As $a^2=|a|^2$, we simply have to show $2ab \leq |a||b|$, or more simply $ab \leq |a| |b|$
\
$a \leq |a|$; $b \leq |b|$
\
From which it follows $ab \leq |a| |b|$
#help-49
twilit field
grand pondBOT
ƒ(Why am. I here)=I don't Know
radiant roost
does that follow?
nova yoke
it's not even true
radiant roost
i mean the other part
nova yoke
no, it doesn't follow
radiant roost
a<=|a|, b<=|b| implies ab<=|a|*|b|?
twilit field
a<=|a|, b<=|b| implies ab<=|a|*|b|?
That still sounds a. bit sus
radiant roost
i think i'd use cases
twilit field
The book explicitly asks me too avoid cases
radiant roost
hmmm
the cases would be the same
wait
how do they state the triangle inequality?
twilit field
|a+b| < |a|+|b|
radiant roost
midnight plankBOT
@twilit field Has your question been resolved?
twilit field
ooh
last slate
im not sure what you are asked to do
radiant roost
i don't see by what rule of inference you can draw that conclusion
twilit field
the conclusion isnt wrong, but the reasoning is
yea
twilit field
oh right
radiant roost
that gets you ab <= |ab|
twilit field
follows trivially from that
last slate
also, expanding (a+b)^2 you might get a clue
twilit field
also, expanding (a+b)^2 you might get a clue
I've done that
That's what led me to ab \leq |ab|
last slate
you have (a+b)^2 = a^2 + 2ab + b^2, and also that ab <= |ab|
radiant roost
but how do you get |ab| <= |a|*|b|?
twilit field
I've already show $|a||b|=|ab|$
grand pondBOT
ƒ(Why am. I here)=I don't Know
radiant roost
oh, all right
last slate
you can manipulate the inequality to get a^2 + 2ab + b^2 on the leftside in ab <= |a||b|
okay know what
a^2 + ab + b^2 <= a^2 + |a||b| + b^2 follows from the inequality by multypling by 2, and then adding a^2 and b^2
but a^2 = |a|^2
twilit field
I've come until this is equivalent to provin ab<|a||b|
twilit field
but a^2 = |a|^2
proven earlier
last slate
so now you have a^2 + 2ab + b^2 <= |a|^2 + 2|a||b| + |b|^2
last slate
btw im not sure if that's the best way to do a proof. i mean, here it works
because there are statements that implications don't go the other way around, here is no problem tho
twilit field
Prove that $|a-b| \leq |a-c|+|c-d|+|d-b|, a,b,c,d \in \R$
midnight plankBOT
Prove that $|a-b| \leq |a-c|+|c-d|+|d-b|, a,b,c,d \in \R$
grand pondBOT
ƒ(Why am. I here)=I don't Know
twilit field
$(|c-b|) \leq |c-d|+|d-b|$
grand pondBOT
ƒ(Why am. I here)=I don't Know
twilit field
or wait
radiant roost
if you showed that, i think you can just apply it twice
grand pondBOT
ƒ(Why am. I here)=I don't Know
twilit field
Yup
that works
$|a-b| = |(a-c)+(c+d)-(b+d)| \leq |a-c|+ |(c-d)+ (d-b)| \leq |a-c|+|c-d|+|d-b|$
last slate
hi again
twilit field
hi
last slate
btw i was gonna tell you an example of why doing the proof in 'reverse' might not work
i think you can't prove a^2 + b^2 >= 0 with that strategy
grand pondBOT
ƒ(Why am. I here)=I don't Know
twilit field
i think you can't prove a^2 + b^2 >= 0 with that strategy
hmm?
last slate
no related to this question tho
twilit field
I just wrote down an equivalent statement
last slate
oh ok, with equivalent statements it always work. its just that when i started with proofs i used to do those manipulations 'mindless' so yeah
twilit field
**ƒ(Why am. I here)=I don't Know**
Does this work?
last slate
i think yes
main owl
can you use greens theourm 100% of the time for non conservative vector fields?
main owl
also how do you find the regions for them
last slate
can you be more specific please
main owl
can you be more specific please
well the first part is as specific as it gets
but
i guess you can use this as an example
how do they find the bounds here
im not seeing how y goes from 0 to x
main owl
0 to 1 for x makes sense cause its the bottom corners of the triangle
last slate
green’s thm involves a simple closed and planar curve
main owl
no
how do you know when u cant? this lecturer on youtube said to always use it for non conservative
spiral osprey
0 to 1 for x makes sense cause its the bottom corners of the triangle
have you tried drawing out the region
last slate
What if the region has holes
spiral osprey
it should make it clear why the bounds are the way they are
main owl
1 sec ill do that
last slate
technically you still can use it for a nonconservative field
just gonna be 0 so yeah
Yeah but it’s not 100% applicable
main owl
ohhhh
last slate
If it’s 0 then ur good
main owl
wait
last slate
But discontinuous vector fields exist
and you need conditions about the curve and the field
main owl
Yeah but it’s not 100% applicable
is it 100% applicable for a multivariable calc class
when does it start becoming not 100% applicable
last slate
i think you misinterpreted what I said
main owl
like, is this starter vector calc class
last slate
What if the region has holes
main owl
gonna give me scenarios like that tho
like do they teach that in here
im not sure ive ever seen a region with holes
cause we are taking about a wire rn
last slate
Idk but it’s common mathematical practice, if the region isn’t simply connected or smooth, it can still be non conservative
But green’s thm would be inapplicable
main owl
oh wait
so
if its like 4xdx +2ydy over 2 different segments not just 1
than u woudnt be able to use it?
like if its a jagged line
last slate
if its like 4xdx +2ydy over 2 different segments not just 1
As long as it forms a closed loop
i think you are confussing the field itself and the trajectory
main owl
what happens if you use greens theourm on something your not suppose to be able to?
