Timeline for What do named "tricks" share?
Current License: CC BY-SA 2.5
11 events
| when toggle format | what | by | license | comment | |
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| Jun 14, 2011 at 18:43 | comment | added | Michael Hardy | A fairly new Wikipedia article: en.wikipedia.org/wiki/Integral_of_the_secant_function Raise your hand if you knew that this was a famous conjecture in the 17th century. Or that it was originally done for the purposes of cartography (I think maybe lots of people know that). Or that Isaac Newton was aware of the conjecture and wrote about it, and Isaac Barrow was the first to prove it. | |
| Jun 14, 2011 at 15:00 | comment | added | Ostap Chervak | I usually integrate this function differently (using something trick-like, but natural one): $$\frac{dx}{\cos(x)} = \frac{\cos(x)dx}{{\cos^2(x)}} = \frac{d(\sin(x)}{1 - {\sin^2(x)}}$$ | |
| Dec 5, 2010 at 2:31 | comment | added | Jeff Strom | I had a student come up with a different way to do this, on the fly, on the final exam. Write $$ \sec(x) = {1\over \sqrt{1 - \sin^2(x)}}, $$ then substitute $u = \sin(x)$ and use partial fractions! | |
| Dec 4, 2010 at 20:37 | comment | added | Michael Hardy | @Thierry: Anyone who likes the tangent-half-angle formula might like this version of it that has some symmetry not found in the versions always seen in present-day texts: $$ \tan\frac{\alpha + \beta}{2} = \frac{\sin\alpha + \sin\beta}{\cos \alpha + \cos\beta}. $$ Could that be useful for anything related to integrals like these? | |
| Dec 4, 2010 at 19:45 | comment | added | Mike Spivey | @Michael Hardy: I asked this question about ways of evaluating $\int \sec x dx$ on math.SE about two months ago. You might be interested in the answers: math.stackexchange.com/questions/6695 | |
| Dec 4, 2010 at 18:42 | comment | added | Michael Hardy | Dammit, I'm being clumsy, and I can't edit a comment here after I've posted it. To atone for my sin, I'll try to get it right: $$ \sec x\ dx = \frac{\sec x(\tan x + \sec x)}{\sec x + \tan x}\ dx = \frac{\sec x \tan x + \sec^2 x}{\sec x + \tan x}\ dx = \frac{du}{u}. $$ | |
| Dec 4, 2010 at 18:18 | comment | added | Thierry Zell | The tan(x/2) formulas can be derived from basic trigonometric identities, and that's the way I learned them in high-school. Needless to say, when I realized the link with the rational parametrization, I was extatic. | |
| Dec 4, 2010 at 18:05 | comment | added | Qiaochu Yuan | Yes, I agree that that argument is trick-like. The rational parameterization of the circle is not. It is a natural and beautiful geometric argument (just taking the slope of a line between two points) and I think it could easily be presented to a first-year calculus class. | |
| Dec 4, 2010 at 17:50 | comment | added | Michael Hardy |
There's the argument often used in calculus texts: $$ \sec x = \frac{\sec x(\tan x + \sec x)}{\sec x + \tan x} = \frac{\sec x \tan x + \sec^2 x}{\sec x + \tan x} = \frac{du}{u} $$ where $u$ is the last denominator above. That seems like a "trick". I've seen other methods but I don't remember what they are right now. One of them seemed to take some of the mystery out of it, but it was still a trick. But I can't show a residue computation to a first-year calculus class. The "Weierstrass trick" (was it Weierstrass?) of $\tan(x/2)$, etc., does seem a bit more like a "method".
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| Dec 4, 2010 at 17:37 | comment | added | Qiaochu Yuan | There is an algorithm to find the integral of any rational function of trigonometric functions: use a rational parameterization (e.g. using tan x/2), then use partial fractions (or a residue computation, etc.). I don't see how this is a "trick" in any sense. It is a direct corollary of the fact that the circle is birational to the projective line. | |
| Dec 4, 2010 at 17:26 | history | answered | Michael Hardy | CC BY-SA 2.5 |