Exploring the Divisor Function

This note is part of the set Exploring the Divisor Function.

In this set, we aim to get a general form for this sum:

n=1d(kn)n2\sum _{n=1}^{\infty}\frac{d\left(kn\right)}{n^2}

Where kk is a positive integer.

So, instead of giving out everything on a note, why not split it up into several problems so that everybody can try it out by themselves?

I will give a clue here, and then you can go ahead to solve the first problem of this set, slowly progressing to the last problem, where you will finally be able to find a general form of the sum. You may skip steps, because your approach might be better than mine. If you do have a better approach, do post it!

Here's the first clue:

If f(n)f(n) is completely multiplicative, that is f(ab)=f(a)f(b)f(ab)=f(a)f(b), then

ff(n)=d(n)f(n)f*f(n)=d(n)f(n)

[n=1f(n)ns]2=n=1f(n)d(n)ns\left[\sum _{n=1}^{ \infty}\frac{f\left(n\right)}{n^s}\right]^2=\sum _{n=1}^{\infty }\frac{f\left(n\right)d\left(n\right)}{n^s}

Where * is the Dirichlet Convolution

and d(n)d(n) counts the number of divisors n.


I would post the solutions for the problems soon.

#NumberTheory

Note by Julian Poon
5 years, 4 months ago

No vote yet
1 vote

  Easy Math Editor

This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

  • Use the emojis to react to an explanation, whether you're congratulating a job well done , or just really confused .
  • Ask specific questions about the challenge or the steps in somebody's explanation. Well-posed questions can add a lot to the discussion, but posting "I don't understand!" doesn't help anyone.
  • Try to contribute something new to the discussion, whether it is an extension, generalization or other idea related to the challenge.
  • Stay on topic — we're all here to learn more about math and science, not to hear about your favorite get-rich-quick scheme or current world events.

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold

- bulleted
- list

  • bulleted
  • list

1. numbered
2. list

  1. numbered
  2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1

paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
    # 4 spaces, and now they show
    # up as a code block.

    print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in \( ... \) or \[ ... \] to ensure proper formatting.
2 \times 3 2×3 2 \times 3
2^{34} 234 2^{34}
a_{i-1} ai1 a_{i-1}
\frac{2}{3} 23 \frac{2}{3}
\sqrt{2} 2 \sqrt{2}
\sum_{i=1}^3 i=13 \sum_{i=1}^3
\sin \theta sinθ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

my mind is blown

Joel Yip - 5 years, 2 months ago

I cannot work it out for 4 as kk Can you help?

Joel Yip - 5 years, 2 months ago

Log in to reply

Oh yeah, thanks for commenting here. I forgot to add the solutions...

If you want to consider k=4, generalise it to k=pak=p^a, where p is prime. I'll post a solution to part 3 of the set by tomorrow, where you can use it for k=4k=4.

You can use part 1 of this set as a clue to part 3, but you'll have to be more creative.

Julian Poon - 5 years, 2 months ago

Log in to reply

thanks!

Joel Yip - 5 years, 2 months ago

Log in to reply

@Joel Yip I have posted a solution to part 3.

Julian Poon - 5 years, 2 months ago
×

Problem Loading...

Note Loading...

Set Loading...