Carl Love

Carl Love

19094 Reputation

24 Badges

7 years, 351 days
Mt Laurel, New Jersey, United States
My name was formerly Carl Devore.

MaplePrimes Activity


These are replies submitted by Carl Love

@Markiyan Hirnyk Yes, but in that example you've made the Asker's original problem into an IVP. For an IVP, dsolve does not leave an indefinite integral in the solution; it leaves an unevaluated definite integral. Other than extra CPU time usage, there is no problem with the numerical evaluation of the latter.

@Markiyan Hirnyk Yes, but in that example you've made the Asker's original problem into an IVP. For an IVP, dsolve does not leave an indefinite integral in the solution; it leaves an unevaluated definite integral. Other than extra CPU time usage, there is no problem with the numerical evaluation of the latter.

@Markiyan Hirnyk The inconvenience of having an indefinite integral in a dsolve solution is that it can't be numerically evaluated in a convenient way. Here's an example of plotting. Here I use the Asker's new version of the ODE, for which dsolve produces essentially the same indefinite integral.

restart;

DQ:= -((D@@2)(y))(r)-2*y(r)/r+y(r) = (2*(2-1/r^2))*r*exp(-r):

Y:= rhs(dsolve(DQ, y(r))):

YY:= simplify(value(subs(Ei(1,-2*r)= -Ei(2*r), Y))):

plot(eval(YY, [_C1= 0, _C2= 0]), r= 1..9);

plot(eval(Y, [_C1= 0, _C2= 0]), r= 1..9);

Warning, unable to evaluate the function to numeric values in the region; see the plotting command's help page to ensure the calling sequence is correct


Download EI_plot.mw

@Markiyan Hirnyk The inconvenience of having an indefinite integral in a dsolve solution is that it can't be numerically evaluated in a convenient way. Here's an example of plotting. Here I use the Asker's new version of the ODE, for which dsolve produces essentially the same indefinite integral.

restart;

DQ:= -((D@@2)(y))(r)-2*y(r)/r+y(r) = (2*(2-1/r^2))*r*exp(-r):

Y:= rhs(dsolve(DQ, y(r))):

YY:= simplify(value(subs(Ei(1,-2*r)= -Ei(2*r), Y))):

plot(eval(YY, [_C1= 0, _C2= 0]), r= 1..9);

plot(eval(Y, [_C1= 0, _C2= 0]), r= 1..9);

Warning, unable to evaluate the function to numeric values in the region; see the plotting command's help page to ensure the calling sequence is correct


Download EI_plot.mw

@matja You asked

...which differs from mathematica in the particular solution, there is on more term, namely: 2*r*exp(-r)*ln(2) which is already covered in the term r exp(-r)*_C2 of the general solution. What could be said to this?

If you put the new differential equation into the worksheet that I posted and run through the steps, then I think you'll see what's going on. The integration generates a term -ln(2*r), which becomes -ln(2) - ln(r) on simplification. One way to think of it is that the indefinite integral produces a constant of integration, but the int command does not supply an arbitrary constant . Since there are already two constants provided by dsolve, it is necessary that this extra constant matches up with a term which already has one of the dsolve-supplied constants. I guess that this can be used as a sort of accuracy check.

Maple's indefinite integration usually, but not always, produces 0 as the constant of integration; it depends on what is convenient. Here's a simple example, pretty close to the case at hand, where it doesn't.

int(ln(2*x)/x, x);

(1/2)*ln(2*x)^2

expand(%);

(1/2)*ln(2)^2+ln(2)*ln(x)+(1/2)*ln(x)^2


Clearly 1/2*ln(2)^2 is a more convenient constant than 0 for the unexpanded form in this case.

Plaese let me know whether this explanation satisfies you.

Download Constant_of_integrat.mw

@matja You asked

...which differs from mathematica in the particular solution, there is on more term, namely: 2*r*exp(-r)*ln(2) which is already covered in the term r exp(-r)*_C2 of the general solution. What could be said to this?

If you put the new differential equation into the worksheet that I posted and run through the steps, then I think you'll see what's going on. The integration generates a term -ln(2*r), which becomes -ln(2) - ln(r) on simplification. One way to think of it is that the indefinite integral produces a constant of integration, but the int command does not supply an arbitrary constant . Since there are already two constants provided by dsolve, it is necessary that this extra constant matches up with a term which already has one of the dsolve-supplied constants. I guess that this can be used as a sort of accuracy check.

