since (a-b)^2 = a^2 - 2ab
#help-10
shadow tangle
no, we did (a+b)^2, which equals a^2**+**2ab+b^2
shadow tangle
we let b=-1, a-b=a+(-b)
viral tree
so 2 - 1 is equal to 2 + ( -1 ) ?
shadow tangle
yep
shadow tangle
yea
shadow tangle
its all good, its partly my fault for using (a+b)^2 instead of (a-b)^2
viral tree
I could have done this with (a-b)^2 as well?
shadow tangle
yep
shadow tangle
let a=sqrt(x+2) and b=1
shadow tangle
let a=sqrt(x+2) and b=1
if were using (a-b)^2
shadow tangle
i mean you can use either one, it just comes down to what values you make a and b
viral tree
so I'll do it as (a-b)^2 now again
shadow tangle
ok, good luck 👍
viral tree
thanks
viral tree
me too
shadow tangle
ok, see ya :)
patent magnet
For the following surface-like objects:
Sketch the object, clearly marking an origin and the +z-direction,
Determine an appropriate coordinate system that could describe the area,
Determine which coordinate(s) could correspond to divergence,
Determine which coordinate(s) could correspond to curl
obtuse pebbleBOT
For the following surface-like objects:
Sketch the object, clearly marking an o...
patent magnet
I'm unsure on what coordinate system to even start this one off with. I took a look around at conical systems perhaps, but that doesn't seem right...Or maybe I'm dumb and it is. Who knows
brave bramble
This is a 2D problem so you only have two real choices
Rectangular vs polar
Polar is the right choice here
patent magnet
Really? There was another one before that was a flat sheet of paper, but I still ended up making it 3D w/ z-axis pointing off the x-y plane
Eh, I'll just run it as 2D and see what happens
.close
obtuse pebbleBOT
winged field
hi
obtuse pebbleBOT
Available help channel!
winged field
i need some help to understand how to factor this: (xy-1)(x-1)(y+1)-xy
obtuse pebbleBOT
i need some help to understand how to factor this: (xy-1)(x-1)(y+1)-xy
slate hamlet
Have you tried?
gaunt walrus
I would prefer to expand it if I were doing the question
winged field
well. i actually have an answer book with th answer, but i1m not compeletely sure on how they get there
gaunt walrus
Send it
slate hamlet
well. i actually have an answer book with th answer, but i1m not compeletely sur...
Did you try solving it on your own?
rich plume
the answer is (xy + z -1) (xy -y -1)
where did 'z' come from?
winged field
sorry, that was a typo, that should be an X
slate hamlet
yep, i tried answering on my own
Can you show your working?
winged field
i got up to (xy-1)/2 + (xy-1) (x-y) -xy
winged field
. reopen
obtuse pebbleBOT
Available help channel!
wise lichen
if f(-1) = 1 and f(3) = 2 and f is linear fnction what is f(5)
obtuse pebbleBOT
if f(-1) = 1 and f(3) = 2 and f is linear fnction what is f(5)
wise lichen
is it just 1/4(5)
fiery shell
it looks like you've calculated the slope, but you're also gonna need the y intercept
hollow dove
you can just use linearity of f
f(a+b) = f(a) + f(b); you want f(5), so try writing 5 as a sum of multiples of -1 and 3
sage geode
I don't think by linear function here means f(x + y) = f(x) + f(y) and f(cx) = cf(x)
fiery shell
that's pretty cool
fiery shell
wait yeah
hollow dove
oh I'm misreading, yeah
wise lichen
i am so lost
hollow dove
my bad
sage geode
Yeah it's f(x) = ax + b
hollow dove
thought we were doing linear algebra here
sage geode
One of the least favourite ambiguities
viral blade
What's your favorite ambiguity?
fiery shell
i am so lost
remember y = mx+b
wise lichen
whats a and b tho is tha tjust -1 and 3
fiery shell
if you can solve for m and b you can solve for any input
wise lichen
remember y = mx+b
right irght
hollow dove
so f(-1) = 1 and f(3) = 2 gives you 2 points, (-1,1) and (3,2), on the graph of f
you have the slope, so find the line with that slope that passes through those points
wise lichen
so if tis slope its 1=(1/4)-1+b?
viral blade
not quite, how did you get that?
fiery shell
so if tis slope its 1=(1/4)-1+b?
x is -1, not 1
wise lichen
x is -1, not 1
mb
wise lichen
not quite, how did you get that?
