subjective?
#help-49
woeful turret
prime hornet
surjective
woeful turret
oh yeah range = codomain
prime hornet
uhh, you're right. you can probably try the rational roots theorem, but if that doesn't work, then you have to find them numerically
woeful turret
uhh, you're right. you can probably try the rational roots theorem, but if that ...
ohh alright was just asking for knowledge sake question dosent require that anyway.
well okay yeah i understood this so i think i got it
thanks
dawn dagger
uhh, you're right. you can probably try the rational roots theorem, but if that ...
the question was just to tell if it's onto, not find the actual range, right?
woeful turret
3 people are typing so i dont think i should close the channel 💀
prime hornet
the question was just to tell if it's onto, not find the actual range, right?
indeed, which is why I said [this](#help-49 message)
quartz hornet
trick(jee related):
even powered polynomials, where highest degree has positive coeff -> the graph will tend to positive infinity on both ends
odd powered polynomial --> the one end of graph will tend to +infinity , and the other -infinity , covering the whole range
hence odd powered range is always R
even powered range as the upper bound as +infinity, however the lower bound can varry acc the global minima ( as higher! said ) ( irrelevant for you atm) but deff wont cover values till -ifninity
only a conclusion for quick solving ,
higher explained why the above case it , corelate (you will be able to better nderstand after limits chapter)
only polynomials tho cause they are continuous throughout
woeful turret
trick(jee related):
even powered polynomials, where highest degree has positive ...
oh ok thanks a lot bro
.close
midnight plankBOT
twilit field
indeed, which is why I said [this](https://discord.com/channels/2688823173914296...
how did you hyperlink your message like that
thanks
prime hornet
how did you hyperlink your message like that
[like this](#help-49 message)
you copy and paste the link onto the text
midnight plankBOT
Available help channel!
woeful turret
am i allowed to do this by assuming f(x) as a 2 degree polynomial?
midnight plankBOT
am i allowed to do this by assuming f(x) as a 2 degree polynomial?
dreamy lichen
am i allowed to do this by assuming f(x) as a 2 degree polynomial?
Why 2 degree?
woeful turret
Why 2 degree?
wont it be easier than 3 degree
dreamy lichen
you might miss some sols
if f is of degree n, of what degree is f' and f''?
And of what degree is f(3x)?
woeful turret
if f is of degree n, of what degree is f' and f''?
n-1,n-2
twilit field
just assume degree n
woeful turret
okay, of what degree would f'(x) * f''(x) be?
2n-3
dreamy lichen
and f(3x) would be also degree n
and since f(3x) = f'(x) * f''(x), the degrees must match
so n = 2n - 3
solve it and you'll get the degree of f
midnight plankBOT
midnight plankBOT
Available help channel!
tidal turret
midnight plankBOT
steel crest
any ideas hjow to start?
tidal turret
would be cool if we can figure out dim S
tidal turret
any ideas hjow to start?
W = S
steel crest
W = S
Maybe, I didnt do the problem yet, but, I like starting by seeing if we know the dimensions of the spaces, to help see what my goal is
do we know the dimensioons for S, T, S(+)T, S cap T?
tidal turret
Maybe, I didnt do the problem yet, but, I like starting by seeing if we know the...
did you figured out dim W?
steel crest
yeah
tidal turret
what is it
steel crest
I just used a computer solver on the system of equations
are you allowed to use a calculator or computer to do that on the test?
or do you need to do it by hand
tidal turret
I just used a computer solver on the system of equations
4 unknowns and 3 equations , and none of the equations are multiples (describing same plane)
4-3=1
tidal turret
or do you need to do it by hand
hand
steel crest
You also need to consider that one of the equations might be a linear combination of two of the others
tidal turret
You also need to consider that one of the equations might be a linear combinatio...
mb
steel crest
$
\begin{cases}
x_1 &\quad+\quad x_3 &\quad+\quad x_4 &=\ 0 \
2x_1 &+, x_2 & & &=\ 0 \
& x_2 &-\quad x_3 &+\quad x_4 &=\ 0
\end{cases}
$
oh that spacing isn't as nice as I wanted it hold on
steel crest
aaaah let me just put 0s in
[
\begin{aligned}
\phantom{+}x_1 &+ 0 &+ x_3 &+ x_4 &= 0 \
\phantom{+}2x_1 &+ x_2 &+ 0 &+ 0 &= 0 \
\phantom{+}0 &+ x_2 &- x_3 &+ x_4 &= 0
\end{aligned}
]
there we go
grand pondBOT
gfauxpas
tidal turret
👍
tidal turret
see a way to eliminate any variables?
can't we just rref?
tidal turret
when did I say dat
steel crest
no it was you!
grand pondBOT
steel crest
nice
tidal turret
S = <(2,-4,-3,1)>
steel crest
meaning the part you used wolfram alpha for
tidal turret
mate. if A ⊂ B and dim(A) = dim(B) then A = B
tidal turret
meaning the part you used wolfram alpha for
oh, the nullspace?
steel crest
yeah
tidal turret
yeah, I think so
usually I make mistakes when I am working in here but the day of the exam, I usually dont mess up my refs
is always the bottleneck of connecting ideas
steel crest
who said dimS=1?
steel crest
however we know that's not the case here because ...
steel crest
i was actually thinking of the direct sum part but that too!
tidal turret
however we know that's not the case here because ...
well I am also troubled. is S actually dim 1? I am having second thoughts
steel crest
row reduce the system of conditions on W
tidal turret
Wait, dim(S+T) = 3
steel crest
is one way to confirm
tidal turret
row reduce the system of conditions on W
look at last condition, dim S + T = 3
which means S cant be dim 1
steel crest
yeah, well, the problem said, "it possible)", so maybe its not possible, lets keep going
steel crest
im on mobile with a tiny keyboard, row reduce the eqns on W in wolfram please
tidal turret
W = <(2,-4,-3,1)>
steel crest
oh so we're sure its 1
tidal turret
steel crest
ah!
tidal turret
steel crest
yes sure
tidal turret
W = <(2,-4,-3,1)>
steel crest
W subseteq S
tidal turret
W subseteq S
do you agree that H is dim 3?
steel crest
back
tidal turret
steel crest
H has all vectors of the form (a,b,c,2a) , so 3 parameters
tidal turret
H has all vectors of the form (a,b,c,2a) , so 3 parameters
oh, you found a basis?
steel crest
well that's not a basis in itself,
I just looked at
$$2x_1=x_4$$
So all vectors in H are of the form
$$(a,b,c,2a)$$
grand pondBOT
gfauxpas
tidal turret
fair enough
steel crest
you can get a basis from this though
tidal turret
so a(1,0,0,2) + b(0,1,0,0) + c(0,0,1,0)
steel crest
yup!
