your biggest challenge is evaluating the other derivative
#help-42
eternal shard
eternal shard
it's alright
well the exponential function with base e has very nice derivative properties
simple musk
,, g'(x) = e^{f(x) - 2} + x \cdot \left((f(x)-2)' e^{f(x)-2}\right)
potent lotusBOT
938c2cc0dcc05f2b68c4287040cfcf71
simple musk
something like this ?
simple musk
f'(x)
eternal shard
pro
simple musk
,, g'(x) = e^{f(x)-2} + xf'(x)e^{f(x)-2}
potent lotusBOT
938c2cc0dcc05f2b68c4287040cfcf71
simple musk
hahah
eternal shard
yup
airy would be proud
Now
you have everything you need
to finish
y = g'(3)(x-3)+g(3)
simple musk
,, g'(3) = e^{f(3)-2} + 3f'(3)e^{f(3)-2}
potent lotusBOT
938c2cc0dcc05f2b68c4287040cfcf71
simple musk
,, g'(3) = 1 -6 = -5
potent lotusBOT
938c2cc0dcc05f2b68c4287040cfcf71
eternal shard
good.
simple musk
y = -5(x-3)+3
eternal shard
yup
eternal shard
nononono
you should thank the other person
i was just acting on their account
i felt bad for how you treated them
simple musk
wdym
well, thank u both
is just that he jumped to solving and used the term gradient to refer to the slope
it was too advanced
eternal shard
like this after 2 paragraphs of trying, it seems to me harsh and disrespectful, like their effor was for nothing
eternal shard
it was too advanced
it is true that gradient in math has a more advanced meaning, but in terms of English gradient and slope are synonymous
so it could also be that they meant what you would undestand as slope
simple musk
I will try to change my behaviour tbh, but idk I really didnt understood anything
eternal shard
that's ok
simple musk
but that is prolly because I didnt know what a tangent line meant
eternal shard
no judges
or grudges
we help
honesty helps and saves time
and if someone ridicules you, you can ping mods
you need to let people know your level in math
thats how people can help at best
and also
adjust
to you
be less advanced more intuitive etc
simple musk
should I say that I am newbie?
simple musk
is just awkward to ask about tangent line if I thought I knew what it was but not really at the same time
like is awkward to say that I dont know
eternal shard
that's why i asked
simple musk
like how much do I know really?
eternal shard
that's agood question
the goal is not to know all but get a good grasp of things
but dont be afraid
if you know something 100 % or not
cause truth is
nobody does
what they say in math
what is a tensor? a tensor is something that behaves... like a tensor
😂
simple musk
lol I didnt knew that one
eternal shard
me neither
but i read so many times now
i really appreciate
how you really try
and do all these tasks
simple musk
wdym?
eternal shard
but that's also no use if there are minor inconsistencies in your understanding
simple musk
yes
simple musk
I agree
eternal shard
when mpl said
you should do it so many times until you understand it
that's partially true
because thing is you also need to understand
doing a thing 1000 times wrong
wont get you also far
simple musk
yes quite precisely
I will try harder to understand tbh I shouldve understoof with what the other helper said
simple musk
it was very rude, I am sorry
eternal shard
no problem
simple musk
.solved
calm coralBOT
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nocturne mist
I need help with limits
calm coralBOT
I need help with limits
strange tinsel
for the first, use $\mathbf{\quad (a-b) \cdot (a^2 + b^2 + ab) = a^3 - b^3} \ with \quad a=(x+6)^{\frac{1}{3}} \quad & \quad b=2$
potent lotusBOT
Milena
calm coralBOT
@nocturne mist Has your question been resolved?
strange tinsel
for the 2nd, i suggest you calculate the 2 limits, when x approaches 3 from the left and when x approaches 3 from the right, separately
in order to be able to remove the absolute value in the denomn
oh, and factor the numerator before that
nocturne mist
thank you i got it already
@strange tinsel can you help with another question>
?
its about looking at a graph
@sly geyser
for a) it would be x= -1 right?
and for b) it should be x=-1 and x=2
west ermine
I need a video or a link on row operations to simplify the matrix using Gauss Jordan, Im kinda lost on how they do it
west ermine
<@&286206848099549185>
calm coralBOT
@west ermine Has your question been resolved?
calm coralBOT
dreamy lance
Please open your own help channel
slate field
Sorry I m new here
just post in one without any name in it
snow roost
Ok
calm coralBOT
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remote mural
sqrt(x^2) = x or mod x ?
calm coralBOT
sqrt(x^2) = x or mod x ?
remote mural
.close
calm coralBOT
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dreamy hull
I am sorry but let me use a help channel for practicing memorization
dreamy hull
Orthogonality properties of Lie algebra, one has lemma
unique jackal
Absta help channel 
dreamy hull
Welp, I am just going to post what I memorized in my Lie algebra for a quiz
Hello higher!
8.3. Orthogonality properties
Lemma 8.3
(1) span(Phi) = H^*
(2) a \in Phi, then -a \in Phi
(3) [xy] = k(x,y) t_a (for x in L_a, y in L_{-a}
(4) [L_a L_{-a}] = F t_a
(5) a(t_a) = k(t_a, t_a) \neq 0
(6) For e_a \in L_a - 0, one has f_a \in L_{-a} so h_a = [e_a f_a], e_a, f_a spans simple Lie alg sl(2).
(7) h_a = 2t_a/k(t_a, t_a), h_(-a) = - h_a
Prop 8.4.
(1) dim L_a = 1,
L_a + L_{-a} + H_a = S_a (simple sl(2)), H_a = [L_a, L_{-a}],
and for each e_a \in L_a, f_a \in L_{-a} s.t. h_a = [e_a f_a]
(2) \Phi \cap F a = {\pm a}
(3) a, b \in \Phi then b(h_a) is integer, b - b(h_a) a \in \Phi
(4) [L_a, L_b] = L_{a+b}
(5) b \neq \pm a. For maximal q, r: b - r a, b + q a \in \Phi, in between b + ia \in \Phi, and b(h_a) = r - q.
(6) L is generated by root spaces L_a.
Thm 8.5
One has \Phi \subset E with pos-def bilin. form (-,-) that
(1) span(\Phi) = E, 0 \notin \Phi
(2) \Phi \cap F a = {\pm a}
(3) b - (b, a)/(a, a) a \in \Phi if a, b \in \Phi, b \neq \pm a
(4) (b, a)/(a, a) is integer (Called Cartan integer)
Lem 8.1
(1) [L_a, L_b] \subset L_{a+b}
(2) x \in L_a is nilpotent.
(3) k(L_a, L_b) = 0 for a + b \neq 0
Cor 8.1
k|_{L_0, L_0} nondegen.
Cor 8.1'
k|_{L_a, L_{-a}} nondegen.
Thm 8.2. H: maximal toral subalgebra, then H = L_0 = C_L(H).
