you can find this constant upper bound by restricting x to lie in an interval centered at 2 where all numbers in this interval lie in a close distance from 2 so let $|x-2|<1$ this restricts x and keeps the close distance from 2
#help-13
dim tiger
so $|x+2|<5$ then $|x^2-4|=|x-2||x+2|<5|x-2|=5\delta=\epsilon$ then $\delta=\frac{\epsilon}{5}$
wraith daggerBOT
calculus is fun
dim tiger
so here you have 2 restrictions on $|x-2|$ which are $|x-2|<1$ and $|x-2|<\frac{\epsilon}{5}$ which led to $\delta=1$ and $\delta=\frac{\epsilon}{5}$ came from trying to bound $|x+2|$ by restricting the domain of x
wraith daggerBOT
calculus is fun
dim tiger
so to satisfy both inequalities you should choose $\delta$ to be the minimum of 1 and $\frac{\epsilon}{5}$ which is written as $\delta =min{1,\frac{\epsilon}{5}}$
wraith daggerBOT
calculus is fun
dim tiger
you didnt continue part 4
cedar kilnBOT
cedar kilnBOT
Available help channel!
bold lagoon
This is a question purely out of curiosity.
Imagine the functions f(x) and g(x) shown below.
How can I prove that for any value k, the graphs of y = f(x) and y = g(x) intersect exactly and only at point (0 , k) ?
wraith daggerBOT
themathboi #2137
junior dome
solve them simultaneously.
bold lagoon
solve them simultaneously.
Could you elaborate?
bold lagoon
just add them
This may sound dumb, but how?
junior dome
do you know how to solve system of equations?
bold lagoon
do you know how to solve system of equations?
Yes, I do actually
junior dome
then just solve
cedar kilnBOT
cedar kilnBOT
Available help channel!
copper oriole
.reopen
white mulch
3a+2b=19
4a+3b=27
cedar kilnBOT
3a+2b=19
4a+3b=27
copper oriole
im trying to type it out
white mulch
3a+2b=19
4a+3b=27
copper oriole
thanks for that
copper oriole
500 times 5 with a protection of 40% adding 25 health per second along with another 75% damage resistance in armor
can you solve that?
maybe this is better
Khei Lapidarist + S.H. + Silver Guard (45% res/1.82x mult)
HP: 486 (+219)
Regen: 5.83/s (+3.03/s)
and add 5x with that
copper oriole
it was closed
candid lotus
#❓how-to-get-help #rules @copper oriole
copper oriole
this was an available math help channel
this was on math help avaialble, i followed these guidlines.
candid lotus
yeah thats the other guy who shouldve opened a new one
copper oriole
since there was no help0
kindred storm
You typed .reopen. You should do that when it was already your channel. When it wasn't and the channel is open, just start off by asking your question.
copper oriole
it didnt specify that so i thought it became mine if i opened the channel
candid lotus
.reopen is when your channel was closed
you just have to type your question and it will open a channel to your name @copper oriole
dont need to type .reopen
copper oriole
while were here can you solve this question? im trying to become the tankiest class in a game?
Khei Lapidarist + S.H. + Silver Guard (45% res/1.82x mult)
HP: 486 (+219)
Regen: 5.83/s (+3.03/s)
and add 5x with that
kindred storm
I don't know the rules of the game you're talking about.
copper oriole
carp i didnt predict that
ill never know the tankiest class in the game unless i ask someone from the game im playing..
kindred storm
Apparently the 45% res adds to your hit points.
candid lotus
what game is it
copper oriole
rogue lineage roblox
roguelike perma death game
the 45% res does add to the hitpoints
i have one instance of someone solving an equation
but its not directly the same question im looking for, but maybe it will help. since it has the same format
you would have to take the Khei Lapidarist + Sigil Helm + Silverguard, divide by 1.25, and add 40 for grindstone, which returns 428.8. Multiplying by 9 for cameo returns 3859.2.
cedar kilnBOT
@white mulch Has your question been resolved?
copper oriole
yes it just had been solved. i just solved it
now how to close this bad boy
.close
damn. life's just never that easy huh?
.close.
i have no value as a closer of this. <@&286206848099549185> channel's done its just collecting dust, you acn go ahead and close this please. ive got all i was looking for
candid lotus
you cant close it since your reopened it
so its not your ticket technically
.close
high coyote
.close
cedar kilnBOT
cedar kilnBOT
Available help channel!
subtle hound
cedar kilnBOT
subtle hound
I'm overthinking this problem
How does adding the two equations find the circle that passes thru intersection
it seems like a random thing to do
cedar kilnBOT
@subtle hound Has your question been resolved?
inland totem
$$ 2x\left(3x+1\right)-3\left(3x+1\right) $$
cedar kilnBOT
$$ 2x\left(3x+1\right)-3\left(3x+1\right) $$
wraith daggerBOT
Black Hat
inland totem
.close
median bear
This is a question about the connection between Lie group/Lie algebra with unitary/hermitian operators in quantum mechanics:
Evolution of a definite state by some operator (eg. Hamiltonian evolving a stationary state) simply multiplies the original state by a phase factor (Complex number of modulus 1). This has the geometric interpretation of the exponential map of the tangent space (imaginary axis / Lie algebra in this case) onto the manifold (Unit circle / Lie group). Energy E is a real number that simply translates the identity up and down the imaginary axis.
