Axel Vogt

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16 years, 304 days
Munich, Germany

MaplePrimes Activity


These are answers submitted by Axel Vogt

Using vv's suggestion to sum up over smaller intervals I get
0.005295 - 0.0009968*I for n=0 and 0.7493 - 0.4808*I for n=1
for epsilon = 1e-4.

My machines needs about 1.5 h for each case.

Find attached a solution using the NAG routines which gives
it in about 20 seconds for each case, giving the same figures.

MP_232754_NAG.mw

Likewise you write down the recursion function in Cents and finally express it in USD

B:=proc(m::nonnegint) # compute in Cents
round( B(m-1)*(1 + 0.0775) + 400000 );
end proc;
B(0):=400000: # initial

B(30)/100.0; # Cents as USD
                           470415.2400

 

 

Writing as iterated integral and handling s first I get
-0.331842471004086e-3 and -.663684942008173e-1

Edit: I have not treated the imaginary part correctly:

-0.331842471004085e-3 -0.331842471004085e-3*I and

-0.0663684942008172 - 0.0663684942008172*I

MP_232562_corrected.mw

MP_232562.mw

Look at plot([0*h[2](x), h[1](x)], x = 0 .. 6, y = -.1 .. .1)

PS: to find possible rounding errors do not use interface(displayprecision = 5), only if you are free of bugs

For Windows you may use its Indexer.

(p+1)*(p-1) = p^2 - 1 is zero modulo 3 by "Little Fermat" (except p=3). Hence one factor divides by 3 and being even (3<=p) it divides by 6.

I once filed Luc Devroye, Non-Uniform Random Variates (1986), having "recipes". May be you can find it as well or find a backup for his site http://cgm.cs.mcgill.ca/~luc/rnbookindex.html through wayback machine

Likewise you may try x = 1+xi, y=x + eta, z=y+zeta (see attached file, it is Maple 2017, I do not have Maple 2021)

MP_232070.mw

I guess you mean "how to prove it", here a suggestion

Task:=Int(diff(u(x, y), x)*diff(u(x, y), x, y) +
  diff(u(x, y), x, x)*diff(u(x, y), y), x);
expand(%);
A,B:=op(1,%), op(2, %);

X:=IntegrationTools:-Parts(A, diff(u(x,y),x));
Y:=IntegrationTools:-Parts(B, diff(u(x,y),y));

Task=X+Y; combine(%);
lhs(%) - rhs(%);
combine(%);

From that it follows obviously.

For example you can save a sheet as html and pick images from there (it is easy to find the according subfolder).

Can one not handle this by providing an (initialized) array and handle it in the code? May be that is oldish.

I think it is 2.695678478*hypergeom([1., 1., 1.645396648],[2., 2.],-1.549434760*exp(-677.0138340*t))*exp(-677.0138340*t)+.95671881+1887.991395*t

find MP_231348.mws attached

 

Added 03. Jan 2021:

Changing coordinates by 1+57154/36887*exp(-342569/506*t)=s instead of 1+57154/36887*exp(-342569/506*t)=1+s gives the solution involving LerchPhi (instead of the more complicated 3F2 hypergeom) or likewise a 2F1, which is valid as well:

SolNum := -.4073509876e-1+62.97977396*t+4.176777934/(exp(677.0138340*t)+1.549434760)^.6453966478*exp(436.9424590*t)*(1.-1.*Re(hypergeom([-.6453966478, 1.],[.3546033522],1.+1.549434760*exp(-677.0138340*t))))

NB: Using Re = real part avoids spurious numerical imaginaries for LerchPhi.

 

Correction 03. Jan 2021

I made an error through copy + paste and took 62.97... instead of 862.97... for the constant.

So 800*t has to be added to both the posted anti-derivatives

@mmcdara 

If you say "expand" then it is written as polynomial and it should work

Plot it, "see" there is no solution and ask yourself whether you ask the correct question

-Pi/2 - arctan(25*x) - Pi +2*Pi/3; plot(%,x, -2*Pi .. 2*Pi);

 


 

 

That does not converge I think:

If you look at your E2 (in your sheet, but without feeding z) it is ~ e0 + rational in exp(z), exp(-z), then squared. For z ---> +- infinity that fraction tends to some constant*exp(-+z), but e0 survives. Thus the integral over the real line can not be finite. Now your z is just a linear function of your actual integration variable t. But a linear transform only changes the magnitude.

Likewise you may use MultiSeries:-asympt for your integrand to find similar reasoning.

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