atomic idol
Alright thanks
#help-27
uncut crow
you can close it if you want
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stable moss
How to go about this $\int tan^{3}(\theta) \cdot sec^{3}(\theta)$
woven radishBOT
inheritance.
devout snowBOT
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1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
marsh trail
**inheritance.**
I think rewriting tan^2 in terms of sec helps as then only one tan will be left and then we can substitute sec as u
devout snowBOT
@stable moss Has your question been resolved?
thorn cloak
why is it when expressing ln(1+x) as a power series you dont have to put +C (when using integration)? or when ln(5+x) you know C=ln5 and so on
devout snowBOT
why is it when expressing ln(1+x) as a power series you dont have to put +C (whe...
young spade
what is ln 1
thorn cloak
no like i get why its not there anymore im just confused why in some functions you still have to write +C instead of evaluating at 0(?)
young spade
the +C comes from the integration itself, has to do with the fact that for differentiable f(x), f(x)+k has the same derivative.
In the particular case of ln(1+x), again, you can procede by evaluating at x = 0
ln(1+0) = C + 0 - 0/2 + + 0/3 ...
ln(1) = C
0 = C
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solid flicker
Can anyone explain the construction in dis
devout snowBOT
Can anyone explain the construction in dis
gloomy aurora
What part of it didn't you get?
small rampart
for questions that wants you to proof a function's continuity or discontinuity are you meant to use epsilon delta proofs most of the times?
devout snowBOT
for questions that wants you to proof a function's continuity or discontinuity a...
sand quarry
for questions that wants you to proof a function's continuity or discontinuity a...
no
rough nova
for questions that wants you to proof a function's continuity or discontinuity a...
It's one of the ways, but not used most of the time
sand quarry
most people avoid epsilon-delta whenever possible
i have never seen an eps-delta proof of discontinuity actually
that just sounds really bad
polar chasm
for questions that wants you to proof a function's continuity or discontinuity a...
Epsilon delta is what defines continuity, but usually they are used mainly to prove some theorems about continuity, which you can then apply more easily
lavish prawn
oh actually using eps delta for discontinuity is really not that bad for simple functions
polar chasm
for example continuity of most common functions follows by the fact that continuity is preserved after addition, multiplication, inverse... and continuity of some basic functions such as e^x
for discontinuity, e.g. the sequential definition of continuity is quite useful - it suffices to find one sequence xn -> x, such that f(xn) doesnt tend to f(x)
lavish prawn
there's like 20 ppl responding to thsi question so i don't wanna add too much more confusion
but i think if you find the epsilon-delta way of proving (dis)continuity very confusing it might be worhtwhile to try and force yourself to apply it in a simple scenario
sand quarry
even before using the sequential criterion doing the regular methods like checking if the limit matches the function, comparing left and rgiht limits is good enough
lavish prawn
like showing that something like f(x) = 1 for x > 0, and f(x) = 0 for x <= 0 is discontinuous in 0 is super easy with eps-delta if you understand what you're doing
but yea as with everything in math, at some point you stop going back to those basics and just use all the easy machinery that has been built from it, i.e. the other answers to this question
small rampart
So what methods are there to proof continuity and discontinuity
other than epsilon delta
lavish prawn
i think MathIsAlwaysRight's answer covers it
polar chasm
even before using the sequential criterion doing the regular methods like checki...
if you have all the limit machinery already, it suffices to check the limits
for some functions such as sin(1/x) as x -> 0, you probably dont have a simple way to conclude that the limit doesnt exist, so thats where you could use the sequential criterion (finding a sequence converging to 0, whose functions values dont converge to f(0))
polar chasm
for example continuity of most common functions follows by the fact that continu...
for proving continuity, you can often use this:
if f and g are continuous, then:
f + g is continuous
fg is continuous
f - g is continuous
c*f is continuous
f / g is continuous, if g isnt equal to 0 anywhere
f^-1 is continuous (if the inverse exists)
f o g is continuous, if well-defined
together with continuity of some basic functions such as
e^x, c, x, trig functions, ...
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graceful monolith
yo how were they able to go from there to there
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yo how were they able to go from there to there
sand quarry
,, \41{\32}x^{\512}=\4{\3x}{\32}=\3{\4x2}
woven radishBOT
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marsh trail
here too
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tidal sable
Y does no 6 have a fractions amount of sides of a polygon
devout snowBOT
Y does no 6 have a fractions amount of sides of a polygon
tidal sable
Ohh ok ty
neon folio
you're welcome
tidal sable
How do u do 13
How do I get the midpoint
I tried making lines from each point of a to a point of b and getting the tangents
They didn’t meet at a pt
neon folio
well
so first you enlarge A by a factor of 3
and then you move it right, then up
<@&268886789983436800>
ebon coyote
or does it have to be 1 step only
what do you think /s
ebon coyote
I tried making lines from each point of a to a point of b and getting the tangen...
tangents?
There're no circles here
Assuming you mean you tried to draw lines connecting corresponding points of A and of B together, did you try extending those lines to see where they met?
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glossy dew
no idea
finite fable
Ig you can use x³+y³+z³ identity
glossy dew
🤨
devout snowBOT
What step are you on?
1. I don't know where to begin.
2. I have begun but got stuck midway.
3. I got an answer but I was told that it's wrong.
4. I got an answer and would like my work checked.
5. I have a question about someone else's work/solution.
6. I have completed the problem and don't need help anymore. Thank you.
7. None of the above
lambda1 + lambda2 + lambda3 = Tr(A) = 6
Lambda31 + Lambda 32 + Lambda33 = Tr(A³) = 36
glossy dew
okk wait so lambda is an eigenvalue?
finite fable
Mhm
Use x²+y²+z² identity
Then the cube one
Do you get what I am saying? Coz I am bad at explaining
I don't want to give the answer. Just hints 😭
glossy dew
ok i understand the steps after that but im already a bit unclear in the entire eigenvalues part
ill just try searching on it then and come later
thanks
.close
upper gazelle
If the function f is defined by f (x) = [ \begin{cases} \text { } 0 \text{if x is rational} \ \text{} 1 \text{if x is irrational} \end{cases} ] then prove that $\lim_{x\to 0} f(x)$ does not exist
woven radishBOT
Toaster
trail eagle
upper gazelle
Contemplating my life choices
wait
But since the irrationals and rationals both have infinite densities then the Dirichlet function is nowhere continuous?
trail eagle
Indeed.
upper gazelle
So if epsilon is 1/3rd then due to the densities of the rationals and irrationals x cannot be 0 or 1 therefore it is undefined?
trail eagle
Yeah well any epsilon < 1 would not work.
