floor function is not continuous for R
#help-13
astral steppe
arctic lava
candid forge
i need help in geometry
cedar kilnBOT
i need help in geometry
scarlet pasture
candid forge
any formulas you're allowed to use?
any is just geometry 1 so idk what to use
thorn lotus
are you familiar with finding equations of a line? Like point-slope form stuff? Does that sound familiar?
candid forge
yes
thorn lotus
ok. then if you want to use the point-slope form of a line, you need to find the slope first
the problem says you're looking for a line that is parallel to the one in the picture
parallel means they have the same slope. so what's the slope of that line
candid forge
that makes sense thank you
crude lagoon
is (xy)^3 simplifyable
cedar kilnBOT
is (xy)^3 simplifyable
wraith daggerBOT
Astral
faint dirge
what are you aiming for?
rare vault
Astral's form is arguably better, but they are about the same - your expression is essentially already simplified
faint dirge
it truly depends on purpose
faint dirge
is (xy)^3 simplifyable
this is more convenient when dealing with substitutions, for instance
faint dirge
**Astral**
but this is easier to get partial derivatives from (which is more common)
atomic obsidian
Prove that for any integer a, 9∤(a^2−3)
cedar kilnBOT
Prove that for any integer a, 9∤(a^2−3)
atomic obsidian
discrete math
so far i have suppose not, that is suppose there exists integer a, 9 | (a^2-3)
then a^2 - 3 = 9r for some integer r
a^2 = 9r + 3
then im stuck
a^2 = 3(3r + 1)
3 | a^2
3 | a
a = 3s
(3s)^2 = 9r + 3
9s^2 = 9r+3
s^2 = (9r+3)/9
anybody??
atomic obsidian
@gleaming zinc
dire geode
Try #elementary-number-theory as well
dire geode
(3s)^2 = 9r + 3
Left side is 0 mod 9 but not the right side
Not sure if that's enough for your contradiction
atomic obsidian
Left side is 0 mod 9 but not the right side
how so?
i dont get what you mean 0 mod 9
dire geode
$9n \equiv 0 \mod 9$ for all integers n
wraith daggerBOT
riemann
dire geode
i dont get what you mean 0 mod 9
Have you learned modular arithmetic
atomic obsidian
kind of i dont really get it
cedar kilnBOT
@atomic obsidian Has your question been resolved?
solid juniper
9s^2 = 9r+3
with a little less modular arithmetic language:
9s^2 is a multiple of 9, but 9r + 3 is not (it leaves 3 remainder when divided by 9)
solid juniper
s^2 = (9r+3)/9
or here, s^2 is an integer, but (9r+3)/9 = r + 3/9 is not
solid juniper
with a little less modular arithmetic language:
9s^2 is a multiple of 9, but 9r ...
i would think more along these lines tho
atomic obsidian
is that the contradiction
that one side is divisible by 9 and the other is not
but theyre supposedly equal
solid juniper
ya, a number that's a multiple of 9 can't be equal to a number that's not a multiple of 9
atomic obsidian
so i can stop 9s^2 = 9r+3 here and just say that
solid juniper
i would think so
atomic obsidian
okay, yea that makes sense to me, hopefully its okay
and the formal way to say it is what riemann said?
solid juniper
with a little less modular arithmetic language:
9s^2 is a multiple of 9, but 9r ...
i would say what riemann said isn't any more formal than this
just uses a more powerful tool
"left side is 0 mod 9 but not the right side" is exactly the same as "left side is divisible by 9 but not the right side"
atomic obsidian
oh okay, tysm!
solid juniper
np ^_^
atomic obsidian
wait is there some formal way to say the rhs isnt divisible by 9
solid juniper
being able to write it as 9r + 3 says it has 3 remainder
idk how formal you wanna get, something with the division algorithm could be said maybe but that's pretty overkill
atomic obsidian
so lhs is 0 mod 9 and rhs is 0 mod 3?
solid juniper
rhs is 0 mod 3 but that's not important
more relevant is rhs is 3 mod 9
a number that's 0 mod 9 can't be equal to one that's 3 mod 9 (i.e. a number that leaves 0 remainder when divided by 9 can't be equal to one that leaves 3 remainder when divided by 9)
solid juniper
so the rhs is both 0 mod 3 and 3 mod 9?
the rhs being 0 mod 3 is irrelevant
atomic obsidian
i get that i was just wondering if those 2 things were equal
atomic obsidian
0 mod 3 and 3 mod 9
solid juniper
asking if "0 mod 3" and "3 mod 9" are equal doesn't make much sense
atomic obsidian
i really just don't understand mod at all tbh and was just trying to gain more insight but that's for another time ig
solid juniper
this might address what you are thinking:
if a number is 3 mod 9, it's also 0 mod 3
not every number that's 3 mod 9 is also 0 mod 3
atomic obsidian
yea that addresses it, tysm
i just have to digest things a bit more
but i think im onto it
solid juniper
haha, gr8
cold briar
if theta is in (0, pi), given that sin theta + cos theta = 1/5, find tan theta.
cold briar
I've started by squaring the sin theta + cos theta = 1/5,
which gives me
sin^2 theta + cos^2 theta + 2 sin theta cos theta = 1/25
1 + sin(2x) = 1/25
sin 2x = -24/25
okay so i have two ways which results in different answers and it starts to vary in this step
first step:
I made a triangle with opposite = -24 and hypotenuse = 25
the adjacent side will be:
sqrt ( 25^2 - (-24)^2) = 7
so tan 2x = -24/7 right?
now i use the tan double angle rule.
