Available help channel!
#help-17
vocal sleetBOT
Available help channel!
river kettle
how can i find the natural domain and range of this where f is a complex function
vocal sleetBOT
how can i find the natural domain and range of this where f is a complex functio...
oak magnet
So f(z) = 1/z² ?
river kettle
yeah
oak magnet
Well domain is ?
alpine wave
vocal sleetBOT
@river kettle Has your question been resolved?
tall hazel
how would i get the z score if only the mean and sd were given without the value of x
vocal sleetBOT
how would i get the z score if only the mean and sd were given without the value...
@tall hazel Has your question been resolved?
manic dawn
how would i get the z score if only the mean and sd were given without the value...
it appears unclear whether the "class mean" and "class standard deviation" refer to the mean and standard deviation of math, science, or english scores
is there no additional context?
tall hazel
there were no additional context, just a question that looked for the grade averages of each student
im assuming the x may be the average but im not sure
manic dawn
yes, it is unclear
if the class mean refers to the average of the average performance of each student on the 3 tests
and the standard deviation refers to the standard deviation of the average performances of each student on the 3 tests
then you can plug in the average performance of the student on the 3 tests as the observed value
but the question does not make it clear what observations the mean and standard deviation were taken of
tall hazel
yeah true, i think im going to ask my teacher about this first
manic dawn
good idea
vocal sleetBOT
vocal sleetBOT
Available help channel!
bleak fossil
vocal sleetBOT
topaz eagle
Is it possible to solve without derivatives
✅
guide: ||1. use basic algebra to isolate some "easy" part. 2. observe symmetry on the "remaining part" 3. apply AM-GM inequality||
vocal sleetBOT
@bleak fossil Has your question been resolved?
manic flax
I don think I understood vin's method, but the way I did it is prove f(x)>0 for all x eleminating B and C, and then checking if f(x1)=f(x2) doesnt necessarily imply x1=x2
manic flax
guide: ||1. use basic algebra to isolate some "easy" part. 2. observe symmetry ...
also could you elaborate on the symmetry and how you applied am gm inequality to solve it/
final rapids
Forgive me–I'm feeling very stupid. Why does this not always generate the correct modular inverse?
vocal sleetBOT
Forgive me–I'm feeling very stupid. Why does this not always generate the correc...
final rapids
The idea is as follows
$$5x \equiv 1 \pmod 6$$
$$\ceil{\frac{6}{5}} = 2$$
$$25x \equiv 2*1 \pmod 6$$
$$4x \equiv 2 \pmod 6$$
$$\ceil{\frac{6}{4}} = 2$$
$$8x \equiv 4 \pmod 6$$
$$2x \equiv 4 \pmod 6$$
$$x \equiv 2 \pmod 6$$
twin meteorBOT
7aman
final rapids
I'm not sure what rules I'd be breaking here, but the answer is incorrect.
$$5 * 2 \equiv 10 \equiv 4 \not \equiv 1 \pmod 6$$
twin meteorBOT
7aman
hard atlas
2x=4 mod 6 does not imply x=2 mod 6
you cant divide by 2 cause 2 is not invertible mod 6
2x=4 mod 6 also has the solution x=5
final rapids
2x=4 mod 6 does not imply x=2 mod 6
for what values of $$a$$ and $$m$$ does $$ax \equiv ay \implies x \equiv y \pmod m$$?
twin meteorBOT
7aman
final rapids
Ah okay
Thank you
I was so confused because this algorithm had like a 20% chance of working lol
hard atlas
you could just keep on multiplying the equation by 5
that would work
the problem here is that you multiplied by 2 which isnt invertible
if you multiplied by 5 (or in general any element that is a power of 5, although that only ends up being 1 or 5 in this case) then you will eventually get x=...
very inefficient way to go about finding the inverse tho
faint pumice
is there a rule for breaking up (xn+x)! into factors
vocal sleetBOT
is there a rule for breaking up (xn+x)! into factors
faint pumice
here is the example
so (3n+3)!
i did 3n! * (n+1) but that is wrong lol, i just kinda applied what (n+1)! does
oh nvm i see it
.close
vocal sleetBOT
faint pumice
.reopen
vocal sleetBOT
✅
faint pumice
for this one
when i do the ratio test and get
((n+1)!)^9
i do ((n+1)n!)^9
how would i simplify the exponent?
or is that the simplest form
paper depot
well you can rewrite that as (n+1)^9 * (n!)^9
vocal sleetBOT
vocal sleetBOT
Available help channel!
viral igloo
vocal sleetBOT
viral igloo
part 3?
winter hawk
twin meteorBOT
ロケットジャンプ
viral igloo
and i can take V' = V = C? so for all z in C, f(exp(z)) = z + a
now what's the contradiction
vocal sleetBOT
@viral igloo Has your question been resolved?
