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2:02 PM
Using Fourier transforms to solve integrals is hard
 
@N3buchadnezzar what do you mean? It sounds like "using integrals to solve integrals is hard"
 
@Nimza INDEED
Say I want to solve
$$ \int_0^\infty \frac{\omega \sin \omega}{\omega^2 + 1} \mathrm{d}\omega $$
Can I define a function $$ f(x) = \frac{\omega \sin \omega x }{\omega^2 + 1} $$ and then find its fourier transform with respect to x?
 
don't just do fourier transforms
 
@N3buchadnezzar not in ordinary sense... It isn't $L_1$ by $x$
 
@N3buchadnezzar is contour integration allowed?
 
2:14 PM
@robjohn No
But I have already figured that if
$$ f(x) = \left\{ \begin{array}{cccc} e^{-x} & \text{if} & > 0 \\ -e^{x} & \text{if} & < 0 \end{array} \right.$$
Then $$ \mathcal{F}(f(x)) = - \sqrt{\frac{2}{\pi}} \frac{i \omega}{\omega^2 + 1} $$
 
Is it possible to have a ring $R$ such that Spec $R$ is connected but reducible?
 
@N3buchadnezzar The answer is $\dfrac{\pi}{2e}$, but I used contour integration.
 
@robjohn Ahh. Thanks.
 
@N3buchadnezzar and Mathematica verifies it :-)
@JayeshBadwaik It is hard to believe that that passed inspection.
 
how does mathematica do it?
 
2:29 PM
@robjohn Yeah, I believe they meant finite interval in the least, probably closed.
 
@cassandra0 I don't know, but perhaps W|A would give the steps.
@JayeshBadwaik My counterexample is for [-1,1]
is that closed and finite enough? :-)
 
@robjohn Ahh. I did not see the example properly, just saw that error. I mean, even a sine function has infinite number of extrema. Hmm.
 
off to the park... bbl
 
later.
 
Maybe I can consider two intersecting curves...
 
2:46 PM
@N3buchadnezzar Out of curiousity, are you already done with a course in analysis?
 
@JayeshBadwaik Not with rudin no, still undergrade.
 
@N3buchadnezzar What is the name of the course you are currently doing then?
 
@N3buchadnezzar hmm. okay. just asked. anyway, going for dinner now, will be back later.
 
3:05 PM
im bored
 
user19161
@cassandra0 Talk to me then.
 
@cassandra0 Help me prepare for my exam ^^
 
how
 
Know anything about Laurent series?
 
3:09 PM
Gosh darnit
 
looks important
> the substitution z=e^{\pi i w} transforms a Laurent series into a Fourier series
that's interesting
or is it trivial?
 
Basically I am given that $$ \frac{1}{1-z-z^2} = \sum_{n=1}^\infty C_n z^n $$
 
why are you starting at n=1?
 
@cassandra0 Because the problem says so ; )
 
not a good answer
 
3:13 PM
@cassandra0 It's trivially true from the shape of the formulas. But it may still be interesting nevertheless, if you find a problem that it helps solve :-)
 
The problem text states the above equation, and asks me to find a formulae for $C_n$
 
@HenningMakholm, thank you
 
I rewrote it as a geometric series and obtained that
 
@N3buchadnezzar, evaluate both sides at z=0
 
$$ \displaystyle \frac{z}{1-z-z^2} = \sum_{n=0}^\infty z^{n+1} (z+1)^n = \sum_{n=1}^\infty (z+1)^{n-1} \cdot z^n $$
 
3:16 PM
I don't think you're really listening to me
 
@cassandra0 I am! I just told you what I had done, I am now looking into what you wrote.
 
@N3buchadnezzar Have you tried partial fraction expansion on the LHS and then geometric series for each term?
 
@HenningMakholm One does not simply partial fraction expand the right hand side.
 
Well, sorry then.
 
@HenningMakholm Just a joke, give me a minute =)
 
3:19 PM
1/(1-z-z^2) = 1 + 1z + 2z^2 + 3z^3 + 5z^4 + 8z^5 + 13z^6 + ... fibonacci numbers
 
Mind you, I haven't actually thought this through. There may be pitfalls down the road.
 
$\varphi$ is the root of $1-z-z^2$.
 
@cassandra0 But there is an ugly $z$ at the top
 
I didn't see it
 
@cassandra0 Yes, of course! N3b, instead of the partial fraction nonsense, multiply both sides by 1-z-z², and then separate like powers of z on the RHS.
 
