I got a solution using DSolve as :
Function[{t}, -((
1. (2. + k (-1. - 2. t) + k^2 (-0.5 + 1. t + 1. t^2)))/k^3) +
E^(-1. k t) C[1]]
What does '.' mean and '()'? For example, in the case of '2.+k(-1.-2.t)', what does it mean?
How to translate it to normal Math notation?
Pretty simple because the first one is regularly used in languages like C and Java too and the second one actually is math-notation. Therefore,
1. just means 1.0. In programming languages this is different from a simple 1 which usually means an integer while 1.0 means a real number. In Mathematica a 1 is a perfect one with infinite precision, while 1.0 has only machine precision.
k(1-t) just means k*(1-t). Parenthesis are used like in normal math too.
Related
Source: Google Code Jam. https://code.google.com/codejam/contest/10224486/dashboard#s=a&a=1
We're asked to calculate Prob(K successes from N trials) where each of the N trials has a known success probability of p_n.
Some Analysis and thoughts on the problem are given after the Code Jam.
They observe that evaluating all possible outcomes of your N trials would take you an exponential time in N, so instead they provide a nice "dynamic programming" style solution that's O(N^2).
Let P(p#q) = Prob(p Successes after the first q Trials)
Then observe the fact that:
Prob(p#q) = Prob(qth trial succeeds)*P(p-1#q-1) + Prob(qth trial fails)*P(p#q-1)
Now we can build up a table of P(i#j) where i<=j, for i = 1...N
That's all lovely - I follow all of this and could implement it.
Then as the last comment, they say:
In practice, in problems like this, one should store the logarithms of
probabilities instead of the actual values, which can become small
enough for floating-point precision errors to matter.
I think I broadly understand the point they're trying to make, but I specifically can't figure out how to use this suggestion.
Taking the above equation, and substuting in some lettered variables:
P = A*B + C*D
If we want to work in Log Space, then we have:
Log(P) = Log(A*B + C*D),
where we have recursively pre-computed Log(B) and Log(D), and A & B are known, easily-handled decimals.
But I don't see any way to calculate Log(P) without just doing e^(Log(B)), etc. which feels like it would defeat to point of working in log space`?
Does anyone understand in better detail what I'm supposed to be doing?
Starting from the initial relation:
P = A⋅B + C⋅D
Due to its symmetry we can assume that B is larger than D, without loss of generality.
The following processing is useful:
log(P) = log(A⋅B + C⋅D) = log(A⋅elog(B) + C⋅elog(D)) = log(elog(B)⋅(A + C⋅elog(D) - log(B))
log(P) = log(B) + log(A + C⋅elog(D) - log(B)).
This is useful because, in this case, log(B) and log(D) are both negative numbers (logarithms of some probabilities). It was assumed that B is larger than D, thus its log is closer to zero. Therefore log(D) - log(B) is still negative, but not as negative as log(D).
So now, instead of needing to perform exponentiation of log(B) and log(D) separately, we only need to perform exponentiation of log(D) - log(B), which is a mildly negative number. So the above processing leads to better numerical behavior than using logarithms and applying exponentiation in the trivial way, or, equivalently, than not using logarithms at all.
I am aware of the fact that the Sieve of Eratosthenes can be implemented so that it finds primes continuosly without an upper bound (the segmented sieve).
My question is, could the Sieve of Atkin/Bernstein be implemented in the same way?
Related question: C#: How to make Sieve of Atkin incremental
However the related question has only 1 answer, which says "It's impossible for all sieves", which is obviously incorrect.
Atkin/Bernstein give a segmented version in Section 5 of their original paper. Presumably Bernstein's primegen program uses that method.
In fact, one can implement an unbounded Sieve of Atkin (SoA) not using segmentation at all as I have done here in F#. Note that this is a pure functional version that doesn't even use arrays to combine the solutions of the quadratic equations and the squaresfree filter and thus is considerably slower than a more imperative approach.
Berstein's optimizations using look up tables for optimum 32-bit ranges would make the code extremely complex and not suitable for presentation here, but it would be quite easy to adapt my F# code so that the sequences start at a set lower limit and are used only over a range in order to implement a segmented version, and/or applying the same techniques to a more imperative approach using arrays.
