Hello all. I was wondering: so we can define the matrix exponential. Is the definition the taylor series for the exponential on say real numbers $e^x$ except replace $x$ with your matrix?
I am wondering in particular if the idea of a taylor series exists for functions defined on domains more general than just scalars
some charlatan even wrote a book that might have a section on other versions of taylor's series. depends on how multivariable he actually got with his mathematics.
ted, you have one of the world's largest supplies of lesliecoin in circulation. i don't know why you would besmirch the good name of a currency that you hold.
Huh. So if you have something like $\mathcal{L}\{e^{-t}\ast e^t \cos t\}$ you'd want to do $\mathcal{L}\{e^{-t}\}\mathcal{L}\{e^t\}$ but what about the $\cos t$ term
if |u| < 1 then sum_{r=1}^infty r u^r = u * sum_{r=1}^infty r u^{r-1} = u * (the derivative, with respect to u, of sum_{r=1}^infty u^r), with that last sum being a geometric series, does that help?
I just read one of the statements the FOX lawyers released already talking about "mismatches in journalistic integrity"..........like give me a F-🤬-in break............
Now I'm stuck on two separate problems where I have the form $F(s) = \frac{1}{s} \cdot \frac{1}{1+\frac{1}{s}}$ and $F(s) = \frac{1}{(s-1)^2} \cdot \frac{1}{1+\frac{1}{s^2}}$
I just wanna take the inverse transform and it's being difficult
@TedShifrin taylor polynomials /ne taylor series ?
Seems pretty clear that a power series in Rudin is not a function of any variable
and is thus an infinite sum of numbers
If your objection is that the taylor series as in the infinite sum is not a polynomial i guess sure.
but my point is that power series as defined in Rudin has no variable dependence whereas taylor series of functions do. but everywhere online people say that taylor series are particular power series
obliv it helps to separate out a few concepts. any function that is N-times differentiable at a point has an Nth taylor polynomial at that point. some functions aren't differentiable to every order at a point, though.
if the function is infinitely differentiable at a point, you can at least form the "taylor series" as a formal thing, but without more information, you don't automatically know that it converges anywhere except the point where you did the expansion.
so "taylor's theorem" as you see it in calculus books often handels a couple of things. it often handles how a finite order taylor polynomial relates to the function that generated it. sometimes it provides estimates of the 'remainder term' (= difference between function and taylor polynomial) in terms of the order of the polynomial.
once you're talking about "representing" a function in terms of a taylor series, you're automatically thinking, infinitely differentiable functions, and also having to think about - gee, what happens to that remainder.
which you can fail to notice on the tenth time through a calculus book because it tends to arrive, if at all, near the end of a term, and because most calculus people don't like thinking about estimates and inequalities.
that's a helpful way of looking at it. if you have any series with a variable in it, and you want to think of it as a 'function' of that variable, you've suddenly got to wonder about questions of convergence a whole lot more.
because you're not just considering convergence of one numerical series, but for a family of them, as your 'variable' varies.
the thing that makes a power series a "power series" is it's sum_n [sequence of numbers depending only on n] x^n, with x the variable, or the same in 'powers' of (x-c) with c fixed. but this thing might or might not represent a function. it might converge only for x = 0 and not be anything at all otherwise.
An interesting group theory exam problem I took an hour ago: Let $F$ be a field with $p^n$ elements, $D$ a diagonal, $U$ strictly upper triangular, $T$ triangular matrices of $\mathrm{GL}_2(F)$. Let $r$ be an integer $1\leq r\leq p^n-1$. Construct an injective group homomorphism $\phi_r:F\rtimes_{\psi_r}F^\times\to T$ such that $\phi_r(F)<U$ and $\phi_r(F^\times)<D$ where $\psi_r:F^\times\to\mathrm{Aut}(F)$ by $\psi_r(\gamma)(a)=\gamma^r\cdot a$ for each $\gamma\in F^\times$ and $a\in F$.
