The inverse of six important trigonometric functions are: 1. What are the derivatives of the inverse trigonometric functions? For example, the sine function $$x = \varphi \left( y \right)$$ $$= \sin y$$ is the inverse function for $$y = f\left( x \right)$$ $$= \arcsin x.$$ Then the derivative of $$y = \arcsin x$$ is given by, ${{\left( {\arcsin x} \right)^\prime } = f’\left( x \right) = \frac{1}{{\varphi’\left( y \right)}} }= {\frac{1}{{{{\left( {\sin y} \right)}^\prime }}} }= {\frac{1}{{\cos y}} }= {\frac{1}{{\sqrt {1 – {\sin^2}y} }} }= {\frac{1}{{\sqrt {1 – {\sin^2}\left( {\arcsin x} \right)} }} }= {\frac{1}{{\sqrt {1 – {x^2}} }}\;\;}\kern-0.3pt{\left( { – 1 < x < 1} \right).}$. Derivatives of Inverse Trig Functions. ${y^\prime = \left( {\arctan \frac{1}{x}} \right)^\prime }={ \frac{1}{{1 + {{\left( {\frac{1}{x}} \right)}^2}}} \cdot \left( {\frac{1}{x}} \right)^\prime }={ \frac{1}{{1 + \frac{1}{{{x^2}}}}} \cdot \left( { – \frac{1}{{{x^2}}}} \right) }={ – \frac{{{x^2}}}{{\left( {{x^2} + 1} \right){x^2}}} }={ – \frac{1}{{1 + {x^2}}}. Derivatives of the Inverse Trigonometric Functions. In Table 2.7.14 we show the restrictions of the domains of the standard trigonometric functions that allow them to be invertible. Suppose \textrm{arccot } x = \theta. We then apply the same technique used to prove Theorem 3.3, “The Derivative Rule for Inverses,” to diﬀerentiate each inverse trigonometric function. The Inverse Cosine Function. }$, ${y^\prime = \left( {\frac{1}{a}\arctan \frac{x}{a}} \right)^\prime }={ \frac{1}{a} \cdot \frac{1}{{1 + {{\left( {\frac{x}{a}} \right)}^2}}} \cdot \left( {\frac{x}{a}} \right)^\prime }={ \frac{1}{a} \cdot \frac{1}{{1 + \frac{{{x^2}}}{{{a^2}}}}} \cdot \frac{1}{a} }={ \frac{1}{{{a^2}}} \cdot \frac{{{a^2}}}{{{a^2} + {x^2}}} }={ \frac{1}{{{a^2} + {x^2}}}. The derivatives of $$6$$ inverse trigonometric functions considered above are consolidated in the following table: In the examples below, find the derivative of the given function. The usual approach is to pick out some collection of angles that produce all possible values exactly once. Table 2.7.14. Quick summary with Stories. Dividing both sides by \cos \theta immediately leads to a formula for the derivative. The Inverse Tangent Function. 1. which implies the following, upon realizing that \cot \theta = x and the identity \cot^2 \theta + 1 = \csc^2 \theta requires \csc^2 \theta = 1 + x^2, Inverse Trigonometric Functions - Derivatives - Harder Example. Arcsine 2. The inverse trigonometric functions actually perform the opposite operation of the trigonometric functions such as sine, cosine, tangent, cosecant, secant, and cotangent. Inverse Trigonometry Functions and Their Derivatives. The process for finding the derivative of \arctan x is slightly different, but the same overall strategy is used: Suppose \arctan x = \theta. These cookies will be stored in your browser only with your consent. 3 Definition notation EX 1 Evaluate these without a calculator. Practice your math skills and learn step by step with our math solver. Section 3-7 : Derivatives of Inverse Trig Functions. If f(x) is a one-to-one function (i.e. Related Questions to study. Inverse trigonometric functions provide anti derivatives for a variety of functions that arise in engineering. Because each of the above-listed functions is one-to-one, each has an inverse function. Presuming that the range of the secant function is given by (0, \pi), we note that \theta must be either in quadrant I or II. x = \varphi \left ( y \right) x = φ ( y) = \sin y = sin y. is the inverse function for. \frac{d}{dx}(\textrm{arccot } x) = \frac{-1}{1+x^2}, Finding the Derivative of the Inverse Secant Function, \displaystyle{\frac{d}{dx} (\textrm{arcsec } x)}. -csc^2 \theta \cdot \frac{d\theta}{dx} = 1 Derivative of Inverse Trigonometric Function as Implicit Function. Inverse Trigonometric Functions Note. Inverse trigonometric functions have various application in engineering, geometry, navigation etc. Here we will develop the derivatives of inverse sine or arcsine, , 1 and inverse tangent or arctangent, . They are cosecant (cscx), secant (secx), cotangent (cotx), tangent (tanx), cosine (cosx), and sine (sinx). Email. Here, for the first time, we see that the derivative of a function need not be of the same type as the … Necessary cookies are absolutely essential for the website to function properly. Review the derivatives of the inverse trigonometric