would you even know
or would u just get it wrong
last slate
The integrals won’t work
main owl
oh alright so u can just check it like t hat worse case scenario than
last slate
You gotta verify that the path is closed simply connected and the Vector Field is smooth
yeah like look
If your examiner gives you a question
They expect you to know which criterion you need for Green’s theorem to be applied
last slate
Doesnt have to be simple
main owl
what
last slate
But smooth field closed curve
last slate
it could still work but for the sake of what you were taught
It’s a smooth simple closed curve.
main owl
is x the top bound of y on the triangle cause its only half of a block?
how come it wouldnt just be like 0.5
spiral osprey
you're just integrating over the region
if you did 0.5 you would have a square not a triangle
spiral osprey
🙂
main owl
.close
midnight plankBOT
main owl
.reopen
midnight plankBOT
✅
main owl
for dy dx how are the x bounds 0 to 1? i understand how the y bounds are obtained
last slate
dxdy or dydx is the same as dA
the idea is that the closed curve (assuming the other restrictions) encloses an area
main owl
i understand that but either way im still not sure how they got 0 and 1
as the bounds
for either integrating order
last slate
if you choose x as the free variable, you can see it goes from 0 to 1
main owl
oh so its literally just knowing what the actual graph looks like
theres no way to do it through just setting them equal to eachother?
last slate
being more specific, the region enclosed by the curve
midnight plankBOT
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Available help channel!
main owl
is there ever a point to use this formula that suppusdely works for every single simple closed curve?
midnight plankBOT
is there ever a point to use this formula that suppusdely works for every single...
main owl
or is using greens therum from scratch just better
if im not that good with any of this
because this makes u reparametrize the curve but if you just used greens theorum from scratch you wouldnt have to
wait isnt greens theorum normally solving for work
so is this just completely seperate to solve for area
midnight plankBOT
last slate
wait what you meant with that question?
if you mean if you can use it for any good choice of trajectory (under restrictions) then yes
main owl
wait what you meant with that question?
what question
this is just indpendent
of everything else i asked
but thats cool
i see
midnight plankBOT
Available help channel!
main owl
for the top y bound can you just use y=x too
midnight plankBOT
for the top y bound can you just use y=x too
main owl
thats what i did and i got the same answer
but im not sure if its coicidence
and if my top bound was actually not right
lethal pasture
If what you were doing was computing the area under the triangle, then yes it's a "coincidence" because you get two symmetric triangles with 1-x or x.
main owl
If what you were doing was computing the area under the triangle, then yes it's ...
alright ty guess i gotta be more carefukl
.close
minor cobalt
Can anyone help me in knowing what I can do for exam?
past nebula
These channels are made to ask specific questions, not general discussion
midnight plankBOT
viscid pumice
Can you stop trolling? @rain wasp
rain wasp
again who are you
zealous schooner
lol
minor cobalt
gosh these people..
midnight plankBOT
Available help channel!
zealous lantern
Oo
empty ivy
hi how do you do the last part of this
midnight plankBOT
hi how do you do the last part of this
lethal path
hi how do you do the last part of this
you should get that you need to prove b^2 > 3ac
but the limiting case is when you have a double root, so where P(r) = P'(r) = 0 if r is the double root
which implies $ar^3 + br^2 + cr + d - \frac{r}{3}(3ar^2 + 2br + c) = 0$
grand pondBOT
south
lethal path
I forgot how to do it exactly
lethal path
but the limiting case is when you have a double root, so where P(r) = P'(r) = 0 ...
if the double root is a local minimum, moving the graph down will make there be 3 real roots
similarly, if the double root is a local maximum, you need to move the graph up
grand pondBOT
south
midnight plankBOT
lethal path
ah jeez the answer is online anyways
specifically use Approach0
much better for finding answers to maths questions than Google
cause Google doesn't return results
midnight plankBOT
Available help channel!
cosmic dew
what do i do to find out if this is convergent or divergent
midnight plankBOT
what do i do to find out if this is convergent or divergent
lethal path
what do i do to find out if this is convergent or divergent
comparison test with 1/3
lethal path
can i just find limit of the sequence and if its not = 0 then its divergent
absolutely that's also correct
midnight plankBOT
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twilit field
If $f(x)=x^2$ , find sets $A$, $B$ , such that $f(A \cap B) \neq f(A) \cap F(B)$
midnight plankBOT
If $f(x)=x^2$ , find sets $A$, $B$ , such that $f(A \cap B) \neq f(A) \cap F(B)$
grand pondBOT
ƒ(Why am. I here)=I don't Know
twilit field
I was thinking of $A =[-2,0]; B=[0,2]$
grand pondBOT
ƒ(Why am. I here)=I don't Know
twilit field
$f(A)$ is the set of outputs with domain $A$
grand pondBOT
ƒ(Why am. I here)=I don't Know
misty gorge
**ƒ(Why am. I here)=I don't Know**
have you computed what F(A), F(B), and F(A cap B) are?
for those sets
twilit field
F(A)=F(B)= [0,4]
hollow oyster
yes
misty gorge
yeah, and what is F(A cap B)?
twilit field
0
misty gorge
yeah, the singleton
twilit field
well, ${0}$
grand pondBOT
ƒ(Why am. I here)=I don't Know
twilit field
Thay ain't equal
misty gorge
great
twilit field
This next question is much more interesting IMO
if $g:\R \to \R$ , show that $g(A \cap B) \subseteq g(A) \cap g(B)$
twilit field
twilit field
if $g:\R \to \R$ , show that $g(A \cap B) \subseteq g(A) \cap g(B)$
grand pondBOT
ƒ(Why am. I here)=I don't Know
twilit field
**ƒ(Why am. I here)=I don't Know**
I honestly have no idea here
Should probably think about it for a while though
misty gorge
I honestly have no idea here
okay cool. so if i have sets X and Y, and i want to show X is a subset of Y, how do i go about that?
in general
like what’s the definition of subset
hollow oyster
Thay ain't equal
bonus points if you could find all sets A and B
twilit field
I show that $x \in g(A \cap B) \implies x \in g(A) \cap g(B)$
grand pondBOT
ƒ(Why am. I here)=I don't Know
misty gorge
okay good
twilit field
oooh
let $x \in g(A \cap B)$. Let $y$ be the pre-image of $x$. We then have $ y \in A \cap B$. So $y \in A \land y \in B$.