Note that even Berstein's implementation of the SoA isn't really faster than the Sieve of Eratosthenes with all possible optimizations as per Kim Walisch's primesieve but is only faster than an equivalently optimized version of the Sieve of Eratosthenes for the selected range of numbers as per his implementation.
EDIT_ADD: For those who do not want to wade through Berstein's pseudo-code and C code, I am adding to this answer to add a pseudo-code method to use the SoA over a range from LOW to HIGH where the delta from LOW to HIGH + 1 might be constrained to an even modulo 60 in order to use the modulo (and potential bit packing to only the entries on the 2,3,5 wheel) optimizations.
This is based on a possible implementation using the SoA quadratics of (4*x^2 + y^), (3*x^2 + y^2), and (3*x^2 -y^2) to be expressed as sequences of numbers with the x value for each sequence fixed to values between one and SQRT((HIGH - 1) / 4), SQRT((HIGH - 1) / 3), and solving the quadratic for 2*x^2 + 2*x - HIGH - 1 = 0 for x = (SQRT(1 + 2 * (HIGH + 1)) - 1) / 2, respectively, with the sequences expressed in my F# code as per the top link. Optimizations to the sequences there use that when sieving for only odd composites, for the "4x" sequences, the y values need only be odd and that the "3x" sequences need only use odd values of y when x is even and vice versa. Further optimization reduce the number of solutions to the quadratic equations (= elements in the sequences) by observing that the modulo patterns over the above sequences repeat over very small ranges of x and also repeat over ranges of y of only 30, which is used in the Berstein code but not (yet) implemented in my F# code.
I also do not include the well known optimizations that could be applied to the prime "squares free" culling to use wheel factorization and the calculations for the starting segment address as I use in my implementations of a segmented SoE.
So for purposes of calculating the sequence starting segment addresses for the "4x", "3x+", and "3x-" (or with "3x+" and "3x-" combined as I do in the F# code), and having calculated the ranges of x for each as per the above, the pseudo-code is as follows:
Calculate the range LOW - FIRST_ELEMENT, where FIRST_ELEMENT is with the lowest applicable value of y for each given value of x or y = x - 1 for the case of the "3x-" sequence.
For the job of calculating how many elements are in this range, this boils down to the question of how many of (y1)^2 + (y2)^2 + (y3)^2... there are where each y number is separated by two, to produce even or odd 'y's as required. As usual in square sequence analysis, we observe that differences between squares have a constant increasing increment as in delta(9 - 1) is 8, delta(25 - 9) is 16 for an increase of 8, delta (49 - 25) is 24 for a further increase of 8, etcetera. so that for n elements the last increment is 8 * n for this example. Expressing the sequence of elements using this, we get it is one (or whatever one chooses as the first element) plus eight times the sequence of something like (1 + 2 + 3 + ...+ n). Now standard reduction of linear sequences applies where this sum is (n + 1) * n / 2 or n^2/2 + n/2. This we can solve for how many n elements there are in the range by solving the quadratic equation n^2/2 + n/2 - range = 0 or n = (SQRT(8*range + 1) - 1) / 2.
Now, if FIRST_ELEMENT + 4 * (n + 1) * n does not equal LOW as the starting address, add one to n and use FIRST_ELEMENT + 4 * (n + 2) * (n + 1) as the starting address. If one uses further optimizations to apply wheel factorization culling to the sequence pattern, look up table arrays can be used to look up the closest value of used n that satisfies the conditions.
The modulus 12 or 60 of the starting element can be calculated directly or can be produced by use of look up tables based on the repeating nature of the modulo sequences.
Each sequence is then used to toggle the composite states up to the HIGH limit. If the additional logic is added to the sequences to jump values between only the applicable elements per sequence, no further use of modulo conditions is necessary.
The above is done for every "4x" sequence followed by the "3x+" and "3x-" sequences (or combine "3x+" and "3x-" into just one set of "3x" sequences) up to the x limits as calculated earlier or as tested per loop.