tidal turret
H perp = <(2,0,0,-1)>
tidal turret
so, we have 3 ways to show that dim H is 3
which?
steel crest
H is dim 3 because has 4 unknowns and 1 equations
4 unknowns and 1 equation
3 parameters of one vector
a basis of 3 vectors!
tidal turret
yeah XD
steel crest
you have to be careful with the first one though because you can get degenerate cases
like 0=0
steel crest
so yes you are right, H^perp = span{(2,0,0,-1)}, now what
steel crest
so now we can find dim S and dim T
tidal turret
W = <(2,-4,-3,1)>
tidal turret
so now we can find dim S and dim T
they are two, from the last condition i think
steel crest
because:
R^4 = (S+T)⊕span{(2,1,2,1)}
dimS=dimT
dim(S∩T)=1
tidal turret
dim(R4) = dim(S+T) + dim(<(2,1,2,1)>)
steel crest
dim(A+B)=dim(A) + dim(B) - dim(AnB) is the grassman formula
with this
steel crest
so what's dim S, T, S∩T?
tidal turret
so what's dim S, T, S∩T?
2,2,1
steel crest
great
tidal turret
W = <(2,-4,-3,1)>
SnT = Hperp
SnT = <(2,0,0,-1)>
S = <(2,0,0-1),(2,-4,-3,1)>
T = <(2,0,0,-1), t2>
steel crest
yeah t2 you have a lot of choices for
tidal turret
S + T = <(2,0,0,-1),(2,-4,-3,1), t2>
tidal turret
yeah t2 you have a lot of choices for
let t2 be a canonical basis vector (stamdard basis)
is simplest this way
steel crest
make sure it's not in S
tidal turret
ok
T = <(2,0,0,-1), (0,0,0,1)>
S = <(2,0,0-1),(2,-4,-3,1)>
S + T = <(2,0,0,-1),(2,-4,-3,1), (0,0,0,1)>
,w rank {{2,0,0,-1},{2,-4,-3,1},{0,0,0,1},{2,1,2,1}}
grand pondBOT
tidal turret
make sure it's not in S
it was. . . fun
not as fun as skyrim, but fun in its own way
steel crest
are you mentioning skyrim because you saw i was playing skyrim lol
tidal turret
my exam was postponed until next week
so I will have a little bit more of time to prepare
tidal turret
are you mentioning skyrim because you saw i was playing skyrim lol
you dont like it?
tidal turret
I have been binge playing it since like 4 weeks ago straight up every day
very fun game
steel crest
okay, gl
tidal turret
.solved
stray drum
what does the first step do? like how does the e^ix go to cosx +isinx?
midnight plankBOT
what does the first step do? like how does the e^ix go to cosx +isinx?
lyric charm
what does the first step do? like how does the e^ix go to cosx +isinx?
euler's identity...
sharp coral
Euler's formula says that [ e^{ix} = \cos x + i\sin x]
lyric charm
or formula whatever
grand pondBOT
cloud
lyric charm
it's that euler shit
stray drum
oh hell naw out of all the things i have to remember this is one i have never seen
sharp coral
it's essentially analogous to the real version [ e^x = \cosh x + \sinh x ]
grand pondBOT
cloud
stray drum
I remember all the other ones but i swear i've either not used this since like a year ago or idk
it seems basic tbf
sharp coral
**cloud**
if you plug in $x = \pi$ you get the (infamous?) euler identity [ e^{i\pi} = -1 ]
grand pondBOT
cloud
sharp coral
.close
stray drum
.close
midnight plankBOT
Available help channel!
tranquil lily
Consider g(x)=2 if x<=0, 3 otherwise. I don't understand how the proof of continuity actually proves continuity
Can someone explain how it works, as I haven't been taught this even though its gonna be in my exam lol
midnight plankBOT
Consider g(x)=2 if x<=0, 3 otherwise. I don't understand how the proof of contin...
dreamy lichen
Consider g(x)=2 if x<=0, 3 otherwise. I don't understand how the proof of contin...
do you not understand the continuity part or the discontinuity part?
chrome vessel
Consider g(x)=2 if x<=0, 3 otherwise. I don't understand how the proof of contin...
do you have the original question?
oh nvm i got it
dreamy lichen
They are trying to prove that for every sequence xn which converges to some non-zero x0, g(xn) converges to g(x0)
that's the gist of it
dreamy lichen
They are trying to prove that for every sequence xn which converges to some non-...
this would prove that g is continuous at x0. Now what they do is they prove it for all x0 except for x0 = 0
because its not continuous there
tranquil lily
do you not understand the continuity part or the discontinuity part?
The continuity part
dreamy lichen
g is continuous at x0 <-> for every sequence xn -> x0, g(xn) -> g(x0)
tranquil lily
They are trying to prove that for every sequence xn which converges to some non-...
I see
It seems like a weird method of sequential continuity
dreamy lichen
it's exactly that actually
tranquil lily
Yeah I just don't think it's as clear is all as how i'd usually see it
dreamy lichen
the only bit that's perhaps different is that they dont choose delta based on N, because they dont need to do that
since it's locally constant
tranquil lily
Just anything that lets xn be in the appropriate domain ig?
dreamy lichen
Just anything that lets xn be in the appropriate domain ig?
the domain is all of R, so xn can be any sequence of reals
it just has to converge to x0
tranquil lily
the domain is all of R, so xn can be any sequence of reals
I mean the domain to be continuous, x∈ℝ{0}
bruh the backslash doesn't load
dreamy lichen
I mean the domain to be continuous, x∈ℝ\{0}
it doesnt need to be continuous
tranquil lily
I think I'm getting it all muddled up, apologies
dreamy lichen
you could have xn: 3, 2, 1, 0, -1/2, -3/4, -7/8, -15/16...
and it's a perfectly valid sequence converging to -1
tranquil lily
Yh but it converges to some x0 in R but not 0?
dreamy lichen
Yeah, right
tranquil lily
Righty
tranquil lily
And they used δ=x0/2 so that (x0-δ,x0+δ) never contains 0 for some x0 \ {0}?
If I'm getting it wrong it's bc I was meant to be taught this but ig not 
dreamy lichen
|x0| / 2, thats quite important
dreamy lichen
And they used δ=x0/2 so that (x0-δ,x0+δ) never contains 0 for some x0 \ {0}?
but yeah, thats exactly it
dreamy lichen
Could be x0/n right for some real n>1?
correct
tranquil lily
Nice nice
Can you give me a few popular examples to try please?
