wait
#help-19
craggy jasper
meager juniper
But 4, 3, and 2 do not need to
craggy jasper
does this question translate to x * 8 in base 9?
more than 100 million
@meager juniper do you agree that we should do it in a help forum here
subtle atlas
this gave no output for some reason
meager juniper
n: int = 1
while True:
n2 = n * 9
s2 = str(n2)
for i in range(2, 10):
s3 = str(n2 * i)
if sorted(s3) == sorted(s2):
print(f"{s2} * {i} = {s3}")
if s3 == s2[::-1]:
print(f"rev! {s2} * {i} = {s3}")
n += 1
if n > 1111111:
break
idk i just copied his code
subtle atlas
```py
n: int = 1
while True:
n2 = n * 9
s2 = str(n2)
for i in range(...
ok this works
meager juniper
n2 = n * 9
I proved that n must always be divisible by 9. So no need to check any other values.
craggy jasper
I proved that n must always be divisible by 9. So no need to check any other val...
oh
meager juniper
Honestly, I should rewrite to use a range
This is just me being lazy and using a phone after copying your old code lol
craggy jasper
ok this works
are you plotting n * 9 = n2 first or the general problem n * i = n2
subtle atlas
are you plotting n * 9 = n2 first or the general problem n * i = n2
what ever you want man
meager juniper
You get the same output either way
craggy jasper
You get the same output either way
?
I think Im misunderstanding what your code does
meager juniper
I'm finding all of the permutations
I'm just checking fewer numbers
Here one sec let me rewrite to be equivalent but perhaps easier to understand.
subtle atlas
```py
n: int = 1
while True:
n2 = n * 9
s2 = str(n2)
for i in range(...
i let the bound of n go till 111111111 instead of 1111111. Its been executing for far longer now
lets see how long this goes on for
craggy jasper
Ill move this to the help forum and close the channel
subtle atlas
good idea
meager juniper
max_val = 10000000
for n in range(9, max_val, 9):
s = str(n)
s_rev = s[::-1]
for i in range(2, 10):
s2 = str(n * i)
if sorted(s) == sorted(s2):
print(f"{s} * {i} = {s2}")
if s_rev == s2:
print(f"rev! {s} * {i} = {s2}")
@craggy jasper ^
odd edgeBOT
odd edgeBOT
Available help channel!
sly drift
odd edgeBOT
split gyro
Yes
You applied by parts which is often applied while dealing with log
Can you tell me more about your problem
!status
odd edgeBOT
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
ripe spire
Monkey
honest turtle
Can't read your answer, but if you work out the second line, and making the logarithms ln2, you should be getting the answer.
odd edgeBOT
@sly drift Has your question been resolved?
odd edgeBOT
@sly drift Has your question been resolved?
odd edgeBOT
@sly drift Has your question been resolved?
obtuse raven
How do I find x here?
obtuse raven
<@&286206848099549185>
obtuse raven
<@&286206848099549185> anyone?
odd edgeBOT
@obtuse raven Has your question been resolved?
obtuse raven
.close
spiral crane
odd edgeBOT
spiral crane
im kind of confused how to obtain --> QR
like
even if i obtain it
would it contribute to finding the answer?
warped glacier
im kind of confused how to obtain --> QR
that means P, Q, R are on the same straight line
cause that's the definition of a scalar multiple of a vector
warped glacier
im kind of confused how to obtain --> QR
okay can you find vector PQ?
spiral crane
PQ
warped glacier
yeah
spiral crane
oh
warped glacier
okay but what's PO in terms of OP?
spiral crane
hold on
spiral crane
3i + 13j
spiral crane
QO + OR?
spiral crane
but we have neither of those
warped glacier
you've found PQ = 3i + 13j already
warped glacier
okay now we use the fact that QR = 4 * PQ
no, use this
warped glacier
which is?
spiral crane
12 i + 52 j
warped glacier
yep
okay and now position vector of R = position vector of Q + QR
basically OR = OQ + QR
spiral crane
21 i + 71 j
this divided by magnitude?
warped glacier
21 i + 71 j
should be 21i + 72j
warped glacier
this divided by magnitude?
yes
warped glacier
yeah and now find the magnitude
spiral crane
warped glacier
yeah and you can simplify that, but that's correct
spiral crane
simplify?
spiral crane
ohhhh
spiral crane
thing is i have non calc paper so will they accept answer if i keep it like this?
spiral crane
^
warped glacier
basically the strategy is to use every single piece of the question
so the first bit of info requires us to find PQ to find QR
then OQ + QR = OR
then we just need to normalise OR so that the length is 1
warped glacier
thing is i have non calc paper so will they accept answer if i keep it like this...
