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I'm reading the book "A First Course in Numerical Methods" by U. Ascher and C. Greif, and in the 2nd chapter it's written that

... we associate $x$ a floating point representation $fl(x)$ of the form similar to that of $x$ but with only t digits, so

$$fl(x) = sing(x) \times (1.d_1d_2d_3 \cdots d_{t-1}d_t) \times 2^e$$

Then it says that

the relative error $$\frac{fl(x) - x}{|x|}$$ is bounded by $n = \frac{1}{2} \times 2^{-t}$.

So for a double-precision floating pointer number (represented using IEEE), which uses $52$ for the mantissa, $n = 2^{-53} \times 1.1 \times 10^{-16}$

Why $-53$ if the mantissa of a double floating point number is $t = 52$?

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  • $\begingroup$ Well I guess you'll get your answer here: stackoverflow.com/questions/4930269/… It's related to what is called the implicit bit. $\endgroup$ – G.Clavier Apr 6 '16 at 10:35
  • $\begingroup$ @G.Clavier I thought it could be because of this, since I'm aware of this implicit digit 1, but I was not sure, so I decided the same to ask the question. $\endgroup$ – nbro Apr 6 '16 at 10:42

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