sure but you have to remember that when you set it equal to 7pi/6 you’re setting
#help-36
soft zealotBOT
knief
rocky tusk
mhm so you’re confused why they set it equal to -pi/6
rocky tusk
we know that t>=0 yes
rocky tusk
what’s messing you up is that you think time can’t be negative and so you’re ignoring negative values yes
can you do me a favor
solve this hold on
flint summit
okie
rocky tusk
$\frac{\pi}{20}(t-10) = -\frac{\pi}{6}$
soft zealotBOT
knief
rocky tusk
you’ll see that just because the rhs is negative doesn’t mean t is negative
rocky tusk
mhm
flint summit
-20/6
+10
but how do i know when to go positive or negative
when grabbing my pi valkues
rocky tusk
well you need to look for the smallest values of t
flint summit
OH WAIT
HOLD UP
do i just grab the two smallest values of sin -(0.5)
which would be - 6 pi and 6 pi
as long as they give me a positive in the final answer im good
rocky tusk
as long as they give me a positive in the final answer im good
yea so long as t >= 0
rocky tusk
well you can look at the initial point which is when t = 0
and that will give you some insight
and the period as well
rocky tusk
you’re welcome
flint summit
.close
final saddleBOT
final saddleBOT
Available help channel!
rapid orchid
What is the approach to this proof? I have proved that the set S cannot be finite, but I just don't see what to do next.
final saddleBOT
What is the approach to this proof? I have proved that the set S cannot be finit...
@rapid orchid Has your question been resolved?
rapid orchid
<@&286206848099549185>
final saddleBOT
@rapid orchid Has your question been resolved?
final saddleBOT
@rapid orchid Has your question been resolved?
final saddleBOT
@rapid orchid Has your question been resolved?
rain scroll
What properties must a generating set of Q satisfy? Let the generating set be:
${\frac{a_1}{b_1},\frac{a_2}{b_2},...}$
soft zealotBOT
LTHMath
rain scroll
What can you determine about the a_i 's and b_i 's?
final saddleBOT
@rapid orchid Has your question been resolved?
robust horizon
how do I proceed further?
final saddleBOT
how do I proceed further?
forest relic
<@&286206848099549185>
final saddleBOT
<@&286206848099549185>
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).
forest relic
<@&286206848099549185>
im stuck in this shi im sure we can do it using l'hôpital rule tho
empty oriole
uhm is that power to the n
forest relic
uhm is that power to the n
x
empty oriole
what
forest relic
sorry for my bad handwriting
empty oriole
there’s n and x?? 😭😭
empty oriole
why u write ur x’s like they’re n’s
forest relic
why u write ur x’s like they’re n’s
im sorry my handwriting is rlly trash 💀
u want me to write this que again?
empty oriole
also I don’t think l’hopital is applicable since the power is in the way
empty oriole
u want me to write this que again?
yes pls.. and turn it sideways
forest relic
yes pls.. and turn it sideways
ight
forest relic
also I don’t think l’hopital is applicable since the power is in the way
ahh i see
@empty oriole
raven marsh
final saddleBOT
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).
forest relic
oh alright
final saddleBOT
@robust horizon Has your question been resolved?
torpid kernel
@robust horizon have you learned the limit as n goes to infinity of sin(n)/n?
final saddleBOT
final saddleBOT
Available help channel!
small fjord
The beginning of the question says find the exact value. My question is why can't I do what o did in the end ? Take the x-4 out
final saddleBOT
The beginning of the question says find the exact value. My question is why can'...
urban burrow
$\int af(x) dx = a \int f(x) dx$ is only for constant $a$
soft zealotBOT
kaue
urban burrow
yeah
final saddleBOT
final saddleBOT
Available help channel!
tranquil pine
hii
muted prairie
what are you confused about?
tranquil pine
if that's right or wrong
muted prairie
you're 3/4
tranquil pine
the empty set?
undone holly
is it true that if A and B are independent, then Pr(A | A or B) = Pr(A)?
from what i gather, one counter-example was P(A | A or A) > Pr(A) when A is independent of A is there any other counter-example cases?
final saddleBOT
is it true that if A and B are independent, then Pr(A | A or B) = Pr(A)?
from w...
@undone holly Has your question been resolved?
hybrid heath
is it true that if A and B are independent, then Pr(A | A or B) = Pr(A)?
from w...
Your counterexample makes no sense. How can A be independent of itself?
undone holly
Your counterexample makes no sense. How can A be independent of itself?
Oh. That is correct. Guess we're looking for other counter-examples then.
gleaming pendant
maybe just let B be the empty event
hybrid heath
maybe just let B be the empty event
empty event trivially depends on A
gleaming pendant
shouldn't it trivially be independent?
hybrid heath
Let $\emptyset$ be the empty event. How are we defining $P(\emptyset)$?
soft zealotBOT
SWR
gleaming pendant
this is by the axioms always 0
hybrid heath
And what is $P(A\text{ and }\emptyset)$?
soft zealotBOT
SWR
gleaming pendant
0
hybrid heath
Oh that was my mistake. I thought it was just P(A)
Wait. What is $P(A\text{ or }\emptyset)$ then?
soft zealotBOT
SWR
gleaming pendant
P(A)
hybrid heath
Then how is empty event a counter example?
gleaming pendant
P(A | A or {}) = P(A | A)
gleaming pendant
let A and B be omega and its true
hybrid heath
okay it's mostly not true then they're not independent
gleaming pendant
P(omega and omega) = P(omega) * P(omega)
final saddleBOT
@undone holly Has your question been resolved?
hybrid heath
@undone holly what is your remaining question?
undone holly
<@723373144164794399> what is your remaining question?
