Evaluating polynomials to 5 significant figures but only 1 sig fig returns - Maple Programming - polynomials

For example, A polynomial is defined as follows:
f := (x, y) -> 333.75y^6 + x^2(11x^2y^2 - y^6 - 12y^4 - 2) + 5.5y^8 + 1/2*x/y
In maple, I look to evaluate this to 5 significant figures like so:
evalf[5](f(77617,33096))
And obtain a value that is: 1*10^32.
Why is this not to 5 sig fig? Why is this not close to a value of 7.878 * 10^29 as you increase the number of sig fig required?
Thanks!


Don't reduce the working precision that low, especially if you are trying to compute an accurate answer (and then round it for convenience).
More importantly, for compound expressions the floating-point working precision (Digits, or the index of an evalf call) is just that: a specification of working precision and not an accuracy request.
By lowering the working precision so much you are seeing greater roundoff error in the floating-point computation.
restart;
f := (x, y) -> 333.75*y^6
+ x^2*(11*x^2*y^2 - y^6 - 12*y^4 - 2)
+ 5.5*y^8 + 1/2*x/y:
for d from 5 to 15 do
evalf[5](evalf[d](f(77617,33096)));
end do;
32
1 10
31
-3 10
30
1 10
29
8 10
29
7.9 10
29
7.88 10
29
7.878 10
29
7.8784 10
29
7.8785 10
29
7.8785 10
29
7.8785 10

Related

How to calculate pow(2,n) when n exceeds 64 in c++?

So, I am new to programming in c++ and i came across this question where i need to calculate pow(2,n)/2 where n>64 ?
i tried using unsigned long long int but as the limit of the c++ is only 2^64. So is there any method to calculate this.
Edit:
1 < n < 10^5
The result of the expression is used in further calculation
The question is asked on online platform.So, i cant use libraries like gmp to handle large numbers.
Question
You are given with an array A of size N. An element Ai is said to be charged if its value (Ai) is greater than or equal to Ki. Ki is the total number of subsets of array A that consist of element Ai.
Total charge value of the array is defined as summation of all charged elements present in the array mod (10^9)+7.
Your task is to output the total charge value of the given array.

An important detail here is that you're not being asked to compute 2n for gigantic n. Instead, you're being asked to compute 2n mod 109 + 7 for large n, and that's a different question.
For example, let's suppose you want to compute 270 mod 109 + 1. Notice that 270 doesn't fit into a 64-bit machine word. However, 270 = 230 · 235, and 235 does fit into a 64-bit machine word. Therefore, we could do this calculation to get 270 mod 109 + 7:
270 (mod 109 + 7)
= 235 · 235 (mod 109 + 7)
= (235 mod 109 + 7) · (235 mod 109 + 7) mod 109 + 7
= (34359738368 mod 109 + 7) · (34359738368 mod 109 + 7) mod 109 + 7
= (359738130 · 359738130) mod 109 + 7
= 129411522175896900 mod 109 + 7
= 270016253
More generally, by using repeated squaring, you can compute 2n mod 109 + 7 for any value of n in a way that nicely fits into a 64-bit integer.
Hope this helps!

The common approach in serious numerical work is to rewrite the formula's. You store log(x) instead of x, and later when you do need x it will typically be in a context where you didn't need all those digits anyway.

Generating variations in C++ - all r-digit numbers among n given digits?

I have a program in which I have to generate all r-digit numbers (if r is 2 - all 2-digit numbers) among n digits (these ought to be all numbers from 1 to n inclusive). My question is how can I do this recursively or iteratively, for example if n = 3 and r = 2, the result should be 12 13 21 23 31 32.