2tan x / 1 - tan^2 x = -24/7
ima just write tan x as u to simplify it
2 u = -24/7 ( 1 - u^2)
14 u = -24(1 - u^2)
0 = 24u^2 - 14u - 24
0 = 12 u^2 - 7u - 12
0 = (4u + 3)(3u - 4)
so u = -3/4 or 4/3
but if you use sin 2x = 2tanx / 1 + tan^2 x you get a different result
-3/4 or -4/3
how....
for the solution
oh... is it because sqrt of 49 and be +- 7...
welp
.close
visual socket
I’m currently working on number 19 but I’m not sure how to do it. Can someone walk me through it?
undone sage
Substitute y²= 1-(x-6)² in the first equation and solve the quadratic formed
visual socket
Substitute y²= 1-(x-6)² in the first equation and solve the quadratic formed
Are you looking at 21?
visual socket
I’m doing 19 and if I substitute -x^2 for y it should be x + (-x)^2 = 1 right?
undone sage
No
visual socket
Oh
undone sage
See the equation properly
visual socket
So I have -y + y^2 = 1
undone sage
Yeah
visual socket
And all you did was subtract the 1?
undone sage
Yeah
long mauve
Why not 4.0λ=0.34 but It's 4.25λ =0.34
cedar kilnBOT
Why not 4.0λ=0.34 but It's 4.25λ =0.34
muted bear
Because theres an extra wave bit at the end
cedar kilnBOT
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visual socket
I’m working on number 29. Can someone help explain to me how to do this?
rustic cove
Hi everyone trying to figure out the subscript vs superscript of an equation
rustic cove
we're looking at the numerator of the division, SUM j P[j,k] - 1
is it (SUM j P[j,k]) - 1 or SUM j (P[j,k] - 1)
cedar kilnBOT
@rustic cove Has your question been resolved?
flat harbor
flat harbor
so the roots of it are 1, cis+-2π/5, cis+-4π/5
cedar kilnBOT
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Available help channel!
severe pasture
cedar kilnBOT
severe pasture
f E --> F
f is a function which all elements from E are defined
ACE and BCF
f^-1(B) = {x€E/f(x)€B}
cedar kilnBOT
@severe pasture Has your question been resolved?
cedar kilnBOT
@severe pasture Has your question been resolved?
peak salmon
cedar kilnBOT
peak salmon
how would solve this?
working out the expression first, then integrating
or applying integration by parts right away?
or maybe something else?
idk
upper ruin
uneven sand
Students from Mrs. Hein’s class are standing in a circle. They are evenly spaced and consecutively labelled using whole numbers starting from 1. The student in place number 3 is standing directly across the student in place number 17. How many students are there in Ms. Hein’s class?
zenith cedar
okay think of it this way
the diameter divides the circle into 2 parts of equal perimeter
this means the number of students between the student number 3 and 17 would be same, if you go right or left
which means there are 13 students in the (say) right side between 3 and 17
then there are another 13 on the left side
so total = 28
uneven sand
then there are another 13 on the left side
Wait how did you get 13 students on the right side again?
uneven sand
yes
zenith cedar
this means 3 and 17 are at extreme opposites of each other
so there are 17-3-1 = 13 students in between them
excluding both of them
uneven sand
so there are 17-3-1 = 13 students in *between* them
Why did u minus 1 again? sorry if im being dumb rn ;-;
uneven sand
2*
uneven sand
Ahh
zenith cedar
and 2 is? 5-2-1
uneven sand
Ahhh alr
uneven sand
Ahh ok thx
cedar kilnBOT
@uneven sand Has your question been resolved?
errant wasp
what does c) mean
cedar kilnBOT
what does c) mean
ebon pulsar
which c?
errant wasp
c)
ebon pulsar
ahh k
errant wasp
i think this is SVD
dawn junco
yeah it is SVD
errant wasp
idk my teacher doesnt actually show us how to do it, he just does one example
its actually really annoying but whatever
dawn junco
it's not like you need a million examples tbh
errant wasp
no but like he doesnt even really show us the steps
its kinda weird
its okay
i need to stop complaining
dawn junco
A = U S V^T, where S has the singular values on the diagonal, U has unit eigenvectors of A^TA, V has unit vectors of AA^T : that's SVD
they're giving you the steps in the question
errant wasp
hmm
dawn junco
so you gotta find the eigenvectors first
and normalize them
is there something specifically hard for you in there ?
dawn junco
eigenvectors are vectors
errant wasp
soryr
dawn junco
not numbers
errant wasp
values
dawn junco
so for [[0 4] [0 0]] ?
errant wasp
i mean one unit eigen vector would have to be [0 1]
hold on no
brb let me fix this
dawn junco
in part b, i got 16, 0 and 1 as eigenvectors
0 and 16 for the eigenvalues of A^TA yeah, i don't see where you get 1
errant wasp
dawn junco
ok sounds good
errant wasp
if you combine the two unit vectors for the left side, you get the identitity matrix
errant wasp
okay you get the same for both
so i guess V is just the identity matrix
im trying to find u
i think im getting the same eigen vals?
