Fie \( f: [0,\infty) \to [0,1] \) o functie monotona. Aratati ca \( \alpha = \lim\limits_{x\to \infty} f(x) \) exista si e finita, iar
\( \lim_{n\to\infty} \sum_{k=1}^n \frac{f(k)}{n}\cdot f \left( \frac{k}{n}\right) =\alpha \int_{0}^1 f(x)dx \)
Functie monotona si suma Riemann
Moderators: Bogdan Posa, Beniamin Bogosel, Marius Dragoi
-
Radu Titiu
- Thales
- Posts: 155
- Joined: Fri Sep 28, 2007 5:05 pm
- Location: Mures \Bucuresti
Functie monotona si suma Riemann
A mathematician is a machine for turning coffee into theorems.
-
Laurentiu Tucaa
- Thales
- Posts: 145
- Joined: Sun Mar 22, 2009 6:22 pm
- Location: Pitesti
Prima cerinta se rezolva stiind faptul ca orice functie monotona are limita si cum functia ia valori in \( [0,1] \) rezulta ca limita \( \alpha\in[0,1] \).
Cum \( \lim_{x\to\infty}f(x)=\alpha\in[0,1] \) rezulta ca \( \forall\eps>0 \) exista \( n_{\eps}\in\mathbb{N} \) a.i. \( \forall n\in\mathbb{N},n\ge n_{\eps},f(n)\in(\alpha-\eps,\alpha+\eps) \). Astfel \( \lim_{n\to\infty}\sum_{k=1}^n \frac{f(k)}{n})f(\frac{k}{n})=\lim_{n\to\infty}\(\sum_{k=1}^{n_{\eps}}\frac{f(n_{\eps})}{n} f(\frac{k}{n})+\sum_{k=n_{\eps+1}}^{n}\frac{f(k)}{n} f(\frac{k}{n})-\sum_{k=1}^{n_{\eps}}\frac{f(k)-f(n_{\eps})}{n}f({\frac{k}{n})\) \)
si de aici totul se rezolva folosind criteriul clestelui si faptul ca limita ultimei sume este 0 (cea de dupa minus).
Cum \( \lim_{x\to\infty}f(x)=\alpha\in[0,1] \) rezulta ca \( \forall\eps>0 \) exista \( n_{\eps}\in\mathbb{N} \) a.i. \( \forall n\in\mathbb{N},n\ge n_{\eps},f(n)\in(\alpha-\eps,\alpha+\eps) \). Astfel \( \lim_{n\to\infty}\sum_{k=1}^n \frac{f(k)}{n})f(\frac{k}{n})=\lim_{n\to\infty}\(\sum_{k=1}^{n_{\eps}}\frac{f(n_{\eps})}{n} f(\frac{k}{n})+\sum_{k=n_{\eps+1}}^{n}\frac{f(k)}{n} f(\frac{k}{n})-\sum_{k=1}^{n_{\eps}}\frac{f(k)-f(n_{\eps})}{n}f({\frac{k}{n})\) \)
si de aici totul se rezolva folosind criteriul clestelui si faptul ca limita ultimei sume este 0 (cea de dupa minus).