Why we need size of the page table should be some fraction of virtual address space

The Page Table should have all virtual page number which are in its logical address space, Why it’s the case?

  1. Is it because we want to access Page Table entry fast just like an array where key is virtual page number i.e. constant time?

Or

  1. Is it due to structure of the process? (I mean Our program uses whole logical space; In general at address 0 we have Code and at address Max we have stack which is variable. Which means can point to any address of logical address space)

Algorithm for Egyptian Fraction

An Egyptian fraction was written as a sum of unit fractions, meaning the numerator is always 1; further, no two denominators can be the same. As easy way to create an Egyptian fraction is to repeatedly take the largest unit fraction that will fit, subtract to find what remains, and repeat until the remainder is a unit fraction, for example:

  1. 7 divided by 15 is less than 1/2 but more than 1/3, so the first unit fraction is 1/3 and the first remainder is 2/15.

  2. Then 2/15 is less than 1/7 but more than 1/8, so the second unit fraction is 1/8 and the second remainder is 1/120.

  3. That’s in unit form, so we are finished: 7 ÷ 15 = 1/3 + 1/8 + 1/120

I’m trying to solve the egyptian fraction problem where I’m using the below greedy method:

def egyptianFraction(nr, dr):   print("The Egyptian Fraction " +       "Representation of {0}/{1} is".              format(nr, dr), end="\n")   # empty list ef to store  # denominator  ef = []   # while loop runs until   # fraction becomes 0 i.e,  # numerator becomes 0  while nr != 0:       # taking ceiling      x = math.ceil(dr / nr)       # storing value in ef list      ef.append(x)       # updating new nr and dr      nr = x * nr - dr      dr = dr * x   # printing the values  for i in range(len(ef)):      if i != len(ef) - 1:          print(" 1/{0} +" .                   format(ef[i]), end = " ")      else:          print(" 1/{0}" .                  format(ef[i]), end = " ")  egyptianFraction(6, 14)  

I need to build an algorithm that guarantees a maximum number of terms or a minimum largest denominator; for instance,

5 ÷ 121 = 1/25 + 1/757 + 1/763309 + 1/873960180913 +  1/1527612795642093418846225  but a simpler rendering of the same number is  1/33 + 1/121 + 1/363. 

I need to convert fraction to hex but there is perhaps a mask [on hold]

I get the following values from a machine. I type the hex value in and it sets the machine to a decimal with two places after the point.

Machine setting hex value
0.1 3DCCCCCD
0.11 3DE147AE
0.12 3DF5C28F
0.13 3E051EB8
0.25 3E800000
0.5 3F000000
0.75 3F400000
1 3F800000
2 40000000

I take 0.31 and I work it out to 0.4F5C28F6 but the machine says 0.31=3E9EB852 How do I get from 0.4F5C28F6 to 3E9EB852 is there some mask I need to apply?

Continued fraction terms in OEIS A215272

The OEIS sequence A215272 has terms: 81, 729, 59049, 43046721, …

I found the terms 81, 729, 59049 by checking the length of the repeating fraction of the product of Primorial/EulerPhi[Primorial], could someone explain the connection? Thanks.

cheers, Jamie

Mathematica code courtesy of a Mathematica stack exchange answer:

Primorial[n_] := Times @@ Prime[Range[n]] pp[i_Integer /; 0 <= i <= 1] := Primorial[i]/EulerPhi[Primorial[i]] pp[i_Integer /; i >= 2] := Primorial[i]/EulerPhi[Primorial[i - 1]]  Table[Length[RealDigits[Product[pp[i], {i, 0, n}]][[1, -1]]], {n, 0, 13}]  {0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 27, 729, 59049, 43046721} 

can’t evaluate integral of the form $\log \left(2+\lambda _+ P+ 2 \sqrt{\lambda _* P^2+\lambda _+ P+1}\right)$ times a fraction

I’m trying to evaluate a bunch of integrals of the form $ $ \int_0^{\infty } \frac{\sum\limits_{i=0}^k (c_i P^i)}{\left(P^2+1\right)^n}\log \left(2+\lambda _+ P+ 2 \sqrt{\lambda _* P^2+\lambda _+ P+1}\right)\, dP$ $ where $ n$ and $ k$ is are integers ($ n\ge 1$ , $ m<2n$ ), $ \lambda_+>0$ , $ \lambda_*>0$ , and the polynomial $ \sum\limits_{i=0}^m(c_i P^i)$ only has even powers of $ P$ ; for example $ \frac{3-2P^2 +47P^4+4P^6}{(1+P^2)^5}$ .

For each particular term of the form $ $ \int_0^{\infty } \frac{1}{\left(P^2+1\right)^k}\log \left(2+\lambda _+ P+ 2 \sqrt{\lambda _* P^2+\lambda _+ P+1}\right)\, dP$ $ parital integration $ \int u \,dv = u v – \int v\,du$ (with $ dv=log(\ldots)$ ) will give me a polynomial of the order $ 2n-4$ divided by $ 1/(1+P^2)^{n-1}$ (which can easily be solved) plus $ $ \int\frac{\arctan(P)}{P}\left(1-\frac{1}{\sqrt{\lambda _* P^2+\lambda _+ P+1}}\right)dP$ $ which I don’t know how to solve (nor does Mathematica apparently). In certain instances (for particular set of $ c_i$ the (indefinite) integral can be solved by partial integration when the $ \arctan(x)$ terms cancel out (for example $ \frac{1-16P^2 +25P^4-6P^6}{(1+P^2)^5}$ ).

The queation is how do I solve the integral when $ \arctan(x)$ terms don’t cancel out? I have tried all the variable substitutions I could think of, and couldn’t figure anything out by trying the residue theorem either.

Any help would be greatly appreciated.