Density of fractions between 2 given numbers

I'm trying to do some analysis over a simple Fraction class and I want some data to compare that type with doubles.
The problem
Right know I'm looking for some good way to get the density of Fractions between 2 numbers. Fractions is basically 2 integers (e.g. pair< long, long>), and the density between s and t is the amount of representable numbers in that range. And it needs to be an exact, or very good approximation done in O(1) or very fast.
To make it a bit simpler, let's say I want all the numbers (not fractions) a/b between s and t, where 0 <= s <= a/b < t <= M, and 0 <= a,b <= M (b > 0, a and b are integers)
Example
If my fractions were of a data type which only count to 6 (M = 6), and I want the density between 0 and 1, the answer would be 12. Those numbers are:
0, 1/6, 1/5, 1/4, 1/3, 2/5, 1/2, 3/5, 2/3, 3/4, 4/5, 5/6.
What I thought already
A very naive approach would be to cycle trough all the possible fractions, and count those which can't be simplified. Something like:
long fractionsIn(double s, double t){
long density = 0;
long M = LONG_MAX;
for(int d = 1; d < floor(M/t); d++){
for(int n = ceil(d*s); n < M; n++){
if( gcd(n,d) == 1 )
density++;
}
}
return density;
}
But gcd() is very slow so it doesn't works. I also try doing some math but i couldn't get to anything good.
Solution
Thanks to #m69 answer, I made this code for Fraction = pair<Long,Long>:
//this should give the density of fractions between first and last, or less.
double fractionsIn(unsigned long long first, unsigned long long last){
double pi = 3.141592653589793238462643383279502884;
double max = LONG_MAX; //i can't use LONG_MAX directly
double zeroToOne = max/pi * max/pi * 3; // = approx. amount of numbers in Farey's secuence of order LONG_MAX.
double res = 0;
if(first == 0){
res = zeroToOne;
first++;
}
for(double i = first; i < last; i++){
res += zeroToOne/(i * i+1);
if(i == i+1)
i = nextafter(i+1, last); //if this happens, i might not count some fractions, but i have no other choice
}
return floor(res);
}
The main change is nextafter, which is important with big numbers (1e17)
The result
As I explain at the begining, I was trying to compare Fractions with double. Here is the result for Fraction = pair<Long,Long> (and here how I got the density of doubles):
Density between 0,1: | 1,2 | 1e6,1e6+1 | 1e14,1e14+1 | 1e15-1,1e15 | 1e17-10,1e17 | 1e19-10000,1e19 | 1e19-1000,1e19
Doubles: 4607182418800017408 | 4503599627370496 | 8589934592 | 64 | 8 | 1 | 5 | 0
Fraction: 2.58584e+37 | 1.29292e+37 | 2.58584e+25 | 2.58584e+09 | 2.58584e+07 | 2585 | 1 | 0