And there you have it: given an appropriate method of dividing the sieve range into segments, which is best used as fixed sizes that are related to the CPU cache sizes for best memory access efficiency, a method of segmenting the SoA just as used by Bernstein but somewhat simpler in expression as it mentions but does not combine the modulo operations and bit packing.
Is there a fast method for taking the modulus of a floating point number?
With integers, there are tricks for Mersenne primes, so that its possible to calculate y = x MOD 2^31-1 without needing division. integer trick
Can any similar tricks be applied for floating point numbers?
Preferably, in a way that can be converted into vector/SIMD operations, or moved into GPGPU code. This rules out using integer calculations on the floating point data.
The primes I'm interested in would be 2^7-1 and 2^31-1, although if there are more efficient ones for floating point numbers, those would be welcome.
One intended use of this algorithm would be to calculate a running "checksum" of input floating point numbers as they are being read into an algorithm. To avoid taking up too much of the calculation capability, I'd like to keep this lightweight.
Apparently a similar technique is used for larger numbers, particularly 2^127 - 1. Unfortunately, the math in the paper is beyond me, and I haven't been able to figure out how to convert it to smaller primes.
Example of floating point MOD 2^127 - 1 - HASH127
I looked at djb's paper, and you have it easier, since 31 bits fits comfortably into the 53-bit precision double significand. Assuming that your checksum consists of some ring operations over Z/(2**31 - 1), it will be easier (and faster) to solve the relaxed problem of computing a small representative of x mod Z/(2**31 - 1); at the end, you can use integer arithmetic to find a canonical one, which is slow but shouldn't happen too often.
The basic reduction step is to replace an integer x = y + 2**31 * z with y + z. The trick that djb uses is to compute w = (x + L) - L, where L is a large integer carefully chosen to provoke roundoff in such a way that z = 2**-31 * w. Then compute y = x - w and output y + z, which will have magnitude at most 2**32. (I apologize if this operation isn't quite enough; if so, please post your checksum algorithm.)
The choice of L involves knowing how precise the significand is. For the modulus 2**31 - 1, we want the unit of least precision (ulp) to be 2**31. For doubles in the range [1.0, 2.0), the ulp is 2**-52, so L should be 2**52 * 2**31. If you were doing this with the modulus 2**7 - 1, then you'd take L = 2**52 * 2**7. As djb notes, this trick depends crucially on intermediate results not being computed in higher precision.
How would I solve:
-x^3 - x - 4 = 0
You can't use quadratic, because it is to the 3rd power, right?
I know I it should come out to ~1.3788 but I'm not sure how I would derive that.
I started with:
f(x) = x + (4/(x^2 + 1)).
Solving for 0, moving the x over to the other side, multiplying by (x^2 + 1) on both sides, I end up with:
-x(x^2 + 1) = 4,
or
-x^3 - x - 4 = 0.
Finding roots of equations by using Newton's method or Fixed point iteration
Algebraically, you want to use Cardano's method:
http://www.math.ucdavis.edu/~kkreith/tutorials/sample.lesson/cardano.html
Using this method, it's about as easy to solve as the quadratic.
Actually, this is possibly clearer:
http://en.wikipedia.org/wiki/Cubic_function#Summary
Find a root by using newton iteration (see link below). Then divide the polynomial by (x-TheRootYouFound). The result will be a quadratic formula that you can plug into your quadratic root finder of your choice.
About Newton Iteration:
http://en.wikipedia.org/wiki/Newton%27s_method
About Polynomial Division
http://en.wikipedia.org/wiki/Polynomial_long_division
This article may be interesting for you as well. It covers more robust ways to solve your problem at the expense of some additional complexity.
http://en.wikipedia.org/wiki/Root-finding_algorithm
It's a cubic function. You're correct, the quadratic formula does not apply.
You gave one root, but in general there are three.
How did you arrive at that single value? Trial and error? That's legitimate. You don't need to "derive" anything.
x^3 + a2*x^2 + a1*x + a0 = 0 can be written as (x-x1)*(x-x2)*(x-x3) = 0, where x1, x2, and x3 are the three roots. If you know that the root you cited is correct, you can divide it out and leave (x-x2)*(x-x3) = 0, which is a quadratic that you can apply the usual techniques to.
