fractal
it is commutative, but it already fails inverses
royal gust
Yeah, weird they didn't just ✅
static fractal
yeah
but if you're checking if its an abelian group or not
and you see that it already fails inverses, then it isn't even a group and you can already say no
weary beacon
can someone please explain the kinematic equation to me and why it uses t^2 and acceleration if you use velocity and dont square it instead
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can someone please explain the kinematic equation to me and why it uses t^2 and ...
steel crescent
can someone please explain the kinematic equation to me and why it uses t^2 and ...
what do you mean?
weary beacon
so basically yk how its 1/2at^2
zenith pollen
you can get it either by integrating or drawing a velocity-time graph
steel crescent
yea
modest hornet
you mean this ? :
1/2at^2 + tv + x0 ?
weary beacon
ive been doing some khan academy stuff they dont square it
should i send ss of problem
steel crescent
should i send ss of problem
sure
weary beacon
why is it v-t
weary beacon
modest hornet
oh you know why ?
because it is using the area under the graph
its basically the same
weary beacon
wdym
modest hornet
area under the graph = displacement
weary beacon
oh yeah
modest hornet
np
modest hornet
its not inital velocity
celest cobalt
- In △ABC, AB = X + 1, BC = X, AC = X − 1. C = 2B, Find X.
modest hornet
its finding the area under the graph
celest cobalt
need helps
modest hornet
1. In △ABC, AB = X + 1, BC = X, AC = X − 1. C = 2B, Find X.
open a new help channel
steel crescent
1. In △ABC, AB = X + 1, BC = X, AC = X − 1. C = 2B, Find X.
please
celest cobalt
?
steel crescent
1. In △ABC, AB = X + 1, BC = X, AC = X − 1. C = 2B, Find X.
type your problem in these channels
any one of these
peak hemlock
Let $f$ be a function from a finite set $A$ to itself. Prove that $f$ is injective if and only if $f$ is surjective. $\$
I understand why this must be true. in the direction injectivity implies surjectivity: $\$
Assume the function is not surjective. This means $|f(A)| < |A|$. $//$
The codomain and domain are the same set, so $| \text{codom} f | = |A|$. Since injectivity implies that $|f(A)| \ge |A |$ and $f(A) \subseteq A$, we know $|f(A)| \le |A|$. Consequently, $|f(A)| = |A|$. This contradicts with $|f(A)| < |A|$, and thus $f$ must be surjective. $\$
I qualitativly understand the reverse, but am even less sure how to write it out. $\$
Is my reasoning ocrrect, and how can I turn it into a proof? Thanks!!
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Let $f$ be a function from a finite set $A$ to itself. Prove that $f$ is inject...
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Nameless
peak hemlock
oh, sorry JWIVY
peak hemlock
I stole your help channel xD
peak hemlock
I pasted in my question, but didn't realize you were typing
pseudo nymph
oh np
peak hemlock
thanks xD
Let $f$ be a function from a finite set $A$ to itself. Prove that $f$ is injective if and only if $f$ is surjective. $\$
I understand why this must be true. in the direction injectivity implies surjectivity: $\$
Assume the function is not surjective. This means $|f(A)| < |A|$. $//$
The codomain and domain are the same set, so $| \text{codom} f | = |A|$. Since injectivity implies that $|f(A)| \ge |A |$ and $f(A) \subseteq A$, we know $|f(A)| \le |A|$. Consequently, $|f(A)| = |A|$. This contradicts with $|f(A)| < |A|$, and thus $f$ must be surjective. $\$
I qualitativly understand the reverse, but am even less sure how to write it out. $\$
Is my reasoning ocrrect, and how can I turn it into a proof? Thanks!!
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Nameless
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peak hemlock
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peak hemlock
NO. IT STILL HASNT. STOP ASKING.
half shard
<@&286206848099549185>
Your reasoning is correct. When you say you qualitatively understand the reverse, are you referring to the if f is surjective then it must be injective part?
peak hemlock
yes, I am
I am struggling with writing a formal proof, though
Saldy I don't have time right now to have a discussion. sorry
half shard
Your first part is fine, you just to make sure it’s clear you’re writing the if injective then surjective part, and state that you’re using pigeonhole principle (becuase of the cardinality stuff)
You should probably also define what not surjective means (there’s some element in the codomain with no element in the domain mapping to it)
And then for the second part, surjective to injective, you can also do a contradiction. So like:
-
If f is surjective then f is injective (Proof by contradiction)
-
Assume f is not injective. So, there are at least two elements in the domain with the same image. This means |f(A)| < |A|
-
Since f is surjective/onto, every element in the codomain is mapped to by at least one element in the domain. This means |f(A)| >= |A|.
