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Bernoulli's resolution of the St Petersburg Paradox - how to find the limit

Keith Woodward, Sydney Uni
Posted 10 years ago
POSTED BY: Keith Woodward
6 Replies
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POSTED BY: Daniel Lichtblau
POSTED BY: Keith Woodward

I just see Daniel's answer. Pretty much the same. well... anyhow...
POSTED BY: Sander Huisman

Well, It is definitely not gonna work when you write 

ln

instead of

Log

!

Here is the correct code to solve this problem:

ClearAll[GetSum]
GetSum[w_?NumberQ,c_?NumberQ]:=NSum[1/2^n*(Log[w+2^n-c]-Log[w]),{n,1,Infinity}]

w=1000000;
Plot[GetSum[w,c],{c,1,30}]

c/.FindRoot[GetSum[w,c],{c,20,15,25}]

enter image description here
POSTED BY: Sander Huisman

Hi Daniel,

Thanks for your reply.

I'm sorry, I mistakenly wrote 100 in my initial wealth rather than 1,000,000 which is what I tried in Mathematica and which didn't give a numerical result.

I changed the value to the correct 1,000,000 five minutes after my initial post, but I guess you only saw the first incorrect version. Sorry about that.

Here is the mathematica code that I tried to run with no numerical result:

w = 1000000;

Solve[Sum[1/2^n*(ln[w + 2^n - c] - ln[w]), {n, 1, Infinity}] == 0, c]

Best regards, Keith
POSTED BY: Keith Woodward

Read the Wikipedia link again. The value of c=10.95 is for a different w.

Also it is better to post actual Mathematica code for purposes of others testing/fixing it.
POSTED BY: Daniel Lichtblau
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