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sudden pivot
cedar kilnBOT
@errant locust Has your question been resolved?
mortal yarrow
Steal one
sharp sage
potent ledge
if p=sin40, how do I express cos140 in terms of p..?
cedar kilnBOT
# if p=sin40, how do I express cos140 in terms of p..?
cedar kilnBOT
@potent ledge Has your question been resolved?
cedar kilnBOT
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eager hemlock
\alpha^T\beta=0, A=E+\alpha\beta^T, how can we know det(A)?
cedar kilnBOT
\alpha^T\beta=0, A=E+\alpha\beta^T, how can we know det(A)?
crimson delta
you can find a few eigenvalues. not sure if you can find all of them
ah yes I think you can. together with the trace
crimson delta
use the properties of the trace
cedar kilnBOT
@eager hemlock Has your question been resolved?
potent ledge
how do 40 and 140 relate to each other?
How to get cos140 from sin40
mental trail
\alpha^T\beta=0, A=E+\alpha\beta^T, how can we know det(A)?
so $\alpha^T\beta=0$, $A=E+\alpha\beta^T$
wraith daggerBOT
rafilou2003
mental trail
E being the identity
eager hemlock
yes
mental trail
what are the eigenvalues of $\alpha\beta^T$
wraith daggerBOT
rafilou2003
eager hemlock
we dont know
mental trail
what is $(\alpha\beta^T)^2$
wraith daggerBOT
rafilou2003
mental trail
I think this should help
eager hemlock
0
mental trail
wait maybe I'm tripping too much wait
ok no I'm not
$(\alpha\beta^T)^2 = \alpha\beta^T\alpha\beta^T$
wraith daggerBOT
rafilou2003
wraith daggerBOT
rafilou2003
mental trail
**rafilou2003**
what's the minimal polynomial of that matrix
eager hemlock
wiat what
eager hemlock
idk
eager hemlock
yep
mental trail
and characteristic polynomial?
eager hemlock
ye
mental trail
and never heard of cayley hamilton theorem?
mental trail
if you've seen characteristic polynomial I don't understand you haven't seen minimal polynomial
mental trail
char poly, EVALUATED in the matrix, is 0
so say $A = \begin{pmatrix}1&2\0&2\end{pmatrix}$
wraith daggerBOT
rafilou2003
mental trail
you can compute char poly as $p_A(\lambda) = (\lambda-1)(\lambda-2)$
wraith daggerBOT
rafilou2003
mental trail
cayley hamilton tells you that $p_A(A) = (A-E)(A-2E) = 0$
wraith daggerBOT
rafilou2003
mental trail
but more importantly
there is a set of polynomials
that "cancel" A
it's all the polynomials $P$ such that $P(A) = 0$
wraith daggerBOT
rafilou2003
eager hemlock
mmm
mental trail
and amongst them
there is a single one, that is unitary, and of minimal degree
we call it the minimal polynomial of A
mental trail
**rafilou2003**
in this example, you can compute the minimal polynomial to be $\pi_A(\lambda) = (\lambda-1)(\lambda-2)$
wraith daggerBOT
rafilou2003
wraith daggerBOT
rafilou2003
mental trail
**rafilou2003**
we found that $(\alpha\beta^T)^2 = 0$
wraith daggerBOT
rafilou2003
mental trail
now unless the matrix is 0 to begin with, its minimal poly is $\pi(\lambda) = \lambda^2$
wraith daggerBOT
rafilou2003
mental trail
the roots of the minimal poly are exactly the eigenvalues of the matrix
there's only one root
0
so the only eigenvalue of the matrix is 0
eager hemlock
ok..
mental trail
if we add the identity to it
all eigenvalues are 1
now for the det
it's the product of eigenvalues
cedar kilnBOT
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full wren
2-2
neat urchin
full wren
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drifting marlin
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misty furnace
Please don't troll
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crimson sedge
is z = 9 - xˆ2 - yˆ2 an equation for an elliptic paraboloid? if not, what kind of surface does it represent?
buoyant latch
Did you try graph it
subtle harbor
think about the level curves
crimson sedge
I did not graph it
teacher said we should try to draw it ourselves
but i was having a hard time imagining the shape
i used wolfram and got the ideia
deep oriole
You could even think about the graph it makes on each set of 2 axes, ie get rid of one variable and see what you get, basically like a projection
subtle harbor
I did not graph it
do you know what level curves mean?
crimson sedge
You could even think about the graph it makes on each set of 2 axes, ie get rid ...
havent got to it yet
subtle harbor
it means fixing z to a certain number, i.e. let z=1, then you have a crossection of the particular surface you're looking at
buoyant latch
Level curves are definitely the way to go here
crimson sedge
its the next topic on the book im using (Stewart)
subtle harbor
z=0, z=-1, etc.
then, you can picture cross sections by fixing x=0 and y=0
and these together give you a natural picture of what the curve looks like
cedar kilnBOT
crimson sedge
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subtle harbor
is z = 9 - xˆ2 - yˆ2 an equation for an elliptic paraboloid? if not, what kind o...
in this case, this is correct, but you should use what was said as your justification
crimson sedge
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cedar kilnBOT
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median jolt
Context: From what I understand, a polynomial is a function) where you want the input to equate to zero so you can tell where the "base" value is. Graphing polynomials via an input vs result graph can provide greater visualization of the data.