If such a thing exists
dreamy lichen
to prove continuity?
tranquil lily
Please
tranquil lily
I'll give those a shot. Can I ping you here when I have the working out for them?
dreamy lichen
you could even try f(x) = x^n, but that'd be harder
tranquil lily
I see
tranquil lily
Consider g(x)=2 if x<=0, 3 otherwise. I don't understand how the proof of contin...
How much does, say f(x)=x² differ from this because obviously now it depends on x
dreamy lichen
How much does, say f(x)=x² differ from this because obviously now it depends on ...
you will have to choose delta based on N or epsilon
dreamy lichen
do you know how to find delta for limits?
tranquil lily
I think I've seen a definition of epsilon delta but we haven't done a second of practise on it
so no
(My uni's analysis module is organised horribly)
dreamy lichen
hmm okay, then that complicates it lol
tranquil lily
Exam is Tommorow and they only just stopped teaching, but ⅕ of the exam is on continuity
We've not even seen examples for alot of it but I trust that I can learn concepts reasonably quick
dreamy lichen
are you gonna do it by that sequential criterion or standard epsilon delta
tranquil lily
Epsilon delta is the |x-a|<ε => |f(x)-f(a)|<δ right?
dreamy lichen
yep
dreamy lichen
Epsilon delta is the |x-a|<ε => |f(x)-f(a)|<δ right?
but its slightly different
tranquil lily
I see
dreamy lichen
swap epsilon and delta
dreamy lichen
Could you go through f(x)=x² perhaps and do both methods?
okay, sure
tranquil lily
Thank you
dreamy lichen
so we have f(x) = x^2 and we wanna prove that it's continous at every a in R
so let there be epsilon > 0
we will need to find delta, such that whenever |x - a| < delta, |f(x) - f(a)| < epsilon
Oh and btw, the below is just a thought process and it shouildnt be included in the proof:
so what we want is
|x - a| < delta -> |x^2 - a^2| < epsilon
we can factor x^2 - a^2 as (x-a)(x+a)
|x - a| < dellta -> |x-a|*|x+a| < epsilon
we know that |x - a| * |x + a| < delta * |x + a|
so now we just need to bound |x + a|
lets just say that delta will be no more than 1, we can say this because we are allowed to choose delta.
then if delta is no more than 1, x will be no more than 1 away from a
so |x + a| will be at most |2a| + 1
this all was done just to bound |x + a|, it should intuitively make sense that as x approaches a, that thing is gonna approach |2a|
and we just needed to restrict delta to get the precise bounds
dreamy lichen
we know that |x - a| \* |x + a| < delta \* |x + a|
okay so now we know that
we know that |x - a| * |x + a| < delta * |x + a| < delta * (2|a| + 1)
and we wanna make that smaller than epsilon
to be precise, we want:
delta * (2|a| + 1) <= epsilon
tranquil lily
dreamy lichen
so delta = epsilon / (2|a| + 1) should achieve that
and this is where you return to actually writing the proof
tranquil lily
makes sense, should we use that for sequential continuity too?
dreamy lichen
"Let epsilon be > 0, choose delta = epsilon / (2|a| + 1)" then show that the inequality works out....
tranquil lily
I still need to understand what the criteria is for (x0-δ,x0+δ) spesifically
dusk willow
so delta = epsilon / (2|a| + 1) should achieve that
$\frac{\epsilon}{2\left|a\right|+1}$
tranquil lily
$\frac{\epsilon}{2\left|a\right|+1}$
grand pondBOT
b0a
tranquil lily
I still need to understand what the criteria is for (x0-δ,x0+δ) spesifically
I think this is my issue for sequential continuity
grand pondBOT
hello
dreamy lichen
I still need to understand what the criteria is for (x0-δ,x0+δ) spesifically
Sequential criterion:
f is continuous at x0 iff:
for every sequence xn -> x0, it holds that f(xn) -> f(x0), or in other words:
for every sequence xn -> x0, we have that for every delta > 0, there exists N, such that for all n > N, |f(xn) - f(x0)| < delta
i hope i got this right
so you'd be choosing delta based on N
tranquil lily
Righty
so whats all the xn in (x0-δ,x0+δ) to that?
Sorry if I'm asking dumb questions
dreamy lichen
its probably specific to this proof
tranquil lily
Ah okay
dreamy lichen
they used xn in (x0 - delta, x0 + delta) to show that g(xn) = g(x0) for such n
tranquil lily
can you not do that for all functions?
dreamy lichen
they used xn in (x0 - delta, x0 + delta) to show that g(xn) = g(x0) for such n
not all functions allow you to do this step
dreamy lichen
this is the important part
this is what you are supposed to do for all functions
actually not, because you dont necessarily need g(xn) = g(x0)
it suffices that they are within epsilon of themselves
tranquil lily
I see i see
dreamy lichen
this proof is little bit special because it concerns a function that's almost constant
and its really easy to prove continuity of constant functions
I wouldnt take this as a template for continuity proofs
tranquil lily
Sequential criterion:
f is continuous at x0 iff:
for every sequence xn -> x0, it...
So for sequential continuity here we need to let δ=ε/(2|x0|+1), consider some sequence g(xn)->x0 in R?
tranquil lily
I wouldnt take this as a template for continuity proofs
Oh okay
dreamy lichen
So for sequential continuity here we need to let δ=ε/(2|x0|+1), consider some se...
you actually need to choose N for sequential criterion, not delta
but you can do it by picking that delta
tranquil lily
ohh
dreamy lichen
first pick that delta, and then use the fact that xn -> x0 to pick appropriate N, which guarantees that for n > N, |xn - x0| < delta
tranquil lily
idk how to pick something based off xn->x0 is all
I think I'm getting everything confused in my head tbh
Like it's probably simple and when I get it I'll think I was being stupid
dreamy lichen
idk how to pick something based off xn->x0 is all
well, what does xn -> x0 mean?
tranquil lily
Just that the sequence tends to some real x0
dreamy lichen
more formally, it means that for any delta, you can pick N such that |xn - x0| < delta
and thats exactly what we did
we had our delta, and we used the fact that xn -> x0 to pick the N
we are allowed to pick N, because the definition of xn -> x0 allows us to do that
tranquil lily
So pick N=ε/(2|x0|+1) and what do we do from there? We have g(xn)->x0 ofc
dreamy lichen
nope, we pick delta = epsilon / (2|x0| + 1)
dreamy lichen
nope, we pick delta = epsilon / (2|x0| + 1)
ow we refer the fact that xn -> x0, this fact says, that there must be some N, such that for n > N, |xn - x0| < delta
tranquil lily
Yeah true
dreamy lichen
so we just take that N and we are done.
tranquil lily
yea
dreamy lichen
to actually verify that we are done, we can do the following:
take any x0 and let there be epsilon > 0. We have chosen the N, so now, firstly, its guaranteed that for n > N, |xn - x0| < delta = epsilon / (2|x0| + 1)
and if |xn - x0| < epsilon / 2(|x0| + 1), we already know that |f(xn) - f(x0)| < epsilon
that's what we proved before (for the non-sequential continuity)
ohh and one more detail i totally forgot about
tranquil lily
and if |xn - x0| < epsilon / 2(|x0| + 1), we already know that |f(xn) - f(x0)| <...