vocal sleetBOT
vocal sleetBOT
Available help channel!
sleek siren
i really can’t do this 😭😭
vocal sleetBOT
i really can’t do this 😭😭
sleek siren
ignore some of my working out it’s probably wrong
but i’ve labelled angles i think i have
i’m just stuck on relating it to angles adg and gdf
mighty idol
i really can’t do this 😭😭
You need to prove that the angle ADG and FDG are half of overall angle ADF
Cus a bisector cuts it into half
sleek siren
okay so am i right in adg being 90 - x or have i messed up
marble wigeon
okay so am i right in adg being 90 - x or have i messed up
that wouldn't be correct since that would say AGD = 90 which it is not
marble wigeon
how I solved it is that say DGF = theta and then work from there
marble wigeon
it's just another variable
sleek siren
okay
marble wigeon
theta is just what I default to when worrying about angles
you can use y if u want
marble wigeon
wanna share your new diagram?
marble wigeon
i got y= 90 - (180-x)
how did you get that?
sleek siren
because EF is a straight line
180 - x is ADF
oh wait
yeah i messed up i did ADF
i don’t think i can halve it yet
arghhh
i’m so stuck
marble wigeon
fill your diagram as much as you can and post it
vocal sleetBOT
@sleek siren Has your question been resolved?
prisma kettle
why is this an ellipse?
vocal sleetBOT
why is this an ellipse?
prisma kettle
bitter pilot
why is this an ellipse?
complete the square and see for yourself
drifting jackal
It's a circle?
vocal sleetBOT
drifting jackal
,w x = x^2 + y^2
drifting jackal
@prisma kettle It's a circle
bitter pilot
<@726460267923505254> It's a circle
a circle is also an ellipse 😛
sage wind
well both of you are correct xD
drifting jackal
Unless wolfram says circle is an ellipse
vast shale
can any1 help me i have a math test tmrw about tan and i don't understand when i use tan and when i use tan-1
oak magnet
,w is a circle an ellipse
twin meteorBOT
bitter pilot
a=b
sage wind
🔥
opal moss
yes a circle is an elipse I'm pretty sure
vocal sleetBOT
Available help channel!
brittle seal
vocal sleetBOT
manic flax
uh haven't you solved it?
brittle seal
uh haven't you solved it?
Im getting wrong answer
manic flax
oh let me try
brittle seal
Okay
manic flax
its alr it happens to all of us
manic flax
e^-8
brittle seal
I mean
manic flax
yess
manic flax
well it should work out now
vocal sleetBOT
vocal sleetBOT
Available help channel!
misty cliff
hello , i’m trying to do a warm up and i’m confused how this would be anything but 17?
vocal sleetBOT
hello , i’m trying to do a warm up and i’m confused how this would be anything b...
misty cliff
i’m using the formula a(a+b) =c(c+d)
this gets me to 5x+5=6x-12
from there i subtract 5
then 6x
to get -1x =- 17x
wary hull
i’m using the formula a(a+b) =c(c+d)
Because of this
misty cliff
sorry i’m a bit confused i was taught to use the same formula for similar questions like this.
how would that formula be different?
wary hull
sorry i’m a bit confused i was taught to use the same formula for similar questi...
That would just be 27*x=45^2
misty cliff
so would x not be 48?
sorry if i’m annoying , im genuinely confused on the last question
wary hull
so would x not be 48?
For which one
wary hull
It should be 5((x+1)+5)=6((x-2)+6), no?
^
misty cliff
It should be 5((x+1)+5)=6((x-2)+6), no?
this would come out to 25x+25=36x-72 right?
wary hull
this would come out to 25x+25=36x-72 right?
No...
$5((x+1)+5)=6((x-2)+6)$, pay attention to parenthesses
twin meteorBOT
mathisfun
misty cliff
5x+30 =6x-24?
wary hull
5x+30 =6x-24?
Change the sign on the 24.
wary hull
so the final answer would be 6=x?
Yes.
misty cliff
thank you for the help
wary hull
thank you for the help
I would prefer to take only the cat.
misty cliff
.close
vocal sleetBOT
vocal sleetBOT
Available help channel!
fresh coyote
vocal sleetBOT
fresh coyote
for item 1
i dont know what to do after this. i tried factoring the denominator out but it doesnt do anything
oh wait
is this correct?
fresh coyote
yeah thats correct
what about item 2?
faint pumice
do you know l'hôpitals ??
fresh coyote
yes but we havent discussed that yet and i wanna solve it the way we were thought just incase
i might get deducted points
fresh coyote
is item 2 correct thooo
faint pumice
hang on
faint pumice
what about item 2?
