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challenge for programmable calculators
12-26-2013, 06:19 AM
Post: #49
RE: challenge for programmable calculators
Or just have a look at the roots: {0.04974, 0.857089, 1.50476, 15.5884}
Since \(f(1)=1\) it's obvious that \(f(n)<0\) for n in {2,...,15}.
To show this for n = 15 with your method isn't possible.
In this case we're lucky that we don't have to do that.
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Messages In This Thread
Proof using number theory - cruff - 12-24-2013, 05:43 PM
RE: challenge for programmable calculators - Thomas Klemm - 12-26-2013 06:19 AM
RE: challenge for programmable calculators - radwilliams - 12-24-2013, 05:57 PM



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