Summer Reading

Monday 11 July 2005 at 5:25 pm | In Articles | 7 Comments

I finally succumbed and bought Martin Gardner’s Mathematical Games: The Entire Collection of His Scientific American Columns. Of course, as it had to be sent from America, the postage wasn’t cheap and it locked up my computer when I first tried it. Turning off CD autorun solved that problem.

Now I can spend the summer reading all those articles I’ve never read or even only half-remembered. The ability to search all the articles is a real boon. Did you know, for example, pi is mentioned 5645 times in the books?

In Fractal Music, Hypercards and More… Martin Gardner refers to the e^\pi>\pi^e problem (which I’ve mentioned before) and says:

    Dozens of proofs have already been published. One of the shortest is based on the fact from elementary calculus that x^{1/x} has a maximum value when x equals e. Hence e^{1/e} is greater than \pi^{1/\pi}}. Multiplying each exponent by \pi^e and canceling yields the inequality e^\pi>\pi^e.

The perils of publishing mathematics :-? Can you spot the error and correct it?

7 Comments »

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  1. Multiplying exponents does not maintain the inequality as it means that you are multiplying each side by different amounts?

    Comment by rohit khetan — Wednesday 13 July 2005 2:21 pm #

  2. You can multiply exponents as a^n>b^m \implies (a^n)^r>(b^m)^r \implies a^{nr}>b^{mr} which means that you have multiplied the exponents by r.

    But you are in the right area – think about what you are multiplying by.

    Comment by Steve — Wednesday 13 July 2005 3:17 pm #

  3. multiply exponents by pi*e rather than pi to the power e.

    Comment by rohit khetan — Friday 15 July 2005 6:04 am #

  4. rohit’s right. i guess the formal(?) way to say it would be to exponentiate both sides by e*pi, right?

    Comment by BL — Friday 15 July 2005 6:09 am #

  5. You got it. I expect somewhere in the publishing process \pi e got transposed to \pi^e. The problem with typing mathematics that such errors, which are easy to make, can invalidate the result.

    Comment by Steve — Friday 15 July 2005 10:31 am #

  6. Yes, but you actually multiply the larger side by a larger amount.

    Comment by guest — Friday 29 July 2005 7:23 am #

  7. I think it best to think of it as raising both sides to the same power. You can say the sides are being multiplied by e^{\pi-1/e} and by \pi^{e-1/\pi} but how do you know one is larger than the other without assuming what you are trying to prove?

    Comment by Steve — Friday 29 July 2005 10:18 am #

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