novel ravine
True
#help-4
vale dockBOT
vale dockBOT
Available help channel!
halcyon tree
where did I go wrong?
vale dockBOT
where did I go wrong?
halcyon tree
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Available help channel!
celest shell
I am confused about how my textbook is introducing and then using this inverse function theorem.
- is the formula saying about the denominator (original'(inverse(x))) or (inverse'(original(x)))
- introduction seems to be saying (original'(inverse(x)))
- the application seems to be saying (inverse'(original(x)))
keen tundra
its original'(inverse)
celest shell
and is it saying that taking the reciprocal of (original'(inverse)) is the derivative of the inverse?
jade ivy
I'm not sure what part of that page is supposed to be inconsistent
keen tundra
and is it saying that taking the reciprocal of (original'(inverse)) is the deriv...
yes
jade ivy
You take the derivative of f and then you plug in the inverse into it
celest shell
the thing that's confusing me are the terms they use in the introduction: f(x) being the original and g(x) being the inverse. and then in the application they use g(x) as the original and f(x) as the inverse
jade ivy
The thing that is making this a nightmare to read is their lack of proper LaTeX making the prime look really ugly.
I read the example as the original function is f and the inverse is g and that they want to find g'
celest shell
ok i'm following that
jade ivy
Whether you call f the inverse of g or g the inverse of f here doesn't really matter
But I agree starting the solution by calling f the inverse of g is confusing
celest shell
ok so i should maybe chop this up to an inconsistency in terms introduced and then terms used... buuut what they are doing is still correct... aaand i should just think of it as: take the reciprocal of the (original'(inverse)) or $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$
rocky lotusBOT
Tubberware
jade ivy
The point is that it is a matter of what your perspective is
Lets say you start with a function f, and you want to find the derivative of a function f^-1. That is the usual "forward" way that one would normally think about the problem
The alternative is that you start with a function g and you want to differentiate it but you don't know how.
One way to then approach the problem is to find the inverse of g and maybe that is easier to differentiate
In practice, you would never think like this because differentiation is too easy in some sense
Actually nevermind, your own book is confusing me. I would never have wrote that solution they way they did in a million years and trying to justify it is just making me say garbage
celest shell
ohh ok i think i see what you mean
jade ivy
The broad point stands that what you call f and what you call f-1 doesn't really matter
celest shell
although i am interested in the point you made. maybe finding the derivative of the inverse would be easier then finding the derivative of the original?
jade ivy
This kind of getting off topic, but while that is a strategy you might use in math, I don't really see it in this situation
The problem is that the derivative, being "just" a function that cares about the behavior locally where h is near 0 is just too easily understood.
Between it being linear and the chain rule, product rule etc, you are just not going to realistically run into a function where you go "how on earth do I differentiate this". The derivative is the least of your worries in most applications
So thinking about the inverse function theorem "backwards" to try to access a way to differentiate something "hard" is just not that useful because nothing is really hard here
celest shell
ok, so its not a tool to differentiate hard problems but this theorem could be thought of as a proof that is used to setup a pattern or a mnemonic that makes finding an inverse function's derivative as easy as taking the reciprocal of the original's derivative? does that sound right?
jade ivy
I just meant that the way they start with g and call f the inverse of g doesn't get you anything helpful
celest shell
gotcha. so is my understanding correct about the symantics of the theorem? is my formula here correct? $(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$ and i'm interpreting that as i can find the derivative of the inverse by taking the original and finding its inverse, and then find the derivative of it, and all of that is reciprocated? or is my formula and process incorrect?
rocky lotusBOT
Tubberware
jade ivy
Yes
celest shell
sweet! thanks for getting in the weeds there with me!
wait i think i still got it wrong. looking at the application example - the piece where they are working on under the 'and' is just the denominator: $f'(g(x)$ in this case f'(x) is the derivative of the inverse of the original function, and then they are plugging the original function into that, and getting $-\frac{x^2}{2}$. aren't they getting that backwards or am i backwards still?
rocky lotusBOT
Tubberware
celest shell
i'm circling back to my original confusing:
- introduction seems to be saying about the denominator: (original'(inverse(x)))
- the application seems to be saying about the denominator: (inverse'(original(x)))
and that process is contradicting my current understanding of the theorem:
$(f^{-1})'(x) = \frac{1}{f'(f^{-1}(x))}$ and i'm interpreting that as i can find the derivative of the inverse by taking the original and finding its inverse, and then find the derivative of it, and all of that is reciprocated
rocky lotusBOT
Tubberware
vale dockBOT
@celest shell Has your question been resolved?
vale dockBOT
@celest shell Has your question been resolved?
vale dockBOT
@celest shell Has your question been resolved?
silent walrus
vale dockBOT
silent walrus
my professer told me to flip the diagram then do the problem but im stil confused on how to get D
shadow moss
consider the x component of velocity
vale dockBOT
Available help channel!
open niche
can someone explain why if x= - 1/2 then this expression equals 0 ?
red tulip
Put -1 /2 into the |2x+1|
open niche
Put -1 /2 into the |2x+1|
and?
dark pewter
what value did you expect the expression to have for x=-1/2
open niche
what value did you expect the expression to have for x=-1/2
2?
whats that gotta do with it
0
so we have |0| * lim (n+1)
but lim (n+1) = infinity
so we have 0 * infinity ?
merry crystal
no, we dont have 0*infinity
hazy pivot
Maybe write out the sequence
Pick n as large as you want
What's the value of the expression (without the limit)
hazy pivot
Do you see what to do now
open niche
that for every ε>0 there is N such that if n>N |an-L|<ε
hazy pivot
can someone explain why if x= - 1/2 then this expression equals 0 ?