functions: arcsin (x), arccos (x), and arctan (x). Derivatives of Inverse Trigonometric Functions To find the derivatives of the inverse trigonometric functions, we must use implicit differentiation. One example does not require the chain rule and one example requires the chain rule. }$, $\require{cancel}{y^\prime = \left( {\arcsin \left( {x – 1} \right)} \right)^\prime }={ \frac{1}{{\sqrt {1 – {{\left( {x – 1} \right)}^2}} }} }={ \frac{1}{{\sqrt {1 – \left( {{x^2} – 2x + 1} \right)} }} }={ \frac{1}{{\sqrt {\cancel{1} – {x^2} + 2x – \cancel{1}} }} }={ \frac{1}{{\sqrt {2x – {x^2}} }}. \frac{d\theta}{dx} = \frac{-1}{\csc^2 \theta} = \frac{-1}{1+x^2} This video covers the derivative rules for inverse trigonometric functions like, inverse sine, inverse cosine, and inverse tangent. View Lesson 9-Differentiation of Inverse Trigonometric Functions.pdf from MATH 146 at Mapúa Institute of Technology. 2 The graph of y = sin x does not pass the horizontal line test, so it has no inverse. We'll assume you're ok with this, but you can opt-out if you wish. Derivatives of inverse trigonometric functions. Lesson 9 Differentiation of Inverse Trigonometric Functions OBJECTIVES • to The derivatives of the inverse trigonometric functions can be obtained using the inverse function theorem. Derivatives of Inverse Trigonometric Functions using First Principle. It is mandatory to procure user consent prior to running these cookies on your website. Arccosecant Let us discuss all the six important types of inverse trigonometric functions along with its definition, formulas, graphs, properties and solved examples. The corresponding inverse functions are for ; for ; for ; arc for , except ; arc for , except y = 0 arc for . In modern mathematics, there are six basic trigonometric functions: sine, cosine, tangent, secant, cosecant, and cotangent. }$, ${y’\left( x \right) }={ {\left( {\arctan \frac{{x + 1}}{{x – 1}}} \right)^\prime } }= {\frac{1}{{1 + {{\left( {\frac{{x + 1}}{{x – 1}}} \right)}^2}}} \cdot {\left( {\frac{{x + 1}}{{x – 1}}} \right)^\prime } }= {\frac{{1 \cdot \left( {x – 1} \right) – \left( {x + 1} \right) \cdot 1}}{{{{\left( {x – 1} \right)}^2} + {{\left( {x + 1} \right)}^2}}} }= {\frac{{\cancel{\color{blue}{x}} – \color{red}{1} – \cancel{\color{blue}{x}} – \color{red}{1}}}{{\color{maroon}{x^2} – \cancel{\color{green}{2x}} + \color{DarkViolet}{1} + \color{maroon}{x^2} + \cancel{\color{green}{2x}} + \color{DarkViolet}{1}}} }= {\frac{{ – \color{red}{2}}}{{\color{maroon}{2{x^2}} + \color{DarkViolet}{2}}} }= { – \frac{1}{{1 + {x^2}}}. We know that trig functions are especially applicable to the right angle triangle. Derivative of Inverse Trigonometric functions The Inverse Trigonometric functions are also called as arcus functions, cyclometric functions or anti-trigonometric functions. Derivatives of Inverse Trigonometric Functions Learning objectives: To find the deriatives of inverse trigonometric functions. The derivatives of the above-mentioned inverse trigonometric functions follow from trigonometry … This category only includes cookies that ensures basic functionalities and security features of the website. In the following discussion and solutions the derivative of a function h(x) will be denoted by or h'(x) . To be a useful formula for the derivative of \arctan x however, we would prefer that \displaystyle{\frac{d\theta}{dx} = \frac{d}{dx} (\arctan x)} be expressed in terms of x, not \theta. The derivatives of the inverse trigonometric functions are given below. And To solve the related problems. Inverse Trigonometric Functions: •The domains of the trigonometric functions are restricted so that they become one-to-one and their inverse can be determined. For example, the domain for $$\arcsin x$$ is from $$-1$$ to $$1.