-
This has reached a contradiction, so f must be injective
When you write it just make sure you write by contradiction at the start of each part and you write a conclusion saying that sonce you proved it both ways, f is injective IFF it’s surjective
drowsy path
Yo was good
final saddleBOT
Yo was good
drowsy path
I’m doing physics project rn
I will send photos
This is what I rearranged the equation as
It’s a simple question really
Has this been rearranged correctly tantheta times delta theta over since theta as cot
Otherwise I think it would be this, but I was lazy to use sintheta for all the values
:
The chain rule states that delta sin theta =cost theta times delta theta
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amber junco
i dont get what there doing here
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i dont get what there doing here
floral nova
They explained that they used that theorem you in the second paragraph.
amber junco
They explained that they used that theorem you in the second paragraph.
but why do they add s0 + s1?
oh wait
floral nova
Probably because the question was phrased so that S_0 wasn't enough
amber junco
they do it until s1
amber junco
Probably because the question was phrased so that S_0 wasn't enough
so they only add up until its enough?
cus s2 could also be enough
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storm void
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storm void
Couldn't find no pttenr
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umbral onyx
mortal bison
Do I solve for right and left limits? The correct answer should be 1/2
mortal bison
the right limit gets me 1/2 but the left one does not
marble agate
you can eliminate absolute value for left side and right side limit
marble agate
no
jusr 2x-2
2x-2>0 2x>2 x>1
and 1 is not close to 2
so you can eliminate for left side too
marble agate
yes
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midnight ginkgo
didn't get any help last time, so I try again.
A box has the shape of a rectangular prism. The side faces have areas of 96 cm², 252 cm², and 168 cm².
a Explore: What dimensions can the box have (length, width, and height)?
b How many liters of rice can fit in the box?
midnight ginkgo
I don't understand what I'm supposed to do in the first challenge, so just explaining it to me would be great!
<@&286206848099549185>
vestal yoke
<@&286206848099549185>
You have given the areas of the faces of the prism and through the given areas you have to find the dimensions. let the dimensions and you’ll get 3 equations
midnight ginkgo
ok, but how do I get the length, width and height from the area?
vestal yoke
Let the dimensions a,b,c and find areas in these form to get 3 equations
And solve these equations to get a,b,c
midnight ginkgo
I don't understand what you are trying to say.
what do you mean by "let the dimensions"
vestal yoke
Let the length = a, let width = b , hight = c
midnight ginkgo
still don't understadnd ;(
vestal yoke
Here, we are assuming these dimensions, so we can get these equations, ab = 252, bc = 96 and ac = 168. Here, through these equations, we can get the values of dimensions (a, b , c) which is asked in the question
vestal yoke
Welcome
lunar forum
$Consider the quotient space $P = P(X)/ \sim$ and let $\pi : P(X) \to P$ be the quotient map. Show that the following defines a metric on $P$:
[
d(\pi(A), \pi(B)) = \mu^(A \Delta B).
]
Show also that
[
|\mu^(A) - \mu^*(B)| \leq d(\pi(A), \pi(B)).
]$
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$Consider the quotient space $P = P(X)/ \sim$ and let $\pi : P(X) \to P$ be the ...
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RealTek
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
lunar forum
im trying to show the inequality on the bottom
$|\mu^(A) - \mu^(B)| &= |\mu^(A) - \mu^(C) + \mu^(C) - \mu^(B)| \
&\leq |\mu^(A) - \mu^(C)| + |\mu^(C) - \mu^(B)| \$
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lunar forum
I understand i need to get some symetric difference here
but i cannot for the life of me see it
ocean lintel
It would certainly be helpful to say what P(X) and ~ are
lunar forum
It would certainly be helpful to say what P(X) and ~ are
yeah i gotta somehow find a symetric difference
ocean lintel
A D B = A \ B + B \ A
Hence mu(A D B) = mu(A) - mu(A n B) + mu(B) - mu(A n B) >= |mu(A) - mu(A n B)| + mu(B) - mu(A n B)| = |mu(A) - mu(A n B)| + |mu(A n B) - mu(B)| >= |mu(A) - mu(A n B) + mu(A n B) - mu(B)| = |mu(A) - mu(B)| ?