Density between 0 and 1
If the integers with which you express the fractions are in the range 0~M, then the density of fractions between the values 0 (inclusive) and 1 (exclusive) is:
M: 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31
0~(1): 1 2 4 6 10 12 18 22 28 32 42 46 58 64 72 80 96 102 120 128 140 150 172 180 200 212 230 242 270 278 308 ...
This is sequence A002088 on OEIS. If you scroll down to the formula section, you'll find information about how to approximate it, e.g.:
Φ(n) = (3 ÷ π2) × n2 + O[n × (ln n)2/3 × (ln ln n)4/3]
(Unfortunately, no more detail is given about the constants involved in the O[x] part. See discussion about the quality of the approximation below.)
Distribution across range
The interval from 0 to 1 contains half of the total number of unique fractions that can be expressed with numbers up to M; e.g. this is the distribution when M = 15 (i.e. 4-bit integers):
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
72 36 12 6 4 2 2 2 1 1 1 1 1 1 1 1
for a total of 144 unique fractions. If you look at the sequence for different values of M, you'll see that the steps in this sequence converge:
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16
1: 1 1
2: 2 1 1
3: 4 2 1 1
4: 6 3 1 1 1
5: 10 5 2 1 1 1
6: 12 6 2 1 1 1 1
7: 18 9 3 2 1 1 1 1
8: 22 11 4 2 1 1 1 1 1
9: 28 14 5 2 2 1 1 1 1 1
10: 32 16 5 3 2 1 1 1 1 1 1
11: 42 21 7 4 2 2 1 1 1 1 1 1
12: 46 23 8 4 2 2 1 1 1 1 1 1 1
13: 58 29 10 5 3 2 2 1 1 1 1 1 1 1
14: 64 32 11 5 4 2 2 1 1 1 1 1 1 1 1
15: 72 36 12 6 4 2 2 2 1 1 1 1 1 1 1 1
Not only is the density between 0 and 1 half of the total number of fractions, but the density between 1 and 2 is a quarter, and the density between 2 and 3 is close to a twelfth, and so on.
As the value of M increases, the distribution of fractions across the ranges 0-1, 1-2, 2-3 ... converges to:
1/2, 1/4, 1/12, 1/24, 1/40, 1/60, 1/84, 1/112, 1/144, 1/180, 1/220, 1/264 ...
This sequence can be calculated by starting with 1/2 and then:
0-1: 1/2 x 1/1 = 1/2
1-2: 1/2 x 1/2 = 1/4
2-3: 1/4 x 1/3 = 1/12
3-4: 1/12 x 2/4 = 1/24
4-5: 1/24 x 3/5 = 1/40
5-6: 1/40 x 4/6 = 1/60
6-7: 1/60 x 5/7 = 1/84
7-8: 1/84 x 6/8 = 1/112
8-9: 1/112 x 7/9 = 1/144 ...
You can of course calculate any of these values directly, without needing the steps inbetween:
0-1: 1/2
6-7: 1/2 x 1/6 x 1/7 = 1/84
(Also note that the second half of the distribution sequence consists of 1's; these are all the integers divided by 1.)
Approximating the density in given interval
Using the formulas provided on the OEIS page, you can calculate or approximate the density in the interval 0-1, and multiplied by 2 this is the total number of unique values that can be expressed as fractions.
Given two values s and t, you can then calculate and sum the densities in the intervals s ~ s+1, s+1 ~ s+2, ... t-1 ~ t, or use an interpolation to get a faster but less precise approximate value.
Example
Let's assume that we're using 10-bit integers, capable of expressing values from 0 to 1023. Using this table linked from the OEIS page, we find that the density between 0~1 is 318452, and the total number of fractions is 636904.
If we wanted to find the density in the interval s~t = 100~105:
100~101: 1/2 x 1/100 x 1/101 = 1/20200 ; 636904/20200 = 31.53
101~102: 1/2 x 1/101 x 1/102 = 1/20604 ; 636904/20604 = 30.91
102~103: 1/2 x 1/102 x 1/103 = 1/21012 ; 636904/21012 = 30.31
103~104: 1/2 x 1/103 x 1/104 = 1/21424 ; 636904/21424 = 29.73
104~105: 1/2 x 1/104 x 1/105 = 1/21840 ; 636904/21840 = 29.16
Rounding these values gives the sum:
32 + 31 + 30 + 30 + 29 = 152
A brute force algorithm gives this result:
32 + 32 + 30 + 28 + 28 = 150
So we're off by 1.33% for this low value of M and small interval with just 5 values. If we had used linear interpolation between the first and last value:
100~101: 31.53
104~105: 29.16
average: 30.345
total: 151.725 -> 152
we'd have arrived at the same value. For larger intervals, the sum of all the densities will probably be closer to the real value, because rounding errors will cancel each other out, but the results of linear interpolation will probably become less accurate. For ever larger values of M, the calculated densities should converge with the actual values.
Quality of approximation of Φ(n)
Using this simplified formula:
Φ(n) = (3 ÷ π2) × n2
the results are almost always smaller than the actual values, but they are within 1% for n ≥ 182, within 0.1% for n ≥ 1880 and within 0.01% for n ≥ 19494. I would suggest hard-coding the lower range (the first 50,000 values can be found here), and then using the simplified formula from the point where the approximation is good enough.