cedar kilnBOT
@errant wasp Has your question been resolved?
errant wasp
the eigenvalues are the same for both AA^T and A^TA yes
gotcha
im going to keep this open for a bit
errant wasp
the eigenvalues are the same for both AA^T and A^TA yes
so basically U and V are both the identity matricies
dawn junco
it would be a bit problematic if that was the case
you need USV^T to be equal to A
if U and V are both identity, that's never gonna happen
errant wasp
if U and V are both identity, that's never gonna happen
can i show my work
errant wasp
so far...
dawn junco
yeah you just have to be coherent when you write U S V^T then
if you choose S as [[4 0] [0 0]], i.e. 4 is on the 1st diagonal spot, you have to put the singular vectors for 4 on the 1st col
it's like when you do diagonalization, you have to put the eigenvectors in the correct order
(coherent with your order of eigenvals)
@errant wasp
errant wasp
shoot am i doing the wrong order
dawn junco
trying to make sure that you're not
errant wasp
trying to make sure that you're not
says in S, they are in descending order
on the diagonal
U is just the anti identity materix
i checked to see that they were in order
dawn junco
yeah that's what you should have @errant wasp
errant wasp
yeah that's what you should have <@448651226007994380>
but then V transpose = U
dawn junco
wait why should V^T = U
errant wasp
i fixed it
dawn junco
V is the identity
cedar kilnBOT
@errant wasp Has your question been resolved?
ivory finch
Does anyone have a way to type latex fast without actually typing latex? I see that typing stuff in Desmos then copying it is one way but I'm wondering if there's actually a dedicated platform to do this.
runic garnet
Maybe ask gpt to do it? Who knows, that may work
neat dune
U mean like wysiwyg editor?
runic garnet
I used to take a math class on a software called canvas, which would convert regular text to latex
ivory finch
Haven't heard of wysiwyg editor
ivory finch
Basically I want to type math like I do in Desmos but have that converted to latex, in a more efficient way than opening Desmos every time
Unless Desmos is the only way to do it
cedar kilnBOT
@ivory finch Has your question been resolved?
crimson meteor
cedar kilnBOT
kindred turtle
when you differentiate a function y with respect to x, you get a function y’ that tells you the gradient of y at any value of x
here differentiating $$x^2$$ with respect to x gives you 2x, which is a generic function
wraith daggerBOT
abhi
kindred turtle
now putting x=2 into that function y’ tells me the gradient of y at the point x=2
crimson meteor
define gradient?
kindred turtle
gradient is rise over run
crimson meteor
slope?
crimson meteor
okay go on
kindred turtle
so the original functiom is $$y=x^2$$
wraith daggerBOT
abhi
kindred turtle
once you differentiate it to get y’ = 2x, you have a gradient (or slope) function that will tell you the slope at any point
crimson meteor
ok i'm following
kindred turtle
if i was to put x=1 for example, it would tell me the slope of y=x^2 at x=1 is 2(1) which is 2
the reason we use derivatives is because for functions like this, the slope is constantly changing
crimson meteor
I get it that we have to find the slope for that point. but why was it the slope of tangent, not the slope of the normal ?
rare vault
is this a question on why derivatives are the way that they are, or a question on what tangent and normal lines are
crimson meteor
I guess the latter?
what made this Mtan? ig?
oh is it because the slope of tangent line is the derivative of the function?
cedar kilnBOT
@crimson meteor Has your question been resolved?
sand patio
Having trouble figuring out where to go from here.
I'm figuring out if it's divergent or convergent
cedar kilnBOT
Having trouble figuring out where to go from here.
I'm figuring out if it's dive...
@sand patio Has your question been resolved?
sharp kettle
Can someone help me write this out for N=2? I don't understand the sums
crimson delta
delta_i doesnt appear in the terms?
crimson delta
uhh
sharp kettle
Sorry for writing bad 😄
sharp kettle
Are u familiar with the partition function=
from the ising model
I am trying to understand the sum for free boundary conditions
crimson delta
no I am not
what are you summing over
are you summing over a set containing some S_i?
crimson delta
or are you summing over all combinations of some S_i ?
sharp kettle
I am plugging the H into the Z and use the property of the exp-function
sigma_i = s_i
I am also trying to write it out for small N
for example N=2
crimson delta
well the sum for H just says $-J ( S_1 S_2 + S_2S_3+\ldots+S_{n-1}S_n)$
wraith daggerBOT
Denascite
crimson delta
so for n=2 just -J S_1 S_2
sharp kettle
But I am not sure if I can write it like that
How am I supposed to sum it all up
Is it a product of sums or something else?