I didn't quite follow the example. So the counter example was with Pr(A | A or {})?
Pr(A | A or {}) != Pr(A)?
And A and {} are independent?
hybrid heath
And A and {} are independent?
Yes
hybrid heath
Pr(A | A or {}) != Pr(A)?
Correct A or {} would be A, and P(A | A) is 1
undone holly
Correct A or {} would be A, and P(A | A) is 1
Independence condition would require Pr(A | {}) = Pr(A), but Pr(A | {}) is undefined because by definition it just means Pr(A & {}) / Pr({}), and Pr({}) is undefined
hybrid heath
Independence condition would require Pr(A | {}) = Pr(A), but Pr(A | {}) is undef...
Independent event requires either $P(A|B)=P(A)$ $\textbf{or}$ $P(A\cap B)=P(A)P(B)$.
soft zealotBOT
SWR
hybrid heath
Because your first condition came out undefined, try the second one
The second one is usually preferred since it won't require a division
undone holly
Pr(A | B) = Pr(A) is logically equivalent to Pr(A & B) = Pr(A) Pr(B)
hybrid heath
Pr(A | B) = Pr(A) is logically equivalent to Pr(A & B) = Pr(A) Pr(B)
Except when P(B)=0
undone holly
Is that true?
soft zealotBOT
SWR
undone holly
Okay I see what you're saying.
Yeah that makes sense.
Okay, the example checks out. Thanks for the contribution!
Is there a different example where A and B aren't empty sets?
hybrid heath
I proved that it basically never works
It only works if P(A)+P(B)=1
So simply let there exist some third independent event with nonzero probality, and it will never work
undone holly
So simply let there exist some third independent event with nonzero probality, a...
So I know that if A and B are mutually exclusive & jointly exhaustive, then Pr(A) + Pr(B) = 1
But does it follow the other way around?
If Pr(A) + Pr(B) = 1, then A and B are mutually exclusive & jointly exhaustive?
hybrid heath
So I know that if A and B are mutually exclusive & jointly exhaustive, then Pr(A...
No. Consider dice roll.
A= roll 1, 2, 3
B= roll 2, 4, 6
If P(A)+P(B)=1 and A and B are jointly exhaustive, then A and B are mutually exclusive
If P(A)+P(B)=1 and A and B are mutually exhuastive, then they are jointly exhaustive
Any two will prove the third
undone holly
I'm not fully following you here
So simply let there exist some third independent event with nonzero probality, and it will never work```It only works if P(A)+P(B)=1 I understand
So simply let there exist some third independent event with nonzero probality, and it will never work
I do not understand how there being a 3rd independent even with non-0 probability would mean Pr(A) + Pr(B) != 1
hybrid heath
yes
Okay, let's say A is you roll a 1, 2, or 3. B is you roll a 4, and C is you roll a 5 or 6
What are P(A), P(B), P(C), P(A or B) and P(A|A or B)?
undone holly
Pr(A) = 3/6, Pr(B) = 1/6, Pr(C)=2/6, Pr(A or B) = 4/6, Pr(A | A or B) = 3 / 4
I'm trying to figure out how you're reasoning from there being a 3rd independent even with non0 prob, to Pr(A) + Pr(B) != 1
hybrid heath
What are P(A), P(B), P(C), P(A or B) and P(A|A or B)?
What is P(A)+P(B) here?
hybrid heath
is it true that if A and B are independent, then Pr(A | A or B) = Pr(A)?
from w...
Going back to your very first question. I am saying that your first question is false whenever there is some third independent event C
undone holly
What is P(A)+P(B) here?
4/6
hybrid heath
4/6
And this is why the answer to your original question is "No". When A+B!=1, then the answer will always be "No". It is only "Yes" when they sum to 1'
undone holly
I agree that it is only yes if Pr(A) + Pr(B) = 1
Gimme a minute to work this all out in my head
undone holly
And this is why the answer to your original question is "No". When A+B!=1, then ...
Okay, so I'm not clear on hwo you're reasoning to the last bit
that Pr(A) + Pr(B) != 1 if there's a 3rd independent non-0 probablity event
With the dice roll example, you're assuming all dice rolls are equally probable to get your estimate.
Let's just suppose all events are not equally probable, is it still true that Pr(A) + Pr(B) != 1 if there's a 3rd independent non-0 probablity event?
hybrid heath
Let's just suppose all events are not equally probable, is it still true that Pr(A) + Pr(B) != 1 if there's a 3rd independent non-0 probablity event?
The events weren't even equally probably in my example
A, B, and C all had different probabilities
undone holly
Oh I meant the outcomes
hybrid heath
Okay
undone holly
The outcomes were equally probable, the events were not
hybrid heath
Let there be 3 outcomes: A, B, C
And let it have these probabilities
then everythng still applies
Let's just suppose all events are not equally probable, is it still true that Pr(A) + Pr(B) != 1 if there's a 3rd independent non-0 probablity event?