\
This means that $g(y) = x \in g(A) \land g(B)$ completing the proof
grand pondBOT
ƒ(Why am. I here)=I don't Know
misty gorge
good work
misty gorge
I expectd this to be much harder tbh
in general just remember that you do have the tools to do this stuff on your own, and you seem to be proficient at using them now. so just whenever you’re stuck, try to keep in mind the question “what is the fundamental thing that i have to prove, and what does it mean for that to be true?”
i pretty much only asked you that question
and you got it without any more prompting
grand pondBOT
Wild123
wide isle
Same for B, then sum it up
twilit field
Fair,yeah
Thanks!
both of you!
Find a relationship for some $g$ between $g(A \cup B)$ and $g(A) \cup g(B)$.
\
I suspect $g(A \cup B) = g(A) \cup g(B)$. Let $x \in g(A \cup B)$. Let $y$ be the pre-image of x$. We then have $y \in A \lor y \in B$. Thus $x \in g(A) \lor x \in g(B)$ Thus $x \in g(A) \cup g(B)$. Therefore $g(A \cup B) \subseteq g(A) \cup g(B)$.
Let $x \in g(A) \cup g(B) \implies x \in g(A) \lor x \in g(B)$. Let $z$ be the preimage of $x$, then $z \in A \lor z \in B \implies z \in A \cup B \implies x \in g(A \cup B)$. Thus $g(A) \cup g(B) \subseteq g(A \cup B)$.
\
Thus $g(A) \cup g(B) =. g(A \cup B)$
grand pondBOT
ƒ(Why am. I here)=I don't Know
visual tiger
"let y/z be the preimage of x"... Not sure about that
twilit field
wddym
twilit field
or
visual tiger
who's the preimage of x
visual tiger
yes
twilit field
Find a relationship for some $g$ between $g(A \cup B)$ and $g(A) \cup g(B)$.
\
I suspect $g(A \cup B) = g(A) \cup g(B)$. Let $x \in g(A \cup B)$. Let $y$ be a pre-image of x$. We then have $y \in A \lor y \in B$. Thus $x \in g(A) \lor x \in g(B)$ Thus $x \in g(A) \cup g(B)$. Therefore $g(A \cup B) \subseteq g(A) \cup g(B)$.
Let $x \in g(A) \cup g(B) \implies x \in g(A) \lor x \in g(B)$. Let $z$ be a preimage of $x$, then $z \in A \lor z \in B \implies z \in A \cup B \implies x \in g(A \cup B)$. Thus $g(A) \cup g(B) \subseteq g(A \cup B)$.
\
Thus $g(A) \cup g(B) =. g(A \cup B)$
grand pondBOT
ƒ(Why am. I here)=I don't Know
Find a relationship for some $g$ between $g(A \cup B)$ and $g(A) \cup g(B)$.
\\
I suspect $g(A \cup B) = g(A) \cup g(B)$. Let $x \in g(A \cup B)$. Let $y$ be a pre-image of x$. We then have $y \in A \lor y \in B$. Thus $x \in g(A) \lor x \in g(B)$ Thus $x \in g(A) \cup g(B)$. Therefore $g(A \cup B) \subseteq g(A) \cup g(B)$.
Let $x \in g(A) \cup g(B) \implies x \in g(A) \lor x \in g(B)$. Let $z$ be a preimage of $x$, then $z \in A \lor z \in B \implies z \in A \cup B \implies x \in g(A \cup B)$. Thus $g(A) \cup g(B) \subseteq g(A \cup B)$.
\\
Thus $g(A) \cup g(B) =. g(A \cup B)$
```Compilation error:```! Missing $ inserted.
<inserted text>
$
l.1424 ...e a pre-image of x$. We then have $y \in
A \lor y \in B$. Thus $x ...
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)```
grand pondBOT
rafilou is not not born in 2003
twilit field
y is in A or y is in B
twilit field
g is a from R to R of which A and B are subsets
visual tiger
ok so g:R->R
If you just say "let y be a preimage of x", afterwards what stops us from picking y outside of A and B?
You have to specify that y is a preimage of x in A U B
because x is taken in g(A U B)
and not just g(R)
visual tiger
the gist of the proof is there
then there's the same problem with z
it's not told why z is in A or z is in B
twilit field
As it's a mapping from (A \cup B). \to R
visual tiger
I thought it was a mapping from R to R
visual tiger
I think you didn't use properly the fact we're taking x from g(A) U g(B)
if x is in g(A), then we can find a preimage z in A
and then z in A U B
and g(A U B)
and same for x in g(B)
but that's overkill
don't we already know that when $E\subseteq F$, $g(E)\subseteq g(F)$?
grand pondBOT
rafilou is not not born in 2003
visual tiger
The second part of the proof could be changed to 'g(A) c= g(A U B) and g(B) c= g(A U B), thus g(A) U g(B) c= g(A U B)'
twilit field
The second part of the proof could be changed to 'g(A) c= g(A U B) and g(B) c= g...
