1, 2, 3 -> 1, 2, 4 .... -> 1, 2, 9 -> 1, 3, 9 -> ... 1, 8, 9 -> 2, 8, 9 -> ... 7, 8, 9
#help-19
kindred inlet
pearl patio
mhm i get that part now
thanks
still doing get the min and max thing though
well i get it but i don't know why u thought of that
or alisia
kindred inlet
not sure what to say really 
kindred inlet
looking at it doesnt it kinda feel like it should be bounded immediately
pearl patio
222(x+y+z) = X
you want all combinations of (x+y+z) that give unique sums
immediately i'm thinking caseworks
not bounding yeah
lol
i wonder if there is a combinatorial argument
pearl patio
looking at it doesnt it kinda feel like it should be bounded immediately
i think i'm seeing it now lol
i think that's all, thanks!!
that was hard
odd edgeBOT
Available help channel!
mystic saffron
odd edgeBOT
mystic saffron
How do I solve this
I don’t know how to do with the division
I always end up with 0
you are on the right track in a sense, although the process you did seems wrong
mystic saffron
I don’t know how to do with the division
let's try and get of both the /3 and /2
can you think of the LCM of both 2 and 3?
LCM standing for least common multiple
6?
yeah
exactly
try multiplying both sides with 6
[
x - \f x3 = \f{x+1}2 \Iff \c b{6\cd}\p{x-\f x3} =\p{\f{x+1}2}\c b{\cd 6}
]
clever fjordBOT
clever fjordBOT
mystic saffron
Oh yeah it’s 6 too
yeah perfect but you don't actually have to distribute it
it'd best to leave it as 6(x+1)
how would you simplify $\ds \f{6x}3$ and $\ds \f{6(x+1)}2$
clever fjordBOT
mystic saffron
2x
And
mhm
3x+3
yeah on point
so now what do you have
5x+3?
I don't see an equality there. I'm asking for like your original expression simplified
mystic saffron
I don't see an equality there. I'm asking for like your original expression simp...
just show me this like
[
6x - 2x = 3x+3
]
is what you have now yes?
clever fjordBOT
mystic saffron
after this
Ohhhhh yeah
that's what I was asking for you to show me
so I know if something went wrong
mystic saffron
anyways uh simplify now. should be straightforward
mhm bingoo
Thank u I understand it much better so it’s just like founding the common thing
mmm wdym by founding the common thing?
Lcm
yeah
Forgot what it was called lol
exactly
Ty
although
Yes
I wanna warn u about something preemptively
[
\f{3x}x = 1
]
clever fjordBOT
mystic saffron
you can multiply both sides by x, right?
it fits what we were saying about the LCM stuff
Yes
so you get 3x = x
so you might think x = 0 because of that because you get 0=0
But isn’t x times x like to the power of
which superficially makes sense
but it actually is incorrect
mystic saffron
you can multiply both sides by x, right?
If we multiply both sides with x
yeah
OHHH
BUT IT CANCELS OUT
[
\c b{x \cd} \f{3x}x = 1 \c b{\cd x}
]
clever fjordBOT
mystic saffron
BUT IT CANCELS OUT
yeahH exactly
so you get 3x = x
you might say x = 0 makes sense right?
Yeah
because you get 0 = 0
right?
Mhm
but it actually is incorrect and you should be very careful about that
Yeah
Then what’s the answer
nothing there is no answer actually
it's because if we let x = 0 in 3x/x we have 0/0
you can check with your calculator but 0/0 has NO solution
it just doesn't exist
so x=0 is something we cannot have
whenever you have an expression in the denominator, you NEVER want it to be equal to 0 because that cannot be evaluated
mystic saffron
so x= 0 is never something we can consider
anyways sorry this is just a warning to the LCM stuff
you can close now if you want to
No thanks I understand now
that's great!
Thank you
alright have a good one
U too
.close
pure flame
Identify minimum/maximum/infimum/supremum
All i wanna know what this says to ger started^^.
Is it x is any number between root of 42 and 69?
regal leaf
x is between $\sqrt{42}$ and $\sqrt{69}$ and a rational number
clever fjordBOT
tobi
regal leaf
root42 and root69 excluded
regal leaf
funny numbers
pure flame
Since its > and not >=
regal leaf
even with = the maximum and minimum wouldnt exist
cold urchin
So theres no minimum and maximum
Yeah but you’d probably want to write out more formally why that is.
!occupied
odd edgeBOT
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pure flame
Yeah but you’d probably want to write out more formally why that is.
Yeah Im about to, my first step is probably doing epsilon delta method.
