It is important to define exactly the concepts of 'derivative' and 'differenziable funtion'. The derivative of a function in is the limit [if a limit exists]...
(1)
A function is differentiable in if we can write...
(2)
... where is independent from and . Now we consider the function...
(3)
... in . On the basis of the (1) the derivative in exists and is but the function isn't differentiable in because we are not able to write a relation like (2)...
Kind regards
Are you sure?
Quoted from wiki: "In calculus (a branch of mathematics), a differentiable function is a function whose derivative exists at each point in its domain."
See Differentiable function - Wikipedia, the free encyclopedia
Im my opinion the 'key' of the question is that the fact that the derivative of a function exists in doesn't mean that is differentiable in . Wiki as other people may have a different opinion but in that case they have to justify the fact that a function like...
(1)
... has the derivatives of all order in and therefore its Taylor expansion around doesn't exist. In fact such Taylor expansion exists only if is differentiable in and that is not the situation...
Kind regards
A confirm of what i posted is in...
http://www.math.jhu.edu/~fspinu/405/...20notation.pdf
2.4 The following statement are true…
a) f is differentiable at a
b) there exist a real number such that
Kind regards
How? I haven't the slightest idea what either of you are talking about.
To say that is differentiable on is to say that exists for all . Thus, one can show that this is true for this function. Moreover, it is infinitely differentiable if it has dervatives of all orders, i.e. if exists for all and all . I showed that. Only after I proved it was differentiable did I make a REMARK about analyticity.
chisigma, I totally fail to understand what you are getting at here. For a start, the statements "the derivative of the function exists at " and " is differentiable at " mean exactly the same thing. Secondly, the function f defined by
is infinitely differentiable (equivalently, it has derivatives of all orders) at every point including . At , every derivative is equal to 0. That does not mean that the Taylor expansion around doesn't exist. The Taylor expansion around does exist, but it is identically zero.
Honestly for me is a little hard to undestand how is possible that this function...
(1)
... that has all its derivatives equal to zero in , can be computed 'somewhere around' with the Taylor expansion...
(2)
If all the are equal to 0 , it is evident that the (2) gives 0 no matter which is x... but y(x) is not equal to the 'nul function'... It seems to me that the situation is 'a little unconfortable'...
Kind regards