ionic finch
and this way i can reduce it down
#help-13
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sand cradle
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proper raptor
I need help with my geometry assignment it’s about minor and major arcs formed by chords secants and tangents
stone mountain
use the secant angle theorem
The measure of an angle formed when two secants
intersect at a point outside the circle
is one-half the difference of the measures of the two
intercepted arcs.
stone mountain
try to relate this example to your problem, and create an equation for it
@proper raptor
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edgy quarry
2^(2n + 1) + 7 * 2^n + 21 = m^2
cedar kilnBOT
2^(2n + 1) + 7 * 2^n + 21 = m^2
edgy quarry
m, n integers
real tree
m, n integers
what exactly are we supposed to do with that equation
edgy quarry
Find m and n
real tree
wait is it possible to find out m and n from just 1 equation
edgy quarry
How?
real tree
im asking
elder sparrow
Find m and n
u cant solve
raven shard
says who?
raven shard
that thinking is not correct
elder sparrow
oh m and n are integers
elder sparrow
2^(2n + 1) + 7 * 2^n + 21 = m^2
trial and error spam by letting m=1, 2, 3, 4, 5, 6 💀
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ocean garnet
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ocean garnet
can someone help me understand the notation of t multiplied by the [1,1]
im not sure what this means
i remember learning it at start of sem but forgot
stiff totem
t[1,1] = [t,t]
vectors multiply with scalars componentwise
but you're parametrising a line here, so it's more useful to have the parameter outside as a scalar
ocean garnet
yeah i took a look at the previous stuff and realized i was just overcomplicating this
thank you
that makes sense
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raw yoke
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raw yoke
so i get the idea is to find a,b and c and then put them back into a cubic function but i think i have the process wrong a little bit?
crimson sedge
At face value your answer seems fine since r(x) seems to agree with the conditions they specified. What do you think is wrong about what you wrote? you substituted the wrong point seemingly
crimson sedge
Actually why did you specify D[3,2] when it was given to be D[3,4]?
crimson sedge
😭
raw yoke
4am gang yay
crimson sedge
wooo
raw yoke
thanks again man ;-;
crimson sedge
it's aight, is the rest fine by you then?
crimson sedge
lmaoo aight take ur time
raw yoke
lies i am not ok ;-;
i thought it'll be a simple; just change out the 2 to the 4 in the last part
wait wait wait
maybe i still got this
raw yoke
mmm nope yea i stuffed it
essentially i went back to
at D[3,4]
r(3) = 27a+(-18-54a) + (20+48a)-8a
2=13a+2
raw yoke
i got an answer of r(x)=−2x^3+5x^2
crimson sedge
isn't correct i assume?
raw yoke
nah :(
raw yoke
okioki
crimson sedge
can y share ur updated notes
crimson sedge
raw yoke
essentially...
r(3) = 27a+(-18-54a) + (20+48a)-8a
2=13a+2
a= 0/13 = 0
subbed that into
b=-2-6a =-2
c=5+12a =5
d=-8a = 0
and then subbed that into the cubic
crimson sedge
aight
crimson sedge
essentially...
r(3) = 27a+(-18-54a) + (20+48a)-8a
2=13a+2
a= 0/13 = 0
subbed th...
I will say, your method will be less error prone if you used like gaussian elimination rather than subbing
raw yoke
we're learning that next week :(
if i dont respond, it's not a you thing; i have actaully just stayed up way too long and im dying
crimson sedge
LOL you should really sleep 😭
and yeah sorry im just eyeballing your work trying to find the algebra mistake u did
crimson sedge
,w rref [[8, 4, 2, 1, 2], [27, 9, 3, 1, 4], [6, 4, 1, 0, -3], [12, 2, 0, 0, -4]]
wraith daggerBOT
crimson sedge
it should match the numbers on the last column here
if i wrote it down correctly anyways 
damn i algebrad off i guess
crimson sedge
cedar kilnBOT
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untold lichen
can I compute a^b quicker than binary exponent if a is 1 digit integer and b is limited to, say natural integer less than 300.
Also I don't need most of the bits, top 32/64 bits is most likely enough
cedar kilnBOT
can I compute a^b quicker than binary exponent if a is 1 digit integer and b is ...
@untold lichen Has your question been resolved?
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@untold lichen Has your question been resolved?
cedar kilnBOT
@untold lichen Has your question been resolved?
main needle
hmmm I don't see much improvement, you can't really truncate numbers/do scientific notation when you do large powers
crimson sedge
what
crimson sedge
can I compute a^b quicker than binary exponent if a is 1 digit integer and b is ...
have you tried repeated squaring or exponentiation by squaring
untold lichen
have you tried repeated squaring or exponentiation by squaring
Binary exponent, what I said in the previous message, is a different name for exponentiation by squaring
So yes
untold lichen
The question is about finding a way to calculating it quicker than "exponentiation by squaring"
In case it wasn't clear, I'm using it for now and I'm looking for improvements
crimson sedge
and i'm just asking how it went
because binary exponentiation is already quite efficient
what about parallelization?
there's specialized libraries too
for exponentiation
recomputed powers might provide some improvement
maybe bit manipulation too
but not marginal, how much improvement are you seeking?
untold lichen
and i'm just asking how it went
I guess it's not the best
That's why I'm here
If that's absolutely impossible to improve with restrictions I said, then I'll stop there maybe
crimson sedge
challenging to significantly improve
bin exp well-suited for O(log b)
you could experiment but plurmorant is right
i recommend you move on
untold lichen
challenging to significantly improve
I mean slight improvement is still appreciated
I have some time to invest to programming overall so...
