Solving Linear System with Noisy Input I have the following triangular system
\begin{equation}  
\begin{pmatrix}
    1   &  &  & & \\
    \mu_1 & 2 &  &  &  \\
    \mu_2 & \mu_1 & 3 & \\
    \vdots & \vdots &\ddots & \ddots & \\
    \mu_{n-1} & \mu_{n-2} & \dots &  \mu_1 & n
  \end{pmatrix}
  \begin{pmatrix}
    c_{n-1} \\
    c_{n-2} \\
    c_{n-3} \\
    \vdots \\
    c_0
  \end{pmatrix}
  =
  \begin{pmatrix}
    - \mu_{1} \\
    - \mu_{2} \\
    - \mu_{3} \\
    \vdots \\
    - \mu_n
  \end{pmatrix}
  \Leftrightarrow A c = b
\end{equation}
For all the values $\mu_j$ I have approximations $\hat{\mu}_j$ such that
$|\mu_j - \hat{\mu_j}| \leq \epsilon$. I want to compute an upper bound for the error in the solution of this system, that is a bound on 
$\| c - \hat{c}\|_{\infty}$. 
Let's assume that we use forward substitution to solve this system. 
Is it true that, assuming that all the operations are done with infinite precision, would give a solution $\hat{c}$ such that $\|c -\hat{c}\|$ would be upper bounded by the forward error of $c$ ?
To bound the forward error of the solution I found the following theorem
in this book.
Let $Ax = b$ and $(A + \Delta A) y = b + \Delta b$, where
  $|\Delta A| \leq \epsilon E$ and $|\Delta b| \leq \epsilon f$, and
  assume that $\epsilon \| |A^{-1}| E \| < 1$, where $\|\cdot\|$ is an absolute
  norm. Then
  \begin{equation}
\frac {\|x-y\|}{\|x\|} \leq
    \frac \epsilon {1 - \epsilon \| |A^{-1}| E\|}
    \frac {\| |A^{-1}|(E|x| + f)\|} {\|x\|}
\end{equation}
  and for the $\infty$-norm this bound is attainable to first order in $\epsilon$.
Since I am not sure that I have fully understood the backward-forward error concept, I am not sure whether using the above theorem with
\begin{equation}
E = \begin{pmatrix}
    0   &  &  & & \\
    1 & 0 &  &  &  \\
    1 & 1 & 0 & \\
    \vdots & \vdots &\ddots & \ddots & \\
    1 & 1 & \dots &  1 & 0
  \end{pmatrix},\qquad
f =   \begin{pmatrix}
    1 \\
    1 \\
    1 \\
    \vdots \\
    1
  \end{pmatrix}
\end{equation}
will give me an upper bound for the error in my noisy system.
 A: You are right. You need the forward error to answer your question. The term $\| x - y \|$ in the inequality from your book corresponds to $\| c - \hat{c} \|$ in your example. Hence, to bound $\| x - y \|$ you first multiply the inequality by $\| x \|$ to get
\begin{equation}
   \|x-y\| \leq
   \frac{\epsilon}{1 - \epsilon \| |A^{-1}| E\|}
    \| |A^{-1}|(E|x| + f)\|
   \,.
\end{equation}  
Then, to bound the infinity norm, you can use the inequality $\| v \|_\infty \le \| v \|_2$. Assuming the inequality from the book uses the $\ell_2$-norm, we have that
 \begin{equation}
   \|x-y\|_\infty \leq
   \frac{\epsilon}{1 - \epsilon \| |A^{-1}| E\|}
    \| |A^{-1}|(E|x| + f)\|
   \,.
\end{equation}
This statement requires, as you mentioned, that all operations are carried out using exact arithmetic.
The problem is, that you cannot assume that the entires of $E$ and $f$ are all one. You have to assume that
\begin{equation}
E = \begin{pmatrix}
    0   &  &  & & \\
    * & 0 &  &  &  \\
    * & * & 0 & \\
    \vdots & \vdots &\ddots & \ddots & \\
    * & * & \dots &  * & 0
  \end{pmatrix},\qquad
f =   \begin{pmatrix}
    * \\
    * \\
    * \\
    \vdots \\
    *
  \end{pmatrix}
\end{equation}
where the stars mark arbitrary numbers with absolute value smaller or equal to one. Using the submultiplicativity of the matrix norm, i.e., $\| A B\| \le \| A \| \| B \|$, however, you can factor out the $\| E \|$. Then, your answer does just depend on $\| E \|$ and $\epsilon$.
