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Fix finite number example

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Tian Cheng (Riccardo) Xia
2023-12-27 17:49:28 +01:00
parent 1479a99c47
commit e0ad5ed846
1 changed files with 3 additions and 2 deletions
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src/statistical-and-mathematical-methods-for-ai/sections/_finite_numbers.tex
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@ -194,11 +194,12 @@ A floating-point operation causes a small rounding error:
However, some operations may be subject to the \textbf{cancellation} problem which causes information loss. However, some operations may be subject to the \textbf{cancellation} problem which causes information loss.
\marginnote{Cancellation} \marginnote{Cancellation}
\begin{example} \begin{example}
Given $x = 1$ and $y = 1 \cdot 10^{-16}$, we want to compute $x + y$ in $\mathcal{F}(10, 16, U, L)$.\\ Given $x = 1$ and $y = 1 \cdot 10^{-17}$, we want to compute $x + y$ in $\mathcal{F}(10, 16, U, L)$.
It is assumed that $U$ and $L$ are sufficient for this example.
\begin{equation*} \begin{equation*}
\begin{split} \begin{split}
z & = \texttt{fl}(x) + \texttt{fl}(y) \\ z & = \texttt{fl}(x) + \texttt{fl}(y) \\
& = 0.1 \cdot 10^1 + 0.1 \cdot 10^{-15} \\ & = 0.1 \cdot 10^1 + 0.1 \cdot 10^{-16} \\
& = (0.1 + 0.\overbrace{0\dots0}^{\mathclap{16\text{ zeros}}}1) \cdot 10^1 \\ & = (0.1 + 0.\overbrace{0\dots0}^{\mathclap{16\text{ zeros}}}1) \cdot 10^1 \\
& = 0.1\overbrace{0\dots0}^{\mathclap{15\text{ zeros}}}1 \cdot 10^1 & = 0.1\overbrace{0\dots0}^{\mathclap{15\text{ zeros}}}1 \cdot 10^1
\end{split} \end{split}