3:20 PM
@N3buchadnezzar maybe you had a typo here
 
seems like a typo then
 
@N3buchadnezzar That looks like the generating function for the Fibonacci sequence
 
this would have been realized earlier had we figured out why the summation starts at 1
 
@cassandra0 But the solution is correct... When I just expanded the function it gives me that the coefficient for z^5 = 5
 
3:23 PM
@N3buchadnezzar $C_n=F_{n+1}$
 
@N3buchadnezzar Makes sense. That is shifted one position from the coefficients cassandra0 gave, because we didn't have the z in the numerator then.
 
@cassandra0 Ah, you already noticed that :-)
 
does anyone have a fast implementation of AKS which actually gives me the polynomial for a composite number?
 
@robjohn But why ?
 
@N3buchadnezzar Take the relation $F_k=F_{k-1}+F_{k-2}$, multiply by $x^k$ and sum...
 
3:28 PM
hi argon!
 
@cassandra0 Hi
 
I am too tired for this it seems
 
@N3buchadnezzar You may have missed this, I didn't ping you: Instead of the partial fraction nonsense, multiply both sides by 1-z-z², and then separate like powers of z on the RHS. This gives you excatly the Fibonacci recurrence between the coefficients.
 
@HenningMakholm Sigh okay, the partial fraction nonsense turned into a horrible mess
 
@Argon, do you know any thing fun we can do with Gamma? I got it's functional equation and robjohn showed that it decays exponentially away from the real line
 
3:29 PM
@N3buchadnezzar Sorry about that.
 
No problems, I learned something from it too ; )
 
@cassandra0 Have you evaluated $\int_0^1 \log \Gamma(x)\, dx?$
 
@N3buchadnezzar $$(f(x)-F(0)-F(1)x)=(f(x)-F(0))x+f(x)x^2$$
 
@cassandra0 It's a fun one. Try what you can, and if you get stumped, I will give you a cool clue.
 
3:31 PM
ok
 
@N3buchadnezzar $$x=f(x)(1-x-x^2)$$
 
@robjohn Then solve for f and compare? =)
 
I can't use the series for log because it doesn't converge in many places
I can't use parts because that would mean I have to x log(Gamma(x)) evaluated from 0 to infinity which diverges
 
morning
 
3:36 PM
@HenningMakholm I must be silly
 
no I can't think what to do
oh wait I got an idea
$$
\Gamma(z)\Gamma(1-z) = \frac{\pi}{\sin{(\pi z)}}\! $$
 
@N3buchadnezzar That could either mean that you suddenly get it, or that you still don't get it.
 
@cassandra0 Yes
 
I was thinking use $x \Gamma(x) = \Gamma(x+1)$
 
You could post it on main...
 
3:38 PM
@HenningMakholm The former,but thanks a bunch for trying to help me =)
$$z = \sum_{n=1}^\infty C_n (z^n - z^{n+1} - z^{n+2})$$
 
@cassandra0 Note that $\int_0^1 \log \Gamma(x)\, dx = \int_0^1 \log \Gamma(1-x)\, dx$
 
shouldn't one be negative?
no :S but why not
 
@cassandra0 AKS Primality? Hmm, this might help
 
@cassandra0, @robjohn: How sure are you that we actually get Fibonacci coefficients? The more I stare at it, the more it looks like it ought to be the reverse Fibonacci sequence.
 
@cassandra0 Think about the transformations.
 
3:39 PM
$$ z + \sum_{n=1}^\infty C_n (z^{n+1} + z^{n+2}) = \sum_{n=1}^\infty C_n z^n$$
 
@JayeshBadwaik, thanks but this programs output doesn't make any sense and I don't see how to edit it to make it print the polynomials
 
@cassandra0 Ahh, so you have already seen it. Hmm.
 
x --> 1-y, dx --> - dy, so integral log gamma (x) dx should go to - log gamma (1-s) ds
 
@HenningMakholm My derivation, recollection, and Mathematica seem to agree with $\frac{x}{1-x-x^2}$ being the generating function of the Fibonacci sequence
 
@cassandra0 What are the bound then?
 
3:41 PM
ohhh
 
:)
 
$\int_0^1 \log \Gamma(x)\, dx = - \int_1^0 \log \Gamma(1-s)\, ds = \int_0^1 \log \Gamma(1-s)\, ds$
 
@cassandra0 we just did the integration of $\log(\sin(x))$
 
ok good now I understand that
 
@robjohn Okay. .. on further thought I may have done something silly such as subtracting exponents where they should have been added.
 