I'm assuming from the use of n that it's a natural number
open niche
it is
open niche
but i dont think it makes any significant difference
open niche
Then for any n, what is a_n
i dont understand the logic here
hazy pivot
It just makes your life easier is all
open niche
why would we be able to just use specific number for n
when we have a limit that says n approaches infinity
hazy pivot
Limits don't exist in a vacuum, they exist in neighbourhoods
dark pewter
||you might as well put the |2x-1|=0 term inside the limit (which means you will be taking the limit of zero) and call it a day x)||
hazy pivot
What's a neighbourhood of infinity
hazy pivot
||you might as well put the |2x-1|=0 term inside the limit (which means you will...
Ehhh 0. Inf is indeterminate
merry crystal
define your sequence f_n (-1/2) and show using the N-epsilon definition that the limit trivially converges to 0 for any epsilon
open niche
||you might as well put the |2x-1|=0 term inside the limit (which means you will...
mhm i see
open niche
Ehhh 0. Inf is indeterminate
well not really
hazy pivot
Ehhh 0. Inf is indeterminate
(I know it isn't for this case but he seems confused)
open niche
cause x isnt dependent on n
hazy pivot
Oh good
open niche
but wait
im still not getting something
if we agree that lim(n+1) = +oo
what is the thing that prevents as from replacing the limit with +oo
hazy pivot
ok but why are we able to use number
The limit as n goes to infinity is basically the value the sequence converges to for large values
hazy pivot
Then for any n, what is a_n
Now answer this
open niche
The limit as n goes to infinity is basically the value the sequence converges to...
no
not for this case
hazy pivot
What the fuck do you mean no
open niche
the limit does not converge
hazy pivot
My brother in christ
open niche
sorry if im mistaken
hazy pivot
Let a_n = that entire expression without the limit
You want the limit of a_n as n goes to inf
Now what is a_n for any n
If x is -1/2
hazy pivot
Good
open niche
but what im asking is
hazy pivot
What do you have after you do that
open niche
0
hardy coral
What's going on here; you can't put the 0 inside the limit
hazy pivot
So every term in the sequence is 0
hazy pivot
What's going on here; you can't put the 0 inside the limit
It's okay, we can in this case
open niche
What's going on here; you can't put the 0 inside the limit
you cant?
hardy coral
No you can't
hazy pivot
What he's talking about makes sense essentially
open niche
its independent to n though?
cause in the previous step
it pulled the 0 out of thelimit
hazy pivot
Anyway
open niche
hazy pivot
So what is the limit as the sequence goes to inf
open niche
So what is the limit as the sequence goes to inf
bro i understood this
but my question is
if we agree that lim (n+1) = +oo
hazy pivot
Yes
open niche
why cant we replace it just like with x
hazy pivot
What
open niche
it just means: grows without bound
ok but what rule makes us unable to replace it
is it because of how we defined the limits
that grow without bound
merry crystal
the definition of a limit to infinity
hazy pivot
Infinity isn't a value your function achieves anywhere
hardy coral
A limit isn't a function though
open niche
the definition of a limit to infinity
so
open niche
A limit isn't a function though
yea was about to say that
so in the definition
saying lim(n+1) = +oo
hazy pivot
A limit isn't a function though
Fair enough
merry crystal
have u seen N-epsilon proofs of limits to infinities?
hardy coral
$$\lim_{x \to p} (f(x)\cdot g(x)) = \lim_{x \to p} f(x) \cdot \lim_{x \to p} g(x)$$
This is only valid if the limits on the right exist (that they are finite values)
open niche
idek how to aask this
rocky lotusBOT
Nel
open niche
**Nel**
was gonna ask this aswell
when saying the lims exist
from my knowledge
we also assume they are converging
thats what it means
including for the identity lim kf = k limf
so how did we pull |2x+1| out in the first place
hardy coral
,w lim_{n to infty} |((n+1)n! (2x+1))/(n!)|
rocky lotusBOT
hazy pivot
,w lim_{n to infty} |((n+1)n! (2x+1))/(n!)|
x is -1/2
open niche
well this disagrees with the page im using
here it says that t for x= -1/2 we have a different case
merry crystal
yes
okay good.
Define a sequence f_n(x) = |2x+1|(n+1). We want to prove the limit is 0
now taking the limit to infinity for f_n(-1/2) means:
for all epsilon>0, there exists N in naturals such that for all n>N,
|f_n(-1/2) - 0| < epsilon
but f_n(-1/2) = 0
So we want to show 0<epsilon which is true by hypothesis
open niche
okay good.
Define a sequence f_n(x) = |2x+1|(n+1). We want to prove the limit i...
yea here i see how its done
when |2x+1| is inside the limit
i was confused when it was outside
hardy coral
x is -1/2
Right sorry, if x = -1/2, this is just 0; but the next line isn't valid in OP's picture
hazy pivot
i was confused when it was outside
For the purposes of a limit of n, 2x+1 is constant
open niche
For the purposes of a limit of n, 2x+1 is constant
how did we get it out of the limit
hardy coral
Your confusion is that what you did here isn't valid
hazy pivot
how did we get it out of the limit
Cuz it's a constant
open niche
Your confusion is that what you did here isn't valid
this isnt what i did
im studying of a page
thats what it said is it not correct?
open niche
Cuz it's a constant
bro
hazy pivot
Your confusion is that what you did here isn't valid
? Why not
hardy coral
Because the limit of the rest doesn't exist (isn't finite)
open niche
here it doesnt exist
open niche
Because the limit of the rest doesn't exist (isn't finite)
so this is a mistake
he should have replaced x inside the limit
hardy coral
Can you show the entire thing?
open niche
Can you show the entire thing?
