Interesting question - Not homework

  • #26
matt grime
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what's wrong with that?
 
  • #27
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Nothing! It has just dawned on me that because I copy and paste TEX through my working that the 2^(2n) was more than likely a mistype otherwise I would not have arrived at my solution!!
 
  • #28
HallsofIvy
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A couple of hours after my post (when I was not near a computer!) it suddenly dawned on me that NewScientist must have meant
[tex]\Sigma log_{2^{2^n}} e[/tex]

That reduces to
[tex]\Sigma \frac{(-1)^n}{ln(2^{2^n}}= \Sigma \frac{(-1)^n}{2^n ln 2}= \frac{1}{ln 2}\Sigma \left(\frac{-1}{2}\right)^n[/tex]
which is, in fact, a simple geometric series.
 
  • #29
benorin
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That's Beautiful.

HallsofIvy said:
A couple of hours after my post (when I was not near a computer!) it suddenly dawned on me that NewScientist must have meant

[tex]\Sigma log_{2^{2^n}} e[/tex]

That reduces to

[tex]\Sigma \frac{(-1)^n}{ln 2^{2^n} }= \Sigma \frac{(-1)^n}{2^n ln 2}= \frac{1}{ln 2}\Sigma \left(-\frac{1}{2}\right) ^n[/tex]
which is, in fact, a simple geometric series.
 
Last edited:

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