Generalizations!

Whenever I solve a problem, I try if I can proceed on with the generalized statement (or expression) for the problem.

I came up with two interesting generalizations today.

  • The straight forward one - The number of ordered n-tuples of integers {xi}i=1n \{ x_i \}_{i=1}^n such that ∑i=1nxi−∏i=1nxi=n−1\sum_{i=1}^n x_i - \prod_{i=1}^n x_i = n-1 is equal to nk−n+1nk-n+1 provided that 1≤xi≤k1 \leq x_i \leq k

  • This one generalizes the type of problems where you need to find sum of binomial coefficients which are at certain gaps. By gaps I mean, suppose you need to find ∑k=07(304k)\sum_{k=0}^7 {30 \choose 4k} as you can find here that binomial coefficients are at certain gaps of 44. (I hope I am able to explain my point clearly). So here's the generalization - for m,n∈Nm,n \in \mathbb{N} such that m≤nm \leq n we have

∑k=0mk≤n(nmk)=2nm∑k=0m−1cos⁡n(πkm)einkmπ\sum_{k=0}^{mk \leq n} {n \choose mk} = \dfrac{2^n}{m} \sum_{k=0}^{m-1} \cos^{n} \left(\frac{\pi k}{m}\right) e^{i\frac{nk}{m} \pi}

Exercise -

  • Evaluate ∑k=110(303k)\sum_{k=1}^{10} {30 \choose 3k}

  • Prove ∑k=0n−1(−1)kcos⁡n(πkn)=n2n−1\sum_{k=0}^{n-1} (-1)^k \cos^n\left(\frac{\pi k}{n}\right) = \frac{n}{2^{n-1}}

  • Prove both the generalizations.

#RootsOfUnity #Induction #Generalization #Exercises

Note by Kishlaya Jaiswal
5 years, 9 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

For "sum of binomial coefficients which are at certain gaps", I think that the best approach is using Roots of Unity instead of trying to prove it by induction.

Calvin Lin Staff - 5 years, 9 months ago

Log in to reply

Yes Sir, indeed that's the method even I used to prove it and henceforth tagged this post with RootsOfUnity filter ;)

Kishlaya Jaiswal - 5 years, 9 months ago

Just say a few words on how do we use roots of unity..

Vishal Yadav - 4 years, 1 month ago
×

Problem Loading...

Note Loading...

Set Loading...