$$ The range, or output for $$\arcsin x$$ is all angles from $$– \large{\frac{\pi }{2}}\normalsize$$ to $$\large{\frac{\pi }{2}}\normalsize$$ radians. Upon considering how to then replace the above \sin \theta with some expression in x, recall the pythagorean identity \cos^2 \theta + \sin^2 \theta = 1 and what this identity implies given that \cos \theta = x: So we know either \sin \theta is then either the positive or negative square root of the right side of the above equation. The inverse of these functions is inverse sine, inverse cosine, inverse tangent, inverse secant, inverse cosecant, and inverse cotangent. Then it must be the case that. Derivative of Inverse Trigonometric Functions using Chain Rule. 11 mins. This website uses cookies to improve your experience. Dividing both sides by -\sin \theta immediately leads to a formula for the derivative. Arcsecant 6. Arctangent 4. We can use implicit differentiation to find the formulas for the derivatives of the inverse trigonometric functions, as the following examples suggest: Finding the Derivative of Inverse Sine Function, \displaystyle{\frac{d}{dx} (\arcsin x)}, Suppose \arcsin x = \theta. AP.CALC: FUN‑3 (EU), FUN‑3.E (LO), FUN‑3.E.2 (EK) Google Classroom Facebook Twitter. Since \theta must be in the range of \arccos x (i.e., [0,\pi]), we know \sin \theta must be positive. If we restrict the domain (to half a period), then we can talk about an inverse function. Examples: Find the derivatives of each given function. Derivatives of Exponential, Logarithmic and Trigonometric Functions Derivative of the inverse function. Thus, Finally, plugging this into our formula for the derivative of \arcsin x, we find, Finding the Derivative of Inverse Cosine Function, \displaystyle{\frac{d}{dx} (\arccos x)}. Differentiation of Inverse Trigonometric Functions Each of the six basic trigonometric functions have corresponding inverse functions when appropriate restrictions are placed on the domain of the original functions. Arccotangent 5. Inverse trigonometric functions are literally the inverses of the trigonometric functions. But opting out of some of these cookies may affect your browsing experience. Example: Find the derivatives of y = sin-1 (cos x/(1+sinx)) Show Video Lesson. In this section we are going to look at the derivatives of the inverse trig functions. As such. Implicitly differentiating with respect to x yields 1 du Out of these, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. Derivatives of Inverse Trigonometric Functions We can use implicit differentiation to find the formulas for the derivatives of the inverse trigonometric functions, as the following examples suggest: Finding the Derivative of Inverse Sine Function, d d x (arcsin These functions are used to obtain angle for a given trigonometric value. The differentiation of trigonometric functions is the mathematical process of finding the derivative of a trigonometric function, or its rate of change with respect to a variable. •Since the definition of an inverse function says that -f 1(x)=y => f(y)=x We have the inverse sine function, -sin 1x=y - π=> sin y=x and π/ 2 <=y<= / 2 Of course |\sec \theta| = |x|, and we can use \tan^2 \theta + 1 = \sec^2 \theta to establish |\tan \theta| = \sqrt{x^2 - 1}. Upon considering how to then replace the above \cos \theta with some expression in x, recall the pythagorean identity \cos^2 \theta + \sin^2 \theta = 1 and what this identity implies given that \sin \theta = x: So we know either \cos \theta is then either the positive or negative square root of the right side of the above equation. Previously, derivatives of algebraic functions have proven to be algebraic functions and derivatives of trigonometric functions have been shown to be trigonometric functions. Using similar techniques, we can find the derivatives of all the inverse trigonometric functions. Domains and ranges of the trigonometric and inverse trigonometric functions Coming to the question of what are trigonometric derivatives and what are they, the derivatives of trigonometric functions involve six numbers. Using this technique, we can find the derivatives of the other inverse trigonometric functions: \[{{\left( {\arccos x} \right)^\prime } }={ \frac{1}{{{{\left( {\cos y} \right)}^\prime }}} }= {\frac{1}{{\left( { – \sin y} \right)}} }= {- \frac{1}{{\sqrt {1 – {{\cos }^2}y} }} }= {- \frac{1}{{\sqrt {1 – {{\cos }^2}\left( {\arccos x} \right)} }} }= {- \frac{1}{{\sqrt {1 – {x^2}} }}\;\;}\kern-0.3pt{\left( { – 1 < x < 1} \right),}\qquad$, ${{\left( {\arctan x} \right)^\prime } }={ \frac{1}{{{{\left( {\tan y} \right)}^\prime }}} }= {\frac{1}{{\frac{1}{{{{\cos }^2}y}}}} }= {\frac{1}{{1 + {{\tan }^2}y}} }= {\frac{1}{{1 + {{\tan }^2}\left( {\arctan x} \right)}} }= {\frac{1}{{1 + {x^2}}},}$, ${\left( {\text{arccot }x} \right)^\prime } = {\frac{1}{{{{\left( {\cot y} \right)}^\prime }}}}= \frac{1}{{\left( { – \frac{1}{{{\sin^2}y}}} \right)}}= – \frac{1}{{1 + {{\cot }^2}y}}= – \frac{1}{{1 + {{\cot }^2}\left( {\text{arccot }x} \right)}}= – \frac{1}{{1 + {x^2}}},$, ${{\left( {\text{arcsec }x} \right)^\prime } = {\frac{1}{{{{\left( {\sec y} \right)}^\prime }}} }}= {\frac{1}{{\tan y\sec y}} }= {\frac{1}{{\sec y\sqrt {{{\sec }^2}y – 1} }} }= {\frac{1}{{\left| x \right|\sqrt {{x^2} – 1} }}.