Base position: I've heard it's very difficult to find the roots of a polynomial via inputting direct inputs as the result can be a fraction.
general solution: People typically solve this by breaking the problem into smaller parts that multiply to create the whole (factoring), as that somehow is "easier" to solve for when finding input values that would make the result of the function) equate to zero.
Q-1: Direct: The information provided does not appear to explain how exactly it would make it "easier or obvious," and how this would make the problem more solvable.
Q-2: How does breaking a function) into pieces that multiply together make it easier to find the input values that would result with zero?
Q-3
Proposals:
1 multiplied results: Perhaps when inputting values into the factored expression you only need to make the parts themselves zero?
2 altered zero position: Perhaps by breaking the problem down like that it moves the zero to a more obvious position (making the input that equates to zero a number closer to one or a whole number).
NARROWING LOGIC: (unused)
median jolt
static whale
what is your actual question? @median jolt
idle rain
so you want to factoring those polynomials I assume?
median jolt
what is your actual question? <@563179988191281222>
lyric narwhal
Because you only need to make the factors individually equal to 0
To find all the roots
$ab=0\implies a=0\lor b=0$
wraith daggerBOT
kheerii
lyric narwhal
the language used here is weird.
Maybe it’s translated
static whale
if P, Q, R are (non-constant) polynomials (with say real coefficients) if P(X) = Q(X)R(X) and we want to solve P(X) = 0 we can solve Q(X) = 0 and R(X) = 0
and Q and R will have a lower 'degree' than P
the degree of a polynomial is the highest power. so for instance x^2 + 3x + 2 is degree 2. x^7 + 5 is degree 7
that is why factoring makes it 'easier'
lyric narwhal
that is why factoring makes it 'easier'
Indeed
static whale
if xy = 0 then x = 0 or y = 0 (or both)
median jolt
1 format:
Lists all current knowledge and information related to the topic and adds additional explanations for clarification.
2 clarification:
So, if I am getting this correctly, when you break a function) into constituent parts, if you can find the inputs that make those parts individually equate to zero they will give you the zeroes when multiplied back together to create the whole?
3 question:
Does this give the zeroes for the factored 'and' the unfactored function?) And, if so, is this because after multiplying them together the real value would be revealed?```consider the quadratic x^2 + 3x + 2, we can factorise this as (x + 2)(x + 1)
so if we want to solve x^2 + 3x + 2 = 0 we have (x + 2)(x + 1) = 0 which says that either x + 2 = 0 (i.e. x = -2) or x + 1 = 0 (x = -1)
a factorisation is just another expression of the same function
so it has all the same values
idle rain
I think for c, you can check it if that is not obvious to you
static whale
x^2 + 3x + 2 = (x + 2)(x + 1) is true for all values of x
median jolt
I think for c, you can check it if that is not obvious to you
this is c?
idle rain
yeah. 3
median jolt
✅
lyric narwhal
The factored and unfactored functions are exactly equivalent, just written in different ways
idle rain
factoring is another way to rewriting things. Like 40 = 4(10)
lyric narwhal
So they must have the exact same roots
cedar kilnBOT
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Available help channel!
weak verge
cedar kilnBOT
idle rain
you multiply both side by -3
so left side, you got your -3 canceled out
right side, you have -3 multiply with x/4
weak verge
you multiply both side by -3
wouldnt i have to multiply the left by -3 and the right by 3 tho
since the left denominator is negative
and the right is positive
idle rain
no. when you deal with equation, you do the same thing both side
if you multiply left side by -3, you multiply right side by -3
Think of it this way: 1 = 1, then 1(-3) = 1(-3)
other way does not make sense
idle rain
yes
cedar kilnBOT
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gusty charm
I need help with this
cedar kilnBOT
I need help with this
near sand
First off, do you know what sections the indicated set is?
gusty charm
No
near sand
Ok so first off A intersect B represents all elements that are both in A and B
So in the diagram this is represented as the intersection of both circles
gusty charm
Ok that makes sense
So that would be 70
What is the P for? And does the ‘ apply to the P?
near sand
sorry, it’s the area in both circles
not just one
what that is is the union which is something else
near sand
What is the P for? And does the ‘ apply to the P?
P represents the probability of a certain event
near sand
Is it still 70?
in this case it would be 15
near sand
exactly
so then there’s the apostrophe symbol
that means the complement of a set
it’s all of the elements not in the given set
gusty charm
So A intersecting B is 15?
near sand
So A intersecting B is 15?
yeah
near sand
So what does the apostrophe apply onto
it applies into (a intersect b)
gusty charm
So it would be everything other than 15?
near sand
yeah
gusty charm
But I leave the 30 alone?
near sand
The thing about complement is that you first have to know what the universal set in this case is
in this case it’s everything in the box
gusty charm
So the answer to the question would be 85.00?
gusty charm
Ohhh alright thanks
near sand
👍
gusty charm
one last question for this, what does the reversed intersection symbol mean
the one that looks like a U
i think it stands for union
but im not sure what union means
fallen root
one last question for this, what does the reversed intersection symbol mean
do you know logic gates in cs?
it's basically or
so the new set contains each value from both sets
gusty charm
I don’t
cedar kilnBOT
wanton sail
Which part of the question are you trying to do rn
Do you want to find the regions where it's increasing/decreasing?
Right, but which part do you want to do first?
