Zorich's "Mathematical Analysis I" Problem 25

  • Thread starter Thread starter Andrei1
  • Start date Start date
  • Tags Tags
    Analysis
AI Thread Summary
The discussion revolves around a problem from Zorich's "Mathematical Analysis I," specifically regarding the representation of numbers on a computer in the form x = ±q^p∑(n=1 to k)(α_n/q^n). The key points include the clarification that n=1 is correct in the summation, as the mantissa must be less than 1, making the first term either 0 or 1/2 when q=2. The order of the number x is defined as the least power p such that q^p exceeds x, with the bounds for p typically being |p| ≤ 1023 and k=53 for a 64-bit IEEE floating point representation, contrasting with the original problem's stated limits of |p| ≤ 64 and k=35. This highlights the importance of understanding the numerical range and representation in computational contexts.
Andrei1
Messages
36
Reaction score
0
Here is a problem from Zorich's "Mathematical analysis I", pg.69.
25. A number $$x$$ is represented on a computer as $$x=\pm q^p\sum_{n=1}^{k}\frac{\alpha_n}{q^n}$$, where $$p$$ is the order of $$x$$ and $$M=\sum_{n=1}^{k}\frac{\alpha_n}{q^n}$$ is the mantissa of the number $$x$$ $$\left(\frac{1}{q}\leqslant M<1\right).$$ Now a computer works only with a certain range of numbers: for $$q=2$$ usually $$|p|\leqslant 64$$, and $$k=35.$$ Evaluate this range in the decimal system.
I suspect this text has misprints: is it correct that $$n=1$$ under $$\sum$$ and why, or it should be $$n=0$$? By order I understand the unique $$p\in\mathbb{Z}$$ such that $$q^{p}\leqslant x<q^{p+1}.$$
 
Technology news on Phys.org
Andrei said:
Here is a problem from Zorich's "Mathematical analysis I", pg.69.

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

Hi Andrei,

No misprint - it should really be $$n=1$$, 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).
 
Thread 'Is this public key encryption?'
I've tried to intuit public key encryption but never quite managed. But this seems to wrap it up in a bow. This seems to be a very elegant way of transmitting a message publicly that only the sender and receiver can decipher. Is this how PKE works? No, it cant be. In the above case, the requester knows the target's "secret" key - because they have his ID, and therefore knows his birthdate.
I tried a web search "the loss of programming ", and found an article saying that all aspects of writing, developing, and testing software programs will one day all be handled through artificial intelligence. One must wonder then, who is responsible. WHO is responsible for any problems, bugs, deficiencies, or whatever malfunctions which the programs make their users endure? Things may work wrong however the "wrong" happens. AI needs to fix the problems for the users. Any way to...
Back
Top