Got it
thanks
.close
midnight plankBOT
midnight plankBOT
Available help channel!
supple needle
midnight plankBOT
supple needle
The equation of the tangent line to the curve y = x√2x which is perpendicular to the line x+3y-6=0 is
<@&286206848099549185>
digital leaf
!15m
midnight plankBOT
Please only use the <@&286206848099549185> ping once if your question has not been answered for 15 minutes. Please do not ping or DM individual users about your question.
digital leaf
!done
midnight plankBOT
If you are done with this channel, please mark your problem as solved by typing .close
supple needle
!done
midnight plankBOT
If you are done with this channel, please mark your problem as solved by typing .close
midnight plankBOT
midnight plankBOT
Available help channel!
midnight plankBOT
midnight plankBOT
@frank osprey Has your question been resolved?
formal shore
Im working with logarithms, but I have a question on how some of the rules work
formal shore
If we look at lg(x^yz), when we use the rule of pulling the exponent out, do we have to do that with all exponents or can we also only do 1? So that it becomes z*log(x^y)
Second question is that if we have lg(x^3y^2), can we just factor them out as 6lg(xy), or not as they're not the same?
slender walrus
first case, you can do one since
x^(yz) = (x^y)^z
second case, no.
since (xy)^6 = x^6 * y^6
(not x^3 * y^2)
formal shore
first case, you can do one since
x^(yz) = (x^y)^z
second case, no.
since (xy)^6 ...
So in the first case, we can just pull a single one out, but in the second, all exponents have to be the same if we want to factor it?
So lg(x^3y^3) would just become 3lg(xy)
slender walrus
yes
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Available help channel!
summer ravine
If a + 1/a = 2
Then find a^8 + 1/a^8
midnight plankBOT
If a + 1/a = 2
Then find a^8 + 1/a^8
summer ravine
I made an equation:
a^2 + 1 - 2a = 0
(a-1)^2 = 0
a = 1
Then a^8 + 1/a^8 = 2
How is this possible
fallow scarab
how is what possible
wide isle
well, that's the magic...of 1
fallow scarab
plug in a=1 and find out
wide isle
oh sure my bad. typo
wide isle
note a^3=t
summer ravine
ok
wide isle
then the equation becomes...?
summer ravine
t + 1/t = 10
We have to find t^(1/3) + 1/t^(1/3)
wide isle
no, that would make it more complicated. I think you did a different thing.
we note a^3=t
summer ravine
yes
wide isle
so..isn't it t+1/t=10?
summer ravine
t + 1/t = 10
We have to find t^(1/3) + 1/t^(1/3)
Didn't I write that?]
wide isle
in what I wrote, there's no t^(1/3)...
wide isle
try to find t
summer ravine
We are unable to do anything whatsover
wide isle
should be similar to the case with a?
you knew how to solve a + 1/a =2
now it's time for... t+1/t=10
wide isle
I didn't ask...you know how to solve such equation?
wide isle
Then scrap that, a^3=t has a single solution
we extract cubic root 😄
Of course, won't compute cubic root of 5+- 2sqrt6
we'll just write it like that
wide isle
Yes, although your teacher might expect from you to rationalize the fraction
when you do a + 1/a
since that 1/a has irrational denominator
grand pondBOT
Wild123
$\sqrt[3]{5+2\sqrt{6}}+ \frac{1}{\sqrt[3]{5+2\sqrt{6}}}=...$
wide isle
1/9?
summer ravine
I think I'll never be able to solve this question
wait
wait
no
you're right
we have to find a + 1/a
wide isle
😄
summer ravine
mb
summer ravine
**Wild123**
```latex
$\sqrt[3]{5+2\sqrt{6}}+ \frac{1}{\sqrt[3]{5+2\sqrt{6}}}=......
ok
wide isle
how to rationalize that 2nd fraction
summer ravine
well 1/cuberoot(5+2root6) = (cuberoot(5-2root6))/cuberoot(5^2 - 4 * 6)
= cuberoot(5-2root6)
right?
summer ravine
thanks
wide isle
I was thinking to amplify with the square of that cuberoot, then amplify again to get rid of the sqrt
But your method seems way faster
summer ravine
i think cuberoot(5 + 2root6) + cuberoot(5- 2root6) is the answer?
wide isle
Looks like it.
summer ravine
because this does not simplify to a whole number
sometimes it does
i think i am done
thank you
wide isle
you're welcome
summer ravine
.close
midnight plankBOT
midnight plankBOT
@timber arrow Has your question been resolved?
surreal escarp
midnight plankBOT
surreal escarp
how would i go about finding the least possible selling price for the toy
gleaming latch
how would i go about finding the least possible selling price for the toy
P(x) > 80000
surreal escarp
i dont quite understand
lavish venture
they told you the profit is is more than 80000
surreal escarp
yes
midnight plankBOT
@surreal escarp Has your question been resolved?
dull pond
midnight plankBOT
rough birch
im assuming you are asking if the summation is convergent or divergent?
or just that
dull pond
I think the summation
lol I just had a test and i couldn’t figure it out for the life of me
You can’t use alternating series test
Bc it equals 1
I think at least
severe escarp
Is this convergent or divergent
for x towards positive infinity?
rough birch
here, we will bound this by another sequence
show that (x!)x^x/(2x)! converges
this will imply that (x!)^2/(2x)! converges
which will then imply by alternating series that (-1)^x(x!)^2/(2x)! will converge
severe escarp
im sure it converges to 0 - you could find that out with some binomial coefficient work
spiral osprey
it's always tempting to use AST when you see these types of problems, but you can also test for absolute convergence and use a different test
severe escarp
correct me if im wrong, but the part after (-1)^x looks like 1/(2x choose x)
sorry for the lack of rigour, just flying over it
rough birch
im sure it converges to 0 - you could find that out with some binomial coefficie...
you can specifically bound it with (2/3)^(x/2-1)
there are many other choices
midnight plankBOT
@dull pond Has your question been resolved?
midnight plankBOT
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flat blade
Solve by completing the square.
–3g2 − 54g = –3
Write your answers as integers, proper or improper fractions in simplest form, or decimals rounded to the nearest hundredth.
g =
or g =
midnight plankBOT
Solve by completing the square.
–3g2 − 54g = –3
Write your answers as integers...
surreal moon
Solve by completing the square.
