x = 0 y = -100
#help-39
jolly parrotBOT
stoic imp
what do u think
cunning comet
the points are right, the graph not. but anyway. you have one root. x= 1. divide your function to to get a polynom of degree 4.
cunning comet
polynomdivsion.
cunning comet
cunning comet
as x = 1 is a root, yes.
jolly parrotBOT
stoic imp
lowkey stuck
ohh I need to keep subtracting x^5 - x^4 I think
and bringing the next term down
a moment
cunning comet
have you ever divided (large) numbers by hand? its just the same method.
stoic imp
I usually use calculator
cunning comet
then read the wiki article carefully, it describes the method very well.
you are at step 3 (from wiki), go on with step 4.
stoic imp
,calc 68-12
jolly parrotBOT
Result:
56
i have 2 qeustions:
- why do you calc this?
- why do you need al calculator for 68-12?
stoic imp
,calc 56 -176
jolly parrotBOT
Result:
-120
Result:
100
jolly parrotBOT
cunning comet
its ok.
cunning comet
draw the graph. start with x = 0, 1, 2, 3, 4, 5, 6
jolly parrotBOT
Result:
25
,calc 2^4
jolly parrotBOT
Result:
16
,calc 16-128+564-120*2+100
jolly parrotBOT
Result:
4
,calc 3^4
jolly parrotBOT
Result:
81
,calc 3^3
jolly parrotBOT
Result:
27
,calc 81-1227+569-120*3+100
jolly parrotBOT
Result:
1
,calc 4^4
jolly parrotBOT
Result:
256
,calc 4^3
jolly parrotBOT
Result:
64
,calc 256-1264+5616-120*4+100
jolly parrotBOT
Result:
4
,calc 5^4
jolly parrotBOT
Result:
625
,calc 5^3
jolly parrotBOT
Result:
125
,calc 625-12125+5625-120*5+100
jolly parrotBOT
Result:
25
you could use
,w x^4 -12x^3 +56x^2-120x +100 where x = 0, which would make it a little bit easier
stoic imp
thats cheating
cunning comet
and then change x = ... value
jolly parrotBOT
Result:
100
x = 0 y = 100
x = 1 y = 25
x = 2 y = 4
x = 3 y = 1
x = 4 y = 4
x = 5 y = 25
x = 6 y = 100
literally no roots
stoic imp
draw the graph. start with x = 0, 1, 2, 3, 4, 5, 6
.
I mean I can do polynom division faster now but no roots so no divisor
wat to do
need hints
cunning comet
you are right, but normally you would have to check this. from just reading the table there could be roots.
but lets skip this check.
its right there are no real roots for this poylnom. lets read your question again.
eternal tulip
what must be true about the quadratic if it has no real roots
what's special about complex roots :3
cunning comet
read your example. do you need real roots or all roots (also the complex one)?
if you need only real roots, you are ready.
stoic imp
what does simple roots mean
cunning comet
assume your polynom would be (x-1)(x-5)^2, so you would call 1 a single root and 5 a double root.
single root and simple root means the same thing.
stoic imp
but what if my polynom has complex roots
cunning comet
that is what i asked you. do you need only real roots, or alos complex roots (the example is ambigous from my point of view). it depends at the level you are.
stoic imp
how do I find the complex roots
cunning comet
in this case we would have to factorize the polynom of degree 4.
in the complex wolrd.
cunning comet
well, look at your table. what do you see?
stoic imp
x = 0 y = 100
x = 1 y = 25
x = 2 y = 4
x = 3 y = 1
x = 4 y = 4
x = 5 y = 25
x = 6 y = 100
cunning comet
i know the values. what do you see?
eternal tulip
polynomial depression :3
cunning comet
x = 0 -> y = 100, x = 6 -> y = 100
x = 1 -> y = 25, y = 5 -> y = 25
stoic imp
yeah there is some repeated outputs, so the function is not bijective
eternal tulip
what are you trying to do
stoic imp
find complex roots of a quartic
cunning comet
x = 0 -> y = 100, x = 6 -> y = 100
x = 1 -> y = 25, y = 5 -> y = 25
@stoic imp what do you see?
cunning comet
the polynom is symmetric by 3, which means x = 3+k and x = 3-k have the same y -values.
stoic imp
damn
rustic gate
x^4 -12x^3 +56x^2-120x +100
there is a general method for simplifying polynomials by removing the term one less power than the leading term
called depression
cunning comet
such a symmetry is not typical for a polynom of degree 4, its typical for a polynom of degree 2.
rustic gate
applied here, you can do it via the substitution x = y + 3
,w x^4 -12x^3 +56x^2-120x +100 where x=y+3
jolly parrotBOT
stoic imp
im lowkey confused where did the x = y + 3 came from, I thought this quartic had 3 complex solutions
cunning comet
im lowkey confused where did the x = y + 3 came from, I thought this quartic had...
thats what i am trying to guide you.
