Currently I have two numpy arrays: x and y of the same size. I would like to write a function (possibly calling numpy/scipy... functions if they exist): def derivative(x, y, n = 1): # somethi... More on stackoverflow.com

As of v1.13, non uniform spacing can be specified using an array as the second argument. See the Examples section of this page. 2019-03-29T02:12:05.913Z+00:00 ... NumPy does not provide general functionality to compute derivatives. More on stackoverflow.com

This is the code that I have but it tells me my answer is incorrect. def dfunc(x): ''' Parameters x: 1D numpy array Returns df: 1D numpy array containing first derivatives wrt x ''' # WRITE YOUR CODE HERE df = np.gradient(x) return df ... Why do you think you are doing something wrong? Can you provide example ... More on stackoverflow.com

One definition of the derivative is f'(x) = (f(x+h)-f(x))/h where h goes to 0. Computers cannot store infinitely small numbers, so they might set h=1e-6 (that is 0.000001). It's a tradeoff because while we want h to be as small as possible, at some point the errors due to computer precision begin to dominate. Given any function that the computer can calculate, it can approximate the derivative. def f(x): return np.sin(x) x = np.arange(-2,2,0.01) y = f(x) dfdx = (f(x+h)-f(x))/h plt.plot(x,y) plt.plot(x,dfdx) plt.show() Assuming that the function is reasonably smooth (i.e. the derivative above exists), another definition of the derivative is f'(x) = (f(x+h)-f(x-h))/(2h) where h goes to 0. Going from x-h to x+h means 2 steps, that's the reason for 2h. Which works just as well. These methods are named finite difference to contrast from the normal derivative definition where h is infinitely small. The first one is the forward difference and the second one is called central difference. The backward difference is (f(x)-f(x-h))/2. Let's assume we want to write a derivative function. It takes a function f and values of x, and gives back f'(x). def f(x): return np.sin(x) def d(fun, x): return (fun(x+h)-fun(x))/h x = np.arange(-2,2,0.01) y = f(x) dfdx = d(f,x) plt.plot(x,y) plt.plot(x,dfdx) plt.show() By passing the function into the function, the derivative function can just call fun wherever it wants/needs to get the derivative. Now things become a bit more inconvenient. For some reason we do not know f. We only know y, i.e. f(x) for some values of x. Let's say that x is evenly spaced as usual. Then our best guess for h is not really tiny but identical to the spacing between neighboring x values. With the forward difference we need to take care at the rightmost value because we cannot just add +h to get a value even further out. Instead we use the backward difference. For values in the middle we decide to use the central difference instead of the forward difference. def f(x): return np.sin(x) def d(y, h=1): dfdx = [(y[1]-y[0])/h] for i in range(1,len(y)-1): dfdx.append((y[i+1]-y[i-1])/2/h) dfdx.append((y[i]-y[i-1])/h) return dfdx h = 0.01 x = np.arange(-2,2,h) y = f(x) dfdx = d(y,h) plt.plot(x,y) plt.plot(x,dfdx) plt.show() The implementation above corresponds to np.gradient in the one-dimensional case where varargs is set to case 1 or 2. The case where varargs is set to 3 or 4 would use x directly in d instead of h. However at that point the formula is more complicated as they mention in the documentation. Effectively any point has a hd (the forward step size) and a hs (the backward step size) and the formula is not just (f(x+hd)-f(x-hs))/(hd+hs) but instead that bigger expression given in the documentation, where the values of hd,hs act as some kind of weights. np.gradient is basically backwards, central and forward difference combined. When you have values like f(1),f(2),f(2+h) and want the derivative at 2, the code notices that 2 and 2+h are very close together and puts greater weight on that (and mostly ignores f(1)). The important part so far is that np.gradient when given a vector with N elements calculates N one-dimensional derivatives, which is not the typical idea of a gradient. np.gradient does support more dimensions which might make things clearer. So in the 1D case, we essentially go through all values from left to right and then consider that value and its direct left and right neighbor to quantify the uptrend or downtrend. In the 2D case, np.gradient still does this, but additionally also walks from top to bottom and does the same. So in 2D it returns 2 arrays, one for left-right and one for top-bottom. The actual definition of the gradient by finite differences is [(f(x+h,y)-f(x,y))/h, (f(x,y+h)-f(x,y))/h] in 2D. These values are indeed returned by np.gradient, the left part is in the first array and the right part in the second array. Say we are in 2D and want the gradient at x=3 and y=0, then we can plug it into