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normal shore
not math related but can anyone explain?
cedar kilnBOT
not math related but can anyone explain?
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scarlet apex
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scarlet apex
okay my gradient to the point is: <-4,4>. How would i find the tangent line?
pastel vault
okay my gradient to the point is: <-4,4>. How would i find the tangent line?
It's the vector perpendicular to <-4, 4>
There's only one direction that is perpendicular in R^2 (look at the graph)
And it also passes through (4, -4)
So you could use point-slope form or y = mx + b etc
cedar kilnBOT
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fading light
how to solve this? This is the solution but idk how to get to it
cedar kilnBOT
how to solve this? This is the solution but idk how to get to it
@fading light Has your question been resolved?
unborn wasp
Try 2x=cos2theta
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peak lynx
2x=sinu or cosu
slim birch
i'm finding it difficult to intuitively understand how laplace's equation is applied - say if i have a square room with walls at a some temperature T1 and the temp at a point in the center T2 (T2 >> T1), how would we expect the heat to diffuse?
slim birch
^ i'm aware of the heat equation, i guess i'm asking if i can picture or visualise the scenario in a less abstract way
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scarlet sand
you would expect the heat to diffuse radially
the rate of change is proportional to the laplacian. Therefore, areas with greater concavity (eg heat spikes) will diffuse the heat quicker.
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slim birch
the rate of change is proportional to the laplacian. Therefore, areas with great...
Thanks for the response! are you saying that we'd expect the heat to diffuse in a linear fashion?
I read that the laplace equation is linear or something to this effect but I wasn't sure I could use that knowledge
scarlet sand
define linearly.
If you are claiming that the heat increases/decreases at a linear rate, this is not true
slim birch
define linearly.
the distance we move away from the center being proportional to the heat decrease
i know it sounded suspicious but i can't really reason in favour of or against this
scarlet sand
unfortunately that is not necessarily true either
perhaps this may help you visualize how the equation works?
karmic terrace
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karmic terrace
The area of ABE : AECD is 2 :3
The length of BE : EC is 3 : 2
What is AB : DC
Idek where to start
silk pecan
Idek where to start
maybe u should start by the area and try to find out the other stuff
the surface is (a+b)*h/2
and u have (BE+EC)²=h²+(b-a)²
and try to find a and b
cedar kilnBOT
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karmic terrace
Still wrapping my head around it
karmic terrace
and u have (BE+EC)²=h²+(b-a)²
Where did h²+(b-a)² come from?
silk pecan
Where did h²+(b-a)² come from?
the triangle one
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amber crest
give an integer solution to the equation 620x+140y =gcd(620,140)
amber crest
am i stupid or does this make no sense
slim birch
am i stupid or does this make no sense
why do you think so?
amber crest
gcd(620,140) is 20
620x + 140y = 20
31x + 7y = 1
how do you get integer solution
slim birch
integers can be negative
plush yarrow
First, you need to find the greatest common divisor (gcd) of 620 and 140.
slim birch
actually scratch that, i was thinking of =0
amber crest
First, you need to find the greatest common divisor (gcd) of 620 and 140.
yeah, its 20
plush yarrow
yeah, its 20
now rewrite the equation as
[ 620x + 140y = 20 ]
plush yarrow
To find an integer solution for ( x ) and ( y ), we can use the Extended Euclidean Algorithm. However, since we know the gcd, we can also find a particular solution by inspection. One such solution is:
[ x = -3, \quad y = 13 ]
wraith daggerBOT
Flamey
plush yarrow
We know that 620 is a multiple of 20, so we can start by setting ( x ) to a negative value that will reduce the total to something close to 20 when multiplied by 620. Similarly, we set ( y ) to a positive value that, when multiplied by 140, will increase the total to 20.
By trial and error, we find that:
[ 620(-3) + 140(13) = -1860 + 1820 = 20 ]
wraith daggerBOT
Flamey
spice phoenix
cedar kilnBOT
spice phoenix
Anybody know? For question b
slim birch
Anybody know? For question b
what would the relationship between $\theta$ and $2 \pi$ rad be?
wraith daggerBOT
jack
slim birch
(or theta and 360 degrees I guess if you're using deg)
slim birch
both are angles
slim birch
are you more comfortable with using degrees instead of radians
slim birch
well think of it this way then
if someone asked you to cut precisely 1/4 of a pie
and you had a protractor handy, how much would you measure
spice phoenix
Pi/2?
slim birch
right because (pi/2)/2pi rad is 1/4
now you are cutting another slice at an angle of $\theta$ rad
wraith daggerBOT
jack
slim birch
how much of the pie is that
spice phoenix
Theta/2pi?
slim birch
you got it
slim birch
Theta/2pi?
now you want to calculate the area of the slice, can you calculate that using this?
