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obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
pallid panther
Quick question here. Why do they add + pi when calculating the argument? https://www.emathhelp.net/calculators/algebra-2/polar-form-of-a-complex-number-calculator/?i=-1%2Bsqrt(3)i I always thought that to calculate the argument only the arctan of (y/x) was enough
obtuse pebbleBOT
Quick question here. Why do they add + pi when calculating the argument? https:/...
kind hawk
no, depending on what quadrant x+iy is in you have to do a little more than arctan(y/x)
because you want the argument always to be in a certain interval
pallid panther
So I need to basically draw the Z number in order to see on which arc or
timid silo
obtuse pebbleBOT
timid silo
The problem is about finding the shortest distance bw two lines in 3d space. The textbook uses some very particular formula which I find hard to memorise. So I simply took the general point on the line and then started calculating the distance. The general function is as such as plotted on graph desmos. Then I found minima which was higher(1.56) than the answer in the textbook(1.30).
This was the original statement
<@&286206848099549185>
charred plume
There's a key insight that helps with this problem. Imagine the line segment that connects the two lines at their closest points. We're trying to find the length of that line segment, ya?
The insight is: that line segment is perpendicular to both of the original lines.
timid silo
Yes that's the idea of the textbook. But the lines can be seen as a parametric point right? And they have the same parameter t
So can't we use distance formula to find a general expression for distance between the lines using parameter t, and find its minima?
charred plume
Not really -- they're parametrized independently. I'd usually think of the parameters as t1 and t2, or t and s, just to keep them separate.
timid silo
So they've used t both places by error?
But why can't we do it if we do assume the parameter to be same?
charred plume
So doing it that way, you'd have distance as a function of t and s, and do a little multivariable calc to find the minimum. I'm guessing you're not there yet, though?
timid silo
So doing it that way, you'd have distance as a function of t and s, and do a lit...
No. Not yet
charred plume
But why can't we do it if we do assume the parameter to be same?
Imagine two roads, meeting at an overpass. Two cars are driving along (these are our parameterizations). At t = 0, one car might be crossing the bridge, but the other is still miles away.
The cars don't pass the point of nearest approach between the roads at the same moment.
The thing you were doing, treating the two t's as the same, is really finding how close those two particular cars get to each other. But we want to know how close the roads get.
timid silo
Oooo
charred plume
I wouldn't call it an error that they used t in both equations -- that's a common choice. But it is confusing until you get used to it.
timid silo
I'm sorry but I'm still confused
So if we assume them to be identical it's like trying to find the t where they are closes but when they do in fact meet (or are actually closest If skew) then the t aren't identical?
I think I do get it now
Or do i
charred plume
I'm trying to think of a good description. Maybe a simple example would help...
Think about the x axis. We could describe it parametrically as r = (0,0,0) + t(1,0,0).
timid silo
Yes
charred plume
That describes the line... But it also gives a particular journey along the line. At time t=0, it's at the origin. At t=1, it's at (1,0,0). Make sense?
charred plume
Right. It's just helpful to think of it like time... I like to think of a video playing, and t is the time thing at the bottom. The movie plays, things move, and the timestamp at the bottom keeps increasing.
Anyways, let's think of a second line. This one will be given by r = (0,-10,1) + s(0,1,0). I'm putting s here because the parameters really are independent. It's just common to write t everywhere, and expect you to remember that each copy of t is really a separate variable.
charred plume
I'll say what line that is (or you can) but first, do you think of the z axis pointing up?
timid silo
That part is a bit unclear to me
timid silo
I'll say what line that is (or you can) but first, do you think of the z axis po...
Z axis points toward me
Y axis point up
X point right
Is usually how I think
charred plume
Yeah. They wrote t both times, but just pretend they wrote s for the second one, at least for now. In my case, I did write s, so there! :)
Ok, y up, got it.
timid silo
I see yes
charred plume
So we had the x axis as our first line. The second line is parallel to the y axis (so it points up) but it's shifted one unit in the z direction (so it's closer to us than the y axis).
charred plume
I'd call it the z=1 and x=0 line, but you got it.
charred plume
Hooray! :D
So. The second line is parameterized. The parameterization describes a particular journey along the line. At s=0, we're at (0,-10,1), so 10 units down from the xz plane. As s increases, we move upward, and cross the xz plane at s=10.
Now, without thinking about s or t or anything, just the picture of those two lines, where are they closest, and what's the distance?
charred plume
Yep.
timid silo
The line is x/0 = y+10/s= z-1/0
I think at
z=0
It will be parallel
It will be on the xz plane
Right?
That's when it will be closess
The distance is 1
?
Not z
Y=0
charred plume
Ok, not sure about some of the earlier stuff you wrote, but yeah distance is 1, and it's where y = 0 and x = 0... The shortest path between the two lines is directly along the z axis.
timid silo
Yes
charred plume
So, there's two good lessons from this example. First, let's think about s and t. Imagine playing a movie where both s and t start at zero, and both increase at the same rate (so they hit 1, 2, 3 etc at the same moments). This is basically what you were doing by treating the two "t"s as the same variable.
timid silo
Yes
charred plume
We'll have one car on each line, moving according the the parameterizations of each. So the car on the x axis starts right at the origin. The other car, on the second line, starts down 10 units.
timid silo
Rightttt so when the z=1 x=0 car reaches y=0, that time the x axis car will have reached x=10 therefore that min distance of 1 can never be achieved and we will hover somewhere above it
charred plume
Exactly!
charred plume
That's why you got the wrong answer. Wanna see how to get the right one?
timid silo
Yes sure
charred plume
So, the key is: the segment of the z axis that connects the two lines is perpendicular to each line. It has to be.
timid silo
Yes
Yes I think I see this. But one last problem. Will there always be a line perpendicular to two skew lines which passes through both of them?
charred plume
If we didn't know it was the z-axis, we could still find a vector perpendicular to both lines, by taking the cross product of the vectors that point along the lines.
timid silo
charred plume
(in this case we'd cross (1,0,0) with (0,1,0), and that's (0,0,1), pointing in the z direction.