Cor 8.2. k|_{H, H} is nondegen.
This gives basis for identifying H = H^* via a = k(t_a, -)
9.1. Reflections
Defines reflection as
sigma_a(l) = l - 2 (l, a)/(a, a) a
Characterizing properties of sigma_a:
- sigma_a (a) = -a
- Fixes P_a (hyperplane orthogonal to a
Lem 8.1.
Finite set \Phi spans E, 0 not in _Phi, each \sigma_a(\Phi) = \Phi.
Suppose \sigma satisfies \sigma(\Phi) = \Phi, \sigma fixes hyperplane P, \sigma(a) = -a. Then, \sigma = \sigma_a, P = P_a
9.2. Root systems
Root system \Phi is defined as
(R1) finite set spanning E, 0 not in \Phi
(R2) \Phi \cap \bR a = {\pm a}
(R3) b, a \in \Phi, b - <b, a> a \in \Phi
(R4) <b, a> \in \bZ for a, b \in \Phi
Has Weyl subgroup, subgp of GL(E) generated by \sigma_a 's.
Lem 9.2'
Isomorphism \phi : E \to E' maps root system \Phi to \Phi', then
\phi \sigma_a \phi^{-1} = \sigma_{\phi(a)}
and so <b, a> = <phi(b), phi(a)>.
Lem9.2. The same can be said for \phi \in GL(E) preserving \Phi.
Hence, isomorphism is just defined by linear iso sending \Phi to \Phi'.
Dual space is given by collecting
a^\vee = 2a/(a,a)
9.4.
Lem 9.4
Suppose a, b \in \Phi, b \noteq \pm a.
- If (a, b) > 0, then b - a \in \Phi.
- If (a, b) < 0, then a + b \in \Phi.
Tf why do I have to memorize all these
7.1.
Decompose module M into h-weight spaces M_l
Lem 7.1.
e M_l \subset M_{l+2}, h M_l \subset M_l, f M_l \subset M_{l-2}
obtuse briar
do the people who know how to help with this even check these help channels
dreamy hull
I am sorry, I know only I can help myself with this memorization. I hope you understand I needed a place to write down stuff..
obtuse briar
I see
dreamy hull
Thanks!
7.2.
For a maximal vector v_0 \in M_l (meaning it is nonzero and e v_0 = 0)
v_i = 1/(i!) f^i v_0
Lem 7.2.
(1) h v_i = (l - 2i) v_i
(2) f v_i = (i+1) v_{i+1}
(3) e v_i = (l - i + 1) v_{i-1}
graceful dust
dreamy hull
Thm 7.3
Irreducible sl(2)-module M.
(1) M = M_m \oplus M_{m-2} \oplus \cdots \oplus M_{-m}, \dim M_i = 1, \dim M = m+1.
(2) It has a unique maximal vector up to scalar, and M has maximal weight of m = \dim M - 1.
(3) For each positive dimension, there is unique irreducible sl(2)-module up to iso.
Cor 7.3
For any finite-dim sl(2)-module M,
(1) Eigenvalues of h on M is integers.
(2) \dim M_i = \dim M_{-i}
(3) M is decomposed into irreducible submodules with \dim M_0 + \dim M_1 summands.
@dreamy hull Has your question been resolved?
calm coralBOT
@dreamy hull Has your question been resolved?
proud vector
calm coralBOT
timid blade
you form cases
1 case where you choose the 2 students, 1 where you dont
and add the answer you get from both the cases for your final answer
can you try that?
calm coralBOT
@proud vector Has your question been resolved?
proud vector
and add the answer you get from both the cases for your final answer
I did the 1st part of the question but I dont know about he last part
timid blade
I did the 1st part of the question but I dont know about he last part
like i said, for the second part you form 2 cases
one case would be where the 2 students are chosen
the other where they are not chosen
timid blade
you add the answers from both the cases
proud vector
like i said, for the second part you form 2 cases
thats the answer for the first part of (b)
the question (b) has two parts
timid blade
oh wait
mb
so you want to find ways of arranging the team
it should be 3!*3! since you can arrange the students and teachers in 3! ways each
proud vector
it should be 3!*3! since you can arrange the students and teachers in 3! ways ea...
how?
there are 10 students and 5 teachers
so I have to choose among 15 of them
timid blade
youve already formed the teams
you now have to arrange them
so its basically arranging 3 teachers and 3 students
in pairs
calm coralBOT
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Available help channel!
latent gulch
hi guys pls help with 14b
calm coralBOT
hi guys pls help with 14b
calm coralBOT
@latent gulch Has your question been resolved?
obtuse briar
ah I see some problems at around the 2nd and 3rd line with parenthesis one sec I'll show what I mean
so for starters, since you're taking natural log I wouldn't even write ln(e) on the left, I would just put 3x+4y
that way you don't have to worry about not putting parenthesis there
for the right side of that second line, the ln is of the whole thing, like this $$\ln(2e^{2x-y}) = \ln(2) + \ln(e^{2x-y})$$
potent lotusBOT
Merosity
obtuse briar
then you can bring the exponent down on the e, since the 2x-y is only an exponent on e, not on (2e)
that make sense? take your time and ping me when you respond or if you have any questions about that so far
remote mural
can an equilateral triangle be constructed inside a square with these conditions?
- atleast one side of the triangle is on one of the sides of the squares (the sides are not required to coincide completely)
- The third vertex lies on the square.
glass heart
with on the square, do you mean on one side of the square?
glass heart
can the side of the triangle be longer than the side of the square that it is on?
glass heart
well you say ofc but clearly if not then how could the third vertex ever reach the opposing side
the height of the triangle is clearly less than the height of the square
calm coralBOT
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latent gulch
- use logarithms to change each of these non-linear equations into the form Y = mX + c
a.) y = a(x^-b)
solution
y = a(x^-b)
ln y = lna * ln(x^-b) => (whole eq times ln)
ln y = -blnx + lna
Y = ln y
m = -b
X = lnx
c = lna
- Solve the inequalities, giving your answer in terms of base 10 logarithms.
a.) 7 x (5/6)^(3-x) < 4
solution:
7 x (5/6)^(3-x) < 4
(5/6)^(3-x) < 4/7 => (whole eq divide by 7)
lg(5/6)^(3-x) < lg(4/7)
3-x lg(5/6) < lg(4/7)
3-x < (lg 4/7)/ (lg 5/6)
-x < [(lg 4/7)/ (lg 5/6)] -3
x > 3 - [(lg 4/7)/ (lg 5/6)]
guys, how do i know i should multiply the whole equation by ln or move the fraction?
latent gulch
for the right side of that second line, the ln is of the whole thing, like this ...
ohhh
latent gulch
that make sense? take your time and ping me when you respond or if you have any ...
