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Mariusz Iwaniuk

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These are answers submitted by Mariusz Iwaniuk

....

Mariusz Iwaniuk 1476
January 20 2018
0 5

If yours integral is:

int(sqrt(ln(x)/x^2-1/x), x)

then Maple can't find it. Possible it has No closed form.

See: https://en.wikipedia.org/wiki/Nonelementary_integral.

Mathematica, Rubi,Axiom,SymPy, Maxima  cannot do it, either.

Second one:

int((ln(x)^(a-1)/x^2-1/x)^(1/2),x)

the same case.

EDITED:

If You want approximate integral by series see worksheet.

Approximate_indefine_integral_by_Series.mw

Use convert....

Mariusz Iwaniuk 1476
January 16 2018
2 7 
convert(exp(x), Sum, dummy = n)

or:

convert(exp(x), FormalPowerSeries)

For Simple function (2 methods):

you can use inverse ztransform:

Sum(x^n*invztrans(eval(exp(x), x = 1/x), x, n), n = 0 .. infinity)#only works if invztrans can find transfrom.

or n-th derivative:

Sum(x^n*(eval(diff(exp(x), x$n), x = 0))/factorial(n), n = 0 .. infinity)#only works if n-th derivative can find.

 

This is not answer Your question....

Mariusz Iwaniuk 1476
January 11 2018
0 1

Only another way to calculate Julian date.

Formula from:http://aa.usno.navy.mil/faq/docs/JD_Formula.php

Compute the JD corresponding to 2018 January 11, 18h30m30s UT.

Substituting Y = 2018, M = 1, D = 11,h = 18, m = 30, s = 30;

restart:
JD := proc (Y, M, D, h, m, s) local x; 
x := evalf(367*Y-trunc((7/4)*Y+(7/4)*trunc((1/12)*M+3/4))+trunc((275/9)*M)+D+1721013.5-
(1/2)*signum(100*Y+M-190002.5)+1/2+(1/24)*h+(1/1440)*m+(1/86400)*s); end proc:

JD(2018, 1, 11, 18, 30, 30);

# 2.458130271*10^6

 

 

....

Mariusz Iwaniuk 1476
December 31 2017
0 0

It may not possible to find anti-derivatives(closed form).

https://math.stackexchange.com/questions/1469123/integral-of-ex2

https://math.stackexchange.com/questions/168860/functions-cannot-be-integrated-as-simple-functions?rq=1

Mathematica gives No solution.

An aproximation only can be done e.g by series:


 

> 

p := sqrt(ln(t)*(1/ln(t)^2))/(t*(1-ln(t)^2/t^2)^(1/4))

(1/ln(t))^(1/2)/(t*(1-ln(t)^2/t^2)^(1/4))

 (1)
> 

with(IntegrationTools):
> 

Int(p, t)

Int((1/ln(t))^(1/2)/(t*(1-ln(t)^2/t^2)^(1/4)), t)

 (2)
> 

Change(Int((1/ln(t))^(1/2)/(t*(1-ln(t)^2/t^2)^(1/4)), t), t = exp(x))

Int(-(1/ln(exp(x)))^(1/2)*(exp(x))^2*(-(ln(exp(x))^2-(exp(x))^2)/(exp(x))^2)^(3/4)/(ln(exp(x))^2-(exp(x))^2), x)

 (3)
> 

`assuming`([simplify(Int(-(1/ln(exp(x)))^(1/2)*(exp(x))^2*(-(ln(exp(x))^2-(exp(x))^2)/(exp(x))^2)^(3/4)/(ln(exp(x))^2-(exp(x))^2), x))], [x > 0])

Int(exp((1/2)*x)/((-x^2+exp(2*x))^(1/4)*x^(1/2)), x)

 (4)
> 

GetIntegrand(Int(exp((1/2)*x)/((-x^2+exp(2*x))^(1/4)*x^(1/2)), x))

exp((1/2)*x)/((-x^2+exp(2*x))^(1/4)*x^(1/2))