3:42 PM
@cassandra0 So can you do it now? :)
 
@HenningMakholm But now I have $F(1) + F(n+1) + F(n+2) = F(n)$ ?
 
that just gives me $$\int_0^1 \log\left(\frac{\pi}{\Gamma(x)\sin(\pi x)}\right)dx$$
 
@cassandra0 What is the sum of the integrals?
 
:D
oh that's so neat
 
@cassandra0 Euler discovered that gem
 
3:44 PM
$$\int_0^1 \log(\Gamma(x)) dx = \frac{1}{2} \int_0^1 \log\left(\frac{\pi}{\sin(\pi x)}\right)dx$$
 
@cassandra0 Right. Now it split it into two integrals!
 
@N3buchadnezzar Hmm, I'd advice ditching the F's in this phase and work directly with C's. You have an equation up there -- for example, if you put everything on one side so it reads 0=x-(polynomial times series), then what is the coefficient of, say, z^7 on the RHS? This must be 0.
 
just $\int_0^1 \log\left(\frac{\pi}{\sin(2 \pi x)}\right)dx$? I can't think what to do now
 
@cassandra0 ^^ I solved it, use $\log(a/b) = \log a - \log b$
 
@cassandra0 What @N3b said!
 
3:50 PM
@Argon Cool integral $$ \int_0^1 \log(\sin x) \, \mathrm{d}x $$ is a sneaky one too.
See 14. here
@HenningMakholm Sorry, but I just dont see it.
$$0 = z- (1-z-z^2) \sum_{n=1}^\infty C_n z^n$$
 
@N3buchadnezzar It is. It is evaluated similarly too!
 
@N3buchadnezzar Yes, and then multiply each term in the series with 1-z-z², and then collect all terms with the same power of z together.
 
$$x^k \cdot \sum_{n=0}^\infty x^n C_n = \sum_{n=0}^\infty x^n C_{n-k}$$
Hence $$(1 - x - x^2) \cdot \sum_{n=0}^\infty x^n C_n = \sum_{n=0}^\infty x^n (C_n - C_{n-1} - C_{n-2})$$
 
I might have finally gotten it, man my head works slow today
@cassandra0 And you do not have to change the indicees, since all the negative terms are zero?
$$ x^k \cdot \sum_{n=0}^\infty x^n C_n = \sum_{n=-k}^\infty x^n C_{n-k} $$? =)
 
$$
\begin{align}
&\int_0^n\log(x)\,\mathrm{d}x+\int_0^1\log(\Gamma(x))\,\mathrm{d}x\\
&\int_0^1\left(\log(x)+\log(x+1)+\dots+\log(x+n-1)\vphantom{\int}\right)\,\mathrm{d}x
+\int_0^1\log(\Gamma(x))\,\mathrm{d}x\\
&=\int_0^1\log(\Gamma(x+n))\,\mathrm{d}x\\
&=\int_0^1\left(\log(\Gamma(n))+x\log(n)+O\left(\frac1n\right)\right)\,\mathrm{d}x\\
&=\log(\Gamma(n))+\frac12\log(n)+O\left(\frac1n\right)
\end{align}
$$
So
$$
n\log(n)-n+\int_0^1\log(\Gamma(x))\,\mathrm{d}x
=\color{#C00000}{n\log(n)-n+\frac12\log(2\pi/n)}+\frac12\log(n)+O\left(\frac1n\right)
Using Stirling and implicitly the log-convexity of $\Gamma$
 
4:03 PM
great!!!
 
@robjohn An interesting way!!
Cool!
 
how should we prove the reflection formula?
maybeI will look at how Euler did
oh it is related to the Beta function
 
@aDangerousIdea Scroll down?
 
$\Gamma(x) \Gamma(1-x) = B(x,1)$
 
@cassandra0 $$\int_{1/4}^{3/4} \log\Gamma(x)\, dx=\frac12 \left(\log\sqrt{2\pi}-\frac{G}{\pi}\right)$$
 
4:09 PM
what's G?
We wish to prove $$B(x,1) = \frac{\pi}{\sin(\pi x)}$$
 
@cassandra0 Catalan's constant
 
yikes
 
Hahaaha!
@aDangerousIdea :)
 
$$ B(x,1) = \int_0^1t^{x-1}\,dt \! $$
 
@cassandra0 ?????
 
4:13 PM
am I wrong?
 