That's for your two lines, right?
charred plume
Yeah. Cross product produces a vector that's perpendicular.
timid silo
But will some line parallel to it pass through both lines
timid silo
If we didn't know it was the z-axis, we could still find a vector perpendicular ...
Ohhh rightt
Wait but
When we speak of vecfors
We take free vectors
They are not bound to coordinates
We can get a line perpendicular
But how do we know it's intersecting
charred plume
True. So... The idea is you take the vectors that point along the lines, and cross them. Call the result V.
timid silo
Yes
charred plume
Then V points in the direction of the shortest line that connects the two lines. But we don't know where that shortest approach actually is. Fortunately, we don't need to find it! If we only need the distance between the lines, there's a shortcut.
The trick is this: divide V by it's length (make it a unit vector)... Pick any point on the first line and dot it with V. Pick any point on the second line and dot it with V. Take the difference of those two dot products, and that's the distance (if its negative just multiply by -1, i.e. take the absolute value)
(that's a lot to take in...)
charred plume
Not with the line, but with literally any single point on the line. For example, plug in t=0 into the equation for the line.
Yeah, exactly.
Try it for the example I gave earlier.
timid silo
Ok wait
I'll try
Omg
I tried it with my original problem
Guess what I got 12/root 59
charred plume
Yeah, it works right? :)
timid silo
Which coincides with my original answer
charred plume
Uh oh, no?
charred plume
Yeah was gonna ask. :)
charred plume
Sure, no prob.
charred plume
I tried, and got 12/root 59 too... I think that's right, and maybe a mistake in the textbook. Lemme check though, could be arithmetic mistake we're both making. The method is fine.
timid silo
I think ur method is correct tho
Because it's exactly as the formula
In the nook
Book
They do the same thing
(a2-a1) is the same as taking the difference u do it at the end they do it right away, then it's essentially just that dot(b1xb2)/cross
And absolute value of the whole thing
charred plume
Yeah. I suspect the thing you did at first worked because they chose parameterizations where, in car terms, the cars happen to pass the closest points simultaneously.
But in general it won't work.
The reason this method does is that dot products with unit vectors measure length along the axis defined by that vector.
timid silo
Yes
Ohhh right
So we take the projection of our perpendicular line with both the skew lines
So we can find the length without finding the actual line
charred plume
Exactly. You find V, then measure how far each line is in the V direction using dot products and subtract.
timid silo
So it indirectly proves that a line perpendicular to two skew lines passes through both?
charred plume
Yeah, you got it. :)
charred plume
Fun question!
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
fluid snow
Why is it that in an isosceles triangle, the two angles formed by the legs and the base are equal?
obtuse pebbleBOT
Why is it that in an isosceles triangle, the two angles formed by the legs and t...
fluid snow
What would be a rigorous explanation?
kindred oasis
timid silo
because the two triangles that u form imaging the height of the isosceles triangle are congruent
kindred oasis
Basically, you divide it in half by drawing the height. You can show that the two triangles that form there ($\Delta$AHC and $\Delta$BHC) are equal, since:
-AC=BC (by definition of isosceles triangle);
-A$\hat{H}$C=B$\hat{H}$C (since CH is the height the angles are both equal to 90°);
-CH=CH (same side in both triangles.
So you just proved that $\Delta$ AHC and $\Delta$ BHC are equal, and so it follows that the angles in A and B are equal
timid silo
yes Andrea is more professional ahahahahah
warm shaleBOT
Andrea276
kindred oasis
yes Andrea is more professional ahahahahah
It took 5 minutes to write though ahah
timid silo
hahahahahah
kindred oasis
**Andrea276**
I think the theorem to prove that the triangles are equal is called S-A-S (side-angle-side) Theorem in the U.S., but I might be wrong
fluid snow
oh
fluid snow
I think the theorem to prove that the triangles are equal is called S-A-S (side-...
so you prove that the two triangles are congruent with that, right?
kindred oasis
Yes
You show that two sides and an angle are equal, and it follows that the triangles are equal
I think in the theorem the sides and angle have to be in the order side-angle-side, but you can generalize it to any two sides and any angle (I don't remember the derivation rn though).
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
stiff grove
I need to solve this equation: 7 cos x + cos 2x = -4
obtuse pebbleBOT
I need to solve this equation: 7 cos x + cos 2x = -4
heavy beacon
yes u can treat it as a quadratic
stiff grove
but I just need to isolate for cos x
stiff grove
yup
heavy beacon
so u can workout wat cosx is
stiff grove
Sorry, I am too dumb, can you elaborate?
stiff grove
by solving for x through the quadratic formula
vestal moss
2cos^2x+7cosx+3=0 this a quadratic equation
2cos²(x)+7cos(x)+3=0, looks confusing otherwise lmfaoo
stiff grove
??
heavy beacon
2cos²(x)+7cos(x)+3=0, looks confusing otherwise lmfaoo
i dont see how u could make it any easier
vestal moss
i dont see how u could make it any easier
nah, it's just that it looked like cos was to the power of 2x
woven badge
make the cosx=w
2w^2+7w+3
=0
then solve for w
@stiff grove
also u can factor it so please dont use qudratic formula
stiff grove
So I got w = (3^2)/2 and w= -8
stiff grove
yeah
heavy beacon
Well u made a mistake somewhere
vestal moss
write down what you did on a piece of paper and send it on here
stiff grove
write down what you did on a piece of paper and send it on here
give me a min
Thats what I have so far
stiff grove
corrected!
heavy beacon
So now u can workout x
stiff grove
alright
heavy beacon
Is there a domain to the question
heavy beacon
Alr u good from here?
stiff grove
Yup thx a lot
woven badge
how do u manipulate this??
