April 23

Plan:

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

Student Questions:

None

Discussion:
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.