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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.
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.
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
\(\Large {1 \over \int_0^1 (1-x^{1/p})^q\;dx} = \binom{p+q}{p} = \binom{p+q}{q} = {1 \over \int_0^1 (1-x^{1/q})^p\;dx} \)
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