I wanted to know how to study P_a (e.g. its convergence and continuity at 0), if you think it has an other form "without product" (or other nice properties).

Marco Hudson 2022-08-09 Answered
Study of an infinite product
During some research, I obtained the following convergent product P a ( x ) := j = 1 cos ( x j a ) ( x R , a > 1 ) ..
Considering how I got it, I know it's convergent and continuous at 0 (for any fixed a), but if I look at P a now, it doesn't seem so obvious for me.
Try: I showed that P a L 1 ( R ), i.e. it is absolutely integrable on R . Indeed, by using the linearization of the cosine function and the inequalities ln ( 1 y ) y (for any y < 1) and 1 cos z z 2 / 2 (for any real z), we obtain
P a ( x ) 2 = j = 1 ( 1 1 cos ( x j a ) 2 ) j > | x | 1 / a ( 1 1 cos ( x j a ) 2 ) exp ( C | x | 1 / a ) , for some absolute constant C > 0. However, I have not been able to use this upper bound to prove continuity at 0 (via uniform convergence for example, if we can).
Question : I wanted to know how to study P a (e.g. its convergence and continuity at 0), if you think it has an other form "without product" (or other nice properties) and finally if anyone has already seen this type of product (in some references/articles), please.
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Answers (1)

Kasey Bird
Answered 2022-08-10 Author has 13 answers
Step 1
May be, what you could do is P a ( x ) = j = 1 cos ( x j a ) log [ P a ( x ) ] = j = 1 log [ cos ( x j a ) ]
Step 2
Now, using Euler polynomials log [ cos ( t ) ] = n = 1 ( 1 ) n 2 2 n 3 ( E 2 n 1 ( 1 ) E 2 n 1 ( 0 ) ) n ( 2 n 1 ) ! t 2 n
log [ cos ( t ) ] = n = 1 2 2 n 3 | E 2 n 1 ( 1 ) E 2 n 1 ( 0 ) | n ( 2 n 1 ) ! t 2 n
log [ cos ( x j a ) ] = n = 1 2 2 n 3 | E 2 n 1 ( 1 ) E 2 n 1 ( 0 ) | n ( 2 n 1 ) ! x 2 n j 2 a n
Step 3
Since all coefficients have the same sign, we can switch the order of summations j = 1 j 2 a n = ζ ( 2 a n )
log [ P a ( x ) ] = 1 8 n = 1 | E 2 n 1 ( 1 ) E 2 n 1 ( 0 ) | ζ ( 2 a n ) n ( 2 n 1 ) ! ( 2 x ) 2 n
is an infinite sum of negative numbers. Now, it remains to prove the convergence (it is and it should be quite fast).
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