A new tag piecewise-continuity has been created by a new user. Currently the tag is used by only 7 questions - all written by the tag creator. All this is typical of a well-meaning user who is not experienced enough about either math in general, or tagging on our site in particular, or both. My...
I'm a little bit confused in piecewise continuity of a function. Say, if we have an odd function like $f(x) = x$ defined on the open interval $(0, \pi)$. We then extend it to a period $2\pi$ function and find its sine Fourier series. Can we say that this function is then piecewise continuous, but...
if I have a piecewise continuously differentiable function. How do I see that on each open interval, where the derivative is continuous, there is a continous extension on the larger closed interval?
Assume $f$ and $g$ are two piecewise continuous functions on an interval $( a , b )$ containing the point $t_0$ . Assume further that $f$ has a jump discontinuity at $t_0$ while $g$ is continuous at $t_0$ .How can i verify that the jump in the product $fg$ at $t_0$ is given by “the jump in $f$ at...
Say we have the piecewise function $f(x) = x^2$ on the interval $0 \le x < 4$; and it equals $x+1$ on the interval $ x \ge 4$. Why is it that, when I take the derivative, the intervals loose their equality and become strictly greater or strictly less than?
Let be $f:\mathbb{R} \rightarrow \mathbb{C}$. Consider the space of piecewise linear curves, with support in the interval [-1,1], sucht that $f(x)= A-|x|$ if $|x|\leq 1$; $f(x)= 0$ otherwise. For this space show that expression $p(f)=\int_{-\infty}^{\infty}|f(x)|^2dx$ is a norm. I am show that $...
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