From this text: Theorem 18.1. Polynomial roots and colleague matrix eigenvalues. The roots of the polynomial $$ p(x) = \sum_{k=0}^n a_k T_k(x),\quad a_n \ne 0 $$ are the eigenvalues of the matrix $$C=\begin{pmatrix} 0&1\\ {1\over 2}&0&{1\over 2}\\ &{1\over 2}&0&{1\over 2}\\ &&\ddots&\ddots&\ddots\\ &&&&&{1\over 2}\\ &&&&{1\over 2}&0 \end{pmatrix} - {1\over 2 a_n} \begin{pmatrix} 0 & 0 & 0 & \dots & 0 \\ \dots & \dots & \dots & \dots & \dots \\ 0 & 0 & 0 & \dots & 0 \\ a_0 & a_1 & a_2 & \dots & a_{n-1}\end{pmatrix} . $$ — Martin Sleziak Nov 13 '12 at 11:26

The problem with $f(x)=\frac{x-3}x$ is the it is not defined for all real numbers. But for example $$f(x)= \begin{cases} \frac{x-3}x & \text{if }x\ne0, \\ 1 & \text{if }x=0. \end{cases} $$ would be a function from $\mathbb R$ to $\mathbb R$. Where is this function continuous and where not? Can you somehow use this example to get $f(x)$ and $g(x)$ as required? — Martin Sleziak Oct 19 '15 at 13:01

You do now have to write math like this: $a$$b$=$p$$x$. You can use $ab=px$ instead. (You simply end the "math block" enclosed by dollars when you want to start to continue with regular text.) — Martin Sleziak Mar 7 '14 at 7:08

I suppose you are referring to the estimate on page 42 saying that $$1+\frac14+\frac19+\dots+\frac1{k^2}+\dots \le 1+\frac13+\frac1{10}+\dots+\frac1{k(k+1)/2} = 2.$$ — Martin Sleziak Apr 1 '15 at 14:01

You know that you do not have to write $$y$$ and you can write $y$, right? The difference should be obvious - the first one is $$y$$ and the second one is $y$. — Martin Sleziak Apr 24 at 9:06

You should probably explain in your question in what type of series you are interested. I guess you do not want something like this: $$f_n(x)= \begin{cases} \frac1{n^2} & \text{if }x\in[-1,1], \\ \frac1n & \text{otherwise}. \end{cases} $$ — Martin Sleziak May 1 at 7:53