Yes. It's an application of all three axioms
undone holly
then everythng still applies
How?
I'm just trying to picture all this
Oh sorry I misread
Yes it would follow that Pr(A) + Pr(B) = 4/6 != 1
But that's not because there's a 3rd independent non-0 probability
That's because you assigned Pr(A) = 3/6 and Pr(B) = 1/6
You could just assign Pr(C) = 0 and it'd still work
I'm just honestly not understanding the deduction here when you said I am saying that your first question is false whenever there is some third independent event C
hybrid heath
That's because you assigned Pr(A) = 3/6 and Pr(B) = 1/6
You could just assign Pr...
Then how is the second axiom of probability satisfied?
undone holly
Okay, let me consider this. My analysis here may be wrong.
acoustic talon
hi everyone
hybrid heath
hi everyone
!redir
final saddleBOT
hi everyone
This channel is only for on-topic discussion. Please take casual conversation to #discussion or #chill.
undone holly
Then how is the second axiom of probability satisfied?
So I jumped the gun here, and didn't catch that you were assuming A, B, C were the only outcomes. The 2nd axiom would get violated there. So I get your example.
I want to clarify here that there is some third independent event C would just mean Pr(C & ~A & ~B) >0?
hybrid heath
So I jumped the gun here, and didn't catch that you were assuming A, B, C were t...
The heart of my argument is that there must exist some third independent event if P(A)+P(B)!=1.
undone holly
So in other words if Pr(C & ~A & ~B) >0, then P(A)+P(B)!=1?
soft zealotBOT
SWR
hybrid heath
If there exists $C$ such that $\emptyset\ne C\subseteq\neg A\cap \neg B$
soft zealotBOT
SWR
undone holly
Or simply $P(\neg A\cap\neg B)>0$
So I'm trying to track the reasoning here. I'm not sure how to complete it
Pr(A) + Pr(B) = 1 = Pr(sample space)
Pr(A or B) = Pr(A) + Pr(B) - Pr(A & B)
so: Pr(A or B) = 1 - Pr(A & B)
how are we getting to Pr(A or B) = 1?
hybrid heath
We are assuming A and B are independent
hybrid heath
So Pr(A & B) = Pr(A) Pr(B)
Oh sorry I was thinking of disjoint events
hybrid heath
Everything here is still correct except that last line
undone holly
What should the last line be instead
undone holly
You were saying previously that this didn't work if there's a 3rd independent condition. Are you saying that's still true here?
hybrid heath
I'm correcting my statement to "It only works if P(A or B) = 1"
That statement applies to independent or dependent events though
undone holly
Like you mean Pr(A) + Pr(B) = 1 only if Pr(A or B) = 1?
hybrid heath
Nah ignore P(A)+P(B)=1 altogether. I was completely wrong about that
Your original question is only valid when P(A or B) = 1
undone holly
Okay. Why's that?
hybrid heath
Because I confused "independent" with "disjoint"
I incorrectly thought P(A and B) was 0 for indepndent events. But no, it's P(A)P(B). It's 0 for disjoint events.
So it screwed up my math
undone holly
Right, I follow you there.
I'm asking ur reasoning for this Your original question is only valid when P(A or B) = 1?
gentle pond
Stepping in here, I think there's something missing from the original question:
I claim that if A and B are independent, then Pr(A | A or B) = Pr(A) * Pr(A or B)
Your original question was missing the last term: if A and B are independent, then Pr(A | A or B) = Pr(A)?
undone holly
No my question was about the formula Pr(A | A or B) = Pr(A)
gentle pond
Oh ok, then my answer is that your formula is incorrect.
We can run through an example of rolling a 6-sided die
lets say A is the event of rolling an even number, and B is the event of rolling a number greater than 4 (so either a 5 or a 6)
first let's confirm that these are independent events right?
Pr(A) is 1/2.
Pr(A | A or B) translates to "what is the chance that i rolled an even number, given that i rolled either a 2, 4, 5, or 6"? And the answer here is 3/4
so right away we see that Pr(A | A or B) = Pr(A) doesn't hold, because in this example it is "1/2 = 3/4"
But if you inject the context of rolling in the set 2, 4, 5, 6, which is P(A or B), into
Pr(A | A or B) = Pr(A) * P(A or B), you get "1/2 = 3/4 * 4/6" which is true
hybrid heath
I'm asking ur reasoning for this `Your original question is only valid when P(A ...
Because of the mathematical definition of independent events
undone holly
Because of the mathematical definition of independent events
Ohh. Okay. Yeah. Makes sense. I don't see how you're using the definition of independent events here, though.
Pr(A & (A or B)) = Pr(A) simply follows because A is logically equivalent to A & (A or B)
gentle pond
wait what
are you using & "and" and | "given" interchangeably?
what is the exact situation you're describing because I'm not quite understanding
undone holly
are you using & "and" and | "given" interchangeably?
the move from right side of line 2 to right side of line 3
hybrid heath
Ohh. Okay. Yeah. Makes sense. I don't see how you're using the definition of ind...