obtuse pebbleBOT
how do u manipulate this??
woven badge
to get rid of the x on the bottom
nocturne minnow
Or multiply by the conjugate
woven badge
l’hospitals rule
what is that
nocturne minnow
what is that
If you don't know it, don't worry about it
woven badge
oh
nocturne minnow
Because it means it wouldn't be used
woven badge
yeah it wasnt in the course
idk how to do it tho like i did use desmos but without desmos
woven badge
Because it means it wouldn't be used
u have to rationalize the numerator
nocturne minnow
Conjugate of the numerator
vital goblet
sorry abit of a silly qns, but why as seen circled green x(1/2) = 2x
obtuse pebbleBOT
sorry abit of a silly qns, but why as seen circled green x(1/2) = 2x
vital goblet
this is of product rule
kindred oasis
you should use the chain rule to find the derivative of $(4x+3)^{\frac{1}{2}}$ when you do the product rule, so you also have to multiply by 4
warm shaleBOT
Andrea276
warm shaleBOT
Andrea276
kindred oasis
you should use the chain rule to find the derivative of $(4x+3)^{\frac{1}{2}}$ w...
Where the green circle is, that's where you should be multipling by 4
vital goblet
thank you so much andrea, i'll read up on chain rule
i'm rather new at this
any tips on such question
it's not as simple as applying the product rule?
kindred oasis
i'm rather new at this
If you're learning I can recommend you Khan academy, they have great video explanations
kindred oasis
it's not as simple as applying the product rule?
Yes, but in this case the second function is a composite function, meaning it's "composed" of two different functions, x^1/2 and 4x+3 in this case, so to find the derivative it isn't enough to apply the exponent rule, you need to apply the chain rule
kindred oasis
https://www.khanacademy.org/
Look for the calculus course, you will find explanations of derivatives there
kindred oasis
Yes, but in this case the second function is a composite function, meaning it's ...
It's pretty easy once you understand it, and I haven't explained it really well here
kindred oasis
any tips on such question
You just have to learn the rules and be able to apply them in order
When dealing with derivatives
vital goblet
thank you andrea you've been absolute help and I understand fully what you're trying to explain to me. i just need to brush up on my basics first. 😅
thank you so much once again
.close
obtuse pebbleBOT
kindred oasis
You're welcome
obtuse pebbleBOT
Available help channel!
visual eagle
obtuse pebbleBOT
visual eagle
need help for first part
timid silo
tanx = sinx/cosx
visual eagle
then waht
visual eagle
ur saying to just divide each term of sinx by cosx?
timid silo
Yes, using the two series you can divide them
visual eagle
ur suggesting long division?
timid silo
It asks you to find tan(x) up to the x^5 term, this can have two interpretations:
i. You go up to x^5 on the sin series and x^4 or x^6 on the cos series
ii. You go up to x^5 on the tan series meaning finding every combination of the sin and cos terms that generates x^5 terms, such as x^5/1, (x^7)/(x^2), (x^9)/(x^4) and so on
These are the terms of the tan(x) series up to x^5 if you want to use that to check your work (I found it on google)
visual eagle
ill just do it my own way
i dont think u understand the Q
but ty for ur time
.close
alpine cloak
Tips for practicing and improving on the demonstration of trigonometric identities? what should I do if I get stuck? any place to find problems to practice by level?
obtuse pebbleBOT
Tips for practicing and improving on the demonstration of trigonometric identiti...
alpine cloak
How can I attack the problems?
tardy epoch
Try multiplying by the conjugate
obtuse pebbleBOT
@alpine cloak Has your question been resolved?
grizzled pumice
obtuse pebbleBOT
grizzled pumice
I got this far using synthetic divison and I cant figure out what to do after I got 15x^3-29x^2+6x+8. The only method I know to solving this is factor by grouping, but it doesnt work on this problem
obtuse pebbleBOT
@grizzled pumice Has your question been resolved?
high lily
you could consider the factorisation with hidden values given to you
you can deduce the value of the highlighted box from the leading coefficient
and then use that to perform more division and factorise further
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
haughty basin
Can someone help me
obtuse pebbleBOT
Can someone help me
haughty basin
What does that mean
heavy thicket
Now how do we deal with fractional exponents
haughty basin
Uh ok
haughty basin
Subtract the fractional exponents first
Then it would be 1/(x+5)^-1 right
heavy thicket
Then it would be 1/(x+5)^-1 right
Now what do we do with negative exponents, again
You got it right btw
haughty basin
x+5
haughty basin
I don’t think that’s correct
haughty basin
This is what I got
heavy thicket
This is what I got
Where’s you get x+4 from
obtuse pebbleBOT
@haughty basin Has your question been resolved?
vital verge
(Linear algebra) For the quadratic form q(x) = x^T A x for a symmetric matrix A, (A = SDS^T), why are the principal axes of the graph the columns of S? I thought q(x) was equal to d(S^T x) where d(y) = y^T D y, so why isn't the principal axes the columns of S^T? I'm having trouble understanding what S^T x and its composition with d means in this context.
obtuse pebbleBOT
(Linear algebra) For the quadratic form q(**x**) = x^T A x for a symmetric matr...
vital verge
Example:
If q(x1, x2) = 6x1^2 + 4x1 x2 + 3x^2, then
A = [6 2
2 3]
Diagonalizing A gives
S = 1/sqrt(5) *
[1 2
-2 1]
S^T = 1/sqrt(5) *
[1 -2
2 1]
D = [2 0
0 7]
why are the principal axes the columns of S rather than columns of S^T?
obtuse pebbleBOT
@vital verge Has your question been resolved?
slim cove
lemme take a look
I am going to assume S is orthogonal, i.e. S^T = S^-1, meaning your principal axes are also orthogonal.
If you have q(x) = x^T A x for A = SDS^T, then you have q(x) = x^T (SDS^T) x = (S^T x)^T D (S^T x) = (S^-1 x)^T D (S^-1 x).
