## Complex partial fraction expansion

I would like to have a tool for partial fraction expansion of polynomial quotient $$\frac{P(z)}{Q(z)},$$ where the order of the polynomial $$P(z)$$ is less than that of $$Q(z)$$.

The output of the function is expected to be the coefficients $$c_{ij}$$ of the expansion: $$\sum_i\sum_{j=1}^{m_i}\frac{c_{ij}}{(z-\zeta_i)^j},$$ where the sum runs over all distinct roots $$\zeta_i$$ (with multiplicity $$m_i$$) of the polynomial $$Q(z)$$.

Is there a built-in function in Mathematica which is suitable for performing the task? For a symbolic computation the list of roots of the polynomial $$Q(z)$$ can be supplied.

## Is simplifying a complete expansion of the right hand side of a trigonometric identity sufficient to prove it?

My first question is, when proving a trigonometric identity for all real values of a variable such as $$x$$, is it sufficient to expand the right hand side into elementary functions and simplify until it equals the left hand side?

I know this is very basic but since I have seen complicated proofs of identities by induction, I want to be sure that this is a sufficient method of proof if the problem allows. My second question is, is this considered a direct proof?

My third question is, assuming that the answers to my first two questions are affirmative, would there be any reason that a trigonometric identity need be proven any way other than the way I specified in my first question? If trigonometric expressions can be rewritten in terms of the elementary trig functions, would there ever be need for an induction proof rather than direct?

## Advice on using this alternative method of finding the Laurent expansion of $\tfrac{1}{z^2(z-1)}$

Say we want to calculate the Laurent series of $$\tfrac{1}{z^2(z-1)}$$ about $$z_0=1.$$ Now I know that one way to do it is to say that $$f(z)=\tfrac{1}{z^2}(\tfrac{1}{z-1})$$ and appy the geometric series expansion to the brackets term. But I wanted to try and do it a different way :

First we split f into partial fractions and compute the Laurent series separately.Now consider the Laurent expansion of $$\tfrac{1}{z^2}$$

We know that $$0$$ is a pole of order 2 which implies that $$\forall n>2,a_{-n}=0$$. Therefore $$\tfrac{1}{z^2}$$ haa series expansion $$\tfrac{a_{-2}}{(z-z_0)^2}+\tfrac{a_{-1}}{z-z_0}+a_0+a_1(z-z_0)+…$$

Now to calculate the the $$a_{-2}$$ coefficient I applied the following trick

$$a_n=\tfrac{1}{2\pi i} \int_{\gamma}\tfrac{f(z)}{(z-z_0)^{n+1} }dz=\tfrac{1}{2\pi i} \int_{\gamma}\tfrac{1}{(z-z_0)^{n+3}}dx=\tfrac{1}{2\pi i} \int_{\gamma}\tfrac{1}{z^{n+3}}dz$$

$$a_{n-2}=\tfrac{1}{2\pi i} \int_{\gamma}\tfrac{1}{(z-z_0)^{n+1} }dz=\tfrac{f^n(z_0)}{n!}$$ But as n $$f(z)=1$$ this implies that $$n$$ must be zero and so $$a_{-2}=1$$

Now when I tried to use the same trick on $$a_{-1}$$ It doesn’t work because now we can’t use Cauchy’s formula. Also when I tried to use u substitution by letting $$u=z-z_0$$ it returns that $$a_{-1}=-\tfrac{1}{z}$$ but I know this is not rue as I know from the method that I mentioned in the first paragraph that $$a_{-1}=1$$ So does anyone have any suggestions on how I can find $$a_{-1}$$ continuing with the method I’m trying to use ?

## The number of distinct terms in this expansion

$$(x^3 + \frac1{x} +1)^{200}$$ What will be the amount of distinct terms? As in each term with $$x$$ will be distinct both my power and by co efficient, hence the result of combination and every terms without $$x$$ will also need to be combined into one.

## Question about the Fourier expansion of adelic Eisenstein series for $\operatorname{GL}_2$

My reference is Daniel Bump’s book, Automorphic Forms and Representations, Chapter 3.7. Let $$k$$ be a number field, $$G = \operatorname{GL}_2$$, $$B$$ and $$T$$ the usual Borel subgroup and maximal torus of $$G$$. For $$\chi$$ an unramified character of $$T(\mathbb A)/T(k)$$, let $$V$$ be the space of “smooth” functions $$f: G(\mathbb A) \rightarrow \mathbb C$$ satisfying $$f(bg) = \chi(b) \delta_B(b)^{\frac{1}{2}}f(g)$$ which are right $$K$$-finite. For $$f \in V$$ and $$g \in G(\mathbb A)$$, define the Eisenstein series

