sqrt(1+i)

[compsystems]

(10-05-2016 07:24 PM)parisse Wrote:  

(10-05-2016 02:59 PM)compsystems Wrote:  I think the rules of simplifications on the ti68k calculators are more powerful in many cases

.... As a counterexample try simplifying 4*atan(1/5)-atan(1/239) on your ti.
The handling of sqrt in Xcas is based on the more general concept of algebraic extension of Q (rootof in Xcas), where the emphasis is on recognizing 0 in a deterministic manner. Internally, there is no embedded sqrt inside Xcas, only a common algebraic extension for all the coefficients. Since displaying rootofs would scare many users, simple algebraic extensions are converted to more standard sqrt after computations, but as a side effect this does not necessarily *look* as simple (especially for embedded sqrt). And you can rework sqrt(1+i): for example if you want real or imaginary part: normal(re(sqrt(1+i))); normal(im(sqrt(1+i)))

ok, but they are more steps (input) to get to the same output,



ti68k:
input (approx mode): √(1.0+i)
output x+y*i: 1.09868411347 + 0.455089860562*i

input (exact mode): √(1+i)
output x+y*i: sqrt(2*(sqrt(2)+1))/2 + (sqrt(2*(sqrt(2)+1))/2)*i


Xcas
input: normal(re(√(1+i))) + normal(im(√(1+i)))*i
output: (√2*√(√2+1)+(1+i)*√(√2+1))/(√2+2) not is a x + y * i form
fabulous if you include a flag in xcas to see the output of a complex expression in the form x + y * i Good Idea?

Xcas (next realease) =)
input (exact mode and new flag): √(1+i)
output: sqrt(2*(sqrt(2)+1))/2 + (sqrt(2*(sqrt(2)+1))/2)*i

another way to transform the output to x+y*i
√(1+i) -> (1+i)*√(√2+1)/(√2)
substituting i -> x
coeff( (1+x)*√(√2+1)/(√2),x) coeff separates the real and complex part
poly1[√2*√(√2+1)/2, √2*√(√2+1)/2]
Ans .* poly1[1, i]
poly1[√2*√(√2+1)/2, √2*√(√2+1)/2*i]
ΣLIST(poly1[√2*√(√2+1)/2, √2*√(√2+1)/2*i])
sqrt(2*(sqrt(2)+1))/2 + (sqrt(2*(sqrt(2)+1))/2)*i


we also need a version of QPI-ROOT cmd

Code:

QPIROOT(normal(re(√(1+i)))+normal(im(√(1+i)))*i)

-> sqrt(2*(sqrt(2)+1))/2 + (sqrt(2*(sqrt(2)+1))/2)*i