ok
hazy pivot
Because the limit of the rest doesn't exist (isn't finite)
Oh right yeah that's weird
hardy coral
he should have replaced x inside the limit
Well yes, doing that makes it trivially 0
open niche
hardy coral
Idk why that second line happens
open niche
Well yes, doing that makes it trivially 0
yes ik i wasnt confused about that
i was confused after he got it out of it
open niche
well it says it converges
and i calculated it without the 2x+1 outside
i just put x = -1/2 so we have lim 0
and since that line is equal to the next
then the next line is also 0
but regardless i dont get how he got |2x+1| out of the limit
merry crystal
is it really true that u cant pull out a constant from a limit if the limit diverges?
im not sure if ive heard that one before
hardy coral
Me neither, but it never says that the final expression equals 0
open niche
is it really true that u cant pull out a constant from a limit if the limit dive...
well the theorem that you are refering to
is lim kf = k lim f correct?
on the top of all these identities in my book it says assuming limits of f and g at xo exist
here it says at xo but i think xo can be infinity too
open niche
I'm pretty sure this is correct even if f is a constant function
no thats a different theorem
lim (kf(x)) = k lim f(x)
hardy coral
No? It's a generalization
hardy coral
Yeah, just make either f(x) = k or g(x) = k, for all x
open niche
in my book it shows as seperate case
but is see how its included
in either case the limits must exist for the theorem to hold
merry crystal
either way i dont think pulling out a constant will ever change the result
hardy coral
This definitely looks sketchy to me. You don't need that last line for the value of L to show that it's 0 when x=-1/2 and +inf otherwise
open niche
This definitely looks sketchy to me. You don't need that last line for the value...
yes you dont
hardy coral
Like I have no issue with anything else in that picture
open niche
either way i dont think pulling out a constant will ever change the result
it doesnt stand though
and this was where my issue came in place
if |2x+1| is outside
and we cant get it inside
it looks like we have 0* (+oo)
which seems wrong to even write down
hardy coral
You should talk to your professor about it
hardy coral
Oh
merry crystal
its meaningless to want to pull out the |2x+1|
i think the standard of ratio tests is to get the x dependent term out
open niche
i think 0* lim that goes to infinity
is something wrong to write
and thats why we cant pull out k if the limit diverges
but im not sure
merry crystal
i think 0* lim that goes to infinity
this is what itches me
merry crystal
if it was something tending to 0 times something tending to infinity then id agree, its more nuanced
broken condor
**Nel**
I mean proof of this that i know assumes that both limits exist. Idk how to generalize that if one limit diverges
open niche
I mean proof of this that i know assumes that both limits exist. Idk how to gene...
i got the proof 1 sec lemme see it
hardy coral
I mean proof of this that i know assumes that both limits exist. Idk how to gene...
I don't think it holds in general if one limit diverges
hardy coral
if it was something tending to 0 times something tending to infinity then id agr...
I don't have the same idea of a limit as you do
hardy coral
A limit isn't something that "tends to" a value
open niche
assuming lim = l
hardy coral
A limit is a value, or doesn't exist (is infinite)
open niche
then you set ε = k ε*
and distribute k inside | |
and you got |kf(x) - kl| < ε
well i got a question though
when we say and define a lim f(x) = + oo
what is +oo
is it something treated as a constant
merry crystal
okay i think i get why 0*divergent lim is not defined
open niche
If you define multiplication with infinity, sure
yes but in this case
i dont think we have different infinities in limits
the infinity from the limit going to infinity
is a
idk im getting stuck again idk how to describe my problem
broken condor
i got the proof 1 sec lemme see it
I meant for general f and g. Not if one is constant.
open niche
I meant for general f and g. Not if one is constant.
in that case
hardy coral
Look if you're working within the domain of the real numbers, multiplication is defined as function of two inputs and one output; all must be real numbers
open niche
Look if you're working within the domain of the real numbers, multiplication is ...
well im not really working with anything
im just exploring math
thats why im asking if 0 * (limf) with limf=+oo is something correct to write down
or if its a mistake
hardy coral
I'm just saying that x * inf is usually not defined
broken condor
I wouldnt define it. For same reason that $0 \cdot \sqrt{-1}$ is not defined in real numbers
rocky lotusBOT
casework
merry crystal
iirc u lose some important field properties if u define it and then ur real numbers arent so good lookin anymore
broken condor
Maybe bad analogy
open niche
I'm just saying that x * inf is usually not defined
we dont have that rn
hardy coral
It's what you just said
open niche
no
thats why im asking if 0 * (limf) with limf=+oo is something correct to write down
hardy coral
0 * (lim f) with lim f = +inf
open niche
yes
hardy coral
That's literally the same thing as 0 * (+inf)
open niche
but we cant replace limf with + inf can we
hardy coral
... how else do you interpret that equals sign
broken condor
... how else do you interpret that equals sign
||As abusing notation?||
hardy coral
Probably a different context, but also probably for the same reasons: it's not a real value
hardy coral
||As abusing notation?||
Yes, that's one way to put it
broken condor
I mean limit doesnt exist. Saying its infinite is just one way of putting the sequence is unbounded , no?
open niche
not to actually use infinity in addition and etc
but yea pretty much this is what im asking
hardy coral
lim f(x) = +inf is an abuse of notation. Formally, you always have some assumption that lim f(x) = L, and then you work with either cases: L is a real value or L doesn't exist
hardy coral
I mean it does, you're just taking one abuse of notation into another
open niche
isnt that a mistake
hardy coral
Writing "with lim f = +inf" is a mistake
broken condor
I mean limit doesnt exist. Saying its infinite is just one way of putting the se...