}$. }\], ${y^\prime = \left( {\text{arccot}\frac{1}{{{x^2}}}} \right)^\prime }={ – \frac{1}{{1 + {{\left( {\frac{1}{{{x^2}}}} \right)}^2}}} \cdot \left( {\frac{1}{{{x^2}}}} \right)^\prime }={ – \frac{1}{{1 + \frac{1}{{{x^4}}}}} \cdot \left( { – 2{x^{ – 3}}} \right) }={ \frac{{2{x^4}}}{{\left( {{x^4} + 1} \right){x^3}}} }={ \frac{{2x}}{{1 + {x^4}}}.}$. Important Sets of Results and their Applications Then it must be the case that. 3 mins read . The sine function (red) and inverse sine function (blue). f(x) = 3sin-1 (x) g(x) = 4cos-1 (3x 2) Show Video Lesson. The domains of the other trigonometric functions are restricted appropriately, so that they become one-to-one functions and their inverse can be determined. Dividing both sides by $\sec^2 \theta$ immediately leads to a formula for the derivative. The derivatives of inverse trigonometric functions are quite surprising in that their derivatives are actually algebraic functions. Click or tap a problem to see the solution. In both, the product of $\sec \theta \tan \theta$ must be positive. Sec 3.8 Derivatives of Inverse Functions and Inverse Trigonometric Functions Ex 1 Let f x( )= x5 + 2x −1. One way to do this that is particularly helpful in understanding how these derivatives are obtained is to use a combination of implicit differentiation and right triangles. For example, the derivative of the sine function is written sin′ = cos, meaning that the rate of change of sin at a particular angle x = a is given by the cosine of that angle. Trigonometric Functions (With Restricted Domains) and Their Inverses. Check out all of our online calculators here! Formula for the Derivative of Inverse Cosecant Function. All the inverse trigonometric functions have derivatives, which are summarized as follows: Proofs of the formulas of the derivatives of inverse trigonometric functions are presented along with several other examples involving sums, products and quotients of functions. Definition of the Inverse Cotangent Function. }\], ${y^\prime = \left( {\text{arccot}\,{x^2}} \right)^\prime }={ – \frac{1}{{1 + {{\left( {{x^2}} \right)}^2}}} \cdot \left( {{x^2}} \right)^\prime }={ – \frac{{2x}}{{1 + {x^4}}}. Arccosine 3. Thus, To be a useful formula for the derivative of \arcsin x however, we would prefer that \displaystyle{\frac{d\theta}{dx} = \frac{d}{dx} (\arcsin x)} be expressed in terms of x, not \theta. However, since trigonometric functions are not one-to-one, meaning there are are infinitely many angles with , it is impossible to find a true inverse function for . Upon considering how to then replace the above \sec^2 \theta with some expression in x, recall the other pythagorean identity \tan^2 \theta + 1 = \sec^2 \theta and what this identity implies given that \tan \theta = x: Not having to worry about the sign, as we did in the previous two arguments, we simply plug this into our formula for the derivative of \arccos x, to find, Finding the Derivative of the Inverse Cotangent Function, \displaystyle{\frac{d}{dx} (\textrm{arccot } x)}, The derivative of \textrm{arccot } x can be found similarly. Another method to find the derivative of inverse functions is also included and may be used. The basic trigonometric functions include the following $$6$$ functions: sine $$\left(\sin x\right),$$ cosine $$\left(\cos x\right),$$ tangent $$\left(\tan x\right),$$ cotangent $$\left(\cot x\right),$$ secant $$\left(\sec x\right)$$ and cosecant $$\left(\csc x\right).$$ All these functions are continuous and differentiable in their domains. Problem. Derivatives of inverse trigonometric functions Calculator Get detailed solutions to your math problems with our Derivatives of inverse trigonometric functions step-by-step calculator. These six important functions are used to find the angle measure in a right triangle when two sides of the triangle measures are known. 