yesss i understand it now, thank you so much! sorry for the late reply
obtuse briar
yesss i understand it now, thank you so much! sorry for the late reply
happy to help!
obtuse briar
loggin off, my bed time goodnight!
latent gulch
loggin off, my bed time goodnight!
okaayy good nightt
upper sparrow
Be a bit careful with the second one, notably here 
(assuming lg is base 10 [oh, you said in the original 
]) you have lg(5/6) being negative, so the inequality actually reverses as you're dividing by a negative 
latent gulch
ohhh yess
latent gulch
(~~assuming~~ lg is base 10 [oh, you said in the original <:catGiggle:6689487177...
thank youuu
upper sparrow
thank youuu
latent gulch
btww how do i know i should multiply the whole equation by ln or move the fraction?
latent gulch
in the first question i multiplied all by ln
latent gulch
in the second picture i divide the 7 first
upper sparrow
in the first question i multiplied all by ln
Aha, on that note! be careful, the line at the top is really "applying ln to both sides", rather than "multiplying both sides by ln", ln is a function (as are all other logs) so you're applying ln to both sides of y = ax^{-b}
So really that should be $\ln(y) = \ln(a\cdot x^{-b})$, and the line below it actually works out as $\ln(y) = \ln(a) - b\ln(x)$ by log rules
potent lotusBOT
@upper sparrow
upper sparrow
Which you seem to have stated further down at least 
upper sparrow
btww how do i know i should multiply the whole equation by ln or move the fracti...
In general though, "it doesn't really matter", you can do either
latent gulch
<@788085606483361802>
ohhh
upper sparrow
So if instead I took the log of both sides for the second one, I would have
[
\lg\qty(7 \qty(\frac56)^{3 - x} ) < \lg(4)
]
but log rules lets you turn the left hand side into
[
\lg(7) + (3 - x) \lg\qty(\frac56) < \lg(4)
]
from where if you subtracted that other log and turned it, using log rules, into
[
(3 - x) \lg\qty(\frac56) < \lg(4) - \lg(7) = \lg\qty(\frac47)
]
it's the same thing!
potent lotusBOT
@upper sparrow
latent gulch
So if instead I took the log of both sides for the second one, I would have
\[
\...
THANK YOU SO MUCHH
upper sparrow
THANK YOU SO MUCHH
Always a pleasure 
calm coralBOT
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vernal plank
i need help to setting up the limit of this function
calm coralBOT
i need help to setting up the limit of this function
vernal plank
c(theta) ?
calm coralBOT
@vernal plank Has your question been resolved?
vernal plank
<@&286206848099549185>
vernal plank
<@&286206848099549185>
vernal plank
<@&286206848099549185>
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@vernal plank Has your question been resolved?
devout zodiac
vernal plank
?
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@vernal plank Has your question been resolved?
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@vernal plank Has your question been resolved?
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@vernal plank Has your question been resolved?
vernal plank
.close
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vital flame
In the figure, angles CDA and DAB are right angles. Calculate the lengths of line segments AB and BE.
calm coralBOT
@vital flame Has your question been resolved?
vital flame
<@&286206848099549185>
remote mural
hello
vital flame
Hi
remote mural
well are you sure if your answer for AB is correct?
vital flame
Alright
remote mural
are you well acquainted with the cos rule?
vital flame
Im not really sure about sin and cos
Im doing sin and cos pretty soon
Havent gotten to it
Is it possible with just tan as thats the chapter its from
remote mural
I am not quite sure if I can think of a solution with tan but I know the solution with sine
so first things first, we would have to figure out the value of angle AEB
,w calculate 180 - 37 - 53
potent lotusBOT
remote mural
,w calculate 90*4
potent lotusBOT
remote mural
as it turns out, angle AEB is also 90 degrees therefore its a right angled triangle and we can use tan now
vital flame
How so
remote mural
alright mate so let's focus in on triangle AEB for now
-
you can see that angle DAB is equal to 90 degrees and angle DAE is equal to 53 degrees. threfore angle EAB should be 90 - 53
-
we know that tan(theta) = opp/adj
therefore tan (angle EAB) = BE / AB
- we already know the value of AB therefore we can write this equation in terms of BE
AB * tan (EAB) = BE
,w calculate (tan (37 degrees)) * 5.4
potent lotusBOT
remote mural
if your calculation for AB is correct, you should obtain the value above as your final answer
vital flame
Which angle do u mean on DAB
vital flame
U said dab equals 90
What do u mean
Isnt ab hypotenusa
Ima bit conufsed as to what u did
Ur first step
calm coralBOT
@vital flame Has your question been resolved?
vital flame
<@&286206848099549185>
calm coralBOT
@vital flame Has your question been resolved?
remote mural
Isnt ab hypotenusa
no AB is not the hypoteneuse
remote mural
U said dab equals 90
in the question, its mentioned quite clearly that that is the case
"Angle DAB is a right angle"
vital flame
Yes but how did u use tan from just angles
Would u not need another side
Unless u used sin nd cos
remote mural
Unless u used sin nd cos
Yeah I think we need side and cos
Atleast I had to use sine and cos to solve this
Best to learn them
calm coralBOT
@vital flame Has your question been resolved?
vital flame
Alri
wind trail
how do i get 63x^9 ?
calm coralBOT
how do i get 63x^9 ?
unkempt drift
how do i get 63x^9 ?
$7x^3 \cdot x^4 \cdot (3x)^2$
potent lotusBOT
south's secret twin brother
unkempt drift
intuitively speaking, (x - 5)^4 is x^4 translated 5 units right
and (3x + 2)^2 is just (3x)^2 translated 2/3 units left
so those 5 and 2/3 units don't make a difference when we are talking about x being plusminus infinity
wind trail
no
unkempt drift
oh, well it's just the y-value something approaches as we take x closer and closer to a certain point
in this case our points are at positive and negative infinity
okay then so you definitely wouldn't be comfortable finding limits algebraically
if you continue with maths for a year I guarantee you'll see this topic
wind trail
oh, well it's just the y-value something approaches as we take x closer and clos...
is that this? just wondering
unkempt drift
is that this? just wondering
yeah!
wind trail
ahh okie
unkempt drift
end behaviour is another way of saying limit as $x \to \infty$
potent lotusBOT
south's secret twin brother
unkempt drift
or negative infinity ofc
so asymptotes are a great way to introduce the concept of a limit
something that approaches a value but never equals that value exactly
wind trail
yes
unkempt drift
that's a commmon thing with limits
they're not always equal to the value of the function
cause the function doesn't exist there
or because our function is discontinuous and does a little jump at the point we are looking at
or because or function oscillates so wildly you can't say it approaches anything
wind trail
my brain will eat your wise words
unkempt drift
yeah there are technical names for these things you'll learn in AP calc or whatever
here you go actually
that's why we need limits
unkempt drift
how do i get 63x^9 ?
cool do you have any more questions?