Another way to see it is that given a point c, then you can always get a sequence x_n -> c of only rationals and a sequence x_n -> c of only irrationals. Both of those would give you different limits for the function.
upper gazelle
Another way to see it is that given a point c, then you can always get a sequenc...
Ahhh I see
ok I have another q
Prove that f is continuous at a if and only if $\lim_{h\to 0} f(a+h) = f(a)$ (only using epsilon delta)
faint hearth
upper gazelle
Ok, for this question I don't understand because for an $\epsilon > 0$ then $\exists\delta > 0$ such that if $0 < |h-0| < \delta$ then $|f(a+h)-f(a)|<\epsilon$ and I have no idea what to do with that
faint hearth
what is your definition of continuity? is it in terms of sequences or limits
upper gazelle
limits
faint hearth
okay, so can you state that definition?
f is continuous at a if ??
upper gazelle
f is continuous at a if $\lim_{x\to a}f(x) = f(a)$
woven radishBOT
Toaster
faint hearth
**Toaster**
okay, now use the epsilon delta definition for limits to expand this definition
upper gazelle
for an $\epsilon > 0, \exists\delta>0$ such that $0 < |x-a| < \delta$ then $|f(x)-f(a)|<\epsilon$
woven radishBOT
Toaster
faint hearth
**Toaster**
now do you see how this is secretly the same as the epsilon delta definition you just wrote?
upper gazelle
now do you see how this is secretly the same as the epsilon delta definition you...
?
faint hearth
lim_{h-> 0 }f(a+h) = f(a) means:
for every eps > 0 there exists delta >0 such that 0<|h| < delta implies |f(a+h) - f(a)| < eps.
the above is the same as:
for every eps > 0 there exists delta >0 such that 0 < |x - a| < delta implies |f(x)-f(a)| < eps
which is the same as:
lim_{x -> a}f(x) = f(a)
which is the same as:
f is continuous at a
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severe prairie
Yo guys i kinda forgot what to do when i get the rankings..
devout snowBOT
Yo guys i kinda forgot what to do when i get the rankings..
severe prairie
I need the test stats right
And then add em snd divide by two
A no
I need the smallest one
Is it correct?
robust bobcat
for my inductive step I have $\frac{2\sqrt{n^2+n} + 1}{\sqrt{n+1}}$. I know that $(n+1)^2 > n^2+n \implies \sqrt{n+1} > {\sqrt{n^2+n}$ but I am unsure how to proceed from here
devout snowBOT
for my inductive step I have $\frac{2\sqrt{n^2+n} + 1}{\sqrt{n+1}}$. I know that...
woven radishBOT
BigBen
Compile Error! Click the 
reaction for more information.
(You may edit your message to recompile.)
jaunty mantle
Do you have the rest of your working
devout snowBOT
@robust bobcat Has your question been resolved?
robust bobcat
Do you have the rest of your working
Yes so induction step is: $\sum_{k=1}^{n+1} k^{-1/2} = \sum_{k=1}^{n} k^{-1/2} + \frac{1}{\sqrt{n+1}} < 2\sqrt{n} + \frac{1}{\sqrt{n+1}}$
woven radishBOT
BigBen
faint gorge
So the idea is to do some "backwards" work basically, you will see the inequality is satisified for every n, so you can use the <= to show why the inequality holds
woven radishBOT
faint gorge
You can also square both sides since both sides are not negative for n>=1
@robust bobcat in case you weren't pinged mb
robust bobcat
so for your right side are you saying that $2n+1 > 2\sqrt{n+1} > 2\sqrt{n}\sqrt{n+1}$
woven radishBOT
BigBen
faint gorge
I just multiplied by sqrt(n+1)
robust bobcat
ye so your saying that if I can show your right side then that implies what we want to show
faint gorge
Yes
robust bobcat
ok so what I was saying is we know $\sqrt{n+1} > \sqrt{n}\sqrt{n+1}$ = $2\sqrt{n+1}> 2\sqrt{n}\sqrt{n+1}$
faint gorge
you have a [ in the begin
woven radishBOT
BigBen
robust bobcat
then we also know that $(2n+1)^2 > 4(n+1) \implies 2n+1 > 2\sqrt{n+1}$ and $2n+1 = 2(n+1)-1$
woven radishBOT
BigBen
faint gorge
**BigBen**
Not quiet sure I can follow
robust bobcat
where did I lose you?
faint gorge
Like for surfficiently large n sqrt(n)sqrt(n+1) will not be less than sqrt(n+1)
It's like quadratic vs linear growth
faint gorge
lmao
woven radishBOT
robust bobcat
We know $n+1> \sqrt{n}\sqrt{n+1} \implies 2n+1 > \sqrt{n}\sqrt{n+1} \implies 2(n+1)-1> \sqrt{n}\sqrt{n+1}$
woven radishBOT
BigBen
faint gorge
From where do we know n+1>sqrt(n)sqrt(n+1)
robust bobcat
well $(n+1)^2 = n^2+2n+1$ and the other one squared is $n^2+n$
faint gorge
Find the backwards steps to get there
faint gorge
**BigBen**
Ok yeah but that ought to be shown as well that this follows from our assumption n>=1
robust bobcat
ok so we just say that $(2n+1)^2> 4(n^2+n)$ square root both sides and then we can solve what you wrote by adding one to the right and dividing by $\sqrt{n+1}$
woven radishBOT
BigBen
faint gorge
Yes, but you start from n>=1 not from (2n+1)^2> 4(n^2+n)
faint gorge
Basically resolving the inequality should yield an equivalence anyway which would include the <= too
so that you start from a trivial truth
like 1>0
robust bobcat
Yes, but you start from n>=1 not from (2n+1)^2> 4(n^2+n)
isn't that a given though?
robust bobcat
wdym the problem asks us for n>=1
faint gorge
yes
robust bobcat
which means I am only looking at those n. Not any n<1
faint gorge
That's not what I meant
(2n+1)^2> 4(n^2+n) what makes you think this was true for n>=1 in the first place
this is the missing step
robust bobcat
you can factor it out
faint gorge
Yes but you need to show that
You can't just claim that's true for n>=1
even if it seems trivial for you
robust bobcat
no I understand where your going. We say that 1>0 add 4n^2+4n to both sides and then were good.
