# Can anyone think of any examples of curves which are smooth but not analytic?

I know there are some, but I can't think of any examples.

I asked my teacher after class but she couldn't think of any either.

Tide
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The Koch curve is an example of a continuous curve but it is nowhere differentiable.

lurflurf
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Tide said:
Defining terms is good!
Here I hope there is not confunsion.
I assume what was meant is
Smooth: A function having derivatives of all orders in some region
ie on (a,b) f(x) exist and f'(x) and f''(x) and so on
Analytic: A function the equals its taylor expansion

The usual example given in introductory analysis texts of a smoth nonanalytic function is something like
f(x)=0 x<=0
f(x)=exp(-1/x) x>0
The taylor expansion about 0 is 0, but the function is not identicaly zero.

The Joys of Calc. 1

Sadly, at my school, this is covered in Calc III

A curve, or space-curve, is smooth everywhere if r'(t) does not equal the 0 vector anywhere on r's domain, and if r'(t) exists everywhere on r's domain.

As far as I remember, an example of the curve r(t) = <t,t^2> is not smooth at t = 0, but as far as I can tell from below, this curve is, infact, smooth at t = 0 because r'(t) exists there and is not 0.

r'(t) = <1,2t>

When t = 0, r'(t) = <1,0>.

Can anyone clarify this for me.

If I reparameterize the curve as follows, r(t) = <t^2 , t^4>, then the curve would not be smooth at t = 0 for the following reason:

r'(t) = <2t, 4t^3>

when t = 0 r'(t) = <0,0>, so the curve r(t) = r(t) = <t^2 , t^4> is not smooth at t = 0.

it is, however, piece-wise smooth, as piece-wise smooth curves are curves that have a finite number of smooth pieces. We never really covered, in detail, the formal definition of what an "analytic" function is, and I am currently in calculus IV, although I know from my own independent studies as well as my current extracurricular involvement in mathematical research.

Any thoughts.

Inquisitively,

Edwin

HallsofIvy
Homework Helper
If by "smooth" you mean infinitely differentiable, the simplest example is

$$f(x)= e^{-\frac{1}{x^2}}$$
if x is not 0, f(0)= 0.

It's easy to show that all derivatives of f are of the form
$$P(x)e^{-\frac{1}{x^2}}$$
,where P(x) is a polynomial, for x not 0, 0 if x= 0 and so the nth derivative exists and is continuous for all n.

However, since the nth derivative of f at 0 is always 0, the Taylor's Polynomial about x= 0 is just the constant 0. That converges for all x, of course, but is not equal to f(x) for any x other than 0 so f is not analytic at x= 0.

Thank you Halls.

By the way, what other definition is there of smooth?

HallsofIvy