Cambridge should be fine with not simplifying fractions
as long as it's correct
including the method ofc so yeah make sure it follows logically
warped glacier
no worries!
spiral crane
.close
odd edgeBOT
odd edgeBOT
Available help channel!
copper urchin
odd edgeBOT
warped glacier
where is this question from?
copper urchin
where is this question from?
my homework
warped glacier
you should sub in u = sqrt(x)
and then multiply everything by u^2 and rearrange to make RHS = 0
warped glacier
yeah it's a 6th degree equation hopefully more than one root will be guessable
or it's related to values of trig functions idk
copper urchin
.close
odd edgeBOT
odd edgeBOT
Available help channel!
glad imp
Im kinda confused how it became 1/(4x+blablabla
odd edgeBOT
Im kinda confused how it became 1/(4x+blablabla
velvet heart
do you know about the chain rule?
glad imp
What is a chain rule
velvet heart
d/dx f(g(x)) = f'(g(x))g'(x)
glad imp
..no
velvet heart
it's a property for taking derivatives, like the product rule but for f(g(x)) instead of f(x)*g(x)
if you apply that to ln(4x + 7) taking f(x) = ln(x) and g(x) = 4x + 7 you get that
glad imp
Im still confused
velvet heart
about?
velvet heart
do you know the derivative of ln(x) ?
glad imp
No
velvet heart
how much do you know about derivatives?
glad imp
Im bad at logarithm
fickle spoke
..no
I would recommend to watch 3 blue 1 brown - essential of calculus. His explanation are really straightforward
glad imp
Okay
manic lotus
hello, how do i show that the series sum of x^(2n) is not uniformly convergent when x in [0, 1[
odd edgeBOT
hello, how do i show that the series sum of x^(2n) is not uniformly convergent w...
@manic lotus Has your question been resolved?
onyx charm
How would you do part 2?
odd edgeBOT
How would you do part 2?
onyx charm
What I've done so far but I'm not sure if it's right
And if what I've done is right how would I go about this one?
odd edgeBOT
@onyx charm Has your question been resolved?
onyx charm
.close
odd edgeBOT
gentle timber
Is this right
odd edgeBOT
Is this right
odd edgeBOT
fluid shoal
?
amber schooner
For some reason I think I did this integral wrong
you forgot to change the limits of integration
odd edgeBOT
Available help channel!
proper plover
odd edgeBOT
proper plover
I have a function a(x) and another function A(x). I want to vary the paramters K and k such that, I minimize the difference between a(x) and A(x)
But I do not know exactly what to do. I have come up with a idea of subtracting the two, integrating the subtraction, then deriving the subtraction with respect to K, then setting that entire integral equal to 0, but it becomes undefined
So something like this I have tried:
But, it doesn't really work because it keeps becoming undefined
Is there anything else I can do to solve for K such that, it minimizes the difference between a(x) and A(x)?
A(x) is my approximation of a(x)
Any advice as to how I find the value of K that makes A(x) closely match a(x)?
odd edgeBOT
@proper plover Has your question been resolved?
wanton bison
I have a function a(x) and another function A(x). I want to vary the paramters K...
If you wanna minimize the error in terms of k and K you can define an error function e(k,K) := |a(k,K)-A(k,K)| and try to find the minimum of that
wanton bison
But the way I see it k is not really doing much except for translating the function
odd edgeBOT
@proper plover Has your question been resolved?
wanton bison
It seems for K = 12 approximately that's the best choice and k = 0
I basically considered the gradient and a solution for x > 0
So something in [12,13]
queen finch
it says its a test 💀
quasi sparrow
Dang honors class and still need to cheat? Smh
mint leaf
tu
viscid flint
.close
odd edgeBOT
odd edgeBOT
Available help channel!
jolly pollen
how do i solve the last multi choice o the bottom?
odd edgeBOT
how do i solve the last multi choice o the bottom?
jolly pollen
i have no idea what a y = x or y = -x reflection is
odd edgeBOT
@jolly pollen Has your question been resolved?
lone cloak
y = x and y = -x are both lines. They are just the same as reflection in the x-axis but in different lines.
odd edgeBOT
@jolly pollen Has your question been resolved?
odd edgeBOT
odd edgeBOT
Available help channel!
odd edgeBOT
@rich sparrow Has your question been resolved?
fallen surge
err they don't use -2cov(x,y), the -2 comes from linearity of expectation
wraith dagger
I'm learning rocket science now somebody help me
velvet heart
<@&268886789983436800>
pulsar elbow
I'm learning rocket science now somebody help me
don’t troll in the help channels please
.close
odd edgeBOT
wraith dagger
.reopen
odd edgeBOT
✅
amber schooner
you’re cooked
wraith dagger
Fine I'm trying to do 0 divided by 0
odd edgeBOT
odd edgeBOT
Available help channel!
swift helm
hello guys, is this correct? or am I wrong about something in here
limpid root
hello guys, is this correct? or am I wrong about something in here
It is correct
odd edgeBOT
odd edgeBOT
Available help channel!
open bramble
In the same way that you simplify the left equation to the equation below it, can I simplify this one too?
velvet heart
In the same way that you simplify the left equation to the equation below it, ca...
apply the formula and check if it simplifies to something nice
open bramble
alr i'll try that
open bramble
apply the formula and check if it simplifies to something nice
it does not look pretty at all
:C
it was worth a shot anyway
.close
feral igloo
I have no idea how to do this problem...
odd edgeBOT
I have no idea how to do this problem...
stable wave
Yes
feral igloo
u meant a_1?
stable wave
Now let's say nth term is the last term
stable wave
u meant a_1?
