My Math Forum Area under a parabolic section (justification)

 Calculus Calculus Math Forum

 December 23rd, 2010, 10:01 AM #1 Member   Joined: Dec 2010 Posts: 51 Thanks: 0 Area under a parabolic section (justification) Hello. I need help with the justification of (n^3)/3 in a particular pair of inequality. In other words, why (n^3)/3 is chosen instead of something else. Also, if possible, I need the proof deduction, not induction (I've already seen it). But first the premise to the question. The sum that under-approximates the area under a parabolic section: $s_n= \frac{b^3}{n^3}[1^2+2^2+...+(n-1)^2]$ The sum that over-approximates the area under a parabolic section: $S_n= \frac{b^3}{n^3}[1^2+2^2+...+n^2]$ We're only interested in the series in each equation. That is: $p= 1^2+2^2+...+(n-1)^2$ $v= 1^2+2^2+...+n^2$ (p and v are arbitrarily picked) Using $(k+1)^3= k^3 + 3k^2 + 3k + 1$ $3k^2 + 3k + 1= (k+1)^3 - k^3$ Let K = 1, 2, ..., (n-1) and manipulating the equation yield: $p= \frac{n^3}{3} - \frac{n^2}{2} + \frac{n}{6}$ $v= \frac{n^3}{3} + \frac{n^2}{2} + \frac{n}{6}$ --- p < m < v, since p under-approximates the area [if multiplied by $\frac{b^3}{n^3}$], v over-approximates the area [if multiplied by $\frac{b^3}{n^3}$], and m is the area [if multiplied by $\frac{b^3}{n^3}$] that is between the two approximations. The pair of inequalities are: $1^2+2^2+...+(n-1)^2 < \frac{n^3}{3} < 1^2+2^2+...+n^2$ Now, here's my questions) why is m set to $\frac{n^3}{3}$? Why is $\frac{n^3}{3}$ chosen as a consequence of p and v? Is it because of the end behavior of p and v? That is, as we partition more and more of the horizontal base (n approaches infinity) we approach $\frac{n^3}{3}$? Thanks in advance! The book I'm reading just shows the inequalities without justification for it.
 December 24th, 2010, 10:57 PM #2 Global Moderator   Joined: Dec 2006 Posts: 19,285 Thanks: 1681 What matters is that as n tends to infinity $s_n$ and $S_n$ approach the same limit, which is the area (since $s_n$ is an underestimate and $S_n$ is an overestimate). The common limit can be worked out without introducing a new variable m at all, but the author happened to prefer to use m. Note that $1^2+2^2+...+(n-1)^2 < \frac{n^3}{3} < 1^2+2^2+...+n^2$ implies $\frac13\,-\,\frac{1}{2n}\,+\,\frac{1}{6n^2}\,<\,\frac13\,<\, \frac13\,+\,\frac{1}{2n}\,+\,\frac{1}{6n^2}.$

 Tags area, justification, parabolic, section

 Thread Tools Display Modes Linear Mode

 Similar Threads Thread Thread Starter Forum Replies Last Post Tau Complex Analysis 6 February 13th, 2014 07:53 AM Albert.Teng Algebra 3 October 30th, 2012 08:04 PM suomik1988 Algebra 3 December 1st, 2009 04:24 PM uri Applied Math 1 October 11th, 2008 12:21 PM Blanshan91 Algebra 2 July 16th, 2008 10:22 AM

 Contact - Home - Forums - Cryptocurrency Forum - Top