However, if what is the geometric interpretation of the action of hermitian operator in the tangent space near the identity. When the unitary operator (eg. Time evolution operator) evolves a quantum state is a non-trivial linear combination of definite states. Where the manifold is formed by the evolution of quantum states.
violet flume
These channels are mostly for undergraduate type questions
you might have more luck if you move to a more specific channel
maybe #advanced-algebra ?
granite silo
that seems complex af
violet flume
I'm not really sure
median bear
you might have more luck if you move to a more specific channel
This is not undergrad?
No worries.
violet flume
maybe that was a poor choice of words
timid gale
This is not undergrad?
I can help
violet flume
I'm just saying people who are good at more advanced stuff sometimes only hang out in those channels, or you can continue to try your luck here
median bear
I can help
Really!
timid gale
This is a question about the connection between Lie group/Lie algebra with unita...
You want to know where manifold is formed right
median bear
Yes
timid gale
You want to know where manifold is formed right
Each point in this manifold corresponds to a unique quantum state that is what you should keep in mind at first
median bear
Yes I know that
I was wondering about the action of the hermitian operator (observable) in the tangent space. And how that reduced to a simple translation in the 1 dimensional case
timid gale
Oooooh
median bear
Sorry, poorly phrased question
timid gale
No worries
timid gale
I was wondering about the action of the hermitian operator (observable) in the t...
This will corresponds to a change in the expectation value of the observable, indicating a potential change in the physical measurement associated with that observable right . The Hermitian operator's action in the tangent space provides insight into how small changes in the parameter (observable value) affect the state in the vicinity of the identity. Does it help ?
timid gale
This will corresponds to a change in the expectation value of the observable, i...
And you should know that In the one-dimensional case, the action of a Hermitian operator on a quantum state near the identity results in a real-valued parameter.
wanton sail
maybe <#1136641179719565342> ?
Not the right channel
#diff-geo-diff-top would be a better place
median bear
And you should know that In the one-dimensional case, the action of a Hermitian ...
Thank you very much. This helps a bit.
wanton sail
Yeah these help channels tend to be more for high-school or early undergraduate questions
Whereas this topic tends to be learned in graduate school
median bear
cool
cedar kilnBOT
@median bear Has your question been resolved?
crimson estuary
how would i do this question
cedar kilnBOT
how would i do this question
crimson estuary
.close
cedar kilnBOT
crimson estuary
.reopen
cedar kilnBOT
✅
crimson estuary
just need to know how to do c
tropic oxide
did you do part b
crimson estuary
yeah i just added it all together
so he paid 105$ in interest
yes
so 105 = 980 x 1 x R
but when i get r it is 3/28 or 10.7% but the answer is 12.5% p.a
but im not sure where that 12.5 comes from
well im just gonna say answers are wrong
how do i do 24?
just c
wraith daggerBOT
guinearW
crimson estuary
oh so the principle was after the deposit
Ok i see
so how would i do 24c
nvm thank you
cedar kilnBOT
@crimson estuary Has your question been resolved?
cedar kilnBOT
cedar kilnBOT
Available help channel!
grave wave
hi
grave wave
i need help
pls
if im switching on a graph from quadrant 1 and the cords are for G (-1,3) and I rotate it 90 degrees clockwise from 0,0 will its cords be (3, 1) for G prime
deep oriole
should do
cedar kilnBOT
@grave wave Has your question been resolved?
normal vale
Can anyone explain why the chain rule is used here?
cedar kilnBOT
Can anyone explain why the chain rule is used here?
patent bridge
yeah so imagine a explicit example
say y= (x-1)^2 right
so y^2 = (x-1)^4
how would you diff that?
normal vale
dy/dx [((x-1)^2)^2] ?
patent bridge
yeah so dy/dx = 2(x-1)
d/dx(y^2) = 4(x-1)^3 = 4(x-1)(x-1)^2
since we know that dy/dx is 2(x-1) and (x-1)^2 = dy/dx and y respectively we can write
d/dx(y^2) = 2dy/dx*y
so that's an explicit example
the thing you wrote in your book is just a general case for y = some function of x
cedar kilnBOT
cedar kilnBOT
Available help channel!
paper pilot
2+2+
junior dome
!15m
cedar kilnBOT
Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.
paper pilot
yes what will i get as ans
dim tiger
cos(-x)=cosx
slate lintel
,tex .rocket trig
paper pilot
ok
wraith daggerBOT
hayley!
paper pilot
prove that cos(pi +x)cos(-x)/sin(pi-x)cos(pi/2+x)=cot^2x
wraith daggerBOT
LE SSERAFIM
versed fulcrum
???
paper pilot
<@&286206848099549185>
paper pilot
prove that cos(pi +x)cos(-x)/sin(pi-x)cos(pi/2+x)=cot^2x
<@&286206848099549185>
cos(pi +x)cos(-x)/sin(pi-x)cos(pi/2+x)=cot^2x
sharp trail
$$\begin{align}\cos(\pi + x) &= -\cos(x)\
\cos(-x) &= \cos(x) \
\sin(\pi -x) &= \sin(x)\
\cos(\frac{\pi}{2}+x) &=-\sin(x)\
\implies \
\frac{\cos(\pi+x)\cos(-x)}{\sin(\pi-x)\cos(\frac{\pi}{2}+x)}&=\frac{-\cos(x)\cdot\cos(x)}{\sin(x)\cdot(-\sin(x))}\
&=\frac{-\cos^2(x)}{-\sin^2(x)}\
&=\cot^2(x)
\end{align}$$
wraith daggerBOT
Drenitor
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
sharp trail
The first four lines are preliminary identities.