Is this right?
except for the second plus sign
sharp kettle
Yea
crimson delta
lets write s=(s1, s2, ..., sn)
then we are summing over all possible values of s
yes?
is that what we are doing?
sharp kettle
I think so yes
wraith daggerBOT
Denascite
crimson delta
and $H(s) = -J \sum_{i=1}^{n-1} s_i s_{i+1}$
wraith daggerBOT
Denascite
crimson delta
is this what we are doing?
sharp kettle
I am not sure about the ${-1, 1}^n$
wraith daggerBOT
Simplex
sharp kettle
I don't know if ${s_i}={-1, 1}^n$ is true
wraith daggerBOT
Simplex
sharp kettle
Because I don't exactly know what $\sum_{{s_i}}$ means
wraith daggerBOT
Simplex
crimson delta
well good
I dont know either
I dont know what {si} is supposed to represent
if the si can only be +-1 then this is just {1, -1} but thats very likely not what it means
sharp kettle
wait, I look at the lecture again
something like this
Not without the J, but that's not important
crimson delta
**Denascite**
that is what this would end up being equal to
crimson delta
so for n=2 it would be e^(1*1) + e^(1*-1) + e^(-1*1) + e^(-1*-1)
ok I forgot beta
but same thing
cedar kilnBOT
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Available help channel!
silent cedar
Hi!
silent cedar
I need help in calculating the error of the Taylor Theroem
y=e^x
a=0
and
e^1/2
cedar kilnBOT
@silent cedar Has your question been resolved?
silent cedar
<@&286206848099549185>
cedar kilnBOT
@silent cedar Has your question been resolved?
cedar kilnBOT
@silent cedar Has your question been resolved?
silent cedar
<@&286206848099549185>
cedar kilnBOT
@silent cedar Has your question been resolved?
thorny rampart
Hey, I need help finding [Y|B]
cedar kilnBOT
Hey, I need help finding [Y|B]
gusty forum
bayes rule
thorny rampart
Okay, how do I apply that
gusty forum
by plugging in the relevant quantities
some you may have to computer from others
thorny rampart
So how do I get B?
In order to do the equation, because I know the intersection is = 0.08
Since this is the equation we use
gusty forum
you have P(B|stuff)with some pesky conditionals
now figure out how to get P(B) without the conditionals
thorny rampart
Can you explain that a little further
gusty forum
eh, at this point i think just reading up law of total probability would be more useful
thorny rampart
Okay, so how do I use the Law?
gusty forum
look at what conditionals B has and figure out the terms you need
thorny rampart
So, Y, A & C are conditionals
cedar kilnBOT
@thorny rampart Has your question been resolved?
cedar kilnBOT
@thorny rampart Has your question been resolved?
livid isle
is there any way to do this without l'hopital-ing like 6 times
livid isle
i forgot how to calculus and im not writing out 6 l'hopitals on the board (x 3)
dire geode
Shouldn't be 6
livid isle
maybe not 6, but more than 3
dire geode
Taylor series expansion of sin(x)/x should also give you the answer
cedar kilnBOT
@livid isle Has your question been resolved?
orchid bison
why is the graph acting this way?
cedar kilnBOT
why is the graph acting this way?
stoic gale
horizontal asymptote is observed as x goes to plus/minus infinity, at x=0, we get that f(x)=0/1=0
orchid bison
horizontal asymptote is observed as x goes to plus/minus infinity, at x=0, we ge...
What?
so there isnt an asymptote?
stoic gale
horizontal asymptotes occur as x goes to infinity or x goes to minus infinity - the asymptote is there, but it does not mean that the asymptote cannot be crossed for some finite value x
the reason it can be crossed is because we are not dividing by zero
on the other hand, vertical asymptotes can never be crossed, as we would be dividing by 0
orchid bison
ohh okay thank you!
stoic gale
nw
orchid bison
.close
cedar kilnBOT
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granite sapphire
can someone help explain this
cedar kilnBOT
can someone help explain this
granite sapphire
remote thunder
cedar kilnBOT
@granite sapphire Has your question been resolved?
cedar kilnBOT
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dapper jewel
cedar kilnBOT
dapper jewel
how do i do this
A is not square, so i can't use gaussian elimination
i thought maybe i could check if the rank without and without b as a column are different to see if there's a nonanswer, but they are the same
crystal raptor
You can do gaussian elimination with any size matrix
dapper jewel
how would i proceed from here
wait the last operation should be this
but still, where do i go
cedar kilnBOT
@dapper jewel Has your question been resolved?
cedar kilnBOT
@dapper jewel Has your question been resolved?
cedar kilnBOT
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young anchor
cedar kilnBOT
young anchor
What am i doing wrong
crimson sedge
What is required?
young anchor
i need to find second derivative
languid mulch
hes trying to find the second derivative ti seems
young anchor
yeah should have stated that from the start my bad
languid mulch
you got the first derivative wrong from what i understand
young anchor
huh
languid mulch
e^x derivative remains as is
young anchor
crimson sedge
It's the first derivative
wheat moss
that is correct for the first derivative
crimson sedge
Yes
young anchor
yeah but i think im messing up the second one
wraith daggerBOT
inf1425
wheat moss
you need to use this
languid mulch
the derivative
of the first of these
with the other one
minus the first one multiplied with the derivative of the other one
not plainly multiply
young anchor
[(xe^x)' * (x+1)^-2] + [xe^x * ((x+1)^-2)' ]
languid mulch
minus not plus
young anchor
but here i rewrote as multiplication
languid mulch
cedar kilnBOT
@young anchor Has your question been resolved?