you forgot somethin here
you didnt multiply the denominators correctly
well, you just forgot the 2*2 part
faint pumice
i think this one is also correct but i js wanna make sure xd
this one is not 0
its DNE
can you guess why ?
fresh coyote
nope. my thought process was 0/5 is 0, and the sqrt(0) is 0
faint pumice
and this one is correct
faint pumice
nope. my thought process was 0/5 is 0, and the sqrt(0) is 0
so its as it approaches 3 from the left
meaning itll never be exactly three, but always slightly smaller
ex 2.999999
fresh coyote
so am i supposed to take the lim of both sides?
faint pumice
so am i supposed to take the lim of both sides?
wym
fresh coyote
oh
faint pumice
which part
#3
it was an example
it is the left of 3 and it very close
since it wont be 3
it could be 1
-38
anything to the left of 3
all of which will result in a negative
fresh coyote
so in the denominator, (3^-)^2 - 9 is negative?
faint pumice
no
the numerator
x^2-9 = (x-3)(x+3)
specifically the x-3 part will be negative
but even if you didnt factor it, your result for 2.999999 will never be 9 exact
leaving you with a negative in the sqrt
does this make sense ?
fresh coyote
oh yeah sorry, i meant the numerator
alright, makes sens. but if there was a way to factor them, would we go with that?
faint pumice
alright, makes sens. but if there was a way to factor them, would we go with tha...
wdym
fresh coyote
wdym
like what if the denominator was a factor of x^2 - 9 and after cancelling them u no longer get a negative sqrt
fresh coyote
alrightt
thanks!
what about this one
thats the floor function right there btw
just incase my handwriting isnt clear
vocal sleetBOT
@fresh coyote Has your question been resolved?
fresh coyote
<@&286206848099549185>
vocal sleetBOT
@fresh coyote Has your question been resolved?
fresh coyote
.close
vocal sleetBOT
vocal sleetBOT
Available help channel!
quick narwhal
Hello, so i was reading the book "Book of Proof Third Edition" by Richard Hammack.
And in chapter 5, question 26. I'm stuck. The question is as follows:
If n = 2ᵏ - 1 for k ∈ ℕ, then every entry in Row n of Pascal's Triangle is odd.
We gotta prove this statement, using direct of contrapositive proof.
Now, what i did was just to put the values.
Proof
Suppose n = 2ᵏ - 1.
Every entry in row = C(n, k) where k ∈ ℕ and 0 <= k <= n.
C(n, k) = (2ᵏ - 1)!/(k!(2ᵏ - 1 - k)!)
= (2ᵏ - 1)(2ᵏ - 2)(2ᵏ - 3)...(2ᵏ - k)/(k!)
now i don't know what to do from here. Like it's the thing i did'nt even knew before i got to prove it.
Btw, it's not any homework or so. I'm a 10th grader and was just reading the book for better understanding of fun and joy of math. Would love if i understood how i can prove it (because it doesn't seem to be able to be proved).
Btw if you want to see the question in the original book then:
https://richardhammack.github.io/BookOfProof/Main.pdf#page=148
The above is the link to it.
vocal sleetBOT
Hello, so i was reading the book "Book of Proof Third Edition" by Richard Hammac...
quick narwhal
Hello, so i was reading the book "Book of Proof Third Edition" by Richard Hammac...
btw i've seen that there are k terms above which is divided by k! but idk what to do still
vocal sleetBOT
@quick narwhal Has your question been resolved?
vocal sleetBOT
vocal sleetBOT
Available help channel!
finite hatch
vocal sleetBOT
outer warren
take limit to infinities
finite hatch
take limit to infinities
whats that mean
is it because x is -1/2
so then we need to force a to be 2
outer warren
wdym x is -1/2
outer warren
is very different from saying x is -1/2
also irrelevant for the horizontal asymptote
outer warren
overcomplicating
finite hatch
how do i do it
outer warren
have you worked with limits before?
finite hatch
umm once
like
i was doing calcoolus
and i was doing f(x+h) - f(x)/h
with lim h -> 0
outer warren
ok.
you want to evaluate
$$\lim_{x \to \infty} \frac{ax+b}{x+c}$$
dividing numerator and denominator by $x$ will help
twin meteorBOT
ℝαμOmeganato5
outer warren
since the degree of the numerator and denomiantor is the same
this will give you the horizontal asymptote
finite hatch
wait
cant i just do 2 = a/1
because theyve given me the horizontal asymptote
OMG
wait i meant a not horizontal
asuymptote
outer warren
if you're allowed to apply pre-established properties,
where you consider the leading coefficients, yes
outer warren
taking limits like above is what justifies it
vocal sleetBOT
vocal sleetBOT
Available help channel!
violet bloom
hi, can someone double check my work?
vocal sleetBOT
hi, can someone double check my work?
bitter pilot
looks good
vocal sleetBOT
vocal sleetBOT
Available help channel!
livid void
Looking for V_ab.
the orange is how I found emf_1
the blue is emf_2
Based on what Ive asked other's I basically am only being taught the calculus part, so just a heads up on that if you explain it on theory level.