assuming you already know the radius of the pie
slim birch
what's the area of a circle
spice phoenix
Pi(r^2)
slim birch
Theta/2pi?
right, and you know that the slice/sector is theta/2pi of the entire pie/circle
so we can say the area of the sector is also theta/2pi of the area of the circle
spice phoenix
Sorry for being slow, this is my first time seeing this ngl
slim birch
no worries 
slim birch
the area of the sector is also theta/2pi of the area of the circle
you inputted the ratio of the sector angle to 2pi rad
but you did not multiply the area
cedar kilnBOT
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edgy coral
I know 4 and 5 is wrong but I don’t know what I did wrong
cedar kilnBOT
I know 4 and 5 is wrong but I don’t know what I did wrong
edgy coral
clear umbra
I know 4 and 5 is wrong but I don’t know what I did wrong
you didnt multiply x(x+2) on 9
clear umbra
same here
edgy coral
I did because I got 9x^2-7x
cedar kilnBOT
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hidden badge
could someone provide some algebraic reasoning for why $\lim_{x\to\infty} \frac{n!}{n^4} = \infty$? i get it intuitively, $n!$ grows WAY faster than $n^4$, but trying to show it algebraically is stumping me a bit. thanks!
$$\frac{n!}{n^4}$$
$$\frac{1\cdot2\cdot3\cdot\ldots\cdot n}{n\cdot n\cdot n\cdot n}$$
$$\frac{1}{n}(\frac{2\cdot3\cdot\ldots\cdot n)}{n\cdot n\cdot n})$$
$$?$$
wraith daggerBOT
declspecl
violet flume
could someone provide some algebraic reasoning for why $\lim_{x\to\infty} \frac{...
maybe you can parse it out this way
say n is appropriately large for this manipulation to make sense
$n! = n(n-1)(n-2)! \leq n \cdot n \cdot (n-2)!$
wraith daggerBOT
jan Niku
violet flume
o, were maybe going the wrong direction 
hmm how about ...
we could expand the factorial 
@hidden badge stil laround 
hidden badge
yeah haha im following
violet flume
so you know what i mean expand the factorial?
$n(n-1)(n-2)(n-3)(n-4)\cdots 2 \cdot 1$
wraith daggerBOT
jan Niku
violet flume
this is a polynomial, for some fixed n
hidden badge
makes sense
hidden badge
**jan Niku**
how did you get $(n - 2)!$ on the right side btw?
wraith daggerBOT
declspecl
violet flume
were going the wrong direction
hidden badge
oh i see
violet flume
by that i mean
i was creating something larger
i think the more helpful thing would be to create something smaller that is more clean to work with and still diverges
so say $n=3$ then $n! = n(n-1)(n-2) = n^3 - 3n^2 + 2n$
wraith daggerBOT
jan Niku
hidden badge
ohh interesting
hidden badge
then couldnt we say that if we expand it infinity, we get $\frac{n^\infty \cdot \ldots}{n^4}$ so its infinitely large on top
wraith daggerBOT
declspecl
violet flume
n^5 plus stuff
violet flume
then couldnt we say that if we expand it infinity, we get $\frac{n^\infty \cdot ...
well its more like
n! is like a polynomial, but the highest degree is growing, too
each time you increase n, you get another degree
violet flume
then couldnt we say that if we expand it infinity, we get $\frac{n^\infty \cdot ...
so yea, maybe something like this
this is why factorial eventually beats out any polynomial in the growth race
because factorial is like a polynomial that keeps adding higher and higher order terms in the front
hidden badge
i see okay that makes sense
violet flume
you can beat it, buts it hard
hidden badge
i think the reasoning is good enough to be able to pass the limit off as true without going further yk
violet flume
you could make a reasonable argument like
a mean a more precise one
let n>4 which is reasonable, as n goes to infinity
then n! > n^m with m=n+1
or something like this 
idk
hidden badge
isnt n^(n + 1) > n! tho 🤔
violet flume
if i was forced to do this
id use binomial expansion
show that each term is bigger than the previous
you could use ratio test, too
well, that would require some doin
hidden badge
id want ratio test too but sadly this came from the root test so im stuck with the limit 😦
i like the binomial exansion idea tho
what do you think about this $$\frac{1}{n} \frac{2}{n} \frac{3}{n} \frac{4}{n} \frac{5 \cdot \ldots \cdot n}{1}$$
wraith daggerBOT
declspecl
hidden badge
this way we see the n in the numerator always escapes the denominator ns
if that makes sense 💀
violet flume
idk are you asking for intuition or how to argue it more cleanly
hidden badge
defiintely cleanly
hidden badge
**declspecl**
i assume this is not cleanly LOL
violet flume
definitely take the ratio of terms in the sequence
$\frac{a_{n+1}}{a_n} = \frac{ (n+1)! \cdot n^4 }{ n! \cdot (n+1)^4 }$
wraith daggerBOT
jan Niku
violet flume
this is $\frac{n^4}{(n+1)^3}$
wraith daggerBOT
jan Niku
violet flume
now, argue that the difference between terms is growing, so it cannot converge
this seems clean enough to me its not a full rigorous proof but is that good enough?
violet flume
have you seen this kind of argument before
hidden badge
**jan Niku**
mm not this part i dont think so
i get that its the ratio test but i wouldnt have thought to use it on a limit
and represent that as its own series
violet flume
as long as you can track the logic i think its fine
but thats just me
idk if you were intended to use something specific here
hidden badge
i think im confused about the fact taht the original limit came from
$$\sum_{n=1}^\infty \frac{(n!)^n}{n^{4n}}$$
so i used the root test to get
$$\lim_{n\to\infty} \frac{n!}{n^4}$$
wraith daggerBOT
declspecl
hidden badge
so why am i allowed to use the ratio test on just a limit and not a sequence?
violet flume
so why am i allowed to use the ratio test on just a limit and not a sequence?