cedar kilnBOT
@paper pilot Has your question been resolved?
vapid scaffold
cedar kilnBOT
vapid scaffold
is this allowed
royal loom
, rotate
wraith daggerBOT
vapid scaffold
royal loom
I'll allow it
royal loom
Way to what
vapid scaffold
to prove this identity
royal loom
yes
royal loom
maybe demonstrate more why the things you rewrite are what they are
like since that is what you are trying to do
why can you go from (cotx)/(tanx) -> 1/tan^2x
like just showing that kinda intermediate steps
vapid scaffold
yeah, thank you for the feedback
i wanted to make sure that cotx/tanx was valid mostly
thanks thanks
.close
modest lily
Could I please have some help with this question. I have no idea how to approach and would appreciate any guidance.
cedar kilnBOT
Could I please have some help with this question. I have no idea how to approach...
@modest lily Has your question been resolved?
modest lily
I got an answer of 673,920,000 but have no clue if thats correct or not
<@&286206848099549185>
modest lily
yeah I have tried to do it with mates and we all got different answers
Ill send my working out
crimson sedge
aight
modest lily
The proof was different question
crimson sedge
hm
have a look
on this
Let's break down the password criteria to calculate the number of possible passwords for this user:
- At least 2 characters that are the same numerical digit and must be placed next to each other.
- At least 3 characters which are different letters.
- The remaining character can be any letter.
For the first criterion:
- There are 10 possible numerical digits (0 to 9).
- There are 10 ways to choose the digit for the first occurrence, and once that's chosen, there are 9 ways to choose the digit for the second occurrence since repetition is allowed.
- There are 2 possible positions for these digits: either the first and second positions or the second and third positions (since they must be next to each other).
So, there are 10 * 9 * 2 = 180 ways to satisfy the first criterion.
For the second criterion:
- There are 26 possible letters (a to z).
- We need to choose 3 different letters out of these 26.
- The order in which these letters are placed doesn't matter, so we'll use combinations.
- The number of ways to choose 3 different letters from 26 is given by the combination formula: C(26, 3) = 26! / (3! * (26 - 3)!) = 2600.
For the third criterion:
- There are 26 possible letters (a to z).
- One character can be chosen in 26 ways.
Now, combining these criteria:
- The first criterion has 180 possibilities.
- The second criterion has 2600 possibilities.
- The third criterion has 26 possibilities.
To calculate the total number of possible passwords, we multiply these counts together:
Total possibilities = 180 * 2600 * 26 = 11,232,000.
So, there are 11,232,000 possible passwords that satisfy the given criteria for this user.
modest lily
Yeah thats a very different style of working then mine
crimson sedge
I see
modest lily
So could you clarify
crimson sedge
yeah
modest lily
a permutation is ordered
crimson sedge
So could you clarify
which part
crimson sedge
wait
yes
that is correct
permutations involve order, while combinations do not.
modest lily
so with that I think the answer is 112,320,000
becuase I was doing 262524 which is 26P3
then multiplying by 5!
but it should already be in order then
modest lily
yeah I think thats what ill do, thanks for the help appreciate it
crimson sedge
no need to thank bro, we both together found the solution
modest lily
.close
crimson estuary
how do i do 30?
cedar kilnBOT
how do i do 30?
crimson estuary
.close
foggy tendon
For some reasons my numerical solution is different from analytical solution by a lot
cedar kilnBOT
For some reasons my numerical solution is different from analytical solution by ...
@foggy tendon Has your question been resolved?
patent bridge
correct me if I'm wrong but are you trying to solve the BVS for y(x) and then subbing in x=1/3,2/3 respectively and comparing your hand derived answers to the computer ones?
foggy tendon
Yes
celest ledge
I don’t know what the issues is. Since y=Ae^(2x)+Be^(5x). Plug in those values you clearly can have A=(3e^3-5)/(e^3-1), B=2/(e^3-1)
foggy tendon
correct me if I'm wrong but are you trying to solve the BVS for y(x) and then su...
Got the answer from wolfram alpha
Google bard got a pretty good approximation
foggy tendon
I know
patent bridge
in ur handwritten notes
sorry but I don't really understand your working out, would it help if I sent my working
foggy tendon
Ok sure
foggy tendon
I don’t know what the issues is. Since y=Ae^(2x)+Be^(5x). Plug in those values y...
Pretty sure that’s the analytical solution, I need to solve it using numerical method
celest ledge
I have no idea what numerical method means
Anyway that is the answer. If one of you method gives you the same answer then that method is correct
The another is wrong if you got different
foggy tendon
Anyway that is the answer. If one of you method gives you the same answer then t...
The final answer is all in numbers, no functions though
Since it’s numerical methods
celest ledge
Online calculators to verify
foggy tendon
Got the answer from wolfram alpha
Which is what I did
patent bridge
nah chips is right
foggy tendon
sorry but I don't really understand your working out, would it help if I sent my...
im using finite difference method here
and the question stated it also, so cannot use linear shooting method
celest ledge
Correct now
y(1/3), I omitted y(2/3)
$y=\frac{e^{2}-1}{e^{2}-e}e^{3x}+\frac{-e+3}{e^{2}-e}e^{4x}+2e^{2x}$
wraith daggerBOT
Cogwheels of the mind
foggy tendon
Got the answer from wolfram alpha
Which is the same thing from what I got from wolfram alpha?
celest ledge
Yeah
I don’t know what this method is called . It’s just a method. I don’t know why I should care about what method I used
foggy tendon
Because the question stated
foggy tendon
I have no idea what numerical method means
Analytical solution, exact answer
Numerical solution, approximation answer
And I have to get the approximation
warm bone
If all description fails you have to show code
Or worse, your manual numerical steps
I'm hoping you're coding here
foggy tendon
Google bard got a pretty good approximation
google bard gave me the code
i know how to do IVP on matlab but not sure about BVP
heres the question again
celest ledge
Oh I see I finally understood your question. You solved y(1/3),y(2/3) by some approximation method without solving the differential equation directly, and got wrong answer.
foggy tendon
yes
warm bone
google bard gave me the code
Uhh, I don't think you should trust codegen from them.