Yes
stable wave
So nth term is 459
feral igloo
i see now! did a similar problem earlier
stable wave
Np
feral igloo
.close
odd edgeBOT
odd edgeBOT
Available help channel!
thick shell
Hello people, i'm working on my internal assessment which is modelling the trajectory of badminton serves. I have a bunch of points and I want to find an equation of the curve that is the line of best fit for those points, are there any ways of doing this without using technology?
grizzled tide
... yes you can do the same thing the computer does by hand, but why?
thick shell
and how would i do that
grizzled tide
although "curve of best fit" is generic enough it might take you too long. look at "least squares" regression
least squares line of best fit
odd edgeBOT
odd edgeBOT
Available help channel!
mystic saffron
How do you do these?
odd edgeBOT
How do you do these?
sonic summit
How do you do these?
just cancel the same terms....
mystic saffron
Wdym by that though?
mighty nimbus
How this notation mean? Rate of convergence or A<B and abs(A-B)<\delta? idk,thanks
sonic summit
How this notation mean? Rate of convergence or A<B and abs(A-B)<\delta? idk,than...
sir ask this in another vacant channel
mystic saffron
How this notation mean? Rate of convergence or A<B and abs(A-B)<\delta? idk,than...
HOLY MOTHER OF SWEET MATH
amber schooner
!occupied
odd edgeBOT
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).
mystic saffron
WHAT
mystic saffron
I did
And I don't get it sorry
Cause I did something else to get a different answer
Oh actually I get it now
We got the same answer
Just written differently
Tysm
Huh?
mystic saffron
It's alright
odd edgeBOT
@mystic saffron Has your question been resolved?
trim venture
hmm, haven't been here for a while lol. At any rate, how do you prove $\sum (\cos n^{-1})^{n^\alpha}$ converges/diverges depending on $\alpha\in\bR$?
clever fjordBOT
DarQ
nocturne belfry
reciprocal of n is arg of cos?
trim venture
I tried applying the root test which gives $(\cos n^{-1})^{n^{\alpha-1}}$
clever fjordBOT
DarQ
trim venture
but I dunno where that converges
nocturne belfry
i wont know ans just asking
trim venture
reciprocal of n is arg of cos?
arg of cos?
nocturne belfry
inside
trim venture
oh yea
my friend told me I could also use the riemann criterion
and that involves calculating $\lim n^\beta (\cos n^{-1})^{n^\alpha}$
clever fjordBOT
DarQ
trim venture
and I dunno where that goes either 😭
I'm especially interested in how you could use the riemann criterion since I havent seen that before
well, interested is the wrong word, I'm forced into learning this nonsense
please ping me if you have an answer!
frigid canopy
trim venture
it's not.
this is supposedly analysis 3 
a very mutilated, abused and disfigured analysis 3
clever fjordBOT
neon
trim venture
oh inchresting, using ln to get rid of the pesky sqrt sign
long tinsel
is this tripping
clever fjordBOT
neon
long tinsel
oh inchresting, using ln to get rid of the pesky sqrt sign
I basically wanted to apply the nth term test
Because at the least this limit needs to be 0
which means n^(α-2) --> -inf
which, is that even possible 
trim venture
which means n^(α-2) --> -inf
why?
I don't think the fraciton involving ln has a limit
I think that's what goes to -inf
long tinsel
,w limit as x goes to 0 ln(1 - x)/x
clever fjordBOT
trim venture
hmm
trim venture
into what?
long tinsel
did you give you a range for alpha
long tinsel
I see
trim venture
(which I think can be done)
long tinsel
let me exp ln it and see what happens
trim venture
let me exp ln it and see what happens
oh, yea, it's literally just the terms of our series but with \alpha = \alpha-1 btw
you don't have to change shit 
trim venture
remember the root test also gives a criterion for divergence, not just convergence
long tinsel
Oh wait less than 1
trim venture
afterwhich we only need to single out the cases where the limit is 1
and treat it separately
long tinsel
hold on ich bin confused
long tinsel
**neon**
we basically have this limit but with alpha - 3 instead of alpha - 2
which has to be less than e this time
long tinsel
isn't this true for all alpha
trim venture
Huh
long tinsel
**neon**
so assuming i'm not estupido it ends up with this
,w sum of (cos(1/n))^(n^2.39) from n = 1 to infinity
clever fjordBOT
long tinsel
why is wolfram alpha so stupid
trim venture
Lol
long tinsel
I think it should just be alpha > 2, and we gotta manually have a look at the boundary
wolfram is known to go caput when trig is involved in sums
trim venture
I don't see the mistake in your calculation tho
trim venture
isn't this true for all alpha
Like, why isn't this true
long tinsel
it should be
the root test says it should converge for all alpha
but we need to make sure the limit at infinity of the term is 0 as well which restricts alpha > 2
trim venture
No, the root test implies that already
trim venture
this is what the root test gave us though <:thonkzoom:533366356536721418>
Yea
Like, if the limit of the root test is less that 1 that you get convergence to 0 for free
long tinsel
do we consider limit at infinity to be legit
long tinsel
hmm
long tinsel
Like, if the limit of the root test is less that 1 that you get convergence to 0...