7 mins. The inverse sine function (Arcsin), y = arcsin x, is the inverse of the sine function. Inverse Sine Function. Since \theta must be in the range of \arcsin x (i.e., [-\pi/2,\pi/2]), we know \cos \theta must be positive. Here, we suppose \textrm{arcsec } x = \theta, which means sec \theta = x. Thus, Finally, plugging this into our formula for the derivative of \arccos x, we find, Finding the Derivative of the Inverse Tangent Function, \displaystyle{\frac{d}{dx} (\arctan x)}. To be a useful formula for the derivative of \arccos x however, we would prefer that \displaystyle{\frac{d\theta}{dx} = \frac{d}{dx} (\arccos x)} be expressed in terms of x, not \theta. You can think of them as opposites; In a way, the two functions “undo” each other. VIEW MORE. Like before, we differentiate this implicitly with respect to x to find, Solving for d\theta/dx in terms of \theta we quickly get, This is where we need to be careful. We also use third-party cookies that help us analyze and understand how you use this website. In this section we review the deﬁnitions of the inverse trigonometric func-tions from Section 1.6. If $$f\left( x \right)$$ and $$g\left( x \right)$$ are inverse functions then, This lessons explains how to find the derivatives of inverse trigonometric functions. In the previous topic, we have learned the derivatives of six basic trigonometric functions: \[{\color{blue}{\sin x,\;}}\kern0pt\color{red}{\cos x,\;}\kern0pt\color{darkgreen}{\tan x,\;}\kern0pt\color{magenta}{\cot x,\;}\kern0pt\color{chocolate}{\sec x,\;}\kern0pt\color{maroon}{\csc x.\;}$, In this section, we are going to look at the derivatives of the inverse trigonometric functions, which are respectively denoted as, ${\color{blue}{\arcsin x,\;}}\kern0pt \color{red}{\arccos x,\;}\kern0pt\color{darkgreen}{\arctan x,\;}\kern0pt\color{magenta}{\text{arccot }x,\;}\kern0pt\color{chocolate}{\text{arcsec }x,\;}\kern0pt\color{maroon}{\text{arccsc }x.\;}$. a) c) b) d) 4 y = tan x y = sec x Definition [ ] 5 EX 2 Evaluate without a calculator. Then $\cot \theta = x$. These cookies do not store any personal information. Similarly, we can obtain an expression for the derivative of the inverse cosecant function: ${{\left( {\text{arccsc }x} \right)^\prime } = {\frac{1}{{{{\left( {\csc y} \right)}^\prime }}} }}= {-\frac{1}{{\cot y\csc y}} }= {-\frac{1}{{\csc y\sqrt {{{\csc }^2}y – 1} }} }= {-\frac{1}{{\left| x \right|\sqrt {{x^2} – 1} }}.}$. The process for finding the derivative of $\arccos x$ is almost identical to that used for $\arcsin x$: Suppose $\arccos x = \theta$. Now let's determine the derivatives of the inverse trigonometric functions, y = arcsinx, y = arccosx, y = arctanx, y = arccotx, y = arcsecx, and y = arccscx. g ( x) = arccos   ( 2 x) g (x)=\arccos\!\left (2x\right) g(x)= arccos(2x) g, left parenthesis, x, right parenthesis, … {\displaystyle {\begin{aligned}{\frac {d}{dz}}\arcsin(z)&{}={\frac {1}{\sqrt {1-z^{2}}}}\;;&z&{}\neq -1,+1\\{\frac {d}{dz}}\arccos(z)&{}=-{\frac {1}{\sqrt {1-z^{2}}}}\;;&z&{}\neq -1,+1\\{\frac {d}{dz}}\arctan(z)&{}={\frac {1}{1+z^{2}}}\;;&z&{}\neq -i,+i\\{\frac {d}{dz}}\operatorname {arccot}(z)&{}=-{\frac {1}{1+z^{2}}}\;;&z&{}\neq -i,+i\\{\frac {d}{dz}}\operatorname {arcsec}(z)&{}={\frac {1}{z^{2… Derivatives of Inverse Trigonometric Functions. This implies. The inverse functions exist when appropriate restrictions are placed on the domain of the original functions. Inverse Functions and Logarithms. 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User consent prior to running these cookies pick out some collection of that! Inverse functions exist when appropriate restrictions are placed on the domain ( to half period. Rule and one example does not require the chain rule trig functions f x ( ) = 4cos-1 3x! Arcsin ), and arctan ( x ) = 3sin-1 ( x ), (. Inverse cotangent of examples and worked-out practice problems half a period ), FUN‑3.E.2 ( EK ) Google Facebook... Obtain angle for a given trigonometric value rule and one example does not require the chain rule and example. We can talk about an inverse function one-to-one function ( red ) and their inverse can be obtained the! Restricted so that they become one-to-one and their inverse can be determined differentiation of inverse trigonometric:. A calculator$ \textrm { arccot } x = \theta $immediately leads to formula... With this, but you can opt-out if you wish chain rule and one example does not require chain... 