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vast hazel
help
vast hazel
i know im being stupid but i like
ig idk exponent rules anymore??? why is it 8 when u plug in ln5 and ln3 what 😭
sand quarry
,, e^{2\2\log(5)} = e^{\log(25)} = 25
woven radishBOT
sand quarry
try to work out the rest on your own
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sand quarry
\textbf{Question:} Let $X$ be a simple graph with $n$ vertices and $e$ edges. If $\lambda$ is an eigenvalue of the adjacency matrix $A$ of $X$, show that $\abs\lambda \le 2e(n-1)$
\medskip
\textbf{My work so far:} It is easy to find a weaker bound of $\abs\lambda \le\3{2e}$ since $\tr(A^2) = 2e$, but I am kinda stuck on how exactly to introduce the vertices to get the same bound in the problem. Hints?
devout snowBOT
\textbf{Question:} Let $X$ be a simple graph with $n$ vertices and $e$ edges. If...
woven radishBOT
devout snowBOT
@sand quarry Has your question been resolved?
worthy raft
Hmm, maybe use the fact that there are n dimensions and then =< n eigenvalues?
sand quarry
uhhh but what exactly does that tell me
wait wait wait wait
I am onto something I reckon
worthy raft
Oh?
sand quarry
,, \lambda + \sum_{i=2}^n\lambda_i = 0 \Implies \sum_{i=2}^n\lambda_i = -\lambda
woven radishBOT
worthy raft
Yea
sand quarry
,, \lambda^2 +\sum_{i=2}^n\lambda_i^2=2e \Implies \sum_{i=2}^n\lambda_i^2 = 2e-\lambda^2
woven radishBOT
sand quarry
aight one million dollar question now
how do we use this to bound it
uhhh
im thinking of cauchy-schwarz bcuz of the sum^2
but idk if that would work
sand quarry
why not i guess
worthy raft
Well if $|\lambda | \ge 1$ then $|\lambda | \le |\lambda^2 |$ right?
sand quarry
i dont think we can say that
worthy raft
Texit come on
sand quarry
remove whitespace around $
woven radishBOT
NotABot
worthy raft
Alright... then what's the most a vector can change?
Entries in the matrix are dependent on the number of edges right? And vectors represent some combination of the vertices
sand quarry
what vectors are we talking about, to be clear
worthy raft
An eigenvector I suppose
But an eigenvector can be written in terms of the vertex representation/basis
sand quarry
well i dont think we need to use the eigenvector to bound this
sand quarry
But an eigenvector can be written in terms of the vertex representation/basis
i dont disagree
but i dont see what you are trying to say
worthy raft
The question is to put a bound on the eigenvalues right?
Eigenvalues measure how much eigenvectors change
devout snowBOT
@sand quarry Has your question been resolved?
spiral raptor
Once, me and my friend was solving a math which had integrals. If I'm not wrong, it was similar to integral of (1-x^2)/(X^4+1), but its probably something different. When I solved it, I got the answer in inverse trignometry. But, when my friend solved it, he got the answer in logarithms. Somehow, both answers were correct. Is there something I am missing, or did one of us make a mistake?
devout snowBOT
Once, me and my friend was solving a math which had integrals. If I'm not wrong,...
stone stump
do you know complex numbers?
spiral raptor
Yes
stone stump
do you know e^ix=cosx+isinx?
spiral raptor
Yes
stone stump
so this basically says that exponentials and trigs are the same. so logs and inverse trigs are also the same
spiral raptor
Ohhh. So its possible to express logarithms as inverse trig functions?
mystic scarab
The other way round, I'd say
spiral raptor
see what happens when you add e^ix and e^-ix
Oh. So you can also express trig functions as only exponential?
stone stump
yes
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proven anchor
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lunar harbor
triangle inequality 
also make sure that you don't reuse n for all three sides
Hello?
faint gorge
We thought u went undercover
proven anchor
I'm waiting for frowny awesome trick
topaz axle
if the largest side is 19, the others are 18+3, 17+4, 16+5... 16 triangles, also 19,19,2 so 17
if the largest side is 18, the others are 17+5, 15+7, 13+9, 11+11... 7 triangles
17: 16+7, 15+8... 10 plus 17,17,6
16: 15+9, 13+11; 4 triangles
15: 14+11, 13+12, 12+13, 11+4 plus 15,15,10
14,13,13
opens trick
it's bruteforce
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vital edge
if the largest side is 19, the others are 18+3, 17+4, 16+5... 16 triangles, also...
!nosols
devout snowBOT
if the largest side is 19, the others are 18+3, 17+4, 16+5... 16 triangles, also...
As a helper, please do not give out answers that could be copied as a homework solution. Have the student work through the problem themselves and guide them along the way.
Available help channel!
gaunt trail
devout snowBOT
gaunt trail
part 5 and 6 i need help with
dull parrot
part 5 and 6 i need help with
You can use vieta's formula for part 5
dull parrot
im lay man
do you know what the sum of the root will be?
dull parrot
alright cool
and what would be the the sum of the products of the roots taken two at a time
gaunt trail
11
dull parrot
nice
now you can use this something like this $\sum z_i^2 = \left( \sum z_i \right)^2 - 2 \left( \sum_{i < j} z_i z_j \right)$
woven radishBOT
MxRgD
gaunt trail
how? there are 6 terms, not three, how can i use this
dull parrot
wdym by three terms
dull parrot
yeah
gaunt trail
what formula you gave me can only be used for (a+b+c)^2 isnt it?
wait
no
ur right
dull parrot
what formula you gave me can only be used for (a+b+c)^2 isnt it?
no it can be used for any amount of terms, the one I gave is more general
gaunt trail
i get it
can you tell me how i can write lie that
the texit thing you did
i found answer 78 then
dull parrot
you can put dollar signs like this $[anything]$
woven radishBOT
MxRgD
dull parrot
$hi$
woven radishBOT
MxRgD
gaunt trail
no i mean the mathematical notations
dull parrot
$\sum$?
woven radishBOT
MxRgD
dull parrot
as an example
usually you put \ for commands
so like $sin(x)$ is different to $\sin(x)$
woven radishBOT
MxRgD
gaunt trail
is there are channel where i can get the list of nomenclature
dull parrot
is there are channel where i can get the list of nomenclature
you mean #latex-testing and #latex-help ?
gaunt trail
$\sum arccot(x)$
woven radishBOT
Heaven Refining Demon Venerable
gaunt trail
you mean <#844681108473118750> and <#840667252793802752> ?
oh thanks
can we move on to part 6
dull parrot
can you explain it to me
basically if a continuous function (like a polynomial) changes sign between two points a and b, there must be a root between those two points
gaunt trail
gaunt trail
basically if a continuous function (like a polynomial) changes sign between two ...
so basically mean value theorem?
how is it different?
oh if it changes sign
o
k
gaunt trail
mvt means if c lies between a and b then f(c) lies btw f(a) and f(b)
for a continuous fn
can anyone else help me??
