weierstrass substitution proof

Note that $$\frac{1}{a+b\cos(2y)}=\frac{1}{a+b(2\cos^2(y)-1)}=\frac{\sec^2(y)}{2b+(a-b)\sec^2(y)}=\frac{\sec^2(y)}{(a+b)+(a-b)\tan^2(y)}.$$ Hence $$\int \frac{dx}{a+b\cos(x)}=\int \frac{\sec^2(y)}{(a+b)+(a-b)\tan^2(y)} \, dy.$$ Now conclude with the substitution $t=\tan(y).$, Kepler found the substitution when he was trying to solve the equation The Weierstrass substitution is an application of Integration by Substitution. must be taken into account. follows is sometimes called the Weierstrass substitution. gives, Taking the quotient of the formulae for sine and cosine yields. , ) Die Weierstra-Substitution (auch unter Halbwinkelmethode bekannt) ist eine Methode aus dem mathematischen Teilgebiet der Analysis. The technique of Weierstrass Substitution is also known as tangent half-angle substitution. Finally, it must be clear that, since \(\text{tan}x\) is undefined for \(\frac{\pi}{2}+k\pi\), \(k\) any integer, the substitution is only meaningful when restricted to intervals that do not contain those values, e.g., for \(-\pi\lt x\lt\pi\). The Weierstrass substitution can also be useful in computing a Grbner basis to eliminate trigonometric functions from a . The singularity (in this case, a vertical asymptote) of By eliminating phi between the directly above and the initial definition of \end{align} (originally defined for ) that is continuous but differentiable only on a set of points of measure zero. We've added a "Necessary cookies only" option to the cookie consent popup, $\displaystyle\int_{0}^{2\pi}\frac{1}{a+ \cos\theta}\,d\theta$. Weierstrass Substitution 24 4. Among these formulas are the following: From these one can derive identities expressing the sine, cosine, and tangent as functions of tangents of half-angles: Using double-angle formulae and the Pythagorean identity \implies & d\theta = (2)'\!\cdot\arctan\left(t\right) + 2\!\cdot\!\big(\arctan\left(t\right)\big)' b f p < / M. We also know that 1 0 p(x)f (x) dx = 0. x Integrating $I=\int^{\pi}_0\frac{x}{1-\cos{\beta}\sin{x}}dx$ without Weierstrass Substitution. The Weierstrass substitution in REDUCE. 2 2.4: The Bolazno-Weierstrass Theorem - Mathematics LibreTexts Date/Time Thumbnail Dimensions User Example 15. Instead of + and , we have only one , at both ends of the real line. The 2 8999. How do I align things in the following tabular environment? Likewise if tanh /2 is a rational number then each of sinh , cosh , tanh , sech , csch , and coth will be a rational number (or be infinite). , Why are physically impossible and logically impossible concepts considered separate in terms of probability? The technique of Weierstrass Substitution is also known as tangent half-angle substitution . Viewed 270 times 2 $\begingroup$ After browsing some topics here, through one post, I discovered the "miraculous" Weierstrass substitutions. $\qquad$ $\endgroup$ - Michael Hardy |Algebra|. Tangent line to a function graph. Sie ist eine Variante der Integration durch Substitution, die auf bestimmte Integranden mit trigonometrischen Funktionen angewendet werden kann. Also, using the angle addition and subtraction formulae for both the sine and cosine one obtains: Pairwise addition of the above four formulae yields: Setting Now consider f is a continuous real-valued function on [0,1]. 