Anyway I can't help with matlab so
foggy tendon
Uhh, I don't think you should trust codegen from them.
Anyway I can't help with ...
the answer is very close to the analytical solution though
patent bridge
I still don't get what the question is
so you're saying that your numerical solution is wrong?
foggy tendon
For some reasons my numerical solution is different from analytical solution by ...
My solution is at y1 and y2, but after comparing it to analytical solution it’s different by a lot
foggy tendon
so you're saying that your numerical solution is wrong?
Yes
Probably I did something wrong when even google bard can do it
patent bridge
is numerical solution the answer you find by hand or by computer
foggy tendon
By hand
patent bridge
but then if its by hand shouldn't you get an exact answer?
or is it in this case we have to use a pre-determined method as dictated by the question to obtain an "approximate" solution?
cus mathematica and matlab both give me numbers with decimals (approx) whilst doing it by hand gives exact solutions (no decimals)
foggy tendon
but then if its by hand shouldn't you get an exact answer?
But it’s different by a lot
If it’s with error of around 5% it’s acceptable, but it’s a lot higher than this
From my hand analysis answer
patent bridge
oh I'm really sorry I don't know how to do finite difference method
I used a different method without knowing you had to use that one
foggy tendon
It’s okay
cedar kilnBOT
@foggy tendon Has your question been resolved?
cedar kilnBOT
@foggy tendon Has your question been resolved?
foggy tendon
.close
cedar kilnBOT
sinful summit
.close
cedar kilnBOT
cedar kilnBOT
Available help channel!
fallow wind
hii, i just wanted to know if my work is correct
cedar kilnBOT
hii, i just wanted to know if my work is correct
fallow wind
this is the problem
cedar kilnBOT
@fallow wind Has your question been resolved?
slate lintel
,calc pi/2
wraith daggerBOT
Result:
1.5707963267949
the first one is right but you'll want to check those bounds for the others
wintry pier
Hello
mighty drift
wraith daggerBOT
Joseph.P
mighty drift
factor
you want the numerator to cancel out the denominator
to prove it's not just a rational, but an integer
wintry pier
Like $(1-k)(1+k)$ ?
wraith daggerBOT
Joseph.P
mighty drift
yes
wintry pier
But that doesn’t simplify the equation?
mighty drift
a priori it doesn't
celest ledge
You write that fraction as n
south tundra
Think about the parity of k given that 1 - k^2 or (1 - k)(1 + k) has to be divisible by 2
celest ledge
k^2+2nk-(1+6n)=0
so k=-n +/- sqrt(n^2+6n+1)
So n^2+6n+1=a^2
Notice that a^2+8=b^2
Where b=n+3
mighty drift
k^2+2nk-(1+6n)=0
what ?
mighty drift
They set (1 - k^2)/(2(k - 3)) = n
ok just turned it into a quadratic in k
celest ledge
Solve corresponding k for each n finally
south tundra
The other way is simpler I think
wintry pier
I don’t understand what a is
wintry pier
Oh finally I understand
celest ledge
Good
wintry pier
Thanks
celest ledge
Np
south tundra
There is another method, want to know?
wintry pier
Ye why not
south tundra
Alternatively you could notice that k has to be an odd number, say k = 2m + 1
Then the expression is simplified to 2m(2m + 2)/2(2m - 2) or just m(m + 1)/(m - 1)
Now, m-1 and m are always coprime, so m-1 must divide m+1
And generally there are two cases, either gcd(m - 1, m + 1) = 1 or gcd(m - 1, m + 1) = 2
In the first case you are forced to say that the denominator has to be 1 or -1 since it doesn't divide the nominator
So m - 1 = 1 or m - 1 = -1
Meaning k = 5 and k = 1 are some of the solutions
celest ledge
Brilliant
south tundra
In the other case you have that m - 1 = 2 meaning k = 7
Uh actually wait I missed something
Ah I see
m - 1 = 2 or m - 1 = -2
Meaning the other pair of solutions is k = 7 or k = -1
So k = -1, 1, 5 or 7
celest ledge
m-1 divides m+1 gives us m-1 divides 2 actually
wintry pier
n^2+6n+1=a^2
How did you find this ?
south tundra
m-1 divides m+1 gives us m-1 divides 2 actually
Yeah so m-1 is either -1, 1, -2 or 2 from there
celest ledge
You understand his first , his is much simpler
celest ledge
How did you find this ?