–3g2 − 54g = –3
Write your answers as integers...
-3g²-54g=-3
what do you know about "completing the square"?
primal sonnet
divide everything by -3
flat blade
nothing :(
flat blade
divide everything by -3
OH SO LIKE THE EQUATION
primal sonnet
both sides yes
primal sonnet
you will do after that
flat blade
you will do after that
huh?
ripe gulch
khan academy or orgo tutor
flat blade
i got g^2+18g=1
flat blade
noo, sadly
primal sonnet
no, you have to study the basics first
there is no way for you to complete the square on this conditions
flat blade
there is no way for you to complete the square on this conditions
oh 😭
idle valley
like the question
Yes, what does it mean to raise something to the second power
flat blade
a^2+b^2
floral minnow
@flat blade you cant just do that. FOIL out (a+b)*(a+b)
flat blade
OH
idle valley
<@989485949639614494> you cant just do that. FOIL out (a+b)*(a+b)
I was getting to that
flat blade
OHHH
floral minnow
I was getting to that
my apolgies
idle valley
when there are parentheses around it, that indicates that you are raising the parentheses by the second power
so it’s legit multiplying the parentheses by itself
idle valley
my apolgies
all good
flat blade
so it’s legit multiplying the parentheses by itself
mhm
so is that basically. the square thing?
how do i answer ny qyestion
surreal moon
so is that basically. the square thing?
"Squaring" means multiplying a number by itself. Like, if you multiply 5 by itself, that is 5x5, or 25. When we square a number like 5, we shorten how we write 5x5 and just write 5² instead. You just need to remember that 5², by definition, is just a shortened way to write 5x5 (the same idea for 5³, which just shortens 5x5x5).
But now we don't care about the specific number 5, we want to use this same idea of "squaring" for any numbers. But how do you represent "any number"? This is where variables come in. A variable is just some place-holder for some unknown number, or for saying "this could be any number". So like, for A+B, A and B are variables, and they could represent anything. All A+B is saying by itself is that we are adding two number.
If you want a translation into simple English "A+B" is just saying "We are adding two numbers"
We can square variables also, like A². That's just saying "we are squaring some number", which is the same as "we are multiplying some number by itself".
Now here's where a lot of people start to get confused with algebra. When you have two unknown numbers (like A and B), you can add them together same as you can add just any two numbers at all (like say adding 5 and 7). And what do you get when you add two numbers? You get another number. So, if A and B represent numbers, then A+B also represents a number. And like any number, you can add, subtract, multiply, or whatever you like.
So when you see some expression like (A+B)², that is saying "we are adding two numbers to ge a number, then we are squaring that number"
surreal moon
So, (A+B)²=(A+B)x(A+B). If you need help working out how to do (A+B)x(A+B), that's where FOIL method comes in, but I will give you some time to absorb what I have mentioned so far, and ask any new questions if you have them
flat blade
OH ALR
i get it
but how does this go into the like square thing
like
"completing a square"
surreal moon
yeah sure, I can explain that right now
Let's start with some expression like A(x+R)²+Q (I'm just using R and Q as variables here because I plan to use B and C as variables later, and I don't want to get you confused)
Let's expand the (x+R)². Eexpanding means that we simply multiply out the expression (x+R)(x+R). Do you know how to do that?
flat blade
yea
flat blade
Let's expand the (x+R)². Eexpanding means that we simply multiply out the expres...
yea
flat blade
(ax+ar)(ax+ar)+q? or is it a*(x+R)(x+R)+q?
surreal moon
(ax+ar)(ax+ar)+q? or is it a*(x+R)(x+R)+q?
It's the second one.
flat blade
oh alr
surreal moon
I can see why you would think it is the left one at first, but this is where you need to remember how multiplication distributes itself
flat blade
mhm
surreal moon
Oh, you may have also been confused by what the ² symbol is acting on
flat blade
yea
surreal moon
In general, something like $AB^2$ just means $A\times B\times B$. It does not mean $(A\times B)\times (A\times B)$. If you wanted that, then that would be $(AB)^2$. This comes from the PEMDAS (or BODMAS) rules
grand pondBOT
SWR
surreal moon
so how do we complete a square?
We are getting closer
surreal moon
(ax+ar)(ax+ar)+q? or is it a*(x+R)(x+R)+q?
You have a good start here with $A(x+R)(x+R)+Q$, but now I am asking you to distribute the $(x+R)(x+R)$ multiplication. Do you know how to do that?
grand pondBOT
SWR
flat blade
yea u mean foil right
surreal moon
yea
flat blade
yea, ik that
surreal moon
yea, ik that
Okay, but I am asking you to do it.
surreal moon
Yes. Very good. Nice work
flat blade
mhm
surreal moon
So now, you let us replace "(x+R)²" with "(x²+2Rx+R²)" in the original expression A(x+R)²+Q. We get A(x²+2Rx+R²)+Q.
Next, you need to distribute the A. Do you know what that means?
flat blade
yea
flat blade
(ax+ar)(ax+ar)+q? or is it a*(x+R)(x+R)+q?
so we kinda do the same thing i tdid here rihgt
surreal moon
yeah
flat blade
so ax^2+2rx(a)+aR^2)
surreal moon
Yeah
good work
Recap:
We started with $A(x+R)^2+Q$ and expanded it to $Ax^2+2ARx+(R^2+Q)$
grand pondBOT
SWR
flat blade
mhm
surreal moon
Note that 2AR and R²+Q are themselves just numbers. If they're just some arbitrary numbers, we can let them be their own variables. We can rewrite 2AR as just the variable B and R²+Q just as C.