I dont really get what you mean here
a limit is a sequence
i mean, maybe thats phrased poorly
violet flume
so why am i allowed to use the ratio test on just a limit and not a sequence?
there are many things you can say about a series just looking at some kind of limit of its summand/sequence
like, you know the nth term test presumably
sometimes its easier to work on just the summand
because we can put restrictions and requirements on it
in order for it to make sense to talk about the series converging at all
does that answer your question?
violet flume
so why am i allowed to use the ratio test on just a limit and not a sequence?
you may need to clarify here if that did not address your issue
hidden badge
your explanations make sense but let me try to write out my confusion 1 second
this is the best way i can put it... i think i have a misunderstanding of the relationship between limits and series maybe
the ratio can be used for :
$$\sum_{n=1}^\infty \frac{n!}{n^4}$$
(sequences)
so why are you able to use ratio test for:
$$\lim_{n\to\infty} \frac{n!}{n^4}$$
(limits)
wraith daggerBOT
declspecl
hidden badge
right because the original problem was
$$\sum_{n=1}^\infty \frac{(n!)^n}{n^{4n}}$$
but that returns
$$\lim_{n\to\infty} \frac{n!}{n^4}$$
not
$$\sum_{n=1}^\infty \frac{n!}{n^4}$$
so why can you use the ratio test for the $\lim$ version?
wraith daggerBOT
declspecl
violet flume
so IDK if 'returns' is fair, you're breaking in the middle of a sequence of logic. We are in the middle of applying the root test.
violet flume
right because the original problem was
$$\sum_{n=1}^\infty \frac{(n!)^n}{n^{4n}}...
what you need to buy is that attributes of the sequence (summand) imply things strongly about the sequence of partial sums
While its a little long, you can actually look at a proof of the root test
it doesnt require any crazy analysis
sometimes proofs are convincing and sometimes not though
violet flume
right because the original problem was
$$\sum_{n=1}^\infty \frac{(n!)^n}{n^{4n}}...
Surely you buy that, at least in theory, there COULD be some relationship between $$(a_1, a_2, a_3, \dots)$$ and $$(a_1, a_1 + a_2, a_1 +a_2 + a_3, \dots)$$
wraith daggerBOT
jan Niku
violet flume
these tests and stuff, theyre usually provided without proof
leaning heavily into your understanding that like
the sum is itself a sequence
hidden badge
right yeah
violet flume
so you say oh okay, there COULD
and they say alright, heres one
but like I said, you can just look at a proof, if you feel unconvinced
theres one here
hidden badge
oh interesting, thats a good way of putting it
yeah i havent seen any of the test proofs so it would help to look into them
violet flume
some are harder than others
and if you havent looked at many proofs, you should know that they can be VERY unhelpful for imparting actual intuition or understanding
hidden badge
uh oh hahaha
violet flume
some are harder than others
like a convincing argument for root test can be done if you know about the geometric series
but lhopitals rule is normally done somewhere in the middle of an introductory real analysis course
hidden badge
ah ok makes sense
yeah im schedule to take discrete math next semester so hopefully i learn it then
that is...
if i pass calc 2 😦
LOL
violet flume
calc ii is a real one lol
it was my first B and absolutely crushed me but i really enjoyed the material
don't worry if youre asked to make some logical leaps during the class just try to latch on to what makes sense
soon enough youll have the tools to go back and prove stuff yourself if it still bothers you
hidden badge
sounds relieving haha
my entire trigonometry knowledge is based on stuff i was told
that i dont understand
hidden badge
yeah i tried to read a trig book but it was like 1200 pages LOL
violet flume
there are some people that can benefit from it but man geometry
too many cool things to do
you dont need that much trig understanding
hidden badge
im scared for linear algebra its like calc 2 + geometry haha
hidden badge
after
violet flume
itll be fine
hidden badge
do you think discrete or linear algebra is harder 🤔
violet flume
depends on the person
i wouldnt worry so much about hard
youre talking about two totally foundational courses
you should worry about understanding the material more than how hard its gonna be
but thats just me
hidden badge
i agree for sure, its just tough trying to understand everything with limited time and when grades are on the line yk
violet flume
are you cs?