rustic gate
for any quartic of the form [ x^4 + 4bx^3 + cx^2 + dx + e,]you can make the substitution $x = y - b$ to depress it
it removes the cubic term
this is a general strategy
jolly parrotBOT
rustic gate
ugh it's 4bx^3 i can't head math
rustic gate
sure but say if you had x^4 + 5x^3, you can still do it by taking b = 5/4
it'll just be less pleasant
,w x^4 + 5x^3 - 2x^2 + 17x where x = y - 5/4
jolly parrotBOT
rustic gate
the cubic term is gone
stoic imp
y = x- 3
rustic gate
ye
so it's ((x - 3)^2 + 1)^2
you can factor the inside using difference of two squares
rustic gate
i mean like (x - 3)^2 + 1 = ((x - 3) + i)((x - 3) - i)
rustic gate
no there's more
stoic imp
but
rustic gate
,, ((x - 3)^2 + 1)^2 = ((x - 3) + i)^2((x - 3) - i)^2
jolly parrotBOT
rustic gate
and then there's the x-1 factor
rustic gate
ye
stoic imp
okay
so roots are
3 + i
3 - i
3 + i
3 - i
1
?
,w solve x^5 -13x^4 +68x^3 -176x^2 + 220x -100
jolly parrotBOT
pearl pondBOT
sharp elbow
.reopen
stoic imp
.reopen
pearl pondBOT
✅
stoic imp
@sharp elbow
little ingot
can someone help me with 10. b. i dont understand how to do composite functions
eternal tulip
!occupied
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sharp elbow
<@1251486076573450301>
?
sharp elbow
....
stoic imp
.close
pearl pondBOT
prime bramble
midnight haven
XD
stoic imp
I reopened for 1 second and everyone trolled me hard
pearl pondBOT
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sharp smelt
how does this prove spans are closed under addition and scalar multiplication
fluid axle
A vector in the span of v1,..vn is a linear combo of v1,..,vn (it seems like that's the defn axler uses)
quartz citrus
the elements of span(V) is the LHS
sharp smelt
yes
quartz citrus
they regroup on the RHS to show the sum is still in span(V)
sharp smelt
I get that
fluid axle
What they're showing is that if you add two lin combos of v1,..,vn, their sum is still a lin combo of v1,..,vn
Like p norm said really
So the sum of two vectors in span v1,..,vn is still in span v1,..,vn
quartz citrus
to show something is closed under addition, we need to show $x,y \in span(V)$ implies $x+y \in span(V)$
jolly parrotBOT
p norm reformed
quartz citrus
the LHS is two elements of span(V)
the RHS is their sum, and it's a linear combination of the vi's
pearl pondBOT
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trail bough
pearl pondBOT
trail bough
How do I solve this last part?
uneven smelt
the climbing wall is the halfway mark
trail bough
So that’s (0.5,2)
uneven smelt
no its given
trail bough
Oh I see
uneven smelt
swing set (given)
climbing wall (given)
parking lot (to find)
they are in that order
trail bough
So swing set is (-3,5)
Climbing wall is (-4,1)
How do I find the parking lot my
Can I use midpoint formula
When I use the midpoint formula I get (-3.5, 3)
@uneven smelt
Oh wait
It’s not the middle of those points
The parking lot isn’t the middle , the climbing wall is
uneven smelt
lets say we have points a b c. b is the mid point of a c. so a-b = b-c. u have a and b. c=2b-a
trail bough
Got it thanks
uneven smelt
welcome
pearl pondBOT
@trail bough Has your question been resolved?
vast portal
Hi guys, what are the rules for using log differentiation? for example, this question looks like chain rule could be used directly but it's actually not
jolly parrotBOT
kheerii
inland ivy
you can differentiate this directly using chain and product rules
although this is equivalent to log differentiation
inland ivy
what?
inland ivy
and what do you plan to do with this
vast portal
like normally thats how chain rule is done
inland ivy
oh you mean the result
vast portal
yeah
inland ivy
that's wrong on many levels
$\frac{d}{dx}(a^u)=a^u\log a\frac{d}{dx}(u)$ is only valid when a is constant wrt x
jolly parrotBOT
kheerii
inland ivy
which it isn't here
vast portal
which it isn't here
so the gist of it is when there is x in the bottom and exponent then we just cant use chain rule
inland ivy
you can
inland ivy
**kheerii**
if you write it like this
but logarithmic differentiation is also the same thing, and it's easier to understand
vast portal
i understand, but like i meant chain rule directly
like if there are x^x then i have to use the log exponent rule first
pearl pondBOT
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frozen solar
pearl pondBOT
heady inlet
if you put a=the thing under the square root and then if you derive that you get what is in the numerator (circled)
wait i'll find you the name in english for the method that is being used
Integration by substitution
heady inlet
if you replace the dx you got here instead of the one on the integral you'll get exactly what they got
the 3x^2 cancels
pearl pondBOT
@frozen solar Has your question been resolved?