Id assume that
Inf = 42root
Sup = 69root
Epsilondelta for sup
69sqroot - e > x > 42sqroot
regal leaf
hint ||Q is dense in R||
pure flame
I looked through all my material and cant find anything helping with the hint^^
About my problem earlier that u helped with, i ended up finding out that they dont require us to prove the derivation for lhospital^^
fiery ermine
take n boxes and k objects, in how many ways can you put the k things into the n boxes?
fiery ermine
You can obviously change this into 1's and 0's
and then you just have n-1 ones and k zeros
and Now the anwser is $\binom{n+k-1}{k}$
clever fjordBOT
bigpufik
fiery ermine
But I dont understand why we can calcuate subsets that include the n-1 ones, since when calcualting subets size k not all will have only zeros
dim thicket
i dont understand your question
dim thicket
take n boxes and k objects, in how many ways can you put the k things into the n...
wouldnt there be $\binom{n}{k}$ ways?
clever fjordBOT
LF
supple mural
But I dont understand why we can calcuate subsets that include the n-1 ones, sin...
What
fiery ermine
No you take n boxes, and you have k things, and you have to put all k things into all boxes, adn the question is in how many ways you can do taht
supple mural
K indistinguishable things, otherwise it would be n^k
supple mural
and Now the anwser is $\binom{n+k-1}{k}$
This is correct
fiery ermine
This is correct
Yeah, but I dont get why
Like okay
I take n-1 ones let say
and k zeros
and we want to find all possible strings
of these ones and zeros
noting the are distingushiable
since otherwise it would be as you said
and I dont understand how can we jsut choose subsets of size k
supple mural
and I dont understand how can we jsut choose subsets of size k
Wdym
We take the 1s as dividers
Like
010110
Corresponds to 4 boxes and 3 balls
supple mural
010110
With 1 ball in the first box, 1 in the second, 0 in the third, and 1 in the fourth
fiery ermine
yes
supple mural
So each string goes to a unique way
fiery ermine
ok yeah
supple mural
That’s why we just have to count the number of strings
supple mural
This is called a bijection
fiery ermine
it made more sense for me if we just chose n+k-1 per n-1
supple mural
Do whatever makes sense for you
fiery ermine
Yeah oaky thanks i hate combinatroics
supple mural
Same lol
supple mural
Same I’ve been looking at pumac, cmimc, and hmmt combo cause it’s my weakest subject and I’m dying
supple mural
Yeah
They make individual problem sets for each subject
So it’s easy to just do combo problems
fiery ermine
oh thats sounds nice tbh, good luck with those
supple mural
Thanks
supple mural
lol np
fiery ermine
.close
odd edgeBOT
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mystic saffron
I'm trying to show that two solutions of the Legendre equation for $\abs x < 1$ are:
\env{align*}{
\m{y_1}x &= \sum_{n = 0}^\infty (-1)^n \f{\prod_{k = 0}^{n - 1} (\alpha - 2k)(\alpha + (2k + 1))}{(2n)!}x^{2n} \
\m{y_2}x &= \sum_{n = 0}^\infty (-1)^n \f{\prod_{k = 0}^{n - 1} (\alpha + 2k)(\alpha - (2k+1))}{(2n+1)!}x^{2n+1}
}
clever fjordBOT
mystic saffron
all i managed to realise is that the Legendre equation has an ordinary point at x=0
but im unable to set up the series solutions
clever fjordBOT
odd edgeBOT
@mystic saffron Has your question been resolved?
mystic saffron
(i will reopen later, mention me in #odes-and-pdes if you have something in mind)
.close
odd edgeBOT
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Available help channel!
leaden light
Did I solve this right ?
leaden light
<@&286206848099549185>
soft zephyr
what ur even trying to do if u can explain
leaden light
is that a differential equation?
yes it's
lean willow
the thing is
it doesnt have an elementary antiderivative 💀
,w y'= 5/sqrt(x+ln(x))^3
leaden light
this equation is cursed isn't it
leaden light
or u made one up yourself
no, I would never made such uncanny equation for myself
lean willow
you sure you want to solve the differential equation? coz it looks like you took derivative of rhs
is it $\frac{5}{\sqrt{x+ln(x)}^{3}}$?
clever fjordBOT
Fractalogist
leaden light
I found it, and yeah we are talking about derivatives. Sorry I just didn't now how to say it English
restive delta
So you've to find the derivative of that?
leaden light
Yes
clever fjordBOT
Lorentz
restive delta
Line 2 should've been this ig
leaden light
so 5 is a part of derivation of complex function ? like in this formula
restive delta
I found it, and yeah we are talking about derivatives. Sorry I just didn't now h...