i don't get what you mean
Oh wait
ah okay
fluid tundra
hmm, haven't been here for a while lol. At any rate, how do you prove $\sum (\co...
is this cos(1/n) or sec(n)
long tinsel
then why did we end up with two separate conditions
long tinsel
**neon**
@trim venture ah alright if alpha > 2 then this turns out to be 0, but if alpha < 2 it turns out to be 1
so alpha cannot be < 2 since we absolutely need that limit to be 0
fluid tundra
I also tried bounding cos x by 1-x^2/2
$a_n = \left(1 - \frac{1}{2n^2} + O\left(\frac{1}{n^4}\right)\right)^{n^{\alpha}} \approx \exp\left(-\frac{n^{\alpha}}{2n^2}\right)$
clever fjordBOT
Mqnic_
fluid tundra
$a_n = \left(1 - \frac{1}{2n^2} + O\left(\frac{1}{n^4}\right)\right)^{n^{\alpha}...
if my understanding of the problem is not wrong then this just completes the problem
leaden karma
**Mqnic\_**
Are you sure we can ignore the O(1/n^4) like that?
long tinsel
you can never be sure
fluid tundra
it should be negligible
long tinsel
here it should be less than 1
yeah alright makes sense then
fluid tundra
since incorporating that into the exp, you just have + O(1/n^{4 - \alpha})
long tinsel
it leads to the same result as me with no approxmations so yeah it should be gucci
fluid tundra
since incorporating that into the exp, you just have + O(1/n^{4 - \alpha})
which is dominated by 1/n^{2 - \alpha} as written above
fluid tundra
for convergence, yes
long tinsel
but then the nth root test needs alpha > 3?
long tinsel
**neon**
this but it's exp 0
I mean you're gonna have to manually have a look for alpha < 3
convergence in jeopardy in (2, 3) and absolutely converget for alpha > 3
trim venture
but then the nth root test needs alpha > 3?
yea
trim venture
$a_n = \left(1 - \frac{1}{2n^2} + O\left(\frac{1}{n^4}\right)\right)^{n^{\alpha}...
I don't understand where this came from tbh
long tinsel
,w sum from k = 1 to infinity (cos(1/k))^(k^(2.2))
fluid tundra
leaden karma
**Mqnic\_**
well yeah actually the series $\sum\exp(-n^{0.5})$ doesn't converge
clever fjordBOT
kheerii
leaden karma
,w sum from n=1 to inf of e^(-n^(0.5))
leaden karma
I was wrong
trim venture
<:hmmcat:831508801646755850>
like, I understand using the O thing or w/e
trim venture
but why is that approximately expo w/e
long tinsel
it converges for 2.4 but not 2.2
fluid tundra
oh i was sending that hmmcat in response to the wolfram
leaden karma
or we deem wolfram unreliable
i think this is the only way
fluid tundra
i plugged it into my browser wolfram for values smaller than 2.4 and it gives me no result
long tinsel
valley is an expert on wolfram being a bitch when it comes to trig in infinite sums
leaden karma
are we sure that $\sum\exp\left(-\frac{n^k}{2}\right)$ converges for all $k>0$
long tinsel
yeah I asked the Wolfram/ChatGPT power ranger
clever fjordBOT
kheerii
long tinsel
yeah that should converge
leaden karma
nth root test?
trim venture
are we sure that $\sum\exp\left(-\frac{n^k}{2}\right)$ converges for all $k>0$
yea, it's less than sum expo(-n/2)
leaden karma
k is alpha-2
leaden karma
it definitely converges for alpha>3
the question is whether it converges between 2 and 3
trim venture
😭
THIS IS SO STUPID
WHAT A STUPID ASS QUESTION
JUST THROW AT IT STUFF AND SEE WHAT STICKS
terrible
fluid tundra
welcome to analysis
trim venture
no dude, this isn't analysis
long tinsel
was this a university level question
long tinsel
💀
trim venture
2nd year too
long tinsel
i'm scared
trim venture
don't be, it's just dumb
leaden karma
wait am I dumb or does the root test literally give you alpha > 3
fluid tundra
are we sure that $\sum\exp\left(-\frac{n^k}{2}\right)$ converges for all $k>0$
i think you show this using integral test
trim venture
this isn't what analysis should look like
long tinsel
wait am I dumb or does the root test literally give you alpha > 3
it confirms it for alpha > 3 but we're unsure for alpha between 2 and 3
fluid tundra
it comes out to gamma function with parameter (3 - alpha)/(alpha - 2)
which is always > -1 for alpha > 2
long tinsel
it comes out to gamma function with parameter (3 - alpha)/(alpha - 2)
wait reflection formula
1 - (alpha - 2)
leaden karma
we don't actually need to solve the integral neon
fluid tundra
lmao.
trim venture
**Mqnic\_**
still confused how you got this btw
how do we know the O(1/n^4) term doesn't mess up stuff
leaden karma
which is dominated by 1/n^{2 - \alpha} as written above
fluid tundra
$\exp\left(-\frac{n^{\alpha}}{2n^2} + n^{\alpha}O\left(\frac{1}{n^4}\right)\right)$
clever fjordBOT
Mqnic_
fluid tundra
taking the O(1/n^4) term into the exp, we have ^
$\exp\left(-\frac{1}{2n^{2 - \alpha}} + O\left(\frac{1}{n^{4 - \alpha}}\right)\right)$
clever fjordBOT
Mqnic_
trim venture
$\exp\left(-\frac{n^{\alpha}}{2n^2} + n^{\alpha}O\left(\frac{1}{n^4}\right)\righ...