Ah okay
yup
OK, so I guess we could start with local extrema
Do you remember how to find local extrema using the derivative?
okay
basically the first thing you want to do is solve f'(x) = 0
At those points, the graph will be flat
No
Points of inflection are when the second derivative changes sign
So you'd want to solve f''(x) = 0 for that
Yes, in English the word for this is a "stationary point"
So, once you find those x-values
You have to check whether it's a maximum, minimum, or neither
And you do that using the second derivative
So you'll plug the x value into f'' to get f''(x)
If it's positive, then it will be a minimum. If negative, then it will be a maximum. If f''(x) = 0, then we can't conclude anything.
No worries
You did a good job translating the terms
yup!
So there's only one stationary point
And then plug that into f''
Yeah
so calculate f''(2)
wanton sail
If it's positive, then it will be a minimum. If negative, then it will be a maxi...
It tells you this ^
whether it's a local minimum, local maximum, or neither
local yea
sharp lotus
so because f'(2) = 0 as you found earlier, that makes it a POTENTIAL max or min, and the second derivative can clarify which of them; this is called second derivative test
if you think about the places where the derivative is 0 in this picture, you can see the function becomes flat (i.e. f'(x) = 0 ) if its changing smoothly (differentiable) and it has a min or max
if it's a minimum (like a valley in the graph) it's always going to be concave up (so second derivative is negative) and if its a max (like the peak of a hill) the function will always be concave down (so second derivative is positive). try to convince yourself of this (or prove this wrong by trying to come up with counter examples, like can you find a minimum that is concave down and has a flat slope?)
yeah so basically second derivative you
- mostly use if you are looking at the concavity of functions and to check for points of inflection
- it is also used if you are investigating local min \ maxes. but first you find the potential ones using first derivative (where it is 0 or undefined). after that you can distinguish between min or max using second derivative like we just talked about (doesn't apply to undefined first derivative though, just if 1st derivative is 0 and the function is differentiable)
yes
yeah, i wasnt here for the start of the problem, but eric had to go, so just real quick, what interval are you finding the absolute min \ max over?
just for whole domain of the function?
ok
and this was only local min \ max?
ok gotcha, so we also need to look at what happens at x = 1 since the function is undefined and when functions are discontinuous they can jump in weird ways, so we need to consider $\lim_{x\to1^-}f(x)$ and $\lim_{x\to1^+}f(x)$
wraith daggerBOT
∫oosh (lemonsaurus appreciator)
sharp lotus
and also need to examine what f(x) does as x-> inf and -inf
shouldnt be too bad, with 1 basically its gonna be a number / 0 so its either -infinity or positive infinity, consider each side from left and right
and with the limits as x -> +- inf the limit will kinda be infinity \ infinity so you can use lhopitals or some such, basically the e^x should overcome the other part of the fraction and be some infinity really you kinda just have to make sure you get the signs right
if you need help with this let me know, but ill let you try to work it out and at least think about it for a couple minutes, brb
ok first 3 look good, for 4th one you got 1?
^^
yes!
yep! so what can we say about absolute min \ maxes?
so actually we don't have an absolute min \ max. the limits at these points dont exist, the +- infinity is more just a description of the function's behavior, sometimes it's said that "the function increases\decreases without bound"
yeah let me see if i can pull up an exact definition
thats just a local minimum yes, but there are other numbers (all the way to -infinity) in the range of the function so it doesn't qualify as an absolute minimum
i think you also had to answer what the range of the function is, speaking of?
yes
thats actually a better word for it
sometimes used interchangeably in some books but sometimes not lol so probably better to avoid the word range altogether
so youre right that theres a gap somewhere there in the middle, how big is the gap? so the function on the x interval (-inf, 1) is taking on what values? and what about on the (1, inf) interval ?
yeah take all the time you need, great that you are working it out : )
well you could build a rough sketch based on all the stuff we found previously, like the 4 limits and the local min, in fact, try to sketch it real quick just using those 5 pieces of information and see what you come up with
those 5 pieces of information should be enough!
(and kinda being aware that it's continuous \ differentiable at all numbers except x = 1)
excellent, that's exactly right
so what else did we have to find
was that it?
were there points of inflection?
what is your thought?
so what is going on with 2nd derivative when x < 1 ?
so there will always be concavity if the function is differentiable
continuity is not enough, for example consider a line: y = mx + b, y ' = m and y '' = 0 so it has a 2nd derivative of 0 everywhere
so it neither concave up or concave down 🤷♂️
but pretty much anything else, if its differentiable, you can think of concavity as "is the slope changing?"
and YES even on your graph, the slope is changing even when it's "nearly flat as x -> -inf" it is changing ever more subtly but it's never a perfectly line
exactly, it's always the same sign, and what sign is this? well you can plug any single test point in the interval (-inf, 1) into this, so let's try 0 because it's as easy as any to calculate:
f''(0) = ?
you're right if you're saying there are no points of inflection, but is the graph concave up or concave down on (-inf, 1) ?
it is
concave up -> positive 2nd derivative
concave down -> negative 2nd derivative
2nd derivative = 0 it means it could be a point of inflection if it happens at a SINGLE point (unlike a straight line where 2nd derivative is 0 everywhere, but no point of inflection) and on either side of the point theres a concave up and a concave down interval
yep, so what's going on with our graph on (-inf, 1) ?
