sqrt(1+i)

09262016, 12:35 PM
(This post was last modified: 05012018 11:16 PM by moonbeam.)
Post: #1




sqrt(1+i)
(deleted by author)


09262016, 06:19 PM
Post: #2




RE: sqrt(1+i)
You must write your own program to do that. I have disabled exact sqrt of complex numbers on the Prime because it could freeze the calculator too easily (computations with algebraic extension becoming much too complicated and difficult to interrupt).


09262016, 08:33 PM
Post: #3




RE: sqrt(1+i)
Can you rewrite as
e^(1/2*LN(1+i)) ? That would be consistent with (1+i)^(1/3), (1+i)^(1/4), etc. which result in e^(1/3*LN(1+i)), e^(1/4*LN(1+i)), etc. Better solution than to force an approximate result in CAS! Thanks. 

09262016, 08:38 PM
Post: #4




RE: sqrt(1+i)
Math is not my favorite hobby all my excuse if the comment does not provide a real alternative...
very strange... With Code: SQRT(1+i) 

09272016, 04:19 AM
Post: #5




RE: sqrt(1+i)
Interesting. Thanks for the tip. The result is exact and correct.
But now try sqrt(sqrt(1+i)) . . . That is probably the reason why Bernard resorted to the approximations. 

10042016, 03:06 PM
Post: #6




RE: sqrt(1+i)
Interesting. If the TI92 can do it, maybe that will provide an incentive for the HP Prime development team


10042016, 03:44 PM
Post: #7




RE: sqrt(1+i)
(10042016 12:41 PM)moonbeam Wrote:(09272016 04:19 AM)Helge Gabert Wrote: But now try sqrt(sqrt(1+i)) . . . That is probably the reason why Bernard resorted to the approximations. Hi, on a HP71B the result for sqr(sqr(1+j)): 1,06955393236 +0,212747504726j No problem with the ancient calculator :) interesting, that ist could be a problem on the newer one  or I misunderstand the question 

10042016, 04:50 PM
Post: #8




RE: sqrt(1+i)
Yes, you are misunderstanding. The idea is to obtain an exact, symbolic result, please see above posts. The 71B does not have a builtin CAS.


10042016, 06:34 PM
Post: #9




RE: sqrt(1+i)  
10042016, 07:39 PM
Post: #10




RE: sqrt(1+i)
(10042016 03:44 PM)Erwin Wrote: on a HP71B the result for sqr(sqr(1+j)): 1,06955393236 +0,212747504726j No problem with the ancient calculator :) interesting, that ist could be a problem on the newer one  or I misunderstand the question It is exclusively when dealing with exact, symbolic results (which keep getting larger and larger the more you apply them). Prime returns the approximated result identically to the 71. TW Although I work for HP, the views and opinions I post here are my own. 

10052016, 05:34 AM
Post: #11




RE: sqrt(1+i)
(10042016 12:41 PM)moonbeam Wrote: Getting intrigued, I tried it on my HP50g. sqrt(1+i) works, sqrt(sqrt(1+i)) crashes it hard. After 11.37 seconds, my 50g returns an exact result for 'sqrt(sqrt(1+i))' EVAL. Perhaps our settings differ? <0ɸ0> Joe 

10052016, 05:38 AM
Post: #12




RE: sqrt(1+i)
(10042016 07:35 PM)moonbeam Wrote:I tried the windows 64, linux 64 and mac os versions without problems. Which version of xcas do you have?(10042016 06:34 PM)parisse Wrote: Strange, the current version of Xcas work with sqrt(1+i) and also sqrt(sqrt(1+i)) (but the answer is very complicated) 

10052016, 12:16 PM
(This post was last modified: 10052016 01:35 PM by roadrunner.)
Post: #13




RE: sqrt(1+i)
I created a little program to return the exact result for nested square roots of a complex number:
Code:
It crashes the emulator for values of n 6 or greater. Example: rootroot(1+i,5) works but rootroot(1+i,6) crashes. However, approx(rootroot(1+i,6)) returns the correct approximate answer, which tells me the program is returning a correct answer but the display can't handle it, and sits there with an hour glass in the corner. What am I doing wrong? road edited a grammatical error 

10052016, 01:52 PM
Post: #14




RE: sqrt(1+i)
It has been reported that the windows 32 bits version of xcas works on windows 10 while the 64 bits version crashes.