wind trail
yes sry i was trying to figure out how to word it
the -11 doesnt have a coeffcient, do i disregard that as well?
not coefficient
the x
$x^5 \cdot (2x) \cdot x^8$
potent lotusBOT
strwbie
wind trail
wild guess
unkempt drift
the -11 doesnt have a coeffcient, do i disregard that as well?
you include the -11
unkempt drift
**strwbie**
so yeah very nearly, it's $-11x^5 (2x) (x^8)$
potent lotusBOT
south's secret twin brother
unkempt drift
ah, another way to think about this is imagine expanding this, then choosing the highest-order term
this method always gives you the highest-order term
wind trail
ah, another way to think about this is imagine expanding this, then choosing the...
what do you mean expanding?
unkempt drift
what do you mean expanding?
like expanding the brackets and multiplying everything to get a polynomial
tulip herald
calm coralBOT
tulip herald
I tried doing $1=(3/p)=(p/3)\equiv p^{(3-1)/2}=p \pmod 3$
potent lotusBOT
somethingwrong
tulip herald
**somethingwrong**
something went wrong here but im not sure what
unkempt drift
oh I see cause it doesn't give you 11 mod 12
tulip herald
I think the only thing im doing is the claim that
(3/p)=(p/3) and
(p/3) is congruent to p^{(3-1)/2} under modulo 3
are either of this not correct
ohh i guess i could try to show that p cannot be congruent to 4,7 and 10 modulo 12
and the statement would technically hold right
ah nvm i got it, (3/p)=(p/3) isn't true
.close
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wise ore
I know how to do partial fraction decomposition, but the other day I wondered why we can do this. When you get to this point, then do the "Let x = 5" for example, are we not kinda dividing by zero? Why does that not matter here?
leaden thunder
x=5 is allowed on the second line, not first
dark heath
I know how to do partial fraction decomposition, but the other day I wondered wh...
you're assigning 5 to x in the second line to isolate B.
you aren't dividing by zero because there's no division in the equation 7x-23=A(x-5)+B(x-2).
wise ore
I suppose, but we use that result in the original equation. How do we know it is valid?
.close
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blazing coyote
calm coralBOT
terse river
huh, interesting
blazing coyote
I feel like this is more of an exercise in induction
because I can use the previous result
just have to prove that for instance using this formula I can obtain a polynomial of degree $3$
potent lotusBOT
A dense set
terse river
just have to prove that for instance using this formula I can obtain a polynomia...
is this true? im not sure
blazing coyote
hmm
blazing coyote
is this true? im not sure
Yeah, every polynomial is of degree m
terse river
right
oh heres an idea
try working the other way
using ''negative degree" or w/e
like the lowest nonzero term
blazing coyote
I mean, I think I should first list out a few terms belonging to this set
terse river
probably helpful
i havent solved it but like, intuitively, what is the only k for which the 0 degree term of p_k is nonzero?
crisp aspen
like the lowest nonzero term
i think this is the way to go, you can probably show that every polynomial will have a unique representation doing this
blazing coyote
I just have to show that $p_m$ is LI
potent lotusBOT
A dense set
blazing coyote
i think this is the way to go, you can probably show that every polynomial will ...
Direct sums?
I doubt direct sums will help here
blazing coyote
finite dim spaces are quite nice
for now 
crisp aspen
Direct sums?
i mean as a linear combination of the basis
terse river
for now <:opencry:586078614865444865>
i mean forever, theyre all just F^n
blazing coyote
I know, just a bad joke
sorry
Wait
oh
I think Here I'd want to work my way from the middle terms
unique jackal
i mean forever, theyre all just F^n
oh wow, I've never seen you in the help channels before
hello henry 
terse river
sometimes i get bored lol
rn im staying up because i have a flight at like 8am and its not worth it to try to salvage 2h of sleep
potent lotusBOT
A dense set
terse river
oh wow, I've never seen you in the help channels before
i used to be here a lot more but i stopped at some point idk
i gave myself the helper role for like an hour and then decided against it
unique jackal
that role does nothing but ping you a lot
unique jackal
mhm. I'll let you continue helping wai
blazing coyote
I think Here I'd want to work my way from the middle terms
Is this idea right
crisp aspen
Is this idea right
can you say more about what you want to do
terse river
personally i would just look at the fact that x^k divides p_k
and what that says about the terms of degree < k
in p_k
blazing coyote
Like I want to show $\exists k st 1\leq k \leq m$ , such that $x^k(1-x)^{m-k}$ has a term that no other polynomial of this form has
crisp aspen
yeah i think you can just show that there is a unique linear combination, hence the set is LI
potent lotusBOT
A dense set
terse river
lmfao the obligatory 10 edits
crisp aspen
**A dense set**
what are you trying to show, with this
terse river
**A dense set**
this isnt actually true i dont think
terse river
like for m = 2 our 3 polynomials are x^2, x^2 - 2x + 1, and -x^2 + x
all of them have an x^2 term
so the first one doesnt have a term that no others have
crisp aspen
only one has a constant term, and only one except that one has a linear term
(insert induction here)
blazing coyote
Is this supposed to be an easy problem 😭
terse river
i would say its not hard but also not trivial, its easy to solve once you see the trick
many such problems
blazing coyote
personally i would just look at the fact that x^k divides p_k
So this is the idea I use, right
terse river
yes
@blazing coyote if you want an interesting linalg problem along these lines i have a good one
blazing coyote
<@796939110815629322> if you want an interesting linalg problem along these line...
Sure, after I finish this?
terse river
yea fs
blazing coyote
$x^k divides p_k$, but not any polynomial of lower degree, thus they cannot be expressed as a LC
potent lotusBOT
A dense set
terse river
a LC of the others you mean?
blazing coyote
a LC of the others you mean?
Alternatively, I could prove that this spans the desired polynomial space
and I'm done
terse river
also true
blazing coyote
That could probably be done via induction
terse river
maybe, i think the LI way is pretty easy tho
its basically the inverse of what you did last problem
instead of working down from x^m you work up from x^0
blazing coyote
$ln(p_1(x)) = ln(x)+(m-1)ln(1-x)$
\
$ln(p_2(x))= 2ln(x)+ (m-2) ln(1-x)$
\
$\vdots $
\
$ln(p_m(x)) = mln(x)$
potent lotusBOT
A dense set
blazing coyote
this is to work out the combination of basis that will give me a desired nth degree polynomial
blazing coyote
i dont think you can use ln on polynomials of an arbitrary field
I'm working over $\R$
potent lotusBOT
A dense set
terse river
i really think doing it the span way is way unnecessary
blazing coyote
i really think doing it the span way is way unnecessary
I'd like to do it this way
if that's fine
terse river
yea its definitely possible, curious to see how you go about it
binomial expansion perhaps?
for 1-x
blazing coyote
no, using logs
terse river
i mean you also run into domain issues right, like x can be negative
blazing coyote
okay, ln(|p(x)|) then
terse river
sure
crisp aspen
dude hello? what is going on
blazing coyote
Trying an alternate method
We wish to prove that $p_0,p_1,\dots , p_m(x)$ span $\mathscr{P}m(\R)$.