using slope formula y=mx+b
fiery shell
so if tis slope its 1=(1/4)-1+b?
this is still wrong (or written confusingly). it's m*b so it should be (1/4)*-1 but it looks like you've written (1/4) - 1
wise lichen
this is still wrong (or written confusingly). it's m*b so it should be (1/4)\*-1...
right right
either way am i supposed to be finding b
fiery shell
yeah
wise lichen
this is still wrong (or written confusingly). it's m*b so it should be (1/4)\*-1...
ty tho
fiery shell
if you know m and b, and set x to 5, you can find y at that point
which is the original question
wise lichen
so f is representing y here right
fiery shell
yes, we're using y instead of f(x)
fiery shell
👍
wise lichen
is it 7/4
fiery shell
that doesn't look right, can you show your work?
wise lichen
wise lichen
i was solving for b there
fiery shell
you wrote 1 = -1/4 + 3/4, which is false
fiery shell
yeah, 5/4 works
wise lichen
5/2 = f(5)
fiery shell
looks good
wise lichen
WOOOOHOOO
fiery shell
:)
wise lichen
b means y intercept right
obtuse pebbleBOT
@wise lichen Has your question been resolved?
grave condor
obtuse pebbleBOT
grave condor
bearings
grave condor
- havent started because i dont know how to
slate hamlet
Bearing as in angle from the ground?
slate hamlet
Try drawing a triangle
slate hamlet
Lets treat N E as y axis and WS as x axis
So 63 degrees would mean 63 degrees from the a axis?
grave condor
wait what
slate hamlet
and 153 would mean 153 from the x axis right?
grave condor
whys ne that
slate hamlet
Is this plane A's path?
grave condor
yeah
slate hamlet
There fore this is B?
grave condor
why is that b
slate hamlet
What would B be then?
slate hamlet
Can you draw A and B's path with the angle
grave condor
do i draw different things
slate hamlet
In the same diagram
grave condor
okay
wait
im gonna revisit this topic \
because i dont think i understand the basics
thankyou for your time
.close
hushed crest
Can someone help me with understanding bearings?
obtuse pebbleBOT
Can someone help me with understanding bearings?
hushed crest
i dont know
hushed crest
new century advanced year 9 math textbook
polar fossil
bearings are just angles measured clockwise where north is 0
hushed crest
oh ok
so if i were to find the bearing of rocky to mulga
would i do 360-320 to find the missing angle on the mulga
and use it like alternating angles?
polar fossil
alternating angles sorta? but you can see that the angle should be obtuse
hushed crest
oh right
polar fossil
think about supplementary angles as well
hushed crest
so what would i do?
gaunt walrus
Can someone help me with understanding bearings?
the keyword from
the concept is turning
hushed crest
ohh
nah i think i understand it now
to get mulga from rocky i use something like supoplementary angles
just find that angle
360-320
i got 40 and then 180-40
i understand it now, thank you
hushed crest
.close
obtuse pebbleBOT
gaunt walrus
bruh, lol
hushed crest
thanks
obtuse pebbleBOT
Available help channel!
frosty river
I'm struggling to understand the following simple geometric proof.
Problem: Let ABCD be a 6cm side rigid square. Let A be raised rigidly by 1 cm, C be raised rigidly by 2cm and let B be fixed. Calculate the height of D after this rigid transformation. The answer is 3cm.
Proof: Since the midpoint of AC and the midpoint of BD are equal (the center of the square) and the height of B is 0cm, then the height of D must be the sum of the height of A and the height of C, i.e., 1cm+2cm=3cm.
obtuse pebbleBOT
I'm struggling to understand the following simple geometric proof.
Problem: Let...
frosty river
The midpoint of AC and BD being the center of the square is trivial. I don't understand the conclusion.
obtuse pebbleBOT
@frosty river Has your question been resolved?
obtuse pebbleBOT
@frosty river Has your question been resolved?
tame dock
the thing is in the question it meant to say that for the centers to remain the same how much should the height of d be increased in a square the intersection of diagonal is same as the center of the square
frosty river
What?
frosty river
<@&286206848099549185>
obtuse pebbleBOT
@frosty river Has your question been resolved?
frosty river
<@&286206848099549185>
timid silo
have u tried to draw the thing?
frosty river
Yes, I have a figure and the solution, but I can't understand it.