I don’t know what that’s called, the b^2-4ac thing for a quadratic
south tundra
Discriminant?
celest ledge
Yeah
wraith daggerBOT
Joseph.P
celest ledge
2m+2 above not 2m+1
celest ledge
You can ignore -
cedar kilnBOT
wintry pier
.reopen
cedar kilnBOT
✅
wintry pier
Alternatively you could notice that k has to be an odd number, say k = 2m + 1
Last question, how did you find that k has to be odd
south tundra
1 - k^2 has to be even
If k was odd, then 1 - k^2 would be even, which is what we need
If k was even, then 1 - k^2 would be odd instead
crimson sedge
translation:
- find the area of the following figure
- use π ≈ 3.14
crimson sedge
no idea how to solve
livid hound
apply formula for area of a circle
crimson sedge
the formula would be Pi right
livid hound
no
crimson sedge
oh what would it be
livid hound
$A_{circle} \redneq \pi$
wraith daggerBOT
ℝam()n()v
livid hound
if you don't know the formula, do a quick search
livid hound
exactly as i said
this question involves finding the area of a circle
formula for that is relevant
first thing to do would be to look that up
if you don't already know it
livid hound
yes, you'd use 3x for the radius
crimson sedge
so it would be
A=formula
A=(pi)•3x
A= answer?
wraith daggerBOT
ℝam()n()v
crimson sedge
i see i see
so i’d put instead 3x instead of r^2, also sorry for asking a lot i’ve just completely forgotten
livid hound
no
crimson sedge
somebody in the other help forum said 28.26x^2
livid hound
r represents the radius
you replace r with whatever the radius is
here you're radius is 3x, replace r with (3x),
parentheses included to maintain the order of operations and ensure your actual radius is being squared
crimson sedge
and then i multiply ?
livid hound
yes
livid hound
(3x)^2 will be 9x^2
long mauve
how to do b) and d)
cedar kilnBOT
how to do b) and d)
long mauve
can I solve like this
cedar kilnBOT
@long mauve Has your question been resolved?
cedar kilnBOT
@long mauve Has your question been resolved?
lilac crane
You could but 90 - 70 is 20
Oh
That's 3
Yeah that all checks
Did you have questions about why the identities are true?
cold briar
btw this channel is closing since you deleted the original message
cant
open a new one
cedar kilnBOT
Available help channel!
upper ruin
.close
cedar kilnBOT
dire geode
Try it
left lodge
Hello I am new
slender meteor
this is wrong tho
slender meteor
wait no i was wrong; i was reading secx as sec^2x
left lodge
My life mission is solving these problems
strong pivot
Inverse function my bad
dire geode
Usual way involves implicit differentiation
crimson sedge
implicit differentiation
celest ledge
y=arcsin(x), x=sin(y). Take derivative in terms of x on both sides cos(y)y’=1, rewrite cos(y) in terms of x
hot sigil
Hello! I am stuck on a discrete calculus problem. Any help would be appreciated!
Suppose P is the unique polynomial of a degree < 10 that satisfies P(k)=2^k for k=0,1,2,…,9
What is P(10)?
hot sigil
I have no idea on how to tackle this problem, I am very new to discrete calculus, but I do have a very good background in regular calculus.
celest ledge
Use Lagrange polynomial
This is the method for all questions like this
It’s on Wikipedia
But on second thought
hot sigil
Oh I don’t know why I didn’t think of that
hot sigil
Could you please explain what does unique mean
celest ledge
But for degree <=9 it’s not unique
I mean. You are able to find a polynomial of degree 9, which satisfies your condition
But you can find more than one polynomial , of degree <=9, that satisfy your condition
Like y=x^2
celest ledge
I mean so your question might be wrong. There might be different answers
celest ledge
Then that is probably correct now.
It said unique, so you can use Lagrange polynomial
hot sigil
Okay I will try it
But I have a question
Why wouldn’t Lagrange Polynomial just give k^2 for the value of p(10) as well
Since it’s just going to extrapolate the values we have and they all follow the same pattern then it should just be 2^10 right
crimson sedge
I need an explanation to why we can do this?
cedar kilnBOT
I need an explanation to why we can do this?
tropic oxide
as in "wtf is going on?" or "i think this is illegal"?
slate lintel
$x\blue\blacksquare - 3\blue\blacksquare = (x-3)\blue\blacksquare$
wraith daggerBOT
hayley!
crimson sedge
.close
cedar kilnBOT
tropic oxide
doorslam
ocean mural
lol
cedar kilnBOT
Available help channel!
summer valley
cedar kilnBOT
summer valley
Not sure where to begin on this problem. Generally for linear combination questions, I've been given W, then asked if it was a linear combination of given vectors.
cedar kilnBOT
@summer valley Has your question been resolved?
summer valley
<@&286206848099549185>
brave wind
Not sure where to begin on this problem. Generally for linear combination questi...
what part are you stuck on?
summer valley
I don't know how to set up W
brave wind
W would be all combinations of (x,y,z) where
2x+6z=10
-x+8y+5z=3
x-2y+z=3
if there are real numbers x,y and z that makes these three equations true at the same time then b is in A
I mean b is in W
summer valley
I see gotcha. For some reason I was under the impression that A was an augmented matrix, not that A and B form an augmented matrix together.
twin heron
I need an explanation to why we can do this?
just wondering, is that Slovene?
brave wind
I see gotcha. For some reason I was under the impression that A was an augmented...
there is also this cool method that I use to solve thees thing
want to know it?
crimson sedge
just wondering, is that Slovene?
serbian although you could probably understand
summer valley
want to know it?
To solve I put the matrix in row echelon form and go from there, but if you have another way sure 🙂
twin heron
guassian elimination?
summer valley
yes I think so
brave wind
To solve I put the matrix in row echelon form and go from there, but if you have...