placid crest
Got a test tmrw of this
pearl pondBOT
Got a test tmrw of this
placid crest
so for 1.
the formula is like
radians = degrees * pi/180
and degrees = radians * 180/pi
so 80 degrees to radians is
80 * pi / 180
80pi/180 = 4pi/9
so for 2
be abc a right triangle in A , as for sin C = 1/2 and AC = 6
find the length of AB
sin 1/2 = 30*
so sin c = ac / cb
sin c = 6/cb
1/2 = 6/cb
cb = 12
so now we using pythagoras
12^2 = sqrt (6^2 + AB^2)
144 = sqrt ( 36 + AB^2)
AB^2 = sqrt ( 144 - 36)
didnt apply the right formula
shouldnt ahve sqrt it already
AB^2 = 144 - 36
so ab^2 = 108
ab = sqrt 108
AB= 108 = 36×3 =6sqrt3
i'd love if someone would ping me when he's available to help me with thsis
basically calculate it
pearl pondBOT
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placid crest
.close
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river aurora
I am having trouble with this double integrals problem. I know that the answer should be 6pi, from calculating the surface area of a cylinder, but I can't seem to get anything other than 0 with double integrals, because I always end up with sqrt(9-y^2), which evaluates to 0.
summer sundial
I am having trouble with this double integrals problem. I know that the answer s...
shouldnt the answer be 12pi by calculating the surface?
like the radius is 3
and height should be 2
river aurora
shouldnt the answer be 12pi by calculating the surface?
Yes but it is only the top half because x is 0 to 2
river aurora
Wait yeah
Sorry
But it is only the z>0 poriton because it asks for above the region, right?
Which is why I put only positive square root, because technically its +-
summer sundial
well technically you have to integrate the area from the equation of the circle first
then use it as a const to integrate it with the x as height var
no need for that much toughness
river aurora
then use it as a const to integrate it with the x as height var
Ohh
river aurora
well technically you have to integrate the area from the equation of the circle ...
Wait but x and y are the bounded variables? Isn't it easier to do it with respect to x and y?
summer sundial
Wait but x and y are the bounded variables? Isn't it easier to do it with respec...
yeah but technically u should check the domain of y
the domain -3 to 3 already
so need to makes cuts n make it harder
wait for surface area u will need perimeter mb
so you would use the perimeter as const instead of area
mb
summer sundial
first i said that u will have to use the area of the circle as const after integrating
but it was wrong because u are calculating surface area
here i got something through which u can write the perimeter of circle in integral form
vital crescent
im pretty sure inside the integral it should be \sqrt{(\frac{d}{dy}\sqrt{9-y^2})^2+1} as it is (dz/dy)^2 not (z/y)^2.
river aurora
So I should change parameter to arc length, basically?
river aurora
im pretty sure inside the integral it should be \sqrt{(\frac{d}{dy}\sqrt{9-y^2})...
I think that is what I did? The square root of the partial derivative of z with respect to y squared plus one
summer sundial
So I should change parameter to arc length, basically?
yeah so one integration u can get as r dtheta and other of x as height
to reach 2pi*r*h
vital crescent
I think that is what I did? The square root of the partial derivative of z with ...
u might be right, I thought u did y^2/z^2. Note your function is not integrable since it is undefined for -3 and 3 (and the improper integral is also undefined)
river aurora
u might be right, I thought u did y^2/z^2. Note your function is not integrable ...
Yeah, I noticed... that's what I am stuck on
Wait why is the improper undefined
summer sundial
Wait why is the improper undefined
y^2 + z^2 = 9 for y = -3 or 3 then z = 0
so y^2/z^2 for y = 3 or -3 isnt defined
vital crescent
Wait why is the improper undefined
you get something like sin^{-1}(x) from 0 to 1 (something like that). the limit of each is undefined
river aurora
Oh wait yeah
vital crescent
so y^2/z^2 for y = 3 or -3 isnt defined
1/\sqrt{x} is undefined for x=0 yet the improper integral from 0 to 1 is well defined so there is something more than that
river aurora
Wait I think I see too
vital crescent
Oh wait yeah
typically you would compute the middle half of the semicircle, so there is no blow up of derivative
you multiply by 2 afterwards
vital crescent
The integral of 1/sqrt(9-y^2) is defined, its arcsin(y/3)
oh wait am i trolling when i said its undefined
yeah i am
the improper integral is well-defined
vital crescent
The integral of 1/sqrt(9-y^2) is defined, its arcsin(y/3)
this should give you the correct answer now
which should be 6pi
summer sundial
which should be 6pi
12pi at last
vital crescent
12pi at last
its 6*pi * height/2=6pi?
summer sundial
its 6*pi \* height/2=6pi?
why /2?
vital crescent
/2 cuz its a semicircle
summer sundial
but y belongs from -3 to 3
summer sundial
domain is equal to given by default
vital crescent
but y belongs from -3 to 3
its above the rectangle formed by 0<=x<=2 and -3<=y<=3
so its only the the stuff above
z>=0
hence a semicircle
vital crescent
🔥
vital crescent
but circle depends on y and z
yeah but above the reactangle is saying z>=0, and z>=0 with y between -3 and 3 yields a semicircle
summer sundial
yeah but above the reactangle is saying z>=0, and z>=0 with y between -3 and 3 y...