Had someone help me with those so got a decent understanding of that now though
livid void
also realizing, do I need to adjust emf_1 at all?
is V_ba to the middle junction right to left?
guessing it includes the top junction cuz of kirchhoffs junctions
Or does it go down from b then split at a? 
<@&286206848099549185> huh, still not getting it i guess
vocal sleetBOT
@livid void Has your question been resolved?
livid void
So im sure I understand then
we just follow the outer loop from b, the center junction isn't included?
hidden gyro
You can use any path
But you just have wrong sign
They are saying Vb - Va
Vb is lower than Va
So sign is negative
vocal sleetBOT
vocal sleetBOT
Available help channel!
hard hill
hi, I am given a Linear programming problem of similar fashion:
find $\beta$ so that $f(x,y) = x+\beta y \rightarrow \text{max} $ has a) one solution, b) infinite sols, c) no sols
the following constraints are given:
$x-y\leq 1$
$x+2y \geq 4$
$x\geq 0$
$y\geq 0$
now the graph shows an unbounded feasible region. how am i supposed to find a beta here? any line we can move it arbitrarily upwards, having this infinitely large optimal solution that maximizes our problem.
twin meteorBOT
sup sup
vocal sleetBOT
@hard hill Has your question been resolved?
wraith python
hi, I am given a Linear programming problem of similar fashion:
find $\beta$ so...
You're answers will be in terms of beta.
vocal sleetBOT
@hard hill Has your question been resolved?
white mesa
I need help with the following problem:
vocal sleetBOT
I need help with the following problem:
white mesa
Prove that every positive integer $n$ has a unique factorial expansion, meaning there exist unique integers $k > 0$ and $f_i$ such that $0 \leq f_i \leq i$ for each $i = 0,1,\dots,k-1$, with $f_k > 0$, and
$n = 1! f_1 + 2! f_2 + \dots + k! f_k$
Given the factorial expansion $(f_1, f_2, f_3, \dots, f_k)$ of
$N = -8! + 16! - 24! + 32! - 40! + 48! - \dots + 2000! - 2008! + 2016!$,
determine the value of
$f_1 - f_2 + f_3 - f_4 + \dots + (-1)^{k-1} f_k.$
twin meteorBOT
WD_Nessuno
vocal sleetBOT
@white mesa Has your question been resolved?
vocal sleetBOT
@white mesa Has your question been resolved?
vocal sleetBOT
vocal sleetBOT
Available help channel!
compact eagle
hello
compact eagle
need some help here
so following this along i get confused at this part
i know that delta = e has to be independent of any x values
but thi get conf
i get confused at the part where they say delta <_ 1
it sounds like they just decided to use it randomly
but how is delta <_ 1 true?
|x+1| < delta <_ 1?
cobalt crypt
well delta = min{1, eps/5}
compact eagle
how is 1 going to be bigger than or equal to |x+1| what if x is like, 10 or something who knows
cobalt crypt
so delta <= 1
compact eagle
im terrible at this
i was crashing out over it last night too
can u go over it step by step
cuz like
idk what i did was use the proof first
silk osprey
do you know what your goal is
cobalt crypt
my steps are gonna be the same as in your picture
compact eagle
that |f(x)-L| < e
silk osprey
solve for |x + 1| essentially
compact eagle
well i assume my goal is to make |f(x)-L|<e equal to |x-a|<delta to like
prove they are equivalent or whatevver
(x^2-2x+3) - 6
x^2-2x-3
factored its |(x+1)(x-3)|<e
but
i dont know where to go after that
silk osprey
what’s your favorite number
compact eagle
7
silk osprey
ok then you can do as they did where you take delta = min(7, some shit in terms of epsilon) where you get the expression in terms of epsilon from |x+1| < epsilon/|x-3|
compact eagle
wha
i dont get it i just choose a random ass number and then what
im just allowed to say this magical number is greater than delta
?