 (5)
> 

series(exp((1/2)*x)/((-x^2+exp(2*x))^(1/4)*x^(1/2)), x = 0, 5)

1/x^(1/2)+(1/4)*x^(3/2)-(1/2)*x^(5/2)+(21/32)*x^(7/2)+O(x^(9/2))

 (6)
> 

Int(exp((1/2)*x)/((-x^2+exp(2*x))^(1/4)*x^(1/2)), x) = int(1/x^(1/2)+(1/4)*x^(3/2)-(1/2)*x^(5/2)+(21/32)*x^(7/2)+O(x^(9/2)), x)

Int(exp((1/2)*x)/((-x^2+exp(2*x))^(1/4)*x^(1/2)), x) = (1/1680)*x^(1/2)*(245*x^4-240*x^3+168*x^2+3360)+O(x^(11/2))

 (7)
> 

``


 

Download Only_aproximation.mw

 

....

Mariusz Iwaniuk 1476
December 30 2017
1 0

Use Maple built function intsolve with method Neumann or with this  paper https://arxiv.org/ftp/arxiv/papers/1309/1309.6311.pdf see another attached file (Fredholm_integral_2ver.)


 

> 

restart
> 

alpha := 1; A := 1
> 

f := proc (k) options operator, arrow; 1-2*k^2/(sqrt(alpha^2+k^2)*(sqrt(alpha^2+k^2)+k)) end proc

proc (k) options operator, arrow; 1-2*k^2/(sqrt(alpha^2+k^2)*(sqrt(alpha^2+k^2)+k)) end proc

 (1)
> 

K := proc (x, t) options operator, arrow; Int(2*f(v)*cos(x*v)*cos(t*v)/Pi, v = 0 .. A) end proc

proc (x, t) options operator, arrow; Int(2*f(v)*cos(x*v)*cos(t*v)/Pi, v = 0 .. A) end proc

 (2)
> 

EQ := h(x) = 4/(Pi*alpha^2)-(Int(h(t)*K(x, t), t = 0 .. 1))

h(x) = 4/Pi-(Int(h(t)*(Int(2*(1-2*v^2/((v^2+1)^(1/2)*((v^2+1)^(1/2)+v)))*cos(x*v)*cos(t*v)/Pi, v = 0 .. 1)), t = 0 .. 1))

 (3)
> 

HT := intsolve(EQ, h(x), method = Neumann, order = 1)

h(x) = 4*(Int(Int(-2*(-v^2+1+(v^2+1)^(1/2)*v)*cos(x*v)*cos(_k1*v)/((v^2+1)^(1/2)*((v^2+1)^(1/2)+v)), v = 0 .. 1), _k1 = 0 .. 1)+Pi)/Pi^2

 (4)
> 

help("intsolve # See Maple Help for more information.")
> 

(1/4)*D/(Pi*mu*U[0]) = int(rhs(HT), x = 0 .. 1, numeric)

(1/4)*D/(Pi*mu*U[0]) = .7334429187

 (5)
> 

NULL


 

Download _Fredholm_integral.mw

_Fredholm_integral_2ver.mw

Expanding with infinte series...

Mariusz Iwaniuk 1476
December 10 2017
0 2

Only for k=1,4,5,6,7 is real solution.


 

> 

restart
> 

sol := allvalues(solve(10*c*x^7-6*x^3+6 = 0, x))

RootOf(5*_Z^7*c-3*_Z^3+3, index = 1), RootOf(5*_Z^7*c-3*_Z^3+3, index = 2), RootOf(5*_Z^7*c-3*_Z^3+3, index = 3), RootOf(5*_Z^7*c-3*_Z^3+3, index = 4), RootOf(5*_Z^7*c-3*_Z^3+3, index = 5), RootOf(5*_Z^7*c-3*_Z^3+3, index = 6), RootOf(5*_Z^7*c-3*_Z^3+3, index = 7)

