Help: symmetric derivative
Ben Logan
I'm having trouble proving that if the derivative of a function
exists, then the symmetric derivative exists.
"The *symmetric derivative* f_s(x) is defined by
f(x+h) - f(x-h)
f_s(x) = lim ---------------------
h-->0 2h
Show that if f'(x) exits, then f_s(x) exists but that the converse is
false."
The problem doesn't seem to ask me to prove that the two derivatives
are *equal*; just that they exist (although it seems that they would
have to be equal).
Thanks in advance for any help.
Ben Logan
Sujit Ray
= lim h->0 [ f(x+h) - f(x-h) ] / 2h
= lim h-> 0 { [ f(x+h)-f(x) / 2h ] + [ f(x)-f(x-h) / 2h ] }
= lim h-> 0 [ f(x+h) - f(x) ] / 2h + lim h->0 [ f(x) - f(x-h) ] / 2h
= 1/2 { lim h->0 [ f(x+h) - f(x) ] / h + lim h->0 [ f(x) - f(x-h)] / h }
= 1/2 { f'(x) + f'(x) }
= f'(x)
So obviously if f'(x) exists f_s(x) also exists, and they are equal. Why
wouldn't the converse be true? It appears to be true.
Ben Logan <wbl...@hotmail.com> wrote in article
<37f27a9a...@news1.lvdi.net>...
Raymond Manzoni
>
> Hi.
>
> I'm having trouble proving that if the derivative of a function
> exists, then the symmetric derivative exists.
>
> "The *symmetric derivative* f_s(x) is defined by
>
> f(x+h) - f(x-h)
> f_s(x) = lim ---------------------
> h-->0 2h
>
> Show that if f'(x) exits, then f_s(x) exists but that the converse is
> false."
>
> The problem doesn't seem to ask me to prove that the two derivatives
> are *equal*; just that they exist (although it seems that they would
> have to be equal).
>
> Thanks in advance for any help.
>
> Ben Logan
Hi,
Try to write f_s(x) as sum of two limits (add f(x)-f(x) at numerator)
For the converse case use, for example, a function symmetric around
x=x0 axis but singular there.
Hoping it was enough,
Raymond
joseph leva
You don't have to be singular. Consider f(x)=abs(x). The symmetric
derivative exists but the function is not differentiable at x=0.
Joe
Raymond Manzoni
>
> You don't have to be singular. Consider f(x)=abs(x). The symmetric
> derivative exists but the function is not differentiable at x=0.
>
> Joe
Right, sorry to be singular ! :-)
Raymond
Ben Logan
Just one more question: according to my calculations, the symmetric
derivative of abs(x) is 0. Now, the derivative of abs(x) exists for
all x except 0 and the symmetric derivative is equal to it for all x
except zero. So, just out of curiosity, wouldn't that mean that the
symmetric derivative of the abs(x) is not a continuous function?
Thanks again,
Ben Logan
Ed Hook
wbl...@hotmail.com (Ben Logan) writes:
|> Hi.
|>
|> I'm having trouble proving that if the derivative of a function
|> exists, then the symmetric derivative exists.
|>
|> "The *symmetric derivative* f_s(x) is defined by
|>
|> f(x+h) - f(x-h)
|> f_s(x) = lim ---------------------
|> h-->0 2h
|>
|>
|> Show that if f'(x) exits, then f_s(x) exists but that the converse is
|> false."
|>
|> The problem doesn't seem to ask me to prove that the two derivatives
|> are *equal*; just that they exist (although it seems that they would
|> have to be equal).
|>
Well, just fooling around a little leads quickly to noticing
that
f(x+h) - f(x-h) 1 [ f(x+h) - f(x) f(x) - f(x-h) ]
----------------- = --- [ --------------- + --------------- ]
2h 2 [ h h ]
1 [ f(x+h) - f(x) f(x) - f(x-h) ]
= --- [ --------------- - --------------- ]
2 [ h -h ]
which ought to prove helpful ...
|> Thanks in advance for any help.
|>
|> Ben Logan
|>
|>
--
Ed Hook | Copula eam, se non posit
MRJ Technology Solutions, Inc. | acceptera jocularum.