2.1.5Theorem (Weierstrass Preparation Theorem)Let U A V A Fn Fbe a neighbourhood of (x;0) and suppose that the holomorphic or real analytic function A . = d Thus there exists a polynomial p p such that f p </M. \end{aligned} \), \( An irreducibe cubic with a flex can be affinely Substitute methods had to be invented to . t The key ingredient is to write $\dfrac1{a+b\cos(x)}$ as a geometric series in $\cos(x)$ and evaluate the integral of the sum by swapping the integral and the summation. . By clicking Accept all cookies, you agree Stack Exchange can store cookies on your device and disclose information in accordance with our Cookie Policy. Introducing a new variable Try to generalize Additional Problem 2. (This is the one-point compactification of the line.) 2 Let \(K\) denote the field we are working in. But here is a proof without words due to Sidney Kung: \(\text{sin}\theta=\frac{AC}{AB}=\frac{2u}{1+u^2}\) and We use the universal trigonometric substitution: Since \(\sin x = {\frac{{2t}}{{1 + {t^2}}}},\) we have. When $a,b=1$ we can just multiply the numerator and denominator by $1-\cos x$ and that solves the problem nicely. as follows: Using the double-angle formulas, introducing denominators equal to one thanks to the Pythagorean theorem, and then dividing numerators and denominators by x Instead of Prohorov's theorem, we prove here a bare-hands substitute for the special case S = R. When doing so, it is convenient to have the following notion of convergence of distribution functions. 2 assume the statement is false). Differentiation: Derivative of a real function. It is just the Chain Rule, written in terms of integration via the undamenFtal Theorem of Calculus. According to Spivak (2006, pp. However, the Bolzano-Weierstrass Theorem (Calculus Deconstructed, Prop. &=\int{\frac{2(1-u^{2})}{2u}du} \\ Let f: [a,b] R be a real valued continuous function. I saw somewhere on Math Stack that there was a way of finding integrals in the form $$\int \frac{dx}{a+b \cos x}$$ without using Weierstrass substitution, which is the usual technique. $$ Check it: The Weierstrass substitution parametrizes the unit circle centered at (0, 0). sin 1 t = q and the natural logarithm: Comparing the hyperbolic identities to the circular ones, one notices that they involve the same functions of t, just permuted. &=\int{(\frac{1}{u}-u)du} \\ 193. Multivariable Calculus Review. In integral calculus, the tangent half-angle substitution is a change of variables used for evaluating integrals, which converts a rational function of trigonometric functions of x What is the correct way to screw wall and ceiling drywalls? Proof. Thus, when Weierstrass found a flaw in Dirichlet's Principle and, in 1869, published his objection, it . = No clculo integral, a substituio tangente do arco metade ou substituio de Weierstrass uma substituio usada para encontrar antiderivadas e, portanto, integrais