floor function and infinite series...

Pls its a request to all the young mathematics mind to publish a note on floor function and how to solve an infinite series problem.....

#Calculus

Note by Sarvesh Dubey
6 years, 2 months ago

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Comments

Here's a simple example to gear you up. Prove that

lim⁡n→∞1n2∑k=1n⌊kx⌋=x2\lim_{n \rightarrow \infty} \frac{1}{n^2}\sum_{k=1}^n \lfloor kx \rfloor = \frac{x}{2}

Proof. By the definition of Greatest Integer Function, x−1<⌊x⌋≤xx-1 < \lfloor x \rfloor \leq x

Thus, we have kx−1<⌊kx⌋≤kxkx-1 < \lfloor kx \rfloor \leq kx ⇒∑k=1nkx−1<∑k=1n⌊kx⌋≤∑k=1nkx\Rightarrow \sum_{k=1}^n kx-1 < \sum_{k=1}^n \lfloor kx \rfloor \leq \sum_{k=1}^n kx ⇒n(n+1)2x−n<∑k=1n⌊kx⌋≤n(n+1)2x\Rightarrow \frac{n(n+1)}{2}x-n < \sum_{k=1}^n \lfloor kx \rfloor \leq \frac{n(n+1)}{2}x ⇒n+12nx−1n<1n2∑k=1n⌊kx⌋≤n+12nx\Rightarrow \frac{n+1}{2n}x-\frac{1}{n} < \frac{1}{n^2}\sum_{k=1}^n \lfloor kx \rfloor \leq \frac{n+1}{2n}x

Taking the limit, we find that lim⁡n→∞n+12nx−1n<lim⁡n→∞1n2∑k=1n⌊kx⌋≤lim⁡n→∞n+12nx\lim_{n \rightarrow \infty} \frac{n+1}{2n}x-\frac{1}{n} < \lim_{n \rightarrow \infty} \frac{1}{n^2}\sum_{k=1}^n \lfloor kx \rfloor \leq \lim_{n \rightarrow \infty} \frac{n+1}{2n}x ⇒x2<lim⁡n→∞1n2∑k=1n⌊kx⌋≤x2\Rightarrow \frac{x}{2} < \lim_{n \rightarrow \infty} \frac{1}{n^2}\sum_{k=1}^n \lfloor kx \rfloor \leq \frac{x}{2}

And therefore by Squeeze Theorem our result follows as

lim⁡n→∞1n2∑k=1n⌊kx⌋=x2□\lim_{n \rightarrow \infty} \frac{1}{n^2}\sum_{k=1}^n \lfloor kx \rfloor = \frac{x}{2} \quad _\square

Also, you can read more about the floor function here

Kishlaya Jaiswal - 6 years, 2 months ago

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So as you see in the above example, we basically make use of inequalities while solving problems related to Floor/Ceil Function. Here are some other inequalities that can help you -

x−1<⌊x⌋≤xx-1 < \lfloor x \rfloor \leq x

⌊x⌋≤x<⌊x⌋+1\lfloor x \rfloor \leq x < \lfloor x \rfloor +1

Also, notice that, any integer can be written as

x=⌊x⌋+{x}x = \lfloor x \rfloor + \{x\}

Since, 0≤{x}<10 \leq \{x\} < 1. That's where we get the inequality ⌊x⌋≤x<⌊x⌋+1\lfloor x \rfloor \leq x < \lfloor x \rfloor +1 and ⌊x⌋≤x\lfloor x \rfloor \leq x

Anyhow, if your concern was regarding any of the well-known series such as Hermite's Identity, then please let me know so that I can add examples regarding the same.

Kishlaya Jaiswal - 6 years, 2 months ago
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