Post by chang on Aug 12, 2021 2:02:08 GMT
It may come as a surprise that Isaac Newton, who invented calculus in the 17th century, actually computed limits and derivatives geometrically. For example, he used similar (congruent) trianges. Here is an example of the enormous power of this old-fashioned (and mostly forgotten) method, showing that the derivative of tan π is 1 + tanΒ²π (= secΒ²π). Quite a powerful technique!
(You can draw a similar picture to show that the derivative of sin π is cos π, and the derivative of cos π is βsin π.)
Here is a very tricky problem from a book by V. I Arnol'd. He remarks that this problem has a quick, easy geometric solution that Newton, Huygens or Hooke would have seen in a heartbeat. I haven't tried the calculus approach (L'Hopital's Rule, i.e., power series) but I have a suspicion it will lead to an endless series of 0/0 expressions. (Note that the functions sin π, tan π, arcsin π and arctan π all look like π when π is close to zero.) Arnol'd remarks that the only person he knows who was capable of "quickly solving" the problem by conventional (non-geometric) methods was 1986 Fields Medal* recipient Gerd Faltings. (*Fields Medal = the mathematics equivalent of the Nobel Prize.)
Here's my geometric solution: Consider π» and π ("well behaved" functions) with slope = 1 at the origin, as shown below. First look at the triangle ABC. In the limit as A approaches the origin, the triangle ABC approaches a 45Β°-45Β°-90Β° triangle, hence |AB|/|BC| β 1.
Now look at the parallelogram BDEC. In the same limit, the angled sides both become 45Β°, so |BC|/|DE| β 1. Hence |AB|/|DE|=|AB|/|D'E'| β 1, and this is precisely the limit to be evaluated. (Make the identification π» = sin(tan π) and π = tan(sin π).)
It's amazing how "old knowledge" can be lost.
(You can draw a similar picture to show that the derivative of sin π is cos π, and the derivative of cos π is βsin π.)
Here is a very tricky problem from a book by V. I Arnol'd. He remarks that this problem has a quick, easy geometric solution that Newton, Huygens or Hooke would have seen in a heartbeat. I haven't tried the calculus approach (L'Hopital's Rule, i.e., power series) but I have a suspicion it will lead to an endless series of 0/0 expressions. (Note that the functions sin π, tan π, arcsin π and arctan π all look like π when π is close to zero.) Arnol'd remarks that the only person he knows who was capable of "quickly solving" the problem by conventional (non-geometric) methods was 1986 Fields Medal* recipient Gerd Faltings. (*Fields Medal = the mathematics equivalent of the Nobel Prize.)
Here's my geometric solution: Consider π» and π ("well behaved" functions) with slope = 1 at the origin, as shown below. First look at the triangle ABC. In the limit as A approaches the origin, the triangle ABC approaches a 45Β°-45Β°-90Β° triangle, hence |AB|/|BC| β 1.
Now look at the parallelogram BDEC. In the same limit, the angled sides both become 45Β°, so |BC|/|DE| β 1. Hence |AB|/|DE|=|AB|/|D'E'| β 1, and this is precisely the limit to be evaluated. (Make the identification π» = sin(tan π) and π = tan(sin π).)
It's amazing how "old knowledge" can be lost.
