- September 1st, 2013, 7:07 pm
- Forum: Numerical Methods Forum
- Topic: Looking for tough integrals
- Replies:
**148** - Views:
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<t>conclusion on cos(257/256*x)/cos(x)^(255/256) ?It was an unfair task, it can not be done in double precision, at least notwithout additional work, as And 2 said. Even the common NAG routines give up. You may tryJ:= n -> Int(cos((1/(2^n)+1)*x)*cos(x)^(1/(2^n)-1), x = 0 .. Pi/2)That should be zero ...

- August 30th, 2013, 10:08 pm
- Forum: Numerical Methods Forum
- Topic: Looking for tough integrals
- Replies:
**148** - Views:
**16955**

<t>I would give up as well, do not mind, here it was for finding limitation.This is where one wants 'hybrid methods', as it is sometimes named in Maplepapers: combining numerical and symbolical methods. It makes sense especiallyfor (numerical) singularities.For example PARI has built in a syntax to ...

- August 30th, 2013, 3:53 pm
- Forum: Numerical Methods Forum
- Topic: Looking for tough integrals
- Replies:
**148** - Views:
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<t>Yes, it follows showing his 'spoiler' is an anti-derivative (use Maple or MMA).It gives evidence what has to be shown: the integral beyond the root towardsPi/2 has to be of same magnitude (while the negative sign is obviuous).Likewise you can shift it to zero to have the singularity in x=0 and st...

- August 29th, 2013, 7:01 pm
- Forum: Numerical Methods Forum
- Topic: Looking for tough integrals
- Replies:
**148** - Views:
**16955**

Ok ... the result is zero.For numerical 'evidence' split the integral at integrand = 0, which is x = 128/257 * Pi

- August 29th, 2013, 5:14 pm
- Forum: Numerical Methods Forum
- Topic: Looking for tough integrals
- Replies:
**148** - Views:
**16955**

wileysw has given an anti derivative and over the interval there is no need tocorrect the imaginary part to have a continuous one.To find out limitations one can use esp = 1/2^n with increasing n and at n=8a usually good numerical integrator gives up with a false result. Else enforce.

- August 29th, 2013, 1:04 pm
- Forum: Numerical Methods Forum
- Topic: Looking for tough integrals
- Replies:
**148** - Views:
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concerning cos ...Yes, it has a formal solution, but it is numerically ugly (though eps = 1/256 is quite"large") it not "mine" and stems from a discussion locating a possible bug / limitationand no, I have no such handbook

- August 28th, 2013, 6:16 pm
- Forum: Numerical Methods Forum
- Topic: Looking for tough integrals
- Replies:
**148** - Views:
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And2,Concerning Gautschi's book: could you name the sections? I never did it usingthe Hankel matrix ... How do you use choose the measure (for the examples) ?PS: what is your result for the numerical ugly integral over 0 .. Pi/2with cos(257/256*x)/cos(x)^(255/256) as integrand ?

- August 26th, 2013, 7:36 pm
- Forum: Numerical Methods Forum
- Topic: Looking for tough integrals
- Replies:
**148** - Views:
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<r>Sounds very interesting and I am curiuos about a (pre)released presentation.For Excel: take care and it's "15 digits curse" for handling and displaying,<URL url="http://axelvogt.de/axalom/Testing%20Excel%202010.pdf">http://axelvogt.de/axalom/Testing%20Excel%202010.pdf</URL> gives a sketch how tob...

- August 25th, 2013, 7:00 pm
- Forum: Numerical Methods Forum
- Topic: Looking for tough integrals
- Replies:
**148** - Views:
**16955**

The tanh-sinh for finite integrals should work: the integrand "is zero" beyond t ~ 26 . D(H)(t) = -t/2 * H(t) has solution const * exp(-1/4*t^2) , so cut off is ~ 52.

- August 25th, 2013, 6:24 pm
- Forum: Numerical Methods Forum
- Topic: Looking for tough integrals
- Replies:
**148** - Views:
**16955**

<t>@And2: yes, that NAG routines are adaptive. In case they also use accelerationmethods. Just sound after decades ....Where it also is possible to compute it by 'tanh-sinh', using brute cut offs.But here is an idea for the Gamma integral, which may work with your framework(without extending it to a...

- August 23rd, 2013, 8:28 pm
- Forum: Numerical Methods Forum
- Topic: Looking for tough integrals
- Replies:
**148** - Views:
**16955**

<t>Seems to be a very nice "framework" :-)PS: do not use Excel for that (it is restricted to 15 decimal placesand you need 18 to check in double precision, so use you privateMMA at home for _relative_ errors).PPS: that recursion(s) is not stable, IIRC. Computing gamma fctvia integrals is done in an ...

- August 23rd, 2013, 6:21 pm
- Forum: Numerical Methods Forum
- Topic: Looking for tough integrals
- Replies:
**148** - Views:
**16955**

I do not understand (and 11 decimals means to loose ~ 1/3 places?):Are that your results for Int(exp(-x^2)*sin(k*x), x=0..infinity) for k = 1and k = 10? Or do I miss a scaling factor?What about the last tasks (especially that for tanh-sinh, 1/sqrt(1-x)) ?

- August 22nd, 2013, 5:39 pm
- Forum: Numerical Methods Forum
- Topic: Looking for tough integrals
- Replies:
**148** - Views:
**16955**

<t>With NAG routines (some come with Maple) it works without complains, 1 <= x.GAMMA(x+1) = Int(exp((-1+y)/y)*y^(-x-2)*(1-y)^x,y = 0 .. 1), t = (1-y)/ymay do for 0 < x in a wide range, at least for the tanh-sinh method (whichmay overflow)Some more ugly integral: cos(257/256*x)/cos(x)^(255/256) over ...

- August 21st, 2013, 7:43 pm
- Forum: Numerical Methods Forum
- Topic: Looking for tough integrals
- Replies:
**148** - Views:
**16955**

<t>Ok. Can wait, sounds very nice.I use Maple. It can not find it with serious help. With its syntax it is -1/2*r*2^(1/2)*(EllipticK(1/2*2^(1/2)*((r-1)/r)^(1/2))-2*EllipticE(1/2*2^(1/2)*((r-1)/r)^(1/2))-EllipticPi((r-1)/r,1/2*2^(1/2)*((r-1)/r)^(1/2)))where cosh(b)=r, b = 1/2Using MMA you may have to...

- August 21st, 2013, 6:59 pm
- Forum: Numerical Methods Forum
- Topic: Looking for tough integrals
- Replies:
**148** - Views:
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@Ant: ok, and I am curiuos about your paper.Fine! I use x = 1 - y^2. BTW: It has a symbolic solution in terms of elliptic functions,but that was not the task.

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