Prove this

12m∑k=12mf(k)k>23. \large\dfrac1{2^m} \sum_{k=1}^{2^m} \dfrac{f(k)} k > \dfrac23.

Let f(n)f(n) be the greatest odd divisor of nn. Prove that for all positive integers mm, the inequality above holds true.

#NumberTheory #Prove

Note by Yash Saxena
5 years, 6 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} ai−1 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 sin⁡θ \sin \theta
\boxed{123} 123 \boxed{123}

Comments

Induct on mm; this is obvious for m=1m = 1. Assume this is true for m=Mm = M, and we will prove this for m=M+1m = M+1.

Note that f(k)=f(2k)f(k) = f(2k) and f(2k−1)=2k−1f(2k-1) = 2k-1. Thus,

12M+1∑k=12M+1f(k)k=12M+1∑k=12M(f(2k−1)2k−1+f(2k)2k)=12M+1∑k=12M(2k−12k−1+f(k)2k)=12M+1∑k=12M(1+12⋅f(k)k)=12M+1∑k=12M1+12M+1∑k=12M12⋅f(k)k=12M+1⋅2M+12M+2∑k=12Mf(k)k>12+14⋅23=23\begin{aligned} \frac{1}{2^{M+1}} \sum_{k=1}^{2^{M+1}} \frac{f(k)}{k} &= \frac{1}{2^{M+1}} \sum_{k=1}^{2^M} \left( \frac{f(2k-1)}{2k-1} + \frac{f(2k)}{2k} \right) \\ &= \frac{1}{2^{M+1}} \sum_{k=1}^{2^M} \left( \frac{2k-1}{2k-1} + \frac{f(k)}{2k} \right) \\ &= \frac{1}{2^{M+1}} \sum_{k=1}^{2^M} \left( 1 + \frac{1}{2} \cdot \frac{f(k)}{k} \right) \\ &= \frac{1}{2^{M+1}} \sum_{k=1}^{2^M} 1 + \frac{1}{2^{M+1}} \sum_{k=1}^{2^M} \frac{1}{2} \cdot \frac{f(k)}{k} \\ &= \frac{1}{2^{M+1}} \cdot 2^M + \frac{1}{2^{M+2}} \sum_{k=1}^{2^M} \frac{f(k)}{k} \\ &> \frac{1}{2} + \frac{1}{4} \cdot \frac{2}{3} \\ &= \frac{2}{3} \end{aligned}

In the first equality, we unroll the summation so we're summing two terms at a time. The second equality uses the aforementioned facts f(k)=f(2k)f(k) = f(2k) and f(2k−1)=2k−1f(2k-1) = 2k-1. The rest should be fairly straightforward; the inequality invokes the inductive hypothesis.

This completes the proof for m=M+1m = M+1, and thus finishes the proof by induction.

Ivan Koswara - 5 years, 6 months ago

Log in to reply

Was writing up a proof, but your method is surely more elegant. Well done!

Jake Lai - 5 years, 6 months ago
×

Problem Loading...

Note Loading...

Set Loading...