Rabi resonance method: Difference between revisions

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[[File:WeierstrassSubstitution.svg|thumb|right|400px|The Weierstrass substitution, here illustrated as [[stereographic projection]] of the circle.]]
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In [[integral calculus]], the '''tangent half-angle substitution''' is a [[integration by substitution|substitution]] used for finding [[antiderivative]]s, and hence definite integrals, of [[rational function]]s of [[trigonometric function]]s. [[Without loss of generality|No generality is lost]] by taking these to be rational functions of the sine and cosine. [[Michael Spivak]] wrote that "The world's sneakiest substitution is undoubtedly" this technique.<ref>Michael Spivak, ''Calculus'', [[Cambridge University Press]], 2006, pages 382–383.</ref>
 
== Euler and Weierstrass ==
 
Various books call this the '''Weierstrass substitution''', after [[Karl Weierstrass]] (1815&nbsp;–&nbsp;1897), without citing any occurrence of the substitution in Weierstrass' writings,<ref>Gerald L. Bradley and Karl J. Smith, ''Calculus'', Prentice Hall, 1995, pages 462, 465, 466</ref><ref>Christof Teuscher, ''Alan Turing: Life and Legacy of a Great Thinker'', Springer, 2004, pages 105–6</ref><ref>James Stewart, ''Calculus: Early Transcendentals'', Brooks/Cole, Apr 1, 1991, page 436</ref> but the technique appears well before Weierstrass was born, in the work of [[Leonhard Euler]] (1707&nbsp;–&nbsp;1783).<ref>Leonhard Euler, ''Institutiionum calculi integralis volumen primum'', 1768, E342, Caput V, paragraph 261. See [http://www.eulerarchive.org/ http://www.eulerarchive.org/]</ref>
 
== The substitution ==
[[File:Stereo.Weierstrass.svg|350px|thumb|right|How the Weierstrass substitution is related to the [[stereographic projection]].]]
One starts with the problem of finding an [[antiderivative]] of a rational function of the sine and cosine, and replaces sin&nbsp;''x'', cos&nbsp;''x'', and the [[Differential (mathematics)|differential]]&nbsp;d''x'' with rational functions of a variable&nbsp;''t'' and the product of a rational function of ''t'' with the differential&nbsp;d''t'', as follows:<ref>James Stewart, ''Calculus: Early Transcendentals'', Brooks/Cole, 1991, page 439</ref>
 
: <math>
\begin{align}
\sin x & = \frac{2t}{1 + t^2} \\[8 pt]
\cos x & = \frac{1 - t^2}{1 + t^2} \\[8 pt]
\mathrm{d}x & = \frac{2 \,\mathrm{d}t}{1 + t^2}.
\end{align}
</math>
 
=== Derivation ===
 
Let
:<math>t = \tan\frac{x}{2}.</math>
 
By the [[double-angle formula]] for the sine function,
:<math>
\begin{align}
\sin x&=2\sin\frac{x}{2}\cos\frac{x}{2}\\[8 pt]
&=2t\cos^2\frac{x}{2}\\[8 pt]
&=\frac{2t}{\sec^2\frac{x}{2}}\\[8 pt]
&=\frac{2t}{1+t^2}.
\end{align}
</math>
 
By the [[double-angle formula]] for the cosine function,
:<math>
\begin{align}
\cos x&=1-2\sin^2\frac{x}{2}\\[8 pt]
&=1-2t^2\cos^2\frac{x}{2}\\[8 pt]
&=1-\frac{2t^2}{\sec^2\frac{x}{2}}\\[8 pt]
&=1-\frac{2t^2}{1+t^2}\\[8 pt]
&=\frac{1-t^2}{1+t^2}.
\end{align}
</math>
 
The differential&nbsp;d''x'' can be calculated as follows:
:<math>
\begin{align}
\frac{\mathrm{d}t}{\mathrm{d}x}&=\frac{1}{2}\sec^2\frac{x}{2}\\[8 pt]
&=\frac{1+t^2}{2}\\[8 pt]
\Rightarrow\mathrm{d}x&=\frac{2\,\mathrm{d}t}{1+t^2}.
\end{align}
</math>
 