I can't see why the midpoints equal implies the height is the sum.
frosty river
<@&286206848099549185>
timid silo
<@&286206848099549185>
help yourself
timid silo
I can't see why the midpoints equal implies the height is the sum.
break the steps calculation. And explain why
it will help
frosty river
<@&286206848099549185>
frosty river
I'm struggling to understand the following simple geometric proof.
Problem: Let ABCD be a 6cm side rigid square. Let A be raised rigidly by 1 cm, C be raised rigidly by 2cm and let B be fixed. Calculate the height of D after this rigid transformation. The answer is 3cm.
Proof: Since the midpoint of AC and the midpoint of BD are equal (the center of the square) and the height of B is 0cm, then the height of D must be the sum of the height of A and the height of C, i.e., 1cm+2cm=3cm.
The midpoint of AC and BD being the center of the square is trivial. I don't understand the conclusion.
trim portal
Stop advertising
frosty river
<@&286206848099549185>
frosty river
<@&286206848099549185>
civic zealot
I'm struggling to understand the following simple geometric proof.
Problem: Let...
the midpoint of A and B would have height 3/2, but that's half the distance from B to D. Since B is at 0, it raises 3/2 half the distance, and then another 3/2. So the height is 3.
frosty river
frosty river
the midpoint of A and B would have height 3/2, but that's half the distance from...
Can I marry you?
THANK YOU! 🫂
civic zealot
np
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
wanton carbon
Hi, so in the diagram we are given, O is the center of the largest circle shown, and that congruent circles A, B, and C are internally tangent to circle O, at R, Q, and P, respectively. I have figured out most of the problem, but I don’t know how to show that ray RA passes through O. Can someone help?
obtuse pebbleBOT
Hi, so in the diagram we are given, O is the center of the largest circle shown,...
oak hatch
Hi, so in the diagram we are given, O is the center of the largest circle shown,...
well the tangent at the point R is perpendicular to A, but that same "line" is also the tangent of the bigger circle, so it is also perpendicular to O
hence it passes through A if you connect R-O
oak hatch
.close
rich marsh
obtuse pebbleBOT
rich marsh
<@&286206848099549185>
pearl kestrel
working on it
rich marsh
I don't know if I took it in the wrong direction..
pulsar quarry
could you explain me what have you done in 3rd step?
you firsst have to take the lcm of 1+x+1/y
and then
reciprocate it
same for all three terms
pulsar quarry
you've done it wrong
you cannot
distribute
the negative power
inside the bracket
rich marsh
ohk! let me try again..
rich marsh
but isn't it -1? i believe we can do it with 1 and -1?
pulsar quarry
no
if its written (x)^-1 then you can write it ass 1/x
but in tthat case
you first have to take lcm of all those 3 terms (inside the bracket one)
reciprocate it
zenith raft
can anyone help me with this? i named the set on the left X_1 and the one on the right X_2 and showing X_1 \subset X_2 was straightforward but X_2 \subset X_1 is giving me trouble
obtuse pebbleBOT
can anyone help me with this? i named the set on the left X_1 and the one on the...
zenith raft
any standard closure definition/equivalent is fair game
frosty river
Take a in X2, a is in every closure and look and how the sets are defined they go from xn in advance each time with one less "xn"
balmy mortar
What does the bar on the right indicate - closure?
zenith raft
What does the bar on the right indicate - closure?
yea
balmy mortar
Take a in X2, a is in every closure and look and how the sets are defined they g...
im sure this will hint you to the answer
balmy mortar
are the accumulation points of the sequence
But i was going to ask if the equality made geometric sense to you, intuitively
zenith raft
i've taken a in X_2 and done some stuff, the straightforward idea does not seem to work
zenith raft
But i was going to ask if the equality made geometric sense to you, intuitively
ehh...
balmy mortar
ehh...
I think it's helpful to
especially since you're stuck now
So for example - do you see why the closure bar is necessary
Without it, what could go wrong
it works with say 0, 1, 0, 1, 0, ...