np.gradient like this: hx = 1e-6 hy = 1e-3 x = [3,3+hx] y = [0,0+hy] xx,yy = np.meshgrid(x,y) def f(x,y): return x**2-2*x*np.sin(y) + 1/x grad = np.gradient(f(xx,yy), y,x) # Note the order. print(grad[1][0,0], grad[0][0,0]) # Note the order. This is dfdx, dfdy. but if the function f can be calculated by a computer, it makes more sense to just use automatic differentiation instead of finite differences. Automatic differentiation has no h that needs to be chosen carefully. It's always as accurate is possible. import torch x = torch.tensor([3.],requires_grad=True) y = torch.tensor([0.],requires_grad=True) z = x**2-2*x*torch.sin(y) + 1/x z.backward() print(x.grad, y.grad) So what's the deal with the Taylor series? It's just a minor piece in the derivation of that more general expression used by np.gradient. We just start by claiming that we can express the gradient by adding together function values in the direct neighborhood. f'(x) = a f(x) + b f(x+hd) + c f(x-hs) Given that finite differences do work out, this approach should work as well and generalize the idea. Expand f(x+hd) and f(x-hs) with their series: f(x+hd) = f(x) + hd f'(x) + hd^2 f''(x)/2 + ... f(x-hs) = f(x) - hs f'(x) + hs^2 f''(x)/2 + ... Then plug it in and reshape: f'(x) = a f(x) + b f(x) + b hd f'(x) + b hd^2 f''(x)/2 + c f(x) - c hs f'(x) + c hs^2 f''(x)/2 = (a+b+c) f(x) + (b hd - c hs) f'(x) + (b hd^2 + c hs^2 )/2 f''(x) 0 = (a+b+c) f(x) + (b hd - c hs - 1) f'(x) + (b hd^2 + c hs^2 )/2 f''(x) The = in the middle is actually more of an approximately equal sign. We won't be able to reach 0 for all f(x) as claimed on the left hand size, but we can get pretty close. We do NOT want to minimize the right-hand-side. We want it to reach 0 (it can go below 0 right now). To turn this into a minimization problem, we square it. This way we get a positive number always and it really becomes a matter of minimization. We COULD also take the absolute value instead of squaring, but it's pain to work this through and the end result are exactly the same parameters anyway. To minimize: E2 with E = (a+b+c) f(x) + (b hd - c hs - 1) f'(x) + (b hd2 + c hs2 )/2 f''(x) One requirement for an optimum is that the gradient is 0. In this case we take the derivatives with respect to a,b,c because we want to find the optimal a,b,c. First a reminder of the chain rule: dE2 /dt = 2E dE/dt for whatever t is. It's optional to do this but a bit less messy than working it through individually. In particular we have dE^2/da = 2E dE/da = 2E f(x) dE^2/db = 2E dE/db = 2E (f(x) + hd f'(x) + hd^2 f''(x)/2) dE^2/dc = 2E dE/dc = 2E (f(x) - hs f'(x) + hs^2 f''(x)/2) We want ALL three of them to be 0 at the same time. This can only happen if E is 0. 0 := (a+b+c) f(x) + (b hd - c hs - 1) f'(x) + (b hd2 + c hs2 )/2 f''(x) and we want this to be 0 for any f, f', f'' for any value of x. The only way for this to happen is if each coefficient is 0, i.e. a+b+c = 0 b hd - c hs = 1 b hd^2 + c hs^2 = 0 We would need to check the second derivative to make sure that this is a minimum, not a maximum, but given the problem it is fairly clear. So why did we stop exactly after f'' in the Taylor series? It's because this way we get exactly 3 unknowns and 3 equations, which is the most convenient to solve. Multiply the second equation by hd then subtract the third from it. (b hd^2 - c hs hd) - (b hd^2 + c hs^2) = hd -c hs^2 - c hs hd = hd c hs (hs + hd) = -hd c = -hd/hs/(hs+hd) = -hd^2 / (hs hd (hs+hd)) where the last step is just so it looks exactly like in np.gradient. Insert c into the second equation. b hd + hd/hs/(hs+hd) hs = 1 b hd + hd/(hs+hd) = 1 b + 1/(hs+hd) = 1/hd b = 1/hd - 1/(hs+hd) b = (hs(hs+hd) - hs hd) / [hs hd (hs+hd)] b = hs^2 / [hs hd (hs+hd)] From the first equation we know that a = -b-c = (hd2 - hs2 )/(hs hd (hs+hd)). So here's your summary: If you have a function that can be calculated by a computer, use torch or tensorflow or any other framework for automatic differentiation. If you have a function that can be calculated by a computer but such a framework is not available, np.gradient is still a bad idea because it is inefficient. Note for the 2D gradient we needed three values, f(x,y), f(x+dx,y), f(x,y+dy). But with np.gradient we would first need to set up arrays where it is almost natural to also include f(x+dx,y+dy) which is not needed for gradient calculations. It's more natural to set up some loop that increments x once, then y once, then z once, and so on. Many solvers in scipy.optimize work with finite differences. If you have a function that cannot be calculated by a computer, np.gradient may be useful. In practice this means that you have data from some experiment. Even there, the concept of a Taylor series plays no role here UNLESS the data was taken on an unevenly spaced grid. More on reddit.com