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This is a follow-up to a much simpler question I asked here, which @PrincessEev answered promptly and perfectly. She showed me how to rewrite the sum $\sum _{i=1}^x \phi (x-i)$ in such a way that Abel Summation could be applied, to obtain

$$\sum _{i=1}^x \phi (x-i) = \lfloor x \rfloor \phi(x-1) - \int_0^{x-1} \lfloor u+1 \rfloor \phi'(u) \, \mathrm{d}u$$

Here is a harder problem - I think. And once again, I don't really know where to start. Let $1 \le j \le x$ be a constant positive integer. Can one apply Abel Summation to the sum $\sum _{i=1}^x \phi (x - i j)$? If so, how?

CLARIFICATION / UPDATE

I initially added follow-up material here asking what I was doing wrong in implementing @junjios's answer. The fault was mine and was a coding error in my use of Mathematica.

Since it has no bearing on @junjios's correct answer, I have deleted those elements of this post, to avoid confusing new readers. If you are puzzled over some of the posts in the comments section, this is the explanation.

Richard Burke
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  • The formula works for me: https://www.desmos.com/calculator/oynnlokdsv – Caleb Briggs Jul 17 '23 at 18:53
  • Sorry, do you want $\phi$ to be $\sin^2(x)$ or $\sin(x^2)$? Just asking, because I don't understand your last equality and also not how Mathematica arrives at the given expression. – junjios Jul 18 '23 at 14:02
  • Apologies @junios, I’m away from my computer until later. I meant $\sin (x^2)$ but until I can get back to Mathematica, I won’t know if I made a typo putting stuff into the software or rewriting it into TeX … – Richard Burke Jul 18 '23 at 17:42
  • Humble apologies, @junjios. The error was entirely mine and was due to poor coding. Mathematica was calculating the derivative $\phi' (x-u j)$ with respect to $$ not $u$. I will change the 'update' section of my post to remove uncertainties, and I will of course mark this as answered. Many thanks. – Richard Burke Jul 19 '23 at 13:32
  • No problem, I'm glad I could help. – junjios Jul 19 '23 at 20:32

1 Answers1

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Let $x\geq 1$ and $j\in\mathbb{R}$ be fixed. (Note: There are no further conditions on $j$!)

We define: $\psi(y):=\phi(x-yj)$. Clearly, $\psi$ is continuously differentiable, so we can apply the Abel summation formula to obtain $$\sum_{i=1}^x\phi(x-ij)=\sum_{i=1}^x\psi(i)=\lfloor x\rfloor\psi(x)-\int_1^x\lfloor u\rfloor\psi'(u)\,\mathrm{d}u\\=\lfloor x\rfloor\phi(x(1-j))+j\int_1^x\lfloor u\rfloor\phi'(x-uj)\,\mathrm{d}u.$$

Generalisation:

More generally, given any (further) continuously differentiable function $f$ (possibly depending on $x$), we may write $$\sum_{i=1}^x\phi(f(i))=\lfloor x\rfloor\phi(f(x))-\int_1^x\lfloor u\rfloor\phi'(f(u))f'(u)\,\mathrm{d}u.$$

junjios
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  • Hi @junjios. This is great - but could you please take a look at my update above? I'm doing something wrong! – Richard Burke Jul 17 '23 at 15:26
  • @RichardBurke-Ward I did and added a comment above. I cannot pinpoint a specific mistake yet, but could you have messed up $\sin^2(x)$ and $\sin(x^2)$ somehow? – junjios Jul 18 '23 at 14:05