This may not help from a programming viewpoint, but from a math perspective...
Note than in this particular cubic function you need to consider imaginary numbers, because when x = i then you have a denominator that is zero (in your original equation). Also, generally speaking you shouldn't multiply or divide by variables (adding and subtracting is fine though) when you move them to the other side of the equation because you'll generally forget of about the condition where the term you multiplied or divided by is zero. Those answers need to be excluded from the solution set.
x = i is an example of an excluded solution in the above cubic. You need to evaluate your excluded solutions before you manipulate the equation at all.
I'm trying to use Taylor series to develop a numerically sound algorithm for solving a function. I've been at it for quite a while, but haven't had any luck yet. I'm not sure what I'm doing wrong.
The function is
f(x)=1 + x - sin(x)/ln(1+x) x~0
Also: why does loss of precision even occur in this function? when x is close to zero, sin(x)/ln(1+x) isn't even close to being the same number as x. I don't see where significance is even being lost.
In order to solve this, I believe that I will need to use the Taylor expansions for sin(x) and ln(1+x), which are
x - x^3/3! + x^5/5! - x^7/7! + ...
and
x - x^2/2 + x^3/3 - x^4/4 + ...
respectively. I have attempted to use like denominators to combine the x and sin(x)/ln(1+x) components, and even to combine all three, but nothing seems to work out correctly in the end. Any help is appreciated.
The loss of precision can come in because when x ~ 0, ln(1+x) is also close to 0, so you wind up dividing by a very small number. Computers aren't very good at that ;-)
If you use the Taylor series for ln(1+x) directly, it's going to be kind of a pain because you'll wind up dividing by an infinite series of terms. For cases like this, I usually prefer to just compute the Taylor series for the entire function as a whole from the definition:
f(x) = f(0) + f'(0) x + f''(0) x/2 + f'''(0) x/6 + ...
from which you'll get
f(x) = 2 + 3x/2 - x^2/4 - x^3/24 - x^4/240 - 23x^5/1440 + 31x^6/2880 ...
(I cheated and plugged it into Mathematica ;-) Like Steve says, this series doesn't converge all that quickly, although I can't think of a better method at the moment.
EDIT: I think I misread the question - if all you're trying to do is find the zeros of the function, there are definitely better ways than using a Taylor series.
As this is homework, I'm just going to try to give a few pointers in the right direction.
Solution 1
Rather than using the Talyor series approximation, try to simply use a root finding algorithm such as the Newton-Raphson method, linear interpolation, or interval bisection (or combine them even). They are very simple to implement, and with an appropiate choice of starting value(s), the root can converge to a precise value quite quickly.
Solution 2
If you really need to use the Taylor series approximation for whatever reason, then just expand the sin(x), ln(x), and whatever else. (Multiplying through by ln(x) to remove the denominator in your case will work). Then you'll need to use some sort of polynomial equation solver. If you want a reasonable degree of accuracy, you'll need to go beyond the 3rd or 4th powers I'd imagine, which means a simple analytical solution is not going to be easy. However, you may want to look into something like the Durand-Kerner method for solving general polynomials of any order. Still, if you need to use high-order terms this approach is just going to lead to complications, so I would definitely recommend solution 1.
Hope that helps...
I think you need to look at what happens to ln(x+1) as x -->0 and you will see why this function does not behave well near x = 0.
I haven't looked into this that closely, but you should be aware that some taylor series converge very, very slowly.
Just compute the Taylor series of f directly.
Maxima gives me (first 4 terms about x=0):
(%i1) f(x):=1 + x - sin(x)/log(1+x);
- sin(x)
(%o1) f(x) := 1 + x + ----------
log(1 + x)
(%i2) taylor(f(x),x,0,4);
2 3 4
x x x x
(%o2)/T/ - + -- + -- + --- + . . .
2 4 24 240
Method used in question is correct - just make sure your calculator is in radians mode.