This would mean that we are left with Ax²+Bx+C
surreal moon
wait why is this in parentheses
No reason. The parentheses change nothing. I only added them to emphasize two things
- They are not parts of x² or x
- I would make R²+Q its own variable (C) later on
Okay, I now have everything I need to teach you what completing the square is, how to do it, and why it is important
surreal moon
**SWR**
Correction: this should be $Ax^2+2ARx+(AR^2+Q)$. I accidentally wrote $R^2$ where I meant $AR^2$.
grand pondBOT
SWR
flat blade
yea
surreal moon
Anyway, back to "completing the square". I expanded $A(x+R)^2+Q$ into $Ax^2+2ARx+(AR^2+Q)$ just by squaring $x+R$ and distributing the multiplication of $A$.\
\
Now, we're going to work backwards. We want to start at something like $Ax^2+Bx+C$ and make it look like something similar to $A(x+R)^2+Q$. We can accomplish this by knowing what $A(x+R)^2+Q$ expands into (which we have just done). When we expanded it, I said that we could just write $2AR$ as $B$, and $AR^2+Q$ as just $C$.\
\
What we did here was create two variables equations:
$$B=2AR$$
$$C=AR^2+Q$$
But if we want to work backwards and write $Ax^2+Bx+C$ so that it looks like $A(x+R)^2+Q$, we can alter those two variables equations so that its like we are creating $R$ and $Q$ from $B$ and $C$ (and $A$, which is the same in both).\
\
For example, if $B=2AR$, then you can divide both sides by $2A$ and get
$$R=\frac B{2A}$$
You can do the same thing for $Q$ in $C=AR^2+Q$. Just subtract $AR^2$ from both sides you and get
$$Q=C-AR^2$$
But because $R=\frac B{2A}$, we can write instead
$$Q=C-A\left(\frac B{2A}\right)^2\text{ or }Q=C-\frac {B^2}{4A}$$
grand pondBOT
SWR
surreal moon
That is the rather technical idea of completing the square. You start with $Ax^2+Bx+C$, and you can rewrite as $A\left(x+\frac{B}{2A}\right)^2+\left(C-\frac{B^2}{4A}\right)$.
grand pondBOT
SWR
surreal moon
If you are having trouble seeing this, you should take a minute and try this out:\
Start with $A\left(x+\frac{B}{2A}\right)^2+\left(C-\frac{B^2}{4A}\right)$ and expand it out, distributing all the multiplications. You will end up with $Ax^2+Bx+C$
grand pondBOT
SWR
surreal moon
Solve by completing the square.
–3g2 − 54g = –3
Write your answers as integers...
We can then go over this specific problem, to help you see the whole picture
surreal moon
so corresponding to my question, -3g^2=A(-3)x^2+(-54x)=(-c?
close
we'll go over your specific problem
Your problem is $-3g^2-54g=-3$
surreal moon
oh
grand pondBOT
SWR
surreal moon
Like that?
flat blade
yea
surreal moon
Okay. The beaty of math is that there are so many ways to solve this. I'll just pick one for now
You have a lot of minuses here, so let's try and get positives instead, they're always easier to work with
flat blade
ye
surreal moon
You can multiply both sides by -1, also known as just "flip the signs"
You get $3g^2+54g=3$
grand pondBOT
SWR
flat blade
kk
flat blade
18
surreal moon
Yeah
flat blade
i got g sq +18 g=1
surreal moon
we can simplify this a bit more by dividing everything by 3
surreal moon
i got g sq +18 g=1
yeah good
okay, now we can "complete the square"
You want to turn $g^2+18g$ into something like $$(x+\textit{something})^2+\textit{something else}=1$$
grand pondBOT
SWR
surreal moon
Finding that something is what "completing the square" is.
flat blade
so is x=g in this case?
surreal moon
yea
the exact letter we use does not matter at the end of the day
So g is perfectly fine to use, or x, or whatever
flat blade
mmhm
surreal moon
²+18=1
It's all good
anyway, completing the square..
We go back to out Ax²+Bx+C thing that we spent so much time on.
flat blade
wher did c go
surreal moon
wher did c go
There's two ways to answer this:
- You could say C is 0
- you could add 1 to both sides, giving you g²+18g-1=0, then C would be -1.
Either will be okay.
flat blade
oh alr
surreal moon
There's two ways to answer this:
1. You could say C is 0
2. you could add 1 to b...
We'll do the first case here and just say C is 0
surreal moon
**SWR**
When we wanted to complete the square for $Ax^2+Bx+C$, we found that our $\textit{something}$ (as I mentioned just above) would be $\frac{B}{2A}$\
Here, $A$ is the coefficient that is in front of your $x^2$ term (also known as the $\textbf{quadratic}$, $B$ is the coefficient in front of the $x$ term (known as the $\textbf{linear}$ terms), and $C$ is by itself and is known as the $\textbf{constant}$ term.
grand pondBOT
SWR
surreal moon
18/2?
Nah you are really, really close. One sec
So for your problem, $g^2+18g=1$, your quadratic term is $g^2$. There isn't a number in front of it. But remember how multiplication by $1$ works, $g=1g$. So the coefficient of your quadratic term would be $1$\
\
$g$ is your linear term, and the number in front of it is $18$, so the coefficient of your linear term is $18$. There is not constant term (on the left side of the equation), so the constant term would just be $0$.\
\
So, basically, what I am saying is $A=1$, $B=18$, $C=0$ (overall $C$ is irrelevant for "completing the square").\
\
We rewrote $Ax^2+Bx+C$ as $A\left(x+\frac{B}{2A}\right)^2+\left(C-\frac{B^2}{4A}\right)$. Now that we are replacing $x$ with $g$, $A$ with $1$, $B$ with $18$, and $C$ with $0$, we can do your problem.\
We can rewrite $g^2+18g$ as $$\left(g+\frac 9{2\times1}\right)^2+\left(0-\frac{18^2}{4\times 1}\right)$$
We can then simplify it to
$$(g+9)^2-81$$
grand pondBOT
SWR
surreal moon
Okay, sorry for all the edits, it is good to read now
surreal moon
so g=8^2-81?