\
We notice the following
$ln(p_1(x)) = ln(x)+(m-1)ln(1-x)$
$ln(p_2(x))= 2ln(x)+ (m-2) ln(1-x)$
$\vdots $
$ln(p_m(x)) = mln(x)$
\
We now suppose we want to obtain a monomial of say degree $l$ , $1 \leq l \leq m$
\
we thus have
$l ln(x) = a_1(ln(x)+(m-1)ln(1-x))+a_2(2ln(x)+(m-2)ln(1-x)) \dots a_m mln(x))$.
\
This gives us
\
$l ln(x) = $(a_1+2a_2 + \dots ma_m)ln(x) + (a_1(m-1) + a_2(m-2) + \dots a{m-1})ln(x-1)$
\
it's evident from this that we want
\
$a_1+2a_2+ \dots + ma_m=l$
\
and
\
$(a_1(m-1) + a_2(m-2) + \dots a_{m-1})=0$
\
It's evident that this system of linear equations has a solution.
\
From which we can conclude that $p_1, p_2,\dots, p_m$ does indeed span $\mathscr{P}_m(\R)$.
Sorry, have to sort out TeX issue
potent lotusBOT
A dense set
We wish to prove that $p_0,p_1,\dots , p_m(x)$ span $\mathscr{P}_m(\R)$.
\\
We notice the following
$ln(p_1(x)) = ln(x)+(m-1)ln(1-x)$
\
$ln(p_2(x))= 2ln(x)+ (m-2) ln(1-x)$
\
$\vdots $
\
$ln(p_m(x)) = mln(x)$
\\
We now suppose we want to obtain a monomial of say degree $l$ , $1 \leq l \leq m$
\\
we thus have
$l ln(x) = a_1(ln(x)+(m-1)ln(1-x))+a_2(2ln(x)+(m-2)ln(1-x)) \dots a_m mln(x))$.
\\
This gives us
\\
$l ln(x) = $(a_1+2a_2 + \dots ma_m)ln(x) + (a_1(m-1) + a_2(m-2) + \dots a_{m-1})ln(x-1)$
\\
it's evident from this that we want
\\
$a_1+2a_2+ \dots + ma_m=l$
\\
and
\\
$(a_1(m-1) + a_2(m-2) + \dots a_{m-1})=0$
\\
It's evident that this system of linear equations has a solution.
\\
From which we can conclude that $p_1, p_2,\dots, p_m$ does indeed span $\mathscr{P}_m(\R)$.
```Compilation error:```! Missing $ inserted.
<inserted text>
$
l.1439 $l ln(x) = $(a_
1+2a_2 + \dots ma_m)ln(x) + (a_1(m-1) + a_2(m-2) + \d...
I've inserted a begin-math/end-math symbol since I think
you left one out. Proceed, with fingers crossed.
Preview: Tightpage -1310720 -1310720 1310720 1310720
[1{/usr/local/texlive/2023/texmf-var/fonts/map/pdftex/updmap/pdftex.map}]
(./796939110815629322.aux)```
blazing coyote
We wish to prove that $p_0, p_1, \dots, p_m(x)$ span $\mathscr{P}_m(\mathbb{R})$.
We notice the following:
[
\ln(p_1(x)) = \ln(x) + (m-1)\ln(1-x)
]
[
\ln(p_2(x)) = 2\ln(x) + (m-2)\ln(1-x)
]
[
\vdots
]
[
\ln(p_m(x)) = m\ln(x)
]
We now suppose we want to obtain a monomial of say degree $l$, $1 \leq l \leq m$.
We thus have:
[
l \ln(x) = a_1(\ln(x) + (m-1)\ln(1-x)) + a_2(2\ln(x) + (m-2)\ln(1-x)) + \dots + a_m m\ln(x)
]
This gives us:
[
l \ln(x) = (a_1 + 2a_2 + \dots + ma_m)\ln(x) + (a_1(m-1) + a_2(m-2) + \dots + a_{m-1})\ln(1-x)
]
It's evident from this that we want:
[
a_1 + 2a_2 + \dots + ma_m = l
]
and
[
a_1(m-1) + a_2(m-2) + \dots + a_{m-1} = 0
]
It's evident that this system of linear equations has a solution.
From which we can conclude that $p_1, p_2, \dots, p_m$ does indeed span $\mathscr{P}_m(\mathbb{R})$.
potent lotusBOT
A dense set
blazing coyote
Had Bard's help for formatting, sorry
blazing coyote
yea its definitely possible, curious to see how you go about it
here
terse river
sorry i cant understand "we thus have" line
terse river
ohh
monomial
my bad
okay sure i buy it
do you want the problem i was talking about earlier?
blazing coyote
Sure! Thanks for bearing with me!
terse river
okay here it is:
let $S(\mathbb{R})$ be the space of all sequences of real numbers (addition and scalar multiplication are defined how you'd expect). Show that the set ${(\alpha, \alpha^2, \alpha^3 .... ) \vert \alpha \in \mathbb{R}, \alpha > 0}$ is linearly independent. Conclude that $\dim S(\mathbb{R})$ is uncountable.
potent lotusBOT
henryduke
blazing coyote
I don't know anything about sequences 
terse river
just consider them like tuples
but theres countably infinitely many
if you want just think of them as power series, like polynomials but infinitely long
blazing coyote
if you want just think of them as power series, like polynomials but infinitely ...
I don't think I can do this, sorry
terse river
basically this set is like
(1, 1, 1, ...), (2, 4, 8, 16 ...), (3, 9, 27, 81 ...)
but for all positive reals
blazing coyote
Like am I proving that if \alpha is fixed, it's sequence is a vector space
tight copper
I don't think I can do this, sorry
In LaTeX?
blazing coyote
or the set of all seqeucnes is a vector space
terse river
the set of all these sequences is LI
tight copper
No, I literaly am struggling to wrap my head around this
oh let me see it
blazing coyote
the set of all these sequences is LI
hmm, I'll save it and do it later
blazing coyote
Hmm, good luck!