@cassandra0 It's right
But you said $$B(x,1) = \frac{\pi}{\sin(\pi x)}$$
 
yes I'd like to show that
then we would have Eulers reflection formula by $$\frac{\Gamma(x)\Gamma(1-x)}{\Gamma(1)} = B(x,1)$$
 
@cassandra0 But you just said about that is is not equal... Look at your integral!
Bye all, see you tonight
 
I don't understand
 
4:15 PM
$$\frac{\pi}{\sin(\pi x)} \neq \int_0^1t^{x-1}\,dt \!$$
 
oh :(
 
later @Argon
 
@aDangerousIdea: I fixed it :-p
 
@robjohn >8(
 
I thought $\frac{\pi}{\sin(\pi x)} = \int_0^1t^{x-1}\,dt \!$ must be true for Eulers formula
 
4:26 PM
Hi
 
4:37 PM
hi
 
For $0< x<1$, we have
$$
\begin{align}
\Gamma(x)\Gamma(1-x)
&=\int_0^\infty e^{-s}s^{x-1}\,\mathrm{d}s\int_0^\infty e^{-t}t^{-x}\,\mathrm{d}t\\
&=\int_0^\infty\int_0^\infty e^{-s-t}s^{x-1}t^{-x}\,\mathrm{d}s\,\mathrm{d}t\\
&=4\int_0^\infty\int_0^\infty e^{-s^2-t^2}s^{2x-1}t^{1-2x}\,\mathrm{d}s\,\mathrm{d}t\\
&=4\int_0^{\pi/2}\int_0^\infty e^{-r^2}\sin^{2x-1}(u)\cos^{1-2x}(u)r\,\mathrm{d}r\,\mathrm{d}u\\
&=2\int_0^{\pi/2}\tan^{2x-1}(u)\,\mathrm{d}u\\
&=2\int_0^\infty v^{2x-1}\,\mathrm{d}\arctan(v)\\
 
@robjohn Did you write this in a 80-column editor and then copy-paste it here? Just curious.
 
im not english, and i got a question about communicating mathematics in english, is that ok for inhere?
 
@JayeshBadwaik No I was translating from the page I had cited before. I just removed the formatting from that page
@JustDanyul sure
 
4:49 PM
@robjohn Ohh, okay. I have got another question. The derivation you just wrote.. Did you look up and write it or were you able to work it out by yourself right now?
 
@robjohn, thanks but this type of trigonometric substitutions are very difficult I didn't really get it
 
@amWhy Hello!
 
@JayeshBadwaik I proved it by myself a while back, and just copied it from the page I cited above: whim.org/nebula/math/eulerreflection.html
 
@robjohn Any tips for finding the laurent series to $$e^{(z+2)/(z-1)}$$ about $z=1$ ? =)
 
@robjohn Oh!! Whim is your own personal insitute!! Aha.. West Hills Institute of Mathematics. Nice. :D
 
4:51 PM
what do you call the set of results of a function, say y = f(x) , so the set of all possible elements (x,y), what are their name? (i know that their super set is called the codomain, but not what the actual set is called)
 
@JayeshBadwaik It is. Thanks
@N3buchadnezzar I would start by using $z=w+1$...
@N3buchadnezzar But wait... $z=1$ is an essential singularity
 
@robjohn $e^{1 + \omega/3}$ then ?
 
Anyone know if there's any difference between $\cong$ and $\equiv$?
 
@JustDanyul range or image.
 
@N3buchadnezzar $e+ew/3+ew^2/18+ew^3/162+\dots$
 
4:59 PM
AH!
@robjohn My book does not cover these series well at all, barely mentions them. Thanks a bunch, sorry for all the silly questions. But this is really helping me
 
@N3buchadnezzar I'm glad. Don't feel silly about asking questions.
 
@JayeshBadwaik thanks for that, it was hard to google for!
 
@Manishearth There is...
 
I always feel bad when I recognize an integral
because it is so hard searching for integrals
 
@robjohn What is it?
 
5:04 PM
@Manishearth $\cong$ means congruence as in geometrical congruence, or sometime "approximately equal to".
@Manishearth $\equiv$ means modular equivalence or "defined as"
 
@robjohn Hmm, really? Then $\simeq$ and $\approx$ are also "approximately equal to"?
 
@robjohn or even "congruent" as in congruent triangles in elementary geometry - isn't maths notation wondrous :)
 
@Manishearth yeah, I'm not sure if there is a generally accepted difference, but within a paper, they may define them more specifically.
 
Ah, makes sense ...
 
