how to get modulo of a value in exponential form - math

Question is about the modulo operator on very large numbers.
For example consider a question where the total number of permutations are to be calculated.
Consider a number of 90 digits with each of the 9 numbers (1 to 9) repeating 10 times
so 90!/(10!)^9) is to be calculated
After reading many answers on StackOverflow I used logarithms to do it.
Now consider the log value to be 1923.32877864.
Now my question is how can I display the answer (i.e. 10 ^ log10(value) ) modulo of "m"?
And is this the best method for calculating the possible number of permutations?
Edit
Got the solution :)
Thanks to duedl0r.
Did it the way you specified using Modular Multiplicative Inverse.Thanks :)

I'm not sure whether this is actually possible and correct, but let me summarize my comments and extend the answer from Miky Dinescu.
As Miky already wrote:
a × b ≣m am × bm
You can use this in your equality:
90! / 10!^9 ≣m x
Calculate each term:
90!m / 10!^9m ≣m x
Then find out your multiplicative inverse from 10!^9m. Then multiplicate the inverse with 90!m.
update
This seems to be correct (at least for this case :)). I checked with wolfram:
(90!/10!^9) mod (10^9+7) = 998551163
This leads to the same result:
90! mod (10^9+7) = 749079870
10!^9 mod (10^9+7) = 220052161
do the inverse:
(220052161 * x) mod(10^9+7) = 1 = 23963055
then:
(749079870*23963055) mod (10^9+7) = 998551163
No proof, but some evidence that it might work :)

I would argue that the way to compute the total number of permutations modulo m, where m is an arbitrary integer (usually chosen to be a large prime number) is to use the following property:
(a * b) % m = ((a % m) * (b % m)) % m
Considering that the total number of permutations of N is N! = 1 * 2 * 3 * .. * N, if you need to compute N! % m, you can essentially apply the property above for multiplication modulo m, and you have:
((((1 * (2 % m)) % m) * (3 % m)) % m) * ..
EDIT
In order to compute the 90! / (10! ^ 9) value you could simplify the factors and then use multiplication modulo m to compute the final result modulo m.
Here's what I'm thinking:
90! = 10! * (11 * 12 * .. * 20) * (21 * 22 * .. * 30) * .. * (81 * 82 * .. * 90)
You can then rewrite the original expression as:
(10! * (11 * 12 * .. * 20) * (21 * 22 * .. * 30) * .. * (81 * 82 * .. * 90)) / (10! * 10! * ... * 10!)
At the numerator, you have a product of 9 factors - considering each expression in parenthesis a factor. The same is true for the denominator (you have 9 factors, each equal to 10!).
The first factor at the denominator is trivial to simplify. After that you still have 8 pairs that need simplification.
So, you can factor each term of the products and simplify the denominator away. For example:
11 * 12 * 13 * 14 * 15 * 16 * 17 * 18 * 19 * 20 <=> 11 * 2 * 2 * 3 * 13 * 2 * 7 * 3 * 5 * 2 * 2 * 2 * 2 * 17 * 2 * 9 * 2 * 2 * 5
The denominator will always be: 2 * 3 * 2 * 2 * 5 * 2 * 3 * 7 * 2 * 2 * 2 * 2 * 3 * 3 * 2 * 5
After the simplification the second pair reduces to : 2 * 2 * 11 * 13 * 17 * 19
The same can be applied to each subsequent pair and you will end up with a simple product that can be computed modulo m using the formula above.
Of course, efficiently implementing the algorithm to perform the simplification will be tricky so ultimately there has to be a better way that eludes me now.

Related

Recursive algorithm that multiplies specific natural numbers

Design a recursive algorithm f(n1, n2, delta) that multiplies specific natural numbers between n1
and n2 (n1 > n2) as follows.
Starting with n1 multiply all natural numbers with a distance of delta from the previous natural
number.
Example: n1 = 35, n2 = 15, delta = 5
f(35, 15, 5) = 35 * 30 * 25 * 20 * 15
Solution:
Input: n1, n2 ∈ ℕ, n1 > n2, ∃ x ∈ ℕ: n2 + (x * delta) = n2
Output: n1 * (n1 - delta ) * (n1 – 2 * delta) * … * n2
f(n2, n2, delta) = n2
f(n1, n2, delta) = n1 * f((n1-delta), n2, delta)
Can someone explain this solution?
This solution implies a recursive algorithm (pseudo-code below). The idea in the solution is that you can "diminish" the problem by decreasing n1 by delta, and the stopping condition is that n1 is smaller than n2.
Note that the explanation you mentioned is a bit inaccurate as it does assumes that after some steps n2 is gotten, while you may reach some value < n2 (for example, if you change the 15 in the call f(35,15,5) to 14).
The pseudo-code:
f(n1, n2, delta):
if n1 < n2:
return 1
return n1 * f(n1 - delta, n2, delta)
Running example on your case (f(n1) instead of f(n1,n2,delta) is used for brevity as only n1 is changes):
f(35) = 35 * f(30) =
35 * 30 * f(25) =
35 * 30 * 25 * f(20) =
35 * 30 * 25 * 20 * f(15) =
35 * 30 * 25 * 20 * 15 * f(10) =
35 * 30 * 25 * 20 * 15 * 1