I'm sorry, from where did the exp come out in the first place? ig it has to do with (1+x/n)^n?
leaden karma
this doesn't seem very rigorous
fluid tundra
oh that's because it's not
fluid tundra
I'm sorry, from where did the exp come out in the first place? ig it has to do w...
$(1 - x)^k \approx e^{-kx}$
clever fjordBOT
Mqnic_
leaden karma
just replace the O(1/n^4) with a <=
fluid tundra
for small enough x and large enough k
trim venture
**Mqnic\_**
bruh
long tinsel
**Mqnic\_**
physics ahh approximation
fluid tundra
it actually works though
leaden karma
$$\begin{aligned}
\sum\left(\cos\frac1{n}\right)^{n^\alpha}&\le\sum\left(1-\frac1{2n^2}\right)^{n^\alpha}\
&=\sum\exp\left(n^\alpha\log\left(1-\frac1{2n^2}\right)\right)\
&\le\sum\exp\left(-\frac{n^{\alpha-2}}{2}\right)\end{aligned}$$
clever fjordBOT
kheerii
leaden karma
I think i cooked
trim venture
(you're free to entertain the question amongst yourselves obviously I just won't be able to answer 😭)
leaden karma
**kheerii**
@fluid tundra we can prove the last one converges by integral test?
long tinsel
**kheerii**
but this doesn't achieve anything we already knew does it
leaden karma
for 2 < alpha < 3
fluid tundra
<@392866251149410305> we can prove the last one converges by integral test?
is that first inequality even accurate...
leaden karma
we already know it converges for alpha >= 3 and diverges for alpha <= 2
leaden karma
is that first inequality even accurate...
i think it is
cosx is bounded below by 1-x^2/2
fluid tundra
thankyu guys for the help!
🎉
long tinsel
,w (cos(1/n))^(n^2.5) < (1 - 1/(2n^2))^(n^(2.5))
clever fjordBOT
Requested by neonperseus
leaden karma
,w plot y=(cos(1/x))^(x^2.5), y=(1-1/(2x^2))^(x^2.5) for 1 <= x <= 5
clever fjordBOT
fluid tundra
ye
leaden karma
wait yeah it's bounded BELOW by 1-x^2/2
im stupid
just take it upto x^4/24 then lol
fluid tundra
problem averted 
long tinsel
,w plot y=(cos(1/x))^(x^2.5), y=(1-1/(2x^2) + x^(-4)/24)^(x^2.5) for 2 <= x <= 3
long tinsel
I think it crosses it at some point
clever fjordBOT
leaden karma
it can't
long tinsel
i cba tbh
leaden karma
this one is definitely right
long tinsel
alright
leaden karma
$\int_0^\infty\exp\left(-\frac{x^{\alpha -2}}{2}\right)\dd{x}$
clever fjordBOT
kheerii
leaden karma
it turns out e^(-x^2/2) >= cos(x) directly
long tinsel
lol
odd edgeBOT
@trim venture Has your question been resolved?
golden marten
Can someone explain how to determine which function has a larger standard deviation in this case?
wild vault
eyeballing it
golden marten
I was under the impression the normal distribution with a lower height would indicate greater standard deviation as hence the data would be more spread
wild vault
I was under the impression the normal distribution with a lower height would ind...
ah, that's how they get u
golden marten
I was under the impression the normal distribution with a lower height would ind...
^ is this an incorrect assumption?
wild vault
the height doesn't actually matter, it's the width of the curve that coutns
but being squashed makes it look wider
golden marten
so height for the normal distribution would be related to number of occurences instead of probability?
wild vault
if u visualise for a bit u see that they're actually the same width
golden marten
because I was thinking in terms of a PDF where area should be 1
wild vault
well, the graph isn't labelled beyond x and y, so we cant say that this is a pdf
golden marten
if u visualise for a bit u see that they're actually the same width
although it says SD(2)>SD(1) as the answer which honestly I can't even see by eyeballing it
golden marten
so it should be just related to width then? That's it?
golden marten
xd
wild vault
but this is saying curve 1 is narrower
golden marten
yeah so SD is related to width then
I'll keep than in mind
and I guess my eyeballing skills are just not up to scratch
wild vault
tbf i thougth they were equal
so
dont be 2 hard on urself
then again i have a fuckin. 80s nerd prescription so
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rich dust
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rich dust
can someone teach me
regal peak
How do i find the minimum and maximum rate of change of an equation and a derivative equation?
tawdry wagon
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regal peak
i dont know how to find rates of change, and maximum and minimum ROC either
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dapper urchin
Hi all! Who likes math over here?? Thumbs up or thumbs down reaction on this message plz!!! BTW I'm new here....