yes
right
you can also think of it....as you move to the right the slope is getting more and more negative
it "starts out" pretty much "flat" from -inf and it gets more and more negative as you move to the right
like if you were skiing down hill it would get more and more steep right?
if youre thinking in terms of moving to the right, in the +x direction
and analytically, we can just plug in any x value into f''(x) thats in the interval (-inf, 1) and we should get a negative number to convince ourselves
yeah, we say concave up and concave down though but yes : )
ok so since we have a portion of the graph that's concave down and one that's concave up, does that mean there should be a point of inflection in between?
uh ok whats the definition of point of inflection? : )
let's look that up and see if it fits
yes that's right
so how do you explain that we can have a concave up part and concave down part but no points of inflection?
if this was a test question in which you had to write out the answer
yeah, and even more fundamentally?
that's also a good go at it, but could you have similar examples where you have no break in domain?
what if i defined g a function to be the same as f, but i say that g(1) = 1 randomly, then suddenly x = 1 is also in the domain
yes, good
so could you have continuous functions where you have a concave up part and another concave down part but no point of inflection?
are you sure? : )
would you bet $100?
so what about this, is it continuous?
yeah that's a good thought so f'(x) is actually what needs to be continuous!
the way we usually phrase that is f(x) needs to be differentiable
i think more intuitively you can think of differentiable as "the slope of f is changing smoothly" rather than thinking of whether f'(x) is continuous which is harder to visualize
sharp lotus
so what about this, is it continuous?
so like in this example the slope is negative, then suddenly it "jerks" to flat immediately then it suddenly it "jerks" again to negative, no smooth change
if we graphed f'(x) there there would be sudden jumps
so yeah all you were saying is right that we have no inflection point because there is an asymptote and whatnot, but the key fundamental property of the graph that affects which is necessary for point of inflection is that the function is differentiable over the relevant interval where its changing concavity
im just trying to get you to think in terms of continuity \ differentiablity which if you can do then you will understand everything in calculus at a more intuitive level : )
when learning new theorems and stuff try to pay attention to the setups of them, like sometimes continuity is necessary (usually for pretty much everything in calculus) sometimes also differentiability, try to consider why the setups of the theorems need one or the other there is usually some geometric interpretation you can come up with that makes you understand the ideas of the theorems intuitively
ok gotta go, hopefully youre all set on that otherwise feel free to open a new channel with more questions
or keep this 1 open
ok
you would find where f'' = 0
and these are "potential" points, kinda like using first derivative for min \ max
then confirm by making sure that on either side of that you have a + and a -
for example, check out y = x^3 real quick, you have y'' = 0 at x = 0 but there is no inflection point because f'' > 0 for all other x
inland solar
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cedar kilnBOT
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cobalt rampart
Using comparison/absolute value criteria for convergence, determine the convergence of
$\int_{0 }^{\infty}\frac{x{^2}e{^-{^x}}dx}{1+x{^2}}$
wraith daggerBOT
Claire
cobalt rampart
my only idea is to somehow prove that x^2 e^-x < 1 so as to use that 1/1+x^2 converges because its arctan(x)
daring kelp
Brilliant idea
daring kelp
Well can you prove x^2e^-x<1
daring kelp
Try multiplying both sides by e^x
daring kelp
it would have to be valid for all X correct?
Only positive x cause that's what the integral is integrating over
daring kelp
The endpoint of an integral doesnt really matter
Only the limit to it
But the inequality is also true for 0 so its no problem
deep oriole
well i mean i dont know how rigorous you have to be, but cant you just say that e^x grows much faster than x^2
the inequality is eventually true
deep oriole
wait actually its true for all x > 0, i thought it was just eventually true
that makes it easier anyway
think of the power series of e^x
what can you say about it
cobalt rampart
hmm
I'm not sure
other than ofc it grows faster
I guess I could say since at 0 0^2 < e^0
deep oriole
actually what i was thinking doesnt work
but surely it must just be simple
gimme a sec
cobalt rampart
I don't think I can take ln on both sides, that would F up the domain at 0
ln(x^2) < ln(e^x) = x --> 2 ln(x) < x
but ln(x) can't have 0 as domain
deep oriole
ok my concern is the x^2/2 in the power series of e^x
it should be easy to prove that the other terms are greater than x^2/2
ok i think i have an idea
but it may be very overcomplicated
consider e^x + e^-x
daring kelp
It's slightly unrigorous but if you divide by x^2 then you get a constant < increasing function, so just check >0 (when considering the power series)
deep oriole
= 1 + x + x^2/2 + x^3/6 + x^4/24 + ... + 1 - x + x^2/2 - x^3/6 + x^4/24 - .... = 2 + x^2 + x^4/12 + .... (all positive after that)
so e^x = x^2 2 - e^-x + x^4/12 + .....
cobalt rampart
okay hold on, I mean
cobalt rampart
how about this
at 0 its true
and the derivative of e^x indicates that the rate of change is greater and thus
for every point onward
its true too
deep oriole
oh tbh you could just do induction for every natural number
is that what you are doing
deep oriole
yeah that should work
daring kelp
so e^x > x^2
Ngl that's really nice
deep oriole
they are both increasing so it works for every number in between
cobalt rampart
x^2 < e^x at 0
and d/dx (x^2) < d/dx (e^x)
is this true tho
2 < e^x
its not
daring kelp
2x<e^x
cobalt rampart
yeah sorry 2x
daring kelp
derivative of x^2 isnt 2
daring kelp
i mean words are fine lol
cobalt rampart
its kind of obvious that x^2 < e^x
daring kelp
it's obvious that x^2<e^x is eventually true
cobalt rampart
yeah but doesn't it have to be true ALWAYS?