I'm not. If A and B are independent, then you get this. The answer to your original question is still "no" regardless.
gentle pond
oh i just realized i mixed up my numbers in my previous comments, but yes i believe the graphic SWR posted is correct
if A and B are independent events, it means that the occurence or presence of one does not affect the probability distribution of the other (hence they are independent of each other)
intersection of A with (A or B) would be the same as A simply because (A or B) doesn't do anything to probability distribution of A
undone holly
I'm not. If A and B are independent, then you get this. The answer to your origi...
Okay, so I understood & agreed your previous derivation that it works only if Pr(A or B) = 1. The graphic you're posting here. I understand how it's being derived. But what is its significance? What is it supposed to show?
hybrid heath
Okay, so I understood & agreed your previous derivation that it works only if P...
That the answer to your original question is "No"
undone holly
Sure. How does it show that?
Eh I guess this is going off from the original question in any case
hybrid heath
I'm not. If A and B are independent, then you get this. The answer to your origi...
Because this is the formula for P(A|A or B), and it is not equal to P(A)
undone holly
Eh I mean do we know that though? There could be some unknown simplification to that or whatever - I don't know.
hybrid heath
Eh I mean do we know that though? There could be some unknown simplification to ...
Yes. Simply assume P(A)+P(B)-P(A)P(B)=1, then solve P(B)=P(A)/(1-P(A))
But P(B)=P(A)/(1-P(A)) is not always going to be true
P(A) and P(B) could be anything
It might be true in some situations, but that was not your original question
undone holly
Oh I see what you're saying. You could just assign them possible values, and then not derive any inconsistency
So long as there are some possible values where that's false, then it does not follow
hybrid heath
correct
undone holly
Yeah. I'm on the same page with you. Thanks for answering my questions! Appreciate the effort
.close
final saddleBOT
final saddleBOT
Available help channel!
candid pulsar
i am once again back with a painful ellipse question\\
If TP and TQ are perpendiculars upon the axes from any point T on the ellipse $x^2/a^2 + y^2/b^2 = 4$ and PQ is always normal to fixed concentric ellipse whose equation is $x^2/A^2 + y^2/B^2 = 1$ where A is?
soft zealotBOT
final saddleBOT
i am once again back with a painful ellipse question\\\\
If TP and TQ are perpen...
candid pulsar
options are some form of $\frac{\pm ab^2}{a^2 \pm b^2}$
soft zealotBOT
final saddleBOT
@candid pulsar Has your question been resolved?
final saddleBOT
@candid pulsar Has your question been resolved?
final saddleBOT
@candid pulsar Has your question been resolved?
final saddleBOT
@candid pulsar Has your question been resolved?
final saddleBOT
@candid pulsar Has your question been resolved?
final saddleBOT
@candid pulsar Has your question been resolved?
normal pagoda
What is the formula for the equation of the pair of tangents to a given circle from a fixed point
final saddleBOT
What is the formula for the equation of the pair of tangents to a given circle f...
normal pagoda
Centre: (α, β)
Point: (x1, y1)
rich tide
final saddleBOT
final saddleBOT
Available help channel!
tranquil pine
I have a question.how do I know when and if I should use sin,cos and tan
soft zealotBOT
fickle bronze
nice you wrote down the adjacent, opposite and hypothenus
tranquil pine
Yea
fickle bronze
use them and use the definition of sin, cos , tan to determine which to use
tranquil pine
I only got the formulas
fickle bronze
yeah
tranquil pine
Ye I hv
In q 1 the opposite and adjacent are missing
Should I use the
Cosine?
What should I use
tranquil pine
I can use either cos and sin?
Ohh
You have to find 1 of the three missing values
--> adjacent
--> hypoteneuse
--> opposite
--> if you have sine --> you can find either opposite or hypoteneuse
--> if you have cosine --> you can find eithe r adjacent or hypoteneuse
--> if you have tan --> you can find either opposite or adjacent
Oo ok
remember the acroynm
SOH CAH TOA
sine --> opposite over hypoteneuse
cosine --> adjacent over hypoteneuse
tangent --> opposite over adjacent
the O stands for opposite
A for adjacent
and H for hypoteneuse
Oh yea I know that soh cah toa
Yeah this is what we use in india
tranquil pine
Yeah this is what we use in india
My teacher taught me this as
Papa Beer Pioge - Haan Haan Beta
I don't even know why I know this one
I have to mentally translate the fact that perpendicular = opposite and base = adjacent
Nah bro
No JEE
I only know of SOH CAH TOA
I am not built for that
tranquil pine
I only know of SOH CAH TOA
yeah stick with that. It's a lot more common than what we are talking about
Anyways have a good day
ciao
Bye bye
Thank u both of u
Have a nice say
.close
sick pulsar
I have a question regarding a theorum
final saddleBOT
I have a question regarding a theorum
sick pulsar
I'm doing series and I'm learning about LCT and I'm reading my packet and it doesnt state that bn has to be greater then an
it just says that they both have to be two positive terms
grim elm
Like what?
final saddleBOT
@sick pulsar Has your question been resolved?
final saddleBOT
final saddleBOT
Available help channel!
