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 May 26th, 2017, 02:53 AM #1 Member   Joined: Apr 2014 From: Greece Posts: 58 Thanks: 0 Solving Riemann Lower sum I'm trying to find the integral of $\displaystyle f(x)=x^{2},f:[0,1]\rightarrow \mathbb{R}$ using Riemann's criterion. I created a partition with equal intervals like this $\displaystyle P_{n}=\{0<\frac{1}{n}<\frac{2}{n}<...<\frac{n-1}{n}=1\}$ and then I started cxalculating the lower sum $\displaystyle L(f,P_{n})=f(0)\frac{1}{n}+f(\frac{1}{n})\frac{1}{ n}+...+f(\frac{n-1}{n})\frac{1}{n}$$\displaystyle =\frac{1}{n}(0+\frac{1}{n^{2}}+...+\frac{(n-1)^{2}}{n^{2}})$$\displaystyle =\frac{1^{2}+2^{2}+...+(n-1)^{2}}{n^{3}}$ I don't know how to proceed in my book it says that this is equal to: $\displaystyle \frac{(n-1)n(2n-1)}{6n^{3}}$ but I don't understand why.. Can anyone explain how we came to the conclusion that these are equal and what is the way to construct a general equation like this? May 26th, 2017, 04:07 AM #2 Senior Member   Joined: May 2016 From: USA Posts: 1,310 Thanks: 551 $\displaystyle L = \left ( \sum_{j=1}^n f \left \{ 0 + \dfrac{j - 1}{n} * (1 - 0) \right \} \dfrac{1}{n} \right ) = \dfrac{1}{n} * \sum_{j=1}^n f \left ( \dfrac{j - 1}{n} \right )^2 = \dfrac{1}{n^3} * \left ( \sum_{j=1}^{n-1}j^2 \right ).$ So you got here: good work. $1^2 = 1 = \dfrac{6}{6} = \dfrac{1 * 2 * 3}{6} = \dfrac{1(1 + 1)(2 * 1 + 1)}{6}.$ $1^2 + 2^2 = 5 = \dfrac{30}{6} = \dfrac{2 * 3 * 5}{6} = \dfrac{2(2 + 1)(2 * 2 +1)}{6}.$ $1^2 + 2^2 + 3^2 = 14 = \dfrac{84}{6} = \dfrac{3 * 4 * 7}{6} = \dfrac{3(3 + 1)(2 * 3 + 1)}{6}$ In general it can be shown that $\displaystyle \sum_{j=1}^mj^2 = \dfrac{m(m + 1)(2m + 1)}{6}.$ So if m = n - 1, we get $\displaystyle L = \dfrac{1}{n^3} * \sum_{j=1}^{m=n-1}j^2 = \dfrac{1}{n^3} * \dfrac{m(m + 1)(2m - 1)}{6} = \dfrac{(n - 1)((n - 1) + 1)(2(n - 1) + 1)}{6n^3} \implies$ $L = \dfrac{(n - 1)n(2n - 1)}{6n^3} = \dfrac{2n^2 - 3n + 1}{6n^2} = \dfrac{1}{3} - \dfrac{1}{2n} + \dfrac{1}{6n^2}.$ Last edited by JeffM1; May 26th, 2017 at 04:18 AM. May 26th, 2017, 04:46 AM #3 Member   Joined: Apr 2014 From: Greece Posts: 58 Thanks: 0 Ok I see what's going on. Can you please tell me how construct such equalities? For example if I want to construct an equality for $\displaystyle \sum_{j=1}^{m}\sqrt{j}$. Is there any specific way of working or do you just use your imagination and intuition and test it to see if it's correct? Tags lower, riemann, solving, sum Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post thekineto Probability and Statistics 1 March 7th, 2015 10:42 PM walter r Real Analysis 2 June 22nd, 2013 07:09 PM Jalaska13 Number Theory 0 May 30th, 2010 03:30 PM cheloniophile Real Analysis 1 November 23rd, 2008 05:30 PM

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