sharp lotus
Have i made any errors up until this point
looks correct
it would probably be easier to simplify in a different way though, both of the large terms of the numerator have a factor of (x+1) and also a factor of e^x if you factor e^x+x^ex (the parenthesis in the first bracket) as e^x(1+x)
$$\left[(e^x+xe^x)(x+1)^2\right]-\left[2xe^x(x+1)\right]$$
$$\left[e^x(x+1)^3\right]-\left[2xe^x(x+1)\right]$$
$$e^x(x+1)\left[(x+1)^2 + 2x\right]$$
wraith daggerBOT
Soosh
sharp lotus
anyway that's just the numerator @young anchor and you can now cancel (x+1) factor with one of the (x+1) factors in the denominator and maybe multiply out the bracket if you like etc.
young anchor
i dont get it
young anchor
the first one
sharp lotus
just factoring the parenthesis (e^x + xe^x) = e^x(1+x) right?
1+ x = x + 1 so thats just another (x+1) factor and we increase the exponent of (x+1)^2 to (x+1)^3
sharp lotus
in last step, both big brackerts have common factors (the red and green ones)
pulling out e^x(x+1) to front, then first factor is juet left with the (x+1)^2 and 2nd is left with 2x
sharp lotus
so you can cancel the (x+1) in the front with (x+1)^4 in the denominator and denom just becomes ^3
and i think thats probably good enough for a final simplified answer, maybe write [(x+1)^2 +2x] as just (x^2 +4x + 1)
!done
cedar kilnBOT
If you are done with this channel, please mark your problem as solved by typing .close
sharp lotus
well im sure you could manipulate things with algebra into that form if you wanted
:p
young anchor
but what about this
sharp lotus
i dont know, i dont really have time to go through everything you've discussed in this channel, i only looked at the last question you had about the simplification
dusty path
Question asks for the general solution of the following. Was not taught this too well, so I’m completely lost. Any help is appreciated.
cedar kilnBOT
Question asks for the general solution of the following. Was not taught this too...
@dusty path Has your question been resolved?
dusty path
<@&286206848099549185>
vernal shell
Question asks for the general solution of the following. Was not taught this too...
Hello. Try isolating first cos(4(x-30°))
dusty path
2 = 6sec(4(x-30))
2/6 = 6/6sec(4(x-30))
1/3 = sec(4(x-30))
3 = cos(4(x-30))
This works right?
vernal shell
2 = 6sec(4(x-30))
2/6 = 6/6sec(4(x-30))
1/3 = sec(4(x-30))
3 = cos(4(x-30))
...
great, now
is that possible? 🤔
3 = cos(....)
dusty path
It would not be possible because cosine oscillates between 1 and -1
dusty path
That was a lot simpler than I thought.
The question was written exactly as such
Thanks for the help. I am still a little confused when a general solution problem goes further than that. I’m having troubles with restrictions (ex: 0 <= theta < 2pi)
I get the values of X but o have troubles applying the restriction to my answer
cedar kilnBOT
@dusty path Has your question been resolved?
vernal shell
Thanks for the help. I am still a little confused when a general solution proble...
ok I see. Well once you have the general solution, you can try values of n from n=0 until you get a number greater than 2pi
thin hull
I had a test now and I didn’t made one question:
f is a function
f:S——>T
A ⊂ S
#A = 2
#f^-1(f(A))=4
Draw the function
thin hull
Is it possible to do?
cedar kilnBOT
@thin hull Has your question been resolved?
dull oxide
So A has two elements?
cedar kilnBOT
@thin hull Has your question been resolved?
cedar kilnBOT
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quasi prism
Not sure where I messed up.
cedar kilnBOT
Not sure where I messed up.
rare vault
Not sure where I messed up.
should be -10(5) in your first line
the idea is right, just flipped the sign
soft cloud
I have a question about cosinus and sinus functions, you can always do 360-the angle but can you also always do 180+ the angle? It seems like that isn’t an option all the time and I’m wondering why
Help would be greatly appreciated as I write my exam tomorrow 
cedar kilnBOT
I have a question about cosinus and sinus functions, you can always do 360-the a...
@soft cloud Has your question been resolved?
crimson sedge
How can I show the convergence of $\int_{0}^{\infty} \frac{\sin(x)}{\sqrt{x}}, dx$
wraith daggerBOT
Şêro
cedar kilnBOT
@crimson sedge Has your question been resolved?
royal finch
Helps if you mention what technique you want to use to even do it. The idea that immediately springs to mind is to just use Complex Analysis
If you use the substitution x = u^2 the interesting part of the integrand becomes sin(u^2) which is the imaginary part of a Gaussian integral
crimson sedge
.close
pine basalt
This might be a bit higher-level math based on a discovery I made.
So the Fibonacci Sequence is defined as F(x) = F(x-1) + F(x-2) with F(1)=0 and F(2)=1.
Let the Squibonacci Sequence be defined as S(x)=S(x-1) + S(x-3) + S(x-4), with F(1)=F(2)=F(3)=0 and F(4)=1.