@OldJohn Yes, that was the geometrical congruence
 
5:07 PM
Thanks!
 
@OldJohn We used $\cong$ for congruence of triangles.
 
@robjohn ah - sorry - missed that! (not fully awake yet!)
 
@OldJohn nothing to be sorry about. just noting
 
@JayeshBadwaik I think we used that for "similar triangles"
 
5:08 PM
@OldJohn For similar triangles we used $\sim$.
 
Both notation and terminology are crazy in maths generally - like "harmonic measure" ... is not actually a measure :)
 
@robjohn Time for another silly question? ^^
Laurent of
$$ f(z) = \frac{z^2 - 2 i z + 3}{(z-2)(z^2+1)} $$ about $z=2$
I see that $z^2 - 2i z + 3 = (z - 3i)(z + i)$ so that
$$f(z) = \frac{z - 3i}{(z-2)(z-i)} $$
$$ f(z) = \frac{1}{5}\left( \frac{4i - 7 }{2-z} - \frac{4i - 7}{i-z}\right) $$
Should I now write $\omega = z + 2$, and write $f(z)$ as a sum of two geometric series?
Complicated! :p
 
5:32 PM
Could anyone point me in the right direction on the topic of quantifying list similarity? I'm comparing two nature conservation sites, and wonder how similar they are in types of animals living there.

Site 1 has the following species: A,B,C,D,E,F,G
Sites 2 has these: A,C,D,H,I

This isn't impossible is it?
 
let P(z) be a degree 2n seperable polynomial, will it have a meromorphic inverse apart from n line segments between the roots?
 
5:53 PM
hello
why do meromorphic functions have consequences in number theory?
 
6:11 PM
 
@OldJohn First comment: the´╗┐ radiators will just be stolen and fuel their viking tribal warfare. They'll have a population explosion and then they'll just need more radiators. By interfering, you're just going to increase the misery. Let them freeze, that's their way of life.
 
@GustavoBandeira wonderful!
Maybe we should teach them how to hunt radiators, rather than just giving them?
 
6:42 PM
Hey guys...so I would post this as a question, but there have already been a few questions to this effect, and they haven't helped. Can anybody help to give me a good geometric intuition to the meaning of the exterior derivative of a differential form?
 
@oldjohn: what version of mint do you have?
 
@Khromonkey 14, I think - downloaded a few days ago
 
is it working fine?
@OldJohn
 
Yep -no problems so far
using on laptop and one of my desktops
 
are you running nadia or cinnamon?
 
6:52 PM
Mate
 
Hmmm
@OldJohn Any tips for rewriting $1/(z-2)$ in such a way that $|z-2|>1$ ? =)
 
I guess that is the same as nadia :)
 
yes, I messed up
thanks
 
@N3buchadnezzar erm - not sure - what are you trying to do?
 
@oldjohn do you teach mathematics?
 
6:54 PM
@Khromonkey I used to -now retired
 
where?
 
also taught computer science for a while
UK
 
cool
 
@OldJohn I need to write $1/(z-2)$ as a series that converges when $|z-2|>1$
Was messing around with things like $$\frac{1}{1 - 1/(z-2)}$$ but could not quite make it work
 
@N3buchadnezzar this might be a silly question, but what is wring with this series with one term: $1/(z-2)$:)
 
6:58 PM
@OldJohn cause the problem asks me to find its laurent series, and later solve a complex integral involving it ;)
 
@N3buchadnezzar OK
let me have a think - not done Laurent series for s couple of decades ...
 
I did start with $$ f(z) = 1/(z-1)(z-2) $$, might be easier to manipulate
 
hi, the math markup is not being rendered here, does it work for you?
 
@aditsu refer to the pinned message on the top of the starboard about LaTeX in Chat.
 
@aditsu points at the starred messages
 
7:01 PM
@N3buchadnezzar are you certain that the series with one term is not actually the Laurent series for that function? not all series are infinte :)
 
I mean everywhere on math.stackexchange, not just in chat
 
@N3buchadnezzar for example, the Taylor series for the function $1+x^2$ is just $1+x^2$
What you are really asking for is a series of the form $\sum a_k(z-2)^k$ where some of the $k$ may be negative - so the series obtained by taking $a_{-1}=1$ and all the other $a_k=0$ would do the job
 
@aditsu it loads OK here -maybe there is a network problem between you and the server? dns problem maybe?
@aditsu can you ping the server from the command line?
 
oh, it's a dns problem
ok, I fixed it, it's working now, thanks
 
7:11 PM
@aditsu great
@N3buchadnezzar do you just need to find the residue of that function at a pole in order to do an integral?
 