definidas, de funes racionais de funes trigonomtricas.Nenhuma generalidade perdida ao considerar que essas so funes racionais do seno e do cosseno. The Weierstrass substitution is the trigonometric substitution which transforms an integral of the form. (1/2) The tangent half-angle substitution relates an angle to the slope of a line. $$y=\frac{a\sqrt{1-e^2}\sin\nu}{1+e\cos\nu}$$But still $$x=\frac{a(1-e^2)\cos\nu}{1+e\cos\nu}$$ one gets, Finally, since 0 This entry briefly describes the history and significance of Alfred North Whitehead and Bertrand Russell's monumental but little read classic of symbolic logic, Principia Mathematica (PM), first published in 1910-1913. . = Remember that f and g are inverses of each other! Here we shall see the proof by using Bernstein Polynomial. Fact: Isomorphic curves over some field \(K\) have the same \(j\)-invariant. Newton potential for Neumann problem on unit disk. = File history. sin eliminates the \(XY\) and \(Y\) terms. ) By similarity of triangles. ( t u The method is known as the Weierstrass substitution. [2] Leonhard Euler used it to evaluate the integral {\textstyle u=\csc x-\cot x,} Use the universal trigonometric substitution: \[dx = d\left( {2\arctan t} \right) = \frac{{2dt}}{{1 + {t^2}}}.\], \[{\cos ^2}x = \frac{1}{{1 + {{\tan }^2}x}} = \frac{1}{{1 + {t^2}}},\;\;\;{\sin ^2}x = \frac{{{{\tan }^2}x}}{{1 + {{\tan }^2}x}} = \frac{{{t^2}}}{{1 + {t^2}}}.\], \[t = \tan \frac{x}{2},\;\; \Rightarrow x = 2\arctan t,\;\;\; dx = \frac{{2dt}}{{1 + {t^2}}}.\], \[\int {\frac{{dx}}{{1 + \sin x}}} = \int {\frac{{\frac{{2dt}}{{1 + {t^2}}}}}{{1 + \frac{{2t}}{{1 + {t^2}}}}}} = \int {\frac{{2dt}}{{1 + {t^2} + 2t}}} = \int {\frac{{2dt}}{{{{\left( {t + 1} \right)}^2}}}} = - \frac{2}{{t + 1}} + C = - \frac{2}{{\tan \frac{x}{2} + 1}} + C.\], \[x = \arctan t,\;\; \sin x = \frac{{2t}}{{1 + {t^2}}},\;\; dx = \frac{{2dt}}{{1 + {t^2}}},\], \[I = \int {\frac{{dx}}{{3 - 2\sin x}}} = \int {\frac{{\frac{{2dt}}{{1 + {t^2}}}}}{{3 - 2 \cdot \frac{{2t}}{{1 + {t^2}}}}}} = \int {\frac{{2dt}}{{3 + 3{t^2} - 4t}}} = \int {\frac{{2dt}}{{3\left( {{t^2} - \frac{4}{3}t + 1} \right)}}} = \frac{2}{3}\int {\frac{{dt}}{{{t^2} - \frac{4}{3}t + 1}}} .\], \[{t^2} - \frac{4}{3}t + 1 = {t^2} - \frac{4}{3}t + {\left( {\frac{2}{3}} \right)^2} - {\left( {\frac{2}{3}} \right)^2} + 1 = {\left( {t - \frac{2}{3}} \right)^2} - \frac{4}{9} + 1 = {\left( {t - \frac{2}{3}} \right)^2} + \frac{5}{9} = {\left( {t - \frac{2}{3}} \right)^2} + {\left( {\frac{{\sqrt 5 }}{3}} \right)^2}.\], \[I = \frac{2}{3}\int {\frac{{dt}}{{{{\left( {t - \frac{2}{3}} \right)}^2} + {{\left( {\frac{{\sqrt 5 }}{3}} \right)}^2}}}} = \frac{2}{3}\int {\frac{{du}}{{{u^2} + {{\left( {\frac{{\sqrt 5 }}{3}} \right)}^2}}}} = \frac{2}{3} \cdot \frac{1}{{\frac{{\sqrt 5 }}{3}}}\arctan \frac{u}{{\frac{{\sqrt 5 }}{3}}} + C = \frac{2}{{\sqrt 5 }}\arctan \frac{{3\left( {t - \frac{2}{3}} \right)}}{{\sqrt 5 }} + C = \frac{2}{{\sqrt 5 }}\arctan \frac{{3t - 2}}{{\sqrt 5 }} + C = \frac{2}{{\sqrt 5 }}\arctan \left( {\frac{{3\tan \frac{x}{2} - 2}}{{\sqrt 5 }}} \right) + C.