The set is {0, 1}
zenith raft
well yea that would be a silly set with no closure bar
balmy mortar
When exactly
zenith raft
it would just be the elements of the sequence that are frequently in the sequence
balmy mortar
yes the ones that appear infinitely often
Which does not necessarily include the accumulation points
zenith raft
yes that is what i mean with frequently
balmy mortar
So now, it seems like you need to work the backwards containment
by using the definition of closure
How exactly are those additional points defined
And show they must be contained in the left
===
The above were my kindof thought processes to convince myself of the statement geometrically
balmy mortar
yes the ones that appear infinitely often
It should be straightforward to show these particular elements appear in the left set
and then you think about the 'extra' ones from the closure
=================================
(nvm, pic i sent was wrong)
zenith raft
that thing you posted being wrong is where the difficulty comes in
balmy mortar
Well this is wrong in general (with random sets)
but im fairly certain it happens to be true here
I think its only true under certain conditions
like a decreasing sequence of sets or something
zenith raft
balmy mortar
X is discrete 
balmy mortar
nvm yeah this is nonsense even with decreasing sequence of sets 
mb
take {1/n}, that intersection will be empty before you take the closure.
balmy mortar
and then you think about the 'extra' ones from the closure
cool, but everything I said before should hopefully be helpful for the direction to go
Take the definition of closure, then take that intersection on all the closures
I guess there are multiple defns with closure you can work with - the one involving sequences will be of most help
hidden compass
whats 3x= x-2
Post the question on another channel among the available ones, since this channel is already occupied, please
gusty token
yes sir
obtuse pebbleBOT
@zenith raft Has your question been resolved?
balmy mortar
if ur still stuck, can show where, idk exactly which defn of closure ur trying
zenith raft
this is how i envision the proof starting
but it is not so obvious to me how to make the subsequence i want
viral blade
if we fix n, we get that for all ε>0 there's an element with index >n less than ε away from a, right
viral blade
An arbitrary one in X_2
zenith raft
why should we care about an arbitrary sequence in X_2?
viral blade
don't we want to show it's in X_1?
zenith raft
you aren't making sense to me 
viral blade
We want to show X_2 is a subset of X_1 right
zenith raft
yes
viral blade
So we can take an arbitrary element of X_2 and show that it's also contained in X_1
zenith raft
don't we want to show it's in X_1?
or wdym "it" here?
viral blade
oh it's limit points, right, my bad
Let a be an element of X2, let x_n be the given sequence
viral blade
if we fix n, we get that for all ε>0 there's an element with index >n less than ...
So this is for the given sequence
zenith raft
yes sure
viral blade
Let's start with n=0, ε=1
zenith raft
i already wrote this, no take backs
viral blade
Some would call that set N
zenith raft
i am not some
viral blade
Let's start with n=0, ε=1
Let k be the resulting index that we get of our element here
Err
viral blade
The first element with index ≥1, less than 1 away from a
zenith raft
go ahead
zenith raft
oh yea n_1 would be better
viral blade
now we look for the first element with index ≥ n_1 + 1, that's less than 1/2 away from a
and call its index n_2
zenith raft
sure
viral blade
then the first element with index ≥ n_2 + 1 less than 1/3 away from a has index n_3
zenith raft
yep, got it
viral blade
And that should give us a subsequence that converges to a
viral blade
it should show X2 is a subset of X1
zenith raft
viral blade
viral blade
ur welcome
ruby vessel
@viral blade is it possible for u to take a look at my question?
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
sage dagger
How can I differentiate ln(3x) and ln(3x^2)
obtuse pebbleBOT
How can I differentiate ln(3x) and ln(3x^2)
fiery shell
does it become 1/3x
how'd you get that?
sage dagger
how'd you get that?
Well ln(x) is 1/x
fiery shell
true
but you can't use 3x isn't x, it's a function of x
so you gotta use the chain rule
wanton hull
,,\frac{d \log y}{dx} = \frac{1}{y} \cdot \frac{dy}{dx}
forest sinew
,,\frac{d \log y}{dx} = \frac{1}{y} \cdot \frac{dy}{dx}
$\dv x$
warm shaleBOT
jan Niku
warm shaleBOT
Bettim
sage dagger
halp
forest sinew
sage dagger
how 2 chain ln(3x)
forest sinew
bettim is tryin to help u
sage dagger
Is 1/3x the derivative
sage dagger
So derivative times derivative?