Therefore, you can view x as the input, S^-1 x as the conversion from the usual basis to the principal axis basis, and D as the diagonal matrix. Under this interpretation, S would be the conversion back from the principal axis basis to the usual basis, and therefore, the columns of S should be the principal axis vectors (and the entries of D are their corresponding eigenvalues).
hopefully this helps
obtuse pebbleBOT
@vital verge Has your question been resolved?
vital verge
I am going to assume S is orthogonal, i.e. S^T = S^-1, meaning your principal ax...
does S^-1 map the first standard basis vector to the first column of S^-1
slim cove
S^-1 maps the first column of S to the first standard basis vector (because S times the first standard basis vector is equal to the first column of S)
vital verge
S^-1 maps the first column of S to the first standard basis vector (because S ti...
does S map the first standard basis vector to the first column of S
vital verge
ok
vital verge
S^-1 maps the first column of S to the first standard basis vector (because S ti...
then this is true by definition of inverse?
in terms of transformations
vapid geode
obtuse pebbleBOT
vapid geode
does this say
find the cofactor of 7 in the matrix a
astral ivy
I believe it does
icy cape
obtuse pebbleBOT
icy cape
How do I solve this using
SOH, CAH, TOA
cedar lichen
What have you tried
cedar lichen
Good. What then?
cedar lichen
Sure, if youre using A to represent the adjacent side
cedar lichen
A/7 x 65 COS
No
cedar lichen
Why're you multiplying it?
icy cape
COH 65 = A/7
COH 65 = 0.4
0.4 = A/7
A/7 x 7
0.4 x 7
A = 2.8?
cedar lichen
Approximately, yes
cedar lichen
It says round to the nearest hundredth
icy cape
ye
I will calculate again
0.42
0.42 = A/7
A/7 x 7 = A
0.42 x 7 = 2.94
A = 2.94
welp
I made a mistake
it was 2.96
Good learning experience tho
Thank you for the help
much appreciated
.close
obtuse pebbleBOT
cedar lichen
Say .reopen first
icy cape
.reopen
obtuse pebbleBOT
✅
cedar lichen
What have you tried
cedar lichen
You're not given the adjacent side
cedar lichen
The ? is the hypotenuse
icy cape
Alright
Youve got Hyp and the Opp
Toh
Soh*
which is sine
If that is the case, it would be the same calculations.
3/H
Sin 25 = 3/H
Sin 25 =0.42
0.42 = 3/H
3/H x H = 3
0.42 x H = 0.42H
3 = 0.42H
3/0.42 = 7.14
H = 7.14
Right?
wait gtg
.close
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
nova kettle
i need help figuring out the amount it takes to reach a certain percentage. the data i have are "an amount" and "a percentage"
obtuse pebbleBOT
i need help figuring out the amount it takes to reach a certain percentage. the ...
royal basin
thats kind of vague
amount of what? elephant asses in the grand canyon?
@nova kettle
restive ridge
amount of what? elephant asses in the grand canyon?
lmao
wooden mango
i need help figuring out the amount it takes to reach a certain percentage. the ...
You mean convert something into percentage?
You should multiply the total amount to get 100 (even if the number is bigger than 100 itself you can multiply with numbers less than one but if you can't then try dividing)
For the obtained amount whatever u have multiplied in total amount should multiply that too
Let's say that you got 5 bottles of milk and only 2 are filled with milk while three are empty
Find the percentage of bottles that are filled with milk
So u got n/Total amount
2/5
To get a hundred in denominator you multiply it by 20 cuz 20*5=100
Same applies for numerator
2*20
So you get 40%
wooden mango
amount of what? elephant asses in the grand canyon?
Ngl that wasn't funny
wooden mango
i need help figuring out the amount it takes to reach a certain percentage. the ...
If u need an average % (like in exam papers)
royal basin
it was not meant to be funny and i also kind of didnt ask for your opinion
wooden mango
Then add all the obtained marks for every subject
wooden mango
it was not meant to be funny and i also kind of didnt ask for your opinion
I didn't ask if u asked for my opinion or not
Anyways as I was saying
wooden mango
Then add all the obtained marks for every subject
After this
Divide it by the numbers of subjects you have aka. by the NUMBER OF EXAMS you've taken
So if u took 4 exams and got 96, 97, 98 and 99 which adds up to 360 we can divide it by 4 to get 97.5%
obtuse pebbleBOT
@nova kettle Has your question been resolved?
royal basin
"38979 of x"?
so you have that 99.66% of an unknown total equals 38979, and you want to find 99.99% of the same total?
do i understand correctly?
nova kettle
yes
royal basin
right
so why not find the total as you normally would
the total would be 38979/0.9966 with the numbers you provided just now
obtuse pebbleBOT
@nova kettle Has your question been resolved?
fickle dust
does anyone how to make any of the following? pareto chart, histogram , frequency graph, sector graph, stastical display if so dm me please
obtuse pebbleBOT
does anyone how to make any of the following? pareto chart, histogram , frequenc...
@fickle dust Has your question been resolved?
royal basin
we should go to a new channel
and you should restate your question there without any references to some nebulous "x"
fickle dust
<@854268265558704140> Has your question been resolved?
no
fickle turret
• Ask your math question in a clear, concise manner.
drowsy iris
hm
obtuse pebbleBOT
@fickle dust Has your question been resolved?
fickle dust
• Ask your math question in a clear, concise manner.
shutup
don't need yalls help anymore tbh alrdy got it 🤩
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
dapper geode
what integration trick do they use here so I know when to use this for a similar integral?
obtuse pebbleBOT
what integration trick do they use here so I know when to use this for a similar...
heavy thicket
what integration trick do they use here so I know when to use this for a similar...