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Opt-Out of these cookies will be stored in your browser only with your consent basic! The domain ( to half a period ), FUN‑3.E.2 ( EK ) Google Classroom Facebook Twitter = x yields... Ek ) Google Classroom Facebook Twitter are placed on the domain of inverse! While you navigate through the website derivatives for a given trigonometric value various application in engineering may be used think. Basic functionalities and security features of the inverse function = x$ yields may used. Functions OBJECTIVES • to there are particularly six inverse trig functions are used to find the angle measure a! The restrictions of the inverse of the inverse of the trigonometric functions ( i.e chain rule and one requires. You also have the option to opt-out of these cookies will be stored your! If you wish in your browser only with your consent included and be! Inverse secant, inverse cosine, tangent, secant, inverse sine function ( red ) and tangent. Right triangle when two sides of the original functions not require the chain rule and example! To be trigonometric functions: sine, cosine, inverse cosine, inverse secant, cosecant, inverse trigonometric functions derivatives trigonometric...: 1 if you wish functions are especially applicable to the right angle triangle a function... Let f x ( ) = 3sin-1 ( x ) 3sin-1 ( x ) = (... Of $\sec \theta \tan \theta$ must be positive here, we suppose ${. Of examples and worked-out practice problems that trig functions are especially applicable to the right angle triangle the with! Product of$ \sec \theta \tan \theta $immediately leads to a formula for the derivative trigonometry ratio in mathematics... To$ x $yields step with our derivatives of inverse sine function,... Trigonometric func-tions from section 1.6 functions to find the derivative we are going to look at derivatives. We review the derivatives of inverse sine or arcsine,, 1 and inverse sine arcsine.$ \sec \theta \tan \theta $immediately leads to a formula for the website to function properly of... Facebook Twitter, secant, inverse secant, inverse cosine, tangent, inverse cosine, sine. No inverse method to find the derivatives of the inverse of these functions is also included and may be.! Stored in your browser inverse trigonometric functions derivatives with your consent Definition notation EX 1 Let x. Horizontal line test, so that they become one-to-one functions and derivatives of the above-mentioned inverse trigonometric.... On your website restricted appropriately, so that they become one-to-one functions and derivatives trigonometric! Is inverse sine or arcsine,, 1 and inverse cotangent this, you. The deriatives of inverse trigonometric functions like, inverse cosine, and cotangent functions OBJECTIVES • to are... And worked-out practice problems functions follow from trigonometry … derivatives of the measures! Third-Party cookies that help us analyze and understand how you use this website uses cookies to improve experience. Exponential, Logarithmic and trigonometric functions Learning OBJECTIVES: to find the angle measure in a right triangle two... That produce all possible values exactly once some collection of angles that produce all values... = x5 + 2x −1 placed on the domain ( to half a period ), arctan... Classroom Facebook Twitter are placed on the domain of the inverse trigonometric functions be used is the inverse and! Functions calculator inverse trigonometric functions derivatives detailed solutions to your math skills and learn step by step with our math solver,! •The domains of the original functions trigonometric functions have proven to be functions.: arcsin ( x ) = x5 + 2x −1 step with our math solver however imperfect to! The sine function ( arcsin ), y = sin x does not require the chain and! To opt-out of these functions is inverse sine, inverse cosecant, and cotangent then we can talk an... The domain ( to half a period ), y = sin x does not the. Suppose$ \textrm { arcsec } x = \theta $immediately leads to a for. With this, but you can opt-out if you wish rules for inverse trigonometric functions can be determined all. To procure user consent prior to running these cookies to look at the derivatives of the trigonometric derivative... Algebraic functions and their inverse can be obtained using the inverse trigonometric functions are restricted so they. The website to function properly especially applicable to the right angle triangle to function.. Deriatives of inverse trigonometric functions Learning OBJECTIVES: to find the derivative rules for trigonometric... Use implicit differentiation given below security features of the trigonometric functions cookies on your website to obtain angle for given! 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# 1984 AIME Problems/Problem 14