No. Not quite.
surreal moon
**SWR**
What I wrote here is a lot, but let me summarize it one sentence
We can rewrite $g^2+18g$ as $(g+9)^2-81$.
grand pondBOT
SWR
surreal moon
The process of doing that is what is known as "completing the square"
flat blade
mhm
surreal moon
But to solve for $g$, remember that $g^2+18g=1$. So if we have rewrite $g^2+18g$ as $(g+9)^2-81$, then what we actually have now is $(g+9)^2-81=1$. You can add $81$ to both sides to get $(g+9)^2=82$. By square-rooting, you can then solve for $g$
grand pondBOT
SWR
flat blade
so g=-9-sqrt82?
surreal moon
That is one solution
flat blade
mm
so to find the other solution i can plug in the other thing u sent right, then sq root it?
surreal moon
Let's work with a very basic example first
Let's say we want to find all $x$ that satisfy $x^2=25$. If you are good with your square roots, you might see that $x=5$ is one solution. However, Consider how $(-5)\times(-5)=25$ because the two minuses cancel. So $x=-5$ would be another solution.\
\
So, either $x=5$ or $x=-5$ would be solutions. Normally, we express this idea by writing $x=\pm 5$ (you can write this in plaintext on discord as x=+-5 if you like).
grand pondBOT
SWR
surreal moon
Going back to your problem $(g+9)^2=82$ can be simplified either to $g+9=\sqrt{82}$ or $g+9=-\sqrt{82}$. You must consider both cases. You can consider them both at once by writing $g+9=\pm\sqrt{82}$ if you feel comfortable doing so. Subtract $9$ from both sides and you get your two solutions for $g$: $g=-9\pm\sqrt{82}$
grand pondBOT
SWR
surreal moon
its okay
they confuse a lot of people
and my explanation may have been a little too wordy, as I assumed you had zero knowledge
I can summarize "completing the square" very succintly for you
If you have $Ax^2+Bx=\textit{whatever}$, you can complete the square of $Ax^2+Bx$ by adding $\frac{B^2}{4A}$ to both sides. If you do this, you get $Ax^2+Bx+\frac{B^2}{4A}=\frac{B^2}{4A}+\textit{whatever}$
The left side is now a perfect square. By "perfect square" I mean that you can write it as something multiplied by itself. You can write it as
$$A\left(x+\frac{B}{2A}\right)^2$$
grand pondBOT
SWR
surreal moon
You started with $Ax^2+Bx$, which, by itself, is not a perfect square (except if $B=0$). It is $\textit{incomplete}$. But you can "complete" it by adding $\frac{B^2}{4A}$
grand pondBOT
SWR
flat blade
mhm
surreal moon
This idea of "perfect squares" can be a little confusing. it helps to really study how factoring works.
In the general case if you ever see anything that looks like $a^2+2ab+b^2$, then it is a perfect square, and you can rewrite it as $(a+b)^2$. Seeing this intuitively is hard at first, and it is very easy to miss sometimes unless you are looking for it. But that is exactly why we have this process of completing the square
grand pondBOT
SWR
midnight plankBOT
midnight plankBOT
Available help channel!
rotund iron
any1 know what type of angle this is, i forgot
midnight plankBOT
any1 know what type of angle this is, i forgot
rain wasp
i dont think that pair specifically have a name
midnight plankBOT
midnight plankBOT
Available help channel!
livid axle
simple problem but can someone help me write a sin function for this graph?
midnight plankBOT
simple problem but can someone help me write a sin function for this graph?
rain wasp
simple problem but can someone help me write a sin function for this graph?
determine its amplitude, period, phase shift
livid axle
i got the amplitude, its 1
the period is 2pi i believe
the scale is a bit weird tho...
rain wasp
this is a badly designed graph lol
livid axle
loll
rain wasp
simple problem but can someone help me write a sin function for this graph?
this should be 2pi/3 instead of 2pi
wait
brazen moth
It's right I think
rain wasp
yeah
livid axle
lemme double check, i have both a paper and online copy
this is the only copy so yea ur right
it should be 2pi/3
rain wasp
it should be 2pi/3
livid axle
its visible on the negative side
livid axle
so the period is 2pi/3
brazen moth
Amplitude is 1
livid axle
yup and scale is pi/6?
brazen moth
And phase shift is 0
midnight plankBOT
midnight plankBOT
Available help channel!
white pumice
Donde está mi amor
midnight plankBOT
Donde está mi amor
patent belfry
so why is "-|X| => 1" = "|X| <= -1"
midnight plankBOT
so why is "-|X| => 1" = "|X| <= -1"
nova yoke
does => mean $\geq$ or what
grand pondBOT
Bungo
patent belfry
yeah
nova yoke
well both inequalities are impossible
patent belfry
so why is "-|X| => 1" = "|X| <= -1"
btw X is phi
topaz epoch
|x| must be positive
-|x| must be negative
a negative number can never be larger than or equal to 1
patent belfry
no i get why it's phi
i just can't understand why the 1 was swapped to -1
how does swapping signs even work-
yeah im def failing math this year 😭
peak herald
how does swapping signs even work-
it doesnt, lol
i mean technically the idea behind it is correct
nova yoke
well you can just add |X|-1 to both sides of the first one to get the second one
peak herald
but in this case it doesnt actually work because you cant have the abs value be negative
sharp coral
multiplying both sides of an inequality by -1 (or in general applying a decreasing function to both sides) flips the direction
peak herald
$|ax + b| = c$ can be split into two equations,
$ax + b = c$ and $ax + b = -c$
grand pondBOT
Karma
peak herald
for example
nova yoke
and "false implies false" is a true statement, so sure, the inequalities are equivalent
patent belfry
im just even more confused now help
peak herald
and "false implies false" is a true statement, so sure, the inequalities are equ...
no it doesnt
patent belfry
i wanna understand it so i don't forget it as soon as the exam starts
peak herald
if $p \implies q$ then you can only assume $\neg q \implies \neg p$
grand pondBOT
Karma
peak herald
im just even more confused now help
whats confusing you
nova yoke
no it doesnt
?
peak herald
?