5? You mean the 5 in the numerator here?
leaden light
yeah
restive delta
Yeah that's a constant
Which is multiplied
So yes it is a part
$\frac{d}{dx} af{(x)} = a\frac{d}{dx} f{(x}), a \in R$
clever fjordBOT
Lorentz
leaden light
i meant, the u (i just don't know how to say it another way) according the formula is a degree of x +ln(x) ?
restive delta
u(v) is (x+lnx)^(-1/5)
u is the function that raises it to power of (-1/5), if that's what you meant
u isn't a degree in itself, because u is a function
leaden light
ok, but why we should raise (x + ln(x)) to -6/5. I thought it just stays without a change
restive delta
Oh
If $x+lnx = t,$
$$\frac{d}{dx} t^{-1/5}= \frac{d}{dt} t^{-1/5} \cdot \frac{dt}{dx}$$
$$\frac{d}{dt} t^{-1/5} = \frac{-1}{5} t^{-6/5}= \frac{-1}{5}{(x+lnx)}^{-6/5}$$
leaden light
oh, okay now I get it
clever fjordBOT
Lorentz
restive delta
Missed the -1/5 at the end
restive delta
oh, okay now I get it
For dt/dx just substitute t=x+lnx back
steady tide
why does it throw math error when i try to calculate $\int_0^3\ln{(x)}dx$ on my calculator when clearly it is defined and solvable?
steady tide
clever fjordBOT
FungusDesu
steady tide
fervent hound
Try to caluculate it manually
solid jungle
Solve it by integrating ln(x).1
That is integration by parts. The calculator is showing zero is because its putting zero, your lower limit, directly in ln(x) which is undefined.
Basically 0 is not in the domain of this function
tall veldt
Yeah your calculator isn't well equipped to compute "improper" integrals
solid jungle
Or you can say it's infinity.
odd edgeBOT
@steady tide Has your question been resolved?
trim kite
can anyone help me with 2)b pls
odd edgeBOT
can anyone help me with 2)b pls
plain badge
can anyone help me with 2)b pls
could you translate it?
odd edgeBOT
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.
@trim kite Has your question been resolved?
odd edgeBOT
@trim kite Has your question been resolved?
toxic monolith
can anyone help me with 2)b pls
$\text{Consider that:}\\\sum_{k=1}^{n}\frac{1}{\left( k+1 \right)^{2}}=\sum_{k=2}^{n+1}\frac{1}{k^{2}}=\\=\sum_{k=1}^{n}\frac{1}{k^{2}}+\frac{1}{\left( n+1 \right)^{2}}-1\\\text{and apply it appropriately in your inequality}$
clever fjordBOT
Joanna Angel
odd edgeBOT
@trim kite Has your question been resolved?
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Available help channel!
fallow thunder
odd edgeBOT
fallow thunder
how is this equal to 5?
wooden python
what does this mean
fallow thunder
limit
fallow thunder
its supposed to be an L, and they wrote it as "Lt" lol
wooden python
how is this equal to 5?
is this a "what is going on" or an "i think it should be something else" situation
fallow thunder
is this a "what is going on" or an "i think it should be something else" situati...
"what is going on", the answer is correct i think
wooden python
you're taking the limit of $\frac{5(x+4)}{(x+1)}$
clever fjordBOT
Ann
wooden python
which can be written in many forms, among others $\frac{5(1 + \frac{4}{x})}{1 + \frac{1}{x}}$
clever fjordBOT
Ann
fallow thunder
oh ok i get it
those 1/x terms become 0
and i get 5
ok understood
thanks!
.close
odd edgeBOT
fallow thunder
.reopen
odd edgeBOT
✅
fallow thunder
err another doubt, gimme a min
@wooden python what do i do here
nvm i got it thank you
.close
frigid canopy
how to find the common tangent to two curves in general?
frigid canopy
Justt wondering
take x^2+y^2=4
and a parabola say (x-5)=y^2
How would I find a common tangent to both of these
crisp wadi
Start by finding their intersection point and verify whether the slopes of the tangents are equal there
The curves that you picked do not intersect by the way
frigid canopy
yeah, my problem is the slope at any point on a circle is given by $-x_1/y_1$
clever fjordBOT
Why am. I here
frigid canopy
and on this parabola $1/2y_2$
clever fjordBOT
Why am. I here
frigid canopy
so then what, ?( This was inspired by KV's question in #help-7｜zen1thxyz , )I'm pretty sure we need points to find a common equation, but that problem has none
crisp wadi
At a point of intersection those would be equal given a common tangent exists
But, again, there is no intersection point in this case
frigid canopy
At a point of intersection those would be equal given a common tangent exists
so no common tangent exists ?
strong prawn
so no common tangent exists ?