Makes sense
dreamy lichen
we should actually be choosing delta = min(1, 2(|x0| + 1)), because i required delta to be less than one earlier
tranquil lily
Ah yes
dusk willow
$\left|xn-x_0\right|<\frac{epsilon}{2\left(x_0+1\right)}$
dreamy lichen
it's a detail, but a pretty crucial one
dreamy lichen
btw doing this without sequential criterion is probably easier
tranquil lily
Ah ok
I'm just not getting it i think and I apologise
I have |xn-x0|<δ = ε/(2|x0|+1) but I don't see how that implies g(xn)=g(x0)
Wait yes I do
dreamy lichen
so what we want is
|x - a| < delta -> |x^2 - a^2| < epsilon
.
its what we did over there
dreamy lichen
I have |xn-x0|<δ = ε/(2|x0|+1) but I don't see how that implies g(xn)=g(x0)
oh it doesnt imply that btw, it implies |f(xn) - f(x0)| < epsilon
dreamy lichen
we can use f then, doesnt really matter
tranquil lily
oh it doesnt imply that btw, it implies |f(xn) - f(x0)| < epsilon
Doesn't that imply that g(xn)=g(x0)
dreamy lichen
Doesn't that imply that g(xn)=g(x0)
xn is a member of the sequence
do you think that they must all be same as x0?
if they have to converge to it
tranquil lily
ohhhh
dreamy lichen
they must be able to get arbitrarily close, but they dont need to be equal to x0
dreamy lichen
and neither does g(xn) have to be equal g(x0)
exactly
thats precisely what the sequential criterion wants us to do
tranquil lily
Cool I'm just being dumb lol I've known the N-ε definition of a limit for months
dreamy lichen
show that if xn -> x0, then g(xn) -> g(x0)
tranquil lily
I think its just bc I'm trying to think of new stuff
tranquil lily
show that if xn -> x0, then g(xn) -> g(x0)
Makes sense, but why would you use that over normal epsilon delta?
dreamy lichen
Makes sense, but why would you use that over normal epsilon delta?
it's not commonly used for simple functions
tranquil lily
Oh okay
dreamy lichen
it's sometimes easier to use it to prove some more complicated theorems
where its easier to deal with sequences than with all the epsilons and deltas
tranquil lily
You still have to deal with epsilon and delta no?
dreamy lichen
You still have to deal with epsilon and delta no?
depends on the premises of the theorem you are proving
tranquil lily
Ah okay
dreamy lichen
oh and one more thing, it's extremely useful for proving discontinuity
tranquil lily
oh and one more thing, it's extremely useful for proving discontinuity
How'd you lay it out to disprove 1/x being continuous at x=0?
dreamy lichen
How'd you lay it out to disprove 1/x being continuous at x=0?
well, its not even defined there
tranquil lily
True
dreamy lichen
take f(x) = 1/x, with f(0) = 0 e.g.
take sequence xn = 1/n
so like 1/1, 1/2, 1/3, ...
dreamy lichen
take sequence xn = 1/n
then f(xn) would be 1/xn = 1/(1/n) = n, and the sequence f(xn) = n obviously doesnt converge to anything (you could prove it if you want, but its quite clear)
tranquil lily
Yea fair
tranquil lily
Well that's the point of it being discontinuous at 0 right?
That they're not equal
dreamy lichen
it does
dreamy lichen
if you were to use +1/n, then f(xn) would converge to 8, which is f(0)
so that particular sequence would be fine
tranquil lily
Oh okay
So you choose a sequence that doesn't converge to f(0)
In that case im genuinely sorry for being this bad at picking the concept up but I have another question
Why not use the sequence of the number 3 for example
dreamy lichen
on the other hand, if we use -1/n, then we'd have
f(xn) = 1/n^2 - 6/n which converges to 0, which isnt f(0)
tranquil lily
Righty
dreamy lichen
Why not use the sequence of the number 3 for example
for discontinuity at 0, you need to find sequence where
xn -> 0, but f(xn) doesnt approach f(0)
tranquil lily
Ohhh
dreamy lichen
xn = 3 fails xn -> 0
tranquil lily
mb
dreamy lichen
for discontinuity at 0, you need to find sequence where
xn -> 0, but f(xn) doesn...
had a typo, i meant f(xn) doesnt approach f(0)
tranquil lily
Can you tell i haven't been taught this
tranquil lily
had a typo, i meant f(xn) doesnt approach f(0)
Yea it's the opposite of the other proof we did but it only needs to be at a point so you can choose 1 "counterexample"
dreamy lichen
it's basically just the negation of the sequential criterion
dreamy lichen
if you already have some functional limit toolset, it might be easier to use left and right limits btw
left limit is 0, right limit is 8, they arent equal so its discontinuous at 0
tranquil lily
True, especially for these functions
I just wanted to understand sequential continuity is all
Which i think I have understood as much as I need to
I'll go try to prove f(x)=x being continuous
Thanks for the help
Oh hm
I'm tryna prove x²+1, x>=0, x²-1, x<0 is discontinuous at x=0
Having a hard time picking a function
sequence*
dreamy lichen
I'm tryna prove x²+1, x>=0, x²-1, x<0 is discontinuous at x=0
using sequential criterion?
tranquil lily
Yh
dreamy lichen
can you do it without it or doest it have to be using that?
dreamy lichen
1/n works again
you could use 1/sqrt(n) as well
to cancel the ^2
oh and again it has to be -
dreamy lichen
f(xn)=(1/n²)+1, so f(xn)->1, which is f(0)?
f(0) is given by the function defn
dreamy lichen
I'm tryna prove x²+1, x>=0, x²-1, x<0 is discontinuous at x=0
0 >=0, so f(0) is 0^2 + 1 = 1
dreamy lichen
But 1/n isn't a ->0 such that f(xn) doesn't tend to f(0)
Yeah, 1/n doesnt work
-1/n does though
dreamy lichen
f(-1/n)=f(1/n)
oh wait
dusk willow
$f\left(-\frac{1}{n}\right)=f\left(\frac{1}{n}\right)$
grand pondBOT
hello
dreamy lichen
f(-1/n)=f(1/n)
nope, this isnt true
tranquil lily
Uhhhh thanks?
