April 23


  • Questions
  • Leibniz - hints, then proof of the fundamental theorem of calculus

Student Questions:


The discussion page can be found here, for observations and remarks.

Today in Class:

  • triangular numbers- number of units that can form a triangle (1, 3, 6, 10, 15, …)
    • $\frac{n(n+1)}{2}$ represents this series
  • The series $1 + \frac{1}{3} + \frac{1}{6} + \frac{1}{10} +...$ converges. We can prove this by comparison to the series $\frac{1}{n^2}$ which Euler discovered converges to $\frac{\pi^2}{6}$
  • The series can be written as $\sum_{n=1}^\infty {\frac{2}{n(n+1)}} = (\frac{2}{1} - \frac{2}{2}) + (\frac{2}{2} - \frac{2}{3}) + (\frac{2}{3} - \frac{2}{4}) + (\frac{2}{4} - \frac{2}{5}) +...$
  • The second fraction of each term in the series cancels with the first fraction of the next which means the sum of the series is $2 + \frac{2}{n}$ where n is approaching $\infty$. This series converges to 2.
  • For any series $(a_1, a_2, a_3....a_n)$, if we are given the sum we can find a function that represents the series such that:
    • $a_i = f(i+1) - f(i)$
    • We used the series $1+3+5+7+9+11 +... (2n+1)$ as an example. Here we are looking for $f$ such that $f(i+1) - f(i) = 2i+1$. The function that meets this requirement is $f(x) = x^2$.
    • Another example is the series $1 + 2 + 3 + 4 +... n$. Here we are looking for $f$ such that $f(i+1) - f(i) = i$.
  • This discussion led us to Leibniz's proof of the Fundamental Theorem of Calculus which is in the text on page 133-134.
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