Not necessarily
strong prawn
Alternatively
You can find the general form of the tangent for both the curves
And then compare the coefficients
Of each term
Cause they should be the same line
crisp wadi
Say there exists a line tangent to x^2 + y^2 = 4 at x1 and tangent to y^2 = x - 5 at x2
Then 1/(2sqrt(x2 - 5)) = -x1/(-sqrt(4 - x1^2)) = x1/sqrt(4 - x1^2)
Or the same thing but with opposite signs in front of the roots
But the equation stays the same
frigid canopy
Then 1/(2sqrt(x2 - 5)) = -x1/(-sqrt(4 - x1^2)) = x1/sqrt(4 - x1^2)
how did you get this?
gaunt stratus
can someone help me analyze this algorithm
odd edgeBOT
can someone help me analyze this algorithm
odd edgeBOT
@gaunt stratus Has your question been resolved?
odd edgeBOT
@gaunt stratus Has your question been resolved?
odd thunder
Can someone prove this?
odd edgeBOT
Can someone prove this?
shy smelt
Can someone prove this?
Do you know the series expansion of e^x?
odd thunder
Yes
mystic saffron
[
\sum_{n=0}^\infty a_n = a_0 + \sum_{n=1}^\infty a_n
]
clever fjordBOT
odd thunder
what should x be to make it similar
2
shy smelt
Yes
odd thunder
But n starts from 0 in maclaurin expansion of e^x whereas n starts with 1 here
shy smelt
Yes, look at what Qylo said
silver pecan
shiii bruh i didnt even realise that mb, good job on understanding it tho!😂
odd thunder
🤣 its ok happens
odd edgeBOT
Available help channel!
gaunt stratus
is this correctly analyzed?
odd edgeBOT
is this correctly analyzed?
quasi sparrow
is this correctly analyzed?
#old-network for CS server
odd edgeBOT
@gaunt stratus Has your question been resolved?
mystic saffron
odd edgeBOT
mystic saffron
I'm getting 2 answers although there are 4
what I did was
I substituted 1/cos^2x for sec^2x
and 2sin^2x/cos^2x for tan^2x
ended up with
$\frac{1-2sin^2x}{cos^2x}$
clever fjordBOT
mj
mystic saffron
unfortunante
clever fjordBOT
noqmilitary
mystic saffron
oh, for some reason I read your message as
"I can't help"
sorry for that lol
I used zero property
to remove the denom
so in reality it's $sinx=\sqrt\frac{1}{2}$
clever fjordBOT
mj
mystic saffron
yep lol
<@&268886789983436800> idk if intentional spam but can you clear this up?
viscid flint
that's weird
clever fjordBOT
mj
mystic saffron
for sin, we must realize that pi-ans puts us in Q2 where it also seems to be positive
giving us just 2 answers
now my question is
is there 4 answers because
cos2(x) has 3 different expanded solutions
and those 3 different expanded solutions would give 2 different answers?
viscid flint
what does cos2(x) have to do with it
clever fjordBOT
mj
viscid flint
please don't ever write cos2(x) for cos(2x) lol
mystic saffron
$2cos^2x-1 = cos(2x)$
clever fjordBOT
mj
mystic saffron
sorry about that lol
viscid flint
will it?
viscid flint
it shouldn't, if they are equal
mystic saffron
pi/4 (again), and 7pi/4 (new answer?)
mystic saffron
it shouldn't, if they are equal
well it makes sense for it to
because you'd have to subtract/add it into different quadrants
than sine for example
viscid flint
,w 1 - 2 * sin(7*pi/4)^2
clever fjordBOT
mystic saffron
Okay, WE
mystic saffron
it's assumed to be 0 to 2pi
we have 3 solns thus far
pi/4, 5pi/4, 7pi/4
not sure where to get the 3pi/4 from though
clever fjordBOT
mj
mystic saffron
$x_{1/2}=\pm\frac{\pi}{4}$
clever fjordBOT
mj
mystic saffron
we have to reject -ve though
since -pi/4 or any -ve number is
outside the range
of the domain
viscid flint
but you have to consider +2pi k
mystic saffron
For sin(x)
$$\frac{\pi}{4}, \frac{3\pi}{4}$$
For cos(x)
$$\frac{\pi}{4}, \frac{7\pi}{4}$$
clever fjordBOT
mj
mystic saffron
but you have to consider +2pi k
yeah, which gives us 7pi/4
ok this is weird
I had 5pi/4, and didn't have 3pi/4
but now i have 3pi/4 and not 5pi/4
mystic saffron
but it'd be impossible, both sinx and cosx can only have 2 solutions
mhm
lets try that for cos
viscid flint
sin and cos should both give you all 4 solutions
mystic saffron
it's already in the +ve quadrant as a negative
adding 2pi to get it back in that very same quadrant as a non-negative is 7pi/4
now for sine
it's in 4th quadrant
we need it in 1st and 2nd
as a positive
the RA is pi/4
so pi/4 in Q1, pi-pi/4 in Q2
3/4 in Q2
the +/- isn't the thing here
we've gotten all the +/- answers
for 5pi/4, I'm trying to think of a scenario
where you'd add the reference angle, pi/4 to pi so you can get it in Q3
(only in tan)
now how about subtract
pi
to get 5pi/4
ah I'm lost lol
clever fjordBOT
viscid flint
$\sqrt{\f12} = \f{\sqrt1}{\sqrt2} = \f1{\sqrt2}$
clever fjordBOT
hayley
viscid flint
it's the -ve solution
it's one of the negative solutions for sine yes
there are two
mystic saffron
which would be in Q3 or Q4
so 2pi-pi/4
or pi+pi/4
ahh i see
this should not have taken this long lol
just put it all in terms of sin/cos then solve the quadratic 😨😨
not to mention it is a difference of squares
2x^2-1
havent worked it out
(x+1)(x-1)
But then u shld get the + and - value of x
2(x+1)(x-1)
and theres 2 values between 0 and 2 pi
yep, got it now
thanks
.close
polar plaza
odd edgeBOT
polar plaza
what do you do with sin(x/2)
quasi sparrow
Use sine trig identities
polar plaza
which one
mental grove
double angle but replace x with x/2?
