Yes.
the variance of the uniform distribution function is 1/12(square of(b-a)) and the mean is 1/2(a+b).
probability density distribution
You integrate the probability distribution function to get the cumulative distribution function (cdf). Then find the value of the random variable for which cdf = 0.5.
The modes of a probability density function might be defined as the (countable) set of points in the domain of the function for which the function achieves local maxima. Since the probability density function for the uniform distribution is constant by definition it has no local maxima, hence no modes. Hence, it cannot be bimodal.
The Fermi energy of a material can be derived from the Fermi-Dirac distribution function, which describes the occupation of energy levels in a system at thermodynamic equilibrium. By setting the distribution function to 0.5 (at the Fermi energy), one can solve for the Fermi energy in terms of material parameters such as the electron concentration.
sss
There are two different kinds of particles (electrons, protons, neutrons, atoms, molecules). Each such particle has a so called "spin" which is quantum mechanical value. Depending on spin particles behave differently in the same conditions and can be described using two different distributions. First one is Bose-Einstein distribution for particles with integer spin. Second on is Fermi-Dirac distribution for particles with spin n/2 (where n is an integer number which can take values starting from 1 and higher).
Fermi-Dirac statistics describe the distribution of particles in a system obeying the Pauli Exclusion Principle, such as electrons in a semiconductor. It states that no two indistinguishable fermions can occupy the same quantum state simultaneously. In a semiconductor, electrons fill energy levels up to the Fermi level according to the Fermi-Dirac distribution function, which determines the probability of occupation for each energy level.
The Fermi level is also known as the electron chemical potential (μ), and is a constant appearing in the Fermi-Dirac distribution formula: F() = 1 / [1 + exp((-μ)/kT)] Even though the gap may not contain any electronic states, there may be some thermally excited holes in the valence band and electrons in the conduction band, with the occupancy given by the Fermi-Dirac (FD) function. By inspecting the FD function, it becomes clear that if a state existed at the Fermi level, it would have an occupancy of 1/[1 + exp(0)] = 1/[1+1] = 1/2. Lastly, do not confuse Fermi level with Fermi energy. One is the chemical potential of electrons, the other is the energy of the highest occupied state in a filled fermionic system. In semiconductor physics, the Fermi energy would coincide with the valence band maximum.
An anyon is a particle which obeys a continum of quantum statistics, of which two are the Bose-Einstein and Fermi-Dirac statistics.
You need to know mu, which is probably in the part of the question you didn't bother to write. Wikipedia has several FD statistics curves; you'll need to decide which one is appropriate for your particular value of mu.
To plot each value of a vector as a dirac impulse, try stem instead of plot.
The signum function is differentiable with derivative 0 everywhere except at 0, where it is not differentiable in the ordinary sense. However, but under the generalised notion of differentiation in distribution theory, the derivative of the signum function is two times the Dirac delta function or twice the unit impulse function.
a pulse (dirac's delta).
The Laplace transform of the unit doublet function is 1.
Enrico Fermi is best known for his development of the first nuclear reactor, which marked a crucial milestone in the field of nuclear physics and paved the way for the development of atomic weapons and nuclear energy. He also made significant contributions to quantum theory and particle physics, and his work on beta decay and the Fermi-Dirac statistics were equally groundbreaking.