Here is a simple numerical approach based upon your formula above. You could improve upon this: derivativeOf takes a function fn and an x-coordinate x and returns a numerical approximation of derivative of fn at x: func derivativeOf(fn: (Double)->Double, atX x: Double) -> Double { let h = 0.0000001 return...

local function Rotate(X, Y, alpha) local c, s = math.cos(math.rad(alpha)), math.sin(math.rad(alpha)) local t1, t2, t3 = X[1]*s, X[2]*s, X[3]*s X[1], X[2], X[3] = X[1]*c+Y[1]*s, X[2]*c+Y[2]*s, X[3]*c+Y[3]*s Y[1], Y[2], Y[3] = Y[1]*c-t1, Y[2]*c-t2, Y[3]*c-t3 end local function convert_rotations(Yaw, Pitch, Roll) local F, L, T = {1,0,0}, {0,1,0}, {0,0,1} Rotate(F, L, Yaw)...

machine-learning,integration,neural-network,implementation,calculus

My personal opinion it is not possible to feed into NN enough rules for integrating. Why? Because NN are good for linear regression ( AKA approximation ) or logical regression ( AKA classification ). Integration is neither of them. It is calculation task according to some strict algorithms. So from...

c,math,recursion,calculus,derivative

I would implement it like this: double derivative(double (*f)(double), double x0, int order) { const double delta = 1.0e-6; double x1 = x0 - delta; double x2 = x0 + delta; if (order == 1) { double y1 = f(x1); double y2 = f(x2); return (y2 - y1) / (x2...

Well David,you can convert this function into one trigonometric function by multiplying and dividing it by √(1^2 + 8) i.e, 3. So your function becomes like this I = 3*(1/3 cos(wt) + √8/3 sin(wt)) = 3* sin(wt + atan(1/√8)) Now, you can easily say its maximum value is I =...

Change 3x to 3*x. (This may be the smallest answer-length-to-question-length ratio I've seen in a long time ;-)...

math,wolfram-mathematica,series,calculus

This should be close: inner[i_, gamma_, k_] := Sum[(-1)^(el[1])/el[1]! Product[(-1)^(el[z] - el[z - 1])/(el[z] - el[z - 1])! ,{z, 2, i}], Evaluate[Sequence @@ ({{el[1], k, gamma}}~Join~Table[ { el[ii], el[ii - 1], gamma }, {ii, 2, i}])]] With[{gamma = 3, k = 1}, Sum[ inner[i, gamma, k], {i, gamma - 1}]]...

c++,algorithm,math,calculus,cybernetics

I integral part is just summation also multiplied by some constant. Analogue integration is done by nonlinear gain and amplifier. Digital integration of first order is just: output += input*dt; second order is: temp += input*dt; output += temp*dt; dt is the duration time of iteration loop (timer or what...

You're then looking for Math.ceil(), not Math.round() Math.ceil(value/ 5000.0) * 5000.0 Compare also nearest integer function with floor and ceiling functions...

See plotting functions and expressions in the Gadfly manual. For your example, something like plot([x->x^2+1, x->(x-4)+(x+3)], -2, 2) should do the trick using anonymous functions. It's not currently possible to do this with multivariate functions as far as I'm aware....

arrays,matlab,vector,matrix,calculus

For two arbitrary lines (e.g. line 1 and line 2) do: RGB_dist(x(1,:), x(2,:)) if you want all the combinations then check out pdist2. If you don't have the stats toolbox (i.e. no pdist2) then use nchoosek to create all possible pairs of rows: I = nchoosek(1:size(x,1),2); D = RGB_dist(x(I(:,1),:), x(I(:,2),:))...

You can use the pracma library, such as: library(pracma) dummy <- function(x) { z <- x[1]; y <- x[2] rez <- (z^2)*(y^3) rez } grad(dummy, c(1,2)) [1] 16 12 hessian(dummy, c(1,2)) [,1] [,2] [1,] 16 24 [2,] 24 12 ...

javascript,math,canvas,calculus

As your code is not time bound (in order for real-time frequency to work) you need to first define how much width represents one cycle (@ 1 Hz, or one second if you will). Lets say the whole canvas width represents one cycle then we can do this: var period...

Assuming that the (x,y) grid is uniform, you can approximate the integral by a 2D-Riemman sum as follows: result = sum(z(:))*delta_x*delta_y; where delta_x, delta_y are the grid spacings in the x and y directions. In your case these can be computed as delta_x = 2*pi/numel(x); %// or 2*pi/(numel(x)-1) delta_y =...

c,math,calculator,integral,calculus

y = 10.0 - (12.0 + (float)x) / 4.0; Followed by y = y+1; This makes sense else you have y uninitialized which leads to undefined behavior because the value of y is undeterminate. During declaration you can initialize y and use += operator. Like float y = 0;...

use sympy >>> from sympy import symbols, diff >>> x, y, z = symbols('x y z', real=True) >>> f = 4*x*y + x*sin(z) + x**3 + z**8*y >>> diff(f, x) 4*y + sin(z) + 3*x**2 ...

sql,database,algebra,relational,calculus

This is the translation to SQL for your formula that starts with {S|∃ D ∈ Doctor (∃ C ∈ Duty (D.Doc_id = C.Doc_id ^ (etc..) ) ) } SELECT * FROM S WHERE EXISTS (SELECT * FROM Doctor D WHERE EXISTS (SELECT * FROM Duty C WHERE (D.Doc_id = C.Doc_id)...