Search results
Results From The WOW.Com Content Network
The function in example 1, a removable discontinuity. Consider the piecewise function = {< = >. The point = is a removable discontinuity.For this kind of discontinuity: The one-sided limit from the negative direction: = and the one-sided limit from the positive direction: + = + at both exist, are finite, and are equal to = = +.
A holomorphic function's singularity is either not really a singularity at all, i.e. a removable singularity, or one of the following two types: In light of Riemann's theorem, given a non-removable singularity, one might ask whether there exists a natural number m {\displaystyle m} such that lim z → a ( z − a ) m + 1 f ( z ) = 0 ...
The above example simply states that the function takes the value () for all x values larger than a. With this, all the forces acting on a beam can be added, with their respective points of action being the value of a. A particular case is the unit step function,
Function without a limit at an essential discontinuity. The function = { < = > has no limit at x 0 = 1 (the left-hand limit does not exist due to the oscillatory nature of the sine function, and the right-hand limit does not exist due to the asymptotic behaviour of the reciprocal function, see picture), but has a limit at every other x ...
Then f is a non-decreasing function on [a, b], which is continuous except for jump discontinuities at x n for n ≥ 1. In the case of finitely many jump discontinuities, f is a step function. The examples above are generalised step functions; they are very special cases of what are called jump functions or saltus-functions. [8] [9]
This page was last edited on 10 January 2015, at 10:07 (UTC).; Text is available under the Creative Commons Attribution-ShareAlike 4.0 License; additional terms may apply.
For example, in the classification of discontinuities: in a removable discontinuity, the distance that the value of the function is off by is the oscillation; in a jump discontinuity, the size of the jump is the oscillation (assuming that the value at the point lies between these limits from the two sides);
This function appears to have a singularity at z = 0, but if one factorizes the denominator and thus writes the function as = it is apparent that the singularity at z = 0 is a removable singularity and then the residue at z = 0 is therefore 0.