 (1)
> 

Matrix([seq([series(sol[k], c = 0, 5)], k = 1 .. 7)])
> 

Matrix([seq([convert(series(sol[k], c = 0, 5), polynom)], k = 1 .. 7)])

Matrix([[1+(5/9)*c+(50/27)*c^2+(19000/2187)*c^3+(934375/19683)*c^4], [-1/2+((1/2)*I)*3^(1/2)-(5/9)*(-1+I*3^(1/2))*c/(I*3^(1/2)+1)+(50/27)*c^2-(38000/2187)*c^3/(I*3^(1/2)+1)+(1868750/19683)*c^4/(-1+I*3^(1/2))], [-1/2-((1/2)*I)*3^(1/2)-(5/9)*(I*3^(1/2)+1)*c/(-1+I*3^(1/2))+(50/27)*c^2+(38000/2187)*c^3/(-1+I*3^(1/2))-(1868750/19683)*c^4/(I*3^(1/2)+1)], [1+(5/9)*c+(50/27)*c^2+(19000/2187)*c^3+(934375/19683)*c^4], [1+(5/9)*c+(50/27)*c^2+(19000/2187)*c^3+(934375/19683)*c^4], [1+(5/9)*c+(50/27)*c^2+(19000/2187)*c^3+(934375/19683)*c^4], [1+(5/9)*c+(50/27)*c^2+(19000/2187)*c^3+(934375/19683)*c^4]])

 (2)
> 

``


 

Download sol.mw

 

Work for me....

Mariusz Iwaniuk 1476
December 07 2017
0 0 
solve(15-(1/100)*m = 5+(1/600)*m, {m})
#{m = 6000/7}

solve(15-1/(100*m) = 5+1/(600*m), {m})
#{m = 7/6000}

Regards,Mariusz

Please read.....

Mariusz Iwaniuk 1476
December 04 2017
2 3

Please read in Maple Help (Ctrl+F1 and put "How Do I,Solve an Ordinary Differential Equation?" in Search->Enter):

Scroll down to:

      -Solving an ODE Numerically

      - Taking Derivatives and Integrals of Numeric Solutions

      -  Why You Should Not Use Int or Diff on a Numeric Solution 

 

Corrected file:question-vers_2.mw

(Edited: 2 times) :P

    
 

Try....

Mariusz Iwaniuk 1476
December 03 2017
3 1


 

> 

restart
> 

with(Student[Calculus1])
> 

solve([-(1.25*y-sqrt(abs(x)))*abs(1, x)/sqrt(abs(x))+2*x, 3.1250*y-2.50*sqrt(abs(x))], [x, y])

[]

 (1)
> 

solX := solve(((125*(1/100))*y-sqrt(abs(x)))^2+x^2-1 = 0, {y})

{y = (4/5)*abs(x)^(1/2)+(4/5)*(-x^2+1)^(1/2)}, {y = (4/5)*abs(x)^(1/2)-(4/5)*(-x^2+1)^(1/2)}

 (2)
> 

c1 := CriticalPoints(rhs(solX[1, 1]), x)

[-1, -(1/12)*((215+12*321^(1/2))^(2/3)-(215+12*321^(1/2))^(1/3)+1)/(215+12*321^(1/2))^(1/3), 0, (1/12)*((215+12*321^(1/2))^(2/3)-(215+12*321^(1/2))^(1/3)+1)/(215+12*321^(1/2))^(1/3), 1]

 (3)
> 

evalf(c1)

[-1., -.5566930951, 0., .5566930951, 1.]