quasi sparrow
But since x/2 = x * 1/2, that's the same thing
quasi sparrow
Oh that works too
polar plaza
ah
clever fjordBOT
riemann
quasi sparrow
Top middle
violet elk
A bit confused on this, I thought it was f (g(-15)
odd edgeBOT
A bit confused on this, I thought it was f (g(-15)
violet elk
You would plug -10 into f(-10)?
So shouldn’t it be 8?
Oh I don’t need help anymore.
.close
odd edgeBOT
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Available help channel!
verbal tusk
Could anyone explain me this solution, primarily the "T has 2^m elements", please.
low locust
well T consists of tuples of length m of which each entry can have 2 values
so 2*2*2*..*2=2^m elements
verbal tusk
so 2\*2\*2\*..\*2=2^m elements
but that would be the number of combinations, for that to be true then all the combinations must be inside T
how can I be sure that all possible combinations are inside T?
low locust
cause T is defined that way
T is the set of all those tuples
the function f doesnt have to hit every element in T
verbal tusk
but like, the tuple (0,0,0,...,0) where the element is inside none of A's is not there
low locust
well first, (0,0,..,0) is not a set
second, why should there be no element x which is in none of the A's
mystic saffron
how do you determine the horizontal asymptote of a function
mystic saffron
I'm aware that if the numerator and denominator have equal highest powers
you would divide the two
LC
ember oak
how do you determine the horizontal asymptote of a function
Depends on the function
mystic saffron
yeah, I wanted to know in the instance they werent the same
maybe the numerator was more or less than the denominator
how would you find it then
wooden python
answered your own question
ember oak
like $\frac{Ax+B}{Cx^2+Dx+E}$?
wooden python
if the fraction is top-heavy there is no HA at all
if it's bottom-heavy the HA is y=0
why the fuck are my messages not sending
clever fjordBOT
SWR
mystic saffron
I also know that there is no horizontal asymptote if the top is one more than the bottom and instead there is an oblique asymptote
mystic saffron
if the fraction is top-heavy there is no HA at all
mm
ember oak
I also know that there is no horizontal asymptote if the top is one more than th...
So where are you confused, exactly?
mystic saffron
so let me get this straight
hexed wigeon
how would i make a eco system similar to chess but for a different game that has more variables affecting it like winrate, average wins in a row, etc
mystic saffron
in p(x)/q(x)
if p(x) is one greater, no HA, but OA
if p(x) >1 difference, HA=0
if p(x)=q(x), LC/LC
ember oak
how would i make a eco system similar to chess but for a different game that has...
!occupied
odd edgeBOT
how would i make a eco system similar to chess but for a different game that has...
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).
mystic saffron
did I get everything?
odd edgeBOT
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).
mystic saffron
in p(x)/q(x)
if p(x) is one greater, no HA, but OA
if p(x) >1 difference, HA=0
i...
what if
p(x)<q(x)
mystic saffron
I used to understand the logic behind them which helped me memorize it better
but is the reason it being 0 because it's approaching 0 the larger the denom is?
ember oak
When looking at p(x)/q(x), look only at the largest coefficients
So this would become $\frac{p(x)}{q(x)}\approx\frac{Ax^n}{Bx^m}$
clever fjordBOT
SWR
mystic saffron
if n>m+1, no asymptote
why is this
mystic saffron
if n<m, then HA=0
I know this is because as m gets infinitely larger, the function approaches 0
or atleast that's my reasoning for it
ember oak
If $n<m$, then $\frac{p(x)}{q(x)}\approx\frac{Ax^n}{Bx^m}=\frac{A}{Bx^{m-n}}$. As $x$ goes toward $\pm\infty$, the fraction approaches $0$.