.close
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proven anchor
z^n=(1+z)^n
devout snowBOT
z^n=(1+z)^n
proven anchor
roots of this equation and what realz will be
rain summit
roots of this equation and what realz will be
you see where both sides have a power of $n$?
woven radishBOT
1 divided by 0 equals Infinity
hasty cargo
roots of this equation and what realz will be
trivial n = z = 0 (if you define 0^0 = 1)
finite fable
trivial n = z = 0 (if you define 0^0 = 1)
but 0^0 aint 1
hasty cargo
but 0^0 aint 1
it is?
according to some definition exponentiation is just counting how many ways a set can map to another set, the empty set to the empty set has only 1 mapping so 0^0 = 1?
finite fable
it is?
see a^0 = 1
and 0^a = 0
so 0^0 suggests two answers
in elementary arithmetic
rain summit
yo guys
finite fable
it is undefined
rain summit
wait for the OP to respond
hasty cargo
z^n=(1+z)^n
there are more solutions tho , z = 1/2 and any even n>2 should work
proven anchor
there are more solutions tho , z = 1/2 and any even n>2 should work
How did you get it
0^0 is undefined
Z=0 should be excluded
hasty cargo
How did you get it
(a^x)^y = a^(xy), so you can factor out the 2 in the even n, then by abusing the fact that squaring n and -n gives the same number we can prove [(1/2)^2] and [(-1/2)^2] gives the same thing
woven radishBOT
1 divided by 0 equals Infinity
hasty cargo
you notice that both sides have a power of $n$?
twin using latex for this is overkill
rain summit
twin using latex for this is overkill
-# ||making discord's server management heavier and heavier anytime||
finite fable
(a^x)^y = a^(xy), so you can factor out the 2 in the even n, then by abusing the...
i was gonna do roots of unity. but this works as well.
proven anchor
n is positive integer
dapper tiger
the only real solution should be z = -1/2 when n is a positive integer
z^n=(1+z)^n
1/2^n = 1/2^n when z = -1/2
1/2^n is not (3/2)^n when z = 1/2 in the first place
hasty cargo
z^n=(1+z)^n
1/2^n = 1/2^n when z = -1/2
1/2^n is not (3/2)^n when z = 1/2 in the...
oof i forgot the -
i misread and thought it was 1-z lmfao
dapper tiger
and I think you can probably prove something like Re(z) is -1/2 in general case so that -1/2 becomes trivially the only real solution
hasty cargo
and I think you can probably prove something like Re(z) is -1/2 in general case ...
tbh i was going off the dividing n by 2 method
not rigorously proving it like this
dapper tiger
I mean if I understand what you intend to do, you are rigourous
it's just a bit of a pain to conclude about 2^n cos(ntheta/2)
kind gust
z^n=(1+z)^n
i think z = 0 is not a solution since 0^n not equal to (1+0)^n
proven anchor
divide n by 2
Buyt why?
proven anchor
if n is not divided by 2 then
could you clarify?
@hasty cargo
z^n=(1+z)^n
z^3=(1+z)^3
Thanks
.close
devout snowBOT
kind gust
Thanks
hold on going through the method its taking a bit of time to do it on paint sorry
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Available help channel!
dapper tiger
oh I found an elegant solution
devout snowBOT
oh I found an elegant solution
kind gust
oh I found an elegant solution
the solution im using is roots of unity
dapper tiger
yeah sure do your thing
rain summit
idk if we're talking about the complex part here
but if in reals then sure $z = \frac{-1}{2}$
woven radishBOT
1 divided by 0 equals Infinity
kind gust
the op asked for the real roots
ohh i didnt see that 😭
hasty cargo
if n is not divided by 2 then
Twin. use n/2 * 2 = n
devout snowBOT
rain summit
oh damn
hasty cargo
not even needed
Pretty sure we are at different levels of math and my logic is certainly not your logic
But thanks
dapper tiger
?? |z| = |1+z| is enough to answer op
proven anchor
.reopen
dapper tiger
not about logic mate, you can solve algebraically from there if you want
proven anchor
?? |z| = |1+z| is enough to answer op
Yes
kind gust
?? |z| = |1+z| is enough to answer op
its probably better then my method actually i think since you get it directly without having to toss through all the other values
kind gust
roots of this equation and what realz will be
do you want the other complex solutions as well or just want reals
dapper tiger
well, good news, the complex ones are conjugate and we already know their real parts
bad news they are a pain to write
kind gust
well, good news, the complex ones are conjugate and we already know their real p...
wait could you check my solution i just want to make sure ive done it right
dapper tiger
I mean I can give a look but I haven't checked them myself since it wasn't asked for
kind gust
alr ty
proven anchor
kind gust
the imaginary part should be of the form cot(pi*k/n)/2
i didnt manage to simplify it down xd
proven anchor
nth roots of unity
kind gust
dapper tiger
I think you should use exp form rather than algebraic
dapper tiger
because from z = w/(1-w) you get e^(itheta)/(1-e^(itheta))
proven anchor
And theta is 2kpi so only cos gives
dapper tiger
the cot comes from there
proven anchor
I got it z=1/(costheta-1)+isintheta
kind gust
because from z = w/(1-w) you get e^(itheta)/(1-e^(itheta))
so you write it like w = 1+1/z
then like you write it back in terms of z
z = w/(1-w)
dapper tiger
yeah
w is nth root of unity
and you have the identity e^(itheta)/(1-e^(itheta)) = -1/2 + i*cot(theta/2)/2
you can prove it by multiplying by e^(itheta/2) on num and denom
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dapper tiger
and then apply it to w
devout snowBOT
and then apply it to w
kind gust
you can prove it by multiplying by e^(itheta/2) on num and denom
so thats why you could just cut off the imaginary bits and get the -1/2
dapper tiger
it's the part that simplifies yeah
dapper tiger
still easier to see from |z| = |1+z| tbh
kind gust
Can we move to <#359052581022203914>
it says no access for me
thick lotus
<id:customize> and choose this?
studying role removes math - discussion as of now
devout snowBOT
kind gust
z^n = (1+z)^n => |z| = |1+z|
this still looks sick
so if i wanted to get the real solution only for a problem like this in general i could just do this?
willow helm
gonna open a new channel for the third time now
sharp yew
Can I ask a question
neon folio
yes
devout snowBOT
Asking the actual question right away is more likely to get responses.
