I would like to calculate the following function for very large number (for example e^e^e^e^10) and would like to know the sign of the following term in general. I tried for some numbers, but it is negative. Is there any m0 such that for all n>m>m0, the following function is positive.
where n is greater than m.
I tied with Mathematica, but it does not compute for the above number. Should I use special package?
thanks
Related
I'm working on a game for which I want deterministic demo playback that is portable between architectures that treat floating point numbers differently. I'm using the Racket language, which conveniently has, as a primitive data type, non-floating-point representations of rational-number fractions. I want to use these to implement an approximately normally-distributed random function that accepts parameters for mean and standard deviation (skewness would be gold-plating).
Because of the limitations I've mentioned, any operation that takes in rational numbers and puts out irrational ones will need to be reimplemented from scratch in a way that produces approximations based on Racket's native fractions, not based on floating points. I've looked around at various algorithms for normal random functions, but of these, even many of the "simplest" ones like the Box-Muller transform involve things like square roots, logarithms, and trig functions. Iterated averaging is easy, so square roots aren't a problem, but I don't want to reinvent any more wheels than I need to here.
What are some algorithms I can use for generating approximately normal random numbers without invoking irrational operations like roots, logarithms, and trig functions?
I settled on a solution after typing up this question but before sending it, so I'll Share My Knowledge Q&A-Style.
After poring over several different SO posts on normally-distributed random numbers, I found that the best solution for my purposes was actually the most naive one: abuse the Central Limit Theorem. Random variables of any distribution, when added up, approximate a normal distribution just fine. In Racket, my solution turned out to be the delightfully concise
(define (random/normal μ σ)
(+ (* (- (for/sum ([i 12])
(random/uniform 0 1))
6)
σ)
μ))
where uniform-random is my function for generating uniformly random rational numbers.
In infix, imperative pseudocode, this means:
Function random_normal(μ, σ):
iterations := 12
sum := 0
for i from 1 to iterations:
sum += random_uniform(0, 1)
sum -= iterations / 2 # center the distribution on 0
return σ * sum + μ
Why 12 iterations?
A few SO answers mention this solution, but don't explain why 12 is a magic number here. When we add up those numbers, we want the standard deviation of that random sum to equal 1 so that we can stretch out or squish down the bell curve by the desired amount in a single multiplicative step.
If you sum a sample of random variables the standard deviation of the approximately normal distribution this creates is equal to
where is the standard deviation of the variables themselves.* The standard deviation of a uniform random distribution from 0 to 1 is equal to † so by substituting this in for we see that what we want is just
which works out easily to
* See "Central Limit Theorem" on Wolfram MathWorld. Equation is given under identity (2), here multiplied by N to give the standard deviation of the sum rather than of the average.
† See "Continuous uniform distribution" on Wikipedia. Table on the right, "variance" square-rooted.
But doesn't this limit your range to ±6 standard deviations?
It does, but the range of your distribution has to be truncated somewhere unless you have infinite memory, and ±6σ is A) almost as good as Box-Muller on a 32-bit machine and B) already huge.
I don't mean a function that generates random numbers, but an algorithm to generate a random function
"High dimension" means the function is multi-variable, e.g. a 100-dim function has 100 different variables.
Let's say the domain is [0,1], we need to generate a function f:[0,1]^n->[0,1]. This function is chosen from a certain class of functions, so that the probability of choosing any of these functions is the same.
(This class of functions can be either all continuous, or K-order derivative, whichever is convenient for the algorithm.)
Since the functions on a closed interval domain are uncountable infinite, we only require the algorithm to be pseudo-random.
Is there a polynomial time algorithm to solve this problem?
I just want to add a possible algorithm to the question(but not feasible due to its exponential time complexity). The algorithm was proposed by the friend who actually brought up this question in the first place:
The algorithm can be simply described as following. First, we assume the dimension d = 1 for example. Consider smooth functions on the interval I = [a; b]. First, we split the domain [a; b] into N small intervals. For each interval Ii, we generate a random number fi living in some specific distributions (Gaussian or uniform distribution). Finally, we do the interpolation of
series (ai; fi), where ai is a characteristic point of Ii (eg, we can choose ai as the middle point of Ii). After interpolation, we gain a smooth curve, which can be regarded as a one dimensional random function construction living in the function space Cm[a; b] (where m depends on the interpolation algorithm we choose).
This is just to say that the algorithm does not need to be that formal and rigorous, but simply to provide something that works.
So if i get it right you need function returning scalar from vector;
The easiest way I see is the use of dot product
for example let n be the dimensionality you need
so create random vector a[n] containing random coefficients in range <0,1>
and the sum of all coefficients is 1
create float a[n]
feed it with positive random numbers (no zeros)
compute the sum of a[i]
divide a[n] by this sum
now the function y=f(x[n]) is simply
y=dot(a[n],x[n])=a[0]*x[0]+a[1]*x[1]+...+a[n-1]*x[n-1]
if I didn't miss something the target range should be <0,1>
if x==(0,0,0,..0) then y=0;
if x==(1,1,1,..1) then y=1;
If you need something more complex use higher order of polynomial
something like y=dot(a0[n],x[n])*dot(a1[n],x[n]^2)*dot(a2[n],x[n]^3)...
where x[n]^2 means (x[0]*x[0],x[1]*x[1],...)