daring kelp
no
cobalt rampart
for the comparison criteria?
daring kelp
you could just split the integral
up to the point where its true
the other will be finite so it doesnt change anything
cobalt rampart
so, just to be sure of what im saying is
deep oriole
tbh you could just show $\lim_{x\to \infty}{\frac{x^2}{e^x}} = 0$ cant you?
wraith daggerBOT
eugene_krabs_has_cake
cobalt rampart
comparison criteria states that for a function 0<F(x) < G(x) if the integral of G(x) converges, then so does F(x) on the same bounds (improper)
daring kelp
so e^x = x^2 2 - e^-x + x^4/12 + .....
. you could also just use this proof
deep oriole
if you prove that this limit is 0, then x^2 is in little o of e^x, so for all c > 0, x^2 < ce^x for large enough x
cobalt rampart
its fine, if the criteria works for "at some point"
I'd like to use that, gives me more flexibility
daring kelp
its fine, if the criteria works for "at some point"
yes, let's say that e^x overtakes x^2 at a then
$\int_{0 }^{\infty}\frac{x{^2}e{^-{^x}}dx}{1+x{^2}}=\int_{0}^a\frac{x{^2}e{^-{^x}}dx}{1+x{^2}}+\int_{a}^{\infty}\frac{x{^2}e{^-{^x}}dx}{1+x{^2}}$
wraith daggerBOT
Daniel
Compile Error! Click the 
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cobalt rampart
and we know integrating from 0 to a converges because ?
deep oriole
its a finite integral
the functions we are dealing with are obviously integrable
so the integral with finite bounds is gonna be finite
cobalt rampart
no point in between breaks it correct?
deep oriole
i dont believe so
cobalt rampart
if we had a point that breaks it which doesn't seem possible here as its whole R2 domain
deep oriole
the only point i would see breaking it would be if 1+x^2 = 0
cobalt rampart
then everything becomes divergent?
deep oriole
which doesnt happen
cobalt rampart
y
deep oriole
so we're good
this integrand should be continuous every where that its being integrated
daring kelp
$\frac{x^2e^{-x}}{1+x^2}<x^2e^{-x}<(a+1)^2$ on (0,a) so the integral converges because its finite and bounded
wraith daggerBOT
Daniel
deep oriole
just a normal discontinuity wont necessarily make the integral unbounded
deep oriole
np
daring kelp
for example this function has a singularity but is still integrable
intuitively, because the singularity "comes and goes" incredibly fast
deep oriole
similar with a step function
cobalt rampart
we were told something about sum zero point or smth
like a single point in the integration makes no difference
daring kelp
yea
cobalt rampart
for it is infinitely small is my guess
daring kelp
that's also why i said the endpoint 0 doesn't matter
it's actually the limit as $x\to 0^+$
wraith daggerBOT
Daniel
cedar kilnBOT
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silver idol
is this correct
cedar kilnBOT
is this correct
silver idol
(a) A' u B'
(b) A ∩ B'
(c) A' u B
silver idol
can i ask how this helps know if these are correct
deep oriole
because i dont know what you think they are
silver idol
math
deep oriole
i cant really help you verify if they are correct if i dont know what expression you are trying to represent with each box
from left to right, what do you think they represent?
silver idol
A′ ∪B ′ represents the union of the complement of set A (denoted as A′ ) and the complement of set B (denoted as B′ ). This set includes all elements that are either not in A or not in B, encompassing elements that are outside both A and B.
hopefully that explains it
upper abyss
The boxes are shaded
deep oriole
hopefully that explains it
ok again completely not what i meant
silver idol
The boxes are shaded
yes
deep oriole
is this one supposed to represent A' U B'?
silver idol
the third boxes represent the final answer
deep oriole
considering there are 3 boxes on each row and one answer
ok then what are the other 2 supposed to be?
deep oriole
i was just wondering since they were wrong, maybe you put it somewhere else idk
my understanding was that the first box would be A', the second would be B' and the 3rd would be A' U B'
but that is not what you have depicted
silver idol
so is it wrong
deep oriole
yes
and i dont understand the steps that lead to the wrong answer
ie i dont understand why those are the steps
silver idol
what i did was
(a) A' u B': Shade all areas that are not in either A or B.
(b) A ∩ B': Shade the area where elements are in A but not in B.
