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

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

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Tide

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

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lurflurf

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Defining terms is good!Tide said:Define your terms!

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

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

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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

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HallsofIvy

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[tex]f(x)= e^{-\frac{1}{x^2}}[/tex]

if x is not 0, f(0)= 0.

It's easy to show that all derivatives of f are of the form

[tex]P(x)e^{-\frac{1}{x^2}}[/tex]

,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.

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Thank you Halls.

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

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

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HallsofIvy

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Oh.

All the texts I have seen have said that smooth means infinitely differentiable.

All the texts I have seen have said that smooth means infinitely differentiable.

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