Actually unbounded is wrong word here. For all M there exists N such that for all n > N we have a_n > M
Unbounded could also be n*sin n
open niche
ok
but like
if we have lim f * lim g
can we make it +oo (limg)
because if its just a symbolization it shouldnt be used in multiplication
broken condor
Whenever you have an expression where some term isnt a real number. I would just call the whole expression undefined (if working with real numbers , which im guessing you are)
hardy coral
If lim f = L_f and lim g = L_g, then lim (fg) = lim f * lim g
open niche
thats not what i reffered to
this is something else
ok lemme do it differently
if we have k * limf
can we say thats equal to k * +oo
is it even defined
open niche
1 sec
i searched up
here it says in the extended real number system
we define c * (+oo) = +oo
broken condor
It would be more accurate to say
$$\lim_{n \to \infty} f(n) \to \infty$$
(Some made up notation)
open niche
if c>0
rocky lotusBOT
casework
hardy coral
here it says in the extended real number system
Right, but that requires working with a new system and understanding its rules
open niche
and the guy writing this made a mistake taking |2x+1| out of limit
hardy coral
Yes
open niche
ok thanks
hardy coral
Maybe it's more intuitive to think about convergence and divergence
open niche
appreciate yalls time helping me and providing your knowledge
yea i got it cleared up what i didnt know
broken condor
How does it fail to be a semigroup?
open niche
thank yall again have a good one imma head to sleep
broken condor
How does it fail to be a semigroup?
It seems to be associative (at least for multiplication)
And has identity 1?
broken condor
I mean for multiplication
hardy coral
Same thing
broken condor
we define c * (+oo) = +oo
Doesnt this hold?
Unless having -oo * +oo messes stuff up
Is it at least a magma
hardy coral
0 * inf is left undefined
0 * (0 * inf) =? (0 * 0) * inf
That's not really an equality
broken condor
Oo so its not even a magma
broken condor
Umm seems kinda useless
broken condor
This is interesting
It could be a magma under addition
Or maybe its a monoid
I dont see why it wouldnt be associative as every addition should be defined
vale dockBOT
@open niche Has your question been resolved?
vale dockBOT
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Available help channel!
jaunty gazelle
vale dockBOT
woeful rover
!status
vale dockBOT
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
woeful rover
Alright
jaunty gazelle
trying to find the moments of each one but i forgot how to compute it
woeful rover
Could you send your work over?
jaunty gazelle
i alr got the 0 points
this is what i got for this part
v is the shear and m is the moment equation cuz its supposed to be the integral of shear
jaunty gazelle
shear 0 points are
.75,4,6.25
uhhh i forgot lemme see what that is
oh yeah
yeah i am
its the 0-2 stuff right at the top its messy
woeful rover
Did you account for the moment caused by the dual 8 kN forces in the middle?
woeful rover
it would jsut be 8kn*m right
16 kNm
8 from top and 8 from bottom
Both working together
woeful rover
.75,4,6.25
Those are correct
woeful rover
1.125 for themoment at .75
Correct
woeful rover
but idk how to move to the other ones
You have 2 options
You can either calculate the sum of areas in the shear diagram
Or you can calculate the sum of moments through the forces
From left of the point
woeful rover
It wouldn't
jaunty gazelle
wait
woeful rover
You sum the areas from x = 0 till x = the position that you want to calculate for
jaunty gazelle
yeah
woeful rover
Positive area is positive and negative area is negative
jaunty gazelle
so why couldnt we intergrate the old equations?
the only true arease would be created by the distrubuted loads right
woeful rover
No
All the point loads and distributed loads and the 16 kNm moment causes moments through areas
jaunty gazelle
yeah
woeful rover
I'll make the equations rn hol on
jaunty gazelle
cuz i got the shear ones but ion think they transfer to moment ones well
4 is 0
6.25 is -6.25?
V=dM
jaunty gazelle
k but how exactly did you integrate it?
woeful rover
jaunty gazelle
cuz mines doesnt do it well
woeful rover
I used macaulay's brackets
Well
Kinda
I defined a function called the ramp function
max(0,x)
jaunty gazelle
interesting
woeful rover
Then I defined the derivative of that function as the unit step function
And I needed to define the derivative of the unit step function as the dirac delta function, but desmos can't recognize it
And even so, it won't make a difference
(This function is needed for the 16 kNm moment at x = 4)
I defined the loads using the unit step function and the dirac delta function load
jaunty gazelle
isnt it just 8 tho if you include the 3kNs too
woeful rover
isnt it just 8 tho if you include the 3kNs too
It's a double 8
woeful rover
It's a tandem moment
jaunty gazelle
cuz they technically all acting at 4m
woeful rover
but for the vertical forces tho
I used a 6 instead of double 3s
woeful rover
$\int \langle x \rangle^n \dd x = \begin{cases} \langle x \rangle^{n+1}, n \le 0 \ \ \frac{1}{n+1} \langle x \rangle ^{n+1} , n > 0 \end{cases}; n \in \mathbb{Z}$
rocky lotusBOT
VulcanOne
jaunty gazelle
ye
woeful rover
Keep in mind that if you graph the bending moment diagram and shear force diagram by hand, you'll find it easier than writing their equations
woeful rover
https://www.desmos.com/calculator/qbdiq7ca6g has the graphs you can check from
v(x) and m(x)
Hope this helps
I gotta now do some errands irl
Good luck :)
vale dockBOT
@jaunty gazelle Has your question been resolved?
lusty bay
Vector B = Vector A / magnitude (A)
So, vector B= the unit vector of A.
how isnt the magnitude of A always bigger than the magnitude of B considering that B literally has the length of 1 and "1" is a "part of" A.
cloud coral
Vector B = Vector A / magnitude (A)
So, vector B= the unit vector of A.
how is...
vector B always has magnitude 1
lusty bay
yes
cloud coral
can vector A have a lower magnitude?