I'm not convinced myself because I make it come out of thin air by whatever shenanigans
Ok nvm it's obvious if you draw it
mu(A) - mu(B) <= mu(A \ B) <= mu(A D B)
Done
And vice versa
lunar forum
Ok nvm it's obvious if you draw it
mu(A) - mu(B) <= mu(A \ B) <= mu(A D B)
Done
okay can you explain to me how its less than equal to the set difference a\b????
whys that not equal
lunar forum
Ok nvm it's obvious if you draw it
mu(A) - mu(B) <= mu(A \ B) <= mu(A D B)
Done
how do you draw something like this? with two circles
ocean lintel
Of course it's circles
Why do potatoes or even worse, not simply connected bounded convex sets when you can ???
ocean lintel
I didn't talk about how to see it with two circles
I said why overcomplicate when you can take 2 circles anyways
lunar forum
damn ive laboured over this for awhile
and its as simple as a -b <= a\b <= a D b = d(a,b)
ocean lintel
Welcome to math I guess
It happens
lunar forum
hah i think im just dumb
anyways thanks for the help
you have a good night 🙂
.close
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dusk mason
hello
dusk mason
it has some python but
can you check this, I feel like I made some logic error
I think it should be higher than 8%
sum just finds the count of true values across the array btw
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dusk mason
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cedar sierra
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cedar sierra
need help
magic sparrow
cedar sierra
i havent tried anything yet because i missed like 3 math classes
so im just confused
magic sparrow
cedar sierra
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mossy patio
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mossy patio
Someone mind helping me out with this question?
Ik the answer, just not how to get there
mossy patio
Mhm
tranquil pine
so you have m*m/2*(m + 1)
mossy patio
So m/2 * m * (m+1) > 1000, right?
tranquil pine
yeah
m^2/2*(m + 1) > 1000
now you can get m^3/2 + m^2/2 > 1000
which can be simplified to m^3 + m^2 > 2000
mossy patio
Mhm
tranquil pine
yeah i didn’t feel like solving a cubic if you want me to ig i will
mossy patio
Nah, that's fine
We haven't covered cubics yet anyway
So
But the answer's weird
Says m >= 14
tranquil pine
oh lol forgot about the domain
because there are m/2 choices for the main course it must be even
or it dosen’t make sense
mossy patio
Ohh okay
tranquil pine
!done
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mossy patio
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tranquil bone
I wonder if the answer to this part for G_2 is 2/3 + 1/3 • 2/3 • 1/3
tranquil bone
Just by considering the probability of going in the right path?
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zenith pollen
I wonder if the answer to this part for G_2 is `2/3 + 1/3 • 2/3 • 1/3`
Not quite, if you try branching out and counting the probabilities you'll have to go on forever. I'd try writing recurrence equations for the middle two points, with variables x=(chance to get to the bottom node from this vertex) and y for the other vertex. Like for the top middle point, there's a 1/3 chance it goes up top, and 2/3 it goes to the y vertex and uses whatever chance that has, for x=1/3*0 + 2/3*y
then you can solve for x and y and the answer will be 2/3+x since you start up top
tranquil bone
Not quite, if you try branching out and counting the probabilities you'll have t...
I’m not really following this. Sorry. How would i go about calculating x and y?
zenith pollen
It's a specific strategy where you look at the next step of a random walk, and write those next step probabilities in terms of the probability of your location.
the idea is that every point has a chance to reach the bottom before the top, and we can look at how they're connected to relate all those chances
for x the path going up immediately hits the top so it fails, 0, and 2/3 of its paths have whatever chance y does
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tranquil pine
can someone help me on bearings?
final saddleBOT
can someone help me on bearings?
tranquil pine
tepid anvil
can someone help me on bearings?
have you drawn a picture
itll be easier to solve if you visualize it
tranquil pine
itll be easier to solve if you visualize it
i tried drawing the diagram
but still can't picture where all the angles are
tranquil pine
i have my diagram, just need that angle
tranquil pine
also i keep seeing that the angle QPR is a right angle
and i don't know why
tranquil pine
did you solve the last side yet?
