# | 53 | 26 | 13 | 6 | 3 | 1 | 0 |
---|---|---|---|---|---|---|---|
r | 1 | 0 | 1 | 0 | 1 | 1 |
So, 101011 $\rightarrow$ reverse the order $\rightarrow$ $\mathbf{110101} = 53$
See Numerical Recipes 3rd Ed. Page 8, section 1.1
Called floating point because the decimal point floats
Addition: 1.57E2 + 2.0E0 is really 157.0E0 + 2.0E0
Floats and Doubles
S (1) | E (11) | M (52) |
---|---|---|
0 | 00000000000 | 0000000000000000000000000000000000000000000000000000 |
From Numerical Recipes: $$S\times M\times b^{E-e}$$
S | E | F | Value |
---|---|---|---|
any | 1-2046 | any | (-1)$^S\times$ 2$^{E-1023}$$\times$ 1.F |
any | 0 | nonzero | (-1)$^S\times$ 2$^{E-1022}$$\times$ 0.F |
0 | 0 | 0 | $+0.0$ |
1 | 0 | 0 | $-0.0$ |
0 | 2047 | 0 | $+\infty$ |
1 | 2047 | 0 | $-\infty$ |
any | 2047 | nonzero | NaN |
See also this site
using Bits
bits(1.5)
<0|011 11111111|1000 00000000 00000000 00000000 00000000 00000000 00000000>
where $|\delta|\le 3\epsilon_{mach}.$
where $|\delta|\le(n-1)\epsilon_{mach}$.
Or
$$x = \frac{2c}{-b\pm\sqrt{b^2-4ac}}.$$What can go wrong in evaluating the solution using either of these two solutions?