Here's a simple code example with the first 182 values of Φ(n) hard-coded. The approximation of the distribution sequence seems to add an error of a similar magnitude as the approximation of Φ(n), so it should be possible to get a decent approximation. The code simply iterates over every integer in the interval s~t and sums the fractions. To speed up the code and still get a good result, you should probably calculate the fractions at several points in the interval, and then use some sort of non-linear interpolation.
function fractions01(M) {
var phi = [0,1,2,4,6,10,12,18,22,28,32,42,46,58,64,72,80,96,102,120,128,140,150,172,180,200,212,230,242,270,278,308,
324,344,360,384,396,432,450,474,490,530,542,584,604,628,650,696,712,754,774,806,830,882,900,940,964,1000,
1028,1086,1102,1162,1192,1228,1260,1308,1328,1394,1426,1470,1494,1564,1588,1660,1696,1736,1772,1832,1856,
1934,1966,2020,2060,2142,2166,2230,2272,2328,2368,2456,2480,2552,2596,2656,2702,2774,2806,2902,2944,3004,
3044,3144,3176,3278,3326,3374,3426,3532,3568,3676,3716,3788,3836,3948,3984,4072,4128,4200,4258,4354,4386,
4496,4556,4636,4696,4796,4832,4958,5022,5106,5154,5284,5324,5432,5498,5570,5634,5770,5814,5952,6000,6092,
6162,6282,6330,6442,6514,6598,6670,6818,6858,7008,7080,7176,7236,7356,7404,7560,7638,7742,7806,7938,7992,
8154,8234,8314,8396,8562,8610,8766,8830,8938,9022,9194,9250,9370,9450,9566,9654,9832,9880,10060];
if (M < 182) return phi[M];
return Math.round(M * M * 0.30396355092701331433 + M / 4); // experimental; see below
}
function fractions(M, s, t) {
var half = fractions01(M);
var frac = (s == 0) ? half : 0;
for (var i = (s == 0) ? 1 : s; i < t && i <= M; i++) {
if (2 * i < M) {
var f = Math.round(half / (i * (i + 1)));
frac += (f < 2) ? 2 : f;
}
else ++frac;
}
return frac;
}
var M = 1023, s = 100, t = 105;
document.write(fractions(M, s, t));
Comparing the approximation of Φ(n) with the list of the 50,000 first values suggests that adding M÷4 is a workable substitute for the second part of the formula; I have not tested this for larger values of n, so use with caution.
Blue: simplified formula. Red: improved simplified formula.
Quality of approximation of distribution
Comparing the results for M=1023 with those of a brute-force algorithm, the errors are small in real terms, never more than -7 or +6, and above the interval 205~206 they are limited to -1 ~ +1. However, a large part of the range (57~1024) has fewer than 100 fractions per integer, and in the interval 171~1024 there are only 10 fractions or fewer per integer. This means that small errors and rounding errors of -1 or +1 can have a large impact on the result, e.g.:
interval: 241 ~ 250
fractions/integer: 6
approximation: 5
total: 50 (instead of 60)
To improve the results for intervals with few fractions per integer, I would suggest combining the method described above with a seperate approach for the last part of the range:
Alternative method for last part of range
As already mentioned, and implemented in the code example, the second half of the range, M÷2 ~ M, has 1 fraction per integer. Also, the interval M÷3 ~ M÷2 has 2; the interval M÷4 ~ M÷3 has 4. This is of course the Φ(n) sequence again:
M/2 ~ M : 1
M/3 ~ M/2: 2
M/4 ~ M/3: 4
M/5 ~ M/4: 6
M/6 ~ M/5: 10
M/7 ~ M/6: 12
M/8 ~ M/7: 18
M/9 ~ M/8: 22
M/10 ~ M/9: 28
M/11 ~ M/10: 32
M/12 ~ M/11: 42
M/13 ~ M/12: 46
M/14 ~ M/13: 58
M/15 ~ M/14: 64
M/16 ~ M/15: 72
M/17 ~ M/16: 80
M/18 ~ M/17: 96
M/19 ~ M/18: 102 ...
Between these intervals, one integer can have a different number of fractions, depending on the exact value of M, e.g.:
interval fractions
202 ~ 203 10
203 ~ 204 10
204 ~ 205 9
205 ~ 206 6
206 ~ 207 6
The interval 204 ~ 205 lies on the edge between intervals, because M ÷ 5 = 204.6; it has 6 + 3 = 9 fractions because M modulo 5 is 3. If M had been 1022 or 1024 instead of 1023, it would have 8 or 10 fractions. (This example is straightforward because 5 is a prime; see below.)
Again, I would suggest using the hard-coded values for Φ(n) to calculate the number of fractions for the last part of the range. If you use the first 17 values as listed above, this covers the part of the range with fewer than 100 fractions per integer, so that would reduce the impact of rounding errors below 1%. The first 56 values would give you 0.1%, the first 182 values 0.01%.
Together with the values of Φ(n), you could hard-code the number of fractions of the edge intervals for each modulo value, e.g.:
modulo: 0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17
M/ 2 1 2
M/ 3 2 3 4
M/ 4 4 5 5 6
M/ 5 6 7 8 9 10
M/ 6 10 11 11 11 11 12
M/ 7 12 13 14 15 16 17 18
M/ 8 18 19 19 20 20 21 21 22
M/ 9 22 23 24 24 25 26 26 27 28
M/10 28 29 29 30 30 30 30 31 31 32
M/11 32 33 34 35 36 37 38 39 40 41 42
M/12 42 43 43 43 43 44 44 45 45 45 45 46
M/13 46 47 48 49 50 51 52 53 54 55 56 57 58
M/14 58 59 59 60 60 61 61 61 61 62 62 63 63 64
M/15 64 65 66 66 67 67 67 68 69 69 69 70 70 71 72
M/16 72 73 73 74 74 75 75 76 76 77 77 78 78 79 79 80
M/17 80 81 82 83 84 85 86 87 88 89 90 91 92 93 94 95 96
M/18 96 97 97 97 97 98 98 99 99 99 99 100 100 101 101 101 101 102