sage dagger
Where does 1/y come from
wanton hull
What is differentiation of ln y
sage dagger
1/y
wanton hull
That's what I've written
wanton hull
Chain rule
wanton hull
What is the dy/dx in this case
Differentiation of y alone, here y is 3x
So differentiate 3x
forest sinew
water you should review the chain rule
sage dagger
I know how to use chain rule just not on logarithms
Cuz I haven’t done it on logs before
forest sinew
i dont think that necessarily makes sense
wanton hull
It works the exact same way it does for other functions
fiery shell
it's a general rule, if you knew it well enough you could use it on logs
fiery shell
<@&268886789983436800>
hot hazel
forest sinew
lol
fiery shell
👍
forest sinew
yes but I haven’t applied chain on logs before
this is like saying you know addition but then saying you cant calculate 7+5 because you havent added anything to 7 before
no offense
sage dagger
ok but I’ve used chain rule on polynomials and trigs before fine but idk how it works on logs
wanton hull
There isn't any difference
sage dagger
Chronically scared of logs
fiery shell
it's possible that they could only understand how to use it in a few set cases if they don't know the general rule
grizzled shore
How would you do d/dx lnx
warm shaleBOT
jan Niku
forest sinew
then at the end, substitute in your functions
sage dagger
This is how I learn;
I need to be exposed to multiple examples and applications of it in every type of problem and then I can use it perfectly fine after
forest sinew
thats okay im sorry if you felt teamed up on 
didnt mean to put you on the defensive
sage dagger
grizzled shore
$\dv{x}\ln(x)$
warm shaleBOT
Frosst
sage dagger
How would you do d/dx lnx
I just know it’s 1/x by memorising
I can’t derive it myself lol
grizzled shore
What about $\dv{x} \ln(2x)$
warm shaleBOT
Frosst
grizzled shore
Yes
grizzled shore
Well
fiery shell
chain rule must be applied
grizzled shore
Ok we’re just trying to get the idea of what goes on the bottom first
fiery shell
ok sure
grizzled shore
The first bit $\dv{x}\ln(f(x))$
warm shaleBOT
Frosst
grizzled shore
Is that the f(x) goes on the denominator
You should notice that it’s the first part of the chain rule
sage dagger
yes
grizzled shore
$\dv{x}f(g(x))=f’(g(x))g’(x)$
warm shaleBOT
Frosst
grizzled shore
That’s the chain rule and you know it right?
sage dagger
Yeah
warm shaleBOT
Frosst
sage dagger
diff outside times inside times diff of inside
fiery shell
diff outside times inside times diff of inside
mmm
derivative of outside with respect to inside, not times inside
grizzled shore
And ln’ is this part $\frac{1}{g(x)}$
warm shaleBOT
Frosst
balmy mortar
The version of the chain rule you should be learning is with the quotients...
In my opinion.
forest sinew
The version of the chain rule you should be learning is with the quotients...
$\dv{y}{x} = \dv{y}{u} \dv{u}{x}$?
balmy mortar
yes.
warm shaleBOT
jan Niku
sage dagger
1/3x • 3x • 3?
grizzled shore
Actually you can just use log laws
sage dagger
isn’t it 1/3x (outer) • inside (3x) • derivative of inside (3)
fiery shell
yes
grizzled shore
ln(3x) = ln(3) + ln(x)
balmy mortar
this inner outer talk.... =...=
grizzled shore
And ln(3x²) = ln(3) + 2ln(x)
sage dagger
ln(3x) = ln(3) + ln(x)
O yeah
forest sinew
isn’t it 1/3x (outer) • inside (3x) • derivative of inside (3)
youre confusing product i think
fiery shell
ln(3x) = ln(3) + ln(x)
yeah but using the chain rule on logs is probably a good thing to learn
grizzled shore
Yes
Certainly
Just figured I’d mention an easier way if you can’t remember the chain rule stuff in an exam
fiery shell
fair enough
sage dagger
1/3x • 3x • 3 
sage dagger
1/3x • [3x]’ then
sage dagger
3
fiery shell
1/3x • [3x]’ • 3 <:hmmCat:831508801646755850> <:hmmCat:831508801646755850> <:hmm...
not (the derivative of 3x) * 3
1 or the other
but yes you are gonna end up with 1/x
sage dagger
So ln(ax) = 1/x
fiery shell
it would appear so
sage dagger
I mean ln(x)
sage dagger
The derivative is
fiery shell
yes
so that's ln(ax) for a = 1
but it won't be so simple for more complicated functions
like d/dx ln(x+1)
fiery shell
true
sage dagger
Oh it is?
sage dagger
ok I was guessing lol
fiery shell
it is kinda funny
grizzled shore
Totally not illegal
sage dagger
So wrong
grizzled shore
>:(
polar fossil
you'll find that mathematicians don't write the log base a lot of times because mostly it doesn't matter
sage dagger
So differentiating ln(3x^2) = is 1/3x^2 • 6x?
grizzled shore
Yes
fiery shell
bingo
grizzled shore
you'll find that mathematicians don't write the log base a lot of times because ...