hidden badge
yep
violet flume
yea i mean you couldnt find two more important classes lol
linear alg is probably more important
violet flume
yea i mean you couldnt find two more important classes lol
at least as far as the math is concerned
hidden badge
i did some basic linear algebra with like graphics libraries but i know it ramps up a lot lol
hidden badge
yeah it does seem really interesting
violet flume
im taking a graduate course in computational linear algebra right now
linear algebra is so completely and utterly pervasive if you do any computational anything you will never escape it
hidden badge
YAYYY 🤣
i saw a gaussian integral proof with linear algebra and it flew so high over my head 😟
violet flume
sounds like a parlor trick
but IDK anything
anyways youll make quick work of discrete math
and people love linear algebra its very easy to find help
i wouldnt worry
hidden badge
ah ok cool thats good to hear
yeah im confident that once i finish calc 2 i'll be more comfortable
it is a doozy though
violet flume
you doing bachelors?
violet flume
take algorithms :p
not that youre askin for advice
but take any database and any algorithms class u can
hidden badge
oh yeah i already took a ds&a and a relational db class
very very fun
they didnt dive into math at all though
i also need to take a theory of computation class which seems cool
like compiler design and optimization
violet flume
i also need to take a theory of computation class which seems cool
nice
cloud plinth
hello can someone explain why there's no +- sign before sin(x)/(1-cos(x) $tan(x/2)=\pm \sqrt{(1-cos(x))/1+cos(x)}=sin(x)/(1+cos(x))$ ?
wraith daggerBOT
Odd_the_Wolf
drifting marlin
where's this from?
cloud plinth
it's half angle formula of tangent
slate lintel
,tex .rocket trig
wraith daggerBOT
"Hayley"
slate lintel
that really doesn't have half angles on it huh
drifting marlin
useless
slate lintel
cloud plinth
so why there's no +- sign? I get sin(x)/(1+cos(x)) part but why it has no +-?
Is it because we multiply $\pm\sqrt{(1-cos(x))/(1-cos(x))}$ when we rationalize(?) denominator?
wraith daggerBOT
Odd_the_Wolf
drifting marlin
i think for the other two the +/- is necessary as the sqrt() term is always positive while sin/cos aren't
however it's not needed as sin(A) and tan(A/2) should have the same sign
don't quote me on that (unless it's right)
uncut quarry
however it's not needed as sin(A) and tan(A/2) should have the same sign
dosen't hold in the 3rd quadrant
drifting marlin
you sure about that?
uncut quarry
in 3rd sin and cos both are negative which gives a positive tan
drifting marlin
we're not concerned with tan(A)
we're concerned with tan(A/2)
if A is in the third quadrant, A/2 is in the second
drifting marlin
what about the 4th?
drifting marlin
however it's not needed as sin(A) and tan(A/2) should have the same sign
ok you may quote me 
cedar kilnBOT
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hollow drift
Can someone guide me through finding the area of the triangle exclosed by the axes and tangent to y=1/x.
I've already found f'(x0) is -1/x0^2 how. This is all the work I've done thus far.
trying to currently find the x intercept of the line to find the length of the bottom of the triangle.
- 0 - 1/x0 = -1/x0^2 (x-x0)
- -1/x0 = (-1/x0^2 * x) + (-1/x0^2 -x0)
- -1/x0 = (-x/x0^2) + 1/x0
What i'm confused about is what to do next here. my lecture notes say that -1/x0 = ((-1/x0^2)*x) + 1/x0 becomes (1/x0^2)x = 2/x0 but I'm a little lost on how they get there What am I supposed to do to simplify the equation to that level?
hollow drift
<@&286206848099549185>
tired flame
<@&286206848099549185>
lucas lee pf?
hollow drift
I guess my lack of intelligence shows
lol
but yeah
i can provide a ss of the lecture notes i'm trying to understand to give a better picture of where I'm confused if that helps
I understand if me typing out my work isn't exactly useful
shut meadow
oh so 1 side is x axis, 1 side is y axis and other side is tangent of y=1/x?
hollow drift
yes
shut meadow
which tangent angle are you using tho
cuz there are many tangents along y=1/x
or is it biggest one
cuz then its probably when the tangent is parallel to y=-x
hollow drift
can i show a picture of the graph
shut meadow
sure
im about to go tho
someone else will probably be able to help u better after i go
hollow drift
I really just need to understand why the 2nd row of equations becomes the 3rd row of equations
shut meadow
uh right undergrad maths, i will be of 0 help ig
hollow drift
darn
shut meadow
oh you simplify
-1/x^20 x to other side
-1/x0 to other side where there is 1/x0
= 2/x0
hollow drift
I don't get it :(
I think it would help to see it done like... step by step. Idk if thats asking a lot but I'm having trouble visualizing what you're saying
cedar kilnBOT
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@hollow drift Has your question been resolved?
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Available help channel!
knotty eagle
are my answers correct?
cedar kilnBOT
are my answers correct?
knotty eagle
this is permutation
stone mountain
seems right
knotty eagle
seems right
ok how bout this one?
cedar kilnBOT
@knotty eagle Has your question been resolved?
slate garden
cedar kilnBOT
slate garden
How do i Find the value for A
Why can i not just solve the quadratic for the 2 values of x?