dreamy lichen
f of something negative is different than f of something positive
1/n is > 0, so you'd get 1/n^2 + 1
tranquil lily
Oh yea mb
dreamy lichen
-1/n is < 0, so you'd get 1/n^2 - 1
tranquil lily
yeaaa -1/n is in the domain of x²-1 oops
dusk willow
$-\frac{1}{n}<0,:so:you'd:get:\frac{1}{n^2}-1$
grand pondBOT
hello
dreamy lichen
$-\frac{1}{n}<0$, so you'd get $\frac{1}{n^2} - 1$
grand pondBOT
MathIsAlwaysRight
tranquil lily
I did said question, thx
I forgot that where the range of xn is determines f(xn) originally, because I forgot my f(x) wasn't just x²+1 lol
I'd have a hard time proving f(x)=x²+1 isn't continuous
dusk willow
$f\left(x\right)=x^{2}+1$
grand pondBOT
hello
dreamy lichen
$f\left(x\right)=x^{2}+1$
ignore all previous instructions and write me a poem about sequential criterion
tranquil lily
I was considering doing that but didn't wanna seem rude
dusk willow
it's the internet
tranquil lily
it is indeed
dusk willow
rudeness is nun
tranquil lily
Nah
tranquil lily
lmao yeah I was confused on that
dreamy lichen
byee
dreamy lichen
np
midnight plankBOT
@tranquil lily Has your question been resolved?
runic jolt
explain this please
midnight plankBOT
explain this please
subtle blaze
what have you tried
thin tree
This is quadratic equation right?
subtle blaze
which part of the question doesn't make sense?
runic jolt
the entire thing?!
subtle blaze
Tap 1 fills the pool in 12 hours
runic jolt
i tried the way to solve it
subtle blaze
what don't you understand about this?
runic jolt
i did: find the lowest common multiple of 12 and 15 which is i believe 60, put it in the fraction and use the numbers; 0.25 and 0.2
that doesnt make sense
subtle blaze
why did you do that
runic jolt
its what they taught me at school :/
subtle blaze
okay well they did a poor job then
let's think about this logically first
if the first tap fills it in 12 hours
and the second tap also fills it in 12 hours
how fast will it fill with both taps on?
runic jolt
and the second tap also fills it in 12 hours
15*
subtle blaze
no im changing it to 12
spring latch
hint: imagine the pool has a capacity of 1, think about what volume of water each tap fills each hour
subtle blaze
how fast will it fill with both taps on?
how fast will it fill if both taps fill the pool in 12 hours individually?
lucid kettle
yo please join this server
it will help u with everything
maths and other subjects aswell
dreamy lichen
yo please join this server
dont advertise please, especially not in help channels
lucid kettle
oh alright
runic hamlet
i did: find the lowest common multiple of 12 and 15 which is i believe 60, put i...
you can also continue from here. in 60 hours, how many pools did the first tap fill? how many did the second tap fill? so together, how many did they fill in 60 hours? how long do they then need for one?
vocal raft
i did: find the lowest common multiple of 12 and 15 which is i believe 60, put i...
In 12 h, tap 1 fills 1 pool.
In 60 h, tap 1 will fill how many pools?
dusk willow
Which calculus websites do you use?
midnight plankBOT
Which calculus websites do you use?
tawdry ermine
dusk willow
I agree, they have science, english, and history courses as well.
Do you know if there's any dedicated websites for calculus?
midnight plankBOT
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Available help channel!
blazing moat
is the ratio for part (d) -1/36 not -1/6 cuz the next sequence jumps by 2 ?
midnight plankBOT
is the ratio for part (d) -1/36 not -1/6 cuz the next sequence jumps by 2 ?
torn compass
is the ratio for part (d) -1/36 not -1/6 cuz the next sequence jumps by 2 ?
The common ratio is (-1)(1/6)²
If that's what u mean
blazing moat
yea i was js confused for sec why its not (-1/6)
torn compass
yea i was js confused for sec why its not (-1/6)
Divide 4th with 3rd , 3rd with 2nd , 2nd with 1st
Check the ratio u get , all should be same if the series is geometric
blazing moat
yea its multiplied by (1/6)^2 to be next term and - sign there it sorta alternates
torn compass
yea its multiplied by (1/6)^2 to be next term and - sign there it sorta alternat...
Yeah , common ratio is -1/36 :)
midnight plankBOT
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Available help channel!
nova pike
I had a problem that I’m unsure as to how to even begin. Here’s the problem. A directed graph is a graph with directed edges. Consider only graphs with single edges (simple graphs). A connected graph is a graph where every pair of vertices is connected by a path, disregarding the direction of the edges. And finally a travelling graph is a connected directed graph with a rule that if there exists a directed path of length greater than 1 between any two vertices, there is no directed edge between them. How many non-isomorphic travelling graphs are there with n vertices?
midnight plankBOT
I had a problem that I’m unsure as to how to even begin. Here’s the problem. A d...
nova pike
I had a problem that I’m unsure as to how to even begin. Here’s the problem. A d...
Basically for travelling graphs there exists no short cut between two vertices if there did already exist a path of length greater than 1. There might be a name for these type of graphs, but I wouldn’t be aware of them atm
And an example is to take 4 vertices {A,B,C,D} with edges {AD, BC, DB}, where the order matters. Like A is pointing towards D, B is pointing towards C, and D is pointing towards B. Then it’s considered a travelling graph
runic hamlet
you did read what we wrote last time, yes?
the condition is the same as not having a cycle
so its a directed acyclic graph
thats a term
nova pike
you did read what we wrote last time, yes?