👍
river aurora
Should I .close? I don't want to cut either of you off lol
summer sundial
its ok
vital crescent
Should I .close? I don't want to cut either of you off lol
yeah its good👍
river aurora
.close
daring juniper
I am having trouble with this problem. I tried to check the solution with various ai tools but I get a different answer / way to solve it every time and I am unsure what the correct answer is.
dusky glen
I am having trouble with this problem. I tried to check the solution with variou...
As a side note
!nogpt
pearl pondBOT
Please do not trust ChatGPT or similar AI tools for mathematical tasks, as they often generate output which "sounds correct" but has numerous factual or logical errors. Use of these AI tools to answer other people's help questions is strictly against server rules (see #rules).
dusky glen
I am having trouble with this problem. I tried to check the solution with variou...
As a hint: notice that the two triangles are similar
prime pond
As a hint: notice that the two triangles are similar
wait
does it say that the line measuring 8 is parallel to the line measured y
i cant read that language so
dusky glen
does it say that the line measuring 8 is parallel to the line measured y
Presumably
daring juniper
The question is to solve for x and y
dusky glen
The text doesn't say, but that's what makes most sense
dusky glen
As a hint: notice that the two triangles are similar
Anyway, use this
prime pond
The question is to solve for x and y
do you know about triangle similarity?
daring juniper
yes I think so
prime pond
can you set up the proportions?
daring juniper
long : long = short : short ?
prime pond
just put the numbers and letters
daring juniper
10 / 8 = 5 / y
dusky glen
10 / 8 = 5 / y
Not quite
dusky glen
5 isn't the length of the whole side
prime pond
is not part of any triangle
daring juniper
I wasnt sure how to solve it so I tried to find the way by using ai tools but it just confused me and I am unsure now how it works. Could you please just tell me how to solve it
dusky glen
I wasnt sure how to solve it so I tried to find the way by using ai tools but it...
As I said above, don't use AI
Regardless, the two triangles are similar and white and blue are corresponding sides
prime pond
I wasnt sure how to solve it so I tried to find the way by using ai tools but it...
lemme help with one letter
then try soplving for th eother
10/10+5 = 8/y
daring juniper
x/x+4 = 8 / 12 ?
daring juniper
x/x+4 = 8 / 12 ?
now ?
daring juniper
x = 8 ?
pearl pondBOT
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fathom orbit
I dont understand why the dot product of the gradient and the vector is representative of the directional derivative, isn't the dot product supposed to tell us how close the two vectors are
I think my main issue here is that I dont understand what the vector w or v are doing to the gradient spatially
fathom orbit
the gradient is the direction of steapest accent at that point
and then your just multiplying a random vector to it to get the directional derivative
?
is it measureing the total change over the entire vector
btw Im working off of this
pearl pondBOT
@fathom orbit Has your question been resolved?
west sapphire
I dont understand why the dot product of the gradient and the vector is represen...
the partial derivative in the x direction is the dot product of the gradient with (1,0), that's clear if you just calculate the dot product
similarly for the partial derivative in the y direction, you dot product the gradient with (0,1)
since the gradient is representing a linear map, this is enough to completely determine what it does to all other vectors - it's just a linear combination of what it does to (1,0) and what it does to (0,1)
pearl pondBOT
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fathom orbit
.close
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heavy snow
im having trouble following this proof
pearl pondBOT
im having trouble following this proof
heavy snow
to prove that every bijection gives the same sum we need to show that
$\forall p : \mathbb{N} \leftrightarrow\mathbb{N}, \lim_{n\to\infty}\sum_{k=1}^{n}a_n=\lim_{n\to\infty}\sum_{k=1}^{n}a_{p(n)}$
jolly parrotBOT
esca (@ with reply)
heavy snow
then this means that if $\lim\sum a_n = s$,
$\forall p, \forall \varepsilon > 0, \exists n_0, |\sum_{k=1}^{n_0}a_{p(k)} - s | < \varepsilon$
jolly parrotBOT
esca (@ with reply)
heavy snow
or i guess in the proof theyre taking s to be the limit of sum a_p(k)
heavy snow
**esca (@ with reply)**
so we have to prove this for a_n
west sapphire
or i guess in the proof theyre taking s to be the limit of sum a_p(k)
you'd think they would at least mention how you know that such an s even exists
i.e. explain why every permutation converges
heavy snow
you'd think they would at least mention how you know that such an s even exists
yeah thats exactly what i was thinking 
but uhh taking that for granted
heavy snow
yeah im starting to see it as i figure out how exactly to phrase my question
heavy snow
im having trouble following this proof
choose n_1 sufficiently large such that the sum over k in K of |a_n| < eps/2 for every finit K subseteq {n | n>=n_1}
this is possible because the sum |a_n| is convergent therefore cauchy?