blazing coyote
oh let me see it
No need
I'm doing other problems now, this is too advanced for now
maybe I''l do it next sem
blazing coyote
I doubt this is a vector space in the first place
nvm
my bad
forgot sets aren't ordered
It's evident that I don't compute $C^6$ here, for that will be , well 
potent lotusBOT
A dense set
terse river
just let C be any 6-th dim space, it works fine
blazing coyote
We suggest the following lemma if $dim(V) = dim (U) \implies dim(V \cap U) \geq 1$
potent lotusBOT
A dense set
blazing coyote
Oh, or even better
Let the basis of $U$ be $e_1,e_2,e_3,e_4$, and that of $W$ be $f_1,f_2,f_3,f_4$.. From this it follows that ${e_1,e_2,e_3,e_4,f_1,f_2,f_3,f_4} \subseteq V+W$. However, as $V+W \subseteq C^6$, it follows that $V+U$ can have at most $6$ linearly independent vectors in a given subset ( lemma 1, proof to be supplied later). It also follows that $dim(V+U) \geq 4$( using lemma 2, proof to be supplied later ). We have already proven the following lemma : $dim(V+U)= dim(V)+dim(U) - dim (V \cap U)$. From this we find that $4 \leq 8 -dim(V \cap U) \leq 6$.We thus find that $4 \geq dim(V \cap U) \geq 2$. From this it follows there are atleast two vectors in $V \cap U$ , such that neither vector is a scalar multiple of the other
potent lotusBOT
A dense set
calm coralBOT
@blazing coyote Has your question been resolved?
blazing coyote
i think you mixed up U, V, W?
Looks like it
terse river
but your idea is correct
and lemma 1 proof is trivial since dim U + W <= dim C6
and also dim U + W >= dim U = 4
useful to remember that U is a subspace of U + W is a subspace of C6
blazing coyote
and lemma 1 proof is trivial since dim U + W <= dim C6
yeah
unique jackal
henry you should go to sleep 
terse river
henry you should go to sleep <:cat_happycry:636407813144576010>
what?
lol i told you earlier
i have a flight at 9
calm coralBOT
calm coralBOT
Available help channel!
vagrant sand
limit x tends to infinity ((sinx)/x)^(1/x)
calm coralBOT
limit x tends to infinity ((sinx)/x)^(1/x)
analog pier
My attempt: take log both sides
L'Hospital log(sin(x)/x)/x a couple of times. Or use Taylor expansitons like sin(x)=x+O(x^3).
vagrant sand
L'Hospital log(sin(x)/x)/x a couple of times. Or use Taylor expansitons like sin...
+?? I guess there should be - sign?
sinx=x-x^3/6+...
analog pier
+?? I guess there should be - sign?
it doesn't matter if you write O(x^3)
analog pier
e^(1/x)log(1-x^2/6+.??)
yeah, and u use ln(1+t) ~ t
vagrant sand
We have -x^2/6?
vagrant sand
e^(-x/6)
analog pier
limit x tends to infinity ((sinx)/x)^(1/x)
Oh, Are you sure that x tends to infinity and not to 0?
vagrant sand
Oh, Are you sure that x tends to infinity and not to 0?
Yes. In the book it is infinity
analog pier
then it is nonsense, as you get negative bases with non-integer powers
sin(x)/x changes sign infinitely many times as x->inf
and you take these negative numbers and raise to powers like 0.0001
what is the value of say (-0.3)^0.001 ? I think it should be x->0 in the statement of the problem
then its ok, you have indeterminate 1^inf
and then everything I said before is correct. You do get like equivalence ~ exp(-x/6) or just exp(O(x)) which is 1 as x->0.
calm coralBOT
@vagrant sand Has your question been resolved?
vagrant sand
and then everything I said before is correct. You do get like equivalence ~ exp...
Yes. I think they we will get e^0=1
calm coralBOT
@vagrant sand Has your question been resolved?
vagrant sand
.close
zinc falcon
anyone can help me understand ?
calm coralBOT
anyone can help me understand ?
remote mural
They're asking you for the magnitude of that complex exponential
zinc falcon
so i rewrite as e^a * e^bi and take the magnitude of that?
no
that wouldn't work
i can rewrite e^bi to cos(b) + isin(b)
remote mural
right
zinc falcon
then i can take the magnitude
remote mural
right
eternal stirrup
that wouldn't work
Why would that not work?
That’s precisely what they probably want you to do
zinc falcon
wouldn't that be sqrt(e^a * e^bi)
remote mural
so...
eternal stirrup
?
remote mural
at least consider how to show the magnitude of e^(bi)
remote mural
eggman, just prove it to yourself
eternal stirrup
what does e^bi describe?
zinc falcon
cos(b) + isin(b)
remote mural
e ^ib=cos(b)+isin(b)
eternal stirrup
Yeah so what’s the distance from the origin in that case?
(It’s a point on a circle with radius 1)
remote mural
what is the magnitude of a complex number x+yi?
eternal stirrup
Don’t really have to fuzz with that
zinc falcon
either my brain is too thick currently or i have no idea what you guys are talking about
eternal stirrup
But if youre somehow in disbelief that a point on a circle with radius 1 has distance 1 then sure go ahead
remote mural
eggman, so let's walk it back. I don't think it's good to assume things because you're being told
I would remove the geometric notion of a complex number from your mind at this point
terse river
in any event, you can use the definition of distance for the complex numbers right
square the real and imaginary components
giving you cos^2(b) + sin^2(b)
zinc falcon
im not too good at seeing the distance thing for absolute values
terse river
and then square root that
zinc falcon
like i was told you can use absolute value as a way to see distance but i don't get it
remote mural
eggman, what is the magnitude of a complex number?
terse river
absolute value is the distance to 0
eternal stirrup
A complex number x + iy is associated with a point (x,y)
How would you calculate the distance for a point? (From the origin)
zinc falcon
sqrt(x^2 + y^2
eternal stirrup
Yes!
terse river
yes
terse river
so sqrt of (a^2 - b^2)?
this would also cause issues because
we want distance to be a real number
remote mural
eggman, I encourage you to understand it in a different way.
eternal stirrup
So that’s the distance formula from the origin to your complex number, i.e |x + iy| = sqrt(x^2 + y^2) like you said above
zinc falcon
okay i think isee it then
eternal stirrup
eggman, I encourage you to understand it in a different way.
What other way?
remote mural
Consider defining the magnitude of a complex number by multiplying it by its complex conjugate
terse river
okay i think isee it then
to see that |e^bi| = 1, refer to this
eternal stirrup
Consider defining the magnitude of a complex number by multiplying it by its com...
How is that helpful lmao
zinc falcon
the conjugate is what you multiply to get the thing equal to 1 right?
terse river
strictly less intuitive lol
terse river
the conjugate is what you multiply to get the thing equal to 1 right?
nope thats the inverse
but its irrelevant
remote mural
Because a lot of complex analysis becomes purely algebraic, and a geometric understanding will become nearly useless
eternal stirrup
They’re probably in high school so that’s not helpful
terse river
Because a lot of complex analysis becomes purely algebraic, and a geometric unde...
sure but in this case it is helpful
and you can view |a + bi| = sqrt(a^2 + b^2) algebraically once you know it
zinc falcon
no im in first year universtiy 😶🌫️
eternal stirrup
Oh sorry
terse river
to see that |e^bi| = 1, refer to this
so we know that |e^bi| = 1 do you agree
zinc falcon
yeah i do
terse river
and e^a+bi = e^a * e^bi ofc
zinc falcon
becasue e^x is a strictly positive function
terse river
yes for real x
remote mural
but why is |e^bi| = 1?
terse river
but why is |e^bi| = 1?
because its sqrt(cos^2(b) + sin^2(b)) = 1
eternal stirrup
why is that true?'