Asking "Can I ask...?" or "Does anyone know about...?" doesn't give people enough information to decide whether they can help, and answering can feel like a promise to help with the actual question, which they might find themselves unable to.
graceful cosmos
This channel just got closed, you want one that's ready for you
neon folio
also in the future don't delete your first message as the channel will close
graceful cosmos
We're gonna get locked out any second
crystal dawn
tbf it wasn't this OP that nuked the initial msg I presume
but yeah time to move, Infinity
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buoyant verge
Hi
buoyant verge
Yesterday I tried to proof that this subset of X was indeed a topology, after sending my work to my professor he cleared to me that I haven't used the properties of this subset to proof if it is a topology. So I tried to reformulate my work and reached here:
I will only send the proofs for the arbitrary union axiom and the finite intersection axiom, which are the parts my professor disliked
Arbitrary union axiom
And this is the proof for the finite intersection:
Are this proofs valid? thanks in advance
atomic osprey
Arbitrary union axiom
This sounds like you're assuming what you're trying to prove; you want to prove that the union U_a C_a is an element of T. Also, I think you're mixing up elements of T and elements of R. You wrote "such an element is in the topology", but an element of U_a C_a is an element of R, not T
sullen island
The main issue is that your proof is fluff, all talk no show
infinite unions of open sets maintain the structure ...
Why tho ? That's exactly the thing you need to show, you’re just saying it's true without justific1tion
Like it's not too hard to give explicitly what A and B tell you that U_aC_a is an open set using the As and Bs of each C_a, do so
buoyant verge
Give me a min and I'll send it
This is for arbitrary union axiom
This are one of my firsts proofs so take it with calm please xD
sullen island
That is essentially the idea yes
Tho you need to show it for arbitrary unions, not just union of 2
buoyant verge
okay
let me do it correctly
btw, thanks for the help
Give me one min and ill send it back
Would this be valid?
sullen island
It’s better but your union still isn’t arbitrary, it’s finite
Like you defined the C_a earlier, use them
buoyant verge
isn't it infinite countable?
buoyant verge
Like you defined the C_a earlier, use them
you mean using the index?
sullen island
isn't it infinite countable?
No, n can grow arbitrarily big sure, it’s still a finite number of things you’re unioning each time tho
And your proof should still work for uncountable unions anyway
buoyant verge
buoyant verge
Up in the paper there is the rest of the proof which hasn't changed since this image
sullen island
And what are Aa and Ba ?
buoyant verge
sullen island
So each Ca = Aa U Ba right ?
buoyant verge
yes sir
sullen island
Write it then
sullen island
What alpha are you talking about on the RHS tho ?
buoyant verge
what is RHS?
sullen island
Right hand side of your equation
buoyant verge
it is the same alpha as in the left side I mean, it's arbitrary right?
sullen island
So idk what that right side is supposed to mean here
buoyant verge
well, the inside of the union is the definition of open set for this topology
sullen island
and?
buoyant verge
good question xD
sullen island
Still stuck in square 1 aren’t we
buoyant verge
Do I have to say that the union preserves this structure?
Do I have to expand the expression?
sullen island
You need to write this whole union as (something in Tu) U (something inside Q)
Essentially yeah
lunar harbor
sullen island
Find A in Tu and B inside Q such that this equation is true
That's how you show that the union fits the def of an open set here
buoyant verge
this is equal to my definition of open set in my topology
sullen island
what
buoyant verge
Find A in Tu and B inside Q such that this equation is true
But, this applies for every A and every B
but A U B like that is my whole topology
I don't know if im following right
sullen island
You want to show that the union of open sets is an open set
buoyant verge
yeah
sullen island
An open set here is a set C such that there exists A in Tu , B in Q with C = A U B
buoyant verge
yeah
sullen island
So if the union is to be an open set you need to find such A and B such that Ua Ca = A U B
That's what it means to be an open set right
These A and B will most certainly depend on the Aa and Ba we defined for each Ca
sullen island
And?
buoyant verge
that i could choose every a and b
i mean, all a and b are valid
i don't have to find any because all satisfy
sullen island
Well no we want A U B to equal the union of Ca
You can't just pick any A and B and expect that to be true
buoyant verge
Well no we want A U B to equal the union of Ca
okay
I see
I was seeing it the other way around
For that to be true we have to pick the biggest open set A which contains all the other open sets which are in Tu
sullen island
So A = all of R ?
buoyant verge
because our topology is A U B where A in Tu and B subset of Q, therefor the union of Ca should be the union of all R and Q?
sullen island
No
sullen island
Yeah
It has to work for whatever Ca you pick
And it has to be an actual equality each time
Not just UC_a included in A U B, that's a copout
So picking A = R, B = Q always doesn’t work, as I said even with only one open set in the union it doesn’t work, there are open sets in T other than R, these are all counterexamples to your claim
buoyant verge
So picking A = R, B = Q always doesn’t work, as I said even with only one open s...
So my claim does not work?
buoyant verge
So my claim does not work?
So this is false
buoyant verge
Okay, I though that you were saying that the statement was wrong
couldn't I pick a open set from the topology and eliminate Q from it
and the same goes the other way?
A = Z in (T\Q)
B = Y in (T\Tu)
?
sullen island
I don’t get why you think these A and B should not depend on the C_a in any way
buoyant verge
You have to know that I'm stupid, thats the case xD
So may I say A = z in (C_a \ Q) and B = y in (C_a \ T_u)
?
sullen island
What does that mean ?
buoyant verge
that are my A and B xD
for the union
or
may A and B just be elements of C_a
_
buoyant verge
sorry, I'm a bit lost with topology as you may see :/
buoyant verge
As in ?
as in ? xD
elements of C_a which is the same as saying elements of the union of (A_a U B_a)?
sullen island
as in ? xD
Which things in Ua Ca go in A and which ones go in B ?
buoyant verge
what do you mean what things in the union go in A and which ones go in B?
Isn't it the opposite
like
you input A and B into C_a?
buoyant verge
Which things in Ua Ca go in A and which ones go in B ?
elements of the usual topology go to A and subsets of Q goes to b
buoyant verge
How would you write that mathematically ?