I always check the det of A first because if it's 0 then there always exist a trio of x,y and z real numbers that solve the equation meaning that the hole R^3 plane is in W
summer valley
i had to google to see if I remembered how to do det of 3x3 matrix, but luckily i remember from calc 3 cross product rule
thats good to know
thank you 🙂
.close
hot sigil
Find the maximum value of P(0) within all polynomials P(x) that satisfy the following conditions:
- P(x) has a degree < 6
- P(1),P(2),…,P(6) are a permutation of the values {1,2,3,4,5,6}
hot sigil
I tried to find the Lagrange Polynomial in terms of P(1),P(2),…,P(6)
And then substitute values for those numbers that will give the maximum value of P(0)
It turned out to be 38 but it’s not the right answer
crimson delta
did you make a mistake with the lagrange polys? quite a bit of multiplying out, easy to make a mistake there
crimson delta
remember that you dont have to multiply all the (x-bla) terms out
you can just plug in x=0
hot sigil
Yeah that’s what I’m doing
celest ledge
did you make a mistake with the lagrange polys? quite a bit of multiplying out, ...
We tried Lagrange polynomial before, he asked a similar question in these channels but the thing is, you can only get maximal for polynomials of degree 5, but sometimes polynomial of lower degrees might give you bigger result, condition is <=5 not =5…
celest ledge
the degree would cancel out with the relevant coefficients
What do you mean cancel out
crimson delta
I don't see what your problem is
the lagrange polys also give polynomials of lower degree
celest ledge
I mean we can find max P(0) for polynomials of degree =5 this way
crimson delta
if you use a "correct" linear combination
celest ledge
But the question asks for maximal P(0) for polynomials of degree <=5
That is how I solved a similar question but smaller than the answer because a lower degree polynomial might give you a bigger result
hot sigil
Yup I think I just had a mistake! The polynomial is 4P(1)-15P(2)+20P(3)-15P(4)+6P(5)-P(6)
The maximum value for this polynomial is 126
celest ledge
if you use a "correct" linear combination
Yes, but how do you rule out polynomials of lower degrees
crimson delta
just because the lagrange polys are of degree 5 does not mean that a linear combination of them is of degree 5
celest ledge
Each term is of degree 5
hot sigil
Yeah I see what you’re saying
crimson delta
x^2 and -x^2+x are of degree 2. their sum is of what degree?
celest ledge
a(6) (x-1)…(x-5)/(6-1)…(6-5) like this
hot sigil
But I don’t think there is any way to check the polynomials of degrees less than 5
celest ledge
1 but
Yeah that’s what I am saying
There exist lower degree polynomial satisfying the condition, for example p(x)=7-x
crimson delta
there is literally no problem here
and 7-x will be found if you plug in the correct coefficients
hot sigil
There exist lower degree polynomial satisfying the condition, for example p(x)=7...
But that’s not the maximum value that P(0) could take
crimson delta
feel free to suffer through all the manipulations
celest ledge
Why
crimson delta
I promise you you will get it
celest ledge
I promise you you will get it
Lagrange polynomial can only get you a polynomial of degree 5 no?
celest ledge
How can you construct a polynomial of degree 4 satisfying the condition?
crimson delta
it will give you the unique poly of degree <=5 that works
by plugging in the correct coefficients
celest ledge
I just want to know, can you construct a polynomial P of degree 4, such that P(1),…,P(6) is permutation of 1,…,6
north bough
What is the problem?
celest ledge
At the beginning
crimson delta
yes I can
well if there is one
I can give you a degree 1 poly even with lagrange
celest ledge
How? Say P(x)=x
hot sigil
I can give you a degree 1 poly even with lagrange
Yeah but idk if you can do it using the conditions
celest ledge
Degree 1 is too trivial can you give me a degree 2 case or something
crimson delta
well I don't know if a degrr 2 poly exists
celest ledge
A degree 2 or 3 or 4 polynomial satisfying the condition
What if it exists, you don’t need to calculate and compare its p(0) to get the answer?
crimson delta
this is pointless
north bough
Question States that p(1) , p(2) and so on are one of the values from the set {1,2,...,6}, am I wrong?
crimson delta
you are suggesting that lagrange interpolation doesn't work
because it can't find polynomials that aren't of maximal degree
celest ledge
I mean Lagrange can only make you calculate max P(0) for degree P=5
crimson delta
no
celest ledge
I don’t know why you are so sure that this equals max P(0) for degree P <=5
crimson delta
lagrange interpolation can find all polynomials of degree <=5 which go through the specified points
celest ledge
Yes. I still don’t know how you can use Lagrange to get lower degree polynomials.
If you construct a degree 4 polynomial using x=1,2,3,4,5, then you can’t guarantee that P satisfies the condition, P(6) could be something else
crimson delta
try adding all lagrange polynomials
just pure sum
which poly do you get
is it of degree 5?
celest ledge
Yes
crimson delta
no
celest ledge
Each term, like (x-1)…(x-5)/(6-1)…(6-5) multiplied by a constant, is of degree 5
celest ledge
And if a lower degree polynomial exists, you are sure that it equals some combination of them?
crimson delta
yes
celest ledge
yes
Shit. So sorry man, I see it now. column vector made up by coefficients of (their product) / (x-x_i), these i vectors are linear independent
celest ledge
I suppose I should have said that earlier
So sorry man, take you so long. I never realized I have always misunderstood Lagrange polynomial method as degree=n while it always is <=n.