How do I prove that (F(x))^2 = (S(2x))?
opal basin
Induction I guess, haven't worked it out
idle tusk
S(x)=S(x-1) + S(x-3) + S(x-4) or S(x)=S(x-1) + S(x-2) + S(x-3)
opal basin
I assume the first one, otherwise S(4) wouldn't have to be specified
idle tusk
okay okay
cedar kilnBOT
@pine basalt Has your question been resolved?
pine basalt
S(x)=S(x-1) + S(x-3) + S(x-4) or S(x)=S(x-1) + S(x-2) + S(x-3)
What I said
idle tusk
okay
pine basalt
(x-1), (x-3), (x-4)
pine basalt
Induction I guess, haven't worked it out
All of these types of problems are induction I s2g
cedar kilnBOT
@pine basalt Has your question been resolved?
cedar kilnBOT
@pine basalt Has your question been resolved?
pine basalt
This might be a bit higher-level math based on a discovery I made.
So the Fibon...
*Initial Problem if somebody can help ;-;
cedar kilnBOT
@pine basalt Has your question been resolved?
cedar kilnBOT
@pine basalt Has your question been resolved?
rapid wigeon
!help
cedar kilnBOT
To ask for mathematics help on this server, please open your own help channel or help thread. See #❓how-to-get-help for instructions.
opal basin
This might be a bit higher-level math based on a discovery I made.
So the Fibon...
Are you sure it's correct actually? Is seem to be getting F(3)^2 = 1^2 = 1 and S(2 * 3) = S(6) = 2
The way you've written down it actually holds that S(x + 2) = F(x), proving that is probably one of the first steps
pine basalt
Are you sure it's correct actually? Is seem to be getting F(3)^2 = 1^2 = 1 and S...
The way I have it, I’m having F1 be zero and F2 be 1. S1, S2, S3 are 0, and S4 is 1
opal basin
Yeah, so F(3) = 1, right?
pine basalt
pine basalt
Yeah, so F(3) = 1, right?
yes. and S(6) (right column) is 1^2=1
I have found some more interesting properties of these two sequences. (The numbers corresponding to "Fibonacci" represents the number of initial values, where the regular Fib seq has two)
opal basin
yeah, you are right, i've written down S incorrectly
pine basalt
yeah, you are right, i've written down S incorrectly
i do understand where one would make that mistake because conventionally, the Fib sequence has 0 at index 0 and 1 at index 1, and here I'm starting from index 1
opal basin
I miscalculated the 6th term
pine basalt
i'm going to get the sheet where I wrote down some other interesting properties, and we can brainstorm how this might help to prove how it works
so S(x)+S(x-1) = {S(x+1)-1 for odd x, S(x+1) for even x
S(2x)-S(2x-1) = sqrt(S(2x)) times F(x-1) = sqrt(S(2x)) times sqrt(S(2x-4))
F(2k)-S(2k)=S(2k+2)=(F(k+1))^2
limit of F(n)/S(n) as n goes to infinity is (phi + 2) = (phi^2) + 1
opal basin
This might be circular, but I think you can turn it into a valid proof
Suppose it holds for x and x+1, then inductive step will become:
2f(x)f(x+1) = s(2x+1) + s(2x+3) - s(2x + 2)
This is the inductive step for proving the previous claim, assuming the previous claim
2f(x+1)(f(x) + f(x+1))= s(2x+3) + s(2x+5) - s(2x + 4)
2f(x+1)f(x+1) + s(2x+1) + s(2x+3) - s(2x + 2) = s(2x+3) + s(2x+5) - s(2x + 4)
2f(x+1)f(x+1) + s(2x+1) - s(2x + 2) = s(2x+5) - s(2x + 4)
2f(x+1)f(x+1) + s(2x+1) - s(2x + 2) = s(2x+1) + s(2x + 2) + s(2x+4) - s(2x + 4)
f(x+1)^2 = s(2x + 2)
So the first claim will hold if this is true, but this was an assumption, so I think we are done
Im on mobile rn, so typing is hard
cedar kilnBOT
@pine basalt Has your question been resolved?
sand cradle
cedar kilnBOT
sand cradle
I used Bezout's identity (line 3 -> 4) and that a = b (mod n) implies n | a - b.
floral arrow
Perhaps I'm not reading this right, but you haven't proven gcd(a,m) | b
sand cradle
Perhaps I'm not reading this right, but you haven't proven gcd(a,m) | b
I've proven b = gcd(ax, m) = gcd(a, m) for x = 1.
b | b holds of course
So gcd(a, m) | b
floral arrow
Right ok, I would've just said "take x=1" at the beginning and gone from a = b mod m
sand cradle
Yeah, we can do that
floral arrow
Also isn't it an equivalence instead of just an implication?
sand cradle
Which step?
floral arrow
a = b mod m iff m | a-b
sand cradle
Yeah, should be an equivalence. Well an equivalence is just a two-sided implication, so I guess it's not wrong to write it like this
We just need the implication here after all
floral arrow
I mean if you just use an equivalence then you don't have to prove the other way
The question just becomes whether you can use that equivalence
sand cradle
I mean if you just use an equivalence then you don't have to prove the other way
Oh
Yeah, true
sand cradle
I mean if you just use an equivalence then you don't have to prove the other way
Alright, then imagine the ==> as a <==>. Would the proof be fine like this, then?
floral arrow
If you can use the equivalence, then yeah I guess so
sand cradle
If you can use the equivalence, then yeah I guess so
Yeah, I've proven the equivalence before, so that step should be fine
Does b = ax - mk_1 mean b = gcd(ax, -m)?