simple oar
#discussion is the "general" kind of chat. This is a help channel that is !occupied
tall pivot
i dont know how to find rates of change, and maximum and minimum ROC either
differentiate wrt time to find rate of change
simple oar
<@823462958079803432>
What's the original question
regal peak
What's the original question
sorry i changed it up, i am confused on how to find growth/decay rates of a function
simple oar
Is this a general question or do you have a specific question that you need help with
regal peak
Is this a general question or do you have a specific question that you need help...
general
simple oar
The derivative expresses the rate of change of some value with respect to another
The rate of change can be seen as growth or decay depending on the function
tall pivot
ohhh alright, i see
+/- growth/decay
tall pivot
ohhh alright, i see
unless you're into product calculus? hehe
analogues of Volterra, geometric and bigeometric integral also exist if you're interested? multiplicative (as opposed to the familiar additive) calc
mystic saffron
OMG
mystic saffron
🤬
lunar hedge
Do you know what a square is
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@mystic saffron Has your question been resolved?
mystic saffron
np, click the above button tho
hi
warped glacier
for part a, you need to use the fact that the angle at the centre is twice the angle at the circumference?
what is the angle at the centre?
mystic saffron
uh idk
warped glacier
can you draw the pentagon?
mystic saffron
odd edgeBOT
warped glacier
if f is odd then we have $f''(x) + f''(-x) + 3f^2 (x) + 3f^2 (-x) = 0$ so $f'' + 3f^2$ is odd
clever fjordBOT
south's secret twin brother
mystic saffron
you got f is odd fn how ?
warped glacier
no I assumed f was odd
mystic saffron
ohh
winter grotto
If f is odd, how can it be monotonic?
warped glacier
If f is odd, how can it be monotonic?
an even function can't be monotonic
winter grotto
oh, sorry
winter grotto
makes sense yes
warped glacier
or even simpler y = x
winter grotto
but not all functions are odd or even
warped glacier
helps to rule something out
mystic saffron
**south's secret twin brother**
what next ?
warped glacier
idk
mystic saffron
okok
actually i made some progress
i made two cases
1st is f is constant and 2nd is f is not constant
for 1st case
f'=0 & f''=0 too
i got f(3+4f+8f^2)=0 after substituting and factorisation
and f =0
but in 2nd case i didnt get anything
winter grotto
Ye
mystic saffron
and in case if f is not const
then f '' = -(4f+3f^2+8f^3)
considering -(4f+3f^2+8f^3)
for large f values (+ or -) both cases we get concave down graph
which implies f cant be monotonic
winter grotto
I mean, the non-constant case is the interesting one
warped glacier
seems like only special kinds of functions work, such as if f is linear, i.e f = ax + b
also searching your question online didn't yield anythig
mystic saffron
but now question is how can we know that f can take large values ?
warped glacier
my point exactly
cause as x tends to plusminus infinity it seems really hard for the expression to equal 0
mystic saffron
but now question is how can we know that f can take large values ?
and if i say function is monotonic so taking large values
then what about arctanx
mystic saffron
i made two cases
this soln is just shit ig
winter grotto
Why that one in particular?
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ruby eagle
The first limit needs to be sqrt3,but idk whats the 2nd one..
ruby eagle
I also tried sm but idk what to do after
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ruby eagle
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steel raven
ill look at the second one also, but for the first i get negative sqrt3 btw
ruby eagle
ill look at the second one also, but for the first i get negative sqrt3 btw
How do you get -sqrt3
steel raven
a_n = (1 + \sqrt{3})^n - b_n \sqrt{3}
citrusmunch
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steel raven
dividing b_n, first term goes to zero and second term is negative, right?
ruby eagle
The limit for the first one is supposed to be sqrt3
steel raven
oh hold up sorry yeah. i'm assuming too much about a and b
ruby eagle
Aaa okok its alright
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@ruby eagle Has your question been resolved?
ruby eagle
Pls
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@ruby eagle Has your question been resolved?
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@ruby eagle Has your question been resolved?
unborn zinc
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unborn zinc
I swr it doesnt give any solutions
mystic saffron
the first one?
mystic saffron
well the equation is sincos=1 right and that means sin=1/cos which never intersect if im not wrong
unborn zinc
Idk
unborn zinc
well the equation is sincos=1 right and that means sin=1/cos which never interse...