but if delta is greater than x-a wouldnt i be like
blocking off a lot of values
idk how it works
cobalt crypt
you get to choose which values of x to inspect
your job is to find a bound on how far x can get from -1 so that f(x) stays within eps of the limit
so you can just arbitrarily say i want x to be within 69 of -1
cuz why not
compact eagle
mhm
but if i want x to be within 69 of -1
well
how owuld i use that to find the epsilon
cobalt crypt
well now -70 < x < 68 right
compact eagle
and at minium 65
cobalt crypt
no
compact eagle
oh
cobalt crypt
at minimum 0
compact eagle
oh right
cobalt crypt
we don't care though
compact eagle
idk what they rlly mean about the delta = min(ab)
cobalt crypt
|(x+1)(x-3)| = |x+1||x-3| is now at most 73|x+1|
so if you want that to be less than epsilon, you need |x+1| to be < eps/73
compact eagle
why is it necessary for that
like just to understand the concept
why do we need to divide
why cant we just say that epsilon is definitely less than 73|x+1|
cobalt crypt
well how do you know
silk osprey
well if delta <= 69 then since |x + 1| < delta you have |x+1| < 69 so
|x-3| <= |x+1| + 4 < 69 + 4 = 73
compact eagle
well i dont know i dont rlly know anything but how do i know thats |x+1|<eps/73
cobalt crypt
well you know you can control how far x is from -1
that is to say you have control over how big |x+1| can get
compact eagle
well if delta <= 69 then since |x + 1| < delta you have |x+1| < 69 so
|x-3| <= ...
|x-3| <= |x+1| + 4 < 69 + 4 = 73
where do u get this
cobalt crypt
so you can declare that |x+1| < eps/73
silk osprey
|x-3| <= |x+1| + 4 < 69 + 4 = 73
triangle
cobalt crypt
that achieves your goal
cobalt crypt
|x-3| <= |x+1| + 4 < 69 + 4 = 73
it's just a rephrasing of what we did
cobalt crypt
so |x-3| is at most 73
right here
compact eagle
unfortunately i dont really understand what we realy did uh
silk osprey
|x - 3| = |(x + 1) - 4|
cobalt crypt
unfortunately i dont really understand what we realy did uh
do you understand that we found that |x-3| is at most 73?
silk osprey
well now -70 < x < 68 right
you can do this as well
cobalt crypt
no we chose 69 as how far x can be from -1
compact eagle
ye so it can be both 69 and -69
so plugging in that
-69-3
wait isnt that -72 then
cobalt crypt
no we're looking at how far x can be from 3 now
if x is at most 69 away from -1, then x is at most 73 away from 3
compact eagle
hlold
x is at most 69 away from the x-value that corresponds to the limit right
cause we chose it to be
cobalt crypt
yes
compact eagle
and per the question
as x > -1
so -1 is that x value
with the limit being 6
isnt 69 basically our delta
cobalt crypt
not yet
cobalt crypt
we're setting up to choose delta
compact eagle
ok so 69 is like a delta but without any limitations and not considering epsilon yet
compact eagle
okay
so our main x value is -1
and delta is going to be 69 or less away from it
on both sides
so it can be -1 -69
and -1+69
68
right? im not sure where 73 comes from\
cobalt crypt
well you want to look at |f(x) - L| = |(x+1)(x-3)|
so you want to know how far x can be from 3
compact eagle
okay so we use the value of -1 which is like the main x value and the maximum bounds produced by it (which is -70 and 69) and then essentially use that bound with (x+1) and (x-3) to realize potentially how far delta can be with those values
cobalt crypt
yes
compact eagle
so like
x+1 for example
or i mean its |x+1|
-70<x+1<69
subtract 1 from every side
-71<x<68
compact eagle
why
cobalt crypt
because we chose delta <= 69
compact eagle
oh
cobalt crypt
delta is how far x can be from -1
compact eagle
oh i see so the -70 and the 69 we kinda put on the side burner its just there to see what the hard limits are
cobalt crypt
that sounds vague enough to not be wrong
compact eagle
what if the values like |x+1| |x+3| cross the limits
does it just mean we ignore everything past it
compact eagle
idk its so hard to explain
whatever lets
go with it
sp
so x+1
69<x+1<69
-1
or sorry
-69
-70<x<68
so for x+1, -70 and 68 is the maximum distance possible from..
from what
cobalt crypt
x is at most 70 distance away from 0
compact eagle
why? couldnt x be like, 265 for example
cobalt crypt
no
compact eagle
oh right
cuz
ye
im dummy
<x<
so x is between -70 and 68
okay
so we plug in the same shit for x-3 then?