Yeah, right?
Bezout's lemma
sand cradle
I should post this to #elementary-number-theory or #proofs-and-logic
Thanks
.close
flat sparrow
I'm pretty sure this is the right one was a little hard to see though ~ am i right?
dapper cairn
I'm pretty sure this is the right one was a little hard to see though ~ am i rig...
second and last one are wrong cuz u cant do ln(0)
floral arrow
I'm pretty sure this is the right one was a little hard to see though ~ am i rig...
No
Hint: ln(1) = 0
And each interval on the axes represents 2
flat sparrow
right so C?
flat sparrow
And each interval on the axes represents 2
yes but i graphed it and still couldn't pick between a and c
floral arrow
Still no
flat sparrow
B
floral arrow
What's f(1)?
flat sparrow
2
floral arrow
So half an interval on the x axis and one full interval on the y axis, right?
floral arrow
...no
Start at the origin, go half an interval right and half an interval up
You should be on the graph
flat sparrow
Start at the origin, go half an interval right and half an interval up
wouldn't this just be A?
I mean i graphed it out
and since yall said it can't b or d
floral arrow
I never said that
flat sparrow
second and last one are wrong cuz u cant do ln(0)
^
floral arrow
Yeah, that's not me
flat sparrow
ok well then its B
flat sparrow
😭 why is it D
it doesn't look like D moved to the right at all
from where it starts
native shuttle
cedar kilnBOT
native shuttle
how can u solve the bottom one?
g(x) = 10
so i thought adding them would work but it didnt
somone said u can factor
into two different integerals
but how would i know that the integeral of -2 -> 5 of 4 dx is?
hollow minnow
Just seperate g(x) and 4 into 2 integrals
hollow minnow
but how would i know that the integeral of -2 -> 5 of 4 dx is?
You should know how to integrate a constant
native shuttle
You should know how to integrate a constant
omg
never did this yet XD
gonna review that now! ty
hollow minnow
Really
hollow minnow
$$\int a dx = ax + c$$
wraith daggerBOT
$Pure$
hollow minnow
Where a is a constant
wraith daggerBOT
Arm
hollow minnow
Np
cedar kilnBOT
If you are done with this channel, please mark your problem as solved by typing .close
hollow minnow
constant times the bounds
Yep you can think of it like an area of a rectangle
Because it’s a constant function and we’re finding the area under it
.close
abstract ice
cedar kilnBOT
cedar kilnBOT
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
4
wraith daggerBOT
hollow minnow
Apparently it’s -2x z
abstract ice
Apparently it’s -2x z
can u tell me where i have made the mistake
oh shi i found it
.close
quiet cypress
how do i find the area of a circle cut across a chord knowing the height of the chord and the radius
glossy inlet
You could do (area of sector) - (area of triangle formed by the radii and chord)
quiet cypress
o true
now i just need to find the angle of theta which should be easy
i can just use pythagoras i think?
this seems overcomplicated.
is there a more elegant solution?
so from this
c=sqrt(r^2-(r-h)^2)
because pythagoras
oh i dont need to find c i can just use cos right?
gritty viper
either one
quiet cypress
this is something im tryna implement computionally btw
preformance doesnt really matter for my applications but but i prefer more elegant solutions
<@&286206848099549185>
cedar kilnBOT
@quiet cypress Has your question been resolved?
cedar kilnBOT
@quiet cypress Has your question been resolved?
silver lance
now i just need to find the angle of theta which should be easy
it's an isosceles triangle
quiet cypress
if you cut it in half it becomes a right triangle
wanton aurora
I am trying to solve this question but I am unsure on how to start it. I started with trying to find the derivative of y first, but I am unsure how to do the chain rule with this and I looked at the solution, but I don't understand how they got this.
wanton aurora
not sure how they got the 2 outside the sinx
dire geode
Do you know chain rule?
wanton aurora
kinda
isnt it inner function' times outerfunction'
(f(u))' times u'?
but the thing is, its to the power of sin2x
so im not sure whether to do the chain rule on the sin(2x) or on the 9^sin(2x)
cedar kilnBOT
@wanton aurora Has your question been resolved?
worthy cairn
so im not sure whether to do the chain rule on the sin(2x) or on the 9^sin(2x)
On 9^sin(2x).
Using (f(g(x)))' = f'(g(x))g'(x), with f(x)=9^x and g(x) = sin(2x)
cedar kilnBOT
Available help channel!
turbid inlet
Whatever this is lol
cedar kilnBOT
Whatever this is lol
cedar kilnBOT
@turbid inlet Has your question been resolved?