 (4)
> 

point1 := evalf(subs(x = 1, rhs(solX[1, 1])))

.8000000000

 (5)
> 

point2 := evalf(subs(x = 0, rhs(solX[2, 1])))

-.8000000000

 (6)
> 

point3 := evalf(subs(x = -.5566930951, rhs(solX[1, 1])))

1.261469543

 (7)
> 

c2 := CriticalPoints(rhs(solX[2, 1]), x)

[-1, 0, 1]

 (8)
> 

with(plots); p1 := pointplot([[evalf(c1)[2], point3], [evalf(c1)[4], point3], [0, point2], [0, point1]], color = [green], symbol = circle, axes = none, symbolsize = 15); p2 := implicitplot(((125*(1/100))*y-sqrt(abs(x)))^2+x^2-1, x = -1 .. 1, y = -1 .. 1.3, axes = normal, gridrefine = 3); display({p1, p2})

 
> 

``


 

Download Critical_Points.mw

solve...

Mariusz Iwaniuk 1476
November 16 2017
1 1 
sol := n > ceil(evalf(solve(product(exp(1/i), i = 1 .. n) = 100, n))); solve(sol, n);

# sol := 56 < n

# RealRange(Open(56), infinity)

......

Mariusz Iwaniuk 1476
November 10 2017
0 1

Calculating point with newton method:

restart;
with(Student[NumericalAnalysis]):
with(plots):
f := x^3-x^2-x-1;
P := Newton(f, x = 2.0, tolerance = 10^(-2));
p1 := pointplot([[P, 0]], color = [blue], symbolsize = 20, symbol = circle, axes = normal):
p2 := plot(f, x = 0 .. 2.2):
display({p1, p2});

 

First iteration:   

evalf(eval(simplify(x-f/(diff(f, x))), x = 2));
# 1.857142857

 

Answer for Legendre wavelets method for ...

Mariusz Iwaniuk 1476
November 09 2017
1 0

Hi

I'm only changed "Equation" to "2*M" , "Solu" to "Solu[1]" and other things.

Regards Mariusz.

Help_v2.mw

 

 I'm not an expert on that topi...

Mariusz Iwaniuk 1476
November 07 2017
0 6

 I'm not an expert on that topic.

With help from: https://www.maplesoft.com/applications/view.aspx?sid=4971  (See: A Numeric Approach).

 

restart;
f := proc (u) options operator, arrow; piecewise(0 <= u and u <= 1, 0, 1 < u and u <= 2, 1) end proc;
ode := diff(y(u), u$2) = 4*Pi^2*(f(u)-e)*y(u); bc := y(0) = 0, y(2) = 0, (D(y))(0) = 1;
Eigen1 := (dsolve({bc, ode}, numeric, range = 0 .. 2, maxmesh = 8192, abserr = 1.*10^(-3), approxsoln = [y(u) = exp(-u), e = 1]))(0)[4];
Eigen2 := (dsolve({bc, ode}, numeric, range = 0 .. 2, maxmesh = 8192, abserr = 1.*10^(-3), approxsoln = [y(u) = u, e = 3]))(0)[4];
Eigen3 := (dsolve({bc, ode}, numeric, range = 0 .. 2, maxmesh = 8192, abserr = 1.*10^(-3), approxsoln = [y(u) = u, e = 6]))(0)[4];

textplot...

Mariusz Iwaniuk 1476
November 06 2017
3 2

In Maple there is not much choice how to do it.

One of the possibilities is textplot.

with(plots):
p1 := pointplot([[3, 8], [-5, 16], [11, 32], [3, -8]], color = [black], symbol = solidbox, axes = none, symbolsize = 12): 
p2 := textplot({[-5, 16+2, "lampart"], [3, (-8)-2, "dog"], [3, 8+3, "cat"], [11, 32+4, "panthera"]}, axes = none):
p3 := implicitplot(y^2 = x^3-43*x+166, x = -40 .. 40, y = -40 .. 40, axes = normal, gridrefine = 2): 
display({p1, p2, p3})

Your formula is very complicated,closed...

Mariusz Iwaniuk 1476
November 05 2017
0 0

Your formula is very complicated,closed form solution may be not exist(Mathematica also can't solve)

Maybe you can try a Numeric Inverse Laplace to solve your problem.

Help.mw

 

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