But why use log over ln when ln has less characters
I still don’t get that
fiery shell
true
grizzled shore
Write your 1’s better
sage dagger
d/dx[ln(ax)] = 1/x sounds just as illegal as dividing by 0
polar fossil
although i usually write log with a loopy l so
sage dagger
ln looks a little too much like 1n
Write cursive
polar fossil
d/dx[ln(ax)] = 1/x sounds just as illegal as dividing by 0 <:NervousSweat:789206...
get desmos and graph y = ln(ax) and vary a
balmy mortar
This is how I learn;
I need to be exposed to multiple examples and application...
@sage dagger example of how id lay out the chain rule. Different from what you've probably seen so far
Let $y = \ln\left(\sin(x^2)\right)$.
Then
\begin{align*}
\dv{y}{x} &= \dv{\left(\ln\left(\sin(x^2)\right)\right)}{x}\\\\
&=\dv{\left(\sin(x^2)\right)}{\left(\sin(x^2)\right)}\dv{\left(\ln\left(\sin(x^2)\right)\right)}{x}\\
&&&\text{(chain rule)}\\
&=\dv{\left(\sin(x^2)\right)}{x}\dv{\left(\ln\left(\sin(x^2)\right)\right)}{\left(\sin(x^2)\right)}\\
&&&\text{(let $u=\sin(x^2)$)}\\
&=\dv{\left(\sin(x^2)\right)}{x}\dv{\left(\ln u\right)}{u}\\\\
&=\dv{\left(\sin(x^2)\right)}{x}u^{-1}\\\\
&=\dv{(x^2)}{(x^2)}\dv{\left(\sin(x^2)\right)}{x}u^{-1}\\
&&&\text{(chain rule)}\\
&=\dv{(x^2)}{x}\dv{\left(\sin(x^2)\right)}{(x^2)}u^{-1}\\\\
&=2x\dv{\left(\sin(x^2)\right)}{(x^2)}u^{-1}\\
&&&\text{(let $w=x^2$)}\\
&=2x\dv{\left(\sin w\right)}{w}u^{-1}\\\\
&=2x\cos(w)u^{-1}\\\\
&=2x\cos(x^2)(\sin(x^2))^{-1}\\\\
&=2x\cot(x^2)
\end{align*}
sage dagger
Oh my
obtuse pebbleBOT
@sage dagger Has your question been resolved?
clear oar
obtuse pebbleBOT
clear oar
I have no clue how to begin
n is the smallest prime in a list of primes
where pi+1 is n or below
we're basically trying to show n | (Pi+1 - Pi)
fossil crag
Hint : you have to show that necessarily d = 0 mod n.
What happens if you suppose not?
clear oar
uhh
wait but
ok holdup
if n is 5
and lets say pi is 11
and pi+1 is 13
however 13-11= 2
but 5| 2is false
fossil crag
well condition (2) is false so we can't apply the proposition here
clear oar
well condition (2) is false so we can't apply the proposition here
wait howso?
balmy mortar
Its worth constructing a simple example that works to see claim actually works for one.
clear oar
n is 5
fossil crag
yes, what are they ?
clear oar
if n is 5 the smallest so p1 to p5 7 11 13
fossil crag
I have an example :
7 37 67 97 127 i think works
clear oar
where n is 7?
balmy mortar
if n is 5 the smallest so p1 to p5 7 11 13
youve written us 3 numbers, not 5?
fossil crag
n is 5
balmy mortar
well condition (2) is false so we can't apply the proposition here
Then check this
fossil crag
if n is 5 you need to find 5 prime numbers where the differences are the same
fossil crag
5 7 11 13 17 19
proposition 2 is false in this one
fossil crag
yes
fossil crag
I have an example :
7 37 67 97 127 i ***think*** works
however check this out for n=5
fossil crag
yes
fossil crag
37-7 = 30
clear oar
prime number?
clear oar
ok...
clear oar
yeah
fossil crag
and we do have 5|30
clear oar
indeed
fossil crag
so that's an example of an instance where the theorem applies
clear oar
so the second condition states thst the differences have to be the same?