$$E(g,f) = \sum\limits_{\gamma \in B(k) \backslash G(k)} f(\gamma g)$$

For suitable $$\chi$$, the series converges absolutely for all $$g \in G(\mathbb A)$$. Now $$E(g,f)$$ has a “Fourier expansion,” which as explained by Bump is gotten as follows: the function $$\Phi: \mathbb A/k \rightarrow \mathbb C$$ $$\Phi(x) = E( \begin{pmatrix} 1 & x \ & 1 \end{pmatrix}g,f)$$ is continuous, hence is in $$L^2(\mathbb A/k)$$, and therefore has a “Fourier expansion” over the characters of $$\mathbb A/k$$. If $$\psi$$ is a fixed character of $$\mathbb A/k$$, then $$\psi_{\alpha}: x \mapsto \psi(\alpha a)$$ comprise the rest of them, for $$\alpha \in k$$. Then

$$\Phi(x) = \sum\limits_{a \in k} c_{\alpha}(g,f) \psi_{\alpha}(x) \tag{1}$$

$$c_{\alpha}(g,f) = \int\limits_{\mathbb A/k} E( \begin{pmatrix} 1 & y \ & 1 \end{pmatrix} g,f) \psi(-\alpha y) dy$$

According to Bump, we may simply set $$x = 0$$, giving us the Fourier expansion for the Eisenstein series

$$E(g,f) = \sum\limits_{\alpha \in k} c_{\alpha}(g,f)$$

My question: Why is this last step valid? The right hand side of equation (1) converges to $$\Phi(x)$$ in the $$L^2$$-norm. As far as I know, this is not an equation of pointwise convergence. In general, the Fourier series of a continuous function need not converge pointwise to that function everywhere (in the classical case $$\mathbb R/\mathbb Z$$, the Fourier series of a continuous function converges pointwise to that function almost everywhere).

## Proof Idea: There are irrational numbers whose decimal expansion cannot be computed

The online lecture I am watching stated a proof idea:

The set of all possible programs is countably infinite, yet the set of irrational numbers is uncountably infinite.

I don’t think this is sufficient, it assumes one program can only return one irrational number. How to show that I can not write one program that returns all irrational numbers?

## Expansion Tile e ListView

Estou tentando criar esse layout para um app, usei um expansion tile dentro de um Card e uma listView abaixo, porém quando eu abro o expansionTile, se bater no fundo da tela acontesse isso e se a listView tiver muitos tiles acontece isso, não estou conseguindo de forma alguma resolver esse problema.

Segue meu código abaixo:

     class HomeState extends State<Teste>{       @override       Widget build(BuildContext context) {         return Scaffold(           appBar: AppBar(             title: Text (""),           ),             body: Container(               padding: EdgeInsets.all(15.0),               child: Column(                 children: <Widget>[                   CardList(),                   MyList()                 ],               ),             ),         );       }     }      Widget CardList(){       return new Flexible(           child: Card(             elevation: 5,             color: Color.fromARGB(255, 0xC8, 0xF4, 0xF2),             child: ExpansionTile(               title: Text("Periodo",                 style: new TextStyle(                 ),textAlign: TextAlign.center,               ),               children: <Widget>[                 ListaPeriodos(10)               ],             ),           )       );     }  Widget MyList(){   return new ListView.builder(     padding: EdgeInsets.all(10.0),     shrinkWrap: true,     itemBuilder: (BuildContext context, int index) {       return new ListTile(         title: new Text(           "Texto",           textAlign: TextAlign.center,           style: new TextStyle(             fontWeight: FontWeight.normal,             color: Colors.black45           ),         ),       );     },     itemCount: 15,   ); } 

## [ Football (American) ] Open Question : What do you think of my NFL expansion and realignment idea?

NFC East New York Giants Dallas Cowboys Philadelphia Eagles Washington Redskins Carolina Panthers Tampa Bay Buccaneers Central Minnesota Vikings Green Bay Packers Chicago Bears Detroit Lions Atlanta Falcons New Orleans Saints West Los Angeles Rams San Francisco 49ers Seattle Seahawks Portland Bulldogs (Expansion Team) Arizona Cardinals Salt Lake City Wildcats (Expansion Team) AFC East New York Jets New England Patriots Buffalo Bills Baltimore Ravens Jacksonville Jaguars Miami Dolphins Central Cleveland Browns Cincinnati Bengals Pittsburgh Steelers St. Louis Stallions (Expansion Team) Indianapolis Colts Tennessee Titans West Los Angeles Generals (Expansion Team) Oakland Raiders San Diego Chargers Las Vegas Blackjacks (Expansion Team) Denver Broncos Kansas City Chiefs Season extended to 18 games Playoffs (3 division winners and 3 wild cards in each conference, lowest division winner plays in the wild card round)