u/X3nion

Hello,

I’ve found this post explaining PWM in this Reddit. Are there more posts like this one here explaining the theory of PWM, how it works etc.?

Best regards,
X3nion

Hey everyone!

Over the last past months, I’ve been thinking about writing a letter to Apple’s CEO motivated by some people here who have posted here that they have written a feedback regarding the issues more and more people experience with PWM and OLEDs, and also with the d-word.

I’ve had a lot of success with personal letters in the past regarding different topics. Of course, in general, not the CEO other company will open the letter, but it will be forwarded to a higher level worker or even technician (so not 1st level / 2nd level etc.) who will then answer you.

My goal is not to achieve an instant improvement. However, my hope is that a letter has much more power compared to an email and maybe I can get in touch with a worker who is willing to convey.

What are your aspects you’ve mentioned when writing those feedbacks? Did you go into detail with modulation depth and so on? And is there maybe a summary here which explains most of the things like PWM, modulation depth, number of “black lines” and so on? I saw such a post some time ago with a great explanation including pictures but wasn’t able to find it anymore.

I’d be very grateful for your assistance!

Best regards,
X3nion

For n ∈ N let S_n = 1 + 2 + 3 +. . . + n = n(n+1)/2 and P_n = 1·2·3···n = n!
Does S_n | P_n hold if

(a) n is odd
(b) n is even and n + 1 is a prime number
(c) n is even and there exists 1 < a < b < n + 1 with n + 1 = ab?

My thoughts are so far:

a) We want to check whether there exists k € Z such that n(n+1)/2 * k = 1 * 2 * …* n, and simplify this to (n+1)/2 * k = 1 * 2 * … * (n-1). As n is odd, n+1 is even and we can write n+1 = 2m and get
2m/2 * k = 1 * 2 * … * (2m-2)
m*k = 1 * 2 * … * (2m-2)

Now I thought about proving by induction that for m >= 2 it holds m <= 1 * 2 * … * (2m-2) and so we would have m | 1 * 2 * … * (2m-2) for m >= 2, and S_n | P_n for n being an odd number 3,5,…
and for n=1 we obviously have S_n | P_n.
What do you think about that, and isn’t there another way to prove this?

b) We take a look at n(n+1)/2 * k = n! <=>
(n+1)/2 * k= (n-1)! <=> 2k = (n-1)! / (n+1).
Now as n+1 is a prime number and >= n-1, there is no term of the form (n+1) in (n-1)!
Hence (n-1)!/(n+1) is a rational number but no integer ( this could only be possible if n+1 appeared in the term (n-1)! ), thus it cannot be written as 2k and in this case S_n does not divide P_n.
What are your thoughts about this?

c) Here I don’t have any clue how to do this, can you give me a tip?

I’d be grateful for every help!

Best regards,
X3nion