i can't without a complete diagram
and i need the angle for that
tepid anvil
ahhh
tranquil pine
mhm
tranquil pine
also i keep seeing that the angle QPR is a right angle
and i don't know why
yk why?
tepid anvil
some dumb geometry rule prob
tranquil pine
some dumb geometry rule prob
that's the thing!
i can't tell
tepid anvil
did they not give you an angle?
tranquil pine
like some alternating angle kind of thing
tranquil pine
maybe the triangle inequality theorem then
which is?
tepid anvil
i think maybe it is 90 degree in between, then you can just solve via Pythagoras
and find the angle that way too
tranquil pine
but i need an exact value
somehow i'm supposed to just use geometry and figure out what the angle QPR is
tranquil pine
i think maybe it is 90 degree in between, then you can just solve via Pythagoras
exactly what i've been seeing
but idk how it's 90
tranquil pine
ok lemme draw it out
thank you!
tepid anvil
i am 99% sure they have a typo here, they didnt give you the bearing from qp and pr because of the big space in between them
also the grammar
tranquil pine
true there is a typo
the bearing of q from p is 030 while the bearing of p from r is 300
tepid anvil
yea there you fo
tranquil pine
so can you get the diagram now?
need to compare yours to what i have
tepid anvil
sum like that
if i read it right
oh that qrp might be wrong
i was just guessing
tranquil pine
shouldn't your bearing of q from r be drawn from the north?
tepid anvil
you have directions?
tepid anvil
where
tranquil pine
you have directions?
idk, js how we were taught
tranquil pine
trying to take a picture of my diagram, but my camera's clapped
tranquil pine
it would just mean my arrow goes in the other direction
i see
tranquil pine
alr
tranquil pine
ping me if when you get it
so this is basically the diagram i have
what would you correct?
tepid anvil
nah you got it
tranquil pine
so what's the angle between the line QR and north?
tepid anvil
would be something like that
after you solve interior angles you can do some algebra to solve for the outside angle
a circle is 360 deg
so 360 - (300 + theta 1) is ur angle
I think
and for the angle P
I applied some angle rules for that 30
we learn that thats the right angle
tranquil pine
once i understand how P is 90 is can move on 😭
just explain your reasoning to how P is 90
tranquil pine
alright
tepid anvil
Bottom angle is 300
We know three quadrants makes 270
300-270 is 30
By interior angle rule the opposite corner must also be 30
For top angle
You know it has 30 degrees
then the angle to east must be 60
The 60 inside plus the 30 inside make 90
Does this make sense?
tepid anvil
👍
tranquil pine
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sonic crystal
How do I start part a?
final saddleBOT
How do I start part a?
chrome hemlock
hi water beam
spare folio
How do I start part a?
take the derivative
4x³-12ax² = 0
x²(4x-12a) = 0
x_1 = 0
x_2 = 12a/4 = 3a
thus, if a ≠ 0, there is two inflection
white tiger
4x³-12ax² = 0
is it not -12ax²
spare folio
yes sorry
sonic crystal
Is this stationary point?
spare folio
yes
these are the points where the derivative of the function is 0
potential minimum points
we can examine which one is the minimum
let's take x = 0
and the function equals b this way
which is the constant of the function
let's take x = 3a
then the function is equal to: (3a)⁴-4a(3a)³+b = 81a⁴-108a⁴+b = -27a⁴+b
so we know that 27a⁴ is always a positive number
then -27a⁴ is always a negative number
so
If we add a negative number from b it will always be less than b
So the minimum point is point 3a
let's look at this
minimum point is point 3a
not zero
ok we have a "derivative = 0" for 0 point, but it's not a minimum
i wrote a graph as an example
sonic crystal
Is the derivative = 0 the point of inflection you’re talking about
spare folio
yes
and a = 6/4 in this example which is 1.5
and when i looked at the second inflection point
it is on x = 4.5 which is 3a
1.5×3
and f(0)-f(4.5) = 27a⁴ ≈ 136.688
spare folio
then -27a⁴ is always a negative number
.
sonic crystal
Hm. So our derivative is 0 and x = 3a with a cannot be 0.
How do we test it’s a minimum if we don’t know it’s positive or negative
spare folio
Hm. So our derivative is 0 and x = 3a with a cannot be 0.