lusty bay
i said no
copper stump
well, small/small can also equal big/big
cloud coral
is 0.5 less than 1
lusty bay
yes
cloud coral
can vector A have magnitude 0.5
lusty bay
yes it can, but- if we take "a part of it" how will it ever be 1 magnitude
cloud coral
run the calculations, you have to divide by 0.5
pick A = (0.5, 0) if you want a 2d vector to try
lusty bay
pick A = (0.5, 0) if you want a 2d vector to try
oh... B is (1 , 0)
okay... A can have magnitude less than 1
and still have its unit vector be 1
cloud coral
yeah
lusty bay
so does that mean that every single vector has a unit vector
stark wedge
yes it can, but- if we take "a part of it" how will it ever be 1 magnitude
division does not always make a number smaller
jovial edge
all we care about vecA is its direction, a vector with length 1 and that direction can be called unit vector of vector A
lusty bay
division does not always make a number smaller
if we take a part of a part we have a part thats bigger than a whole part 🤩 i love math
yeah dividing two fractions sometimes gives numbers that are bigger than 1
ig that makes sense
lusty bay
all we care about vecA is its direction, a vector with length 1 and that directi...
mhm
okay ty people
.close
vale dockBOT
vale dockBOT
Available help channel!
dark pond
\sin \left(\frac{\theta }{\cos \left(\pi \right)^{20}}\right)^2=\log _b\left(a\right)
jovial edge
you forgot the $ sign
dark pond
can u re type it im new
jovial edge
$\sin \left(\frac{\theta }{\cos \left(\pi \right)^{20}}\right)^2=\log _b\left(a\right)$
rocky lotusBOT
Alexis_Fx
dark pond
=$\log _b$\left(a\right)
rocky lotusBOT
theonlygardener
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jovial edge
$=\log _b\left(a\right)$ ?
rocky lotusBOT
Alexis_Fx
jovial edge
what's the question?
ashen prawn
$\sin \left(\frac{\theta }{\cos \left(\pi \right)^{20}}\right)^2=\log _b(a)$
rocky lotusBOT
Dhairya
jovial edge
\sin \left(\frac{\theta }{\cos \left(\pi \right)^{20}}\right)^2=\log _b\left(a\r...
what's the question? since I'm seeing 3 variables here
dark pond
sole for a
jovial edge
so express a in term of theta and b ?
dark pond
yes
jovial edge
what have you tried
jovial edge
so you want us to check the ans?
dark pond
yes
jovial edge
yeah that seems correct
dark pond
what should it be for the lim
jovial edge
wdym
dark pond
i need it
jovial edge
!original
vale dockBOT
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
dark pond
jovial edge
so the question is express a in term of theta and b ? and Nothing else
dark pond
yes
jovial edge
you just did it
dark pond
wait now try
\sin \left(\frac{\theta :}{\cos \left(\pi \right)^a}\right)^2=\log _b\left(a\right)
jovial edge
$\sin \left(\frac{\theta :}{\cos \left(\pi \right)^a}\right)^2=\log _b\left(a\right)$
rocky lotusBOT
Alexis_Fx
jovial edge
what
could you send the whole question?
You clearly do more than just express here
jovial edge
what does $(\theta:)$ mean
rocky lotusBOT
Alexis_Fx
dark pond
no :
dark pond
b=e^{\frac{\ln \left(a\right)}{\sin ^2\left(\frac{θ}{\left(-1\right)^a}\right)}}
jovial edge
$b=e^{\frac{\ln \left(a\right)}{\sin ^2\left(\frac{θ}{\left(-1\right)^a}\right)}}$
rocky lotusBOT
Alexis_Fx
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dark pond
wait
b=e^{\frac{\ln \left(a\right)}{\sin ^2\left(\frac{\theta}{\left(-1\right)^a}\right)}}
jovial edge
$b=e^{\frac{\ln \left(a\right)}{\sin ^2\left(\frac{\theta}{\left(-1\right)^a}\right)}}$
rocky lotusBOT
Alexis_Fx
jovial edge
how did you get this
dark pond
pause
jovial edge
isn't $b=a^{-\sin^2{\left((-1)^a\theta\right)}}$
rocky lotusBOT
Alexis_Fx
rocky lotusBOT
Alexis_Fx
dark pond
what if we change
it to sin(b/theta) to th e power of pi
rocky lotusBOT
Alexis_Fx
dark pond
\sin ^2\left(\frac{θ}{\cos ^a\left(\pi \right)}\right)=\frac{\ln \left(a\right)}{\ln \left(b\right)}
\sin ^2\left(\frac{θ}{\cos ^a\left(\pi \right)}\right)\ln \left(b\right)=\frac{\ln \left(a\right)}{\ln \left(b\right)}\ln \left(b\right)
ashen prawn
\sin ^2\left(\frac{θ}{\cos ^a\left(\pi \right)}\right)\ln \left(b\right)=\frac{\...
$\sin ^2\left(\frac{θ}{\cos ^a\left(\pi \right)}\right)\ln \left(b\right)=\frac{\ln \left(a\right)}{\ln \left(b\right)}\ln \left(b\right)$
rocky lotusBOT
Dhairya
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dark pond
\sin ^2\left(\frac{θ}{\left(-1\right)^a}\right)\ln \left(b\right)=\ln \left(a\right)
ashen prawn
ah heck
dark pond
\ln \left(b\right)=\frac{\ln \left(a\right)}{\sin ^2\left(\frac{θ}{\left(-1\right)^a}\right)}
b=e^{\frac{\ln \left(a\right)}{\sin ^2\left(\frac{θ}{\left(-1\right)^a}\right)}}
$b=e^{\frac{\ln \left(a\right)}{\sin ^2\left(\frac{θ}{\left(-1\right)^a}\right)}}$
rocky lotusBOT
theonlygardener
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ashen prawn
$\sin ^2\left(\frac{\theta}{\left(-1\right)^a}\right)\ln (b)=\ln(a)$
dark pond
@jovial edge where u at
rocky lotusBOT
Dhairya
ashen prawn
$b=e^{\frac{\ln \left(a\right)}{\sin ^2\left(\frac{θ}{\left(-1\right)^a}\right)}...