This is exactly the same as: (Sum of phi(k)) where m <= k <= M where phi(k) is the Euler Totient Function and with phi(0) = 1 (as defined by the problem). There is no known closed form for this sum. However there are many optimizations known as mentioned in the wiki link. This is known as the Totient Summatory Function in Wolfram. The same website also links to the series: A002088 and provides a few asymptotic approximations.
The reasoning is this: consider the number of values of the form {1/M, 2/M, ...., (M-1)/M, M/M}. All those fractions that will be reducible to a smaller value will not be counted in phi(M) because they are not relatively prime. They will appear in the summation of another totient.
For example, phi(6) = 12 and you have 1 + phi(6), since you also count the 0.

sas generate 5 digit id code that first 3 must be letters and last 2 numbers

How can I generate in SAS and ID code with 5 digits(letters & Numbers)? Where the first 3 must be letters and last 2 must be numbers.

You can create a unique mapping of the integers from 0 to 26^3 * 10^2 - 1 to a string of the format AAA00. This wikipedia page introduces the concept of different numerical bases quite well.
Your map would look something like this
value = 100 * (X * 26^2 + Y * 26^1 + Z * 26^0) + a * 10^1 + b * 10^0
where X, Y & Z are integers between 0 and 25 (which can be represented as the letters of the alphabet), and a & b are integers between 0 and 9.
As an example:
47416 = 100 * (0 * 26^2 + 18 * 26^1 + 6 * 26^0) + 1 * 10^1 + 6 * 10^0
Using:
0 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25
A B C D E F G H I J K L M N O P Q R S T U V W X Y Z
You get:
47416 -> [0] [18] [6] (1) (6)
A S G 1 6
So 47416 can be represented as ASG16.
To do this programatically you will need to step through your number splitting it into quotient and remainder through division by your bases (10 and 26), storing the remainder as part of your output and using the quotient for the next iteration.
you will probably want to use these functions:
mod() Modulo function to get the remainder from division
floor() Flooring function which returns the rounded down integer part of a real numer
A couple of similar (but slightly simpler) examples to get you started can be found here.
Have a go, and if you get stuck post a new question. You will probably get the best response from SO if you provide a detailed question, code showing your progress, a description of where and why you are stuck, any errors or warnings you are getting and some sample data.

max and min values that can be represented with a 5-digit

How do I find max and min values that can be represented with
a 5-digit number that is in base 13 assuming only positive integers
are represented? then the answer needs to be in base 10.
does 5 digit number mean 5bits? Isn't the smallest number that
can be represented a zero and largest is 2^(N-1)?

This sounds like homework, but I'll bite anyway :)
5 digits probably means 5 digits, as in 12345.
Base 13 means there are 13 possible digits where we as humans are used to calculate with 10.
We could represent the extra 3 digits with A, B, C so that the full range of possible digits is 0123456789ABC. With this representation, it's clear that the smallest 5-digit value is 00000 and the largest CCCCC.
To convert CCCCC in base 13 to base 10 you do
((((C * 13) + C) * 13 + C) * 13 + C) * 13 + C
=
((((12 * 13) + 12 ) * 13 + 12 ) * 13 + 12 ) * 13 + 12
=
371,292
00000 is of course zero in any base.