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Method of Exhuastion for the area of a parabolic segment

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  1. Sep 5, 2015 #1
    Hello everyone,

    I'm using the book Apostol- Calculus Vol. 1 for self-studying to get a better understanding of proof based Calculus. They said this book was good for self studying, but I am already stuck in the first chapter. I'm trying to understand how he got the identity:

    12+22+...+n(2)= n3/ 3 + n2/2 + n/ 6
     
  2. jcsd
  3. Sep 5, 2015 #2

    Ssnow

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  4. Sep 5, 2015 #3

    mathman

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    You could assume it is a cubic polynomial and set up a system of 4 linear equations by using n=1,2,3,4.
     
  5. Sep 7, 2015 #4

    mathwonk

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    there is a footnote on page 27 of courant's calculus book that explains how all siuch formulas can be found by recursion. it was when i saw that in week one of freshman year in college that i knew i wasn't in kansas anymore, to paraphrase Dorothy of wizard of oz fame. to make the derivation easier, notice that all you really need to know is the lead term, i.e. that the formula is n^3/3 + a quadratic in n.


    https://www.ime.usp.br/~gorodski/ps/Courant-DifferentialIntegralCalculusVolI.pdf
     
  6. Sep 10, 2015 #5

    Svein

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    Actually, it is straight-forward induction. The formula is obviously correct for n=1, so assume that it holds for n: [itex] \sum_{i=1}^{n}i^{2}=\frac{n^{3}}{3}+\frac{n^{2}}{2}+\frac{n}{6}[/itex]. We want to verify the formula for n+1, so we add (n+1)2. Now observe:
    1. [itex]\frac{(n+1)^{3}}{3}-\frac{n^{3}}{3}=n^{2}+n+\frac{1}{3} [/itex]
    2. [itex]\frac{(n+1)^{2}}{2}-\frac{n^{2}}{2}=n+\frac{1}{2} [/itex]
    3. [itex](n+1)^{2}=n^{2}+2n+1=(n^{2}+n+\frac{1}{3})+(n+\frac{1}{2})+\frac{1}{6} [/itex]
    Now, adding the three expressions in parentheses in (3) to the corresponding elements in the formula, we end up with [itex] \sum_{i=1}^{n+1}i^{2}=\frac{(n+1)^{3}}{3}+\frac{(n+1)^{2}}{2}+\frac{(n+1)}{6}[/itex]. Therefore, the formula is (by induction) proven for all n.
     
  7. Sep 11, 2015 #6
    Basic Idea: You can find ##\displaystyle\sum_{k=1}^n k^{p+1}## if you know ##\displaystyle\sum_{k=1}^n k^{p}## , for ##p\in\{0,1,2,\ldots\}##.

    Procedure: (for ##p=1##)

    We know that (Why?) $$\sum_{k=1}^n k^1=\frac{n(n+1)}2.\tag1$$ We also know that $$(k+1)^3-n^3=3n^2+3n+1.$$ Writing this formula for ##k=1,2,\ldots,n## we get:
    $$
    \begin{array}{rcl}
    2^3-1^3 & = & 3\cdot1^2+3\cdot1+1\qquad \\
    3^3-2^3 & = & 3\cdot2^2+3\cdot2+1 \\
    \cdot\,\ \quad & & \\
    \cdot\,\ \quad & & \\
    \cdot\,\ \quad & & \\
    \quad(n+1)^3-n^3 & = & 3\cdot n^2+3\cdot n+1 \\ \hline
    (n+1)^3-1^3 &=&3[1^2+2^2+\cdots+n^2]+3[1+2+\cdots+n]\\&&\qquad\qquad+[\underbrace{1+1+\cdots+1}_{n\,\text{times}}]
    \end{array}$$ This last equality can be simplified using ##(1)## to get: $$(n+1)^3-1=3\sum_{k=1}^n k^2+3\frac{n(n+1)}2+n.$$ We need to solve this equation for ##\displaystyle\sum_{k=1}^n k^2## to get our desired result: $$\begin{align*}
    (n+1)^3-1=3\sum_{k=1}^n k^2+3\frac{n(n+1)}2+n&\iff n^3+3n^2+3n=3\sum_{k=1}^n k^2+3\frac{n(n+1)}2+n\\
    &\iff n^3+3n^2+3n-3\frac{n(n+1)}2-n=3\sum_{k=1}^n k^2\\
    &\iff n^3+3n^2+2n-\frac{3n(n+1)}2=3\sum_{k=1}^n k^2\\
    &\iff \frac{2n^3+6n^2+4n}{2}-\frac{3n^2+3n}2=3\sum_{k=1}^n k^2\\
    &\iff\frac{2n^3+3n^2+n}2=3\sum_{k=1}^n k^2\\
    &\iff\frac{2n^3+3n^2+n}{2}\frac13=\sum_{k=1}^n k^2\\
    &\iff\frac{2n^3+3n^2+n}{6}=\sum_{k=1}^n k^2\\
    &\iff\sum_{k=1}^n k^2=\frac{n^3}3+\frac{n^2}2+\frac n6\\
    \end{align*}$$

    Now substitute ##n## with ##n-1## to get your desired formula. :wink:

    N.B: This is (overall) the same method Apostol uses to evaluate ##\displaystyle\sum_{k=1}^n k^{2}## , you just had to look further in the text :woot:, see Calculus vol.I, pp.5-6.

    Exercise: Find ##\displaystyle\sum_{k=1}^n k^{3}##, ##\displaystyle\sum_{k=1}^n k^{4}## and ##\displaystyle\sum_{k=1}^n \frac1{k(k+1)}## using this method. Good luck! :oldtongue:
     
    Last edited: Sep 11, 2015
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