@Nimza, I found this people.math.gatech.edu/~ecroot/riemann_notes.pdf some moredetails
about inverse mellin transform
 
@OldJohn I was able to solve it :D:D:D
 
@N3buchadnezzar was I right about it being its own Laurent series?
 
For the completion of this hazardous task I obtained large quantities of females and adult soda.
@OldJohn Noooo
 
@N3buchadnezzar really? - so what is the Laurent series for $1/(z-2)$ about 2 ???
 
7:21 PM
@OldJohn $$f(z) = \frac{z-3}{(z-2)^2} \sum_{n=0}^\infty (z-2)^{-n}$$ when $|z-2|>0$
 
that cannot be the Laurent series $1/(z-2)$ that series is zero when $z=3$, but the function is not!
 
But it works for most values, I can not really see where my mistake lies then
 
@N3buchadnezzar OK - my logic is faulty - the series does not converge for $z=3$
 
I know
@OldJohn |z-2|>1
$z=3$, does not satisfy this criteria ;)
 
indeed
 
7:33 PM
hi anon
 
hey
 
how is it going
 
pretty good. reorganized my hard disk and finished a long sylow theory assignment.
 
cool
can we derive $$
\Gamma(z) = \frac{e^{-\gamma z}}{z} \prod_{k=1}^\infty \left(1 + \frac{z}{k}\right)^{-1} e^{z/k} $$ without using the limit form of gamma?
I don't really understand why $e^{z/k}$ is there
I know why there is e^{something} there but not z/k
 
ah, Weierstrass factorization
the z/k is so that exp(z/k) factors prevent the (1+z/k)^-1 factors from diverging
If we view proofs as paths in proposition space, I feel that any proof of that formula would be homotopic to the one with the limit definition of gamma.
 
7:45 PM
yeah I was thinking that because of the definition of $\gamma$
but Idon't really like that proof...
 
whyzat?
 
$$ \Gamma(z) = \lim_{n \to \infty} \frac{n! \; n^z}{z \; (z+1)\cdots(z+n)} = \frac{1}{z} \prod_{n=1}^\infty \frac{\left(1+\frac{1}{n}\right)^z}{1+\frac{z}{n}}$$
I don't know how someone would find this limit version
the denominator makes sense
 
take the & out
 
maybe we could the integral somehow: $$\Gamma(z) = \sum \frac{(-1)^n}{(z+n)\cdot n!} + \int_1^\infty e^{-t} t^{z-1}$$
I did take the & out but I think there's some lag, is it displaying ok now?
 
@N3buchadnezzar That function you gave earlier - it might be a series which converges to $1/(z-2)$ ... but it is definitely not a Laurent series!
 
7:49 PM
@cassandra0 weird, I had to refresh to see the edit
@OldJohn it's more or less Laurent around z=2 innit?
 
maybe I would be better imitating the proof of that formula rather than using the result
 
erm, arbitrarily negative powers of z-2, nevermind
 
@anon yes - but a Laurent series does not have a factor $(z-3)$ sticking out the front :)
 
@OldJohn that's just (z-2)-1, and $(w-1)\sum a_nw^n$ is pretty close to being in form
 
I am still convinced that the Laurent series for $1/(z-2)$ is just ... $1/(z-2)$ for $|z-2|>1$
 
7:51 PM
modulo breaking apart and reindexing
 
Hi!
 
just like the Taylor series for $z^2+1$ is just ... $z^2+1$
 
@OldJohn right. Laurent series are unique, that is a valid Laurent series, and n3bu's series is not Laurent because it has arbitrarily large negative powers
 
hello
 
@cassandra0 Hello!
 
7:54 PM
hey whats going on
 
@cassandra0 i'm studying linear algebra....
 
@anon Yes - it was the uniqueness bit that was worrying me - my "one term" series seemed to be a well-definied Laurent series, and I couldn't see why his wasn't
 
cool
 
@cassandra0 symmetric operators...hmm.. having some problems...
 
what's a symmetric operator
 
7:56 PM
operator equal to its adjoint (there needs to be a given basis and therefore isomorphism between the space and its dual)
so A=A^T
 
@anon I did not understand that invariant subspace thing...
 
I never said anything about invariant subspaces..
 
@anon i know, that's why i'm asking
 
well, then what is "that invariant subspace thing"?
 

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