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Integration? - Printable Version +- Tetration Forum (https://math.eretrandre.org/tetrationforum) +-- Forum: Tetration and Related Topics (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=1) +--- Forum: Mathematical and General Discussion (https://math.eretrandre.org/tetrationforum/forumdisplay.php?fid=3) +--- Thread: Integration? (/showthread.php?tid=465) Pages:
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Integration? - 73939 - 06/28/2010 Hello! I tried integrating the function x^^n for any nonnegative integer n. I wanted to share my method and results, so that you can check fofr accuracy (if I've done it correctly, it'd be a rare occurance ![]() First we write ![]() Then we can expand on the indefinite integral in the parentheses. ![]() Once we have calculated this out, we can use substitution. ![]() ![]() If we repeat this process n-1 times, then we get ![]() Notice that I defined i_0 = 1. This merely simplifies the notation. I haven't yet completed that general integral, but what matters is that it can be completed on a case-by-case basis through integration by parts. This is demonstrated for n=2... ![]() Integrating by parts, ![]() If we do this indefinitely, we end up with ![]() Plugging in, ![]() Bernoulli's integral, the so-called "Sophomore's Dream", is calculated as ![]() Note that 0^0 is 1, so the second summation only has a non-zero value when n=k. If anyone has any corrections, or if this has been done before, I'd love to know. ![]() Edit: Ugh. Just noticed that in the last few equations, I wrote the summations wrong. Rest assured, it's from k=0 to infinity. RE: Integration? - bo198214 - 07/01/2010 Nice done. So you reduce the integral to a series containing x and ln(x). Another way to do it would be to just integrate the Taylor expansion of RE: Integration? - 73939 - 07/02/2010 Unfortunately, I have no idea what the Taylor series is. What you linked to uses x and ln(x), while a true Taylor series would only involve x. It's be cool to see how to calculate the nth derivative of On another note, I decided to tackle ![]() which I'm not going to bother TEXing. Unfortunately, all those summations make calculating even RE: Integration? - bo198214 - 07/03/2010 (07/02/2010, 09:35 PM)73939 Wrote: Unfortunately, I have no idea what the Taylor series is. What you linked to uses x and ln(x), while a true Taylor series would only involve x. You should read the whole thread not just the first post. RE: Integration? - bo198214 - 07/03/2010 (07/02/2010, 09:35 PM)73939 Wrote: Unfortunately, all those summations make calculating even I use sage with my formal_powerseries package. With a code like this Code: sage: load formal_powerseries.py we get the Taylor coefficients of We can now integrate this: Code: sage: pt3i = pt3.integral() I always include the derivatives which is the powerseries coefficients multiplied by n! because this is an integer sequence, which you possibly can look up at Sloane's Online Encypclopedia of Integer Sequences. And now I see that the sequence for the 3rd powertower (or its integral which is just the series shifted to the right by one) isnt found there (the second powertower is of course contained). I will filed a new submission. And now we can compute the integral pt3i(1-1) - pt3i(0-1) = -pt3i(-1) by truncating the infinite series, say to 100. To have an idea about the accuracy we also calculate the 101-truncation: Code: sage: p = pt3i.polynomial(100) This means Edit: If we improve precision a bit taking a polynomial with 200 terms which takes a bit longer to compute we see that the last digits are not accurate: Code: sage: p3 = pt3i.polynomial(200) This time a bit more careful with the last digits I would say: Caution: actually one need to be careful with this kind of evaluation of a powerseries. Because we know that the convergence radius of the powertowers are at most 1. That means that -1 is at boundary of the disk of convergence of pt3, so it may or may not converge. And I guess the same is true for pt3i. RE: Integration? - Gottfried - 07/04/2010 (07/03/2010, 11:34 AM)bo198214 Wrote: [quote='73939' pid='4939' dateline='1278102954'] Pari/GP gives using numerical integration and internal float-precision of 200 digits gp> intnum(x=-1,0,(1+x)^(1+x)^(1+x)) %1 = 0.573121567043619582758830 [update] I also had tried it with my matrix-formulation and my simple possibilities of convergence-acceleration using Euler- et al summation. Having the same taylor-series with which Henryk worked, an improvement would only be possible, if such a convergence-acceleration were effective. But what I see if I look at the terms of the powerseries of the integral with the powers of x(=0 and =-1) mutiplied in, is that the summands decrease slowly to zero (?) but not with alternating sign - so common convergence-accelerators should not work easily. (The extreme case would be to try to sum the harmonic series, where such convergence acceleration would even be contradictory). So with all my usual tools I come not nearer than four to six digits (using 64 terms) like Henryk, - thus, to apply this model of powerseries at all, would require some more consideration before ... [/update] Gottfried RE: Integration? - Ztolk - 07/05/2010 I have also worked out an integral for x^x. Mine involves the incomplete gamma function. RE: Integration? - 73939 - 07/05/2010 That's pretty interesting. I'd certainly like to see that. Could you post it? RE: Integration? - 73939 - 07/05/2010 On another note concerning integration: Could the function RE: Integration? - Ansus - 07/08/2010 It may be useful that In other words, |