-69<x-3<69
cobalt crypt
no
you don't really care about what the bounds on x are
you care about |(x+1)(x-3)| = |x+1||x-3|
so we need bounds on |x+1| and |x-3|
cobalt crypt
well yes
but you decided to figure out the bounds on x
which we don't need
we need bounds on |x-3|
cobalt crypt
no
silk osprey
isnt that what were finding
you choose the bound on |x + 1| from your choice of delta then with that bound you can bound |x - 3|
cobalt crypt
we start with -69 < x+1 < 69
|x+1| is the only thing we have direct control over
by setting delta
everything else must be deduced
compact eagle
well i dont have a delta yet though right
cobalt crypt
but we declared delta <= 69
silk osprey
you pick your favorite number
compact eagle
so delta is less than or equal to 69
cobalt crypt
but we declared delta <= 69
this is how we conclude that -69 < x+1 < 69
silk osprey
so delta is less than or equal to 69
yes but you’re not done
compact eagle
|x+1|<delta<=69
cobalt crypt
sure
compact eagle
|x+1|<=69
silk osprey
<
compact eagle
why
silk osprey
because |x + 1| < delta
compact eagle
but delta<=69
silk osprey
right so |x + 1| < 69
compact eagle
x+1 is less than delta which is less than or equal to 69
silk osprey
69 < 420 <= 420 —> 69 < 420
compact eagle
uhh
cobalt crypt
|x+1| can't be equal to 69
|x+1| is always strictly less than delta
the most delta can be is 69
compact eagle
oh right
right
|x+1|<69, so 69 has to be the bigger value meaning x in this equation can be at most like
less than 68
silk osprey
we need bounds on |x-3|
^
compact eagle
why 69
compact eagle
i mean like if u plug in a x value tho
cobalt crypt
okay sure
compact eagle
idk
cobalt crypt
if you mean x then sure
compact eagle
yes
cobalt crypt
but again we don't care about x
compact eagle
okay
cobalt crypt
we care about |x-3|
compact eagle
why do we care about that
cobalt crypt
so we need bounds on |x+1| and |x-3|
.
silk osprey
so you can get |x+1| < some expression of epsilon
cobalt crypt
you care about both |x+1| and |x-3|
silk osprey
then from there reverse to get f —> 6
cobalt crypt
they appear in the expression you're trying to bound
how could you not care about them
compact eagle
i feel like im missing some sort of basic piece of this puzzle to just understand
the textbook didnt explain any of this shit it just threw random examples at me
fuck
compact eagle
true but that doesnt necessarily make it not confusing
okay
we have
|x+1||x-3|<e
cobalt crypt
compact eagle
and |x+1|<delta
in the simplest of terms
when u see those 2 values
what does it mean that needs to be done
cobalt crypt
you need to make both |x+1| and |x-3| small
and you do it by controlling how big |x+1| can be
compact eagle
epsilon has gotta be more than |x+1||x-3| and delta has to be more than |x+1|
cobalt crypt
sure
compact eagle
delta and epsilon are like dependent on each other right
i mean ig of course they are
cobalt crypt
by the definition of the limit
silk osprey
why is epsilon fixed and delta isnt
do you remember the statement of convergence
cobalt crypt
epsilon is the challenge your enemy gives you
compact eagle
no
cobalt crypt
you need to satisfy the challenge with delta
compact eagle
so im kinda looking for a delta value that satisfies the boundaries of epsilon
cobalt crypt
yes
your enemy is like: i challenge you to make |f(x) - L| < epsilon
and your response is: okay i choose |x - a| < delta for that to happen
silk osprey
"for any epsilon there exists some delta" so you let epsilon be some arbitrary positive number then based on that epsilon you need to find a delta so that if x is within a distance delta from your limit point then you know the value of your function is within a distance epsilon from the claimed limit
compact eagle
the output of the function minus the limit
is less than epsilon
epsilon being the total units of distance possible for the output of the function to extend from the limit
compact eagle
thanks :(
your enemy is like: i challenge you to make |f(x) - L| < epsilon
and your response is: okay i choose |x - a| < delta for that to happen
silk osprey
"for any epsilon there exists some delta" so you let epsilon be some arbitrary p...
look up a visual or something
compact eagle
so we need the distances away from the limit to be less than epsilon
epsilon being an unknown number basically
silk osprey
your enemy is like: i challenge you to make |f(x) - L| < epsilon
and your respon...
yep
cobalt crypt
yes your enemy is evil, so they can give you any epsilon > 0
you must be ready for all
compact eagle
so because epsilon can whatever the fuck they want i have to somehow find a value of delta, which is the maximum distance from the input of the function that results in the output of the limit, that meets the bounds of epsilon
cobalt crypt
yes
compact eagle
and i can do that by proving that
|x-a|<delta and |fx-l|<epsilon
well
its more like
|fx-l|<delta no?
or well
i guess if i had to visualize it
its like |x-a|<delta all just based on the x axis
cobalt crypt
no you must prove that
- if |x-a|<delta
- then |f(x) - L| < eps
compact eagle
and it has to kinda match the values on the y axis which is |fx-l|<epsilon
cuz a and L are connected
input a, you get L
so if the distance from a is less than some delta value, we know that the distance from L is less than some epsilon value
and we basically need to find a delta value that meets the absolute bound that will make the statement true
cobalt crypt
yes
cobalt crypt
then you're done
cobalt crypt
you choose delta
compact eagle
i pick a delta value and then i have to
find out where in that chosen delta value it will meet the boundaries for epsilon?