Are you asking why the trig identity is true?
zinc falcon
its the absolute value of e^bi which is cos(b) + isin(b)
terse river
bro
remote mural
you haven't proven it, it's the point of the problem
terse river
from the definition of the euclidian norm
remote mural
squared powers of trigonometric functions?
eternal stirrup
squared powers of trigonometric functions?
?
terse river
because |a + bi| = sqrt(a^2 + b^2)
remote mural
henry, I'm not asking you
terse river
where is he coming up with squared powers?
who are you asking then?
remote mural
eggman
zinc falcon
i guess me
terse river
then why use the third person for the second person lol
zinc falcon
i don't really get what you are asking for though
the squares came from the equation of the absolute value
remote mural
so how did they come up with that equation
eternal stirrup
That’s a definition
terse river
its literally the definition yea
remote mural
they're missing something very important in the definition
eternal stirrup
??
zinc falcon
bro teasing us
remote mural
which is why I challenged you to understand this on an algebraic basis to redefine your understanding of the absolute value of a complex number
remote mural
so.... z = x+i*y
what then is a number I can multiply by, to come up with squared values
which still isn't the magnitude precisely
one of the first concepts that they setup is the notion of a complex conjugate
terse river
can we just .close lmao
remote mural
I can't do the latex here, but if you define z* = x-iy, and (z)(z) = (x+iy)(x-iy), then you can get your magnitude defined as Asian suggested
eternal stirrup
My point is that how is that helpful in their case?
They’ve already been given the definition
zinc falcon
alright yeah imma close
remote mural
it was a challenge to define it, I know they don't understand the fundamental concepts surrounding complex numbers
jovial linden
can anyone help me or give me guides on solving this? i am really stuck on that and i appreciate your help on me
pseudo socket
sin(x)^2+cos(x)^2 is always equal to 1
jovial linden
then?
pseudo socket
sin(x)^2+cos(x)^2 is always equal to 1
Since we know this we should make use of it somehow
jovial linden
so sin²x + cos²x = 1
pseudo socket
Yes this is the proof
jovial linden
then how does it become such that we proceed from that addition thing
pseudo socket
Maybe try (sina+cosa)^2 cuz we know what's sina+cosa
fallow crow
can anyone help me or give me guides on solving this? i am really stuck on that ...
squaring both sides?
jovial linden
Maybe try (sina+cosa)^2 cuz we know what's sina+cosa
then its sin²x + 2sinxcosx + cos²x = 1/16
pseudo socket
Yes
jovial linden
sin²x + cos²x + 2sinxcosx = 1/16
since sin²x + cos²x = 1
so we subtitute that
it becomes 1 + 2sinxcosx= 1/16
2sinxcosx = 1/16 - 1
2sinxcosx = -15/16
divide both sides by 2
sinxcosx = -15/32
am i right?
fallow crow
yeah but i think theres a formula for this
jovial linden
what?
calm coralBOT
@jovial linden Has your question been resolved?
remote mural
It is to be proved to be divisible by 24 for every value of n.
2 * 7^n + 3 * 5^n - 5 this expression
remote mural
by induction
vagrant oak
Have you tried anything?
vagrant oak
i see, have you done the base case already?
remote mural
wdym
remote mural
isn't it just sub?
vagrant oak
yeah
remote mural
yes
vagrant oak
that needs to be included in your proof though
remote mural
Ik
vagrant oak
Alright, so now for the inductive step
we can assume 2 * 7^n + 3 * 5^n - 5 is divisible by 24
and we need to prove that it holds for n + 1 as well
remote mural
mhm
vagrant oak
$2 \cdot 7^{n+1} + 3 \cdot 5^{n+1} - 5$
potent lotusBOT
MæthIsAlwaysRight
vagrant oak
we need to prove that this is divisible by 24
calm coralBOT
@remote mural Has your question been resolved?
calm coralBOT
calm coralBOT
Available help channel!
blazing coyote
blazing coyote
To do so , we consider a basis of $U$ to be $\beta_1= {e_1,e_2 \dots e_m}.$We then proceed to extended this basis to one of $V$. Such a basis is $\beta_2 {e_1,e_2,\dots,e_m, f_1,f_2,\dots, f_{n-m}}$. In order to get a vector space of dimension $n-1$, we need to remove one of the basis vectors, this can be done in $^{m+n}C_1$ ways. This tells us that there are $m+n$ vector subspaces
potent lotusBOT
A dense set
blazing coyote
To do so , we consider a basis of $U$ to be $\beta_1= {e_1,e_2 \dots e_m}.$We then proceed to extended this basis to one of $V$. Such a basis is $\beta_2 ={e_1,e_2,\dots,e_m, f_1,f_2,\dots, f_{n-m}}$. In order to get a vector space of dimension $n-1$, we need to remove one of the basis vectors tha aren't in $U$ this can be done in $^{n-m}C_1$ ways. This tells us that there are $n-m$ vector subspaces, with dimension $n-1$
How does my proof look so far?
potent lotusBOT
A dense set
blazing coyote
To do so , we consider a basis of $U$ to be $\beta_1= {e_1,e_2 \dots e_m}.$We then proceed to extended this basis to one of $V$. Such a basis is $\beta_2 ={e_1,e_2,\dots,e_m, f_1,f_2,\dots, f_{n-m}}$. In order to get a vector space of dimension $n-1$, we need to remove one of the basis vectors that aren't in $U$ this can be done in $^{n-m}C_1$ ways. This tells us that there are $n-m$ vector subspaces, with dimension $n-1$.
\
Let each of these subspaces be given by $W_i$. We now attempt to find $\bigcap_{i=1}^{n-m} W_i$. As $\beta_1 \in W_i \forall 1 \leq i \leq n-m$, it follows that $span(\beta_1) \subseteq W_i \forall 1 \leq i \leq n-m$. It follows that $U \subseteq \bigcap_{i=1}^{n-m} W_i$. We now attempt to prove no other element can lie in this vector space. This follows from the fact in each $W_i$, we're removing one $f_j$. Thus $span( \beta_2. \setminus \beta_1) \not \subseteq \bigcap_{i=1}^{n-m} W_i$.
\
From this it follows that $U= \bigcap_{i=1}^{n-m} W_i$.