How do we test it’s a...
i explained above
f(0) = b always
do you see this
and f(3a) = b-27a⁴ when you plug in 3a instead of x
27a⁴ is always a positive number so when you positive number from ANY number, is always less than the first number
b-27a⁴<b
then
f(3a)<f(0)
Therefore point 3a is the minimum point
For there to be a minimum point, the derivative must be 0
derivative became zero in two places
x = 0 and x = 3a
but we learned that the value of 3a is smaller
so 3a is the minimum, there is no other option
sonic crystal
f(0) = b always
So it doesn’t matter if b is like 1000 or something?
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glacial torrent
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karmic kindle
So
What arccos(-1)
Equal to
Should be angle measurmet
Helpful graphic
To know the quadrant restrictions of inverse of trig functions
@glacial torrent
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tired knot
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tired knot
Is there a specific reason on why they put the y's in brackets
What is the purpose
Or is it just good in general to have Y's in a bracket
or doesnt rlly matter
Not that its difficult but i just wanna know if it helps you later on in life
yk
ionic rose
tired knot
3y^2 +3
Into 3(1 +y^2)
I have just started implicit differentiation
and i was wondering if it serves a purpose later on
foggy minnow
just to factor out 3
foggy minnow
it's good to simplify
foggy minnow
product rule maybe
tired knot
what i understand its
(xy) -> 4xy^2 ->
4x dy/dx + y^2
y^2 = ny^n-1 dy/dx
=
4x dy/dx + 2y dy/dx
= 4 + 2y dy/dx
???
Idk can someone check this for me
foggy minnow
this is what i got
tired knot
how?
tired knot
so
foggy minnow
i don't know whats up with the rules you were given
foggy minnow
dv = y^2 dy/dx
tired knot
oh?
foggy minnow
right?
tired knot
Oh ye
srry my bad
u = 4x v = y^2
du =4 dv = 2y dy/dx
4x . 2y (dy/dx) y^2 x 4
so 8xy (dy/dx) + y^2 x 4
8y + 4y^2
HUHHHHHHHHh
LOL
tired knot
stuff is crazy
me too
im going to watch some youtube
ty anyways! @foggy minnow
.close
final saddleBOT
tired knot
good day to you
final saddleBOT
Available help channel!
crude axle
I've used the quadratic formula and found
final saddleBOT
I've used the quadratic formula and found
spare folio
now how to proceed to express it in the cartesian form ?
what is cartesian form actually?
this is an imaginary number
Re(z) = 0
right?
spare folio
okay i understand
a little complicated
i dont know actually but let me examine it more
there must be a way to write √(2+i) in terms of i
like √i
√i = (i+1)/√2
actually
but i dont know about √(i+2)
still worth to thinking
But i'll have to go soon
crude axle
yes, the problem looked straight forward at the beginning, then this 😄
spare folio
xD
maybe it has
it has a real part
probably yes
because even sqrt(i) has real part
we shouldn't miss this
tiny nova
That looks a lot like a quadratic formula
Ohhhhhh
That's because you used it
That makes sense
Hehehehehe
crude axle
yeah, that was the easy bit
spare folio
wait,
...
i+2 = i(1+2/i)?
no
yes but
it doesn't seem like its getting anywhere
what can we do
crude axle
I'm wondering if I should have solved the equation straight away substituting z = x +iy
crude axle
can separate Re and Im and then isolate x in one of the two and sub in the other
but still get a pol grade 4
tiny nova
I suggest you try to remove the root
So x = i+- I*√(2+I) right?
So what if you make it a single fraction, then square both sides
spare folio
hm that's good i think
tiny nova
Then simplify
tiny nova
Well actually there is a cool thing you can do
tiny nova
√x = x/√x
Oh
Okay
The above thing just means squareroot of anything
Meaning you can rewrite the √2+i
Then give the i you are adding/subtracting common denominator
spare folio
Meaning you can rewrite the √2+i
how
tiny nova
Well we turned √2+I into a fraction right?
spare folio
not yet
tiny nova
What is that
tiny nova
I mean pascals triangle could be useful
crude axle
any grade polynomial divisions
tiny nova
Yeah
spare folio
i heard i think but
tiny nova
The coefficients
tiny nova
What if we try making it one fraction then squaring it to remove the roots
Also if it is one fraction we don't have to do foil
spare folio
how can we do it
tiny nova
So √2+I =(2+I)/√2+I right?
