Question

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Answer 1

So, what ideas have you had for this problem so far?

I don't know how much help I can be for this, but I'll try my best. :)

Answer 2

thank you

Answer 3

From what I have been looking into: http://en.wikipedia.org/wiki/Subharmonic_function#Subharmonic_functions_in_the_complex_plane
A function \(\varphi\) of a complex variable defined on a set \(G \subset \mathbb{C} \) is subharmonic if and only if for any closed disk \( D(z, r) \subset G \) of center z and radius r one has
\( \displaystyle \varphi \left(z \right) \le \frac{1}{2\pi} \int_{0}^{2 \pi} \varphi \left( z + r e^{it} \right) \ dt \).
This bears striking resemblance to our current question I believe. Does this seem familiar / useful to you?

Answer 4

i know this but how to solve it this is what i want
any way thank you accessdenied

Answer 5

If we have:
\( \displaystyle u(z) \le \dfrac{1}{2 \pi} \int_{0}^{2 \pi} u(z + re^{it} ) \ dt \)
This gives on the right
\( \displaystyle \dfrac{1}{2 \pi} \int_{0}^{2 \pi} \left( \sum_{n=1}^{\infty} \frac{\ln |z+re^{it} - 2^n |}{2^n} \right) \ dt \)
and I believe we can interchange the integration and summation here so that the interior begins to match the original integral, right? That's what I am thinking so far.

Answer 6

\( \displaystyle \sum_{n=1}^{\infty} \left( \frac{1}{2^n} \times \frac{1}{2 \pi} \int_{0}^{2 \pi} \ln |re^{it} + z - 2^n| \ dt \right) \)
The integral on the inside matches the form of the original integral so that we could evaluate it using that statement.
This puts \(\zeta = 2^n - z \) and we should then check for each case \(r \le \ln |\zeta| \) and \(r > \ln |\zeta| \)

Answer 7

Were you able to figure it out? :)