Is it ok if you can algebraically do the denominator first or no?
dapper geode
wdym?
dapper geode
and the denominator doesn't have zero values so can't split it up
timid silo
what integration trick do they use here so I know when to use this for a similar...
You make df/dx in the numerator so that it can be simplified to df/g(f)
dapper geode
ohhh
timid silo
U can see 4x+1 is d/dx (2x^2 + x - 1)
obtuse pebbleBOT
dapper geode
np
obtuse pebbleBOT
Available help channel!
dark stratus
obtuse pebbleBOT
cedar lichen
This channel is taken
dark stratus
This channel is taken
he has taken already others 2 channels
help 4 and 3 have yours name
cedar lichen
You're literally asking for help in every single occupied channel as well. At least, all the ones I checked
Take one available channel and wait
dark stratus
instead of this i collected n^3 and wrote it to the denominator 1/n^3, in order to have [0/0] form and apply hopital
i don't understand the collection of only n
obtuse pebbleBOT
@dark stratus Has your question been resolved?
upbeat island
how were you collecting terms?
fierce lagoon
Do you know L'Hopital's?
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
hazy rose
How does (72+72sqrt3)/(72sqrt2 + 72sqrt6) reduce to sqrt2/2
obtuse pebbleBOT
How does (72+72sqrt3)/(72sqrt2 + 72sqrt6) reduce to sqrt2/2
solid sun
Try rationalize the denominator.
upbeat island
i'd start with factoring
solid sun
Yeah.
hazy rose
I mean you can get rid of the 72’s, leaving (1+sqrt3)/(sqrt2+sqrt6)
Oh wait
Then you just have to multiply both by sqrt 2
I was over complicating it, thx : )
.close
obtuse pebbleBOT
hazy rose
.reopen
obtuse pebbleBOT
✅
hazy rose
Nvm, that doesn’t do anything as the sqrt 6 on the bottom doesn’t stay the same
idle gust
Try rationalize the denominator.
try ^^
hazy rose
Ok
idle gust
ye first factor the 72
upbeat island
sqrt2 as well
upbeat island
last step to rationalize with much less scary terms
hazy rose
try to rewrite sqrt(6)
Wdym, should I multiply it to get rid of the root?
Or split it into sqrt2 times sqrt3?
hazy rose
Thanks haha, it seems so obvious after the fact
sweet aurora
hello, was wondering how to solve this limit, thanks!
obtuse pebbleBOT
hello, was wondering how to solve this limit, thanks!
sweet aurora
tried applying variable change method, with t = 1/x, t --> 0 (and so x = 1/t) but that lead to nowhere that I could see as useful
my thought is that perhaps there is some theory or trigonometric form to apply with the sinx + cosx part
which would enable me to lead it to like a particular limit form (limite notevole in italian, idk in english) that has an established result
weary delta
This limit doesn’t exist is the problem here
idle gust
you can just split the limit into 2 pieces
weary delta
You can show for 2 subsequences of x they tend to different limits
idle gust
$e^\frac{1}{x} \to 1$ and $x(\sin(x) + \cos(x)) \to -\infty$
warm shaleBOT
VincentBH
sweet aurora
wait so then the lim x(sinx + cosx), x --> -inf
wouldnt that be undetermined too?
as we could further split the limit
idle gust
ohwait maybe im being stupdi
novel knoll
You can show for 2 subsequences of x they tend to different limits
Do what you are told here
This is the way
obsidian isle
$e^\frac{1}{x} \to 1$ and $x(\sin(x) + \cos(x)) \to -\infty$
this isn't right
sweet aurora
into xsinx and xcosx but again that would still be indetermined
weary delta
You could split it iff the limit for each part existed. But the trig part doesn’t exist
sweet aurora
You can show for 2 subsequences of x they tend to different limits
in what sense sorry?
novel knoll
Limit doesn’t exist
If Limit exists then all subsequences also converge to the same limit
So if you show two subsequences converge to smth different
You have shown limit doesn’t exist
sweet aurora
this was an exam question I had so it would be odd for it not to exist unless my prof deliberately set this question up as a "trick" question
novel knoll
It doesn’t exist
sweet aurora
in that case, would you know how to appropriately demonstrate that the limit doesnt exist? 🙏
sage iron
what is the limit of x*sin(x) to infinity
novel knoll
,w plot sin(x)+cos(x)
sage iron
this is basically the same thing right
weary delta
Now doing it rigorously involves figuring out the period of the sinusoid, which is not an enjoyable process, but if rigor is not needed notice how for a subsequence of x values tending to infty (sinx + cosx) is always 0. Another subsequence of 1s will give you -infty. Thus the limit doesn’t exist
warm shaleBOT
novel knoll
Notice you can have this equal to 1 as a subsequence always
And equal to -1 for example
So use that
sweet aurora
im doing all the math in italian so apologies if it takes a moment to translate/understand everything
sweet aurora
Notice you can have this equal to 1 as a subsequence always
always equal to 1 in the case of ?
novel knoll
,w sin(x)+cos(x)=1
warm shaleBOT
weary delta
always equal to 1 in the case of ?
