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Let X be a binomial random variable with mean Np and variance Np(1−p).
Show that the ratio X/N also has a binomial distribution with mean p and
variance p(1 − p)/N.
Let r = X/N. Since X has a binomial distribution, r also has the
same distribution. The mean and variance for r can be computed as follows:
Mean , E[r] = E[X/N] = E[X]/N = (N p)/N = p .
Variance , E[(r − E[r])2] = E[(X/N − E[X/N])2] = E[(X − E[X])2]/N2 = N p(1 − p)/N2 = p(1 − p)/N
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I know that it may seem a duplicate question, but I could not find my answer in previous questions.
I mean how to write a log base 10 function by simple loops and not using built in log function in c++.
The easiest way is to calculate the natural logarithm (ln) with a Taylor series. Once you have found the natural logarithm, just divide it by ln(10) and you get the base-10 log.
The Taylor series is quite simple to implement in C. If z is the number for which you are seeking the log, you just have to loop a few iterations multiplying an accumulator by (z-1) each time. To a limit, the more iterations you run, the more accurate your result will be. Check it a few times against the libC log10() version until you are happy with the precision.
This is a "numeric approach". There are other numeric solutions to finding the logarithm of a number which can give more accurate results. Some of them can be found in that Wikipedia link I gave you.
Assuming by "log base 10" you mean "the number of times n can be divided by 10 before resulting in a value < 10":
log = 0;
// Assume n has initial value N
while ( n >= 10 ) {
// Invariant: N = n * 10^log
n /= 10;
log += 1;
}
You'll get faster convergence with Newton's Method. Use something like this (hand written not compiled or tested uses f(r) = 2**r - x to compute log2(x) ):
double next(double r, double x) {
static double one_over_ln2 = 1.4426950408889634;
return r - one_over_ln2 * (1 - x / (1 << static_cast<int>(r)));
double log2(double x) {
static double epsilon = 0.000000001; // change this to change accuracy
double r = x / 2;. // better first guesses converge faster
double r2 = next(r, x);
double delta = r - r2;
while (delta * delta > epsilon) {
r = r2;
r2 = next(r, x);
delta = r - r2
}
return r2;
}
double log10(double x) {
static double log2_10 = log2(10);
return log2(x) / log2_10;
}
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I have a 2D point A inside the [0,1]² square.
The square is divided into 9 subsquares (of equal dimensions)
http://www.noelshack.com/2015-23-1433689273-capture.png
I want to know which subsquare the point A belongs to.
I can do a if elseif else on the first coordinate, then inside each branch, another if else if else on the second coordinate.
There is a lot of code repeating (the check on the second coordinate)
Is there a better way ?
The trouble is, it is not clear what your 2D point is, what you want the sub-square value as, etc. So a definitive answer is difficult. But anyway, I will make some assumptions and see if I am right about what you are asking...
Assuming you had a point A with a coordinate such as:
float point[2] = {0.1224, 0.4553}
Then to work out where it is inside the square you can do some simple maths:
float x = point[0] * 3;
float y = point[1] * 3; //Multiply both by 3
int xIdx = floor(x);
int yIdx = floor(y); //Floor the result - this gives a number 0 to 2
int cell = yIdx * 3 + xIdx + 1; // Calculate the cell index (based on your diagram)
Now you could generalise this for any point - for example the point might be (1.23343, 2.6768) - by simply removing the integer part from the point - leaving a number between 0 and 1. The integer part would be the super cell, and the fractional part would be converted into the sub cell as above.
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Is it possible to simplify (a+b)xor(c+b)? What is the contribution of b to the final result?
Note that I'm mixing boolean algebra with arithmetic, xor is a bitwise exclusive or on corresponding bits and + is a standard addition on 8 bits, that wraps around when overflown.
a, b, c are unsigned char;
We can use an SMT solver to test our hypothesis that your formula can be simplified. You can head over to http://rise4fun.com:
x = BitVec('x', 8)
y = BitVec('y', 8)
z = BitVec('z', 8)
print simplify((x + z) ^ (y + z))
and the result, anticlimactically, is:
x + z ^ y + z
Which means your formula cannot be further simplified.
(a+b)xor(c+b)
--------------
=((not(a+b))*(c+b))+((a+b)*(not(c+b)))
-----------------------
=((not a)*(not b)*(c+b))+((a+b)*(not c)*(not b))
----
=((not a)(not b)*c) + (a*(not c)(not b))
----
=(not b)((not a)c + a(not c))
----
=(not b)(a xor c)
----
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can anyone tell me how to generate a 2d gaussian filter kernel using the gaussian filter equation? how does the x and y value vary?
ref: http://en.wikipedia.org/wiki/Gaussian_function
To generate the kernel is quite simple. If your problem is in applying the kernel, you need to update the question.
The kernel is simply a square matrix of values, generally an odd number size so that there's a clearly defined center. To fill it, the x and y values go from -(n-1)/2 to (n-1)/2 where n is the size of the matrix.
double half_n = (n - 1) / 2.0;
for (i = 0; i < n; ++i)
{
double x = i - half_n;
for (j = 0; j < n; ++j)
{
double y = j - half_n;
kernel[i][j] = // use formula with x and y here
}
}
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I am trying to solve T(n) = 2T(n/2) + log n
substituted n = 2^k
T(2^k) = 2T(2^(k-1)) + k
T(2^k) = 2^2 T(2^(k-1)) + 2(k-1) + k
after k steps
T(2^k) = 2^k T(1) + 2^(k-1) + 2 * (2^(k-2)) +....+k
So basically I need to sum a term of i*2^i where i = 1 to log n - 1.
And I could not find an easy way to sum these terms. Am I doing something wrong ? Is there any other way to solve this recursion ? would master theorem work her ? if yes than how ?
Thanks.
first you should define a recursive export,say T(1)
then:
since T(2^k) = 2T(2^(k-1)) + k; *
we define g(k) = T(2^k)/2^k;
then * come into:
g(k) = g(k-1) + k/2^k = g(1) + sum(i/2^i); i=2,3,4...k
where g(1) = T(1)/2 = c;
where you could then unfold the sum expression and define it = y;
then unfold the expression of y/2;
y-y/2 is a geometric progression, so youcan solve it
as I worked out, sum = 3/2 - (k+2)/2^k;
so T(n) = 2^k * g(k) = (3/2+c)*n - (2+logn)
Wolfram|Alpha gives a closed form solution:
for a constant c_1 that is fixed by the initial condition.
By the way, log(n)/log(2) = lg(n), where lg is the base two logarithm.