NAS, NASA Ames Research Center | I can barely speak for myself, much
Internet: ho...@nas.nasa.gov | less for my employer
G.E. Ivey
Yep, that's what it means. Nothing at all unusual about that. The
"symmetric derivative" of |x| is 1 if x>0, 0 for x= 0 and -1 if x<0.
The regular derivative is also not continuous at 0. The regular
derivative of |x| is 1 if x>0, -1 if x< 0 and does not exist for x=0.
Dave L. Renfro
wrote
Elementary calculus texts aside (which do this for
pedagogical reasons, I suspect), one typically says that
the derivative of |x| has no continuous extension to x=0,
rather than that it fails to have a derivative at x=0.
There are a number of results known about the set of
points where a function can be symmetrically differentiable,
but not differentiable.
Let I be an open interval and f : I --> reals. Denote by
D(f) the set of points in I at which f has a finite derivative
and denote by SD(f) the set of points in I at which f has
a finite symmetric derivative.
1. As several others have observed, D(f) is a subset
of SD(f).
2. If f is measurable, then the relative complement of D(f)
in SD(f), SD(f) - D(f), has measure zero.
A. Khintchine, "Recherches sur la structure des fonctions
measurables", Fund. Math. 9 (1927), 212-279.
3. The continuum hypothesis implies that there exists
a measurable function f such that SD(f) - D(f) is
residual in I. [Residual in I means "has a first
(Baire) category complement in I".]
Let G be an additive subgroup of the reals that has
second category in every interval [Erdos constructs
such a group using the continuum hypothesis in
"Some remarks on subgroups of real numbers", Colloq.
Math. 42 (1979), 119-120.], and let f be the
characteristic function of G. Then it is not difficult
to check that SD(f) = G and D(f) = empty set.
I don't know if there exists in ZFC a function f
such that SD(f) - D(f) is residual. My guess would
be "yes". (The answer may already be known.)
4. If the set of points at which f is continuous is dense in I,
then SD(f) - D(f) is both first category and measure zero.
In fact, the set is sigma-porous, a notion strictly
stronger than "first category and measure zero".
C. L. Belna, Michael J. Evans, and Paul D. Humke,
"Symmetric and ordinary differentiation", Proc. Amer.
Math. Soc. 72 (1978), 261-267.
5. There exist functions f continuous everywhere on I
such that SD(f) - D(f) contains a nonempty perfect
set. [In other words, even for everywhere continuous
functions, SD(f) - D(f) can be uncountable.]
James Foran, "The symmetric and ordinary derivative",
Real Analysis Exchange 2 (1977), 105-108.
Foran asks in this paper whether the set can have positive
Hausdorff dimension. As far as I know, this is still open.
6. If the points at which f is continuous is dense in I,
then SD(f) - D(f) is a countable union of uniformly
symmetrically porous sets, where the uniformity
constant for each set in this countable union can
be uniformly chosen larger than any fixed number
less than 1.
It is known that such sets can still have Hausdorff
dimension 1. Apparently, it is not known whether or not
SD(f) - D(f) MUST have Hausdorff dimension less than 1,
nor whether or not SD(f) - D(f) CAN have Hausdorff
dimension greater than 0. (However, this may no longer
be the case.)
Ludek Zajicek, "A note on the symmetric and ordinary
derivative", Atti Sem. Mat. Fis Univ. Modena 41 (1993),
263-267. [Proved the result for f continuous everywhere
on I.]
Michael J. Evans, "A note on symmetric and ordinary
differentiation", Real Analysis Exchange 17 (1991-92),
820-826. [Generalized Zajicek's result for functions
whose points of continuity are dense in I.]
7. The most complete reference available on these and
other related matters is
Brian S. Thomson, SYMMETRIC PROPERTIES OF REAL FUNCTIONS,
Monographs and Textbooks in Pure and Applied Mathematics
183, Marcel Dekker, 1994, xiii + 447 pages.
<http://www.dekker.com/e/p.pl/9230-0>
<http://www.amazon.com/exec/obidos/ASIN/0824792300/qid=938891045/sr=1-4/002-3040240-8137458>
<http://www.math.sfu.ca/mast/grad_info/brochure_59.html>
I don't have a copy of this book, but I feel certain
that everything above (except possibly #8) can be found
in it.
Dave L. Renfro