Hi. Im so stuck on this chapter.
Also, it seems there may be an error in the theory section under Higher-order Trotterization.
Hi! thanks for this question.
I’m so glad you are working through the Codebook.
I think you found a mistake, but the solution is rather s_{2k}=1/(1-4^{1/(2k+1)}). I found the correct formula in a reference.
I will fix this both in the theory and the Codercises. Your codercise solution will be almost correct, you just need to change s_2k to the correct value that we just discussed.
Check back in a few days and things should be working properly.
Thank you again for spotting these mistakes and reporting them.
I see. Thanks! I was losing my head over this.
No worries, i really enjoy doing the exercises. Even though sometimes im losing my mind over the math, im really having a blast learning something this exciting
I think i see the issue the formula for s_2k is 1/(4-4^(1/(2k+1))), but the k is for the 2k order, as in when k = 2, the order is 4, therefore the s used is 1/(4-4^(1/(2(k-1)+1))) which equals to 1/(4-4^(1/(2k-1))).
This
would be more accurate if we are to use the correct formula in the reference as you mentioned.
For U_2k, s_2(k-1) = 1/(4-4^(1/(2(k-1)+1))) = 1/(4-4^(1/(2k-1))), which when k=2, s_2 = 1/(4-4^(1/(3)))
I also found this module, let alone codercise challenging. Like @cyyoung, I had implemented
s_2k = 1 / (4 - 4**(1 / (2*k - 1)))
but I’m still incorrect.. I’ve implemented the same lines of recursion as well.
Hi!
Thanks for the feedback. Is there anything in particular that you think we can improve to make it more understandable?
Regarding your question, the formula both in the theory and the codercise is s_2k = 1 / (4 - 4**(1 / (2*k + 1))). There is a plus one and not minus one. This one should work, let me know.
Oh I see now.. I got confused between the discussions of s_2k but needed to pay attention to whether it was + or - for the correct formula. The small details are killer!
Thank you Daniela!