(c) A' u B: Shade all areas that are not in A, and B.
deep oriole
well you should think about what union and intersection represents in venn diagrams
my advice would be to draw each of the 2 parts to each expression
deep oriole
my understanding was that the first box would be A', the second would be B' and ...
ie do this
but for each of them
if the question wants a union, shade everything that is shaded in either of the previous 2 boxes
if its intersection, only shade the part that is commonly shaded in both of the previous 2 boxes
silver idol
ok i think i might have got it
forgot to shade in the third box for the first one but you get the idea
deep oriole
its wrong but i still dont see where you get that from
silver idol
i looked at examples online
deep oriole
gimme a sec, lemme show you my method for doing this kind of question
this is what i would recommend
in the first 2 boxes putting each part of the expression
and then joining those 2 things to get the answer
here it wants the union of these 2 shaded parts, so shade every part thats shaded in at least one of the 2 boxes
silver idol
Ok I gotcha
silver idol
how about A ∩B'
deep oriole
well draw A and then draw B', and since its intersection, only shade the parts that are shaded in both A and in B'
silver idol
deep oriole
well if we take the second one, its also not right
and again idrk what the first 2 boxes are trying to represent
silver idol
what exactly did i do wrong this time
deep oriole
well if this is using the method i suggested, the first one doesnt represent A, the second doesnt represent B'
if its not using the method i suggested, idk what is being done to try to get the answer
it just seems like random shading
silver idol
i literally did what you asked wdym
deep oriole
ok then its done incorrectly if you did
this doesnt represent A
if you were representing A you would need the middle aswell
deep oriole
this doesnt represent B as the middle is actually part of B, you need the parts not included in B, ie what is shaded here, and the outside
silver idol
can you do another example drawing then
deep oriole
ok here are the first 2 boxes
i'll leave you to do the 3rd
remember only shade the parts that are shaded in both boxes
if its only shaded in 1 of them, dont shade it
silver idol
this is for A ∩ B' right not A' u B
deep oriole
yeah this is for the second row
deep oriole
of this
deep oriole
ok youve shaded too many things in that 3rd box
for example the middle
the middle is only shaded in 1 of the 2 boxes before it, ie the first one
its not shaded in the second one
hence its not part of the intersection
that same explanation goes for one more part that you shaded in the 3rd box
silver idol
ok youve shaded too many things in that 3rd box
i mean i have to show the final answer
deep oriole
wdym?
deep oriole
yeah?
im just saying that the final answer is wrong
because youve shaded too many things
silver idol
ok so what do i unshade from the 3rd box
deep oriole
well the middle part for 1 thing
i explained why above
theres one more part
the outside
do you understand why?
silver idol
the outside
just to clarify, you want me to erase the union
deep oriole
the middle part yeah
i'll try and explain it a bit better
see here the middle is only shaded on the left box, not the middle
so its not supposed to be shaded in the right box
as another example, the outside is only shaded in the middle box, not the left box, so again you dont shade it in the right box
but this one is different
this part is shaded in both the left and middle box, so you would shade it on the right box
the only bit thats left to consider is this part
but this isnt shaded in either of the left or middle boxes, so again you wouldnt shade it in the right box
silver idol
so im just gonna clarify again, you only want me to leave it like this
deep oriole
yeah
deep oriole
would you know how to draw A' or B by themselves as 2 seperate venn diagrams?
silver idol
no
deep oriole
if someone asked you to just draw B, what do you think that would look like?
silver idol
a box with a circle inside it
deep oriole
well if you were to say that something is in B, what circle would that thing go into?
silver idol
idk B, the intersection
silver idol
deep oriole
you are close with the middle box
and you have the correct idea for the 3rd
in the middle box, the middle is also included
deep oriole
so it would be this
silver idol
by middle you mean the intersection
deep oriole
yeah
since if something is in the intersection, its in A and its in B, so by definition its in B
so the intersection is part of B
now for drawing A', just think of it as drawing A, then unshade everything thats shaded, and shade everything thats unshaded
basically draw the complete opposite of A
silver idol
so for Box 1 of C. unshade circle A and shade circle B
deep oriole
well lets say first draw A quickly
you would do the exact same thing that you did when drawing B, but ofc switch A and B around
should make it easier to see how to draw A'
silver idol
so for Box 1 of C. unshade circle A and shade circle B
ok so yes
silver idol
deep oriole
well i mean you changed the bottom right box from when you initially had it
deep oriole
if the first 2 boxes were correct here, then the 3rd would also be correct
so just do what you did before, after changing the first 2 boxes
silver idol
which was what
deep oriole
im saying you had the right idea here
any part thats shaded in at least one of the first 2 boxes gets shaded in the 3rd
so apply that idea to this
silver idol
can you do a drawing like you did here while i work on getting the final answer
deep oriole
so apply that idea to this
well you have both the drawings correct there all you need to do is imagine the 3rd box is completely unshaded, look at each part of the box and think about whether that same part has been shaded in either of the 2 boxes here, if so then shade it, if not then dont
i want you to do this part on your own, since this is literally the last part
it will give you more of an understanding compared to if i just drew it
deep oriole
ok ill draw it just because i feel bad
but i want to know how you drew the 3rd box then if you dont understand this
silver idol
i just took A and B and put them together in the 3rd box nothing hard
deep oriole
exactly
deep oriole
but thats literally what you do to get this
so how could you do it before but couldnt do it there?
silver idol
like i said i just connected the first two boxes together
jovial anchor
interesting to see how people struggle with this kind of thing
oak bluff
I need help on question 9, idk how to get d
cedar kilnBOT
I need help on question 9, idk how to get d
cerulean star
I need help on question 9, idk how to get d
Use the formula
a_n = a_1 + d(n-1)
Solve for d
No idea why they changed from s to t
Typo I think
oak bluff
it means sum vs term i think
cedar kilnBOT
@oak bluff Has your question been resolved?
trail flare
The shape given by the area bounded between y=x^2 and y=sqrt(x) inside of the first quadrant (x>=0) has a density of p(x)=2x+x^2, what is it's mass and center of mass in terms of x
trail flare
how would I find it's mass?
buoyant latch
You’re looking for the volume under the surface
dull oxide
$M=\int_V dm$
wraith daggerBOT
SWR
buoyant latch
The shape given by the area bounded between y=x^2 and y=sqrt(x) inside of the fi...