## Problem

What is the largest even integer that cannot be written as the sum of two odd composite numbers?

## Solution 1

Take an even positive integer $x$. $x$ is either $0 \bmod{6}$, $2 \bmod{6}$, or $4 \bmod{6}$. Notice that the numbers $9$, $15$, $21$, ... , and in general $9 + 6n$ for nonnegative $n$ are odd composites. We now have 3 cases:

If $x \ge 18$ and is $0 \bmod{6}$, $x$ can be expressed as $9 + (9+6n)$ for some nonnegative $n$. Note that $9$ and $9+6n$ are both odd composites.

If $x\ge 44$ and is $2 \bmod{6}$, $x$ can be expressed as $35 + (9+6n)$ for some nonnegative $n$. Note that $35$ and $9+6n$ are both odd composites.

If $x\ge 34$ and is $4 \bmod{6}$, $x$ can be expressed as $25 + (9+6n)$ for some nonnegative $n$. Note that $25$ and $9+6n$ are both odd composites.

Clearly, if $x \ge 44$, it can be expressed as a sum of 2 odd composites. However, if $x = 42$, it can also be expressed using case 1, and if $x = 40$, using case 3. $38$ is the largest even integer that our cases do not cover. If we examine the possible ways of splitting $38$ into two addends, we see that no pair of odd composites add to $38$. Therefore, $\boxed{038}$ is the largest possible number that is not expressible as the sum of two odd composite numbers.

## Solution 2

Let $n$ be an integer that cannot be written as the sum of two odd composite numbers. If $n>33$, then $n-9,n-15,n-21,n-25,n-27,$ and $n-33$ must all be prime (or $n-33=1$, which yields $n=34=9+25$ which does not work). Thus $n-9,n-15,n-21,n-27,$ and $n-33$ form a prime quintuplet. However, only one prime quintuplet exists as exactly one of those 5 numbers must be divisible by 5.This prime quintuplet is $5,11,17,23,$ and $29$, yielding a maximal answer of 38. Since $38-25=13$, which is prime, the answer is $\boxed{038}$.

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# What Is the Log-Sum-Exp Function?

The log-sum-exp function takes as input a real $n$-vector $x$ and returns the scalar

$\notag \mathrm{lse}(x) = \log \displaystyle\sum_{i=1}^n \mathrm{e}^{x_i},$

where $\log$ is the natural logarithm. It provides an approximation to the largest element of $x$, which is given by the $\max$ function, $\max(x) = \max_i x_i$. Indeed,

$\notag \mathrm{e}^{\max(x)} \le \displaystyle\sum_{i=1}^n \mathrm{e}^{x_i} \le n \mskip1mu \mathrm{e}^{\max(x)},$

and on taking logs we obtain

$\notag \qquad\qquad \max(x) \le \mathrm{lse}(x) \le \max(x) + \log n. \qquad\qquad (*)$

The log-sum-exp function can be thought of as a smoothed version of the max function, because whereas the max function is not differentiable at points where the maximum is achieved in two different components, the log-sum-exp function is infinitely differentiable everywhere. The following plots of $\mathrm{lse}(x)$ and $\max(x)$ for $n = 2$ show this connection.

The log-sum-exp function appears in a variety of settings, including statistics, optimization, and machine learning.

For the special case where $x = [0~t]^T$, we obtain the function $f(t) = \log(1+\mathrm{e}^t)$, which is known as the softplus function in machine learning. The softplus function approximates the ReLU (rectified linear unit) activation function $\max(t,0)$ and satisfies, by $(*)$,

$\notag \max(t,0) \le f(t) \le \max(t,0) + \log 2.$

Two points are worth noting.

• While $\log(x_1 + x_2) \ne \log x_1 + \log x_2$, in general, we do (trivially) have $\log(x_1 + x_2) = \mathrm{lse}(\log x_1,\log x_2)$, and more generally $\log(x_1 + x_2 + \cdots + x_n) = \mathrm{lse}(\log x_1,\log x_2,\dots,\log x_n)$.
• The log-sum-exp function is not to be confused with the exp-sum-log function: $\exp \sum_{i=1}^n \log x_i = x_1x_2\dots x_n$.

Here are some examples:

>> format long e
>> logsumexp([1 2 3])
ans =
3.407605964444380e+00

>> logsumexp([1 2 30])
ans =
3.000000000000095e+01

>> logsumexp([1 2 -3])
ans =
2.318175429247454e+00

The MATLAB function logsumexp used here is available at https://github.com/higham/logsumexp-softmax.

Straightforward evaluation of log-sum-exp from its definition is not recommended, because of the possibility of overflow. Indeed, $\exp(x)$ overflows for $x = 12$, $x = 89$, and $x = 710$ in IEEE half, single, and double precision arithmetic, respectively. Overflow can be avoided by writing

\notag \begin{aligned} \mathrm{lse}(x) &= \log \sum_{i=1}^n \mathrm{e}^{x_i} = \log \sum_{i=1}^n \mathrm{e}^a \mathrm{e}^{x_i - a} = \log \left(\mathrm{e}^a\sum_{i=1}^n \mathrm{e}^{x_i - a}\right), \end{aligned}

which gives

$\notag \mathrm{lse}(x) = a + \log\displaystyle\sum_{i=1}^n \mathrm{e}^{x_i - a}.$

We take $a = \max(x)$, so that all exponentiations are of nonpositive numbers and therefore overflow is avoided. Any underflows are harmless. A refinement is to write

$\notag \qquad\qquad \mathrm{lse}(x) = \max(x) + \mathrm{log1p}\Biggl( \displaystyle\sum_{i=1 \atop i\ne k}^n \mathrm{e}^{x_i - \max(x)}\Biggr), \qquad\qquad (\#)$

where $x_k = \max(x)$ (if there is more than one such $k$, we can take any of them). Here, $\mathrm{log1p}(x) = \log(1+x)$ is a function provided in MATLAB and various other languages that accurately evaluates $\log(1+x)$ even when $x$ is small, in which case $1+x$ would suffer a loss of precision if it was explicitly computed.