clever fjordBOT
SWR
mystic saffron
ah okay
m-n
m is larger and larger
until n doesn't matter anymore
or am I getting confused
ember oak
Yeah you basically have it
Also, if $n>m+1$, then $Ax^{n-m}$ is always a polynomial asymptote
clever fjordBOT
SWR
ember oak
example:
mystic saffron
I guess I just need a bit more practice with these lol
thanks for clarifying though
ember oak
This example is more of a "neat thought" than something they may end up teaching you
mystic saffron
ah
clever fjordBOT
SWR
mystic saffron
lol alright
ember oak
But yeah, neat stuff
Yeah you don't need to remember this fact, just sharing something cool
.close
eternal sorrel
odd edgeBOT
eternal sorrel
why is (CD,AB)=π/2 +kπ where k is an integer
shouldnt it be π/2 +2kπ or am i missing something
upper onyx
for the same direction you need 2pi (a full circle), for (co)linearity you need only pi (half circle)
thats what i said.
eternal sorrel
for the same direction you need 2pi (a full circle), for (co)linearity you need ...
direction of the 2 vectors CD and AB ?
it is the same
bro what are you talking about
wdym by 2 vectors for CD
this doesnt make sense ig
cinder elk
does this question require any integration @eternal sorrel ? from the beginning?
eternal sorrel
i mean this sentence doesnt make any sense
no this is just a question concerning arguments of complex numbers
cinder elk
why?
eternal sorrel
bro integrate what
upper onyx
never ever.
eternal sorrel
if i have 4 points A,B,C and D in the complex plane with affixes a,b,c and d respectively
bro there is no integration here
cinder elk
based off another person that was helped and this conversation I find it hard to believe that Rick knows what they are doing and just trolling
or guessing
eternal sorrel
based off another person that was helped and this conversation I find it hard to...
ig this fits
upper onyx
if i have 4 points A,B,C and D in the complex plane with affixes a,b,c and d res...
sorry, my "what" was the reply to another post.
cinder elk
<@&268886789983436800>
eternal sorrel
sorry, my "what" was the reply to another post.
dw np
warped grove
Do the question before you try to help others with it
eternal sorrel
tysm
so my question is how do ik when to add kπ and when to add 2kπ where k is an integer
eternal sorrel
upper onyx
colinearity is when the are on one line, so pi is enough.
cinder elk
Quick question, what is the name of this math and the like "unit/chapter name"?
eternal sorrel
colinearity is when the are on one line, so pi is enough.
do i have to check if both vectors are collinear or no before finding the angles/arguments of the corresponding complex numbers ?
and even if they are collinear this isnt enough
i would have to check their sense too
upper onyx
you asked why its pi and not 2pi. my argument was: for the same direction you would need 2pi, but for colinearity you need only pi.
eternal sorrel
how do ik when they are collinear algebraically
how do ik when do i add kπ and when to add 2kπ
without having a geometrical reference such as a figure
upper onyx
vectors are a geometrical reference.
eternal sorrel
i mean without a figure
upper onyx
i mean you are posting a screenshot with the title "geometric representation":
in this screenshot its mod 2pi because its about the same direction, so its 2pi. arg z is the angle, and again here you need the same direction therefore 2pi.
but for colinearity ...
you need only pi.
eternal sorrel
ok
i was told before that if i have z=x+yi then arg(z)=arctan(y/x)+2kπ or arg(z)=arctan(y/x)+π+2kπ
depending on the sign of x
but why does it only depend on the sign of x
i mean doesnt the sign of y also play a role
upper onyx
well, assume |y|=2, |x|=1, and now try all combinations with + and - -> you will get 4 vectors. make a sketch, and think about arctan(y/x)., then you should see, why its +pi in one case.
ore more technically:
arctan gives you values in (-pi/2, pi/2) which means x > 0.
so you need a case for x < 0.
eternal sorrel
let arg(z)=arctan(y/x)+2kπ=α+2kπ where x>0 and y>0
then arctan(-y/x)+2kπ=-α+2kπ
so from here where does +kπ come from for x<0
upper onyx
as said before: arctan gives you values in (-pi/2,pi/2) which means x > 0. so you need an extra case for x < 0. as tan has a periodicity pi you need the +pi case.
upper onyx
because of the definition of arctan. arctan gives you values in (-pi/2,pi/2) which means x > 0.
eternal sorrel
x here is referring to the input of arctan am i wrong
eternal sorrel
because of the definition of arctan. arctan gives you values in (-pi/2,pi/2) whi...
so you are talking about arctan(x) here
but for arctan(y/x) you cant simply ignore y and look at the sign of x bc y changes the sign too
upper onyx
you started with an x+iy. why do you change the meaning of x now?