Yh first part i got
.close
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sleek galleon
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sleek galleon
So I already studied this in school, but a year and a half ago, forgot lots of things
so I tried writting the expression ^-1
and integrating by parts
but didnt thelp
mystic saffron
do you remember u-substitutions?
mystic saffron
what could I substitute?
well, the fact that (ln(x))^2 is in the denominator is an incentive to be rid of it
so you should consider u = ln(x)
mystic saffron
why not ln^2 ?
because one of them has a convenient du/dx
recall that you need $\int f(u)\frac{du}{dx}dx$ for a u-sub
clever fjordBOT
Micabo
clever fjordBOT
Chuti | Argentina
sleek galleon
mystic saffron
well, it looks correct to my uninformed eye
sleek galleon
heheeh
cool
and this?
is there like an integration by parts for divisions ? 🤣
I feel like it's super easy but I just dont remember
sleek galleon
$\int \frac{3x-2}{:\left(x-1\right)\left(x+2\right)}$
clever fjordBOT
Chuti | Argentina
sleek galleon
hmmm I dont see how can I move on 💀
mystic saffron
now split it into the form A/(x-1) + B/(x+2)
mystic saffron
A/(x-1) + B/(x+2) = (3x-2)/(x-1)(x+2) and equate coefficients
mystic saffron
I've 2 unknowns
the x-coefficient must be the same on both sides, and so does the constant term
it’s actually two equations in disguise
sleek galleon
mystic saffron
start by multiplying both sides by (x-1)(x+2)
half swallow
I've 2 unknowns
if you have Ax+B=Cx+D, than we know A=C and B=D
half swallow
they're both in the form of the integral that =ln(x)
sleek galleon
rihgttttt
half swallow
I hated partial fractions 😭
but what helped me wasn't remembering the formula, but trying to recall it backwards, so we know how to add 2 fractions, partial fraction decomp is just doing that in reverse
mystic saffron
alr so no I have two different integrals
yes
mystic saffron
but what helped me wasn't remembering the formula, but trying to recall it backw...
if it works, it works
sleek galleon
ok got the result
half swallow
rihgttttt
so 1/(x-2)=ln(x-2), but be careful, this works since d/dx(x-2)=1, remember your chain rule
sleek galleon
yup
I got this
$\frac{1}{3}\cdot ln\left(3x-3\right)+8\cdot 3\cdot ln\left(3x-6\right)+C$
clever fjordBOT
Chuti | Argentina
sleek galleon
and now time for dinner
mystic saffron
that seems wrong
half swallow
what lol
mystic saffron
for one, it should be 3x+6 at any rate
and the factors of 3 within the brackets of ln are unnecessary although not incorrect
i’ll check your coefficients
sleek galleon
my A and B?
mystic saffron
yes
clever fjordBOT
sleek galleon
my coeffs are right, maybe I didnt something wrong afterwards
mystic saffron
oh, should it be 8/3 and not 8•3?
half swallow
oh, should it be 8/3 and not 8•3?
no since it's 8x3xln(3x+6) since the chain rule
only thing wrong is the -6 sign
mystic saffron
no since it's 8x3xln(3x+6) since the chain rule
what?
half swallow
oh yeah I made the same mistake, reverse chain rule
should be divide not multiply
mystic saffron
8/[3(x+2)] integrates to 8ln(x+2)/3, unless i am very mistaken
integrates
sorry
half swallow
yeah you're right
mystic saffron
just treat 8/3 as a coefficient and integrate 1/(x+2)
no need for reverse chain rule in this case
half swallow
true
odd edgeBOT
@sleek galleon Has your question been resolved?
zealous sable
I have a problem where I would like to find the volume of the solid bounded by z = 1 - x^2 and y = 1 - x in the first octant. If I want to write the triple integral in the order dxdydz or dxdzdy, it makes sense to me that we have to split the integral up, but how do we know what the dividing curve between the two regions in the zy plane is?
wanton bison
I have a problem where I would like to find the volume of the solid bounded by z...
what is a dividing curve
if you mean how to find the intersection of these two planes, you can solve y=1-x for x and plug that into the other equation
zealous sable
Yeah, that's what I meant, but how do we know that this curve is necessarily the boundary? I drew an image in an attempt to clarify what I mean; wouldn't the intersection of the two planes give you like a curve that is tilted with respect to the y-z plane?
odd edgeBOT
@zealous sable Has your question been resolved?
odd edgeBOT
wanton bison
.reopen
odd edgeBOT
✅
wanton bison
Yeah, that's what I meant, but how do we know that this curve is necessarily the...
It is a bit difficult to explain, how to determine the boundary based on the intersection
wanton bison
Yeah, that's what I meant, but how do we know that this curve is necessarily the...
This is why drawing will always help, but basically if you took a look at what you wanna actually calculate the region is simply the upper part that is bounded by z = 1 and z = 1-(y-1)² the parabola basically
and you would also need two integrals if you do it dx dA way because you will basically miss out on one part of the whole volume due to this region, dz dA or dy dA are far better, since they require only one integral
zealous sable
Ah, I see. So even though technically the two planes intersect at an angle wrt the y-z plane if we push that intersection curve onto the y-z plane we just get z = 1-(y-1)^2, setting the two z's equal?
Thanks for reopening the thread by the way
wanton bison
Ah, I see. So even though technically the two planes intersect at an angle wrt t...
Yes, this is also called projection, you project their intersection in 3D onto a plane, here yz plane. But notice it is not plainly z = 1-(y-1)^2 but also z = 1 (upper bound) because otherwise you would end up with a different kind of volume.