tranquil pine
final saddleBOT
tranquil pine
I require assistance with the a part of this question
I would like to get my working checked:
$$\lambda_n = \frac{4L}{n}, \quad n \in { 2p + 1 \mid p \in \mathbb{Z}, p \geq 0 }
$$
soft zealotBOT
Edmund Cloudsley (Hello CHAT)
tranquil pine
using this information, we can say that
$$v = f\lambda$$
$$v = 427 \times \frac{4L}{1}$$
I am using $n = 1$ here because I guess they are trying to imply that this is the first time the loud sound is heard
soft zealotBOT
Edmund Cloudsley (Hello CHAT)
tranquil pine
substituting information for length
$$v = 427 \times \frac{4 \times 2 \times 10^{-2}}{1}$$
soft zealotBOT
Edmund Cloudsley (Hello CHAT)
tranquil pine
,w calculate 427 * 4 * 2 * 10^(-2)
soft zealotBOT
tranquil pine
yeah
Why am I getting the speed of sound as 34.16?
when the actual value if around 330 ms
if anyone knows, feel free to ping me
thanks in advance
yeah
but a lot of people here do physics
or have alteast done a little bit of highschool physics
and I think it should be fine to post HS physics questions? Not sure if the mods mind?
What is the value of lambda?
no that's L
but lambda is 4L/n according to the formula booklet where L is the length of the medium :(
let's take your value into account
then the speed which is
$$v = f \lambda$$
it should be
soft zealotBOT
Edmund Cloudsley (Hello CHAT)
tranquil pine
,w calculate 20 * 10^(-2) * 427
soft zealotBOT
tranquil pine
85.4
Still not the speed of sound
why does the answer change everytime
Cause all of these are incorrect
I need the answer to be ~330
,w calculate 20 * 10^(-2) * 4
soft zealotBOT
tranquil pine
,w calculate 427 * 0.800
soft zealotBOT
tranquil pine
okay close
ahhhhhhhhhhhhhhhhhhhhhhhhhh
I just realised my mistake
I took the air column as 2 cm instead of 20 cm
Sorry for wasting your time guys
😭i didnt even notice that LOL
Thanks @brazen relic and @tranquil pine
i didnt do anything
but yw!
Thanks for attempting to help
I mean thanks for spending time on this help channel iykwim
anyways have a good day regardless
.close
final saddleBOT
final saddleBOT
Available help channel!
sinful nexus
sup
sinful nexus
- Please create a list containing the following information.
- Describe the steps in numbered points and describe their purpose/function and how they help make solving easier/solve the question.
- Give a basic description of quadratics and polynomials aswell as factoring and how it helps with quadratics.
- nil
I am in need of this as due to great prior stress it is hard for me to format it all, remember, and then perform synthesis to pursue the answer; therefore, it would be of great assistance to me if someone were to do this for me.
cyan kayak
I'm not certain what you are asking for here.
sinful nexus
I'm not certain what you are asking for here.
Please describe the steps taken to solve quadratics via factoring and how each step is useful in making the problem easier to solve and why it is used. I also request that you number the steps as I did in my prompt.
I need someone to do this as due to prior stress with math and this problem aswell as very large time expendature it became hard mentally for me to think about this clearly; therefore, it would be greatly helpful to me if you were to format the questions and explanation in such a manner.
cyan kayak
well, I don't currently have time to give you an entire lecture over discord. May I instead recommend Khan Academy?
sinful nexus
well, I don't currently have time to give you an entire lecture over discord. M...
Alright, thank you for the suggestion of utilizing readily available documentation that may list the solutions as described per my prompt as I had not considered this as I've just returned to the problem after a while.
thank you
.close
final saddleBOT
final saddleBOT
Available help channel!
manic tusk
Can the following function be thought of as a polynomial or expressed as one (for computer science)? $2n^2 \cdot \log_2(n)$
final saddleBOT
Can the following function be thought of as a polynomial or expressed as one (fo...
soft zealotBOT
shadox
final saddleBOT
@manic tusk Has your question been resolved?
manic tusk
<@&286206848099549185>
Because it's a stupid question or because it's too complex xd?
manic tusk
Taylor series?
Haven't learned about that yet. But afaik, logarithms can't be polynomials, but my TA says otherwise.
manic tusk
Where are you from, country?
manic tusk
Like an assistant for the professor xd
tulip root
Haven't learned about that yet. But afaik, logarithms can't be polynomials, but ...
Like if you want you can even expand ln(x) near x=0 using puiseux series
manic tusk
But say, $2n^2 \log_2(n) < n^3$ , since it's a polynomial of 3'rd degree, we can say there's 3 solutions/intersections. Specifically, n = 0,2,4
soft zealotBOT
shadox
manic tusk
We haven't learned about that yet. I'm kind of fresh outta high school
tulip root
Though I study in Europe temporarily, they have very unique tutoring systems called TA
Puiseux series?
That’s just Taylor though
manic tusk
I just wanted to know if we can treat the function involving log as a polynomial to conclude there's 3 solutions
manic tusk
That’s just Taylor though
Haven't learned that neither D:
long flame
**shadox**
Correct me if I am wrong here
I don't think this is always true
There will be 3 solutions where r will be equal
It will like from 0 to2 and from 4 to infinity
manic tusk
Yes indeed
manic tusk
Yes. But can I treat 2n^2 log2n as a polynomial compared with n^3?
manic tusk
Hm
tulip root
What’s the problem
manic tusk
Say I put them equal to each other, then tried some sample values from 1,2,3,4,5
long flame
Okay.....
manic tusk
Then concluded with, since n^3 leads, from n > 4. n^3 will dominate the other function?
tulip root
What’s function bc
long flame
What’s the problem
We trying to figure that out
long flame
Then concluded with, since n^3 leads, from n > 4. n^3 will dominate the other fu...
