Axel Vogt

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15 years, 315 days
Munich, Germany

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These are answers submitted by Axel Vogt

Expand the results to see that are the same

Yet another way, using "type":


 

restart;

f:=x^6-3*x-5;
L:=[solve(%)];
select(type, L, 'realcons');
evalf(%);

f := x^6-3*x-5

L := [RootOf(_Z^6-3*_Z-5, index = 1), RootOf(_Z^6-3*_Z-5, index = 2), RootOf(_Z^6-3*_Z-5, index = 3), RootOf(_Z^6-3*_Z-5, index = 4), RootOf(_Z^6-3*_Z-5, index = 5), RootOf(_Z^6-3*_Z-5, index = 6)]

[RootOf(_Z^6-3*_Z-5, index = 1), RootOf(_Z^6-3*_Z-5, index = 4)]

[1.45157146456211, -1.09429273656731]

(1)

 


 

Download MP_228473_real_roots.mws

Besides 'adding' to fourier rules you may use

inttrans:-fourier(erf(x),x,k); convert(%, Int, only=fourier);
value(%); # Maple 2017

It seems you zeroes are 10*Pi*k*(1/3), 5*Pi*(2*k+1)*(1/3), k an integer

I converted to 'rational', used asymptotics to find a scaling and final get the above (find an ugly sheet attached)

MP_228037.mw

no problem in Maple 2017.3

210867-Evaluating-A-Sum.pdf

Your task has a result through the bivariate cumulative normal. Maple does not have it built in.

2*exp(1/4*tp^2*delta^2+sigma)*tp*Pi^(1/2)*
  cdfN2(-1/2*(delta*tp^2-2*tf)*2^(1/2)/tp,-1/2*delta*tp+lambda,1/2*2^(1/2))

MP227459_-_short.mws

 

PS 15-07-2019: in case of questions please post them here (I do not want privat emails if that involves gmail = Google).

Writing r = E__fv and feeding data it is obvious that numerically almost nothing is contributed to r (if being not close to zero) within the exp term (depends on settings for Digits). The equation boils down to

'.239735220903707e26*(ln(1+.241440919407021e21*exp(r-.3012136e-20 * 0))+ln(1+.241440919407021e21*exp(r-.1185628e-19 * 0)))/Pi = .3e25';

Which is solved immediately

I used upper case Int and Digits:=15 to get 0.187286133249768

Int(exp(-b*x)/(exp(-2*b*x)*a+exp(-4*x)),x):
IntegrationTools:-Change(%, x=ln(t),t): simplify(%) assuming 0<t; #simplify(%, symbolic);
value(%):
simplify(%) assuming  b::real, 0<t;
#convert(%, hypergeom);
                         /      (-b + 3)
                        |      t
                        |  ----------------- dt
                        |   (-2 b + 4)
                       /   t           a + 1


             (-b + 4)            (-2 b + 4)        b - 4
            t         LerchPhi(-t           a, 1, --------)
                                                  -4 + 2 b
          - -----------------------------------------------
                               -4 + 2 b

 

You may try the following to study the task:

#Eq1:=Eq1*omega__n^5:
#Eq2:=Eq2*omega__n^5:
sys := { Eq1 , Eq2 };

eliminate(%, [p]): map(simplify, %);

I think that your integral is sol := 2*R/Vzero*InverseJacobiAM(1/2*arccos(1/3*(2*R*g+Vzero^2)/g/R),2*I/Vzero*R^(1/2)*g^(1/2))

I do not have a reasonable idea for the limit or series in Vzero, but plotting shows that it explodes

MP_226316_plot.mws
 

# https://www.mapleprimes.com/questions/226316-Why-Does-This-Integral-Evaluate-To-Infinity

restart; interface(version);

`Classic Worksheet Interface, Maple 2017.3, Windows, Sep 27 2017, Build ID 1265877`

(1)

sol := 2*R/Vzero*InverseJacobiAM(1/2*arccos(1/3*(2*R*g+Vzero^2)/g/R),2*I/Vzero*R^(1/2)*g^(1/2));

sol := 2*R*InverseJacobiAM((1/2)*arccos((2*R*g+Vzero^2)/(3*g*R)), (2*I)*R^(1/2)*g^(1/2)/Vzero)/Vzero

(2)

Sol:=subs(g=ScientificConstants:-GetValue(ScientificConstants:-Constant(g)), R=3, sol);
plot(Sol,  Vzero = 10^(-4) .. 1.5*10^(-3) );

Sol := 6*InverseJacobiAM((1/2)*arccos(0.113301801441992e-1*Vzero^2+.666666666666668), (6.26311424133394*I)*3^(1/2)/Vzero)/Vzero

 

plot(Sol,  Vzero = 0 .. 1.5*10^(-3), smartview=false);

 

 


 

Download MP_226316_plot.mws

 

I use 'rationalize' and instead of op(...) I write ln = ln@rationalize (so first rationalize and then take log). Similar for the second and more complicated example

eval(f, ln = ln@rationalize); #completesquare(%, x);
                                       2
                                (x - 1)
                             ln(--------)
                                   4

 

Find a suggestion appended for your first task: you can "eliminate" the variables Io and Rs, now solving for n only (though it needs higher Digits).

Note that it is a bit questionable since you have only 4 - 5 decimals in your input and I do not want to check sensitivity (it is a must here for practical reasons, I guess).

n ~ 0.9941, Io ~ 0.4431e-11, Rs ~ 1.160

MP_225728.mws

I depends how Maple has implemented the function, for the (extreme) tails one may use erfc. That is quite fast and exact.

1-erf(x): convert(%, erfc) =%;
                         erfc(x) = 1 - erf(x)

Digits:=32;             # or higher
erfc(x)=1e-20;          # complementary
#ln(erfc(x))=ln(1e-20); # not needed here
b:=fsolve(%, x=6);

                b := 6.6015806223551425615163916324187

 

# using Maple 2017

(1-x^floor(u))/((1-x)*u^2); eval(%, x=-1);
 plot(%, u = 1 .. 21);

suggests

1/2*(1-(-1)^floor(u))/u^2; #eval(%, u=k);
Int(%, u= 2*k-1 .. 2*k);
value(%) assuming k::posint;
Sum(%, k=1 .. infinity);
value(%);

giving ln(2)

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