fossil crag
yes
fossil crag
wanna try to prove it?
fossil crag
When we don't know, proof by contradiction is god sent
(or math sent if you're atheist)
fossil crag
yes
clear oar
uhh
clear oar
yes but because
fossil crag
p1 is prime
fossil crag
yes
fossil crag
yes, and n cannot divide any pi
clear oar
yeah
fossil crag
However, what can we say about all the integers of the form "p1 + k*d" (for k between 0 and n-1) ?
fossil crag
it's alright, this is the part that's a bit confusing
clear oar
it gives us the next value?
fossil crag
yes, we can notice that p1 + k*d = p_{k+1}
clear oar
yeah
fossil crag
but this is not over yet!
since n doesn't divide d, and n is prime, what can we say about gcd(n,d) ?
clear oar
gcd(n d ) = 1
clear oar
uhh
fossil crag
It becomes more intuitive when we go through an example
clear oar
mhm
fossil crag
I'll do it with you
fossil crag
yes
fossil crag
yep
so here's k*d mod n for values of k from 1 to 10 :
2
4
1
3
0
2
4
1
3
0
notice a pattern ?
fossil crag
the important thing to notice is that it spans through all the possible remainders mod n
clear oar
mhm
fossil crag
there is something to do with it yes !
clear oar
ok...
fossil crag
let r be the remainder of p1 mod n
then there must be some k between 0 and n-1 such that k*d = -p1 mod n right? since k*d cycles through all possible remainders
fossil crag
ok so going with this example : n=5, d=8
clear oar
yeah
fossil crag
we have "primes" 7, 7+8, 7+2*8, ..., 7+4*8
clear oar
mhm
fossil crag
p1 = 2 mod n
clear oar
when you say = do u mean ≡
fossil crag
yes
fossil crag
so we want k such that kd ≡ -p1 mod n
so kd ≡ 3 mod n
it's easy here because we can take k =1
and 8 ≡ 3 mod n
but then look : p1 + kd ≡ 0 mod n
clear oar
right...
fossil crag
so $p_{k+1}$ is not prime
warm shaleBOT
rafilou2003
fossil crag
yes
clear oar
and we get n | p1 + remainder
sorry can you explain how we got here again im just a little lost
fossil crag
Intuitively, since k*d cycles through all possible remainders mod n, it eventually lands on the opposite remainder of p1
clear oar
opposite remainder?
fossil crag
basically ≡ -p1 mod n
clear oar
uhh p1 ≡ -p1 mod n?
fossil crag
nonono
clear oar
uhh
fossil crag
a ≡ b mod n means n|(a-b)
clear oar
yeah
fossil crag
meaning they have the same remainder
clear oar
yeah
fossil crag
so it's 0 mod 5
clear oar
mhm
fossil crag
and subtracting 2, we get -2 ≡ 3 mod 5
clear oar
yes
fossil crag
uhh p1 ≡ -p1 mod n?
so be careful, p1 is not congruent to -p1
fossil crag
uhh p1 ≡ -p1 mod n?
no it was just to show you this was not true
clear oar
oh alright
fossil crag
but we can go back to the example of n=5, d=8
clear oar
right
fossil crag
If you remember, k*d cycles through all possible remainders mod n
clear oar
mhm
fossil crag
and since -p1 is a possible remainder mod n
k*d has to land on this remainder at some point
fossil crag
I took 7,7+8,...,7+4*8 as examples of "primes"
clear oar
yeah
fossil crag
so -p1 would be -7, which is 3 mod n
clear oar
it cycles to -7?
fossil crag
it cycles to -7?
it cycles to the same remainder
clear oar
ohhh
fossil crag
at some point, -7 and kd will have the same remainder
and so kd+7 will have a remainder of 0 mod n
clear oar
right
fossil crag
but kd+7 is the k+1-th prime
clear oar
right
fossil crag
and that's a contradiction
clear oar
so from the beginning we assume the contradiction
and then since n is prime and p1 .... pn is prime
or rather their gcd is 1
then they cant divide each other
n cannot divide any Pi
fossil crag
yes
clear oar
and then since we know P1 + k*d gives us the Pk+1 term
when k*d mod n
we know we get the remainder?