$b=e^{\frac{\ln (a)}{\sin ^2\left(\frac{\theta}{\left(-1\right)^a}\right)}}$
dark pond
wann a be friends
rocky lotusBOT
Dhairya
dark pond
my final q
$\sin \left(\frac{\theta :}{\cos \left(\pi \right)^a}\right)^2=\log _b\left(a\right)=\sum _{n=0}^{\infty }:\sqrt[3]{pi}$
rocky lotusBOT
theonlygardener
jovial edge
huh?
dark pond
ya its in my text book
jovial edge
the last equation doesn't even has n anywhere in side the sum
jovial edge
ya its in my text book
could you send a pic
dark pond
me
thinking
sorry wrong q
$\sin \left(\frac{\theta :}{\cos \left(\pi \right)^{a^n}}\right)^n=\log _b\left(n\right)=\sum _{n=0}^{\infty }:\sqrt[3]{pi}$
rocky lotusBOT
theonlygardener
dark pond
@jovial edge u there
jovial edge
nah
vale dockBOT
Please show the original problem, exactly as it was stated to you, with the entire original context. A picture or screenshot is best. If the original problem is not in English, then post it anyway! The additional context might still be helpful. Do your best to provide a translation.
white adder
this guy is a troll
why even bother
he's just writing whatever into latex
and posting it
dark pond
no
jovial edge
this guy is a troll
I was bored, but now I'm not 
white adder
yes u are
ashen prawn
this guy is a troll
how dare you say that
dark pond
$\sin \left(4θ\right)-\frac{\sqrt{3}}{2}=0,:\forall :0\le :θ<:2\pi$
ashen prawn
try it
rocky lotusBOT
theonlygardener
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white adder
cuz look at what this guy is posting
its nonsense
he's just typing nonsense and wasting everyone's time
white adder
**theonlygardener**
convince me this is not a troll
ashen prawn
convince me this is not a troll
who are you to determine who is a troll btw?
fine leave it
i do not inted to create a lengthy discussion
wheat garnet
can someone pls call with me to help me with some complex numbers and how to write them down in different forms, I am relearning it and am confused about it for some reason
verbal badger
We don't do calls. Ask your question here and someone will help
wheat garnet
alright
this bit confuses me since I believe 3. point should be + pi
or am I wrong? if so pls tell me why, I'm struggling
lyric sundial
If both x and y (i.e. real and imaginary parts, respectively) are negative, it means that your number is in the third quadrant of the complex plane. Hence, the angle will be between π and 3π/2, if you follow the convention of having the principal angle in the interval [0, 2π). Otherwise, the more typical convention is having the principal angle in [-π, π). In this case (which is the one followed by the picture you sent), the third quadrant has angles between -π and -π/2
All in all, it just depends on the convention.
vale dockBOT
@wheat garnet Has your question been resolved?
trail aspen
Help
trail aspen
Definition of addition
light saddle
wdym
river shale
vale dockBOT
@trail aspen Has your question been resolved?
vale dockBOT
turbid valve
hushed carbon
https://tenor.com/view/furina-absolutecinema-furinaabsolutecinema-gif-1488900600...
sorry I though this was #discussion 💀
vale dockBOT
Available help channel!
civic umbra
vale dockBOT
civic umbra
ngl i got no clue how to go about this
ig if you integrate f'(x) from like ax to bx you get the numerator
of the integrand
and then maybe put some bounds into the integral
stark wedge
thats kind of the idea but not quite
stark wedge
try to think of how you could integrate something from a to b specifically to get f(bx) - f(ax)
civic umbra
f'(tx) from a to b?
integrate over t?
oh yeh that works actually
oh and then you integrate that again
over x
from like c to d and c tends to 0 and d tends to infintiy
oohhhhh
oh
fringe niche
looks like you didn't need much help
fringe niche
i didn't do anything but you're welcome
civic umbra
.close
vale dockBOT
fringe niche
you also need to show the integral exists
civic umbra
exists coz integral from 1 to infinity 1/x f(x) exists
so it tends to 0
yeh
and the other way round too
for the tending to 0
fringe niche
why does integral from 0 to 1 f(x)/x exist?
civic umbra
i don't think it necessarily exists
but it doesn't matter coz the integral from 0 to 1 of (f(bx)-f(ax))/x exists
mint meteor
im so confused 😔
civic umbra
of f(bx) and f(ax)
fringe niche
f(x) =f(0) + xf'(0)?
civic umbra
f(x) =f(0) + xf'(0)?
ye
jaunty gazelle
just need a double check
pretty sure its all right but wanna make sure
its moment and shear stuff
next vortex
could you send an image instead of a PDF? 
late creek
don't upload files..
vale dockBOT
sweet sinew
hi, if someone could check these for me that would be much appreciated
jade ivy
In c) you are claiming that there is a real number x such that no matter what number you plug in for y, you always have x = 27^y
vale dockBOT
@sweet sinew Has your question been resolved?
sweet sinew
In c) you are claiming that there is a real number x such that no matter what nu...
oh right
for c then
im assuming this would make it work?
sweet sinew
does the last one not work for the trig one?
jade ivy
No it's fine, I just suspect you are wasting it on Q
and this kind of problem is a bit of a headache when there are multiple correct answers, so you may end up flopping all your answers in a daisy chain when one doesn't work
sweet sinew
hmm true
jade ivy
If you can find a set of solutions that are correct for all of them simultaneously without changing it, then there is no problem
In particular, because of that |x|, the fact that they specified Q > 0 is essentially useless here, and maybe the problem is tight enough that you can't use sin here, or maybe it's okay.