Hmm I wonder what $$\mathcal{L}^{-1}\left\{\frac{1}{(s-1)^2}\cdot \frac{s^2}{s^2 - 1}\right\}$$ is.. since I can put it in the form $\frac{te^t}{2}\ast \cos t \ast s$ Does that look right?
Go back to Spivak one variable calculus for a basic explanation on series, then return to Rudin.......if it "must" be RUdin. but many have said Rudin ain't it..
@D.C.theIII These are all phrases which we should strive to eliminate from our vocabulary. Definitely from our writing, and hopefully from our spoken language, as well.
@D.C.theIII Yeah, I got called to task for it when I started student teaching. I did not get along well with my supervisor, but she had some good advice sometimes.
I just found out about a 12 yr old who teaches part-time at MIT and part-time in India. But I can’t see that he got any graduate degrees or wrote a thesis. Fascinating.
@leslietownes Now I can understand if this is your starting definition for a power series. But my claim is that this is not Rudin's definition for power series.
"For example, a vibrating airplane wing could be struck by lightning, a mass on a spring could be given a sharp blow by a ball peen hammer, and a ball (baseball, golf ball, tennis ball) could be sent soaring when struck violently by some kind of club (baseball bat, golf club, tennis racket)."
$\sum_{n=0}^{\infty} a_n(z-z_0) ^n is a power series centered at z_0 . If z, z_0, (a_n) $ are complex numbers, we call complex power series and of z_0, z, (a_n) are real numbers then we call real power series.
Though we simply called power series and (real Or complex)should be clear from the context.
@SillyGoose f(z) =... is valid only inside the disk of convergence.
$f(z) =\sum_{n=0}^{\infty}a_n(z-z_0) ^n $ for all $z$ with $|z-z_0|<R$
For an example $f(z) =\sum_{n=1}^{\infty} \frac{z^n}{n}$ is not valid at $z=1$
I guess I do not see what text in the definition 3.38 that tells me $z$ should be treated like a variable. I read it like "choose a z, then the power series is given by (19)" like that
@SillyGoose As I have told earlier, f(z) is valid only inside the disk of convergence, writing f(z) =... automatically implies z inside the disk of convergence.
which to me is different than saying we say that the power series parameterized by $z$ converges when $z$ is in the disk of convergence
like this above statement seems like a generalization of my last last message?
i mean if definition 3.38 is literally talking about a power series of a single complex variable ($z$) and if that is the unanimous opinion, then I can accept that. But I just want to make sure that I am understanding what everyone is saying. In particular, I would like to make sure that (19) is literally talking about a power series of a single complex variable ($z$), not that it implies what a power series of a single complex variable looks like.
BTW I wouldn't consider this a soft-question - but I did not remove the tag. It seems that there are already some answer - and some useful comments from Dave L. Renfro.
You have to consider all $n$-th roots of unity $\zeta_k = e^{2k\pi i/n}$ with $k = 0,\dots,n-1$.
$V = \{e^{it} \mid -1 < t < 1\}$ is an open neigborhood of $1$. Then $U_k = \{ e^{it} \mid (2k\pi - 1)/n < t < (2k\pi + 1)/n\}$ is an open neigborhood of $\zeta_k$ which is mapped by $f$ homeomorphica...
@SillyGoose I would suggest that you read Tao's post on the levels of rigour in mathematics. You are getting really caught up on a technicality that (1) doesn't seem like a problem at all, and (2) probably isn't actually a problem, anyway.
I think that it takes lot of time in computing cellular homology so not computationally friendly. And another thing is that this works only for CW complexes (right?). But singular works for any space.
Physicists like to pretend that $\delta$ is a normal function, and that $\delta(0) = \infty$, but this is not how mathematicians generally think about it.
Really, it would be better to drop the integral notation entirely, and think about $\delta$ as a dual object (in the sense of linear algebra / functional analysis).
There are a few ways to think about Dirac's $\delta$. If you want to think of it as a "function", the appropriate idea is probably to think of it as a limit of "nice" functions.
