Mathematica Asked by meraj on April 27, 2021
The expression
|log(1+x) – log(1 – x)| <= |x-y| /{1+|x-y|}, 0<x<1, 0<y<1
is valid.
How we can prove using Mathematica (either graphically or using construct with x = y and x # y)
Your claim is invalid.
Look at your expression:
Abs[Log[1 + x] - Log[1 - x]] <= Abs[x - y]/(1 + Abs[x - y])
Assume x==y
, then the right hand side is zero. The left hand side is e.g. for x == y == 0.5
Abs[Log[1 + x] - Log[1 - x]]/. {x -> .5, y -> .5}
(* 1.09861 *)
Answered by Daniel Huber on April 27, 2021
Clear["Global`*"]
The inequality is not valid over the specified ranges.
ineq = Abs[Log[1 + x] - Log[1 - x]] <= Abs[x - y]/(1 + Abs[x - y]);
Plot3D[Evaluate[List @@ ineq],
{x, 0, 1}, {y, 0, 1},
PlotLegends -> "Expressions", ClippingStyle -> None,
AxesLabel -> Automatic]
ineq /. {x -> 1/2, y -> 1/2}
(* False *)
Answered by Bob Hanlon on April 27, 2021
Get help from others!
Recent Answers
Recent Questions
© 2024 TransWikia.com. All rights reserved. Sites we Love: PCI Database, UKBizDB, Menu Kuliner, Sharing RPP