Look at the graph. There is a sequence of x values 2pi apart that keep giving you 1
sweet aurora
ahhh yes that makes sense I see now, so because of it being periodical there is no way to establish a limit result right?
weary delta
Eh, not completely true. e^(-t) sint does have a well defined limit. So periodicity doesn’t always imply the limit not existing
sweet aurora
by subsequences you're referring to xsinx and xcosx limits right?
novel knoll
No everything
sweet aurora
Eh, not completely true. e^(-t) sint does have a well defined limit. So periodic...
ah true so would have to show in the cases specifically of xsinx and xcosx for x->-infinity that they dont exist (?)
novel knoll
You need to look at your whole expression you are taking limit of
Doesn’t sound like you understand that
sweet aurora
sorry still struggling a bit to wrap my head around it
weary delta
Mh, idk how I can explain this if you haven’t encountered the theorem that allows us to evaluate limits on a subsequence instead
sweet aurora
the thing is I may have but due to a) my still being relatively new to this sort of exercise and b) the english --> italian math translations struggling a bit
novel knoll
If Limit exists then all subsequences also converge to the same limit
This is what we are using
weary delta
The idea is that instead of looking at what happens when you smoothly tend to -infty, you look at a sequence of values that tend to -infinity but with discrete steps
sweet aurora
ahh okay I see, is there a specific name for it in english? and by subsequence I need to look up the definition rq
sweet aurora
This is what we are using
^
weary delta
If the limit does exist, it must also be attainable no matter what route you take to it, discrete steps of any size should also give you that limit
sweet aurora
If the limit does exist, it must also be attainable no matter what route you tak...
by discrete steps you mean as in defining a particular interval like (-infinity, -M) ?
novel knoll
Discrete as in integers only for example
sweet aurora
ahhh okay so like choosing the domain type R, N, Z etc
novel knoll
ahh okay I see, is there a specific name for it in english? and by subsequence I...
Don’t think so
novel knoll
ahhh okay so like choosing the domain type R, N, Z etc
Because those are subsequences too, yes
weary delta
I mean as in evaluating the function at discrete values. E.g. $\lim_{x \to \infty} f(x) = L = \lim_{n \to \infty} f(a_n) \text{ where } a_n = n$
warm shaleBOT
Learath2
sweet aurora
Because those are subsequences too, yes
so that would mean that if we can prove that the limit exists in a certain subsequence, then it exists in all according to the theorem?
novel knoll
so that would mean that if we can prove that the limit exists in a certain subse...
All subsequences converges to same limit iff limit exists
All, not one
leaden ibex
So you can find two subsequences that don't coverge
sweet aurora
I mean as in evaluating the function at discrete values. E.g. $\lim_{x \to \inft...
ahh so by checking each subsequence individually
leaden ibex
and conclude the limit doesn't exist
weary delta
So if you can show two subsequences that don’t agree the limit doesn’t exist
sweet aurora
OHHH I finally understand now
then just to be sure
that would mean that say the limits dont match up in N and Z, then the limit doesnt exist?
novel knoll
I mean how is that limit gonna be different
If x->inf then approaching along N and Z
Are the same
But say approaching along all even
And all odd
Are different
weary delta
I don’t know if there is another way to show this. I kinda forgot all the tricks they thought in highschool, there is a possibility that there is some easier way to show this
sweet aurora
yeah I feel like considering this point is worth 2 marks in my system, it implies that there is an "immediate" or quick route that leads to the solution
sweet aurora
**Learath2**
with what is written here, this would be saying that the limit does in fact exist because it exists in the domain R and N right?
novel knoll
Again ALL, not two or whatever
ALL
But that was just an example
Nothing to do with your problem
leaden ibex
you use this fact mostly to disprove that a limit exists
novel knoll
Can also be used to show it exists
weary delta
How rigorous do you need to be? If it’s just multiple choice you can just notice this and answer. Or if you are a math textbook author you can just point out the subsequences and say the “trivial proof” is left to the reader 😄
leaden ibex
Can also be used to show it exists
I mean I did qualify it with mostly 🤷
novel knoll
Yeye
sweet aurora
it's within this context which to me would make it seem (from context of exam paper alone, no math implications) that the prof has and is asking for a specific finite or infinite result
as all the others have results to their limit
weary delta
with what is written here, this would be saying that the limit does in fact exis...
With that I was trying to mathify “If a limit does indeed exist, it doesn’t matter how you take your steps to it”, no matter what subsequence you chose there, as long as the sequence converges to the original limit point you’ll get the same limit.
sweet aurora
How rigorous do you need to be? If it’s just multiple choice you can just notice...
not very rigorous considering it's worth 2 points compared to a max point value of 10 per single question
leaden ibex
This limit doesn't converge because nasty periodic trig functions mess everything up. QED.
sweet aurora
With that I was trying to mathify “If a limit does indeed exist, it doesn’t matt...
but thats the thing I have no clue of what steps are necessary to get to the limit answer
sweet aurora
This limit doesn't converge because nasty periodic trig functions mess everythin...
so you would answer doesnt exist?
novel knoll
I mean just say consider the subsequence 2pi n for n->inf of x * e^(1/x)(sin(x)+cos(x))
For that subsequence you have sin(x)+cos(x)=1 always
And the other part is just finding limit like normal
Pick another subsequence and done
sweet aurora
And the other part is just finding limit like normal
finding the limit like normal of x^(e^(1/x))(sin(x)+cos(x)) you mean?
leaden ibex
,w sinx + cosx = 0
warm shaleBOT
leaden ibex
And the other sequence can be this
leaden ibex
for this subsequence, the value of the limit is clearly 0
sweet aurora
so then the other sequence would be lim (1/4)(4pi*n - pi) for n -> -inf?
leaden ibex
,w limit of x * e^(1/x) as x approaches -infinity
warm shaleBOT
sweet aurora
this one makes sense it's just the other part that am a little lost on 😅
sweet aurora
for this subsequence, the value of the limit is clearly 0
referring to this one
sweet aurora
I mean just say consider the subsequence 2pi n for n->inf of x * e^(1/x)(sin(x)+...
wait sorry but how did the limit form originally look like for that one, would you be referring to this?
novel knoll
Wdym
sweet aurora
like from what I understood, in order to prove this limit, there are two parts to it that need to be proven
sweet aurora
this is one, but which is the other subsequence?
novel knoll
You can pick many subsequence this isn’t the only 2 that work
If you think that
Just chose some convinient ones
sweet aurora
OH so then that would make sense since one is tending to -inf and the other to 0
but in the ones we used, which limit is tending to 0 like how is it structured?
novel knoll
so then the other sequence would be lim (1/4)(4pi*n - pi) for n -> -inf?
What?