But if you can't find something that works for all of them, you should find another place for that one
jade ivy
I think that is the most natural choice for the sin one yeah
but to be clear, this is just an intuitive hunch, your answer for d was correct too
sweet sinew
the first one js doesnt work now
sweet sinew
but to be clear, this is just an intuitive hunch, your answer for d was correct ...
yea ik
jade ivy
I think it is a good idea to pair 3, 1/3 with that one
because 27^(1/3) = 3
but it definitely doesn't work for (a)
This is kind of an obnoxious problem in my opinion, because you already demonstrated you know how to solve the problem, and now you are spending extra time fiddling with their poorly thought-out choices for options
sweet sinew
This is kind of an obnoxious problem in my opinion, because you already demonstr...
yeah lol i agree
idk tbh i tried swapping a) and c) but c) still wont work
jade ivy
I don't have a great answer for you, this is just a lot of choices combinatorically speaking
It was really mean of them to make all the "for alls" and "there exists" different. You are matching 12 options in disguise here
sweet sinew
It was really mean of them to make all the "for alls" and "there exists" differe...
i still cant seem to get it 😭
do u by chance know the answer
jade ivy
I haven't done the problem no
You should look at the options and find the most restrictive ones, for example in that picture, grouping c with R and xy in S is probably the best possible choice
vale dockBOT
Available help channel!
cyan dock
how many possible ways are there to move from the bottom left corner to the top right corner in a 7 x 5 grid. However 1 move to the left must be included and you cannot leave the grid at any given moment.
if the move to the left is not required the solution simply comes to be 12 c 5.
if you can leave the grid the soultion is 14 c 5, but if we consider faulty cases we have 2 (first left before first right, and first left after all 8 right).
So will the solution simply be 14 c 5 * 7/9??
dapper dagger
how many possible ways are there to move from the bottom left corner to the top ...
do you mean to include the restriction that all moves must be rightward or upward (except for one left)?
cyan dock
do you mean to include the restriction that all moves must be rightward or upwar...
yeah. sorry
dapper dagger
is it also valid to go left, but then immediately right?
cyan dock
is it also valid to go left, but then immediately right?
no because you exit the grid
if u do it before first right
but on any other turn yes
dapper dagger
for example, right-right-up-left-right-...
cyan dock
for example, right-right-up-left-right-...
valid
dapper dagger
sure
then I guess you'd have to account for the occurence of the first rightward move
dapper dagger
I was imagining a similar idea with the easier version of the same problem, wherein the number of moves is fixed, and you choose which of the moves go right, so that the upwards moves are forced
except this time you must include a left, so there are 14 moves total, but the first horizontal move cannot be left
cyan dock
except this time you must include a left, so there are 14 moves total, but the f...
and the last cant be left either right
dapper dagger
hence you take special interest in the first right
cyan dock
since ureach the target before
dapper dagger
yeah, also true, good point
cyan dock
so i thought
that since there are 14 moves
and 5 ups
we can do 14 c 5 to find all combinations
and then multiply by 7/9 to exclude the 2 exceptions for horizontal moves?
dapper dagger
but does multiplying by 7/9 really remove those as options faithfully?
cyan dock
but does multiplying by 7/9 really remove those as options faithfully?
not sure, which is why i wanted to ask this question. to figure out if im removing cases correctly
actually
it might be 14 c 5 * 7
since if we could exit grid it would be * 9
right?
dapper dagger
I agree, yeah. you're choosing which of the 9 remaining moves are left, but it can't be first or last
vale dockBOT
@cyan dock Has your question been resolved?
hardy coral
You must do 8 right moves in total, and somewhere in that you need to include a left move. It must be in between two right moves, so there are 7 choices. Then, you have 9 horizontal moves and 5 vertical moves to mix together, so 14 choose 5. Total is 7 * 14C5, you are correct
vale dockBOT
@cyan dock Has your question been resolved?
vale dockBOT
vale dockBOT
Available help channel!
tulip barn
When I’m writing the domain and range for let’s say y=x^2 would this be right?
Domain: { x | -infinity, infinity, x all real numbers}
Range: { y | 0 < or equal to y < or equal to infinity, y is all real numbers}
Sorry if I typed it really weird
stark wedge
you can write the domain in just one symbol:
R
yes btw you did type it weird. if you can't type the symbols ≥ and ≤ you should replace them with >= and <=
{y in R | y >= 0}
or even {y | y >= 0}
vale dockBOT
vale dockBOT
Available help channel!
fallow lake
each side of the square has 10 cm, and the point E is in the middle of the square ABCD, i need to calculate the full area of the new polygon without the overlap
fallow lake
I thought about rotating the bottom square so that the overlap would form a 5x5 square, and then the overlapping area would be 25.
Total = 200
200 - 25 = 175
Can someone confirm if my reasoning makes sense?
heavy juniper
each side of the square has 10 cm, and the point E is in the middle of the squar...
I'm assuming the angle at which it is hinged, does not affect the area of overlap
fallow lake
the question doesn't bring any angle
heavy juniper
Yeah exactly
heavy juniper
I thought about rotating the bottom square so that the overlap would form a 5x5 ...
I mean, the question is, if you rotate, does the area of overlap remain the same
fallow lake
yeah, thats the point, it didnt says nothing
so i think how this is the only way to answer, it might be correct
heavy juniper
You will need to prove that the area remains the same under rotation action
fallow lake
sorry, but i dont know how do that
i dont know if that helps, but it has 5 alternatives
a) 125 cm²
b) 145 cm²
c) 150 cm²
d) 165 cm²
e) 175 cm²
.closed
.close
vale dockBOT
copper stump
.reopen
vale dockBOT
✅
copper stump
sorry, but i dont know how do that
its easy enough to prove this
Instead of a single square at E, you can draw four squares in its place, and its easy to pove that the shapes are contruent to each other
you can prove in multiple ways that the overlapped areas 1,2,3,4 are from congruent shapes, and have the same area.
