I know that the derivative is 0 for all x but can somebody explain this to me. I understand derivatives but I have been having a lot of trouble here. I know that $ceil(x)=(x+1/2) - (arctan(tan(pi*(x+1/2))))/(pi)$ for all non integer x so I differentiated that on wolfram alpha https://www.wolframalpha.com/input/?i=derivative+of+(x%2B1%2F2)+-+(arctan(tan(pi(x%2B1%2F2))))%2Fpi and got $1-(csc^2(pi*x)/(cot^2(pi*x)+1)$ and using the desmos graphing calculator to graph this I got https://www.desmos.com/calculator/mmzdqg6nba which is 0 for all non-integer x. can anybody show me an easier way to solve this problem.
Asked
Active
Viewed 448 times
2
-
4if $x$ is not an integer then there is a neighborhood of $x$ where ceil($x$) is constant – J. W. Tanner Jun 23 '19 at 19:36
-
If $f=g$ on some open set containing $x$ and $g$ is differentiable, then $f'(x)=g'(x)$. – copper.hat Jun 23 '19 at 19:45
2 Answers
2
The ceiling function is piecewise constant and the derivative of a constant is zero.
The ceiling function is discontinuous at integer values and is non-differentiable there.
0
$x\le \lceil x\rceil<x+1$.
Let $\epsilon=\min(\lceil x_0\rceil - x_0,x_0+1-\lceil x_0 \rceil$). If $x_0$ is not an integer, then $x_0<\lceil x_0\rceil, $ so $\epsilon>0$.
Then for all $x$ in $(x_0-\epsilon, x_0+\epsilon)$, $\lceil x\rceil =\lceil x_0\rceil,$ i.e., the ceiling function is constant on this interval,
so its derivative is $0$ at $x_0$.
J. W. Tanner
- 60,406
-
To be pedantic, note that if $x<x_0+\epsilon$ then $x<x_0+\lceil x_0\rceil-x_0=\lceil x_0\rceil$ and if $x>x_0-\epsilon$ then $x>x_0-(x_0+1-\lceil x_0\rceil)=\lceil x_0\rceil -1$ so if $x_0-\epsilon<x<x_0+\epsilon$ then $x<\lceil x_0\rceil<x+1$ – J. W. Tanner Jun 23 '19 at 20:13