Yes
For n>4 n^3 leads
tulip root
Like I have been confusing what’s expressed to that
manic tusk
The problem is: Determine a problem size $\Tilde{n}$ such that for all $n > \Tilde{n}$, $f(n) = 2n^2 \cdot \log_2(n)$ grows slower than $n^3$
tulip root
Asymptotic analysis
long flame
.-.
soft zealotBOT
shadox
long flame
Differentiate
tulip root
Can’t you just make it into a fraction
And then lhopital to infinity
That’s basically what lhopital good for
This is straightforward with lhopital
manic tusk
uhm i don't know, since we use base 2 log
soft zealotBOT
meow
tulip root
Basic log identity
manic tusk
But $\lim _{x\to \infty }\left(\frac{2log_2\left(x\right)}{x}\right) = 0$
soft zealotBOT
shadox
long flame
$\frac{\2ln(n)}{\ln2*n}$
soft zealotBOT
Lex
$\frac{\2ln(n)}{\ln2*n}$
```Compilation error:```! Undefined control sequence.
<argument> \2
ln(n)
l.49 $\frac{\2ln(n)}{\ln2*n}
$
The control sequence at the end of the top line
of your error message was never \def'ed. If you have
misspelled it (e.g., `\hobx'), type `I' and the correct
spelling (e.g., `I\hbox'). Otherwise just continue,
and I'll forget about whatever was undefined.```
manic tusk
x = n same
tulip root
You are comparing with n^3
manic tusk
Simplifying gives 2log_2(n) = n
tulip root
2n^2ln(n)/ln2
long flame
2n^2ln(n)/ln2
That's one of the function
manic tusk
It evaluates to 0
manic tusk
But it's the same xd?
$\lim _{n\to \infty }\left(\frac{2n^2\cdot :ln\left(n\right)}{n^3ln\left(2\right)}\right) = 0$
soft zealotBOT
shadox
manic tusk
$\lim _{n\to \infty }\left(\frac{2log_2\left(n\right)}{n}\right) = \lim _{n\to \infty }\left(\frac{2n^2\cdot ln\left(n\right)}{n^3ln\left(2\right)}\right) = 0$
long flame
You are right
soft zealotBOT
shadox
manic tusk
xd im dumb rn
what does that evaluation tell us? that n is the dominating term, or n^3 * ln(2) ?
tulip root
Then if take reciprocal it goes to infinity
manic tusk
Yes, but I need to show that the solutions I got, i.e., n = 2 and n = 4 are the only solutions available. So when n > 4, n^3 starts to dominate
I tried to input some sample values, like 1, 2 etc... and the only time those two are equal to each other is when n = 2,4
long flame
And 0
manic tusk
Yes, but I don't think we count is as a solution because it's neither negative nor positive
tulip root
Let them equal
manic tusk
I'm not sure though
long flame
The functions are equal at 3 points
tulip root
N<2
manic tusk
tulip root
N>4
3 points it’s easy
Since always faster
Than there must be two points where before after it will be faster for n^3
manic tusk
Simplifying untill I cant, so that fg becomes n = 2^n/2, then -n on both sides
manic tusk
The thing is
I can't solve, 2n^2*log2(n) = n^3
Even if I put 2n^2*log2(n) - n^3 = 0
But can I say, since they're both polynomials, the leading degree dominates, which is n^3, so there's 3 solutions?
So for n = 2, n = 4, they're equal, but after n > 4, n^3 dominates?
long flame
We can just simplify the equation to an extent
To give n^2 = 2^n
I mean then it's just hit an trial
long flame
So for n = 2, n = 4, they're equal, but after n > 4, n^3 dominates?
Yes
Keyword being always dominates
Honestly
I would've just differentiated
But your choice @manic tusk
manic tusk
I don't know 
tulip root
I will answer 2 and 4 though
tulip root
Really?
long flame
And I don't think we resolved his issue
tulip root
Tbh i am still confused what’s asked
tulip root
3
I got it
So there’s only one point that is the same possible growth rate
Since n^3 dominates
Symmetrically speaking
We have 3 roots
manic tusk
Yes, but how do I show/prove that there's only 3 roots
tulip root
So basically they are either decelerating or accelerating
…
N^3 is dominating I guess
Using delta epsilon to show that it approaches to 0 or infinity
Which is stupid
I don’t know
manic tusk
It's the only solution I've got. Because I haven't learned about Taylor series etc.
