The error I get from the calculation
When I try to calculate the square root of a fraction using a calculator, I get a mathematical error. (see link to image) Why is this an apparently invalid operation?
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The error I get from the calculation
When I try to calculate the square root of a fraction using a calculator, I get a mathematical error. (see link to image) Why is this an apparently invalid operation?
The Page Table should have all virtual page number which are in its logical address space, Why it’s the case?
Or
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:
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.
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.
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 believe following is a continued fraction. I’m stumped on how to solve for x
$ x = \frac{a}{b} = \frac{b}{a/3}$
I know it can be re-written as
$ x = \frac{a}{b} = \frac{3b}{a}$
$ x = a^2 = 3b^2$
I’m unsure where to go from here.
Below are the choices
$ x=9$
$ x=\frac{1}{3}$
$ x=3$
$ x=\frac{1}{\sqrt3}$
$ x={\sqrt3}$
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?
I am currently using custom formatting to get an improper fraction (###/###), however, I will need the integers to be 7 instead of 7/1. How do I achieve both improper fraction OR integer (if it can be simplified)?
Question no 3 , kindly explain the relation btw lagrange and partial fraction](https://i.stack.imgur.com/EYVIk.jpg)
I am unable to figure out the relation btw lagrange interpolation formula and partial fractions.
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}
How to solve using perturbation method for small e?
$ \frac{d}{dx}((1-\frac{ey}{y+1})\frac{dy}{dx})=0$
$ x=0,y=0$
$ x=1,y=1$
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.