if |x+1|<delta
then |(x+1)(x-3)| < eps
whatever the fuck x+1 is it has to be less than delta
and whatever the fuck x+1*x-3 is has to be less than epsilon
cobalt crypt
yes
compact eagle
because we dont know whatever the fuck epsilon is
we pick a delta
why can we pick any delta we want
how can we use any picked delta to find the real delta
cobalt crypt
well it doesn't really matter ultimately
you just have to find one that works
your enemy is out to get you, you'll do anything you can
compact eagle
so i have to just assume a random delta because i have no clue what epsilon could be and because i have no clue what epsilon could be really in a way any delta value could work
potentially anyway
any delta value could work
cobalt crypt
well the idea is like
you start with a crude estimate
until you get a better idea of what's going on
so that you can make a better estimate
it's not that you must start with the crude estimate
if you were all knowing you could start with an estimate that works immediately
compact eagle
so the delta i start with isnt rlly gonna be the delta i end with
compact eagle
we use the starting delta to fine tune it down to a delta that will
make it match the epsilon
cobalt crypt
yes
compact eagle
imma write it down 1 min
righto
so basically i start with a delta, find out the bounds of that delta given a,
and then
idk magic
so with this problem
my a value is -1
ill do a crude estimate of delta being 2
so...
|x+1| has to be less than or equal to 2
right?
i mean
sorry
compact eagle
why cant it equal 2
cobalt crypt
it doesn't matter
cobalt crypt
whether you need the = sign is the least of your concerns
compact eagle
im saying delta is at MOST 2 right?
cobalt crypt
yeah
cobalt crypt
-3<x<1
cobalt crypt
no
cobalt crypt
a is -1 here you're not doing anything with it
compact eagle
x is the real delta?
compact eagle
so any value between -3 and 1 works for my imaginary possible delta
how do we use these values now
compact eagle
so what is -3<x<1 telling us
compact eagle
wait
-3<x<1, x is between -3 and 1
if |x+1|<delta
then |(x+1)(x-3)| < eps
so
-3<x+1<1 and -3<x-3<1
?
why do we skip x+1
oh
we skip it cuz we already did it
right
compact eagle
so using my imaginary delta of 2 for -1
cobalt crypt
yes
compact eagle
-3 and 1
cobalt crypt
that's for x
compact eagle
so thats for x alone but now we wanna figure out the boundaries of this imaginary delta at x-3
cobalt crypt
weird wording but okay
cobalt crypt
you want to bound |x-3|
compact eagle
what does bound mean
cobalt crypt
a bound for something is how large it can get
you say |x-3| is bounded by B if |x-3| <= B
compact eagle
B in our case is 2
compact eagle
u mean x+1 right
cobalt crypt
yes
compact eagle
i control x+1 cuz the proof says it has to be less than delta, and right now i say that delta has to be less than or equal to 2
saying that the delta has to be less than or equal to 2 means that the x value can only be between -1 and 3
or uh
i forgot what it was
cobalt crypt
-3 and 1
compact eagle
-3 and 1
right
well now we know that x can only be between -3 and 1
so would it mean that we use that?
-3<x-3<1
?
compact eagle
why does -3<x-3<1 not work (as in i add 3 to all sides)
oh
well i guess then it breaks our definition
0<x<2
compact eagle
so from -2<x+1<2 i now know its -3 and 1
for the boundaries of x alone
and now to find it for x-3
i subtract 3 in -3<x<1
cobalt crypt
yes
compact eagle
ig the reason i thought we could use -3 < x-3 < 1 is for some reason my brain is thinking that like
the -3 and 1 are hard points
i dont know
cobalt crypt
you're pattern matching on the wrong things
cobalt crypt
knowns: delta <= 2
|x+1| < 2
-2 < x+1 < 2
-3 < x < 1
you have to produce the inequality for x-3 using things which you know
compact eagle
so for |(x+1)(x-3)|<e, we already know that x+1 is definitely between -2 and 2
compact eagle
wait
-3<x<1
subtract 3 on everything
right
-6<x-3<-2
wait
why is x-3 less than 6
isnt it more?
|x-3| is at least more than -6 and less than -2
ok no im definitely confusing something
cuz absolute value
cobalt crypt
yeah
compact eagle
what do i do after determing -6<x-3<-2
do we just make it all absolute value
|-6<x-3<-2|
idk aa
i suck with the absolute value part of it idk how to plug it in
cobalt crypt
like
how far can x-3 be from the origin
well it's gonna be a distance of between 2 to 6
so 2 < |x-3| < 6
compact eagle
ok right x-3 can be a maximum of 6 away or 2 away
compact eagle
|x-3|<6
cobalt crypt
so in |(x+1)(x-3)|, by choosing delta <= 2, we've taken back control over both terms
we know |x+1| < 2 and |x-3| < 6
compact eagle
alrigjt
we know that 6 is the maximum distance achievable for x-3, and 2 is the maximum distance away achievable for x+1
cobalt crypt
so now you need to keep control over |x-3|, and make |x+1| very small by tuning delta
compact eagle
how do we tune delta? wdym?
