Re: [sage-support] query on PowerSeriesRing(CC)

2016-05-25 Thread kbriggs
Thanks and sorry - yes, I realized my mistake, K On Wednesday, 25 May 2016 15:03:35 UTC+1, John Cremona wrote: > > R.gen() is z, not the square root of minus 1, so your f is 4*z^2. > > With I = CC.gen() you get > > sage: print(f.dict()) > {1: 2.00*I, 2: 2.00} > > > On 2

Re: [sage-support] query on PowerSeriesRing(CC)

2016-05-25 Thread kbriggs
On Wednesday, 25 May 2016 15:34:30 UTC+1, kbriggs wrote: > > Thanks and sorry - yes, I realized my mistake, > K > > On Wednesday, 25 May 2016 15:03:35 UTC+1, John Cremona wrote: >> >> R.gen() is z, not the square root of minus 1, so your f is 4*z^2. >> >> With I = CC.gen() you get >> >> sage: p

Re: [sage-support] query on PowerSeriesRing(CC)

2016-05-25 Thread John Cremona
R.gen() is z, not the square root of minus 1, so your f is 4*z^2. With I = CC.gen() you get sage: print(f.dict()) {1: 2.00*I, 2: 2.00} On 25 May 2016 at 14:13, wrote: > Why do I get this? > > > R.=PowerSeriesRing(CC) > > I=R.gen() > f=2*I*z+2*z^2 > > print(f.dict()) >

[sage-support] query on PowerSeriesRing(CC)

2016-05-25 Thread keith.briggs
Why do I get this? R.=PowerSeriesRing(CC) I=R.gen() f=2*I*z+2*z^2 print(f.dict()) > {2: 4.00} Thanks, Keith -- You received this message because you are subscribed to the Google Groups "sage-support" group. To unsubscribe from this group and stop receiving emails from it, s