west sapphire
or simply because you can write the following:
$$\sum_{n=1}^\infty |a_n| = \sum_{n=1}^N |a_n| + \sum_{n=N}^\infty |a_n|$$
jolly parrotBOT
Bungo
heavy snow
okok
west sapphire
rearrange to get:
$$\sum_{n=N}^\infty |a_n| = \sum_{n=1}^\infty |a_n| - \sum_{n=1}^N |a_n|$$
heavy snow
i can see how the proof follows from the last line
west sapphire
god damn copy and paste haha
jolly parrotBOT
Bungo
west sapphire
now the limit of the RHS exists and equals zero
heavy snow
im having trouble following this proof
so i guess the logic here is to choose this sum k in K such that it fills in the gaps of the sum p(k)
west sapphire
hence same is true of the LHS
heavy snow
now the limit of the RHS exists and equals zero
mhm
west sapphire
yea you want the K sum to be small, so to do that you take n large enough that all the terms that are big are in the n0 sum
(loosely speaking)
heavy snow
so to confirm:
we choose $K = {1,\dots,n} - {p(1),\dots,p(n_0)}$ in order to make $\sum_{k=1}^{n_0}a_{p(k)}+\sum_{k\in K}a_k = \sum_{k=1}^na_k$ because ${p(1),\dots,p(n_0)}\cup K = {1,\dots,n}$. we also need to ensure that $\sum_{k\in K}|a_n| < \frac{\varepsilon}{2}$ is actually true, so we make sure its not including any n < n1 by setting $n_0$ large enough that all of ${1,\dots,n_1}\subseteq{p(1),\dots,p(n)}$.
jolly parrotBOT
esca (@ with reply)
west sapphire
yea correct
heavy snow
ok great thanks
heavy snow
yeah
west sapphire
but yea you got it
heavy snow
ok great thanks
yeah it was just chasing around the n0 and n1 that was confusing me
.close
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fading frigate
How to do rref on [A, I]?
I want to keep track the operations done on A, and apply it to I
A = 1 2 -2
2 5 -4
4 9 -8
I tried rref([A, I]) in Octave.
it gives me
1.0000 0 -2.0000 0 -4.5000 2.5000
0 1.0000 0 0 2.0000 -1.0000
0 0 0 1.0000 0.5000 -0.5000
but in fact, I want
pearl pondBOT
How to do `rref` on [A, I]?
I want to keep track the operations done on `A`, an...
@fading frigate Has your question been resolved?
pearl pondBOT
@fading frigate Has your question been resolved?
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@fading frigate Has your question been resolved?
warm roost
can u give a step-by-step simplification of whole cubing both sides
warm roost
unkempt yacht
$x^3=(\sqrt[3]{2}+3\sqrt[3]{4})^3$
jolly parrotBOT
unkempt yacht
then apply (a+b)^3 formula
warm roost
then apply (a+b)^3 formula
thats what im asking for
a step by step simplification of that formula
unkempt yacht
$(a+b)^3=(a+b)(a+b)(a+b)$
jolly parrotBOT
jolly parrotBOT
unkempt yacht
do the same, multiply each term by term and add them all up
eventually you will obtain [
(a+b)^3=a^3+3a^2b+3ab^2+b^3
]
jolly parrotBOT
unkempt yacht
the process isnt hard, its just a bit arduous
warm roost
wait im not talking about that aswell
(sorry for not being understandable)
i used this formula
my question is, how did 3ab(a+b) lead to this
warm roost
x = a + b
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@warm roost Has your question been resolved?
pearl pondBOT
@warm roost Has your question been resolved?
warm roost
yes
jolly parrotBOT
faiyrose
warm roost
2
jolly parrotBOT
faiyrose
jolly parrotBOT
faiyrose
jolly parrotBOT
faiyrose
warm roost
i accidentally took 3 cube as 9
warm roost
**faiyrose**
108
jolly parrotBOT
faiyrose
warm roost
ohh
it will be 2
yes
so u get this after solving the other part aswell?
ohhh
i get it now
thanks!
rotund wyvern
i need some help
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kind elbow
Derive a generating function for this expression
pearl pondBOT
Derive a generating function for this expression
@kind elbow Has your question been resolved?
pearl pondBOT
@kind elbow Has your question been resolved?
mental verge
can someone tell me why i keep getting the signs of these wrong for example here in this question i got 2sinx+1 * 2sinx-3 and they got 2sinx-1 * 2sinx+3
mental verge
this happened in the previous question as well
let me share the questions
this is my channel
dont interrup please
for 14.19 i did this
fast shuttle
you can't divide the angle insidee
fast shuttle
like you do 2x=4y > x=2y
mental verge
so is sin theta = cos 2 theta wrong?