10052016, 05:54 PM
(This post was last modified: 07042017 02:46 PM by DedeBarre.)
Post: #15




RE: sqrt(1+i)
Hi..i am a new user here. In my case it crashes the emulator for values of n 6 or greater. rootroot(1+i,5) works but rootroot(1+i,6) crashes.However, approx(rootroot(1+i,6)) returns the correct approximate answer, which tells me the program is returning a correct answer but the display can't handle it, and sits there with an hour glass in the corner.
pcb assembly usa 

10052016, 07:24 PM
(This post was last modified: 10052016 07:34 PM by parisse.)
Post: #16




RE: sqrt(1+i)
(10052016 02:59 PM)compsystems Wrote: I think the rules of simplifications on the ti68k calculators are more powerful in many casesThis is your opinion, I don't share it. 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))) 

10062016, 01:03 PM
(This post was last modified: 04172017 12:53 AM by compsystems.)
Post: #17




RE: sqrt(1+i)
(10052016 07:24 PM)parisse Wrote:(10052016 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. 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 QPIROOT cmd Code: QPIROOT(normal(re(√(1+i)))+normal(im(√(1+i)))*i) 

07042021, 03:50 PM
Post: #18




RE: sqrt(1+i)
(10052016 12:16 PM)roadrunner Wrote: Code is wrong. Think polar form (cis(θ) = cos(θ) + i*sin(θ)) z = z * cis(arg(z)) √z = √z * cis(arg(z)/2) With arg(z) = [pi,pi], sign(arg(z)) = sign(sin(arg(z)) → sign(im(z)) = sign(im(√z)) CAS> csqrt(z) := √((abs(z)+re(z))/2) + sign(im(z))*i*√((abs(z)re(z))/2) CAS> csqrt(1+i) \(\displaystyle \sqrt{\frac{1}{2} \cdot (\sqrt{2}+1)}+ i \sqrt{\frac{1}{2} \cdot (\sqrt{2}1)}\) CAS> csqrt(Ans) \(\displaystyle i \sqrt{\frac{1}{2} \cdot (\sqrt{\frac{1}{2} \cdot (\sqrt{2}+1)}+2^{\frac{1}{4}})}+\sqrt{\frac{1}{2} \cdot (\sqrt{\frac{1}{2} \cdot (\sqrt{2}+1)}+2^{\frac{1}{4}})}\) CAS> approx(Ans) → 1.06955393236+0.212747504727*i CAS> Ans^4 → 1+i Also, with radical not simplified, mess exploded with nested square root. CAS> rootroot(1+i, 1) \(\displaystyle \frac{i \sqrt{\sqrt{2}1}}{\sqrt{2}}+\frac{\sqrt{2} \cdot \frac{1}{2}}{\sqrt{\sqrt{2}1}}\) CAS> rootroot(Ans, 1) \(\displaystyle \frac{1}{2} \cdot \frac{\sqrt{\sqrt{2}1}}{\sqrt{\frac{1}{2} \cdot (2 \sqrt{\frac{1}{4} \cdot (\left(\frac{\sqrt{2}}{\sqrt{\sqrt{2}1}}\right)^{2}+4 \left(\frac{\sqrt{\sqrt{2}1}}{\sqrt{2}}\right)^{2})}\frac{\sqrt{2}}{\sqrt{\sqrt{2}1}})}}+\frac{i \sqrt{\frac{1}{2} \cdot (2 \sqrt{\frac{1}{4} \cdot (\left(\frac{\sqrt{2}}{\sqrt{\sqrt{2}1}}\right)^{2}+4 \left(\frac{\sqrt{\sqrt{2}1}}{\sqrt{2}}\right)^{2})}\frac{\sqrt{2}}{\sqrt{\sqrt{2}1}})}}{\sqrt{2}}\) CAS> approx(Ans) → 1.06955393236+0.212747504727*i 

07052021, 05:31 PM
Post: #19




RE: sqrt(1+i)
I compared the calculated expressions √√...√(1+5×i) on the afx2.0 and tiv200 calculators.
When the number of roots increases, afx2.0 outputs a formula with trigonometric functions within 1.2_sec for any number of roots. Tiv200 gives the result only with numbers and arithmetic operations. When the number of roots is eight ( √√√√√√√√ (1+5×i) ), v200 calculates the example for about 3 minutes 40 seconds (too long). In HPPrime, it would be worth implementing automatic switching of the calculation method depending on the complexity of the final result and on the availability of free space in RAM (not only in this example). 

07072021, 01:35 PM
(This post was last modified: 07072021 01:36 PM by roadrunner.)
Post: #20




RE: sqrt(1+i)
(07042021 03:50 PM)Albert Chan Wrote: Code is wrong. Think polar form (cis(θ) = cos(θ) + i*sin(θ)) That works a lot better. Revision B of rootroot: Code: #pragma mode( separator(.,;) integer(h32) ) Thank you sir! road 

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