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March 22nd, 2017, 03:44 PM  #11  
Senior Member Joined: Jul 2008 From: Western Canada Posts: 3,319 Thanks: 39 
I'll have a look at Pari. In fact, I think I already have it on my computer. I recall downloading it quite a while back. Quote:
Thanks for pointing it out.  
March 27th, 2017, 12:47 AM  #12 
Senior Member Joined: Jul 2008 From: Western Canada Posts: 3,319 Thanks: 39 
billymac00, There seems to be something corrupted in my user settings, which makes private messages unpredictable. So, I thought it best to mention that I did get your PM, and hopefully you got my reply. If not, just let me know here in this thread. I haven't yet tried your Pari code, because I'll have to figure out where I actually installed Pari, and also because I decided to pursue studiot's suggestion of the Second Euler Summation Formula. This led to other things, which in turn, led to even more things (branches going off in all directions), all of which turned out to be very interesting. Regarding EulerMaclaurin, I did some research. Looking at an article on the EulerMaclaurin formula, I had one of those epiphany moments, and realized that—for a completely unrelated problem that I'm also working on—this would give the solution for a finite summation that I'd been trying to solve. So, I solved one problem, but not the one that I originally posed here. On the subject of the Second Euler Summation Formula for the original problem, from what I had read, there didn't appear to be a huge difference between this and the basic trapezoidal rule. Further reading of some interesting articles on the trapezoidal rule indicate that for cyclical functions, the trapezoidal rule may be optimum, giving exponential convergence. I thought I'd tried the trapezoidal rule back in the early days when I first worked on this problem. But I decided to revisit it. I confirmed that it didn't converge nearly as fast as my customized Simpson's rule solution, and I wondered why. The reason why the trapezoidal rule is supposed to converge so fast for cyclical functions is that the approximation error cancels out more quickly than the worst case error formulae would predict. The trapezoidal rule underestimates where the function is concave downwards, and overestimates where the function is concave upwards. Since a cyclical function alternates between concave up and down, much of the error cancels, giving very fast convergence. But curiously, not in my case. I decided that I should have another look at the function I'm trying to integrate. And on that note, I'll point out that I made a typo when I posted the function. I had a $cos^2$ term in the numerator, and it should have been just a first order $cos$. This is the corrected expression: $f(x)=\dfrac{k_1 cos(xk_2)+k_3}{\sqrt{k_4(k_5k_5 cos(xk_2))+k_6x}}$ When I first started working on this problem about 5 years ago, I graphed the function for a particular set of $k_n$ values, and nothing appeared amiss. It looked like a typical decaying cosine function. However, I've now realized that with certain values of $k_n$, this can change so that the graph looks like a series of extremely narrow spikes occurring at intervals of $2π$. This explained why the concave up and concave down won't cancel. I realized that to solve this efficiently, I need to concentrate the function samples around multiples of $2π$. I decided to split the integral into subintegrals with limits $0..\pi, \pi..3\pi, 3\pi..5\pi, 5\pi..7\pi$, etc., so that the spike always occurs in the centre of the integration limits (except for the first, and possibly last interval). Then I applied the double exponential method for each subintegral (which from my past experience has been most efficient for these kinds of integrals). The result is that I'm seeing a reduction in the required number of function evaluations by a factor of about 10 or more to achieve the same error. Considering what the graph of the function looks like, I suspect I won't get much better than this. Also of note (since I'm not integrating off to infinity, fortunately): As can be seen from the formula above, the spikes decay very slowly as a function of $1/\sqrt{x}$, which means that I have to count every one, which is 400 in the case that I'm currently working on. Anyway, I'd like to say that I appreciate the suggestions you've all given me. They may not necessarily have led to what you expected, but it's had very positive results. I still intend to pursue some of the suggestions that I haven't yet followed. I'm already thinking of a couple tangents to veer off on. 

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algorithm, integration, numerical 
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