Menu Close

Is-this-solvable-y-f-x-floor-function-dy-dx-

Tinku Tara 0 Comments
Pin




Question Number 3928 by Filup last updated on 24/Dec/15
Is this solvable: y=⌊f(x)⌋ floor function (dy/dx)=???
$$\mathrm{Is}\:\mathrm{this}\:\mathrm{solvable}: \\ $$$$ \\ $$$${y}=\lfloor{f}\left({x}\right)\rfloor\:\:\:\:\:\:\:\:\:\mathrm{floor}\:\mathrm{function} \\ $$$$\frac{{dy}}{{dx}}=??? \\ $$
Commented by Yozzii last updated on 25/Dec/15
∃f(x)∣(dy/dx) exists. Let f(x)=x , x∈R. For n<x<n+1, n∈Z y=⌊f(x)⌋=⌊x⌋=n ⇒ (dy/dx)=0. Suppose we wished to determine the value of lim_(x→n) y. We observe that, for the left hand limit, y=⌊x⌋=n−1 since for 0≤n−1<x<n, ⌊x⌋=n−1. ∴ lim_(x→n^− ) y=lim_(x→n^− ) (n−1)=n−1. The right hand limit yields y→n, i.e lim_(x→n^+ ) y=n≠n−1=lim_(x→n^− ) y. Therefore, a jump discontinuity exists at all points x=n>0, n∈Z If n<0, then n−1<n<x<n+1≤0. ∴ lim_(x→n^− ) y=lim_(x→n^− ) n−1=n−1 but lim_(x→n^+ ) y=lim_(x→n^+ ) n+1=n+1 or lim_(x→n^+ ) y=0 0≥n+1>x>n⇒⌊x⌋=n+1 or 0. Again, lim_(x→n^− ) y≠lim_(x→n^+ ) y so jump discontinuities exist at x=n<0, n∈Z. At points of discontuity, derivatives do not exist⇒(dy/dx) is undefined.
$$\exists{f}\left({x}\right)\mid\frac{{dy}}{{dx}}\:\:{exists}.\:{Let}\:{f}\left({x}\right)={x}\:,\:\:{x}\in\mathbb{R}. \\ $$$${For}\:{n}<{x}<{n}+\mathrm{1},\:{n}\in\mathbb{Z}\:{y}=\lfloor{f}\left({x}\right)\rfloor=\lfloor{x}\rfloor={n} \\ $$$$\Rightarrow\:\frac{{dy}}{{dx}}=\mathrm{0}.\: \\ $$$${Suppose}\:{we}\:{wished}\:{to}\:{determine}\:{the} \\ $$$${value}\:{of}\:\underset{{x}\rightarrow{n}} {\mathrm{lim}}{y}.\:{We}\:{observe}\:{that}, \\ $$$${for}\:{the}\:{left}\:{hand}\:{limit},\:{y}=\lfloor{x}\rfloor={n}−\mathrm{1} \\ $$$${since}\:{for}\:\mathrm{0}\leqslant{n}−\mathrm{1}<{x}<{n},\:\lfloor{x}\rfloor={n}−\mathrm{1}. \\ $$$$\therefore\:\underset{{x}\rightarrow{n}^{−} } {\mathrm{lim}}{y}=\underset{{x}\rightarrow{n}^{−} } {\mathrm{lim}}\left({n}−\mathrm{1}\right)={n}−\mathrm{1}. \\ $$$${The}\:{right}\:{hand}\:{limit}\:{yields}\:{y}\rightarrow{n}, \\ $$$${i}.{e}\:\underset{{x}\rightarrow{n}^{+} } {\mathrm{lim}}{y}={n}\neq{n}−\mathrm{1}=\underset{{x}\rightarrow{n}^{−} } {\mathrm{lim}}{y}. \\ $$$${Therefore},\:{a}\:{jump}\:{discontinuity}\: \\ $$$${exists}\:{at}\:{all}\:{points}\:{x}={n}>\mathrm{0},\:{n}\in\mathbb{Z} \\ $$$${If}\:{n}<\mathrm{0},\:{then}\:{n}−\mathrm{1}<{n}<{x}<{n}+\mathrm{1}\leqslant\mathrm{0}. \\ $$$$\therefore\:\underset{{x}\rightarrow{n}^{−} } {\mathrm{lim}}{y}=\underset{{x}\rightarrow{n}^{−} } {\mathrm{lim}}{n}−\mathrm{1}={n}−\mathrm{1} \\ $$$${but}\:\underset{{x}\rightarrow{n}^{+} } {\mathrm{lim}}{y}=\underset{{x}\rightarrow{n}^{+} } {\mathrm{lim}}{n}+\mathrm{1}={n}+\mathrm{1}\:{or}\:\underset{{x}\rightarrow{n}^{+} } {\mathrm{lim}}{y}=\mathrm{0} \\ $$$$\mathrm{0}\geqslant{n}+\mathrm{1}>{x}>{n}\Rightarrow\lfloor{x}\rfloor={n}+\mathrm{1}\:{or}\:\mathrm{0}. \\ $$$${Again},\:\underset{{x}\rightarrow{n}^{−} } {\mathrm{lim}}{y}\neq\underset{{x}\rightarrow{n}^{+} } {\mathrm{lim}}{y}\:{so}\:{jump}\:{discontinuities} \\ $$$${exist}\:{at}\:{x}={n}<\mathrm{0},\:{n}\in\mathbb{Z}. \\ $$$${At}\:{points}\:{of}\:{discontuity},\:{derivatives} \\ $$$${do}\:{not}\:{exist}\Rightarrow\frac{{dy}}{{dx}}\:{is}\:{undefined}. \\ $$$$ \\ $$$$ \\ $$$$ \\ $$$$ \\ $$
Commented by 123456 last updated on 26/Dec/15

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *