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Exercise 7.3
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Joined: 04 Feb 2009, 15:52
Posts: 8
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 with z=const. Then, and I'll write the Maclaurin serie of g(p):  Now, let z=variable, p=const, and let's multiply all by f(z): and then (without rigorous justification), But if we can use Cauchy formula in then 'origin shifted', that is , then And the integral is the 'definition' of the k-th derivative at the origin, so Now, we take p=variable and rename it as 'z': 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 Supporter Joined: 07 Jun 2008, 08:21
Posts: 235
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 with z=const. Then, and I'll write the Maclaurin series for g(p):  Now, let z=variable, p=const, and let's multiply everything by f(z): and then (without rigorous justification), But if we use the Cauchy formula in the 'origin shifted' form, that is , then And the integral is the 'definition' of the k-th derivative at the origin, so Now, we take p=variable and rename it as 'z': 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
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 with z=const. Then, and I'll write the Maclaurin series for g(p):  Now, let z=variable, p=const, and let's multiply everything by f(z): and then (without rigorous justification), But if we use the Cauchy formula in the 'origin shifted' form, that is , then And if the integral exist we just wrote a power series of f(p), like this: where Now, we take p=variable and rename it as 'z': 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" :  , , therefore f is at origin.

¿Is this proof right?

12 Feb 2009, 17:05 Page 1 of 1 [ 3 posts ] 