== Examples ==
[[File:Weierstrass substitution.png|thumb|right|400px|The tangent half-angle formula relates an angle to the slope of a line.]]
=== First example ===
(See also [[integral of the secant function]].)
:<math>
\begin{align}
\int\csc x\,\mathrm{d}x&=\int\frac{\mathrm{d}x}{\sin x}&\\
&=\int\frac{\mathrm{d}t}{t}&t=\tan\frac{x}{2}\\
&=\ln t+C\\
&=\ln \tan\frac{x}{2}+C.
\end{align}
</math>
 
=== Second example: a definite integral ===
:<math>
\begin{align}
\int_0^{2\pi}\frac{\mathrm{d}x}{2+\cos x}&=\int_{x=0}^{x=\pi}\frac{\mathrm{d}x}{2+\cos x}+\int_{x=\pi}^{x=2\pi}\frac{\mathrm{d}x}{2+\cos x}&&\\
&=\int_{t=0}^{t=\infty}\frac{\mathrm{d}x}{2+\cos x}+\int_{t=-\infty}^{t=0}\frac{\mathrm{d}x}{2+\cos x}&t&=\tan\frac{x}{2}\\
&=\int_{t=-\infty}^{t=\infty}\frac{\mathrm{d}x}{2+\cos x}&&\\
&=\int_{-\infty}^{\infty}\frac{2\,\mathrm{d}t}{3+t^2}&&\\
&=\frac{2}{\sqrt 3}\int_{-\infty}^{\infty}\frac{\mathrm{d}u}{1+u^2}&t&=u\sqrt 3\\
&=\frac{2\pi}{\sqrt 3}.&&
\end{align}
</math>
In the first line, one does not simply substitute <math>t=0</math> for both [[limits of integration]]. The [[Mathematical singularity|singularity]] (in this case, a [[Asymptote#Vertical asymptotes|vertical asymptote]]) of <math>t=\tan\frac{x}{2}</math> at <math>x=\pi</math> must be taken into account.
 
== Geometry ==
[[File:Weierstrass.substitution.svg|400px|thumb|right|The Weierstrass substitution parametrizes the [[unit circle]] centered at&nbsp;(0,&nbsp;0).  Instead of +&infin; and &minus;&infin;, we have only one &infin;, at both ends of the real line.  That is often appropriate when dealing with rational functions and with trigonometric functions.  (This is the [[one-point compactification]] of the line.)]]
 
As ''x'' varies, the point (cos&nbsp;''x'',&nbsp;sin&nbsp;''x'') winds repeatedly around the [[unit circle]] centered at&nbsp;(0,&nbsp;0).  The point
 
: <math> \left(\frac{1-t^2}{1+t^2}, \frac{2t}{1+t^2}\right) </math>
 
goes only once around the circle as ''t'' goes from &minus;&infin; to&nbsp;+&infin;, and never reaches the point&nbsp;(&minus;1,&nbsp;0), which is approached as a limit as ''t'' approaches&nbsp;±&infin;.  As ''t'' goes from &minus;&infin; to &minus;1, the point determined by ''t'' goes through the part of the circle in the third quadrant, from (&minus;1,&nbsp;0) to&nbsp;(0,&nbsp;&minus;1).  As ''t'' goes from &minus;1 to&nbsp;0, the point follows the part of the circle in the fourth quadrant from (0,&nbsp;&minus;1) to&nbsp;(1,&nbsp;0).  As ''t'' goes from 0 to 1, the point follows the part of the circle in the first quadrant from (1,&nbsp;0) to&nbsp;(0,&nbsp;1).  Finally, as ''t'' goes from 1 to&nbsp;+&infin;, the point follows the part of the circle in the second quadrant from (0,&nbsp;1) to&nbsp;(&minus;1,&nbsp;0).
 
Here is another geometric point of view.  Draw the unit circle, and let ''P'' be the point {{nowrap|(&minus;1, 0)}}.  A line through ''P'' (except the vertical line) is determined by its slope.  Furthermore, each of the lines (except the vertical line) intersects the unit circle in exactly two points, one of which is ''P''.  This determines a function from points on the unit circle to slopes.  The trigonometric functions determine a function from angles to points on the unit circle, and by combining these two functions we have a function from angles to slopes.
 
==See also==
*[[Rational curve]]
*[[Stereographic projection]]
 
== Notes and references ==
{{reflist}}
 
== External links ==
* [http://planetmath.org/encyclopedia/WeierstrassSubstitutionFormulas.html Weierstrass substitution formulas] at [[PlanetMath]]
 
[[Category:Integral calculus]]

Latest revision as of 19:09, 28 October 2014

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