## 370 Reputation

10 years, 126 days

## @mmcdara Shouldn't the expanded seri...

@mmcdara Shouldn't the expanded series and the product coincide? In particular as you showed that the product converges...

## @Kitonum I did, but are you trying to te...

@Kitonum I did, but are you trying to tell me that there is a point y € Y which is mapped onto at least twice?

Which one?

Did you understand my question 3. ?

## @Kitonum 2. Are you sure? Because the ma...

@Kitonum 2. Are you sure? Because the mapping

r=sqrt(x*y)

t=ln(sqrt(x/y))

seems fine to me...

3. I mean:

seq([x, t, t = 0 .. 1], x = 0.1 .. 1, 0.5) gives

[0.1, t, t = 0 .. 1], [0.6, t, t = 0 .. 1]

Now commands like linspace(0.1,1,3) in Matlab or Julia specify that the interval [0.1,1] in the above example is sliced in 3 points, namely [0.1,0.55,1]. For the map in 2. I have to omit the point zero, but I don't want to omit the point 1.

## @Kitonum 2. So what is the reason that o...

@Kitonum 2. So what is the reason that one side is a straight line? And what do you mean by "it is easy to verify that it is not straight" ??

Do you mean the not straight line is given by r=1, so

plot([exp(t), exp(-t), t = 0 .. 1])

is curved?

So is that map bugged?

3. No I really mean that in

x=r*exp(t)

y=r*exp(-t)

x€[0,1] and y€[0,1], so

r^2=xy

exp(2t)=x/y

However how do you proceed if you cannot solve for r and t ?

Is there some build in stuff or do you need to use RootOf?

4. Is there something like

linspace(0,1,10) ?

using

seq([x, t, t = 0 .. 1], x = 0.1e-1 .. 1, 0.5e-1)

does not have the other endpoint at 1, because the intervals do not match up to 1.

## @Kitonum When I do you old version F(S) ...

@Kitonum When I do your old version F(S) seems to work the same as F(plots:-display(S))

Actually I just noticed that you started with (r,t) € [0,1] x [0,1]

However here: (x,y) € [0,1] x [0,1], so I would have to take the inverse first which is sort of doable, but what if I can not solve for (r,t) ???

## @Kitonum Thank you.I'm wondering abo...

@Kitonum Thank you.

why did you write F(plots:-display(S))

and not F(S)

which also seems to work.

PS: Why does you new plot not map onto a triangle anymore?

## @vv " infolevel[`evalf/int`]:=5; &q...

@vv `infolevel[`evalf/int`]:=5;`

Thanks. I always forget about that.

## @Mariusz Iwaniuk What is this method _d0...

@Mariusz Iwaniuk What is this method _d01ajc  precisely?

Also what does maxintervals=300000 mean?

Setting your epsilon doesn't it determine the number of intervals in [0,1] ??

## @vv Yes but how do you know it must have...

@vv Yes but how do you know it must have done that? Does it check everytime for convergence but since there is a period of 1 where nothing changes it assumes convergence??

Did you just try it out with 20 ?

## @vv How do you have this insight?...

@vv How do you have this insight?

## @vv If necessary I will rewrite the...

@vv If necessary I will rewrite the problem tomorrow or so using different names, though I dont think it will change the essence.

## Is it not clear what I mean? If so, can...

Is it not clear what I mean?

If so, can somebody explain to me what information I‘m missing?

## @Preben Alsholm  "The freezing...

@Preben Alsholm  "The freezing that goes on is just a Maple technicality, which doesn't go to the essence of what is really going on. "

I'm aware of that. I just don't see why it is done here in particular as it seems he is freezing some arbitrary constants such as

_csgn(-(2*I)*RootOf(_Z^2+1, index = 1))

## @Carl Love Why does he deem it necessary...

@Carl Love Why does he deem it necessary to freeze these objects anyhow?

## @Preben Alsholm When using infolevel[int...

@Preben Alsholm When using infolevel[int]:=5 then there seems to be lots of freezing going on which is however hidden.

For example this list

{_th[1] = ln(t-(1/2)*RootOf(_Z^2+1, index = 1)), _th[2] = ln(t+(1/2)*RootOf(_Z^2+1, index = 1)), _th[3] = exp(-(1/2)*s*(-2*ln(2)+ln(t-(1/2)*RootOf(_Z^2+1, index = 1))+ln(t+(1/2)*RootOf(_Z^2+1, index = 1))-(1/2)*RootOf(_Z^2+1, index = 1)*Pi*`freeze/R13`*(-`freeze/R13`+`freeze/R9`)*(-`freeze/R13`+`freeze/R10`))), _th[4] = exp(-(1/2)*s*(ln(-2*RootOf(_Z^2+1, index = 1))+ln(t+(1/2)*RootOf(_Z^2+1, index = 1))-(1/2)*RootOf(_Z^2+1, index = 1)*Pi*`freeze/R12`*(-`freeze/R12`+`freeze/R8`)*(-`freeze/R12`+`freeze/R10`)-ln(2*RootOf(_Z^2+1, index = 1))-ln(t-(1/2)*RootOf(_Z^2+1, index = 1))+(1/2)*RootOf(_Z^2+1, index = 1)*Pi*`freeze/R14`*(-`freeze/R14`+`freeze/R11`)*(-`freeze/R14`+`freeze/R9`)))}

is not very enlightening. Is it possible to show what is behind the frozen objects here?

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