.close
vale dockBOT
vale dockBOT
Available help channel!
north scarab
hey I need some help with this, last time I had to leave suddenly but I understand ramonov
north scarab
I'm trying to find the the value of b basically
fringe lava
I'm trying to find the the value of b basically
did you solve f'(b.x) = 8.38 for b.x
tardy orbit
hey I need some help with this, last time I had to leave suddenly but I understa...
what is the derivative of f(x)
north scarab
b.x?
north scarab
what is the derivative of f(x)
one second
north scarab
what is the derivative of f(x)
f'(x)=3(x-2)^2
tardy orbit
you can set f'(x) = 8.38
glass kelp
Yes
rocky lotusBOT
verbal hound
my final answer is 721/100
seems good to me
verbal hound
I just want to confirm with others real quick
its fine
vale dockBOT
vale dockBOT
Available help channel!
dense spoke
how do i find the inverse function of 3sinh(x+2) in terms of logs
stark wedge
do you know the definition of sinh itself in terms of exponentials, and/or the formula for sinh^-1 in terms of logs?
dense spoke
im aware of the definition of sinh itself yes
not the formula for sinh^-1 in terms of logs although
stark wedge
ok well there's two ways here
stark wedge
a) you can either figure out the formula for sinh^-1 (without the 3 and +2) separately & then apply it to your question
b) you can rawdog it for this problem specifically just applying the exponential defn of sinh and going from there
stark wedge
I need to learn trigonometry
!occupied
vale dockBOT
I need to learn trigonometry
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).
tight stag
..
dense spoke
a) you can either figure out the formula for sinh^-1 (without the 3 and +2) sepa...
hmm
ill try rawdogging
see what happens
.close
vale dockBOT
dense spoke
.reopen
vale dockBOT
✅
dense spoke
ok how do i find the inverse in terms of logs for y=sinh^-1(x)
stark wedge
$\frac{e^y-e^{-y}}{2} = x$, solve for $y$.
rocky lotusBOT
Ann
stark wedge
it'll help to multiply both sides by e^y at some point & then recast the thing as a quadratic in e^y.
dense spoke
hmm
dense spoke
$\frac{e^y-e^{-y}}{2} = x$, solve for $y$.
so x=sinh(y) is the same as x=sinh^-1(y)
stark wedge
no, rather we're starting with sinh(y) = x, which is equivalent to y = sinh^-1(x).
dense spoke
why are we doing it the other way?
stark wedge
you want to know how to find sinh^-1(x)
dense spoke
oh my bad
i meant x=sinh^-1(y)
woops
😭
oopsies
y=sinh(x), find inverse function in logarithm, basically
stark wedge
that's... yes that's what i directed you to do
dense spoke
ok so
i got
x=ln(y+-(sqrt(y+1)))
okie dokie wat now
hmmmmm
actually
that gives me an idea
.ckise
.close
frozen tree
could i have help with volumes of revolutions around the y axis
frozen tree
so ive tried putting it in cartesian form
but didnt work out
so i used the parametric method
and i keep ending up in circles
vale dockBOT
@frozen tree Has your question been resolved?
midnight pier
hmmm we eliminate t
y=sin(2t) -> 2sintcost = 2sqrt(1-cos^2 t) cost -> y^2 = 4x^2(1-x^2)
for y e [-1,1] the curve intersetcs the horisontal line y = const at two positive radi
x^2 = (1+- sqrt(1-y^2))/2
now you do R^2(y) r^2(y) and find the volume V = pi integral (1 -1) (R^2(y) - r^2(y)) dy
frozen tree
hmmm we eliminate t
y=sin(2t) -> 2sintcost = 2sqrt(1-cos^2 t) cost -> y^2 = 4x^2...
but doesnt the equation have to be interms of y?
oh wait ye
midnight pier
in other words when we eliminate t we see that (-1 < _ y < - 1, -1 < _ x < _ 1)
this depends on x^2 so the curve is summetric about the y
we intersect a horisontal line with the curve
y e [-1, 1]
solve y^2 = 4x^2(1-x^2) for x^2
x^2 = (1+- sqrt(1-y^2))/2
since 0 < _ sqrt(1-y^2) < _ 1 both are > _ 0 and ordered
i can't type the equal or less than lol
so:
(1-sqrt(1-y^2))/2 lessequal (1+sqrt(1-y^2))/2
horizontal line y = const cuts the curve at
x = +- r(y) +- R(y)
R (1+sqrt(1-y^2))/2
r (1-sqrt(1-y^2))/2
frozen tree
wait sorry how did u hget this? x^2 = (1+- sqrt(1-y^2))/2
midnight pier
quadratic
if we say that x^2 = v for example
v e [0,1]
y^2 = 4v - 4v^2
we make this quadratic equation
4v^2 - 4v + y^2 = 0
and you solve
and get x^2
or v
or whatever you wanna call it
frozen tree
ok
midnight pier
this requires |y| lessequal 1
it's the range of y = sin(2t)
the cross-section area is pi(R^2(y) - r^2(y))
if we replace R and r from above
we can solve it
lemme do it
or?
you can do it as well
then you integrate that solution for -1 1
do you understand where x^2 = (1+- sqrt(1-y^2))/2 comes from now?
frozen tree
um
midnight pier
it's when you do y= sin(2t)
you eliminate t and come to y^2=4x^2(1-x^2)