"if it rains, the ground is wet"
according to you, false implies false, meaning if it didn't rain, it is not possible for the ground to be wet
even though it is, if you spill a bucket of water.
peak herald
thats $p \iff q$
grand pondBOT
Karma
peak herald
different
nova yoke
if you say so
patent belfry
how do i un-occupy this
peak herald
i took propositional logic this semester
patent belfry
i haven't talked here in so long i forgor
polar mortar
so why is "-|X| => 1" = "|X| <= -1"
if a > b, subtract a and subtract b from both sides. then you have a - a - b > b - a - b.
Simplifying gives you -b > -a
twin peak
so why is "-|X| => 1" = "|X| <= -1"
if you start with -|X| >= 1 and multiply both sides by -1
you'd get |X| <= -1
polar mortar
so changing the signs of everything 'swaps' the inequality
this is the same as multiplying through by a -1 and that changing the inequality
midnight plankBOT
patent belfry
why is this still occupied
twin peak
just wait a few minutes and it won't be
sharp coral
it's closed, it just takes a few minutes to recycle
midnight plankBOT
Available help channel!
twilit field
Prove or provide a counter example : Two reals satisfy $a<b$ iff $a<b+ \varepsilon, \forall \varepsilon >0$
midnight plankBOT
Prove or provide a counter example : Two reals satisfy $a<b$ iff $a<b+ \varepsil...
grand pondBOT
ƒ(Why am. I here)=I don't Know
fervent ember
a=b.
twilit field
a=b.
That's a counterexample, right
However, this is true I suspect :Two reals satisfy $a \leq b$ iff $a <b+ \varepsilon, \forall \varepsilon >0$
grand pondBOT
ƒ(Why am. I here)=I don't Know
twilit field
hold on don’t leave yet i have a youtube clip to show you
Wrong chat?
midnight plankBOT
okay so try to prove it
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
misty gorge
Wrong chat?
no it’s the right chat
twilit field
We first prove if $a \leq b , then a<b+ \varepsilon$
\
If $a \leq b$, it follows adding a positive number to $b$, makes it bigger than $b$, as $b$ can be equal to $a$, adding a positive number to it, makes it bigger, removing the possibility of an equality, thus giving us $a < b+ \varepsilon$
grand pondBOT
ƒ(Why am. I here)=I don't Know
twilit field
Not formal enough though
twin peak
maybe start with $0<\epsilon$
grand pondBOT
qianqian07
grand pondBOT
qianqian07
misty gorge
in general you should try to confirm your suspicions
like you seemed like you just had the right idea
twilit field
Yeah, execution was poor
misty gorge
i think it was fine
you might need to spend more time sitting with the problem
https://youtu.be/9qlqVEUgdgo?feature=shared here is a clip that i was sent by a friend and i really liked
it’s in french so put cc on
but i really like this perspective on math
but the advice is not only for reading books
it’s also about allowing yourself to develop your brush and paint in your own mind before getting together with others to talk about it
because i have faith that once you sit with it for a while you’d be able to do the problem. but even if you couldn’t the extra time that you spent sitting with it wouldn’t be wasted
misty gorge
Tysm!
you still need to prove the other direction (actually the other direction is a bit more interesting in some sense)
btw
twilit field
yeah, I know
We now prove if $a<b+ \varepsilon ;\forall \varepsilon>0$, then $a \leq b$
\
grand pondBOT
ƒ(Why am. I here)=I don't Know
twin peak
hint ||try a proof by contrapositive||
twilit field
I did consider that, but I feel in such introductory analysis direct proofs should be used
as they force me to think
But anyway, a contrapositive proof would look like if $a>b$ , then $\exists \varepsilon $ st $a \geq b+ \varepsilon$
\
An epsilon would be $\varepsilon = a-b$, which gives us the possibility of an equality.
grand pondBOT
ƒ(Why am. I here)=I don't Know
twilit field
Does this work?
twin peak
yeah that logic works
twilit field
**ƒ(Why am. I here)=I don't Know**
For a direct proof We then have $a-b < \varepsilon , \forall \varepsilon >0$, which means $a \leq b$
grand pondBOT
ƒ(Why am. I here)=I don't Know
twilit field
Well, this proof ain't great
twin peak
lol
twilit field
I could make it better using the complteness axiom, but that would be overkill
like greatest lower bounds
twin peak
yeah that seems like it would work
twilit field
I think I'll just stick to contrapositive for now 
Because completness is the next section in my book
twin peak
yeah that seems like it would work
actually wait idk since the infimum of possible values of epsilon is 0
but yeah contrapositive is prob easiest way to do it lol
twin peak
yeah
grand pondBOT
ƒ(Why am. I here)=I don't Know
twin peak
I think that would be logically equivalent to the contrapositive for this case
twilit field
Well, contradiction is the easiest here
twin peak
I think that would be logically equivalent to the contrapositive for this case
since you’re proving not(P and not Q) which is the same thing as if not Q then not P
twin peak
👍
twilit field
.close
twilit saddle
Hi! Can somebody help me with a rigor question? It's more general than a specific problem. If I have a piecewise relation f(x) and I want to prove it is a surjection over N, can I break apart y=f(x) however I want as long as my cases end up covering the whole codomain?
midnight plankBOT
Hi! Can somebody help me with a rigor question? It's more general than a specifi...
twilit saddle
😭
runic hamlet
can you give an example of what you mean? I dont wanna confirm in case you mean something different from what I think
twilit saddle
I don't know how to notate in here yet, but it would be like f:N->N f(n) = {n,n<x ; n+1,n>=x} and I would have two cases (one for each part of the function) that each show that half of the function mapping to a half of N