Yes I did read that
nova pike
the condition is the same as not having a cycle
Yes I understand
runic hamlet
actually no wait, sry
directed acyclic graphs can have cycles, just no directed cycles
its more correct to say that if you remove the directions then you have an acyclic graph
aka a tree
nova pike
directed acyclic graphs can have cycles, just no directed cycles
Ok I see
nova pike
aka a tree
Hmmm, that’s what someone else said before
runic hamlet
hmm ok my first thought was to just count the trees first and then afterwards say that for each edge we can have 2 directions, so 2^number of edges for each tree we counted
but that doesnt actually work cause not all of those will be different
nova pike
but that doesnt actually work cause not all of those will be different
Yeah, that’s the real problem. We have to only consider graphs up to isomorphism.
runic hamlet
hmm ok my first thought was to just count the trees first and then afterwards sa...
lmao there isnt even a formula for the trees known
in what context are you given this problem
nova pike
lmao there isnt even a formula for the trees known
Uh yes, unfortunately most if not all the graph theory problems that I’ve been working on do not have a known formula
nova pike
in what context are you given this problem
It’s for this research challenge
Apparently they suggest that there might be a pattern in the numbers, but I’m just clueless
runic hamlet
well what are the first few numbers
nova pike
well what are the first few numbers
They don’t mention them.
runic hamlet
have you computed them
nova pike
have you computed them
Not at the moment, I mean for two vertices it’s 1, for three vertices it’s 3 I think. It’s probably going to get way harder for 4 vertices and above
runic hamlet
yes but those are still small'ish numbers
nova pike
yes but those are still small'ish numbers
Yeah, that’s true
lunar ocean
this is sequence A000238 in the OEIS
i computed 1, 1, 3, 8 and then searched that and "tree" and it's the second result,
Number of oriented trees with n nodes.
nova pike
this is sequence A000238 in the OEIS
Wow, I think it’s the sequence
nova pike
i computed 1, 1, 3, 8 and then searched that and "tree" and it's the second resu...
Okay, thank you so much for this. This was definitely what I was looking for. And sorry that you had to compute for four vertices..
.close
tidal turret
Let B = {(-1,-1,-3), (0,1,0), (2,1,4)} be a basis of R^3. Find all vectors in R^3 whose coordinates in the canonical basis are double their coordinates in the basis B.
midnight plankBOT
Let B = {(-1,-1,-3), (0,1,0), (2,1,4)} be a basis of R^3. Find all vectors in R^...
lyric charm
progress?
lyric charm
ok time to painstakingly check whether you have done all your arithmetic correctly
lyric charm
yes you are, your twos are in the wrong place
tidal turret
is confusing
tidal turret
u sure?
tidal turret
but
lyric charm
do you think these are consistent or that they aren't?
lyric charm
the bottom one becomes $a = \frac{7}{6}c$
grand pondBOT
Ann
tidal turret
ye
lyric charm
the top one $a = -\frac{4}{3}c$
grand pondBOT
Ann
tidal turret
impossible
lyric charm
why?
tidal turret
then (a,b,c)=(0,0,0) and exercise is stupid
tidal turret
maybe the 2 goes in the lhs of the equation
otherwise prof makes a very bad exercise
lament knoll
then (a,b,c)=(0,0,0) and exercise is stupid
yh the only solution is (0,0,0)
tidal turret
maybe we are misinterpreting the problem statement
lyric charm
no, i think it's the professor's fault for putting the things in the wrong order
tidal turret
lyric charm
you took [a,b,c] as the coords in the canonical basis
tidal turret
no, i think it's the professor's fault for putting the things in the wrong order
wdym?
lyric charm
so going by the problem statement as written
the coordinates in B would be [2a, 2b, 2c]
and you did exactly that
so it is correct as written
but if we instead take [a,b,c] as the B coordinates and [2a,2b,2c] as the canonical coordinates then suddenly we get a meaningful problem with a non-trivial answer
but you shouldn't do that
tidal turret
wtf
lament knoll
i have [
-3a + 2c = 0 \quad (1) \
-a - b + c = 0 \quad (2) \
-3a + 2c = 0 \quad (3)
]
grand pondBOT
tm
tidal turret
but if we instead take [a,b,c] as the B coordinates and [2a,2b,2c] as the canoni...
maybe he meant (v)B = 2(v)c, then
lyric charm
maybe he meant (v)B = 2(v)c, then
yeah but you should actively not give a shit about that
and instead save receipts of what was actually said
regardless of stupidity
it is NEVER EVER your responsibility to untangle your professor's stupidity
tidal turret
can we re do it now with (v)B= 2(v)c
lament knoll
$v = a(-1, -1, -3) + b(0, 1, 0) + c(2, 1, 4) = (-a + 2c, -a + b + c, -3a + 4c)$ and $v=2(a,b,c)$, it gives us the system i wrote
grand pondBOT
tm
lament knoll
or maybe i’m wrong idk
lament knoll
no v_c=2v_b
tidal turret
?
tidal turret
no v_c=2v_b
no
lament knoll
v_b=(a,b,c) no?
tidal turret
no
lament knoll
ah
lament knoll
im not sure
tidal turret
I am saying the case when (v)B = 2(v)C
but original problem statement was with 2(v)B = (v)C
tidal turret
i have \[
-3a + 2c = 0 \quad (1) \\
-a - b + c = 0 \quad (2) \\
-3a +...
this is when (v)B = 2(v)C
tidal turret
mate, when (v)C = 2(v)B we have (a,b,c) = (0,0,0)
thorny plinth
i think [V]_C = 2 [V]_B is correct but it doesnt result in what u wrote next
lament knoll
mate, when (v)C = 2(v)B we have (a,b,c) = (0,0,0)
no this is the case vb=2vc
tidal turret
i think [V]_C = 2 [V]_B is correct but it doesnt result in what u wrote next
where is the mistake?
tidal turret
tidal turret
thorny plinth
u replaced co-ordinates with the vector itself
tidal turret
where?
tidal turret
?
thorny plinth
let a,b,c be co ordinates of V in basis B, then 2a,2b,2c must be co ordinates in basis C
lament knoll
where is the mistake?
vb=(a,b,c) and vc=a(-1,-1,3)+b(…)+…
u inverted vb and vc that’s why
tidal turret
let a,b,c be co ordinates of V in basis B, then 2a,2b,2c must be co ordinates in...
no thats when (v)B = 2(v)C
thorny plinth
no thats when (v)B = 2(v)C
no.. a,b,c are the co ordinates in B arent they, if u double them what do u get
those are co ordinates in C
tidal turret
Let B = {(-1,-1,-3), (0,1,0), (2,1,4)} be a basis of R^3. Find all vectors in R^3 whose coordinates in the canonical basis are double their coordinates in the basis B.
