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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.

(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.

(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.

My G2 didn't crash; it behaved exactly like the simulator. Perhaps you have x defined somewhere?

(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