eternal sorrel
arg(z)=arctan(y/x)+ either kπ or 2kπ where k is an integer and my question was when do i add kπ and when 2kπ
eternal sorrel
yes how did i change the meaning of x
i am saying that i am looking at arctan(y/x)
so x<0 or >0 isnt the only thing that we should be looking at
we should look for the sign of y too
why is this claim wrong
upper onyx
so, i will try it a fourth (and last) time. arctan gives you values in (-pi/2,pi/2) which means x > 0.
so you need a case for x < 0.
eternal sorrel
ah wait
eternal sorrel
so, i will try it a fourth (and last) time. arctan gives you values in (-pi/2,pi...
bc of x=rcosα and y=rsinα then for α in (-π/2,π/2) x>0 necessarily is this the reason ?
upper onyx
eternal sorrel
here x is always positive but y changes sign
so we need to cover the cases when x<0 to cover all possible values
to reach the region of the complex plane where x<0 we need to add an integer odd multiple of π to the angle between the positive x-axis and the vector/line joining the origin of the system and the point with coordinates (x,y)
thats why for x<0 we have arg(z)= arctan(y/x)+(2k+1)π instead of arctan(y/x)+2kπ
but if x<0 then we can have the case arg(z)= -arctan(y/x)+(2k+1)π no?
oh wait no bc sin(π-α)=sinα and cos(π-α)=cosα so tan(π-α) neq tanα
tysm and sorry i wasted your time bc of my stupidity i shouldve tried using trig form
.close
odd edgeBOT
odd edgeBOT
Available help channel!
fallow thunder
odd edgeBOT
fallow thunder
can someone give me an example of how to use tangent function to represent dot product/ cross product?
exotic bobcat
I need to find those weights w,1,2,3 such that on the interval [0,1] all polynomials of degree 2 (or lower) are integrated exactly by the given formula
exotic bobcat
How do I even go about that? I don't even understand where/how to start with this
(it's numerical integration)
mortal trench
,w \int_0^1 (ax^2 + bx + c)
clever fjordBOT
mortal trench
f(0) = c
f(1/3) = a/9 + b/3 + c
f(2/3) = 4a/9 + 2b/3 + c
then you have a system of linear eqns
exotic bobcat
oh so I write down the form of a 2nd deg. polynomial in general and then integrate its general form (with constants, in this case a,b,c)
and then just resolve the linear equations
mortal trench
y
exotic bobcat
thanks!
I assume I could also do this for larger degrees using basically the same idea?
mortal trench
yes. but you will need as many points as you have coefficients
notice that we have three weights for three coefficients (3 points needed to define a second degree poly)
exotic bobcat
does it matter which points so long as the system of equations is not underdetermined?
mortal trench
as long as the system is fully determined you are good
odd edgeBOT
odd edgeBOT
Available help channel!
frigid canopy
Is there any way to do this without finding p and q explicitly? More specifically, could somoeone give me a hint on how to find pq
frigid canopy
p+q is pretty easy
I've tried expressing $p^4+q^4$ as $ (p^2+q^2)^2-2p^2q^2$, but I don't know how to find that either
clever fjordBOT
Why am. I here
mystic saffron
yes you are on the right track
calm raft
you can try factorising it
that's one way I guess
and maybe form linear equations?
idk if there's a more efficient way tho
mystic saffron
[
(p^2+q^2)^2-2p^2q^2 = 272 \Implies \p{(p+q)^2-2pq}^2-2p^2q^2
]
clever fjordBOT
signal oar
I've tried expressing $p^4+q^4$ as $ (p^2+q^2)^2-2p^2q^2$, but I don't know how ...
Vieta's formula is very useful here
mystic saffron
you can expand the right parentheses
frigid canopy
One of its statements are that the product of the roots is equal to c
thanks
frigid canopy
you can expand the right parentheses
will try, thanks
mystic saffron
do some algebra and you get [
2p^2q^2 -16pq +16 = 272
]
then what kepe is talking about applies
clever fjordBOT
signal oar
\[
(p^2+q^2)^2-2p^2q^2 = 272 \Implies \p{(p+q)^2-2pq}^2-2p^2q^2
\]
What about doing one more cycle of what he did above
$\big((p + q)^2 - 2pq\big)^2 - 2p^2q^2$
plugging in A), we have $p + q = 2$ and $pq = 8$ from Vieta's formulas
Just check if that gives us 272
Similarly for all 4
frigid canopy
where did you get the 8 from? Sorry, I'm not very familiar with vieta's formula, just know the relation for squares and cubes
signal oar
where did you get the 8 from? Sorry, I'm not very familiar with vieta's formula,...