Solve sigma example

I have a sigma example:
And I don't have any idea how to solve it. Can you help me with the code, please?
(Code pascal, java or c++)
Expanding the inner term, you get m^3 - 3m^2n + 3mn^2 - n^3, which yields a double summation of m^5, -3m^4n, 3m^3n^2 and -m^2n^3. These summations are separable, meaning that they are the product of a sum on m of a power of m and a sum on n of a power of n.
You can evaluate these sums by means of the Faulhaber formulas up to degree five, which are polynomial expressions. Evaluate them by Horner's method.
int F1(int n) { return (n + 1) * n / 2; }
int F2(int n) { return ((2 * n + 3) * n + 1) * n / 6; }
int F3(int n) { return ((n + 2) * n + 1) * n * n / 4; }
...
int S= F5(20) * 30 - 3 * F4(20) * F1(30) + 3 * F3(20) * F2(30) - F2(20) * F3(30);
Using the obvious method of summation, the inner loop will evaluate 30 cubes of a difference, for a total of 60 additions and 60 multiplications, and the outer loop will repeat this 20 times, with extra multiplications and additions, for a total of 1220 + and 1240 *.
Compare to the above method, performing 18 +, 30 * and 7 divisions in total (independently of the values of m and n).

How to find where an equation equals zero

Say I have a function and I find the second derivative like so:
xyr <- D(expression(14252/(1+exp((-1/274.5315)*(x-893)))), 'x')
D2 <- D(xyr, 'x')
it gives me back as, typeof 'language':
-(14252 * (exp((-1/274.5315) * (x - 893)) * (-1/274.5315) * (-1/274.5315))/(1 +
exp((-1/274.5315) * (x - 893)))^2 - 14252 * (exp((-1/274.5315) *
(x - 893)) * (-1/274.5315)) * (2 * (exp((-1/274.5315) * (x -
893)) * (-1/274.5315) * (1 + exp((-1/274.5315) * (x - 893)))))/((1 +
exp((-1/274.5315) * (x - 893)))^2)^2)
how do I find where this is equal to 0?
A little bit clumsy to use a graph/solver for this, since your initial function as the form:
f(x) = c / ( 1 + exp(ax+b) )
You derive twice and solve for f''(x) = 0 :
f''(x) = c * a^2 * exp(ax+b) * (1+exp(ax+b)) * [-1 + exp(ax+b)] / ((1+exp(ax+b))^3)
Which is equivalent that the numerator equals 0 - since a, c, exp() and 1+exp() are always positive the only term which can be equal to zero is:
exp(ax+b) - 1 = 0
So:
x = -b/a
Here a =-1/274.5315, b=a*(-893). So x=893.
Just maths ;)
++:
from applied mathematician point of view, it's always better to have closed form/semi-closed form solution than using solver or optimization. You gain in speed and in accuracy.
from pur mathematician point of view, it's more elegant!
You can use uniroot after having created a function from your derivative expression:
f = function(x) eval(D2)
uniroot(f,c(0,1000)) # The second argument is the interval over which you want to search roots.
#Result:
#$root
#[1] 893
#$f.root
#[1] -2.203307e-13
#$iter
#[1] 7
#$init.it
#[1] NA
#$estim.prec
#[1] 6.103516e-05