fast shuttle
yees
mental verge
howw
fast shuttle
mental verge
ohh right
fast shuttle
its not dividing, its changing the equation 😉
mental verge
actually it doesnt make senseif u think it in terms of algebra but it makes sense when u do the trig lol
fast shuttle
use identities
mental verge
for 14.20 i did this and they did reverse the signs and idk where im doin wrong
fast shuttle
which is what you implied
mental verge
sin(60) != 2sin(30)
oh right
mental verge
i got 2 solutions there
check the last one
where i didnt do any of r = things and directly multiplied by r
fast shuttle
for 14.20 i did this and they did reverse the signs and idk where im doin wrong
i am refering to this
right?
mental verge
i am refering to this
yes
(4 + 4 sin theta) is r
starting from there i did another solution
fast shuttle
4sin(x)+4sin^2(x)= 3
sin(x) = x
4x^2+4x-3=0
x1 = 0.5
x2 = -3\2
sin(x) = 0.5
sin(x) = -3/2
second solution is impossible since -1<sin(x)<1
mental verge
oofff i got it
fast shuttle
which means sin(x) = 0.5
fast shuttle
x= pi/6
fast shuttle
np
mental verge
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unborn cosmos
For sigma notation double/multiple sums in general, does the order in which I place the sigma's matter? [ \sum_{i=1}^{n} \sum_{j=1}^{n} a_{ij} ]
jolly parrotBOT
Kakaka
unborn cosmos
i.e., is it that [ \sum_{i=1}^{n} \sum_{j=1}^{n} a_{ij} =\sum_{j=1}^{n} \sum_{i=1}^{n} a_{ij} ]
jolly parrotBOT
Kakaka
unborn cosmos
for finite sums: no, for infinite sums: yes
I'm a bit confused since in programming the order of nested loops matter
i.e., for each i, sum up all j's
but in maths, there's not really a concept of "point in time" is there
rustic gate
this isn't a matter of programming or not
if you do this with computer program
you will get the same total at the end
unborn cosmos
so is the double sum is really just saying "sum over all pairs of indices (i,j)"
rustic gate
ultimately yes
for finite sums only
for infinite sums, the issue is more complicated
rustic gate
there are convergence issues
rustic gate
well i do but im very lazy about typing it up rn
unborn cosmos
hopefully it'll make sense then
rustic gate
lemme see if i can give you an idea of what i mean
1 -1 1 -1 1 -1 ...
0 1 -1 1 -1 1 ...
0 0 1 -1 1 -1 ...
0 0 0 1 -1 1 ...
consider an infinite grid like this
summing rowise first, you might naively think you get all rows are 0 cuz each pair of 1 and -1 cancel
but summing columnwise first, all columns except the first are 0 by the same reasoning
while the first column is a 1
rowise first you get 0, columnwise first you get 1
rustic gate
yes
anyway the example i gave you is probably bad
the convergence is all over the place
unborn cosmos
**Kakaka**
wait
this is only true if the summation range for each of the sigma's is the same right
otherwise
you could get problems if you exchange the ordering
rustic gate
well
steady crescent
for infinite sums, the issue is more complicated
Ooh, Riemann rearrangement theorem fun
rustic gate
,, \sum_{i = 1}^m \sum_{j = 1}^n a_{ij} = \sum_{j = 1}^n \sum_{i = 1}^m a_{ij}
jolly parrotBOT
unborn cosmos
true in general for finite n,m?
steady crescent
Yes
unborn cosmos
,, \sum_{i = 1}^m \sum_{j = 1}^n a_{ij} = \sum_{j = 1}^n \sum_{i = 1}^m a_{ij}
wait so the order still doesn't matter?
steady crescent
Yes, it’s kinda just like summing the terms in a mxn matrix, A. Obviously the sum of A = A^T
rustic gate
wait so the order still doesn't matter?
you're still summing the same terms
rustic gate
👍
pearl pondBOT
@unborn cosmos Has your question been resolved?
unborn cosmos
How do they get the final step (green highlighted part)?
pearl pondBOT
How do they get the final step (green highlighted part)?
pearl pondBOT
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limber remnant
I think it's an axiom of a norm that $\norm{x} =0 \leftrightarrow x = 0$. But my LinAlg has almost decayed, unfortunately.
jolly parrotBOT
poypoyan
grave swan
in baby rudin's proof that Q is dense in R. where did the line $m-1 \le nx < m$ come from?
jolly parrotBOT
babario
pearl pondBOT
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grave swan
<@&286206848099549185>
muted shale
i think thats the archimedian property
cunning comet
m1 > nx. look at m1-1 if this is m1-1 <= nx, you are ready. if not, do the same again.as you know there is a -m2 which is smaller than nx this process is finite.
grave swan
huh
grave swan
m1 > nx. look at m1-1 if this is m1-1 <= nx, you are ready. if not, do the same ...
i dont understand
cunning comet
what exactly?
grave swan
what exactly?