\
Therefore there exist $n-m$ subspaces of $V$ , each of dimension $n-1$ , such that $\bigcap_{i=1}^{n-m} W_i =U $.
blazing coyote
now show the intersection is U
How does this look?
potent lotusBOT
A dense set
blazing coyote
<@&286206848099549185>
mortal orbit
**A dense set**
this is not what you're trying to prove
you're trying to prove a stronger statement
not even that
you want to prove U = intersection(Wi)
blazing coyote
Yes
mortal orbit
so once you proved U is a subset of the intersection
you have to prove the intersection is a subset of U
mortal orbit
this is not what you're trying to prove
this doesn't hold the info you want
you just said "there exists some guy in span(b2 \ b1) that isn't in the intersection
blazing coyote
you have to prove the intersection is a subset of U
I feel like an argument would be circular
*any
<@&286206848099549185>
glass heart
lets say that n-m=2. so you just have two spaces W_i
take one vector in the intersection
then it is a vector of W_1 and of W_2
lets call it x
we can write it as x=c1e1+...+cmem+af1 and as x=d1e1+...+dmem+bf2
can you conclude from there that x is in U?
blazing coyote
No?
blazing coyote
lets say that n-m=2. so you just have two spaces W_i
We can't , I suppose?
blazing coyote
this is an example aimed at understanding stuff
Hmm?
glass heart
I made the problem simpler
I want you to solve this simplified version
afterwards you can go back to the original problem
prove that x in U from the setup I gave
blazing coyote
hmm
blazing coyote
prove that x in U from the setup I gave
I think there's a mistake
for e_i and f_i are LI
glass heart
and?
blazing coyote
thus $x$ cannot be reprsented both in terms of $e_i$ and $f_i$
potent lotusBOT
A dense set
blazing coyote
Why is my conclusion wrong?
glass heart
things can be zero
blazing coyote
things can be zero
Can I have another hint
glass heart
c1e1+...+cmem+af1=d1e1+...+dmem+bf2
you already mentioned linear independence
what does lin independence mean
blazing coyote
that the sum can only be zero if the coefficents of the terms are all 0 too
glass heart
so how can you use that here
blazing coyote
From that we find that $(c_1-d_1)e_1+(c_2-d_2)e_2 \dots + (c_m-d_m)e_m -bf_2-af_1=0$
We now set the left hand side to be 0
this suggests that bf_2=af_1
which is obviously absurd
blazing coyote
ah , nvm
potent lotusBOT
A dense set
blazing coyote
The only way this is possible is if all the ceofficents are $0$
potent lotusBOT
A dense set
glass heart
yes
blazing coyote
so that tells me that all the vectors are linear combinations of $e_is$
potent lotusBOT
A dense set
glass heart
vectors?
blazing coyote
Al the vectors in $U$
potent lotusBOT
A dense set
glass heart
what did I want you to show?
glass heart
Al the vectors in $U$
well thats obvious, the e_i are a basis for U
glass heart
and how is that coming along?
blazing coyote
I think From this we can conlude that x in U
Anyway, it's past my bedtime, is it fine if I turn in?
Sorry
and thanks a alot!
.close
quiet island
Assume that ( A_i \subseteq X_i ) for every ( i \in I ). Prove that
[
\prod_{i \in I} X_i \setminus \prod_{i \in I} A_i = \bigcup_{i \in I} \prod_{j \in I} B_{ij},
]
where ( B_{ii} = X_i \setminus A_i ) and ( B_{ij} = X_j ) if ( i \neq j ).
potent lotusBOT
Halex
quiet island
Struggling with the other inclusion
Let $f \in \bigcup_{i \in I} \prod_{j \in I} B_{ij}$. By generalized union definition, there exists an $i \in I$ such that $f \in \prod_{j \in I} B_{ij}$, which implies $f: I \to \bigcup_{j \in I} X_j$ is such that for all $j \in I$ we have $f(j) \in B_{ij}$.
potent lotusBOT
Halex
quiet island
Now since for all $j \in I, f(j) \in B_{ij}$ means that $f(j) \in X_i \setminus A_i$
potent lotusBOT
Halex
quiet island
Struggling with the other inclusion
@mortal orbit may we discuss the problem we were in yesterday?
calm coralBOT
@quiet island Has your question been resolved?
quiet island
Let $f \in \bigcup_{i \in I} \prod_{j \in I} B_{ij}$. By generalized union definition, there exists an $i \in I$ such that $f \in \prod_{j \in I} B_{ij}$, which implies $f: I \to \bigcup_{j \in I} X_j$ is such that for all $j \in I$ we have $f(j) \in B_{ij}$.
Now since for all $j \in I, f(j) \in B_{ij}$ it follows that $f(i) \in X_i \setminus A_i$ and $f(j) \in X_j$. Thus $f \in \prod_{i \in I} X_i \setminus \prod_{i \in I} A_i$.
potent lotusBOT
Halex
calm coralBOT
@quiet island Has your question been resolved?
calm coralBOT
@quiet island Has your question been resolved?
bright sentinel
hello guys what am I doing wrong. B should be 2, but I got 4 instead
bright sentinel
the green text I already calculated (which is good), but there is a mistake in red
I can't find the mistake
😦
tranquil wasp
nice integral diddy
bright sentinel
nice integral diddy
thanks what does that mean
tranquil wasp
thanks what does that mean
diddy kong from donkey kong
urban relic
why is the +2C term 0?
bright sentinel
no problem my friend
urban relic
if A is 3 do you even have enough x^2 terms on the right to actually make the polynomials the same
urban relic
(Bx+C)(x+1)
well that fixes what i pointed out
eternal shard
sorry for stealing the show
bright sentinel
(Bx+C)(x+1)
wait I dont understand
so (B(1)+C)(2) and then?
ohhh
2B + 2C
?
eternal shard
its okay bacc ill steal it from you next time
you cant steal something from a peasant like me
bright sentinel
this is coming up on the test thanks for saving @urban relic @eternal shard
.close
twilit bramble
does dot product apply to complex numbers?
calm coralBOT
does dot product apply to complex numbers?
twilit bramble
for example, if z1z2 = 0, is the angle between them 90 degrees?
glass heart
do you mean treating the complex numbers as vectors and actually doing the dot product?
or do you mean the normal complex number multiplication?
in the second case, just like you are used from real numbers, if ab=0, then a=0 or b=0
twilit bramble
do you mean treating the complex numbers as vectors and actually doing the dot p...
yea i mean that
can u not do that?
glass heart
well you can but then you really have no reason to call them complex numbers anymore. the interesting part about complex numbers is the complex multiplication
swift laurel
you can actually get the "dot product" by taking [ \operatorname{Re}(z_1 \bar{z_2}) ]
glass heart
(grossly oversimplified)
potent lotusBOT
cloud
glass heart
if you are treating complex numbers as vectors then its really just like vectors
twilit bramble
if you are treating complex numbers as vectors then its really just like vectors
how would i prove this
,rcce
,rccw
potent lotusBOT