Do I always have to complete the square
for these types of question
to see if a solution gets rejected?
cedar kilnBOT
@slate garden Has your question been resolved?
slate garden
wait actually
cedar kilnBOT
wait actually
slate garden
shouldnt that mean
its 4+4root 2
how can the answer be negative
if thats the domain
mint musk
I'd like to post my working to get an answer but I'm not sure how to get an answer. What's the method that one would use to get the integral of ∫cot(3x)dx ? happy to just find a video that explains the general method and I can work it out myself hopefully :)
crimson delta
do you know u-sub?
otherwise, for something simple as that, you can also try differentiating cos(3x) a few times to see if you notice a pattern, and then try to reverse that pattern
mint musk
oh sorry sorry I meant cot
mb, I do know about integration by subsitution and inspection
remote salmon
oh sorry sorry I meant cot
you should use u-sub here too
mint musk
I'm not sure how that would get me to the answer, but let me try it and I'll get back to this
crimson delta
write cot as cos/sin
mint musk
royal loom
yes but we're doing u-sub
wraith daggerBOT
Austin
royal loom
we want to pick u=f(x)
I suggest you use something other than f(x)=3x
For example, is there something you could pick, some u=g(x)
such that
du=cos(3x)dx
crimson delta
well u=3x works as an initial simplification. but you need a second u-sub afterwards
mint musk
For example, is there something you could pick, some u=g(x)
I've gotten the answer by multiplying it by 3/3 to get f'(x)/f(x) so that I can put in ln, but I'm unsure what your reffering to above?
royal loom
I've gotten the answer by multiplying it by 3/3 to get f'(x)/f(x) so that I can ...
I have no clue why you are talking about ln
mint musk
I have no clue why you are talking about ln
?
royal loom
Just because the integral of 1/x is ln|x|
does not mean that the integral of g(x)/f(x) = ln(f(x))
which is what it seems like you have written
you're not doing this integral correctly, I suggest a different method
mint musk
no but is the integral of f'(x)/f(x) not ln(f(x))?
mint musk
For example, is there something you could pick, some u=g(x)
I seem to have gotten the correct answer, I'm just curious what method your reffering to here?
mint musk
well u=3x works as an initial simplification. but you need a second u-sub afterw...
and below here?
I'm a bit confused
royal loom
I seem to have gotten the correct answer, I'm just curious what method your reff...
I just meant to instead begin by u=sin(3x)
royal loom
I seem to have gotten the correct answer, I'm just curious what method your reff...
you're missing absolute values
mint musk
I just meant to instead begin by u=sin(3x)
ahh okay
maybe it would have been simpler yeah
royal loom
and below here?
here denascite means to then take v=3sin(u)
royal loom
maybe it would have been simpler yeah
it would've been
mint musk
so I've done it correctly, but had I taken a different subsitution I would have gotten the answer more efficiently? Am I correct in this statement?
royal loom
yes other than that you are missing absolute values
and you could show a bit more work
royal loom
no what I mean is
you wrote
ln(3sin(3x))
but ln is only defined for inputs > 0
so
it is actually
ln(|3sin(3x)|)
crimson delta
multiplying by 3/3 didnt change anything here
mint musk
ahh okay yes, this is something that they would accept in my specification and would expect the domain to be defined outside of the function
mint musk
multiplying by 3/3 didnt change anything here
is that not what I did?
crimson delta
the 3 in the ln just gets converted to +ln3 by log laws and then that goes into the +C
mint musk
ahh okay
mint musk
the 3 in the ln just gets converted to +ln3 by log laws and then that goes into ...
it seems like that gives 3cot(3x) though?
crimson delta
did you forget the 1/3 in the front?
mint musk
?
didn't you say multiplying by 3/3 made no difference? sorry I'm just confused
OHHH I know what you mean, I think that 3 is there because of subsituting U
actually with the way I did it there was no need to sub U I guess, only need to sub U if sin(3x) = u
I think that's closed though, thanks!
.closed
.close
cedar kilnBOT
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Available help channel!
ebon oxide
cedar kilnBOT
ebon oxide
the options are surjective only, injective only, bijective, or none of the above
south tundra
What do you think?
ebon oxide
im thinking none of the above
its not injective since you can have the same imaginary part for different values of b
its not surjective since there is at least one Im(z) value which you dont have a value of b to get
south tundra
Would be great to provide counterexamples
ebon oxide
assume a=0 for all these examples that im gonna give
f(-1i)=2+0i=2
f(i)=2+0i=2
but -1=/=1 so its not injective
to test surjectivity:
the possible imaginery parts of the function are -1 to infinity
wrong place lol
slate lintel
!occupied
cedar kilnBOT
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).
south tundra
Yeah, that proves it's none of the above
tribal jewel
Sorry about that
slate lintel
the possible imaginery parts of the function are -1 to infinity
yep those look good to me
ebon oxide
i got this from finding b^2-1
the range ^^
this is an issue though because 2-5i is a complex number but there is no value of b to obtain this output
therefore its not surjective and hence its neither
is that right?
slate lintel
yep
cedar kilnBOT
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verbal prairie
Question is to evaulate this limit explicitly
cedar kilnBOT
Question is to evaulate this limit explicitly
dusty hazel
Looks correct.
wraith daggerBOT
! What the hell am I doing here?
verbal prairie
but i squared the top and bottom
verbal prairie
oh
dusty hazel
This time it works cause it's 1
dusty hazel
Correct.
dusty hazel
That's what this server is about.
verbal prairie
so you just to the reverse operation depending on what operation you need when you initially manipulate the expressio
so you basically change nothing but make the form the expression such that you can take x to infinity and get the limit if it is a number?
dusty hazel
so you just to the reverse operation depending on what operation you need when y...
