[ 3 posts ] 
 Exercise 7.3 
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Joined: 04 Feb 2009, 15:52
Posts: 8
Post Exercise 7.3
I’m sorry, but my english is very bad. I’m engineer but I’m working as teacher at high school in Spain.
I’m not sure about my proof, but here it is.

Let’s g(p)=\frac{1}{z-p} with z=const. Then,

g^{(n)}(p)=\frac{n!}{(z-p)^{n+1}} and I'll write the Maclaurin serie of g(p):

g(p)=g(0)+\frac{g'(0)}{1!}.p+\frac{g''(0)}{2!}.p^2+...+\frac{g^{(n)}(0)}{n!}.p^n+...


g(p)=\frac{1}{z-p}=\frac{1}{z}+\frac{1}{z^2}.p+\frac{1}{z^3}.p^2+...+\frac{1}{z^{n+1}}.p^n+...


Now, let z=variable, p=const, and let's multiply all by f(z):

\frac{f(z)}{z-p}=\frac{f(z)}{z}+\frac{f(z)}{z^2}.p+\frac{f(z)}{z^3}.p^2+...+\frac{f(z)}{z^{n+1}}.p^n+... and then (without rigorous justification),


\frac{1}{2\pi i}\oint\frac{f(z)}{z-p}dz = \frac{1}{2\pi i}.(\oint\frac{f(z)}{z}dz+\oint\frac{f(z)}{z^2}dz.p +...+\oint\frac{f(z)}{z^{n+1}}dz.p^n +...)


But if we can use Cauchy formula in then 'origin shifted', that is \frac{1}{2\pi i}\oint\frac{f(z)}{z-p}dz=f(p), then


f(p)=\sum^\infty_{k=0}\frac{1}{k!}\frac{k!}{2\pi i.}\oint\frac{f(z)}{z^{k+1}}dz.p^k


And the integral \frac{k!}{2\pi i}\oint\frac{f(z)}{z^{k+1}}dz is the 'definition' of the k-th derivative at the origin, so


f(p)=\sum^\infty_{k=0}\frac{f^{(k)}(0)}{k!}.p^k


Now, we take p=variable and rename it as 'z':


f(z)=f(0)+\frac{f'(0)}{1!}.z+\frac{f''(0)}{2!}.z^2+...+\frac{f^{(n)}(0)}{n!}.z^n+...

So, f(z) is analytic Q.E.D.

Please, could you correct my grammaticals errors and tell me if this proof is right?!
Thank you.


05 Feb 2009, 15:43
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Joined: 07 Jun 2008, 08:21
Posts: 235
Post Re: Exercise 7.3
Pibe wrote:
I’m sorry, but my english is very bad.

Actually what you wrote is pretty good. I haven't checked your proof yet - that might take a bit longer, and someone else might check it before me.
However here's a version with the English corrected:

I’m an engineer but I’m working as a teacher at a high school in Spain.
I’m not sure about my proof, but here it is.

Let g(p)=\frac{1}{z-p} with z=const. Then,

g^{(n)}(p)=\frac{n!}{(z-p)^{n+1}} and I'll write the Maclaurin series for g(p):

g(p)=g(0)+\frac{g'(0)}{1!}.p+\frac{g''(0)}{2!}.p^2+...+\frac{g^{(n)}(0)}{n!}.p^n+...


g(p)=\frac{1}{z-p}=\frac{1}{z}+\frac{1}{z^2}.p+\frac{1}{z^3}.p^2+...+\frac{1}{z^{n+1}}.p^n+...


Now, let z=variable, p=const, and let's multiply everything by f(z):

\frac{f(z)}{z-p}=\frac{f(z)}{z}+\frac{f(z)}{z^2}.p+\frac{f(z)}{z^3}.p^2+...+\frac{f(z)}{z^{n+1}}.p^n+... and then (without rigorous justification),


\frac{1}{2\pi i}\oint\frac{f(z)}{z-p}dz = \frac{1}{2\pi i}.(\oint\frac{f(z)}{z}dz+\oint\frac{f(z)}{z^2}dz.p +...+\oint\frac{f(z)}{z^{n+1}}dz.p^n +...)