spare folio
hm
tiny nova
2+I= √2+I *√2+i
spare folio
yes
tiny nova
(2+I)/√2+I = √2+i
spare folio
yes
tiny nova
So we can replace the √2+I with the fraction, then give a common denominator to the I we are adding then combine the fraction
Then square both sides
spare folio
and then?
tiny nova
I am just thinking of a way to remove the roots
tiny nova
I hope so
spare folio
XD
tiny nova
Hehehehehe
crude axle
I guess I rest my head 5 min, feeling like I'm in the rabbit hole rn
tiny nova
Hehehe
crude axle
yeah, it's getting too hairy
tiny nova
Too many coefficients
Hmmm
Well right now we are trying to combine the two into one fraction
Then square it
Because squaring applies to the numerator and denominator
tiny nova
What did you get
spare folio
again we have
tiny nova
Send a picture
tiny nova
Did you square it
spare folio
i didn't write it exactly but from my mind, you know
spare folio
there is -4i term i can see
at the bottom
(a+b+c)² = a²+b²+c²+2ab+2bc+2ac
and 2bc is -4i
here
which doesnt gives anything
i guess
tiny nova
Where is the -4i
spare folio
i mean when you squared it
tiny nova
Ohh okay
So now I am trying to figure out if we can cancel out the squareroot
Wait
If you take out the i you get i +- 2+I/√2+i
tiny nova
Hmmmmmm
tiny nova
What do you mean
spare folio
then z = i(x/i + y)
tiny nova
I guess
tiny nova
So we plug in the values so far?
tiny nova
Okay
tiny nova
Well we can't really say it's 1
spare folio
and x/i = √(2+i)
spare folio
yes
spare folio
yes
tiny nova
That makes it way easier
spare folio
but there is plus sign before than that
tiny nova
It's fine
spare folio
yes
tiny nova
Okay
spare folio
looks like it should be like that
tiny nova
Well isn't √2+I = y
spare folio
and x is a real number
tiny nova
Hmmmm
spare folio
yes
tiny nova
But the thing is
In the equation we are not multiplying the entire solution by i
Only multiplying y
So that complicates things
I guess you made up for it by placing a x/i
But it's still messes up the format I think
So it's harder to work with
Wait
Uh oh
spare folio
Only multiplying y
there was no y at the first
tiny nova
I think I know what you meant
Wasn't it x+iy
Then you made it I(x/I+y)
The weird thing is though
Is that x has to be a real number
spare folio
yes
tiny nova
But according to that it is i
spare folio
it is i times √(i+2)
tiny nova
But we still add/subtract i
spare folio
yes but
sometimes
i makes real ones
xD
okay let me just
calculate it
for now
tiny nova
Okay
tiny nova
So like complex
spare folio
5 is complex too
complex is general definiton
whatever
it's not just real
as i see
tiny nova
Yeah
spare folio
so it's wrong
tiny nova
What is wrong
spare folio
but
if z = x+iy = i+i√(2+i)
then i(x/i+y) = i(1+√(2+i))
and
x/i+y = 1+√(2+i)
okay yes but
y ≠ 1
tiny nova
Well that still makes x= i
tiny nova
Well what if you kept it
spare folio
here
√(2+i) has real part too
so Re(√(2+i))+1 = y
but the question is
what's the √(2+i) actually
so we have a trouble still
it doesnt works
tiny nova
What if you try to simplify the root
tiny nova
Hehehe
spare folio
there must be a way to write √(2+i) in terms of i and real
tiny nova
Wait
spare folio
like √i
tiny nova
I have an idea
spare folio
√i = (i+1)/√2
actually
what an amazing thing
but it is like that
and looks beautiful
because
we can rewrite it as:
1/√2+i/√2
which is cos45+isin45
or π/4
which is e^(iπ/4)
and we can easily seperate this
tiny nova
I guess
spare folio
as you can see
tiny nova
Well take natural log?
spare folio
but what about √(i+2)
spare folio
Well take natural log?
then?
tiny nova
I have an idea
spare folio
itheta you have
tiny nova
Move the I*pi/4 out of the log
spare folio
okay
tiny nova
Well actually taking the natural log is useless
spare folio
just iπ/4
spare folio
lne = 1
tiny nova
Yeah
But that useless
But I had an idea
If we just assume √2+I = w for example
Then square both sides
spare folio
okay
spare folio
yes