Booth approaches results in function with the same "direction"
if any x[i] rises then y rises too
if you want to change that then you have to allow also negative values for a[]
but to make that work you need to add some offset to y shifting from negative values ...
and the a[] normalization process will be a bit more complex
because you need to seek the min,max values ...
easier option is to add random flag vector m[n] to process
m[i] will flag if 1-x[i] should be used instead of x[i]
this way all above stays as is ...
you can create more types of mapping to make it even more vaiable
This might not only be hard, but impossible if you actually want to be able to generate every continuous function.
For the one-dimensional case you might be able to create a useful approximation by looking into the Faber-Schauder-System (also see wiki). This gives you a Schauder-basis for continuous functions on an interval. This kind of basis only covers the whole vectorspace if you include infinite linear combinations of basisvectors. Thus you can create some random functions by building random linear combinations from this basis, but in general you won't be able to create functions that are actually represented by an infinite amount of basisvectors this way.
Edit in response to your update:
It seems like choosing a random polynomial function of order K (for the class of K-times differentiable functions) might be sufficient for you since any of these functions can be approximated (around a given point) by one of those (see taylor's theorem). Choosing a random polynomial function is easy, since you can just pick K random real numbers as coefficients for your polynom. (Note that this will for example not return functions similar to abs(x))
I am working with large amounts of probabilities that I multiply so i quickly obtain very small numbers. But it seems that python finally store the final result as zero.
to overpass this difficutly, I decided to sum the logs of these probabilities (instead of directly multiplying the probabilities). This strategy returns a negative number (call it c) as expected.
But then, if I want to apply the exponential on c (to come back on the real value of my product of probabilities), I obtain the value zero because c is too largely negative (something like -123445,4).
How could I overpass this problem?
If you are going to use numbers of that magnitude you should use a specialized library which can handle arbitrary floating point precision. Check out mpmath or bigfloat package for example.
Computers natively only support number down to approximately exp(-300). Alternatively, you could restrict your code to store only the exponent and never convert it in a decimal representation.
I have a list of numbers x obtained from measurements. I need to calculate the quantity
where a is a positive number. Unfortunately, the individual terms of the sum can be very large and the sign of each term is determined by k and the associated value of x. This leads to cancellation and loss of precision.
I have found a couple of approaches such as compensated summation but wanted to check whether I am on the right path and whether there are better alternatives.
I have a very large positive integer number (million digits). I need represent it with the smallest possible function, this number is variable, it means, I need an algorithm that generates the smallest possible function to get the given number.
Example: For the number 29512665430652752148753480226197736314359272517043832886063884637676943433478020332709411004889 the algorithm must return "9^99". It must be able to analyze numbers and always return a math function that represent the number. Example the number 21847450052839212624230656502990235142567050104912751880812823948662932355202 must return "9^5^16+1".
Heard of Kolmogorov complexity?
To answer your question: unless you restrict yourself to some specific set of functions, it's impossible.
EDIT: Even in your example, how do you know that the shortest representation of 21847450052839212624230656502990235142567050104912751880812823948662932355202 is actually 9^5^16+1? Isn't it a quite hard to prove even in this specific case?
If you restrict yourself to some set of functions then you can use the following algorithm:
For i = 1 to n
enumerate all strings s of length i
if s represents a valid expression according to rules chosen a priori,
and evaluates to the number in the input,
return s
It is guaranteed to halt because on the last iteration of the outer loop (i = n) you will get eventually to a string contains the input verbatim.
Of course, this is not very efficient. Specifically O(bn) where n is the length of the input and b is the size of the alphabet.
Expanding on #ybungalobill's terse answer, your function is equivalent to a function that computes the Kolmogorov complexity of an arbitrary string. (The equivalence is obvious if you treat each digit of your very large numbers as characters, and the numbers as sequences of characters.)
According to the Wikipedia page on Kolmogorov complexity, the K(s) function that gives the complexity of a string s is not a computable function. (The page includes a proof.)
In other words, the algorithm you want simply does not exist.
#BlueRaja - Danny Pflughoeft: yes, it is. I'm trying to create some compression that uses this algorithm, but by the way this is impossible.
That's because it's technically impossible to compress arbitrary data, for the same reason, but that doesn't stop us from doing it :)
There are much better ways of compressing data, however. Take a look at, for instance, LZ. It is so ubiquitous that you can almost certainly find a library to do the compression for you, regardless of what language you're writing in. DEFLATE is another popular one.
Hope that helps!
If you're not looking for optimality, just a reasonably good job, then there are a bunch of heuristics you can use. For example, try to decompose n using all of the following
n = a^k + b
for k = 2, 3, ..., log n, and pick the one with the smallest a + b, say. You can compute a and b using a = floor(n^(1/k)) and b = n-a^k. Then recurse on a and b.
Of course, this uses only exponentiation and addition to find a good compression. If you allow subtraction as well, use a=round(n^(1/k)) instead and let b be negative.
Allowing multiplication as well makes it quite a bit harder because you would probably need to factor n.