Piecewise smooth functions..

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- Oct 15th 2012, 12:12 PMsarideli18P.S
Piecewise smooth functions..

- Oct 15th 2012, 01:25 PMTheEmptySetRe: I need help with piecewise smooth functions
The definition of peicewise smooth is that a function is differentable on the domain except at a finite number of points. Also at these points both the left and right derviatve must exist.

So for i)

h(x) is continous and differentable everywhere except at x=0. For simplicity note that h(x) is an even function so we only have to test the derviative on one side.

$\displaystyle \lim_{h \to 0}\frac{(0+h)^2\ln(0+h)}{h} = \lim_{h \to 0}h\ln(h)$

This limit can be resolved using L'hosipitals

by L.H

$\displaystyle \frac{\ln(h)}{\frac{1}{h}}$

So now this is the form infinity divided by infinity so we take the derivative to get

$\displaystyle \frac{\frac{1}{h}}{\frac{-1}{h^2}}=-h$

So the limit is 0 and both the left and right derviative exist at that point. Also the function can be made continous if we define h(0)=0.

Now try the 2nd one. - Oct 15th 2012, 01:46 PMsarideli18Re: I need help with piecewise smooth functions
Well, I just found that the first and fourth ones are piecewise smooth, and the second one is not piecewise smooth. But I am stuck on the third one. Since the function itself (2/(1-x)) and 0) and its derivative (2/(x-1)^2 and 0) are continuous I think it is piecewise smooth. But there is a jump at x=1. That confuses me.

- Oct 15th 2012, 01:49 PMsarideli18Re: I need help with piecewise smooth functions
I mean, the limit from left is infinite and the limit from right is 0. It is the same case as the derivative of the function. Therefore, I think it is not piecewise smooth, but I am not sure.

- Oct 15th 2012, 01:57 PMTheEmptySetRe: I need help with piecewise smooth functions
- Apr 12th 2013, 11:41 PMaddictionsharyRe: I need help with piecewise smooth functions
I just found that the first and fourth ones are piecewise smooth, and the second one is not piecewise smooth.

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