\], \[t = \tan \frac{x}{4},\;\; \Rightarrow d\left( {\frac{x}{2}} \right) = \frac{{2dt}}{{1 + {t^2}}},\;\; \Rightarrow \cos \frac{x}{2} = \frac{{1 - {t^2}}}{{1 + {t^2}}}.\], \[\int {\frac{{dx}}{{1 + \cos \frac{x}{2}}}} = \int {\frac{{d\left( {\frac{x}{2}} \right)}}{{1 + \cos \frac{x}{2}}}} = 2\int {\frac{{\frac{{2dt}}{{1 + {t^2}}}}}{{1 + \frac{{1 - {t^2}}}{{1 + {t^2}}}}}} = 4\int {\frac{{dt}}{{1 + \cancel{t^2} + 1 - \cancel{t^2}}}} = 2\int {dt} = 2t + C = 2\tan \frac{x}{4} + C.\], \[t = \tan x,\;\; \Rightarrow x = \arctan t,\;\; \Rightarrow dx = \frac{{dt}}{{1 + {t^2}}},\;\; \Rightarrow \cos 2x = \frac{{1 - {t^2}}}{{1 + {t^2}}},\], \[\int {\frac{{dx}}{{1 + \cos 2x}}} = \int {\frac{{\frac{{dt}}{{1 + {t^2}}}}}{{1 + \frac{{1 - {t^2}}}{{1 + {t^2}}}}}} = \int {\frac{{dt}}{{1 + \cancel{t^2} + 1 - \cancel{t^2}}}} = \int {\frac{{dt}}{2}} = \frac{t}{2} + C = \frac{1}{2}\tan x + C.\], \[t = \tan \frac{x}{4},\;\; \Rightarrow x = 4\arctan t,\;\; dx = \frac{{4dt}}{{1 + {t^2}}},\;\; \cos \frac{x}{2} = \frac{{1 - {t^2}}}{{1 + {t^2}}}.\], \[\int {\frac{{dx}}{{4 + 5\cos \frac{x}{2}}}} = \int {\frac{{\frac{{4dt}}{{1 + {t^2}}}}}{{4 + 5 \cdot \frac{{1 - {t^2}}}{{1 + {t^2}}}}}} = \int {\frac{{4dt}}{{4\left( {1 + {t^2}} \right) + 5\left( {1 - {t^2}} \right)}}} = 4\int {\frac{{dt}}{{4 + 4{t^2} + 5 - 5{t^2}}}} = 4\int {\frac{{dt}}{{{3^2} - {t^2}}}} = 4 \cdot \frac{1}{{2 \cdot 3}}\ln \left| {\frac{{3 + t}}{{3 - t}}} \right| + C = \frac{2}{3}\ln \left| {\frac{{3 + \tan \frac{x}{4}}}{{3 - \tan \frac{x}{4}}}} \right| + C.\], \[\int {\frac{{dx}}{{\sin x + \cos x}}} = \int {\frac{{\frac{{2dt}}{{1 + {t^2}}}}}{{\frac{{2t}}{{1 + {t^2}}} + \frac{{1 - {t^2}}}{{1 + {t^2}}}}}} = \int {\frac{{2dt}}{{2t + 1 - {t^2}}}} = 2\int {\frac{{dt}}{{1 - \left( {{t^2} - 2t} \right)}}} = 2\int {\frac{{dt}}{{1 - \left( {{t^2} - 2t + 1 - 1} \right)}}} = 2\int {\frac{{dt}}{{2 - {{\left( {t - 1} \right)}^2}}}} = 2\int {\frac{{d\left( {t - 1} \right)}}{{{{\left( {\sqrt 2 } \right)}^2} - {{\left( {t - 1} \right)}^2}}}} = 2 \cdot \frac{1}{{2\sqrt 2 }}\ln \left| {\frac{{\sqrt 2 + \left( {t - 1} \right)}}{{\sqrt 2 - \left( {t - 1} \right)}}} \right| + C = \frac{1}{{\sqrt 2 }}\ln \left| {\frac{{\sqrt 2 - 1 + \tan \frac{x}{2}}}{{\sqrt 2 + 1 - \tan \frac{x}{2}}}} \right| + C.\], \[t = \tan \frac{x}{2},\;\; \Rightarrow x = 2\arctan t,\;\; dx = \frac{{2dt}}{{1 + {t^2}}},\;\; \sin x = \frac{{2t}}{{1 + {t^2}}},\;\; \cos x = \frac{{1 - {t^2}}}{{1 + {t^2}}}.\], \[\int {\frac{{dx}}{{\sin x + \cos x + 1}}} = \int {\frac{{\frac{{2dt}}{{1 + {t^2}}}}}{{\frac{{2t}}{{1 + {t^2}}} + \frac{{1 - {t^2}}}{{1 + {t^2}}} + 1}}} = \int {\frac{{\frac{{2dt}}{{1 + {t^2}}}}}{{\frac{{2t + 1 - {t^2} + 1 + {t^2}}}{{1 + {t^2}}}}}} = \int {\frac{{2dt}}{{2t + 2}}} = \int {\frac{{dt}}{{t + 1}}} = \ln \left| {t + 1} \right| + C = \ln \left| {\tan \frac{x}{2} + 1} \right| + C.\], \[I = \int {\frac{{dx}}{{\sec x + 1}}} = \int {\frac{{dx}}{{\frac{1}{{\cos x}} + 1}}} = \int {\frac{{\cos xdx}}{{1 + \cos x}}} .