But if we use the Cauchy formula in the 'origin shifted' form, that is \frac{1}{2\pi i}\oint\frac{f(z)}{z-p}dz=f(p), then


f(p)=\sum^\infty_{k=0}\frac{1}{k!}\frac{k!}{2\pi i.}\oint\frac{f(z)}{z^{k+1}}dz.p^k


And the integral \frac{k!}{2\pi i}\oint\frac{f(z)}{z^{k+1}}dz is the 'definition' of the k-th derivative at the origin, so


f(p)=\sum^\infty_{k=0}\frac{f^{(k)}(0)}{k!}.p^k


Now, we take p=variable and rename it as 'z':


f(z)=f(0)+\frac{f'(0)}{1!}.z+\frac{f''(0)}{2!}.z^2+...+\frac{f^{(n)}(0)}{n!}.z^n+...

So, f(z) is analytic Q.E.D.

Please, could you correct my grammatical errors and tell me if this proof is right?!
Thank you.


07 Feb 2009, 16:49

Joined: 04 Feb 2009, 15:52
Posts: 8
Post Re: Exercise 7.3
Thank you very much for your help, vasco.
I reflect on my proof, and I think "I went too far..."
I think the proof is better like this:

Let g(p)=\frac{1}{z-p} with z=const. Then,

g^{(n)}(p)=\frac{n!}{(z-p)^{n+1}} and I'll write the Maclaurin series for g(p):

g(p)=g(0)+\frac{g'(0)}{1!}.p+\frac{g''(0)}{2!}.p^2+...+\frac{g^{(n)}(0)}{n!}.p^n+...


g(p)=\frac{1}{z-p}=\frac{1}{z}+\frac{1}{z^2}.p+\frac{1}{z^3}.p^2+...+\frac{1}{z^{n+1}}.p^n+...


Now, let z=variable, p=const, and let's multiply everything by f(z):

\frac{f(z)}{z-p}=\frac{f(z)}{z}+\frac{f(z)}{z^2}.p+\frac{f(z)}{z^3}.p^2+...+\frac{f(z)}{z^{n+1}}.p^n+... and then (without rigorous justification),


\frac{1}{2\pi i}\oint\frac{f(z)}{z-p}dz = \frac{1}{2\pi i}.(\oint\frac{f(z)}{z}dz+\oint\frac{f(z)}{z^2}dz.p +...+\oint\frac{f(z)}{z^{n+1}}dz.p^n +...)


But if we use the Cauchy formula in the 'origin shifted' form, that is \frac{1}{2\pi i}\oint\frac{f(z)}{z-p}dz=f(p), then


f(p)=\sum^\infty_{k=0}\frac{1}{k!}\frac{k!}{2\pi i}\oint\frac{f(z)}{z^{k+1}}dz.p^k


And if the integral \frac{k!}{2\pi i}\oint\frac{f(z)}{z^{k+1}}dz exist \forall k we just wrote a power series of f(p), like this:


f(p)=\sum^\infty_{k=0}a_k.p^k where a_k=\frac{1}{2\pi i}\oint\frac{f(z)}{z^{k+1}}dz


Now, we take p=variable and rename it as 'z':


f(z)=a_0+a_1.z+a_2.z^2+...+a_n.z^n+...

So, f(z) is analytic at origin, Q.E.D.

But moreover, this power series ¿must be? the Maclaurin series, so we can use this fact to "define" f^{(k)}(0):

a_k=\frac{1}{2\pi i}\oint\frac{f(z)}{z^{k+1}}dz=\frac{f^{(k)}(0)}{k!}

f^{(k)}(0)=\frac{k!}{2\pi i}\oint\frac{f(z)}{z^{k+1}}dz, \forall k, therefore f is C^\infty-smooth at origin.

¿Is this proof right?


12 Feb 2009, 17:05
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