If you imagine the density function as a surface over the x-y plane
trail flare
could I not just do area (ΔyΔx) multiplied by the density equation?
and derive them
so that it's
idk how ot write the integral bu 0-->1 (2x+x^2)(sqrt(x)-x^2)dx?
buoyant latch
It’s $mass = \iint_D \rho(x, y), dxdy$
wraith daggerBOT
frosst
buoyant latch
Where D is the area enclosed by the 2 curves
trail flare
oh jeez we haven't learned that yet
buoyant latch
You’ll need to think about if you want to do dx or dy first, and what the bounds are
trail flare
huh ok
buoyant latch
You haven’t learned this yet?
trail flare
yeah we havent covered the double integralso r whatever
buoyant latch
Huh…I’m not sure how you’d formulate this without double integrals
trail flare
I also havent seen notation with a isngular bound on the bottom
buoyant latch
Seeing as you want to find the mass of a 2D object
Given by the density at each point on the object
trail flare
my thought process was that you take slices in terms of Δx, since the density changed along the x-axis and not the y-axis, and then you multiply the density by the area of the 2d object
and the area is ΔxΔy, and since we're integrating by Δx, that'll be our dx, and the Δy would be the top minus the bottom of the shape
so sqrt(x)-x^2
would that technically... work?
i'll write it down on some itegral writer or smth so I can show it properly
buoyant latch
The thing is, you can do $\int_a^b\rho(x)\int_c^d,dy,dx$
wraith daggerBOT
frosst
muted bear
yeah, $\int \rho(x)(f(x)-g(x)) dx$ should work
wraith daggerBOT
jeeby eff
buoyant latch
But you’d still need the bounds no?
trail flare
yeah the bounds would be from 0 to 1 since that's when they converge
buoyant latch
**jeeby eff**
Ohhh I see
muted bear
intersect, not converge
buoyant latch
Thanks garlic
trail flare
lmao
oh sick I thin kI got htat right
sorry for the typing..
but what about center of mass?
i'm not too sure how to go about that
oh and the bounds would be x=0 to x=1
btw the equation I got for mass is
muted bear
im assuming center of mass in terms of x just means the x coordinate of center of mass
trail flare
yeah
muted bear
ok remember that center of mass means the if you cut the piece through the center of mass, you cut the piece into two pieces of equal mass
trail flare
do I not jsut use the center of mass equation?
muted bear
that being?
trail flare
[mass*Δx]/mass
trail flare
yeah its jsut my class was only taught using that
muted bear
yeah you can use that
cedar kilnBOT
cedar kilnBOT
Available help channel!
foggy blaze
cedar kilnBOT
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unreal folio
how did he calculate $8.42 for the interest of the 2nd month?
drifting marlin
,calc (844.2+840)*.005
wraith daggerBOT
Result:
8.421
so the balance at the start of august (844.20) plus the deposit (840) all earns the .5% interest over the month
cedar kilnBOT
@unreal folio Has your question been resolved?
unreal folio
so the balance at the start of august (844.20) plus the deposit (840) all earns ...
oh 💀 thanks, im confused on how the 3rd interest is $12.66 then
plucky owl
It's the same process, take the balance from the previous row add the deposit times the interest
So the 12.66 is from (1692.62 + 840) * .005
,calc (1692.62 + 840) * .005
wraith daggerBOT
Result:
12.6631
thank you
nimble estuary
cedar kilnBOT
nimble estuary
how to approach this problem
i tried rewriting this as $\sum_{k=0}^{2n}\binom{4n}{2k}(-1)^k$
wraith daggerBOT
pirateking0723
nimble estuary
this sum is the same as $2(\binom{4n}{0}+\binom{4n}{2}+\cdot\cdot\cdot+\binom{4n}{2n})=2\sum_{k=0}^{n}\binom{4n}{2k}(-1)^k$
wraith daggerBOT
pirateking0723
nimble estuary
what can i do now
nimble estuary
**pirateking0723**
does this even help
tidal turtle
通过动画讲解二重积分的换次序积分
random shale
You could use complex numbers
Expand $(1+i)^{4n}$, the real part is the sum that they have given you
wraith daggerBOT
Max
random shale
Write this in modulus argument form. Then find the real part 👍
nimble estuary
**Max**
i got what you mean all of the terms with even powers on the i will be real
these even powers will have the coefficients which i am trying to sum
nimble estuary
so i rewrite this as (2i)^(2n)=(-2)^(2n)=(-4)^n
but then isnt this the same as (1+i)^(4n)
so it also includes the imaginary parts too
cedar kilnBOT
@nimble estuary Has your question been resolved?
nimble estuary
so it also includes the imaginary parts too
right ?