Thank you so much
cedar kilnBOT
@hot sigil Has your question been resolved?
cedar kilnBOT
cedar kilnBOT
Available help channel!
random kelp
The limit here is 0, however cos(1/theta) would just become cos(inf). Shouldnt it make the whole solution DNE as a result?
slate lintel
yes, the expression is undefined at 0
but you wouldn't have a problem evaluating $\lim_{x\to0} \frac1x\cdot x$ i assume
wraith daggerBOT
hayley!
random kelp
but you wouldn't have a problem evaluating $\lim_{x\to0} \frac1x\cdot x$ i assum...
hmmm
it would be indeterminate
(inf)(0)
slate lintel
yep it would be an indeterminate form
but that doesn't mean the limit doesn't exist
random kelp
that is true
slate lintel
it just means we need to like simplify or something
random kelp
there are other ways to evaluate this
for example, we could take x/x right?
then it would just become 1
random kelp
yes, x/x = 1 as long as x ≠ 0
do I have to state that?
slate lintel
nah
random kelp
alright
random kelp
The limit here is 0, however cos(1/theta) would just become cos(inf). Shouldnt i...
so about this problem
we have the same dilemma
(0)(inf)
slate lintel
(0) * cos(inf) really
slate lintel
like as in take the limit of the two factors separately? that only works if both limits are finite
and exist
random kelp
like as in take the limit of the two factors separately? that only works if both...
thats interesting because
here they're getting me to split it up
when lim x=> 2^- ( 1/x-2 ) = inf
slate lintel
okay yeah it works with one finite nonzero and one infinite as well
you should look at your book's examples of using the squeeze theorem
random kelp
okay yeah it works with one finite *nonzero* and one infinite as well
so infinity is allowed then?
slate lintel
paired with something that has a finite, nonzero limit, yes that would work
random kelp
gahhh
and everyone was telling me 'it must be 2 finite values or it doesnt work 🤬 🤬 🤬 🤬 '
are there any other exceptions I should be aware of?
slate lintel
vagrant elbow
and everyone was telling me 'it must be 2 finite values or it doesnt work 🤬 🤬 ...
They probably meant the limit exists only if the two limits are finite and exist
In your example here, the limit does not exist
random kelp
In your example here, the limit does not exist
the one from the textbook I gave?
vagrant elbow
yes
random kelp
are you saying that if the limit is infinity, it is counted as DNE/not existing?
vagrant elbow
yes
slate lintel
are you saying that if the limit is infinity, it is counted as DNE/not existing?
that's how a lot of people define that yeah
random kelp
thats very very interesting
slate lintel
however, the mathexchange link i sent gives a good overview of how to treat inf
vagrant elbow
wdym
slate lintel
For Limits
0 • ∞ is indeterminate
∞ • ∞ = ∞
x • ∞ = ∞
x • 0 = 0
-x • ∞ = -∞
∞ • -∞ = -∞
random kelp
this is basically saying that the product of the limits both have to be finite numbers
vagrant elbow
just think of infinity as a large number and not some almighty last number
That should do it
sand cradle
The limit here is 0, however cos(1/theta) would just become cos(inf). Shouldnt i...
theta will never actually become 0, a limit is just what it approaches as theta goes to 0
random kelp
theta will never actually become 0, a limit is just what it approaches as theta ...
i understand
random kelp
__For Limits__
0 • ∞ is indeterminate
∞ • ∞ = ∞
x • ∞ = ∞
x • 0 = 0
-x • ∞ = -∞
...
so when an unknown is multiplied by infinity it is infinity?
However that doesnt help when the limit approaches 0
slate lintel
correct, 0 • ∞ remains indeterminate
this means you need to use your brain a little bit
random kelp
what I know is that I cant split it up
so I'll have to use the squeeze theorem like the question said
slate lintel
yes you will
random kelp
if that applies, (theta)^3 on both sides will become 0
slate lintel
theta^3 would approach 0 as theta approaches 0 yes
you should look at the examples in your book on the squeeze theorem
random kelp
you should look at the examples in your book on the squeeze theorem
yeah I just did
random kelp
__For Limits__
0 • ∞ is indeterminate
∞ • ∞ = ∞
x • ∞ = ∞
x • 0 = 0
-x • ∞ = -∞
...
but for this, if i've got x* ∞, I still cant split it up and apply l'hopitals rule right?
because infinity isnt finite
sand cradle
Though this is a bit different from how you would usually apply it, since I say that the inequality holds for x > 1
Usually, when applying the squeeze theorem, it holds for all x in R
but it here doesn't matter, since x is approaching 2 anyways
random kelp
however, what hayley said was still correct right, where you can have 1 finite nonezero and 1 infinite value
sand cradle
however, what hayley said was still correct right, where you can have 1 finite n...
Yes, for finite x > 0, x * infinity = infinity
sand cradle
The limit here is 0, however cos(1/theta) would just become cos(inf). Shouldnt i...
You won't really have anything like x * infinity here, though.
