What are application of numerical methods in engineering?
There’s so many uses for numerical methods, it is impossible to list them all. But essentially, we can cover first the basic math problems they can be used for, which are often:Computing integrals and derivativesSolving differential equationsBuilding models based on data, be it through interpolation, Least Square, or other methodsRoot finding and numerical optimizationEstimating the solution to a set of linear and nonlinear equationsComputational geometryThere’s other areas I haven’t listed, but that’s some of the common fundamental uses. With respect to real world problems, here are some examples where numerical methods are used:Development and computation of optimal control algorithmsDevelopment of high fidelity simulations to model viscous flow around a race car to see if the wing designs generate sufficient downforceMachine learning algorithms, like estimating optimal weights of parametric models using only subsets of the full dataset (like stochastic gradient descent)Photorealistic rendererDesign optimization based on simulation and multi-objective optimization formulationsGame EnginesFiltering of noisy data based on an approximately expected model of the dynamics (Kalman Filter, Particle Filter, etc.)There are many more uses for numerical methods out there, but this will hopefully show a range of areas to prove its uses are broad.
Help with numerical methods question?
The relative error is (x - sin x)/sin x = (x/sin x) - 1 so we require (x/sin x) - 1 = 1/2 x 10^-4 = 5 x 10^-5 x/sin x = 1.00005 When x = 0.01732, x/sin x = approx. 1.000049999 When x = 0.017325, x/sin x = approx. 1.000050028 And sin (-x) = - sin x So x is accurate to a relative accuracy of 1/2 x 10^-4 when - 0.01732 < x < 0.01732 (to the nearest 4 significant figures)
Numerical solutions help please (iteration)?
(i) Directly follows from the formulas for the area of the triangle by 2 sides (r) and angle (α) in-between: (1/2)*r^2*sinα, the area of a sector is (1/2)*r^2*α, so (1/2)*r^2*sinα = (1/2)*(1/2)*r^2*α, or sinα = (1/2)α or α = 2sinα; (ii) The function f(x) = x - 2sinx for x belonging to [π/2, 2π/3] is continuous, f(π/2) = π/2 - 2 < 0, f(2π/3) = 2π/3 - sqrt(3) > 0, so there is a root of the equation x - 2sinx = 0 in the given interval /ONLY 1 root, because f'(x) = 1 + 2cos(x) > 0, so f is increasing!/; (iii) Let L = lim x_{n} when n -> infinity /it exists according the condition!/, where recurrence relationship is: x_{n+1} = (1/3)(x_{n} + 4 sin x_{n}), then, having n -> infinity we obtain L = (1/3)(L + 4 sin L) or 3L = L + 4 sin L, or 2L = 4 sinL or L = 2sinL, i.e. L = α from (i).