"look at m1-1 if this is m1-1 <= nx, you are ready. if not, do the same again.as you know there is a -m2 which is smaller than nx this process is finite."
pretty much every thing
pearl pondBOT
@grave swan Has your question been resolved?
grave swan
m1 > nx is given, ok?
yea n then all the others i dont understand
wdym m1-1 until m1-1 <= nx
why does that help?
cunning comet
now look at m1-1 and compare this with nx.
there are two cases m1-1 <= nx or nx < m1-1, ok?
grave swan
yea
cunning comet
ok, if m1-1 <= nx, then we are ready. we found our m such that m - 1 <= nx < m (in this case m = m1), ok?
grave swan
ok
cunning comet
so for the case nx < m1 - 1, we no subtract again 1, which means we compare m1-2 with nx. again we have to cases: m1-2 <= nx or nx < m1-2, ok?
grave swan
yes so we do this again right until our objective is reached, but why is that objective so important?
cunning comet
we do this again and again, the only thing you have to check is that this process ends. which means you can find some m1-k with m1-k <= nx. as you know that -m2 <= nx you know this process is fnite.
so at the end you have found a m, such that m-1 <= nx < m (and -m2 < m <= m1).
grave swan
so.. they let m-1<=nx<m, and we know that -m2<=m<=m1 since the process of m1-k until m1-k<=nx is finite because -m2<nx
cunning comet
yes
grave swan
how about the next line $nx<m\le1+nx<ny$ where does the $m<=1+nx$ come from
jolly parrotBOT
babario
cunning comet
- nx < m is just the right part of the line above
- m <= 1+ nx is the left part of the line above (when you add 1 at both sides)
- 1 + nx < ny is just the start n(y-x)>1
cunning comet
youre welcome
grave swan
.close
pearl pondBOT
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mental verge
pearl pondBOT
mental verge
i mean 14.22 here in the solution i shared
but i couldnt do 14.21 either so any help would be appreciated
light helm
1 sec, misinterpreted the stuff in red
why's 2cos(30) not equal to cos(60) being written there
light helm
i mean its true, but not really relevant
mental verge
2cos(30) not being equal to cos(60) isn't mentioned anywhere in the solution
this was an examplw
of mine
it says 2 arccos 1/2
arccos 1/2 = B
nvm i got it
the took the cos of it as cos 2b and i was confused there but then i remembered that we did take cos of both sides after we multiplied 2 and b
anyways no need for help
thanks for trying though
btw can u do 14.21?
i dont get the first part there what does it being pos or negativ has to do with the solution
did this for 14.21
light helm
by deducing the sign of x, they reduce the number of cases/signs to consider
pearl pondBOT
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primal pawn
primal pawn
Hello, I am in need of assistance with a simple question.
sharp smelt
sharp smelt
,w (2x-5)(3x+1)
muted shale
expand the right
primal pawn
It appears as though there are far too many to easily figure out a solution, unless there are multiple solutions.
primal pawn
expand the right
muted shale
ohh, it shows that the left side is equal to the right side
are you asking how you find these?
primal pawn
Correct.
primal pawn
are you asking how you find these?
Yes, as I have been unable to determine how exactly one would find one solution out of seemingly thousands.
I assume there are ways you can narrow it down? Or, perhaps there are multiple solutions?
muted shale
you can use the quadratic formula, or abc formula, or pq formula or whatever you call it
primal pawn
Thank you, I will go learn that now.
muted shale
but for quite simple stuff like this you can also do it another way
muted shale
$(x-a)(x-b)=x^2-ax-bx+ab=x^2-(a+b)x+ab$
jolly parrotBOT
Flappie
muted shale
so if you get something like $x^2-4x+3$ you need to find a and b such that a+b=4 and a*b=3
jolly parrotBOT
Flappie
tranquil badger
muted shale
so here, -5+4=-1 which is the x term and -5*4=-20 which is the constant term
but this is only really doable with integer solutions
primal pawn
Are A and B factors in this case? And, assuming this is true, could you simply manipulate the values of the rest of the expression to get the values you want?
muted shale
for anything else and this is almost impossible by eye
primal pawn
LIke, for example, how they turned 13x into one? Could you continue decreasing or increasing the value incrementally until you got the value you wanted?
muted shale
$x^2-2x+1$ find a and b such that $a+b=-2$ and $ab=1$
jolly parrotBOT
Flappie
primal pawn
That might take a second.
tranquil badger
mild shadow
$x^2-2x+1$ find a and b such that $a+b=-2$ and $ab=1$
I cant see the question properly
mild shadow
**Flappie**
Where are a and b
muted shale
Where are a and b
the a and b you have to find
primal pawn
No, I cannot solve that as of now, as it would take a lot of refining my understanding of the multiplication.
muted shale
well, a=b=-1
primal pawn
Even though I am perfect (metaflow 2nd half), I still haven't done enough processing yet.
jolly parrotBOT
Flappie
muted shale
,w solve a+b=-2,a*b=1
jolly parrotBOT
primal pawn
This is going to take alot.