$A \in T_u$ , $B \subseteq \mathbb{Q}$
I don't know how to write what you are asking me xD
woven radishBOT
S0S4 - Feel free to ping
buoyant verge
like
the things in Ua Ca that goes in A and in B
I think there is a language misunderstanding on my part
i am not a native english speaker
sullen island
Ok i’ll be more explicit
From all the Ca, you have a whole bunch of Aa and Ba
Can you write suitable A and B using these Aa and Ba ?
buoyant verge
mmmm
sullen island
Like the answer is just repackaging these Aa and Ba inside A and B
woven radishBOT
S0S4 - Feel free to ping
sullen island
Something a bit like that yes
woven radishBOT
S0S4 - Feel free to ping
sullen island
Indeed
sullen island
So just union all of these Aa to get A
Cause Tu is a topology and the Aa are all in Tu, then Ua Aa is in Tu
Do the same with the Ba, B = Ua Ba
Then A U B is indeed Ua Ca, we’ve just shuffled the Aa and Ba around
buoyant verge
sullen island
It’s sufficient if the person reading your answers in the exam isn’t a hardass but you never know
Cover your ass as ppl say
buoyant verge
so should I just specify that A is a element of Tu?
and the same for B?
oh weell
i say it in the proof
this is the fool proof
sullen island
Ok right with the first paragraph it’s understandable
buoyant verge
okay
but this is just for the first axiom
omg
xD
Now we need to go for the other
im going to kms
sullen island
sullen island
At least for the intersections you can just pick two open sets
buoyant verge
for the intersection axiom is it valid what I did before?
yeah
or say, given a n in the naturals
?
sullen island
Yeah but intersection of 2 implies intersection for all nats
Cause induction yada yada
sullen island
And doing it for arbitrary finite sounds like a pain in the ass here
So let’s stick with 2
buoyant verge
So let’s stick with 2
stick two in the ass?
joking
lets see xD
mods pls dont ban me
let me see how I do this
sullen island
If you were banned for saying ass, I’d be banned a thousand times
buoyant verge
btw, how common is that you get headaches when doing proofs? xD
sullen island
If you were banned for saying ass, I’d be banned a thousand times
Apparently only 69 times looking at my whole record here
nice
sullen island
btw, how common is that you get headaches when doing proofs? xD
Do you tend to get headaches for other things ?
buoyant verge
no xD
just proofs
like, proving makes me use the 150% of my brain for a long period of time
as you can see a stupid proof takes me hours
and thats having help
buoyant verge
Maybe there's been some studies done
"Monkeys get headaches after doing basic proofs"
sullen island
Monkeys have written all of shakespeare on a typewriter, now they're fking tired
buoyant verge
nice reference over there
buoyant verge
no fucking way
sullen island
Well we take two open sets C0 = A0 U B0 and C1 = A1 U B1
And we wanna find A and B such that C0 cap C1 = A U B
(cap is the intersection symbol)
echo zodiac
wat is 1+1
buoyant verge
And we wanna find A and B such that C0 cap C1 = A U B
this is like the last time right?
sullen island
Nope it’s different you screwed up the last line
It's unions in the parentheses
Not caps
buoyant verge
sullen island
Now the "refactoring" doesn’t work
buoyant verge
im going to kms
sullen island
That's why I suggested going with 2
sullen island
It’s not as easy as the first proof
Ua Aa and Ua Ba are pretty obvious choices in retrospect
sullen island
Here it’s gonna be less obvious
One nice thing to use here is that intersections distribute over unions
1% now
sullen island
(a+b)(c+d) = ab + ac + bc + bd
Same thing here
Replace + by union
And products are intersections
They work the same
@buoyant verge
My time prolly comes to an end
sullen island
You’re a big boy
I pushed you through the first thing
Hopefully you can emulate that for the second part
buoyant verge
Well, thank you very much, I'll try my best
sullen island
(a+b)(c+d) = ab + ac + bc + bd
And this is the trickiest step really
buoyant verge
Ill go have some ramen then came again to fight this proof
sullen island
If you use that you’re pretty much done
sullen island
bai
buoyant verge
.close
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static temple
Hello, I'm turning what feels like it should be almost an immediately-done function into a power series,
f(x) = [ -5 / (1 - 10x) ], and finding the interval of convergence.
I might be done, but I could use a check.
devout snowBOT
Hello, I'm turning what feels like it should be almost an immediately-done funct...
static temple
It feels too simple, I'm wondering if I got lazy with how I'm looking at r and too-quickly thought it looks like both the bounds of the convergence interval (-1/10 and 1/10) converge by the geometric series test, but I'm having trouble thinking of a reason why they wouldn't.
lunar harbor
It feels too simple, I'm wondering if I got lazy with how I'm looking at r and t...
Your common ratio is 10x, not x.
static temple
Your common ratio is 10x, not x.
I've assumed that by the ratio test, I want to have a value of | r | that's less than 1. is it wrong to divide my r < 1 by 10 on each side?
lunar harbor
I've assumed that by the ratio test, I want to have a value of | r | that's less...
My point is that you identified $r=10x$. But when testing convergence at the endpoints, you used a common ratio of $x$ instead.
woven radishBOT
Civil Service Pigeon
lunar harbor
Also why are you using the ratio test here
The geometric series test tells you the series converges if and only if $|10x|<1$, no need to test endpoints (why?)
woven radishBOT
Civil Service Pigeon
static temple
trying to think of how to word this, one sec
static temple
I think I was misremembering the thing with the ratio test, sorry, it probably just got used a lot in the last problems I've done finding the covnergence window for other series.
static temple
When I did those, I had to test the convergence of the endpoints because the ratio test is inconclusive at p = 1, so I think I mixed things up.
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If you are done with this channel, please mark your problem as solved by typing .close
static temple
Juust to be sure I'm all the way there, in this case I should be putting 10x in as the r for the geometric series?
lunar harbor
mhm
static temple
/asking "is the sky blue" voice
lunar harbor
Your common ratio is 10x, not x.
lunar harbor
/asking "is the sky blue" voice
eh I'm naturally quite terse in general lol
devout snowBOT
lunar harbor
No problem! This helps a lot. Have a good one
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Available help channel!
thick flicker
This is the first assignment my teacher assigned me and I'm completely lost
thick flicker
I would appreciate the help
gloomy aurora
well for starters, the domain is the set of all values of x such that f(x) is defined
Look at the first one.
when is f(x) undefined?
