Wikipedia:Reference desk/Archives/Mathematics/2015 February 12

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February 12

Diffusion of innovations: logit or probit?

For the diffusion of innovations scenarios to which Rogers' model is applicable, does innovation utilization where possible (e.g. Bernoulli trial = IP request, success = IPv6 request) empirically follow a logit model or probit model more closely? Does the choice of curve change when a non-negligible number of users revert from the innovation to the previous technology (e.g. IPv6 to IPv4) compared to when they don't (e.g. electric lighting to oil lamps)? NeonMerlin 02:45, 12 February 2015 (UTC)

I suspect a rigorous answer would constitute WP:OR, and might even be publishable :) That being said, probit is just a limiting case of logit. So you have to ask yourself: is there anything intrinsic that requires strictly discrete outputs? If you do a logistic fit, the data will tell you how steep the switching is. If it's very steep, then a probit is likely just as good, and may have some benefits in terms of further analyses available or computation time needed to fit. SemanticMantis (talk) 15:15, 13 February 2015 (UTC)

peculiar integer factorization

Hello, I've been playing around with maxima and noticed that $10^{60}+1=73\times 137\times 1676321\times 99990001\times 5964848081\times 100009999999899989999000000010001$. Two of the factors are suspiciously close to powers of 10. Is this part of a pattern? Is this significant? Robinh (talk) 08:05, 12 February 2015 (UTC)

More generally, x⁶⁰+1=(x⁴+1)(x⁸-x⁴+1)(x¹⁶-x¹²-x⁸-x⁴+1)(x³²+x²⁸-x²⁰-x¹⁶-x¹²+x⁴+1). 10⁴+1 = 73⋅137 and 10¹⁶-10¹²-10⁸-10⁴+1 = 1676321⋅5964848081 is where the other factors come in. Cyclotomic polynomial has related information. --RDBury (talk) 09:41, 12 February 2015 (UTC)

Some computer algebra systems can do the work for you. Using the free PARI/GP, just start the program with gp.exe, enter factor(x^60+1) and you immediately get:

[                                   x^4 + 1 1]

[                             x^8 - x^4 + 1 1]

[               x^16 - x^12 + x^8 - x^4 + 1 1]

[x^32 + x^28 - x^20 - x^16 - x^12 + x^4 + 1 1]

The "1" at the end is the power of the factor. [x^4 + 1 2] would have meant (x^4 + 1)^2. Wolfram Alpha at http://www.wolframalpha.com/ can do it online. PrimeHunter (talk) 14:28, 12 February 2015 (UTC)

By the way, all 4 factors are prime for x = 2, 46, 91872, 132930, 136054, 265512, 638798, 744168, ... PrimeHunter (talk) 14:47, 12 February 2015 (UTC)

(OP) This also explains the observation that there are "too many" nines and zeros and ones in the two factors. thanks! Robinh (talk) 20:10, 12 February 2015 (UTC)

Resolved