acc wajt
clear oar
when k*d mod n
lost a tad from here
fossil crag
we know that k*d cycles through every possible remainder mod n for k between 0 and n-1
clear oar
k*d mod n cycles through each remainder
for example
n = 5 d = 8
where 7, 7+8, ... 7 + 4*8
4*8 mod 5
gives 2
fossil crag
yes
clear oar
so from here
uhh
letting r to be the remainder of p1 mod n
sorry can you define -p1
fossil crag
have you seen negative integers?
clear oar
have you seen negative integers?
im just a tad lost do you mean that since we're doing pi+1 - pi
fossil crag
no, as we saw, 2 mod 5 is different from (-2) mod 5 ≡ 3 mod 5
clear oar
yes
fossil crag
ok, so since gcd(d,n) = 1, d has an inverse mod n, let's say "u"
and so we take k = -p1u mod n
clear oar
mhm
fossil crag
yes
clear oar
in our case
p1 is 7
so -p1 is -7
so k*d mod n cycles for the remainders until we get the remainder between -7 mod n?
fossil crag
yes
fossil crag
if d=8, we were lucky to reach it at k=1
-7 congruent to -7 + 2*5 = 3
mod 5
10-7 = 3 mod 5
clear oar
right
fossil crag
so after that :
since kd congruent to -p1 mod n, kd+p1 is divisible by n
but that means that p_{k+1} is not prime
which is impossible
fossil crag
I'll go to sleep, if you need someone to reexplain ping helpers, have a good day/night
clear oar
aight tyy
old glacier
hi, I've known the e is 1(1+1/n)^n, but how to understand ,,5000e^{0.024(8)}=6058.35, 5,000 invested for 8 years in an account that pays 2.4% interest continually.
why the e is used as a base, and interest multiple years is the exponent
obtuse pebbleBOT
hi, I've known the e is 1(1+1/n)^n, but how to understand ,,5000e^{0.024(8)}=605...
nocturne minnow
hi, I've known the e is 1(1+1/n)^n, but how to understand ,,5000e^{0.024(8)}=605...
https://math.stackexchange.com/questions/1140023/why-does-pert-work
See if that helps
worn gale
obtuse pebbleBOT
worn gale
is it just me that the question looks weird?
in the first sentence
it says measured from July 1 to June 30...
how is time going backwards
im so lost
forest sinew
theyre different years
worn gale
is that so
forest sinew
yea
worn gale
i mean thought of that too
i gues
that makes sense
oh i never knew that february has only 28 days
wow lack of common sense i guess
worn gale
i was like
why do they say 28th
since there is also
29th and 30th
but then i realized
february is only til 28th
thanks i solved that provlem
blem
but now i don't understand how i got this wrong
isn't the answer 2/3
nocturne minnow
Why would it be 2/3
nocturne minnow
Why are you finding slope
nocturne minnow
You're overcomplicating it, by a lot
It's asking for the domain of that graph
Aka the bounds of that graph
Like what's the highest value x could be?
worn gale
oh
wait
i thought they used
() these
lol..
like coordinates
you know what i mean
ok i get it now
thanks thanks
oh wait
as soon as i uploadedit
ithink i figured out
why it's wrong
because there's no decimal months irl
it wouldbe just 9 months
lmao
like no one counts months in decimals
obtuse pebbleBOT
@worn gale Has your question been resolved?
obtuse pebbleBOT
obtuse pebbleBOT
Available help channel!
crude ice
can someone help me with this qn
obtuse pebbleBOT
can someone help me with this qn
crude ice
these are my steps
royal basin
royal basin
NEVER, EVER, EVER USE CHATGPT FOR MATH!
crude ice
these were my other qns
crude ice
these were my other qns
oh woah
okay but i thought i had like wolfram alpha plugin i thoough ti would be more accurate haha
royal basin
CHATGPT CANNOT DO MATH, AND SHOULD NOT BE USED TO DO MATH OR TO CHECK YOUR WORK. IT PRODUCES WORK THAT LOOKS ACCURATE BUT OFTEN ISN'T.
if you don't know better, you can't tell the genuine stuff from the bullshit.
and if you do, then you don't need chatgpt.
,w sum[k=5, n] (1/4^k + 3k^2)
warm shaleBOT
royal basin
ok yeah your #2 checks out.
crude ice
okay
crude ice
ok yeah your #2 checks out.
fr? i checked with my friends and they had different answers
royal basin
and how do you know your friends are infallible?
royal basin
out of energy sorry
crude ice
out of energy sorry
what about a quick glance
you know how you can kinda tell if it looks right
royal basin
quick glance says it looks reasonableish