For $x^2 + bx + c = 0$, Vieta's formulas state that \begin{enumerate} \item $p + q = -b$ \item $pq = c$\end{enumerate}
Where p and q are the roots
Wait, did I mess something up
Doesn't seem so, @plain badge?
plain badge
For $ax^2 + bx + c = 0$, Vieta's formulas state that \begin{enumerate} \item $p + q = -{b/a}$ \item $pq = {c/a}$\end{enumerate}
signal oar
For $ax^2 + bx + c = 0$, Vieta's formulas state that \begin{enumerate} \item $p ...
See how a is 1 there
If you had ax^2, sure
I stated it for x^2 + ... though
plain badge
See how a is 1 there
but formula is -b/a and c/a
clever fjordBOT
Spacephoenix
signal oar
but formula is -b/a and c/a
Yes, for $ax^2 + bx + c = 0$. If you have $x^2 + bx + c = 0$, $a$ is equal to $1$
mystic saffron
btw uh
mystic saffron
there is no need to overcomplicate this that much i guess
signal oar
👍
Yeah. In this case, we always have x^2, there is never a coefficient
So we just have
(I agree that for ax^2 + bx + c = 0, Vieta's formulas state what you wrote, but I just mean the a = 1 case!)
plain badge
Ig its better if we give the complete information to the person who has doubt as they will be able to deduce the later..
and it would help
signal oar
Ig its better if we give the complete information to the person who has doubt as...
Sure
frigid canopy
ah, I see, that's it? Thanks!
signal oar
Well, now just deduce pq of all of them using this
odd edgeBOT
signal oar
For the first one, pq = 8, for the second 16, ...
Check if the expression becomes 272
odd edgeBOT
Available help channel!
warm gate
help
warm gate
noble forge
which one?
warm gate
all of em but lets start with #1, the explanations are too short for me
noble forge
are you aware of change of bases rule?
warm gate
can you give an example
noble forge
$\log_{a}b=\frac{\log_{c}b}{\log_{c}a}$
warm gate
Im sorry Im new to some terms
clever fjordBOT
B-eard
warm gate
oh not rlly
noble forge
sure
noble forge
**B\-eard**
using this formula can you bring log base 10 on num and well as denom?
for #1
clever fjordBOT
B-eard
noble forge
now break 20 into a multiple of 10
Also, are you aware of this: $\log_{c}ab=\log_{c}a+\log_{c}b$?
clever fjordBOT
B-eard
warm gate
yes
noble forge
$\log_{2}20=\frac{\log_{10}20}{\log_{10}2}=\frac{\log_{10}\left(10\cdot2\right)}{\log_{10}2}$
Do you see what to do from here?
clever fjordBOT
B-eard
noble forge
$\log_{2}20=\frac{\log_{10}20}{\log_{10}2}=\frac{\log_{10}\left(10\cdot2\right)}{\log_{10}2}=\frac{\log_{10}10+\log_{10}2}{\log_{10}2}$
clever fjordBOT
B-eard
noble forge
Well, from here on you should be able to get to the answer
noble forge
yep
warm gate
okay thanks
odd edgeBOT
@warm gate Has your question been resolved?
main totem
log base changes?
main totem
seems right to me I think
stuck igloo
yeah its right
warm gate
warm gate
who're you asking? you are the one who asked the question?
i didnt mean to put a question mark lol mb
stuck igloo
write 56 as 7x8 and 14 as 2x7
warm gate
kk
wait last
what exactly was done here
I dont get how it got to 4
<@&286206848099549185>
light raven
hello
toxic rose
,rccw
clever fjordBOT
plucky mountain
light raven
normal simplification of logarithmic expression @warm gate
like
log_3(2) + log_3(2) = log_3(2*2) = log_3(4)
morever,
-2log_3(2)
= - log_3(2^2)
= -log_3(4)
log_3(4) + 3 + 1 - log_3(4)
= 4
@warm gate
light raven
sure
express 27 and 18 on terms of multiplication of prime numbers or rather on terms of the power of the base
then simplify further
light raven
keep it going
u r good to go
calm raft
yw
warm gate
thanks a lott
calm raft
there're literally countless helpers lmao no need to worry
there are like 100+ Math Professors in this server lol
and 10000+ university students
im probably only better at maths than 40% of the population
warm gate
warm gate
ohh
okay okayy I get it
maybe its just because im not so familiar with the rules of log
calm raft
maybe its just because im not so familiar with the rules of log
there's really not a lot of rules in log
warm gate
do you guys have a list of those rules
clever fjordBOT
lebesgue
warm gate
uhh
calm raft
they haven't teach you yet?
calm raft
thats one of the rules
calm raft
this isnt a lesson for me, im forced to self learn
Ohhhh I see
are you preparing for a specific exam?
light raven
do you guys have a list of those rules
warm gate
thanks
tepid acorn
Oops