So, for example, define $\eta_1(t) = C \mathrm{e}^{-1/(1-|t|^2)}$ on the interval $(-1,1)$ (and zero elsewhere). Note that $\eta_1$ is smooth, and supported on that interval. Choose $C$ so that $\int \eta_1 = 1$.
For each $\varepsilon > 0$, define $\eta_{\varepsilon}(t) = \varepsilon \eta_1(\varepsilon t)$. Note that $\eta_{\varepsilon}$ is smooth, supported on $(-\varepsilon, \varepsilon)$, and integrates to $1$.
Assuming that I have put the pieces together correctly (I'm not taking the time to be super careful here), $$\lim_{\varepsilon\to 0} \int_{\mathbb{R}} f(t) \eta_{\varepsilon}(t) \,\mathrm{d}t = f(0). $$
The problem is that there is no function $\delta$ (e.g. no $L^1$ function, or smooth function with compact support, or whatever) which has the property that $$ \int f(t) \delta(t) \,\mathrm{d}t = f(0). $$
@Obliv You want compact support for a number of reasons, which have to do with more advanced topics in functional analysis / PDE than I am prepared to address here.
In any event, you can kind of write $\delta(t) = \lim_{\varepsilon\to 0} \eta_{\varepsilon}(t)$. This isn't quite right, since this limit doesn't really make sense (it is zero everywhere except $0$, and at $0$ the limit is infinite), but this is, I think, the way physicists generally think about Dirac's $\delta$.
@Obliv You should compute these derivatives and see what you get.
So I'd have to define it in terms of unit step functions.. with $$\delta(t-t_0) = \begin{cases} 0, & t > t_0+a \\ \frac{1}{2a}, & t_0 - a \leq t \leq t_0 + a \\ 0, & t < t_0 - a\end{cases}$$ I think
That seems like nonsense. You could define $\eta_a$ in this way, and then take a limit...
Reread your text.
They are not defining $\delta$ via the piecewise function you have given.
They are defining $\delta_a$ in this manner, and then defining $\delta$ as a limit of these $\delta_a$, in the same way that I defined it for $\eta_\varepsilon$.
I defined $\delta$ to be, more or less, the limit of a bunch of smooth bump functions. Your book is defining it to be the limit of a bunch of step functions.
Note that in both cases, the functions being defined are (1) supported only on a small interval near zero, and (b) integrate to $1$.
To compute the derivative of $\delta$, I like the following approach: $$ \int_{-\infty}^{\infty} f(t)\delta'(t) \,\mathrm{d} t = \bigl[ f(t) \delta(t) \bigr|_{t=\infty}^{\infty} - \int_{-\infty}^{\infty} f'(t)\delta(t)\,\mathrm{d}t = f'(0). $$
Note that the $f(t)\delta(t)$ is zero, since $\delta$ is "compactly supported" (i.e. if $|t|$ is large, $\delta(t) "=" 0$; really, the approximations are compactly supported and we want to be careful-ish about taking the limit).
So $\delta'$ is the "function" which, when you integrate a function against it, give you the derivative of that function at zero.
Note that I am using integration by parts here, which means that $f$ needs to be at least $C^1$, and I am assuming that $\delta'$ "makes sense". I kind of don't like your book's approach, since their $\delta_a$s are not differentiable.
I'm taking the limit of \frac{1}{c} \int_a^b f'(x)cos(cx)dx as c goes to infinity. I know that both f' and cos are continuous so the integral exists. In this case can I treat the integral as just a constant and conclude that the limit is zero?
@Obliv I actually started with $$\int_a^b f(x)\sin(cx)dx$$ and got the integral by using integration by parts. All I know for the question is that $f'$ is continuous so that's what I tried to get.
I think the key is to make $\sin(cx) = dv$ so you have $\frac{-1}{c}\cos(cx)f(x) + \frac{1}{c}\int f'(x)\cos(cx)$ I think that's what you did, which you can probably conclude it is 0
