Aristotle wrote one of the earliest systemizations of logic, his syllogisms. Unfortunately, he made two tiny oversights. An easy and fun logic lesson with students is to use Venn diagrams to check each syllogism, like so:
Thanks to Keith Devlin for the idea.
Essays on education, debate, and math instruction; neat math problems; and whatever else I get around to.
Saturday, June 20, 2009
Thursday, June 11, 2009
A neat application for derivatives
Whoa! May was crazy. But I survived. Here's a fun problem.
The function
gives a Bell curve. The function
gives a downward-opening parabola that fits exactly under the Bell curve, like so:

In other words, f(x) is a lower boundary to g(x):
How to go about proving this? Algebraically, simplifying the above doesn't do much. It turns out, though, that looking at the derivatives does prove this.
Using the Chain Rule for compounded functions on g(x) reveals the similarity between the two functions. Here's a graph of the two derivatives:

On the lower right-hand side, quadrant IV, f(x) has a more negative slope than g(x), which is easy enough to prove:
On the upper left-hand side, quadrant II, f(x) has a more positive slope than g(x), which has virtually the same proof:
(Just remember that the xs in these inequalities are negative, so the inequality signs are flipped twice, back to their original position.)
What does this prove? One final fact: at x = 0, f(0) = g(0).
As x increases, the slope of f(x) is always more negative than the slope of g(x) -- they start equal at x = 0, but f(x) decreases more rapidly, so f(x) must always be less than g(x) as x increases. On the other side, the logic is simply reversed.
The function
gives a Bell curve. The function
gives a downward-opening parabola that fits exactly under the Bell curve, like so:
In other words, f(x) is a lower boundary to g(x):
How to go about proving this? Algebraically, simplifying the above doesn't do much. It turns out, though, that looking at the derivatives does prove this.
Using the Chain Rule for compounded functions on g(x) reveals the similarity between the two functions. Here's a graph of the two derivatives:
On the lower right-hand side, quadrant IV, f(x) has a more negative slope than g(x), which is easy enough to prove:
On the upper left-hand side, quadrant II, f(x) has a more positive slope than g(x), which has virtually the same proof:
(Just remember that the xs in these inequalities are negative, so the inequality signs are flipped twice, back to their original position.)What does this prove? One final fact: at x = 0, f(0) = g(0).
As x increases, the slope of f(x) is always more negative than the slope of g(x) -- they start equal at x = 0, but f(x) decreases more rapidly, so f(x) must always be less than g(x) as x increases. On the other side, the logic is simply reversed.
Subscribe to:
Posts (Atom)