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Integral hangs G2 - lrdheat - 03-29-2020 06:38 PM The integral from 0 to 1 of 3Root(1-x^7) - 7Root(1-x^3) hangs the G2, requiring a reset. The virtual unit gives a warning message, pressing enter returns original integral (which one can then ~ans). Virtual and G2 work in Home...almost produce the correct answer of zero. RE: Integral hangs G2 - Albert Chan - 03-29-2020 09:56 PM (03-29-2020 06:38 PM)lrdheat Wrote: The integral from 0 to 1 of 3Root(1-x^7) - 7Root(1-x^3) ... correct answer of zero. Last month, we had a thread about Wallis' product exploration. Googled "wallis product", I found this: Mathematical Analysis and the Mathematics of Computation, p.404, eqn. 7.87 \(\Large {1 \over \int_0^1 (1-x^{1/p})^q\;dx} = \binom{p+q}{p} = \binom{p+q}{q} = {1 \over \int_0^1 (1-x^{1/q})^p\;dx} \) This explained why your integral had area of zero. P.S. I do not know how above is derived. Any insight is appreciated. RE: Integral hangs G2 - toml_12953 - 03-30-2020 12:16 AM (03-29-2020 09:56 PM)Albert Chan Wrote:(03-29-2020 06:38 PM)lrdheat Wrote: The integral from 0 to 1 of 3Root(1-x^7) - 7Root(1-x^3) ... correct answer of zero. The G2 shouldn't crash over it, though. RE: Integral hangs G2 - roadrunner - 03-30-2020 12:27 AM My G2 didn't crash; it behaved exactly like the simulator. Perhaps you have x defined somewhere? RE: Integral hangs G2 - Albert Chan - 03-30-2020 05:16 PM (03-29-2020 09:56 PM)Albert Chan Wrote: Googled "wallis product", I found this: Mathematical Analysis and the Mathematics of Computation, p.404, eqn. 7.87 Just realized the integral is simply the beta function (see equation 18) Let \((1-u)^p = x\quad →\quad -p(1-u)^{p-1}\;du=dx\) \(\large \int_0^1 (1-x^{1/p})^q\;dx = p \int_0^1 u^q (1-u)^{p-1}\;du = p\;B(q+1,p) = {\Gamma(p+1)\Gamma(q+1) \over \Gamma(p+q+1)} = 1 / \binom{p+q}{p} \) Update: We can prove u integral is beta function by induction (via integration by parts) Foundations of Combinatorics with Applications: Appendix A, Example A.4 |