$\lim _{n\to \infty :}\left(\frac{2\cdot :::log_2\left(n\right)}{n^3}\right) = 0$ and $\lim _{n\to \infty }\left(\frac{n^3}{2\cdot :log_2\left(n\right)}\right) = \infty $
soft zealotBOT
shadox
final saddleBOT
final saddleBOT
Available help channel!
unreal hornet
can someone explain why the first image is true but also why the answer d doesnt follow this?
final saddleBOT
can someone explain why the first image is true but also why the answer d doesnt...
final saddleBOT
@unreal hornet Has your question been resolved?
final saddleBOT
@unreal hornet Has your question been resolved?
final saddleBOT
@unreal hornet Has your question been resolved?
plucky minnow
i dont how how to do this
final saddleBOT
i dont how how to do this
plucky minnow
i have tried all day
final saddleBOT
Show your work, and if possible, explain where you are stuck.
plucky minnow
the only way i think i can do it is through the quadratic formula
but i keep getting diff answers everytime
final saddleBOT
Show your work, and if possible, explain where you are stuck.
hybrid mantle
can you use a calculator
flat saddle
but i keep getting diff answers everytime
if you don't show your work, there is no way to help you
final saddleBOT
plucky minnow
.close
final saddleBOT
Available help channel!
river oasis
is this a trick question
tulip coyote
river oasis
like isnt it 16
urban burrow
you need to know $\sqrt[n]{x} = x^{\frac{1}{n}}$
urban burrow
like isnt it 16
it's taking the square root twice
soft zealotBOT
kaue
urban burrow
no, like you take the square root of 256, and then you take the square root again
river oasis
its 256^1/4
urban burrow
yeah
urban burrow
are there alternatives or you need to write down the answer?
soft zealotBOT
kaue
urban burrow
yeah
river oasis
thats correct?
urban burrow
yes
urban burrow
if by 3√ you mean cube root then yeah
final saddleBOT
final saddleBOT
Available help channel!
hollow shard
yo if i have xy+zf
final saddleBOT
yo if i have xy+zf
hollow shard
how would i solve for x/f?
verbal viper
hollow shard
is that not possible
ashen shadow
yo if i have xy+zf
You just have an expression not a relation , you can't do anything with it
hollow shard
You just have an expression not a relation , you can't do anything with it
2x=3y
same thing?
visual marlin
what do u mean you're not given anything bro
hollow shard
like solve for x/y?
hollow shard
what do u mean you're not given anything bro
like i’m just thinking about it
is there a way to solve for x/y
visual marlin
are u given an equation
visual marlin
^
final saddleBOT
final saddleBOT
Available help channel!
cerulean jackal
hello
cerulean jackal
im integrating this by substitution
i used 1-x^2 as my substitution
let that = u
differentiated it
got -2x
du/dx = -2x
dx = du/(-2x)
so now im trying to solve
tranquil pine
so now im trying to solve
I got an idea
cerulean jackal
i have the integral of 2x/rt(u) - du/2x
2x's cancel
and im left with the integral of u^1/2 - du
tranquil pine
2x's cancel
after the 1st 3 steps yeah
cerulean jackal
now that i have this
if i integrate it
the answer is wrong
so where did i go wrong in my method
blissful meadow
and im left with the integral of u^1/2 - du
It should be -u^{-1/2}
tranquil pine
if i integrate it
u sure its wrong with the integration?
I think u went wrong with ur substitution
cerulean jackal
It should be -u^{-1/2}
why this?
so @blissful meadow i have the integral of 2x/rt(u) - du/2x
this line is correct?
blissful meadow
$\frac{1}{\sqrt{u}} = u^{\frac{-1}{2}}$
soft zealotBOT
Azyrashacorki
cerulean jackal
**Azyrashacorki**
Oh shi
i see where i messed up
when i canceled the 2x's
i didn't treat it as 1/sqrt(u) just sqrt(u)
alr
so
cerulean jackal
It should be -u^{-1/2}
where did this other minus sign come from
i understand u^(-1/2)
slender rampart
soft zealotBOT
Azyrashacorki
cerulean jackal
You had a - out front
o so
you pull the minus sign from the du after you cancel the 2x's
and put it infront of u?
blissful meadow
I mean you can put it wherever
You can bring it outside the integral if you want
cerulean jackal
I mean you can put it wherever
alr
i understand now
so now i integrate that i get
-u^(1/2)/(1/2)
so i get
-2u^(1/2)
- c
then just plug back in u
so
-2 sqrt(1-x^2) + c
?
final saddleBOT
@cerulean jackal Has your question been resolved?
unique estuary
yo can someone help me, im not sure if my answer is correct
final saddleBOT
yo can someone help me, im not sure if my answer is correct
tranquil pine
what's your answer/?
peak parrot
yo can someone help me, im not sure if my answer is correct
-2 and 0
unique estuary
what's your answer/?
(0,-4) and (-2,0)
peak parrot
(0,-4) and (-2,0)
its right
tranquil pine
the integrate it from 0 to 2
unique estuary
yeah i got 4
unique estuary
so basically my teacher is wrong
tranquil pine
unique estuary
im confused
tranquil pine
wait a second isn't it 19/2
why / how?
unique estuary
yeah
visual marlin
ok
unique estuary
integrate and get (x^4)/4 +x^3 -4x
for the interval from 0 to 1 i got -11/4
for the interval from 1 to 2 i got 27/4
tranquil pine
why are you integrating them separately?
long flame
why are you integrating them separately?