Differentiation using r

I'm new to using R or any type of programming and I'm trying to differentiate 3xcos(xy) with respect to x. I've tried four different ways and was wondering which one is best/correct.
D(expression(3*x*cos(xy)),"x")
D(expression(3*x*cos*(xy)),"x")
D(expression(3*x*cos*(xy)),"x")
D(expression(3*x*cos*(x*y)),"x")
thanks in advance
Shane
None of those.
This is the correct expression:
D(expression(3*x*cos(x*y)),"x")
#3 * cos(x * y) - 3 * x * (sin(x * y) * y)
This treats xy as one variable:
D(expression(3*x*cos(xy)),"x")
#3 * cos(xy)
This treats xy as one variable and cos as a variable (and not a function):
D(expression(3*x*cos*(xy)),"x")
#3 * cos * (xy)
This treats cos as a variable:
D(expression(3*x*cos*(x*y)),"x")
#3 * cos * (x * y) + 3 * x * cos * y

Maths help with exponentials [closed]

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Improve this question
Evaluate
(z x^-1 y)^5 y^5
~~~~~~~~~~~~~~~~~~~~~~~ OVER
x^-4 z^-4
How would I evaluate this if X = 10, y = -3 and z = 3? I would like a step-by-step solution to help me fully understand it.
Numerator evaluates as (z*y*x^-1)^5 * y^5
further rewriting ((z^5*y^5)*y^5)/x^5
Denominator ((1/x^4)*(1/z^4))
Final Answer would be ((y^10)*(z^9))/x
as per your values it (3^19)/10
Exponentials have higher priority in most computer languages, so adding parentheses like this should make it clearer. I'm assuming that you're dividing the first polynomial by the second. It's simple algebra.
(z x^-1 y)^5 y^5
---------------- =
x^-4 z^-4
(y^10)(z^9)/x
You substitute the numbers.
Start with:
((z * x^-1 * y)^5 * y^5)/(x^-4 * z^-4)
Commute the exponent to the z factor: (A * B)^N => A^N * B^N
(z^5 * (x^-1 * y)^5 * y^5)/(x^-4 * z^-4)
Commute the exponent to the x and y factors: (A * B)^N => A^N * B^N
(z^5 * (x^-1)^5 * y^5 * y^5)/(x^-4 * z^-4)
Simplify the exponenet on the x factor: (A^N)^M => A^(N*M)
(z^5 * x^-5 * y^5 * y^5)/(x^-4 * z^-4)
Combine the y factors: A^N * A^M => A^(N+M)
(z^5 * x^-5 * y^10)/(x^-4 * z^-4)
Remove the negative exponent on x: 1/A^-N => A^N
(z^5 * x^-5 * y^10 * x^4) / (z^-4)
Remove the negative exponenet on z: 1/A^-N => A^N
z^5 * x^-5 * y^10 * x^4 * z^4
Combine the z factors: A^N * A^M => A^(N+M)
z^9 * x^-5 * y^10 * x^4
Combine the x factors: A^N * A^M => A^(N+M)
z^9 * x^-1 * y^10
Remove the negative exponent on x: A^(-N) => 1/A^N
(z^9 * y^10)/(x^1)
Simplify the x factor: A^1 => A
(z^9 * y^10)/(x)
And that's the algebraic form of your answer.
Next, subsitute the values:
3^9 * (-3)^10 / 10
Factor the exponents:
(3^3)^3 * (-3)^10 / 10
(3^3)^3 * ((-3)^2)^5 / 10
Evalutate the innermost exponents:
(3 * 3 * 3)^3 * ((-3)^2)^5 / 10
(9 * 3)^3 * ((-3)^2)^5 / 10
27^3 * ((-3)^2)^5 / 10
27^3 * 9^5 / 10
Continue evaluation exponents, breaking them down for simplicity:
27 * 27 * 27 * 9^5 / 10
27 * 27 * 27 * 9^5 / 10
729 * 27 * 9^5 / 10
19683 * 9^5 / 10
19683 * 9^2 * 9^2 * 9 / 10
19683 * 81 * 81 * 9 / 10
Then multiply the factors:
19683 * 81 * 729 / 10
19683 * 59049 / 10
1162261467 / 10
116226146.7
And there's your final answer.
You could also take advantage of the fact that X^N = (-X)^N if N is even by replacing -3 with 3 since 10 is even.
3^9 * (-3)^10 / 10
3^9 * 3^10 / 10
3^19 / 10
3 * 3^18 / 10
3 * (3^9)^2 / 10
3 * (3 * 3^8)^2 / 10
3 * (3 * (3^2)^4)^2 / 10
3 * (3 * ((3^2)^2)^2)^2 / 10
3 * (3 * (9^2)^2)^2 / 10
3 * (3 * 81^2)^2 / 10
3 * (3 * 6561)^2 / 10
3 * (19683)^2 / 10
3 * 387420489 / 10
1162261467 / 10
116226146.7

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