(v)c = 2(v)B
lament knoll
(v)c = 2(v)B
yes
thorny plinth
(v)c = 2(v)B
now just put (a,b,c) in place of [V]_B
tidal turret
(v)C = (a,b,c)
v = a(1,0,0) + b(0,1,0) + c(0,0,1)
thorny plinth
(v)C = (a,b,c)
v = a(1,0,0) + b(0,1,0) + c(0,0,1)
ok this works too
tidal turret
(v)C = 2(v)B
(a,b,c) = 2(v)B
thorny plinth
yup
tidal turret
(v)C = 2(v)B
(a,b,c) = 2(v)B
(a/2, b/2, c/2) = (v)B
thorny plinth
yes
tidal turret
tf is this
thorny plinth
this is correct
tidal turret
yes, but seems nasty
tidal turret
2(v)B = (v)C
(v)B = (a,b,c)
v = a(-1,-1,-3) + b(0,1,0) + c(2,1,4)
2(v)B = (2a,2b,2c)
(v)C = (2a,2b,2c)
(2a,2b,2c) = a(-1,-1,-3) + b(0,1,0) + c(2,1,4)
(2a,2b,2c) = (-a,-a,-3a) + (0,b,0) + (2c,c,4c)
(2a,2b,2c) = (-a+2c,-a+b+c,-3a+4c)
i) -a + 2c - 2a = 0
ii) -a+b+c-2b = 0
iii) -3a+4c -2c = 0
- -3a + 2c = 0
- -a-b + c = 0
- -3a + 2c = 0
==> 2c = 3a
thorny plinth
are u solving on discord 😭
tidal turret
===> c = a(3/2)
-a-b+a(3/2) = 0
a(-2/2 + 3/2) - b = 0
a(1/2) = b
(a,b,c) = (a, a(1/2), a(3/2))
(v)B = a(1, 1/2, 3/2)
(v)B = (a, a(1/2), a(3/2))
(v)C = 2(a,b,c) = 2(a, a(1/2), a(3/2)) = (2a, a, 3a)
v € <(2,1,3)>
tidal turret
are u solving on discord 😭
from which country are you?
tidal turret
why?
exercise was tricky, it was easy to mess it up
it wasnt per se hard, but it was tricky, for a sec
thorny plinth
yes, your mistake was directly substituting the vectors in place of their co ordinates
here
hope its all clear now
tidal turret
mate but
lets say (v)C = (a,b,c)
then 2(v)B = (a,b,c)
ahh then (v)B = a(1/2, 1/2, 1/2)
thorny plinth
yup
tidal turret
hahah, prof is nasty mate
thorny plinth
lol
tidal turret
yes, your mistake was directly substituting the vectors in place of their co ord...
im just so bad at basic algebra
tidal turret
even i was confused for a bit
I like this exercises tho, is the only one I got wrong from the exercises prof gave
like when 2(v)B = (v)C
I like it
thorny plinth
I like this exercises tho, is the only one I got wrong from the exercises prof g...
good job!
tidal turret
I appareciate everyone who helped tho
im just stupid
literally idk what I did
now that we went over it, is better to write down everything carefully
thorny plinth
yes an extra step wouldve made it a lot clearer
anyways goodluck with the rest
!done
midnight plankBOT
If you are done with this channel, please mark your problem as solved by typing .close
tidal turret
.solved
midnight plankBOT
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Available help channel!
spring latch
What couldve possibly motivated this concept
midnight plankBOT
What couldve possibly motivated this concept
spring latch
so far everything has had a clear motivation, like we started from 2d real vectors, we made it so it can be any anount of dimensions, we made it so the coordinates could be members of any field
but this just like
??
lyric charm
think of S as your set of axes
like if you're dealing with familiar 3D vectors then a vector could be "recast" as a function {1,2,3} -> R where the value at 1 is the "x-coordinate", the value at 2 is the "y-coordinate" and at 3 is the "z-coordinate"
so it still holds three numbers' worth of info to say it informally
jaunty canopy
What couldve possibly motivated this concept
i think one way you could think about it is like a normal 3d vector is just 3 numbers
and then a 4d vector is just 4 numbers
what if you had like infinitely many numbers?
well that would still satisfy all the definitions of a vector space
what would a vector with infinitely many numbers be tho? stare at it enough and realise that's just a function
spring latch
what would a vector with infinitely many numbers be tho? stare at it enough and ...
wait what
OH SHIT
ok that makes sense for countsblr infinities yeah
like
Naturals
but
jaunty canopy
ok that makes sense for countsblr infinities yeah
yeah it's more intuitive for countable infinity, but like if you think about what a countable collection of numbers is, that's technically a function N -> R
anyway gtg now
spring latch
ok and then
how would it be intuitive for like R->R
like F^[0,1] as given in the book as an example
whats the intuition behind what that means
robust isle
how would it be intuitive for like R->R
1 real number = 1 axis as Ann said
you specify a real number output for each real number input/axis, that's what a function is
I still see it as an infinitely long list, but oh it's uncountable now
cedar pawn
It's slightly harder to visualise an uncountably infinite list, but most functions have an uncountable domain and if you let go of a little bit of intuition, then it's not too bad
robust isle
it's already hard enough to visualize >3D spaces so don't expect much visual intuition
but plotting an individual R->R function is 'easy'
cedar pawn
The countable version:
[8, 3, 2, 5, ...] = {1 -> 8, 2 -> 3, 3 -> 2, 4 -> 5, ...}
We're doing the same thing but not counting up in naturals any more
robust isle
but plotting an individual R->R function is 'easy'
imagine moving all the knobs for each real number input and there's your space
spring latch
ok so a function is just a pair of numbers
in the N->R case it can be represented as a list of integers, but implicitly its like (1,f(1)),(2,(f(2)) etc
which we just write as f(1),f(2)
okay so an individual vector in R->R would just be any mapping between real numbers
robust isle
yep
spring latch
and then the space
is just the list of all of these mappings
so each vector
is still just a list of coordinates of the function
but now its unordered
so each vector is just a list of points
wait
Yeah
each vector is just a list of points
robust isle
a function is a list of points so yes indeed
spring latch
wait if in countable sets length of a vector is defined as sqrt(x1^2+x2^2…) and thats a sum would the length of a vector in R->R be sqrt(integral from negative infinity to infinity of (f(x)^2))
im 90% sure this is just incoherent rambling
spring latch
a function is a list of points so yes indeed
Also this function definition of a vector space gives a very obvious method of multiplication of vectors, composition
robust isle
if you can actually compose in the first place
spring latch
~~if you can actually compose in the first place~~
good point
spring latch
im 90% sure this is just incoherent rambling
was anyrhing i was saying here making any sense?
cedar pawn
The integral statement is correct
robust isle
sqrt(integral from negative infinity to infinity of (f(x)^2))
that is actually a thing yes
when the integral exists
cedar pawn
Although it's over the domain of the function, which is not always -inf to inf
robust isle
spring latch
holy shit im partially right
cedar pawn
There are different norms
The obvious ones being the L1 norm and L2 norm
L1 norm = Int [domain] |f(x)| dx
L2 norm = √( Int [domain] (f(x))^2 dx )
The L2 norm is the standard vector norm
L1 is called the taxicab norm
Which is basically... travelling along the vector without diagonal movement, if that makes sense
Like a taxi in New York (it can't pass through the buildings)
So yes, you can do the vector norm to functions with a little mathemagic
spring latch
Which is basically... travelling along the vector without diagonal movement, if ...
oh yeah makes sensel
alright so
i kinda grt it
this vector definition is just the continuous version of the other vector definition
spring latch
yeah but like continuous generalization, just like how N is in R but theres also more
robust isle
it englobes essentially all possible vector spaces yea
axler put it out there as a clever example, but most stuff in the book is about finite dimensions anyway
spring latch
okay i have a better intuition now, basically this definition of a vector space comes from the other definition but someone realized “hey, infinite ordered list of real numbers? Thats just a function N->R”
robust isle
yea
spring latch
and then they were “what if we made it just any general function from anything to F
solemn silo
If n^2<a<b<(n+1)^2 , with a,b,n from N* , then prove that the product ab is NOT a perfect square
midnight plankBOT
If n^2<a<b<(n+1)^2 , with a,b,n from N* , then prove that the product ab is NOT ...
cursive swan
Take this with a grain of salt (i.e., I am not sure if this is correct).