digital cliff
.close
winged jewel
If I have a function with multiple variables, for example x = cos(wt), if I differentiate dx/dt, why would I treat the w in cos(wt) as a constant and get x = -wsin(wt), because in implicit differentiation before whenever I differentiated a variable with respect to a different variable e.g. y with respect to x I would have to write dy/dx
livid hound
depends on what you're told about w and whether it's related to t
winged jewel
Ahh ok, would the fact they are in the same equation not show they are related? Because I thought you could rewrite any equation in terms of either of the variables in it, and then technically that variable would be dependent on the others
I'm sorry if I'm looking too far into this I'm just genuinely confused by it
livid hound
you could apply chain rule if you want, leading to components with
dw/dt,
if w doesn't change with t, those components will be 0
livid hound
depends on the question, how the variable is defined, info you're given
winged jewel
Ok, so in the question I gave I assume it's not obvious whether it does change with respect to t or not
@livid hound ok I've looked it up further and for that equation (it represents simple harmonic motion), omega typically is assumed to be constant to make analysis easier
So in that example I would treat it like a constant
So I guess as a rule of thumb if the variable isn't given as a function of what I'm differentiating with respect to, I should treat it as a constant
Thank you for your help! I really appreciate it
.close
steel jewel
cedar kilnBOT
steel jewel
I wrote the function as $A = 2xy = 8x\cos{x/2}$
wraith daggerBOT
lunchbox
digital cliff
alrighty
steel jewel
and I took the derivative => $A' = 8\cos{x/2} - 4x\sin{x/2}$
wraith daggerBOT
lunchbox
steel jewel
hmm
then i multipled by $2\cos(x/2) + x\sin{x/2}$ which turns the $\cos{x/2}$ into $\cos^2{x/2}$ but I found it difficult to solve the $4x\sin{x/2}$ part
wraith daggerBOT
lunchbox
steel jewel
my current equation is $4(4-(4+x^2)\sin^2{x/2})=0$
wraith daggerBOT
lunchbox
brazen wigeon
im assuming the recangle cannot extend below y = -4?
steel jewel
um
the rectangle should be contained between the x axis and y = 4cos(x/2)
oh i see what you're saying
brazen wigeon
the rectangle should be contained between the x axis and y = 4cos(x/2)
does that mean the rectangle cannot have corners at $-\pi \l x \l \pi$
wraith daggerBOT
@brazen wigeon
steel jewel
it should look smth like this
brazen wigeon
it should look smth like this
OHHHH
ok
alright
the width of the rectangle is 2x
assuming (x, y) is the top right corner of the rectangle, and the rectangle is symmetrical about the y-axis
brazen wigeon
the width of the rectangle is 2x
assuming (x, y) is the top right corner of the ...
@steel jewel do you a) understand and b) agree?
steel jewel
yeah
brazen wigeon
ok
now make a function h(x). h(x) is the height of the rectangle, and x is the location of the top right corner
steel jewel
h(x) is 4cos(x/2) right
steel jewel
**lunchbox**
after that i made the area function ^
brazen wigeon
**lunchbox**
yep, this looks correct
are you trying to find the maximums of the function with the derivatives?
steel jewel
yeah
brazen wigeon
ok one sec
steel jewel
i set derivative equal to 0
brazen wigeon
**lunchbox**
,tex $A' = cos(\frac{x}{2}) - x \times sin(\frac{x}{2}) \times \frac{1}{2}$
There's 3 rules needed:
product rule
chain rule
der of cos
steel jewel
yeah
brazen wigeon
ok
steel jewel
der of cos is -sin so i think it's -x * sin(x/2)
wraith daggerBOT
@brazen wigeon
brazen wigeon
,tex $0 = cos(\frac{x}{2}) - x \times sin(\frac{x}{2}) \times \frac{1}{2}$
steel jewel
👍
wraith daggerBOT
@brazen wigeon
brazen wigeon
<@655263219459293210>
i did what u said, and set A' = 0
brazen wigeon
<@655263219459293210>
now ill add the second part of the right equation to the left side
,tex $\frac{x \times sin( \frac{x}{2})}{2} = cos(\frac{x}{2})$
wraith daggerBOT
@brazen wigeon
steel jewel
yes
brazen wigeon
hmmm
steel jewel
i think i have to use newtons to approximate the root
brazen wigeon
i think i have to use newtons to approximate the root
i have no clue what newtons is 😭
brazen wigeon
sry its +
steel jewel
yep
brazen wigeon
<@655263219459293210>
,tex $\frac{x \times sin( \frac{x}{2})}{2} = sin(\frac{x}{2} + \frac{\pi}{2})$
wraith daggerBOT
@brazen wigeon
steel jewel
ok
steel jewel
:0
brazen wigeon
,tex $sin(\alpha + \beta) = sin(\alpha) \times cos(\beta) + cos(\alpha) \times sin(\beta)$
wraith daggerBOT
@brazen wigeon
steel jewel
yeah
steel jewel
lol
steel jewel
you basically get back cos(x/2)
brazen wigeon
fuck
steel jewel
😔 math is so hard
brazen wigeon
😔 math is so hard
alr bro W idea. substitue cos(x/2) with sin(x/2 + pi/2) /s
brazen wigeon
wait this time i actually think i got it
steel jewel
👀
brazen wigeon
<@655263219459293210>
,tex $\frac{x}{2} = \frac{ cos(\frac{x}{2} )} {sin (\frac{x}{2})}$
wraith daggerBOT
@brazen wigeon
steel jewel
ok
brazen wigeon
,tex $\frac{2}{x} = tan(\frac{x}{2})$
wraith daggerBOT
@brazen wigeon
steel jewel
its $\cot$ i think
wraith daggerBOT
lunchbox
brazen wigeon
its $\cot$ i think
i also inverted the x/2
brazen wigeon
because tan is easier to work with than cot (in my opinion)
steel jewel
ok