On Wednesday, September 16, 2015 at 1:54:40 AM UTC-7, Mandeep Singh wrote:
>
> Can we declare lambda as a variable? I want to use it in eigenvalues
> (linear algebra).
> like:
> l = var("lambda");
> It'll give error. So how can I declare it as a variable.
>
This works. I don't know if it will kee
On Thursday, May 28, 2009 at 9:31:46 PM UTC+5:30, Jason Grout wrote:
Mike Hansen wrote:
> Hello,
>
> On Thu, May 28, 2009 at 8:38 AM, Paul Sargent > wrote:
>> # Subs for lambda (have to use "lambda", but that's a keyword)
lambda is a reserved keyword in Python, which means you can't use it
On Thu, May 28, 2009 at 9:06 AM, Paul Sargent wrote:
> Yes I'd found the dictionary form in the docstring, but didn't know
> that it took the python variable rather than the symbol name (if you
> see what I mean), and so was still having the same problem.
> The docstring suffers from the fact tha
On 28 May 2009, at 16:49, Mike Hansen wrote:
> sage: sage: e1.subs({l:3})
> theta == 9
Yes I'd found the dictionary form in the docstring, but didn't know
that it took the python variable rather than the symbol name (if you
see what I mean), and so was still having the same problem.
The doc
Mike Hansen wrote:
> Hello,
>
> On Thu, May 28, 2009 at 8:38 AM, Paul Sargent wrote:
>> # Subs for lambda (have to use "lambda", but that's a keyword)
>>
>> sage: e1.subs(lambda = 3)
>>
>> File "", line 1
>> e1.subs(lambda = Inte
Hello,
On Thu, May 28, 2009 at 8:38 AM, Paul Sargent wrote:
> # Subs for lambda (have to use "lambda", but that's a keyword)
>
> sage: e1.subs(lambda = 3)
>
> File "", line 1
> e1.subs(lambda = Integer(3))
> ^