upper laurel
does $C^a_b$ mean $\binom ba$ here
wraith daggerBOT
mtt
wicked mantle
通过动画讲解二重积分的换次序积分
I don’t think it’s related to calculus
upper laurel
sounds like an unrelated question
I swear there should be a youtube animation or 10 on that but finding it wont be easy
upper laurel
but then isnt this the same as (1+i)^(4n)
(1 + i)^(4n) = (2i)^(2n) = (2i * 2i)^n = (-4)^n
similarly (1 + i)^(4n) = ((1 + i)^4)^n = (-4)^n
(1 + i)^4 just is -4
upper laurel
so it also includes the imaginary parts too
the imaginary parts are 0 anyways so it doesnt matter
cedar kilnBOT
@nimble estuary Has your question been resolved?
nimble estuary
the imaginary parts are 0 anyways so it doesnt matter
how are they always 0
upper laurel
(1 + i)^(4n) = (2i)^(2n) = (2i * 2i)^n = (-4)^n
.
nimble estuary
i wrote that they sum to 0
upper laurel
they do
nimble estuary
yes
upper laurel
what you see here is how
1 + i has angle pi/4
(1 + i)^4 has angle pi/4 * 4
or angle pi
which means (1 + i)^4 is real
its -4
nimble estuary
but that doesnt mean that the imaginary parts in each term are 0
upper laurel
can you name me a term
nimble estuary
$\binom{4n}{6}i^{4n-3}$
wraith daggerBOT
pirateking0723
upper laurel
does that go against what I said?
upper laurel
the imaginary parts are 0 anyways so it doesnt matter
oh wait theres a typo in this statement here
nimble estuary
i^(-3)=i
upper laurel
the imaginary parts add to 0 anyways so it doesnt matter
thats likely what you meant
nimble estuary
the imaginary parts add to 0 anyways so it doesnt matter
yes thats what i meant
upper laurel
there we go
as a precaution you say you choose the real part
here we didnt have to throw any terms away
since all the imaginary parts happen to add to 0
nimble estuary
ah yes because the imaginary part is already 0 after summing
upper laurel
in the rush of saying that, I accidentally say the imaginary parts are 0 instead which isnt true
nimble estuary
np dw about that
nimble estuary
yes because this statement can mean that the sum of the imaginary part is 0
you can interpret it like this
tysm and sorry if i wasted your time
but i have a question
how can i find about (1+i)^(4n)
upper laurel
I was wondering why you didnt say that before
nimble estuary
is it something that can only come from experience
upper laurel
the clue you can take is that the signs alternate
1, -1, 1, -1, 1, -1...
but the powers involved are all even like:
0, 2, 4, 6, 8, 10...
now if you remember powers of i, they go:
1, i, -1, -i, 1, i, -1...
0, 1, 2, 3, 4, 5, 6...
you can see there the even parts suggest ideas
thats the first clue
nimble estuary
ah yes this relats to i^2=-1,i^4=1....
but how can i realise (1+i) why not 1+2i for example
nvm
because the coefficients will not just be the binomials
tysm have a great day
.close
cedar kilnBOT
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Available help channel!
tidal fable
cedar kilnBOT
cedar kilnBOT
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
olive wind
you need to use the log rules
tidal fable
can u just give me answer my friend sent me this saying i couldn't solve this question
olive wind
!noans
cedar kilnBOT
The purpose of this server is to help you learn, not to hand out answers. Do not ask someone to give you the answer directly.
split socket
With that attitude he was obviously right
tidal fable
ok
olive wind
can u just give me answer my friend sent me this saying i couldn't solve this qu...
I guess he was right
olive wind
so two ways to start trying to do this
based on the form of the options
one of these should get you to the answer
tidal fable
ok
tidal fable
why this so hard
olive wind
xlog(a)+ylog(b)+zlog(c)=log(a^x)+log(b^y)+log(c^z)=log(a^xb^yc^z)
olive wind
why this so hard
It's ok. It was not designed with the average eight grader in mind
no need to feel bad
olive wind
I'm not doing the question myself
it has a bunch of algebraic manipulations which while not technically challenging are annoying
olive wind
good idea
mighty mango
a hint is to look at the maximum degree vertex
cedar kilnBOT
a hint is to look at the maximum degree vertex
@mighty mango Has your question been resolved?
cedar kilnBOT
@mighty mango Has your question been resolved?
mighty mango
.close
cedar kilnBOT
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Available help channel!
wooden socket
Can I get help
cedar kilnBOT
Can I get help
twilit escarp
For the first one you factor by 2^x on top
dull oxide
,rccw
wraith daggerBOT
rare schooner
Factorise common factors from the top
wooden socket
Ooh
wooden socket
Where did the other 2x go
shell talon
the denominator you js multiply w the power
shell talon
Where did the other 2x go
oh waitt sorry
deep oriole
Are you talking about part a?
shell talon
2^x(2^1 + 1)
should be this
shell talon
Are you talking about part a?
yes
deep oriole
should be this
Ok I was just wondering
wooden socket
How did the 1 come from
deep oriole
But you’ve corrected it
wooden socket
2^x(2^1 + 1)
Ohhh
shell talon
so after u factorise it you get 2^x(2^1 + 1)
wooden socket
So we separate the terms and multiplied them
shell talon
the one can be ignored
wooden socket
Ok
shell talon
So we separate the terms and multiplied them
and factorised them
first step --> look at factorising it
wooden socket
Ok
shell talon
after factorising simplify the brackets
wooden socket
Ok
shell talon
and then simplify the whole eqn
wooden socket
Thanks 🙏
wooden socket
.close
inland solar
how would I solve this?
cedar kilnBOT
how would I solve this?
whole wigeon
