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RichE

Yes they do or you have a different definition of analytic - analytic means defined, or capable of being defined, by a power series

So isn't $f(x)=x^{-\frac{1}{2}}$ analytic? It can be represented by a power series, but $f(-1)$ does not belong in the set of Real numbers.

LennonMcCartney

So isn't $f(x)=x^{-\frac{1}{2}}$ analytic? It can be represented by a power series, but $f(-1)$ does not belong in the set of Real numbers.

but -1 won't be in the domain of any power series you choose to use to define x^(-1/2) - the domain being defined as within the radius of convergence of the power series

Anyway, I thought analytic could also mean infinitely differentiable.

RichE

but -1 won't be in the domain of any power series you choose to use to define x^(-1/2) - the domain being defined as within the radius of convergence of the power series

Ah yes good point. What about f(-0.5)?

RichE

the term for that is $C^\infty$ or "smooth"

Ohhhh.... so smooth and analytic aren't identical properties of a function; now I get it. Thanks for clearing that up.

RichE

well -0.5 wouldn't be in the domain either

what power series do you have in mind?

what power series do you have in mind?

And again a good point. I'm just being silly here.

LennonMcCartney

Ohhhh.... so smooth and analytic aren't identical properties of a function; now I get it. Thanks for clearing that up.

for a complex function to have a derivative on an open set is enough for it to be analytic (or holomorphic)

this isn't the case, even for smooth functions, with the real numbers

the function

f(x) = exp(-1/x^2) if x =/= 0 and f(0)=0

has derivatives of all orders at 0 and they are all 0

clearly it isn't analytic though because its Taylor series defines the zero function

Good shout. Cheers mate.

That's shiny's random comment of the day (:

Bezza

i'm gonna hope that was a type from chloé - if not, she should be glad we've got no more lectures this year! :vmad:

maybe it's a sign of affection....or some other freudian slip

And nobody has answered my standard error query But I think it is that the standard error is the standard deviation divided by Root(N-1). But then what exactly is the root mean square deviation?

Bezza

i'm gonna hope that was a type from chloé - if not, she should be glad we've got no more lectures this year! :vmad:

Sorry!!!! It most certainly was a typo (rep coming your way!) as I was posting before I had to go to a tute (which has now been rearranged for tomorrow). Sorry NICK; I know how annoying it is as people keep calling me Cholé which is incrediably annoying..however possibly calling you a girl is worse! Sorry!!!

Hoofbeat

Sorry!!!! It most certainly was a typo (rep coming your way!) as I was posting before I had to go to a tute (which has now been rearranged for tomorrow). Sorry NICK; I know how annoying it is as people keep calling me Cholé which is incrediably annoying..however possibly calling you a girl is worse! Sorry!!!

i didnt even name Cholé was a name? Or is that another typo (in which case you've completely confussled me)?

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