cobalt crypt
so at this point, we can conclude that
|(x+1)(x-3)| = |x+1||x-3| < 6|x+1|
our goal is to make all of that < epsilon
compact eagle
why do we multiply the 6 by x+1
cobalt crypt
|x+1||x-3|
|x+1|6
compact eagle
so because |x-3| is definitely going to be less than 6 as per our current estimate for delta,
cobalt crypt
yes
so you will certainly not multiply |x+1| by anything more than 6
xo |x+1||x-3| is bounded by 6|x+1|
compact eagle
we start with the biggest possible value which is going to be 6
cobalt crypt
yes
compact eagle
and cuz |x+1| and |x-3| are multiplied together, we can basically just say that |x+1| is being multiplied by 6 (assuming |x-3| is the largest possible number)
cobalt crypt
thats not a good way to phrase it
compact eagle
how do u phrase
cobalt crypt
you say that |x+1||x-3| is bounded by 6|x+1|
or that |x+1||x-3| is less than 6|x+1|
compact eagle
ok to phrase it better we are saying that |x+1| can be multipled by at MOST 6 for it to be < epsilon
cobalt crypt
the end there is also not good phrasing
we are trying to make |x+1||x-3| < epsilon
we can go through an intermediate stage to achieve that goal
if we can show that 6|x+1| < epsilon, then certainly |x+1||x-1| < 6|x+1| < epsilon
compact eagle
|x+1||x-3|< epsilon, and by our findings, |x-3| is at most going to be 6, so for the sake of narrowing it all down we will say that |x-3| IS 6 so we can find the....
compact eagle
but were multiplying |x+1| by 6 right?
cobalt crypt
yes
which is an upper bound for |x-3||x+1|
you are not setting |x-3| = 6
you are saying that |x-3||x+1| < 6|x+1|
compact eagle
oh right cuz 6 is the maximum value were basically saying that whatever |x+1||x-3| ACTUALLY is is definitely going to be samller than 6|x+1|
cobalt crypt
yes
compact eagle
and 6|x+1| is also smaller than epsilon
cobalt crypt
okay so
we've shifted our goal
it used to be proving that |(x+1)(x-3)| < epsilon
but now, we've said that when delta <= 2, we also have |x-3| < 6
so its enough to prove that (|(x+1)(x-3)| <) 6|x+1| < epsilon
compact eagle
when delta is less than or equal to 2
cobalt crypt
that is what we chose
compact eagle
(|(x+1)(x-3)| <) 6|x+1| < epsilon
cobalt crypt
the () is just there to remind you
its not part of the goal
we only need to show 6|x+1| < epsilon now
compact eagle
okay so our goal is to prove that 6|x+1|<epsiloin
we can do that by dividing by 6?
cause then that forms |x+1|<epsilon/6, and we know that |x+1|<delta
compact eagle
lemme just get it fully down in our brain
cobalt crypt
there is one hiccup
which is that we also required delta <= 2 for us to make the claim that |x+1||x-3| < 6|x+1|
compact eagle
|x+1|<delta, meaning that if |x+1|<epsilon, epsilon and delta are equivalent
cobalt crypt
no thats bad phrasing
we are required to show that |x+1| < epsilon/6
we can choose a delta such that |x+1| < delta
choosing delta = epsilon/6 completes the goal
compact eagle
so if we now choose epsilon/6 to be our delta to put in the place of |x+1|<delta
it just looks like |x+1|<e/6
cobalt crypt
yes
compact eagle
so does this mean we proved it
cobalt crypt
no
cobalt crypt
there is one hiccup
.
we did all the work previously to show that |x-3| < 6
this only works under the condition that delta <= 2
compact eagle
but now delta is epsilon/6
cobalt crypt
so if your enemy gave you epsilon = 1000000
compact eagle
so does this mean we need to prove epsilon/6 <=2
compact eagle
oh right
cobalt crypt
but we have control over delta
notice that we don't need to choose delta = epsilon/6 for our goal to be achieved
any delta < epsilon/6 would also work
we would just be overachieving
cobalt crypt
ye
precisely speaking, it should be delta should be at most epsilon/6
delta <= epsilon/6
compact eagle
|x+1|<delta
cobalt crypt
|x+1| < delta <= epsilon/6 -> |x+1| < epsilon/6