\], \[I = \int {\frac{{\cos xdx}}{{1 + \cos x}}} = \int {\frac{{\frac{{1 - {t^2}}}{{1 + {t^2}}} \cdot \frac{{2dt}}{{1 + {t^2}}}}}{{1 + \frac{{1 - {t^2}}}{{1 + {t^2}}}}}} = 2\int {\frac{{\frac{{1 - {t^2}}}{{{{\left( {1 + {t^2}} \right)}^2}}}dt}}{{\frac{{1 + {t^2} + 1 - {t^2}}}{{1 + {t^2}}}}}} = \int {\frac{{1 - {t^2}}}{{1 + {t^2}}}dt} = - \int {\frac{{1 + {t^2} - 2}}{{1 + {t^2}}}dt} = - \int {1dt} + 2\int {\frac{{dt}}{{1 + {t^2}}}} = - t + 2\arctan t + C = - \tan \frac{x}{2} + 2\arctan \left( {\tan \frac{x}{2}} \right) + C = x - \tan \frac{x}{2} + C.\], Trigonometric and Hyperbolic Substitutions. Proof Chasles Theorem and Euler's Theorem Derivation . = 0 + 2\,\frac{dt}{1 + t^{2}} This allows us to write the latter as rational functions of t (solutions are given below). Polynomial functions are simple functions that even computers can easily process, hence the Weierstrass Approximation theorem has great practical as well as theoretical utility. An irreducibe cubic with a flex can be affinely transformed into a Weierstrass equation: Y 2 + a 1 X Y + a 3 Y = X 3 + a 2 X 2 + a 4 X + a 6. . \text{cos}x&=\frac{1-u^2}{1+u^2} \\ The Weierstrass Approximation theorem = @robjohn : No, it's not "really the Weierstrass" since call the tangent half-angle substitution "the Weierstrass substitution" is incorrect. \end{align} The Weierstrass substitution is an application of Integration by Substitution . [1] 6. My question is, from that chapter, can someone please explain to me how algebraically the $\frac{\theta}{2}$ angle is derived? {\textstyle \csc x-\cot x=\tan {\tfrac {x}{2}}\colon }. Finally, since t=tan(x2), solving for x yields that x=2arctant. dx&=\frac{2du}{1+u^2} rev2023.3.3.43278. {\displaystyle a={\tfrac {1}{2}}(p+q)} sin Retrieved 2020-04-01. Then we have. Why do academics stay as adjuncts for years rather than move around? (2/2) The tangent half-angle substitution illustrated as stereographic projection of the circle. and Definition of Bernstein Polynomial: If f is a real valued function defined on [0, 1], then for n N, the nth Bernstein Polynomial of f is defined as, Proof: To prove the theorem on closed intervals [a,b], without loss of generality we can take the closed interval as [0, 1]. Site design / logo 2023 Stack Exchange Inc; user contributions licensed under CC BY-SA. {\textstyle t=\tan {\tfrac {x}{2}}} A point on (the right branch of) a hyperbola is given by(cosh , sinh ). . As t goes from 1 to0, the point follows the part of the circle in the fourth quadrant from (0,1) to(1,0). WEIERSTRASS APPROXIMATION THEOREM TL welll kroorn Neiendsaas . {\textstyle t=\tan {\tfrac {x}{2}},} \begin{align*} Note sur l'intgration de la fonction, https://archive.org/details/coursdanalysedel01hermuoft/page/320/, https://archive.org/details/anelementarytre00johngoog/page/n66, https://archive.org/details/traitdanalyse03picagoog/page/77, https://archive.org/details/courseinmathemat01gouruoft/page/236, https://archive.org/details/advancedcalculus00wils/page/21/, https://archive.org/details/treatiseonintegr01edwauoft/page/188, https://archive.org/details/ost-math-courant-differentialintegralcalculusvoli/page/n250, https://archive.org/details/elementsofcalcul00pete/page/201/, https://archive.org/details/calculus0000apos/page/264/, https://archive.org/details/calculuswithanal02edswok/page/482, https://archive.org/details/calculusofsingle00lars/page/520, https://books.google.com/books?id=rn4paEb8izYC&pg=PA435, https://books.google.com/books?id=R-1ZEAAAQBAJ&pg=PA409, "The