As a slight hint, think about cos(x)
cedar kilnBOT
@random kelp Has your question been resolved?
sand cradle
Do you want another hint, @random kelp?
random kelp
Do you want another hint, <@287899997251436545>?
you gotta apply the squeeze theorem
sand cradle
Yes
sand cradle
you gotta apply the squeeze theorem
To do that, think about cos(x) and its range
random kelp
To do that, think about cos(x) and its range
-1, 1
sand cradle
Yes
random kelp
oh ive already solved it btw 😭
sand cradle
You "squeeze" the middle term inbetween them, while the left and right terms approach the same value
So the middle term also needs to be the limit of the left and right terms
umbral dew
$\log_{3\sqrt{3}}{81}$
cedar kilnBOT
$\log_{3\sqrt{3}}{81}$
wraith daggerBOT
yamdoot
umbral dew
ik this is a super stupid ques
but i really dont know what to do
i dont know what happened to me
why i cant solve this
mental trail
You have to solve the equation $(3\sqrt{3})^x = 81$
wraith daggerBOT
rafilou2003
umbral dew
i can only see writing 81 as 3^4
mental trail
i can only see writing 81 as 3^4
This is great already
mental trail
And $3\sqrt{3} = 3^?$
wraith daggerBOT
rafilou2003
umbral dew
😭
i seriouly cant find out
what?
pls help
what should be there on the place of ?
deep oriole
what is sqrt(3) as a power of 3
umbral dew
$3^{3/2}$
wraith daggerBOT
yamdoot
umbral dew
?
mental trail
Correct !
deep oriole
yeah
crimson sedge
**rafilou2003**
substitute
deep oriole
you now have log_3^3/2 (3^4)
mental trail
So $(3^{\frac{3}{2}})^x = 3^4$
wraith daggerBOT
rafilou2003
crimson sedge
yep
mental trail
And do you see how to finish ?
umbral dew
8/3?
crimson sedge
yep
deep oriole
yeah
deep oriole
bot is broken
umbral dew
it broke again lol
deep oriole
dont worry
umbral dew
lol
dire geode
lucky number 13 in perpetuity
mental trail
Everybody wants to .close
sinful lark
mental trail
.close
cedar kilnBOT
cedar kilnBOT
Available help channel!
shrewd rampart
Determine the sum of two given powers and record the process. 2^3 and 2^4. How do I do this help pls
runic sage
add 2^3 and 2^4
surreal cave
Determine the sum of two given powers and record the process. 2^3 and 2^4. How d...
$2^3+2^4$?
wraith daggerBOT
MrFancy
shrewd rampart
2^7?
runic sage
what is 2^3
shrewd rampart
what is 2^3
8
shrewd rampart
exponent rules very important you should memorise >:)
They are important?
crimson sedge
They are important?
Yea
shrewd rampart
good, now what is 2^4
16
wraith daggerBOT
MrFancy
shrewd rampart
Thank you
How do I do Determine the difference of two given powers and record the process. Find the difference between 5^7 and 5^3.
Nvm ik
.close
carmine tundra
where did the -4 from the 2nd row come from? g'(1) is -1
cedar kilnBOT
where did the -4 from the 2nd row come from? g'(1) is -1
carmine tundra
This is increasing and decreasing intervals topic btw
safe ermine
wanton heron
Yes
glad dagger
Hi. Best learning apps.
cedar kilnBOT
Hi. Best learning apps.
glad dagger
For math.
13-15 year old math.
Especially algebra
And creating algebra equation
Example like this
The price of a theater ticket is 64$ more than a cinema ticket. A total of 400$ was paid for 4 theater tickets and 5 cinema tickets
Calculate the price of one theater ticket. Save calculations.
I wanna learn how to do these
I suck at these
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obsidian coral
Example like this
The price of a theater ticket is 64$ more than a cinema tick...
Khan academy
cedar kilnBOT
@glad dagger Has your question been resolved?
crimson sedge
Especially algebra
i watched a lot of black pen red pen lmao
cedar kilnBOT
cedar kilnBOT
Available help channel!
jolly sable
cedar kilnBOT
jolly sable
I need to find eigen vectors here
So first i did det(a-lembda i)=0
But here lembda seems complicated
tropic oxide
,w det {{1-λ, -3, 3}, {3, -5-λ, 3}, {6, -6, 4-λ}}
tropic oxide
working out the charpoly of that matrix is a minefield for arithmetic errors but apparently it has nice roots still.
cedar kilnBOT
@jolly sable Has your question been resolved?
jolly sable
L=-2 satisfied
Let me find 2 more roots
I got -2,-4
Wow
I shouldn't leave it between. I thought it is complicated
,rotate
wraith daggerBOT
jolly sable
After solving i got this
x=y-z for lembda=-2
<@&286206848099549185>
How to get eigen vector here
,rotate
wraith daggerBOT
jolly sable
Eigen values are lembda=-2 and 4
For lembda =-2 how to find eigen vectors
<@&286206848099549185>
,w det {{-3, -3, 3}, {3, -6, 3}, {6, -6, 0}} find eigen vectors
wraith daggerBOT
crimson sedge
To say a preposition has a truth value of T
cedar kilnBOT
To say a preposition has a truth value of T
wraith daggerBOT
proven fable
The equation of AD = y = -2x+18 AD and BC intersect at P which is equal to (5,8) It is given that the surface of PCD is 44. How to find C?
fallen moat
i think you'll have to use the area od triangle PCD
cedar kilnBOT
The equation of AD = y = -2x+18 AD and BC intersect at P which is equal to (5,8)...
Show your work, and if possible, explain where you are stuck.
proven fable
i think you'll have to use the area od triangle PCD
what is the base of the triangle
fallen moat
that's what we have to find
like finding the coordinates of D
since the height is given
fallen moat
and the height is given in P
proven fable
and the height is given in P
yes but what to do now?
fallen moat
yes but what to do now?
you have the height, you have the area, then you can find base length
proven fable
the height is 8
fallen moat
yea
fallen moat
area of triangle is ?
fallen moat
hint: it is not base times height