The problem for me is how it isn't so simple as finding an answer, it's also having to run a bunch of post-calculations on that number, and then sifting through a sea of possible answers to find one needle-in-a-haystack.
I'm assuming you must memorize and format the answer mentally in such a way they become instantaneous?
muted shale
consider (x-a)(x-b)
if we expand it, we get x^2-ax-bx+ab
correct?
$(x-a)(x-b)=x^2+(-a-b)x+ab$
jolly parrotBOT
Flappie
muted shale
so, if we get some kind of quadratic equation such as x^2+8x+12
we get -a-b=8 and ab=12
correct?
you can also see this as, we need to sum a and b to get the term infront of x and we need to multiply a and b to get the constant term
but if thats confusing, ignore the statement
primal pawn
I believe so, I'm still processing it.
gleaming moss
I believe so, I'm still processing it.
I suggest you start by writing down the coefficients a,b,c then write down the factors of a*c
Then just choose the pair of factors that sum to make b
pearl pondBOT
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devout heath
f:g:h g is 60 percent more than h , f is a third of g , simplify f:g:h
scarlet mirage
not me failing at latex
so g is 60% more than h
we can write g * 160% = h
right?
scarlet mirage
from there try to find g : h
scarlet mirage
i do have to go soon though its the end of my shift
ill explain some of it here for when you get back
scarlet mirage
we substitute 160/100 instead of 160% and get 160g/100 = h
multiplying both sides by 100 we get 160g = 100h
from there simplify and get the ratio between g and h
do the same method for f : g and continue from there
pearl pondBOT
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boreal igloo
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muted shale
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What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
fleet oriole
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opaque compass
Could someone help me on how to find this?
pearl pondBOT
Could someone help me on how to find this?
opaque compass
I need to find X, Y and Z
random ermine
pythagoreas
opaque compass
pythagoreas
so 3^2 + x^2 = y^2?
random ermine
yes
opaque compass
but how do i solve it with two unknowns
opaque compass
wdym
random ermine
so 3^2 + x^2 = y^2?
did you come up with this
or did someone tell you
if you came up with that it should be obvious what to do
just use pythagoreas again for other triangles
opaque compass
did you come up with this
i cam up with it
Triangle 1: 3^2 + x^2 = y^2
Triangle 2: z^2 + x^3 = 12^2
Triangle 3: y^2 + z^2 = 15^2
pearl pondBOT
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tranquil badger
tranquil badger
tranquil badger
Still confused on this.
But I think I know what it is now.
If I am correct, are they basically just narrowing it down, as I stated prior?
I should be able to understand most explanations now via contextualizing so I shouldn't have trouble with that...
tranquil badger
yea, especially that last question
6x^2 - 13x - 5
if I'm right, they basically just narrow down the answers by figuring out what would multiply into 3 and then add to get 13, yes?
also, does adding and subtracting constants affect terms?
Was I right? Were they basically just narrowing it down, without using any particular method?
leaden wadi
There are several methods for finding the linear factors of quadratic polynomials.
tranquil badger
List their names.
tranquil badger
Do most of them involve just writing out the factors and doing the usual multiplication and adding already shown in the problem?
leaden wadi
Quadratic Formula
For these examples, the Quadratic Formula would be the most straightforward method.
tranquil badger
to get this
what i dont get
is i dont see how they couldve got this
bro just tell me
i asked earlier
when using the guessing method are narrowing methods usually employed?
like, you'd know the answer must be in a certain range, which is why they'd be able to find it despite hundreds of possible answers existing.
or no?
leaden wadi
tranquil badger
..
leaden wadi
Here is QF which is the best method for finding the linear factors of a quadratic equation.
The PQ Method is better for any polynomial equation of a degree higher than 2.
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vernal grove
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vernal grove
I dont understand how F(theta - epsilon) = 0
pearl pondBOT
vernal grove
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vernal grove
wait actually
vernal grove
what is theta-epsilon
so we know y <= theta implies 0 but theta - epsilon could be smaller thn y no?
muted shale
so we know y <= theta implies 0 but theta - epsilon could be smaller thn y no?
theta-epsilon is smaller than theta, so its 0
muted shale
so we know y <= theta implies 0 but theta - epsilon could be smaller thn y no?
theta-epsilon is y
pearl pondBOT
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midnight haven
Question 19
pearl pondBOT
Question 19
muted shale
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What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
2
I did something with fx first
Multiplied and divided by 2 first
Yeh I know that, but I am dumb to know how to use it
midnight haven
I did something with fx first
Multiplied and divided by 2 first
Wrote fx in terms of single terms
You know what I mean
Now I need to find the value of x from that GIF function
Not sure how to do it
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midnight haven
...
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