Also you are not setting the two z's equal, otherwise you cannot express the region you integrate the surface over in terms of y and z.
zealous sable
Ah yes, sorry, you want to use the two equations to get z in terms of y only rather than set the two z's equal. Is this how projection works in general if you want to project the intersection onto one of the planes formed by two axes (i.e. just eliminate one of the variables)?
wanton bison
Ah yes, sorry, you want to use the two equations to get z in terms of y only rat...
yes
zealous sable
Ah, thanks! What did you mean when you said that "otherwise you would end up with a different kind of volume"?
wanton bison
We want basically to integrate over yellow region, not the red
So if we now change order of integration we would have to integrate over the blue region
which is z = 1-(y-1)^2 to z = 1 not z = 0 to z = 1-(y-1)^2 because that one would be the red
it's kinda difficult to explain, so that's why i shared the link so you can see for yourself
modest skiff
do u know answer to this bro?
zealous sable
it's kinda difficult to explain, so that's why i shared the link so you can see ...
Hmmm that somewhat makes sense. Shouldn't one of the integral still have bounds from z = 0 to z = 1-(y-1)^2 though?
wanton bison
Hmmm that somewhat makes sense. Shouldn't one of the integral still have bounds ...
yes lol
wanton bison
Hmmm that somewhat makes sense. Shouldn't one of the integral still have bounds ...
you meant that piece
right
if so yea absolutely
it was basically luck that it worked out with the other region as well because of symmetry i assume
zealous sable
Oh, yeah, I was a little confused because I thought it was a little off but the answer was the same which is an interesting coincidence; so to summarize the the 0 to 1 - (y - 1)^2 would only be wrong in the second integral because we'd be integrating on the wrong side of the y = 1-x plane, right?
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chilly scaffold
is this right?
odd edgeBOT
is this right?
chilly scaffold
yes 3/2
wheat bramble
The only thing you're missing is a +C since it's an indefinite integral 
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shrewd helm
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@shrewd helm Has your question been resolved?
fallen surge
can you visualize or draw the region they are talking about
shrewd helm
yeah its a plane with a normal vector <2, 5, 1>
fallen surge
yeah its a plane with a normal vector <2, 5, 1>
yeah, but the part contained in the octant
shrewd helm
looks like a triangle
fallen surge
ok, so how would you find the area of the triangle
shrewd helm
base times one half height
thanks for the approach
i shouldve been more specific sry about that. theres a way to do it through integration that i dont understand
it uses parameterization, notable turning x and y into u and v respectively and leaving z in terms of u and v
fallen surge
i shouldve been more specific sry about that. theres a way to do it through inte...
can you send it somehow
fallen surge
alright, then what are you confused about
shrewd helm
the answer to the question is D also but i just dont get how they arrived at the anwer
like how do they use this approach and get D
fallen surge
err well what part don't you understand
do you understand the transformation they did?
shrewd helm
yes i want to know the bounds for the integration and the thing that is being integrated
odd edgeBOT
fallen surge
well for the bounds of integration, just look at what is the range of x and y
bruh
@shrewd helm type .reopen
shrewd helm
.reopen
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fallen surge
err wait
this isn't the same example
this one
so here, if r(u,v) = (x,y,z) = (u,v, 6-3u-2v)
we need to find the range of x and y
shrewd helm
o i got it tthx
fallen surge
err, ok so are you confused about anything else
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ripe sluice
hey can someone help me with this word problem. i keep getting confused: kian and yak are standing 6240 feet apart on a straight road. they both spot a hot air balloon floating in the air above the road between them. kian looks up at the balloon and sees it at an angle of 65 degrees, while zak looks up and sees it at a steeper angle of 72 degrees. find the height of the balloon
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@ripe sluice Has your question been resolved?
rain junco
odd edgeBOT
rain junco
Q3 Q4 Q8 Q16 Q21 Q30 Q35 Q37
I just need answers not explanation i have exam soon of this exact thing and google don't got answer of those
Someone pls help
<@&286206848099549185>
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vapid halo
Q.35
x+y=11
xy=42
x²+y²..?
(x+y)²=x²+2xy+y²
(11)²=x²+y²+2(42)
121=x²+y²+84
x²+y²=121-84
x²+y²=37
steep hare
i just spent probably 6 hours on this, not a single person nor youtube video had any help on this whatsoever and even a tutor told me he had no idea. im out of options.
ashen flax
steep hare
how so?
its an optimization problem, im in calc 1 but ive never seen one like this
usually they ask you to find the area not the other way around
latent scaffold
What is the area of one of the rectangles on the figure?
steep hare
x*y
latent scaffold
Ok, and then in the statement they tell you that each rectangle should have 1000 ft^2 of area
What equation does that give?
steep hare
2xy=2000
steep hare
would x = 1000/y and y = 1000/x?
latent scaffold
That'll be used later. Just concentrate on the total length of fence used in the figure
latent scaffold
Yes
See you want to minimize 4x + 3y, but there's two variables in there, and you don't have the tools (yet) to deal with that.
However, you know that, say, y=1000/x.
steep hare
uh huhh
latent scaffold
So can you use that information to make the perimeter in terms of only one variable?
latent scaffold
Yep!
So now, we know that the length of fence needed, given the value of x, is
L(x) = 4x + 3000/x.
And we want to minimize L(x)
steep hare
do we take the derivative now?
latent scaffold
Yes
steep hare
so L’(x) = 4 + 3000/x^2