Because it says area
tranquil pine
oh yeah
tranquil pine
So take abs of the negative area and then add to the positive
unique estuary
why thank you
long flame
@unique estuary
unique estuary
yeah?
unique estuary
so it is 19/2?
long flame
Yup
long flame
Good job!
unique estuary
!close
long flame
.close
final saddleBOT
final saddleBOT
Available help channel!
native valve
yo wtf
tranquil pine
PD is coming 2.5
!show
final saddleBOT
PD is coming 2.5
Show your work, and if possible, explain where you are stuck.
mortal berry
!show
final saddleBOT
Show your work, and if possible, explain where you are stuck.
mortal berry
@native valve
native valve
hey
mortal berry
did you read the warnings?
native valve
why?
robust fulcrum
show what you tried
native valve
did you read the warnings?
hey mate, theres no way i can click an image with my broken cam, but i swear i solved it
i proved them similar
but pd is coming 2.5
tranquil pine
mortal berry
hey mate, theres no way i can click an image with my broken cam, but i swear i s...
What ?
mortal berry
@native valve But shouldn't you be the main character since you're a giga chad?
native valve
<@994574861886308473> But shouldn't you be the main character since you're a gi...
real
native valve
<@994574861886308473> But shouldn't you be the main character since you're a gi...
help bro
mortal berry
Ok, this is my help then
Don't let others intervene
@native valve
Hey
where did you go?
native valve
hey
native valve
Why you get 2.5? What did you do?
see
DB/AC=DP/PA=PB/PC
DP/3=6/2.4
DP=18/2.4
DP=2.5
OH
WTF
final saddleBOT
Available help channel!
buoyant holly
my math teacher never taught this and i have an assignment due at midnight, pls help
final saddleBOT
my math teacher never taught this and i have an assignment due at midnight, pls ...
buoyant holly
usually he links videos but he did not for like 1/3 of the questions
i feel like it shouldnt be hard but i dont even know where to start
visual girder
Did he teach you about linear approximations?
buoyant holly
that is in a different homework assignment i think
er no it is this one
but i find most of this unit very difficult
visual girder
Yeah cuz I think the intended way is to use a linear approximation
buoyant holly
what do i plug in for f(t)
visual girder
Just 2 cuz 2 grams is the weight, and f(t) represents the weight
visual girder
Yes, but you have to be careful with the language. That equation represents rate of grams gained per second, which is -0.4. Rate of grams lost per lost per second is 0.4
buoyant holly
i see
so how do i estimate t=3
ohhhhh with -0.4
so at t=3 my weight would be 1.6?
and i plug that in for f(t)?
with the final answer being -17.92/60?
visual girder
That would imply a negative weight, so something went wrong
buoyant holly
but is asking for weight
visual girder
so at t=3 my weight would be 1.6?
I meant this one
buoyant holly
oh
oh it is
thank you
do u still have time by chance
i have a few more questions...
he gave us this and i have never seen it before
kind pelican
calculus
buoyant holly
yes
buoyant holly
:3
kind pelican
like whether the derivative is positive or negative
buoyant holly
i see
kind pelican
whether the second derivative is positive or negative
like this lol
but just with derivatives
buoyant holly
so there is no way for me to graph this is there
kind pelican
there's no way to get an exact graph
but with the sign graphs you should be able to tell generally how the function looks
buoyant holly
but everything i need is in the diagram?
kind pelican
in terms of how it curves
buoyant holly
i see
kind pelican
but everything i need is in the diagram?
yes
for example g increases when the first derivative is positive
so there's your answer to A
buoyant holly
so the t values where g is decreasing is (-oo,0)?
buoyant holly
i think i have to input it in a weird format tho
buoyant holly
but 0-5 is positive no?
buoyant holly
why doesnt it extend to -infinity
kind pelican
take this number line
buoyant holly
wha
kind pelican
it's saying for t values between -5 and 5, the G'(T) value is negative
buoyant holly
yes
buoyant holly
so if g is increasing/decreasing we talk abt the derivitave?
kind pelican
yes because it measures rate of change
buoyant holly
second deri is like acceleration?
kind pelican
the value of g(t) increases when the derivative is positive, because that means we have a positive rate of cahnge
buoyant holly
i see i see
kind pelican
it'll be important for the concave up convave down problems
buoyant holly
so g increasing would be (-oo,-5)U(5,oo))
kind pelican
yes
buoyant holly
those are in the question too
kind pelican
mhm
kind pelican
buoyant holly
m
kind pelican
it's when the first derivative is DECREASING
aka when the second derivative is negative (since the second derivative is the rate of change of the first derivative)
buoyant holly
so for that i look at second deri
kind pelican
yep
buoyant holly
so its concave down at (-5,0)U(5,oo)?
kind pelican
yuh
buoyant holly
excellent
thanks
how do i find relative minimum
is that between - and positive
of first deri?
because thats when it curves
kind pelican
there's gonna be a minimum when
-
the first derivative is zero (because that means the slope is 0, which is what you need at a minimum)
-
the graph is concave UP (if it's concave down then it's a MAXIMUM)
robust fulcrum
Do you mean the relative minima? f'x = 0 AND f"x > 0
kind pelican
Do you mean the relative minima? f'x = 0 AND f"x > 0
minima is plural of minimum so lowkey that diagram is wrong
buoyant holly
the sign diagram
from positive to negative first deri is concave down aka maximum, correct?
buoyant holly
so maximum would be -5 and min would be 5
kind pelican
well
robust fulcrum
If the graph is ↓ then ↑ →→ minimum
↑ then ↓ →→ maximum
buoyant holly
dawg what
robust fulcrum
sorry I'm just playing osu!
buoyant holly
hahaha mayuu can u help someone else pls