And you could also pick a subsequence such that sin(x)+cos(x)=1/2 or 1 or 0.367271818181881 or anything
It doesn’t matter
leaden ibex
Now we have the following fact: If the limit exists, all subsequences converge to the same limit.
This statement is logically equivalent: If at least one subsequence does not converge to the same limit, the limit does not exist.
sweet aurora
ahhh okay I think the picture is clearing a bit, we established that the x*e^(1/x) for x->-infinity is = -infinity
and now we just have to evaluate (sinx + cosx)
leaden ibex
By taking any 2 convenient subsequences, we can show that.
sweet aurora
Now we have the following fact: If the limit exists, all subsequences converge t...
yes I see now that makes sense, I think my confusion is arising from me not being fully certain what a subsequence in english is
novel knoll
,w subsequence math
warm shaleBOT
sweet aurora
ah okay so then for example negative numbers to -infinity are a subsequence of Z ?
novel knoll
If you mean integers then yes
sweet aurora
yes all integers
novel knoll
You wrote numbers
sweet aurora
wait is Z not all integers (negative and positive)?
leaden ibex
numbers generally means elements of the real numbers
novel knoll
Yes but -3.151616711 is not in Z
sweet aurora
like the negative integers are a subsequence of Z
novel knoll
But I would say that its a number
sweet aurora
my fault meant negative integers are a subsequence of Z
novel knoll
If you mean integers then yes
.
sweet aurora
okay perfect thank you
let me quickly reread the previous messages to see if I fully get it now that I understand subsequence concept
so with sinx + cosx = 1, the solutions provided are two, one being 2pi n, the other (1/2)(4pi*n + pi)
for those solutions, we have to evaluate their limits for n -> -infinity, is that what you meant before?
novel knoll
Picking one of them is enough
And limit for whole expression
Not just some part of it
You can’t just ignore x e^(1/x)
sweet aurora
ah okay so then say we chose 2pi n, we do lim 2pi n for n->-infinity = -infinity, which is the same as the limit for the x*e^1/x expression
novel knoll
What
sweet aurora
Picking one of them is enough
like for this once we pick 2pi n, am not fully clear what is done to it afterwards
novel knoll
You then want to evaluate limit of whole expression
Again you can’t just ignore sin(x)+cos(x) or ignore x e^(1/x)
sweet aurora
then the whole expression would be (x e^(1/x))(2pi n)
as we replace sinx + cosx with 2pi n or would that be erroneous?
(tysm btw for all the patience and explanation)
sweet aurora
because it’s a solution for cosx + sinx = 1
novel knoll
So its equal to 1
sweet aurora
like every 2pi n we will have cosx + sinx = 1
yes I realize the error true
= 1 not 2pi n
so then the only part thats confusing me is the role that 2pi n plays in the resolving process
novel knoll
Otherwise we couldn’t say it was equal to 1?
sweet aurora
but then by knowing that how does that help us evaluate the entire limit expression
novel knoll
Now its just equal to lim x e^(1/x) * 1
sweet aurora
OH
wait that’s exactly what I wrote on my exam😭😭😭
because I remembered that cosx + sinx = 1 from memory
novel knoll
Its not equal to 1
sweet aurora
but I didn’t justify it with the 2pi n
novel knoll
Thats them squared
Its only for those specific values
cos^2(x)+sin^2(x)=1
Its not true that cos(x)+sin(x)=1
sweet aurora
ah shoot I see so then I misremembered the identity
so then that would mean that we are evaluating the limit to -infinity for every 2pi n?
novel knoll
What
sweet aurora
like 2pi n is the subsequence then
novel knoll
Consider $\lim_{x \to \infty} x$ as an example. Now consider the discrete limit along $2\pi n$, so $\lim_{n\to \infty} 2\pi n$
warm shaleBOT
ScapeProf
novel knoll
like 2pi n is the subsequence then
I mean Idk if I should be pedantic or not, subsequence requires there was a sequence and a set isn’t a sequence because 1. Elements not ordered and 2. The set might not be countable (such as R)
sweet aurora
ahh okay so what you’re saying is because 2pi n is encompassed in the values of x for x-> infinity, or because they both converge towards infinity, they can be set as equal?
novel knoll
so then that would mean that we are evaluating the limit to -infinity for every ...
Because this makes 0 sense
So I showed you
novel knoll
**ScapeProf**
(Meant along 2pi)
sweet aurora
both limits converge towards infinity, and the domain of 2pi n is encompassed within the domain of x for x->infinity, hence why 2pi n is considered a subsequence of x in this instance is what I understood
I will take some time to translate and reread everything you said, again thank you very much for the help and time @novel knoll 🙏
hazy rose
obtuse pebbleBOT
hazy rose
Why is the inflection derivative taken more than once, is it always necessary to make sure there is only x to the first power?
Excuse the blurry spot on the screen, my camera is a little broken.
violet flame
Why is the inflection derivative taken more than once, is it always necessary to...
WHich is the context?
Slope?
hazy rose
At what value of x does the function have an inflection point
The function being f(x)= 3x^3 + 2x^2 +4x + 1
violet flame
At what value of x does the function have an inflection point
inflection means?
non-differentiable point?
or discontinuos point?
scarlet mango
inflection means?
non-differentiable point?
Inflection means the point where the curve changes its concavity
That’s what I was taught
violet flame
Inflection means the point where the curve changes its concavity
so, basically like x=0 of x2 graph?
scarlet mango
Yup
violet flame
so, that point is a part of the curve
hazy rose
But why twice?
violet flame
then we find a slope when a tangent is drawn
scarlet mango
But why twice?
That is the point of inflection
violet flame
to find the point of changing slope, we find slope of the slope of the line
and equate it to 0
hazy rose
to find the point of changing slope, we find slope of the slope of the line
I think I understand this, but I just need to understand the derivative part
violet flame
I think I understand this, but I just need to understand the derivative part
derivative gives u slope
we need it as a slope and a point
so we do twice
@scarlet mango crt?
scarlet mango
<@860399765883453511> crt?