(hint its a fraction, so the denominator should be ..........)
thick flicker
Umm
gloomy aurora
the denominator cannot be zero, right?
thick flicker
yes
thick flicker
4+2x=0?
thick flicker
Its -2?
thick flicker
Anything except -2?
gloomy aurora
yes but we need to write that in a bit better format
gloomy aurora
Anything except -2?
does your school accept answers of this sort?
what did you do in class?
thick flicker
No its a bit informal
gloomy aurora
hm so i assume you know the basics of set theory and a bit of notation?
thick flicker
Yes
gloomy aurora
try to write it as the difference of two sets, A-B
,texsp$\mathbb{R} - {-2}$
oh wait mb
yes now it is correct
you must envelope 2 in a bracket
we can only subtract sets
not elements.
sand dove
-2 and not 2
woven radishBOT
Annie Maqionde
thick flicker
- -?
Is the correct answer supposed to be - (-)?
Oh wait
Im sorry
Got a little confused
There was 2 images
gloomy aurora
In case you don't know, I'll provide a list for what you need to check given a function to find its domain.\\
- If there is a fraction, check when the denominator is zero.\
- If there is a root/a term of the form $a^{1/n}$ where n is even, check always for $a>0$. Exclude values of $x$ such that $a<0$.\
- If there is a logarithm $log_a{N}$ then always $N>0$, $a=1$, $a>0$. Exclude those values of x for which this doesnt hold.\\
Here ofc, a and N may or may not depend on x(they usually do in given problems)
woven radishBOT
Annie Maqionde
thick flicker
Ill try working on it myself, thanks for the help!
gloomy aurora
!done
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thick flicker
.close
devout snowBOT
thick flicker
Appreciate it
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dry saffron
How does a Koch snowflake has infinite parameters the parameter increase gets smaller and smaller right? So it should be a number?
devout snowBOT
How does a Koch snowflake has infinite parameters the parameter increase gets sm...
gloomy aurora
if i remember correctly $P_n = 3 \times (\frac43)^n$
woven radishBOT
Annie Maqionde
gloomy aurora
the perimeter doesnt get smaller
polar chasm
How does a Koch snowflake has infinite parameters the parameter increase gets sm...
adding infinitely many decreasing changes doesnt mean that the result is gonna be finite
and yeah, the changes actually do get bigger i think. Youre tweaking it on a small scale, but you're tweaking lots of things at once (there are many corners)
dry saffron
I by 2. 1 by 4 1 by 8.... Sum is 2 right? So
polar chasm
I by 2. 1 by 4 1 by 8.... Sum is 2 right? So
Thats one example.
1 + 1/2 + 1/3 + 1/4 + 1/5 + ... = infty
dry saffron
Thats one example.
1 + 1/2 + 1/3 + 1/4 + 1/5 + ... = infty
Why for that it works but not the other one I infinitely it should reach a really smaller point?
gloomy aurora
A very good exercise on this would be to analyze the criteria for convergence of a series and where your argument falls.
dry saffron
Why for that it works but not the other one I infinitely it should reach a reall...
Basically 1 by 2 1 by 4 1 by 8. Is finite but not 1 by 2 1by 3 1by 4... Sum
sand dove
I by 2. 1 by 4 1 by 8.... Sum is 2 right? So
And also, the example of the Koch snowflake is one where each size of new sides you add may decrease exponentially to 0, but the amount of new sides you add increases exponentially to infinity
so it's not even 1 + 1/2 + 1/4...
more like 1 + 1/2 * 2 + 1/4 * 4 + ...
which is 1 + 1 + 1...
gloomy aurora
Basically 1 by 2 1 by 4 1 by 8. Is finite but not 1 by 2 1by 3 1by 4... Sum
Do you know how to check the convergence of a series?
sand dove
even worse
dry saffron
Do you know how to check the convergence of a series?
I just want to know why sum decreasing summation series is a number while the other one is infinity even thoug it's decreasing I mean if I sum numbers infinity that r decreasing I should be able to reach a really really small number that doesn't matter any more right?
Like how it is for 1by 2 1 by 4 1 by 8...
sand dove
The best definition we have for a series is the limit of its partial sums
while 1 + 1/2 + 1/4 + ... + 1/2^n is 2 - 1/2^n
dry saffron
Can u dumb down ur explanation to like a 13 year old?
sand dove
which goes to 2
dry saffron
Lots of unknown words
gloomy aurora
Can u dumb down ur explanation to like a 13 year old?
So do you know what a series is?
dry saffron
So do you know what a series is?
Series summation yes
gloomy aurora
do you understand the notation i used?
gloomy aurora
um so a partial sum $S_n$ is a sum of the form $S_n = a_1+......a_n$
woven radishBOT
Annie Maqionde
sand dove
No
Basically, while a series would be summing an infinite amount of terms in some order
In order to define that, we need to start at looking what happens when we sum the first "n" terms
gloomy aurora
A partial sum is a sum of only a finite number of terms.
sand dove
(Since it's only a part of the entire summation)
gloomy aurora
Now let's say we increase $n$. The value that $S_n$ approaches to can be regarded as the sum $a_1+ a_2+..........$ For simplifications, we'll assume $a_1>0, a_2>0,...$.\
What do I mean by approaches? As I increase $n$, the value $S_n$ becomes closer and closer to a certain value. This 'certain value' is called the limit of the partial sums of the series. If this 'certain value' is finite, the limit of the partial sums is said to exist\
Now observe one thing. Let's say the limit of the partial sums of the series is $S$. Then, you can see that $S_n = a_1+a_2+....a_n < a_1+a_2+....a_n+a_{n+1}+.......... = S$. So $S_n < S$. And now let's say some real number $k>S$ then $S_n < k$, right? So one very important rule is that:\\
If the limit of the partial sums exists, then the partial sums must be less than some finite value $k$
woven radishBOT
Annie Maqionde
gloomy aurora
If we show that for a particular sequence $a_1...$, no $k$ exists such that $S_n<k$ for ALL $n$, then our job is done.
woven radishBOT
Annie Maqionde
gloomy aurora
I'm sorry for the textwalls but you wanted a 13yo explanation, and unfortunately explaining this to a 13yo will involve a lot of text
polar chasm
As for the reason why 1 + 1/2 + 1/3 + ... specifically goes to infty:
1 + 1/2 + 1/3 + 1/4 + 1/5 + 1/6 + 1/7 + 1/8 + ... is certainly larger than
1 + 1/2 + 1/4 + 1/4 + 1/8 + 1/8 + 1/8 + 1/8 + ... now group the like-terms
1 + 1/2 + 2(1/4) + 4(1/8) + 8(1/16) + ... and this is
1 + 1/2 + 1/2 + 1/2 + 1/2 + ... which goes to infty
gloomy aurora
Well its the same thing: for the series $1+1/2+....$, there is NO such $k$ such that $S_n<k$ for all n
woven radishBOT
Annie Maqionde
dry saffron
I'm sorry for the textwalls but you wanted a 13yo explanation, and unfortunately...
