Oxbridge Physics Prelims Revision

LennonMcCartney

Yes but analytic functions of real variables don't always map into the real numbers. That's my point, which is what you said below.

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

Reply 202

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.

Reply 203

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

Reply 204

Anyway, I thought analytic could also mean infinitely differentiable.

Reply 205

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)?

Reply 206

LennonMcCartney

Anyway, I thought analytic could also mean infinitely differentiable.

the term for that is

C^\infty

or "smooth"

Reply 207

LennonMcCartney

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

well -0.5 wouldn't be in the domain either

what power series do you have in mind?

Reply 208

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.

Reply 209

RichE

well -0.5 wouldn't be in the domain either

what power series do you have in mind?

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

Reply 210

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

Reply 211

Good shout. Cheers mate.

Reply 212

LennonMcCartney

Good shout. Cheers mate.

de rien - really I should apolologise for mathecising a physics thread :rolleyes:

Reply 213

shiny

12

LennonMcCartney

No quaternions at all in our physics.

Fooking quaternions! :eek:

Reply 214

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

Reply 215