Zorich's "Mathematical Analysis I" Problem 25

[Andrei1]

Here is a problem from Zorich's "Mathematical analysis I", pg.69.

25. A number \(\displaystyle x\) is represented on a computer as \(\displaystyle x=\pm q^p\sum_{n=1}^{k}\frac{\alpha_n}{q^n}\), where \(\displaystyle p\) is the order of \(\displaystyle x\) and \(\displaystyle M=\sum_{n=1}^{k}\frac{\alpha_n}{q^n}\) is the mantissa of the number \(\displaystyle x\) \(\displaystyle \left(\frac{1}{q}\leqslant M<1\right).\) Now a computer works only with a certain range of numbers: for \(\displaystyle q=2\) usually \(\displaystyle |p|\leqslant 64\), and \(\displaystyle k=35.\) Evaluate this range in the decimal system.

I suspect this text has misprints: is it correct that \(\displaystyle n=1\) under \(\displaystyle \sum\) and why, or it should be \(\displaystyle n=0\)? By order I understand the unique \(\displaystyle p\in\mathbb{Z}\) such that \(\displaystyle q^{p}\leqslant x<q^{p+1}.\)

 

[I like Serena]

Here is a problem from Zorich's "Mathematical analysis I", pg.69.

I suspect this text has misprints: is it correct that \(\displaystyle n=1\) under \(\displaystyle \sum\) and why, or it should be \(\displaystyle n=0\)? By order I understand the unique \(\displaystyle p\in\mathbb{Z}\) such that \(\displaystyle q^{p}\leqslant x<q^{p+1}.\)

Hi Andrei,

No misprint - it should really be \(\displaystyle n=1\), since a mantissa is always less than 1, so with $q=2$ the first term is either $\frac 0 2$ or $\frac 1 2$.

The order, as used here, would be the least power of $p$ that is greater than the number. That is:
$$q^{p-1}\le |x| < q^p$$

I don't know where those bounds for $p$ and $k$ are coming from, but $|p| \le 1023$ and $k=53$ is about the most common, belonging to a 64-bit IEEE floating point number (see wiki).