News, Views & Insights
Mathematics
Seeing beyond a Theorem
Mathematics is a notoriously technical subject that prizes exactingly precise statements. The square of the hypotenuse of a right triangle is the sum of the squares of the legs, not the sum of their cubes, nor the difference of their squares. Such precision produces the clarity that makes the subject so powerful, but occasionally it comes at the cost of easy understanding. Indeed, more-complicated mathematical statements often sound bewildering upon first reading. Take the following theorem in plane geometry (deep breath...):
Let ABC be a triangle. Let DEF be parallel to AC with D on AB and E on BC. Let FGH be parallel to AB with G on BC and H on AC. Let r, r1, r2 and r3 be the radii of the incircles O, O1, O2 and O3 of the triangles ABC, DBE, EFG and HGC, respectively. If F is outside of ABC, then r = r1 + r2 + r3. Got it? Many theorems of mathematics, including this one, are easier to communicate by picture than by words. Here’s the scenario described in the theorem (images in this post are produced by slightly modified versions of the code for the Demonstration “The Radii of Four Incircles,” which is one of nearly 200 Demonstrations about theorems in plane geometry written by Jay Warendorff for the Wolfram Demonstrations Project):Differential Geometry Carved in Stone
John Wheeler (1911-2008)
| Gravitation versus Curved Spacetime | Zonohedron Turned Inside Out |
| Particle Moving around Two Extreme Black Holes | Schwarzschild Space-Time Embedding Diagram |
Pi Day
A New Look and New Features for MathWorld
Mathematica and the Fundamental Theorem of Calculus
Most calculus students might think that if one could compute indefinite integrals, it would always be easy to compute definite ones. After all, they might think, the fundamental theorem of calculus says that one just has to subtract the values of the indefinite integral at the end points to get the definite integral.
So how come inside Mathematica there are thousands of pages of code devoted to working out definite integrals---beyond just subtracting indefinite ones? The answer, as is often the case, is that in the real world of mathematical computation, things are more complicated than one learns in basic mathematics courses. And to get the correct answer one needs to be considerably more sophisticated. In a simple case, subtracting indefinite integrals works just fine. Consider computing the area under a sine curve, which equals