evaluation of trigonometric integrals avoiding spurious discontinuities", "A Note on the History of Trigonometric Functions", https://en.wikipedia.org/w/index.php?title=Tangent_half-angle_substitution&oldid=1137371172, This page was last edited on 4 February 2023, at 07:50. &= \frac{\sec^2 \frac{x}{2}}{(a + b) + (a - b) \tan^2 \frac{x}{2}}, Weierstrass' preparation theorem. A geometric proof of the Weierstrass substitution In various applications of trigonometry , it is useful to rewrite the trigonometric functions (such as sine and cosine ) in terms of rational functions of a new variable t {\displaystyle t} . [4], The substitution is described in most integral calculus textbooks since the late 19th century, usually without any special name. ( the sum of the first n odds is n square proof by induction. ) The integral on the left is $-\cot x$ and the one on the right is an easy $u$-sub with $u=\sin x$. 2 answers Score on last attempt: \( \quad 1 \) out of 3 Score in gradebook: 1 out of 3 At the beginning of 2000 , Miguel's house was worth 238 thousand dollars and Kyle's house was worth 126 thousand dollars. The secant integral may be evaluated in a similar manner. \text{tan}x&=\frac{2u}{1-u^2} \\ csc As t goes from 0 to 1, the point follows the part of the circle in the first quadrant from (1,0) to(0,1). Draw the unit circle, and let P be the point (1, 0). Alternatives for evaluating $ \int \frac { 1 } { 5 + 4 \cos x} \ dx $ ?? ) Let M = ||f|| exists as f is a continuous function on a compact set [0, 1]. Mathematica GuideBook for Symbolics. Now, fix [0, 1]. tanh 1 derivatives are zero). x d the \(X^2\) term (whereas if \(\mathrm{char} K = 3\) we can eliminate either the \(X^2\) Other sources refer to them merely as the half-angle formulas or half-angle formulae . 2 So you are integrating sum from 0 to infinity of (-1) n * t 2n / (2n+1) dt which is equal to the sum form 0 to infinity of (-1) n *t 2n+1 / (2n+1) 2 . cos We show how to obtain the difference function of the Weierstrass zeta function very directly, by choosing an appropriate order of summation in the series defining this function. ( Assume \(\mathrm{char} K \ne 3\) (otherwise the curve is the same as \((X + Y)^3 = 1\)). by setting p.431. $$\int\frac{dx}{a+b\cos x}=\frac1a\int\frac{dx}{1+\frac ba\cos x}=\frac1a\int\frac{d\nu}{1+\left|\frac ba\right|\cos\nu}$$ Now, add and subtract $b^2$ to the denominator and group the $+b^2$ with $-b^2\cos^2x$. , differentiation rules imply. tan \(\text{cos}\theta=\frac{BC}{AB}=\frac{1-u^2}{1+u^2}\). (1) F(x) = R x2 1 tdt. 2.3.8), which is an effective substitute for the Completeness Axiom, can easily be extended from sequences of numbers to sequences of points: Proposition 2.3.7 (Bolzano-Weierstrass Theorem). . {\textstyle \int d\psi \,H(\sin \psi ,\cos \psi ){\big /}{\sqrt {G(\sin \psi ,\cos \psi )}}} \(\Delta = -b_2^2 b_8 - 8b_4^3 - 27b_4^2 + 9b_2 b_4 b_6\).