digerdiga

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12 years, 151 days

MaplePrimes Activity


These are replies submitted by digerdiga

sorry for a third question:

But how can I do this:

Let xs be the numerical solutions as above:

xs:= dsolve({ics, ode}, numeric);
plots:-odeplot(xs, t= 0..3);

works without any problems

how can one plot:

plots:-odeplot(f(xs), t= 0..3);

with some function f

it says:

Error, (in plots/odeplot) input is not a valid dsolve/numeric solution

I have a second question:

with another differential eqation:

ode3 := diff(y(t), t, t) = y(t)*sqrt((1-y(t))^2-1)/(1-y(t))

ics3 := y(0) = -2, (D(y))(0) = 0

ys := dsolve({ics3, ode3}, numeric)

I can ode plot this as the first one.

But my next step was to parametrically plot both solutions like

plots:-odeplot([xs(t), ys(t), t = 0 .. 3])

plots:-odeplot([xs, ys, t = 0 .. 3])

plots:-odeplot([x(t), y(t), t = 0 .. 3])

which results in an error:

Error, (in plots/odeplot) input is not a valid dsolve/numeric solution

I have a second question:

with another differential eqation:

ode3 := diff(y(t), t, t) = y(t)*sqrt((1-y(t))^2-1)/(1-y(t))

ics3 := y(0) = -2, (D(y))(0) = 0

ys := dsolve({ics3, ode3}, numeric)

I can ode plot this as the first one.

But my next step was to parametrically plot both solutions like

plots:-odeplot([xs(t), ys(t), t = 0 .. 3])

plots:-odeplot([xs, ys, t = 0 .. 3])

plots:-odeplot([x(t), y(t), t = 0 .. 3])

which results in an error:

Error, (in plots/odeplot) input is not a valid dsolve/numeric solution

so I dont actually need to asign as I did with xs??

in ordinary circumstances plot(eval(x(t),t=0..3) would be sufficient?

could I similarly use eval(xs,x(t)) ??

originally 1^2 was omega^2, but I replaced it later on to a numerical value to eliminate potential error sources

so I dont actually need to asign as I did with xs??

in ordinary circumstances plot(eval(x(t),t=0..3) would be sufficient?

could I similarly use eval(xs,x(t)) ??

originally 1^2 was omega^2, but I replaced it later on to a numerical value to eliminate potential error sources

@PatrickT I do know what radnormal is...

the question was what the line:

Normalizer:=radnormal;

does???

While doing everything the same the only difference to my approach is this line.

without this line after certain assumptions the limit is given by 0.

then restarting the maple kernel and writing this line, miraculously maple does not let the limit vanish...

if I would not have restarted the maple server the limit would still be 0

so there must be some other background..

@PatrickT I do know what radnormal is...

the question was what the line:

Normalizer:=radnormal;

does???

While doing everything the same the only difference to my approach is this line.

without this line after certain assumptions the limit is given by 0.

then restarting the maple kernel and writing this line, miraculously maple does not let the limit vanish...

if I would not have restarted the maple server the limit would still be 0

so there must be some other background..

what exactly does this do?:

Normalizer:=radnormal;

what exactly does this do?:

Normalizer:=radnormal;

-.- second order ;)

sry^^

-.- second order ;)

sry^^

well then I dont get whats wrong with this:

tanh(1/2*arccosh(1/s)-beta*u)

=(exp(arccosh(1/s))-exp(2*beta*u))/(exp(arccosh(1/s))+exp(2*beta*u))

= sqrt((1-s)/(1+s))*(1-2*beta*u*s/sqrt(1-s^2))+O(u^2)

 

the sign which is not shown is a +

so

2*arctanh(x)=ln((1+x)/(1-x))

=ln((2-2*beta*u*s/sqrt(1-s^2))/(2*beta*u*s/sqrt(1-s^2)))

=ln(sqrt(1-s^2)/(beta*u*s)-1)

the other integral is just

ln(beta*u)

thus:

ln(sqrt(1-s^2)/(beta*u*s)-1) + ln(beta*u) = ln(sqrt(1-s^2)/s-beta*u)

so the upper limit is just the zero order of the expansion but the first order contains the s

ln(sqrt(1-s^2)/s)-ln(sqrt(1-s^2)/s-beta*u)

=-ln(1-beta*u*s/sqrt(1-s^2))

where is the error?

well then I dont get whats wrong with this:

tanh(1/2*arccosh(1/s)-beta*u)

=(exp(arccosh(1/s))-exp(2*beta*u))/(exp(arccosh(1/s))+exp(2*beta*u))

= sqrt((1-s)/(1+s))*(1-2*beta*u*s/sqrt(1-s^2))+O(u^2)

 

the sign which is not shown is a +

so

2*arctanh(x)=ln((1+x)/(1-x))

=ln((2-2*beta*u*s/sqrt(1-s^2))/(2*beta*u*s/sqrt(1-s^2)))

=ln(sqrt(1-s^2)/(beta*u*s)-1)

the other integral is just

ln(beta*u)

thus:

ln(sqrt(1-s^2)/(beta*u*s)-1) + ln(beta*u) = ln(sqrt(1-s^2)/s-beta*u)

so the upper limit is just the zero order of the expansion but the first order contains the s

ln(sqrt(1-s^2)/s)-ln(sqrt(1-s^2)/s-beta*u)

=-ln(1-beta*u*s/sqrt(1-s^2))

where is the error?

ok this works...

ok this works...

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