Maple's indefinite integration usually, but not always, produces 0 as the constant of integration; it depends on what is convenient. Here's a simple example, pretty close to the case at hand, where it doesn't.

int(ln(2*x)/x, x);

(1/2)*ln(2*x)^2

expand(%);

(1/2)*ln(2)^2+ln(2)*ln(x)+(1/2)*ln(x)^2


Clearly 1/2*ln(2)^2 is a more convenient constant than 0 for the unexpanded form in this case.

Plaese let me know whether this explanation satisfies you.

Download Constant_of_integrat.mw

@Markiyan Hirnyk Your timing is not valid because of remember tables. You have to test on different numbers. Here's a better test. Actually, doing the integration symbolically first is about 25 times faster using 512 evaluation points. 

restart;

DQ:= -((D@@2)(y))(r)-2*y(r)/r+y(r) = ((4/9-exp(-r)/r)*2)*r*exp(-r):

Y:= rhs(dsolve({DQ, y(1)=1, D(y)(1)=2}, y(r))):

L1,L2:= seq(RandomTools:-Generate(list(float(range= 1..9), 2^9)), k= 1..2):

st:= time():

Y1:= simplify(Y):

CodeTools:-Usage(assign('EY1', [seq](evalf(eval(Y1, r= x)), x= L1))):

memory used=2.49GiB, alloc change=60.01MiB, cpu time=25.38s, real time=24.55s

t1:= time()-st;

25.406

st:= time():

Y2:= simplify(value(applyrule(Ei(1,_X::algebraic)= -Ei(-_X), Y1))):

CodeTools:-Usage(assign('EY2', [seq](evalf(eval(Y2, r= x)), x= L2))):

memory used=81.38MiB, alloc change=0 bytes, cpu time=860.00ms, real time=809.00ms

t2:= time()-st;

1.016

t1/t2;

25.0059055118110


Download Ei_integration_timin.mw

@Markiyan Hirnyk Your timing is not valid because of remember tables. You have to test on different numbers. Here's a better test. Actually, doing the integration symbolically first is about 25 times faster using 512 evaluation points. 

restart;

DQ:= -((D@@2)(y))(r)-2*y(r)/r+y(r) = ((4/9-exp(-r)/r)*2)*r*exp(-r):

Y:= rhs(dsolve({DQ, y(1)=1, D(y)(1)=2}, y(r))):

L1,L2:= seq(RandomTools:-Generate(list(float(range= 1..9), 2^9)), k= 1..2):

st:= time():

Y1:= simplify(Y):

CodeTools:-Usage(assign('EY1', [seq](evalf(eval(Y1, r= x)), x= L1))):

memory used=2.49GiB, alloc change=60.01MiB, cpu time=25.38s, real time=24.55s

t1:= time()-st;

25.406

st:= time():

Y2:= simplify(value(applyrule(Ei(1,_X::algebraic)= -Ei(-_X), Y1))):

CodeTools:-Usage(assign('EY2', [seq](evalf(eval(Y2, r= x)), x= L2))):

memory used=81.38MiB, alloc change=0 bytes, cpu time=860.00ms, real time=809.00ms

t2:= time()-st;

1.016

t1/t2;

25.0059055118110


Download Ei_integration_timin.mw

If you're trying to solve with fsolve, then you need to drastically reduce the value of Digits.

If you're trying to solve with fsolve, then you need to drastically reduce the value of Digits.

Your g(x) reduces to g(x) = A, because of most-well-known trig identity: cos^2 + sin^2 = 1. So, there is not really any parameter C.

Nicely detailed plot!

I don't how clock faces usually appear in your culture, but I think that in most of the Western World the position that you have labeled as 0 is labeled as 12 (even though mathematicians and especially computer programmers must wish it was 0). It is so culturally ingrained that we refer to an distant object directly in front of us as "at 12 o'clock" and a distant overhead object as being "12 o'clock high".

@Christopher2222 Like I said, bytes and alt-codes are not the same for me, except that the regular keyboard characters 32-127 are the same. Maybe it has something to do with what language keyboard you're using. I'm uaing USA Standard English. What are you using?

@Christopher2222 Like I said, bytes and alt-codes are not the same for me, except that the regular keyboard characters 32-127 are the same. Maybe it has something to do with what language keyboard you're using. I'm uaing USA Standard English. What are you using?

@Joe Riel You're right, the Output Display can be set to 2D Math Notation or Typseset. However, I still need Typesetting Level set to Standard, not Extended. Which begs the question, What is Extended Typesetting for?

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