So the infinity of 1 by 2 1by 3 1by 4..... Is a bigger infinity than 1by 2 1by 4 1by 8.....?
Like the first one will always match the 2nd series numbers but more late. And it's not enough. So the first one isn't fast enought and the numbers still big so it's infinity? So it matters how fast it's decreasing?
A lot of unknown words in that text so didn't understand can u just do English? I think that would be better
Chatgpt said some r infinity cuz it's not decreasing fast enough. Is it like gillbards hotel in veritasium video
gloomy aurora
So the infinity of 1 by 2 1by 3 1by 4..... Is a bigger infinity than 1by 2 1by 4...
- $1/2+ 1/4+ 1/8+.........$ converges. no infinity here
woven radishBOT
Annie Maqionde
gloomy aurora
idk what you're talking about
- that method of logic is extremely fallacious.
we don't 'compare' infinitie's' like that
polar chasm
And also, the example of the Koch snowflake is one where each size of new sides ...
Btw it needs to be said that this whole 1 + 1/2 + 1/3 + ... thing is quite a detour, jsyk. I only brought it up to correct one of the 2 gaps in your reasoning - that sum of increasingly smaller numbers needs to be finite
The reason why koch's snowflakes perimeter goes to infty is much simpler, look at rafi's explanation.
dry saffron
1) $1/2+ 1/4+ 1/8+.........$ converges. no infinity here
I meant the series is infinite
gloomy aurora
I meant the series is infinite
yes, so?
im going to use VERY vague language here, but both 'infinities' in this case are equal.
dry saffron
Like let's say the series of 1 by 2 1by 3 1by 4. Has all the numbers of 1by 2 1by 14 1by 8. But late so the the first series is a bigger infinite series than 1by 4 1by 8 one
gloomy aurora
Like let's say the series of 1 by 2 1by 3 1by 4. Has all the numbers of 1by 2 1b...
no.
what do you define by bigger infinite series?
if you say the number of terms, the number of terms are uhm uhm 'equal'(vague language but works)
polar chasm
So the infinity of 1 by 2 1by 3 1by 4..... Is a bigger infinity than 1by 2 1by 4...
Here is the simplest explanation of why some decreasing series go to 0 and some to inf i can give:
Take a stick of length 1. Go at 0 and move 1/2 to the right, then move half of the remaining distance and repeat. Obviously, this will result in a series that starts like 1/2 + 1/4 + 1/8 + ... and its sum will be finite - it will be 1
Now take a never-ending stick. Again start at 0 and walk a distance of 1m. Once you're at the 1m, decrease your step size to 1/2 and keep walking until you cover a distance of 1m (that's 2 steps). Now decrease your step size to 1/3 and keep walking until you cover a distance 1m, so 3 steps. And continue like this. Obviously, this way youll get as far as you wish - if you do enough steps.
This results in the sum 1 + 1/2 + 1/2 + 1/3 + 1/3 + 1/3 + 1/4 + 1/4 + 1/4 + 1/4 + ... and the sum of this series will be infinite - because it cant be finite, we can cover arbitrary distance.
gloomy aurora
thats actually offtopic here
dry saffron
*thats actually offtopic here*
I understand the Koch one the original one should I go to another channel for this question or continue?
gloomy aurora
you may continue ig. (idk)
dry saffron
Here is the simplest explanation of why some decreasing series go to 0 and some ...
But we are just assuming the sticks r finite and infinite from the start
polar chasm
But we are just assuming the sticks r finite and infinite from the start
in the first one, i always walked half of the remaining distance to the end of the stick
that means that id never go beyond that stick
so finite stick sufficed
you could probably remove the sticks completely
Lets just say that in the first case, I always walk half the distance to 1 and in the second one, i always keep walking at steps of size 1/n until i reach 1 meter and then I change n by 1, so my new step size is 1/(n+1)
dry saffron
Lets just say that in the first case, I always walk half the distance to 1 and i...
the 2nd series will always have the numbers of that halfing series so the 2nd series should also reach finite number? Cuz the nplus 1 series is just halfing series plus all the other numbers ? It will be a bigger finite number result ?
polar chasm
the 2nd series will always have the numbers of that halfing series so the 2nd s...
how big do you think could it be?
maybe it could be around a 100 lets say
first I walk 1
Then 1/2 + 1/2
then 1/3 + 1/3 + 1/3
1/4 + 1/4 + 1/4 + 1/4
1/5 + 1/5 + 1/5 + 1/5 + 1/5
at this point, ive walked 5 meters
if i did it all the way to 100 rows, i would be at 100 already
and I wouldnt be done!
so it certainly has to converge to something greater than 100
and you could do the same for any finite number
ill always surpass it, if i do enough steps
1 +
1/2 + 1/2 +
1/3 + 1/3 + 1/3 +
1/4 + 1/4 + 1/4 + 1/4 +
1/5 + 1/5 + 1/5 + 1/5 + 1/5 +
...
Notice that each row is literally 1. So im just summing infinitely many 1's, ofc thats not gonna be finite
dry saffron
first I walk 1
Then 1/2 + 1/2
then 1/3 + 1/3 + 1/3
1/4 + 1/4 + 1/4 + 1/4
1/5 + 1...
But it's not 1 by n into n. It's just 1 by n summation.
I just don't understand if 1 by n series sum reaches a really small number infinitely then it shouldn't matter if it increases or like 1 by 100 quintillion so then we can say it's just a finite number right? Just like how we do with the halfing one?
polar chasm
But it's not 1 by n into n. It's just 1 by n summation.
I just don't understa...
It absolutely does matter
polar chasm
As for the reason why 1 + 1/2 + 1/3 + ... specifically goes to infty:
1 + 1/2 + ...
for the 1/n summation, see this
there is no reason to expect that summing infinitely many, albeit very small numbers will always get us a finite result
just as there is no reason to expect that summing infinitely many very small numbers will always get us an infinite result
you need a better argument than "it seems like it should be finite"
devout snowBOT
@dry saffron Has your question been resolved?
jaunty mantle
So I’m trying to do this problem
devout snowBOT
So I’m trying to do this problem
vital edge
Any 1-D lie alg is abelian isnt it