I mean even though square root is reverse for squaring, you shouldn't usually use that.
verbal prairie
i should raise to the half as you said?
dusty hazel
If the limit were something like -1, you'd first square to get 1 and once you square root that, you don't know if it's 1 or -1.
verbal prairie
oh okay i see
dusty hazel
Raising to 1/2 isn't always that simple either. Imagine if x -> -∞
Then the limit would be -1 and you'd raise to 1/2 still, but differently.
dusty hazel
so you basically change nothing but make the form the expression such that you c...
You usually try simpler stuff like multiplication and division by the same number. Or rationalisation.
verbal prairie
okay
so in the case of a a whole argument in the limit being raised to an exponent less than 1 its the same as the limit of the argument raised to that exponent less than 1
thats the step they did here
dusty hazel
yeah
verbal prairie
thanks
while i have you here can i ask a question unrelated to this topic
is there general situation that can explain what is 2x2 matrix and larger
idk how to learn this theory
and exams are in a month
i understand that you use a matrix to transform a vector or a function but i dont know how to do that from scratch
dusty hazel
What about matrices do you have to learn?
dusty hazel
i understand that you use a matrix to transform a vector or a function but i don...
Oh.
You should just watch yt videos or something.
If you want to study from scratch.
verbal prairie
to be able to answer these kind of questions:
verbal prairie
If you want to study from scratch.
nah i mean like to demonstrate the theory
dusty hazel
Well ok most of this involves matrix multiplication/ linear equation but using inverse.
verbal prairie
from first principles without using axioms
dusty hazel
There's only a few things you need to know to be able to do all of this.
I'd still recommend yt
opaque root
That is because u chose a bad u sub
Think of another sub
Also you have to so something before
worldly obsidian
What do you have to do before?
raven shard
That is because u chose a bad u sub
i'm unsure of what this means
opaque root
i'm unsure of what this means
It means that there is a much easier way to do this
raven shard
i would be interested in hearing about that so i will wait a bit and see what happens then
opaque root
I am gonna let him think a bit more about it, i will jusy say that you can rewrite the integral in other form
worldly obsidian
I really don't have a clue. I've never done one of these before. Been looking at it over an hour
opaque root
You can do something with (2+2e^x)^8
worldly obsidian
Could you show me the process please? I'm not getting it. I was hoping to see how it's solved then I can try other examples until it clicks
opaque root
You can do something with (2+2e^x)^8
You can rewrite this as (2(e^x+1))^8
And this as 2^8 (e^x+1)^8
Do you know what to do with 2^8 now?
@worldly obsidian
worldly obsidian
I don't sorry. I'm completely new to this and have no clue except for I'm supposed to rewrite things in terms of U.
opaque root
You can rewrite this as (2(e^x+1))^8
Did you understand this?
worldly obsidian
You took 2 out as a factor but I don't see where it gets me.
opaque root
And can you see that after doing that i get 256 * (e^x+1)^8?
worldly obsidian
Sure.
opaque root
And i am sure that you remember what to do with that 256
Can you take out from the integral?
worldly obsidian
Yeah
wraith daggerBOT
Samuel
opaque root
Now what u sub can you choose? This time it will be easier to handle it
worldly obsidian
E^x +1 so I can have u^8?
worldly obsidian
So, int of u-1(u-9)u^8 ?
opaque root
Show what you wrote
worldly obsidian
Just what I said. I used 1+e^x as U and subbed it in
I appreciate you trying to help man but I don't know the process. Can you show me how to solve this one then once I've seen the steps I'll have a better idea. I'll be guessing for hours like this.
worldly obsidian
E^x
worldly obsidian
Ok
worldly obsidian
Du/e^x ?
worldly obsidian
U-1
opaque root
**Samuel**
Now use all that information to make the correct subs here
worldly obsidian
I'm gonna say the same as before lol
U-1(u-9)u^8 but with du/e^X at the end?
worldly obsidian
Ok
opaque root
I'm gonna say the same as before lol
U-1(u-9)u^8 but with du/e^X at the end?
Also remember to put u-1 with brackets
(u-1)
And what did we say it is u-1?
U made a mistake here
u-9
Remember e^x is u-1
So e^x-10 is
u-1-10
Which is u-11 not u-9
worldly obsidian
Ok sorry
opaque root
Ok so
((u-1)(u-11)u^8)e^x
But u-1 is e^x so
(e^x(u-11)u^8)/e^x
Can u simplify here?
worldly obsidian
Why is there an e^x at the end sorry?
opaque root
Because dx was du/e^x
worldly obsidian
So what has happened to get it from that to just the e^x part?
opaque root
In the numerator you mean?
worldly obsidian
Ok sorry. So it becomes (u-11)u^8 ?
I guess I am confused how the dx = du/e^x results in the sub you gave.
I think I'm missing a step
worldly obsidian
I'll play with it. I'm still not solid on how we got here though sorry.
