Codercise HS.1.3

Codebook
1

Hi. Im so stuck on this chapter.

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Could I get some help, I don’t know what im doing wrongly

Also, it seems there may be an error in the theory section under Higher-order Trotterization.

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for s_2, it seems that k =1, but the given value for s_2 takes k = 2. Should the formula for s_2k be 1/(4-4^1/(2(2k)-1))?

2

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.

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3

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

4

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
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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)))

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