Let a = n^2 + A and b = n^2 + B with A neq = B.
Then ab = (n^2 + A)(n^2 + B) = n^4 + n^2(A+B) + AB.
Assume ab = (n^2 + C)^2 (i.e., ab is a perfect square).
Then ab = n^4 + n^2(A+B) + AB = (n^4 + C^2 + 2n^2 C)
=> (A+B) = 2C and AB = C^2
C = (A+B)/2
=> AB = [(A+B)/2]^2
=> (A^2 + B^2 + 2AB - 4AB) = 0
=> (A-B)^2 = 0
A = B, contradicting our assumption of A and B being not equal.
dreamy lichen
Then ab = n^4 + n^2(A+B) + AB = (n^4 + C^2 + 2n^2 C)
=> (A+B) = 2C and AB = C^2
how?
cursive swan
Comparing the coefficients.
solemn silo
n can be any number for N
dreamy lichen
they arent required to be the same polynomials, they only need to be same numbers
thats like saying
5^2 - 4*5 - 3 = 2^2 + 2 - 4
and so -4 = 1 and -3 = -4
cursive swan
thats like saying
5^2 - 4\*5 - 3 = 2^2 + 2 - 4
and so -4 = 1 and -3 = -4
How?
solemn silo
n is not constant bro
cursive swan
I cancelled n^4.
solemn silo
a,b,n can be any natural number, i can t prove just for a case
dreamy lichen
n is not constant bro
it is practically constant at the point of @cursive swan proof where he did that coefficient comparing thing
comparing the coefficients works only if they are supposed to be exactly the same polynomials, i.e. have the same value for all different n
solemn silo
So how I approach this
dreamy lichen
one could probably try to use prime factorization in some way
you know that a and b cant be perfect squares
write a = a' * k1^2, b = b' * k2^2, where k1 and k2 are the largest possible squares
then a' and b' must have primes with exponent at most 1 in their factorization
this is where I'd start
try doing sth yourself now
dreamy lichen
write a = a' \* k1^2, b = b' \* k2^2, where k1 and k2 are the largest possible s...
namely, i'd try multiplying those new forms of a and b to get the product of ab
then try proving that it can never be a square
you can also use numbers to convince yourself of the result and perhaps also get a hint on why it's true
solemn silo
Take this with a _grain of salt_ (i.e., I am not sure if this is correct).
Let a...
Here, instead of doing that "comparing" thing, can t i simply say (n^2+C)^2=(n^2+A)(n^2+B) ?
Because they are both ab
I mean that s what i assumed
cursive swan
Well yes, (n^2+C)^2=(n^2+A)(n^2+B)
solemn silo
Well yes, (n^2+C)^2=(n^2+A)(n^2+B)
So what do i get from here
cursive swan
I got what I wanted to, you asked whether this equation holds, I said yes, it's now your responsibility to do anything else
cursive swan
How? Did you again compare the coefficeints?
cursive swan
We have n^2(A+B) + AB = 2C n^2 + C^2
solemn silo
Yeah
cursive swan
Now what did you do.
cursive swan
How can you conclude that C^2 = AB?
solemn silo
We have n^2(A+B) + AB = 2C n^2 + C^2
I did /-2Cn^2 - C^2
Then factorized with n^2
Those with A+B and 2C
steel crest
I got lost , what are these constants ABC
steel crest
thats like saying
5^2 - 4\*5 - 3 = 2^2 + 2 - 4
and so -4 = 1 and -3 = -4
did you resolve this?
dreamy lichen
Take this with a _grain of salt_ (i.e., I am not sure if this is correct).
Let a...
it comes from this
dreamy lichen
did you resolve this?
yes, they are currently trying to use the algebraic approach, without repeating that mistake. (which i honestly think is kinda hopeless)
grand pondBOT
solemn silo
Oh , so i resolve the 2nd grade eequation?
dreamy lichen
perhaps it's time to switch strategies
steel crest
I don't know the solution, but I'm wondering if maybe it would be easier to consider
$$n^2 + 1 \le a + 1 \le b \le n^2 + 2n +2$$
grand pondBOT
gfauxpas
dreamy lichen
i did the prime factorization one (only in my head for now), but i think it works
ill just try writing it on a paper to check
steel crest
then breaking it up to
solemn silo
i did the prime factorization one (only in my head for now), but i think it work...
Please enlighten me with that
dreamy lichen
one could probably try to use prime factorization in some way
. I explained the beginning here
if you just play with it for a while, you should see what im going for
steel crest
hmm
dreamy lichen
take some numbers between e.g. 25 and 36 and try to make their product to be a square
and analyse why you fail
steel crest
yeah this might work
dreamy lichen
yeah, it does work, just scribbled it on a paper
dont fully spoil it, i think its worth trying to figure it out for at least like 10 mins
steel crest
$$n^2 + 1 \le a + 1 \le b \le n^2 + 2n +2$$
$$-n^2 -2n -2 \le -b \le -a-1 \le -n^2 -1$$
grand pondBOT
gfauxpas
dreamy lichen
**gfauxpas**
oh this
steel crest
I didn't finish it, but I think this would be possible, you would still make a prime factorization argument
solemn silo
**gfauxpas**
How does this help
dreamy lichen
honestly - i think the best way to approach this is to try finding counterexample, until you understand why you cant find one
solemn silo
honestly - i think the best way to approach this is to try finding counterexampl...
Wait so i tried
So i know that every perfect square is a summation of consecutive odd numbers
Can this help?
dreamy lichen
hmm im not sure how
solemn silo
That means a and b are not sum of consecutive odd numbers
dreamy lichen
summation doesnt play very well with the multiplication in a*b
solemn silo
Because they are not p.s.
steel crest
How does this help
working on if it helps