I wouldn't naturally get to
((u-1)(u-11)u^8)e^x
I'd have wrote du at the end instead of e^x
I guess I need to play around with that
opaque root
Ok, just have in mind that when u are doing u=e^x+1 you gotta play with everythiing here
Finding du, dx, u, e^x
And after you find everything u start subtituting
u=e^x+1
du=e^x dx
dx=du/e^x
e^x=u-1
(u-1)(u-11)u^8 du/e^x
((u-1)(u-11)u^8 du)/e^x
(e^x(u-11)u^8 du)/e^x
((u-11)u^8 du)
worldly obsidian
Oh Ok. That makes more sense. I'll give it a try. Thanks
opaque root
u^9-11u^8
Integral u^9 - integral 11u^8
Usub back
Dont forget also that those integrals are multipied but 256
worldly obsidian
Thanks for your patience.
Missed a month of class for surgery. Now gotta practice for exams and I'm lost lol.
When the topic started we were told the du dx etc were all just terminology you leave alone. Now they're being used to manipulate the equation. not used to it at all. Im not sure what it really means so tracking is hard. I'll keep at it.
opaque root
No prob, just keep doing them and u will get everything
hot crag
there's an online website that explains calc really well, hang on
this is basically all the calc u need in the words of someone whose name idr
worldly obsidian
Thanks. I'll take a look!
cedar kilnBOT
@worldly obsidian Has your question been resolved?
peak salmon
cedar kilnBOT
mighty shuttle
no, it;s not one-one
peak salmon
why?
i tried doing it this way too
but from the graph it looks like for each x value theres a unique y value, no?
peak salmon
no, it;s not one-one
could you explain why, please?
oh wait its actually obviopus from the graph why its not injective
im an idiot
but in that case, how would you solve this algebraically?
sand cradle
but in that case, how would you solve this algebraically?
Wdym solve algebraically
peak salmon
Wdym solve algebraically
as in if you got a question like "prove that 1 / (x^2 + 1) is injective"
youd do suppose f(x1) = f(x2), right?
sand cradle
Yeah
peak salmon
like in the screen shot above
but the last line looks weird
what is it meant to say?
sand cradle
We would have to conclude x1 = x2
But you can't arrive at that here
Because it's not injective
sand cradle
what is it meant to say?
|x_2| = |x_1| is probably a clearer way to write it
peak salmon
We would have to conclude x1 = x2
so then we did arrive at this then?...
peak salmon
|x_2| = |x_1| is probably a clearer way to write it
alright yeh, what im trying to ask is how is this different to that
sand cradle
alright yeh, what im trying to ask is how is this different to that
Well this doesn't necessarily mean x1 = x2
We could have x1 = -2
and x2 = 2
And this would hold
cedar kilnBOT
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Available help channel!
quartz holly
can someone help me with these two problems i dont get where to go once i get to this point
quartz holly
the first one i think is done but im not sure if i did it right and the second one i just got to that point and feel stuck
finite raven
send first image again
quartz holly
There is a better pic
finite raven
Did chat gpt write the first one?
quartz holly
i tried but no
that first one ive done like 2 or 3 times but that all get to that point of 5 not equaling 2 so i restart
finite raven
ok
Assume m² +1 = 2^n, so m² = 2^n -1. So m is odd. Now take m = 2k+1 and reach a contradiction
quartz holly
do i have show even ? since im showing odd?
finite raven
"do i have show even" what does this mean
quartz holly
usually whenever our professor shows something as odd it'll use proof by cases and case 1 is even and case 2 is odd
so case 1 is like m = 2k and case 2 is m = 2k + 1
finite raven
Ok, do this then
quartz holly
And also the picture of the second one is kinda different now I got to this point
cedar kilnBOT
@quartz holly Has your question been resolved?
quartz holly
still waiting on one and am working on the current one
ashen palm
cedar kilnBOT
ashen palm
Can someone help me figure out where the -sin(2x) went in the solution?
This is an example from my professor's sample exam
I can follow the whole variation of parameters to the end and get to an answer where u_1 contains that -sin(2x) term
I just can't see where it goes in the y=y_c+y_p part
cedar kilnBOT
@ashen palm Has your question been resolved?
ashen palm
<@&286206848099549185> If you wouldn't mind please
cedar kilnBOT
@ashen palm Has your question been resolved?
lone burrow
Confused about the bottom of the page
cedar kilnBOT
Confused about the bottom of the page
cedar kilnBOT
@lone burrow Has your question been resolved?
oak ginkgo
struggling with this one with the quadratic on the bottom
cedar kilnBOT
struggling with this one with the quadratic on the bottom
junior dome
try completing the square
crimson sedge
take u = x² -6x + 2
junior dome
or this ^
cedar kilnBOT
@oak ginkgo Has your question been resolved?
crimson sedge
make the numerator 2x-6
crimson sedge
replace the bottom with u and dx = du/2x-6 ?
well it works if you just do that as well but it won't everytime
crimson sedge
what do you get after factoring 2 out of the denominator
cedar kilnBOT
@oak ginkgo Has your question been resolved?
junior dome
so you set dx = du/2x-6 ?
yeah
cedar kilnBOT
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bold mauve
I am having difficulty understanding the "coordinate vectors relative to" part
cedar kilnBOT
@bold mauve Has your question been resolved?
cedar kilnBOT