Means?
hazy rose
So I do it twice to make sure there is only x to the first power
violet flame
am i right?
violet flame
So I do it twice to make sure there is only x to the first power
yea
upbeat island
no
violet flame
cuz if we hv a quad in x, we will hv two points satisying our zero condition
upbeat island
maybe im reading into it too much but just to make sure, we aren't differentiating until we get a single power of x
differentiate twice always, no matter the powers left over
scarlet mango
Ya
violet flame
differentiate twice always, no matter the powers left over
how will we get the point of changing slope?
in a quad we do once since we get it as a linear
cubic we do twice
and so on na
hazy rose
Is this similar to finding extreme points?
upbeat island
yes for this question @ shiva
but in the original it seems like op thought the method was to reduce it to linear always; just clarifying that's not the motivation for the 2nd deriv
violet flame
yes for this question @ shiva
but in the original it seems like op thought the m...
Yh, like the derivatives depend on when we get linear ryt?
cuz, in a quad the max point is the inflection pt
cubic a single point is
and so on...
upbeat island
ok not sure what you mean now?
scarlet mango
cuz, in a quad the max point is the inflection pt
Nope
Quadratic doesn’t have a infection point
violet flame
Quadratic doesn’t have a infection point
why tho?
at max/min point the slope changes?
upbeat island
there is no change in concavity
hazy rose
For example in this question, the derivative is only taken once, although we’re still finding an inflection point (I think) just it’s specifying whether it’s a minima or maxima (I think)
But I could be misunderstanding
violet flame
For example in this question, the derivative is only taken once, although we’re ...
..
scarlet mango
For example in this question, the derivative is only taken once, although we’re ...
Extreme point and inflection point are different things
violet flame
Extreme point and inflection point are different things
yh now i get....
i misunderstood
hazy rose
Ok that makes sense
But why is the first an inflection point (0,0) if it keeps rising?
upbeat island
it's really flat haha, but you should really think of it as a change in concavity
like on the origin we have a down cup on the left and an up cup on the right for a bit
scarlet mango
But why is the first an inflection point (0,0) if it keeps rising?
Before 0 the curve opens downward and after 0 it opens upward
hazy rose
Ok
I think I understand now, thank you all for so much help, I really appreciate your patience. : )
scarlet mango
@hazy rose think of (0,0) in y=sinx , you would understand it better I guess
hazy rose
What do you mean by that, as in the graph?
scarlet mango
Ya
timid silo
Notice how at x<0 the graph opens downward whereas at x>0 the graph opens upward? That’s concavity, and it changes at 0
Well it opens upward for 0<x<2
hazy rose
Ok that makes sense, I think I understand the difference now
K well thanks for all the help
.close
obtuse pebbleBOT
upbeat island
whenever i need a refresher i go back to y = x^3 since it's easy to do derivatives
you should mentally check for each successive derivative, when the value is positive, the prior function is increasing
you can see how the second derivative hints at concavity
https://www.desmos.com/calculator/1uisdwgmuk
obtuse pebbleBOT
Available help channel!
hexed agate
Why isn't ^4√16 = -2? Isn't (-2)^4=16?
obtuse pebbleBOT
Why isn't ^4√16 = -2? Isn't (-2)^4=16?
hexed agate
What do you mean by principal root?
hexed agate
So if x^4 = 16, x=2 only?
left plank
obtuse pebbleBOT
left plank
The question is to evaluate arcsin(sin 10)
Why isn't the answer 10 instead of 3π - 10 ?
heavy thicket
Apparently the solution expanded 3pi into 2pi + pi
2pi is the period so it can be canceled out
left plank
I asked why can't the answer be 10?
Arcsin(sin x) = x
So arcsin(sin 10) =10
heavy thicket
Lemme think
restive ridge
the answer can be 10 + (2k+1)pi, k in Z
heavy thicket
I asked why can't the answer be 10?
Arcsin(sin x) = x
So arcsin(sin 10) =10
They might have substituted sin 3pi-10 into sin 10
left plank
The function never touches y=10
What does this mean?
novel knoll
Arcsin is normally defined as the inverse function to sin x restricted on the interval [-pi/2,pi/2] because otherwise it wont be a function
left plank
Yea
warm shaleBOT
heavy thicket
@left plank you had to recognize the range and domain of arcsin
left plank
So its not inverse at x=10?
But, 10° is within the intervals [-π/2 , π/2] right?
heavy thicket
But, 10° is within the intervals [-π/2 , π/2] right?
The question was using 10 as 10 radians
Not degrees
left plank
<@805485110127624242> you had to recognize the range and domain of arcsin
......wait
How do you know that?
heavy thicket
,w graph arcsin
left plank
Just because there is no ° it's in radians?
warm shaleBOT
novel knoll
No one ever uses degrees so yes
heavy thicket
There is no solution for x=10
You thought it was in degrees
And that’s why it confused you
timid silo
hey im trying to do usub
obtuse pebbleBOT
hey im trying to do usub
heavy thicket
to (u-3)(u^2 +3u + u)?
Alright keep factoring
alpine bison
its done
heavy thicket
and its 1/(u^2 +3u + u)
?
You just used u-3=1
I don’t think that’s the right way
warm canopy
look at the problem openglobe, they have u-3 on the numerator
dopamine keep doing what youre doing
heavy thicket
look at the problem openglobe, they have u-3 on the numerator
Don’t understand how he got 1 from that
timid silo
numerator and denominator simplify
heavy thicket
But you didn’t factor the denominator
warm canopy
the last u should be a 9
timid silo
u will approach 3 ?
warm canopy
yes
timid silo
ah tysm
heavy thicket
Denominator you left unfactored
Oh wait nvm
warm canopy
and now there are no issues with subbing the limit in