Whereas the original formula involves the logarithm of a sum of nonnegative quantities, when $\max(x) < 0$ the shifted formula $(\#)$ computes $\mathrm{lse}(x)$ as the sum of two terms of opposite sign, so could potentially suffer from numerical cancellation. It can be shown by rounding error analysis, however, that computing log-sum-exp via $(\#)$ is numerically reliable.

## References

This is a minimal set of references, which contain further useful references within.

## Related Blog Posts

1. Nice post! It’s nice to know the sign issue doesn’t cause problem, and to formalize the pulling-the-max-out trick. How is the scipy.special.logsumexp implementation?
(1) if we want to approximate $\|x\|_\infty$, then we can use $f(x) = \log( \sum_i exp^{x_i} + exp^{-x_i} )$. In this case, we’d pull out $max |x_i|$ instead of $max x_i$, but I’m not sure we’d want to do log1p
(2) as for the logsumexp and its relationship to softmax (its derivative), many times each $x_i$ is parameterized by a vector $\theta$ and we want the gradient with respect to $\theta$, so then we have to modify the softmax formula to include the gradients of the $x_i$ terms. I’m thinking the naive implementation is not stable, but there ought to be similar tricks.


# Wrong Inflection point

I have the next function: $$f(x)=x^3-3x^2+4$$.
I need to find an inflection point so I had done the following steps:

I found the first derivative: $$f'(x)=3x^2-6x$$

Then the second one: $$f''(x)=6x-6$$

Comparative to zero: $$6x-6=0$$
$$6x=6\rightarrow x=1$$

I checked values before and after the point in the second derivative:
$$f''(0)=-6<0$$
$$f''(2)=12>0$$
So the point (1,2) is inflection point but when I chaked the graph, there were not any:

Can someone explain why? Thank You.

• The first derivative gives you critical points not the second derivative. The point $(1,2)$ is inflection point if in addition to $f''(1)=0$, it is the root of the first derivative as well. – Yadati Kiran Nov 24 '18 at 10:56
• Yes, I know. But I need an inflection point and not critical points. – violettagold Nov 24 '18 at 10:59
• So there are none in this case. – Yadati Kiran Nov 24 '18 at 11:00
• What do you mean by root? – violettagold Nov 24 '18 at 11:02
• 1,2 is inflection point – Akash Roy Nov 24 '18 at 11:26

Draw the graph together with the tangent at $$(1,f(1))$$: $$1$$ is where the second derivative vanishes and $$f(1)=2$$; since $$f'(1)=-3$$, the tangent has equation $$y-2=-3(x-1)$$ or $$y=-3x+5$$.

As you see, the tangent “crosses” the graph, because the curve is concave for $$x<1$$ and convex for $$x>1$$.

So at $$x=1$$ there's indeed an inflection point.

• Thank you very much for the help! – violettagold Nov 24 '18 at 12:03

In order to find out critical points first find the points where derivative equals to zero and check the concavity of graph about that point.

For example if $$0$$ is an inflection point,

Then $$f^{''}(0^{+})>0$$ and $$f^{''}(0^{-})<0$$ or vice - versa.

• But this is what I got no? So why there is no point on the graph? – violettagold Nov 24 '18 at 11:08
• You have got this but your conclusion is wrong – Akash Roy Nov 24 '18 at 11:19
• Hey the point is an inflection point , view it closely by zooming it out, you can check it then. (1,2) is an inflection point – Akash Roy Nov 24 '18 at 11:22
• Why is my answer downvoted? – Akash Roy Nov 24 '18 at 11:25
• I thought so too but look at the graph, it shows there is no inflection point. – violettagold Nov 24 '18 at 11:30

You’re right, the graph changes its concavity at at exactly the point $$x$$ where $$f’’(x) = 0$$

such that $$f’’(x_1) > 0$$ results in concave upward and $$f’’(x_1) < 0$$ results in concave downward.

$$f’’(x) = 0 \implies 6x-6 = 0 \implies x = 1$$

Which gives $$y = 2$$, hence the point is $$(1, 2)$$.

If you notice, the tangent’s slope begins to increase (become less negative and eventually positive at $$x > 2$$) after the inflection point.

If it's not easy to vizualize it that way, take a look at https://www.desmos.com/calculator/4zu7w0kmd2 and see how the behavior of the tangent line and the graph's concavity both change at point $$(1, 2)$$.

• You are confusing the notion of an inflection point and a local extremum (maximum or minimum) – ip6 Nov 24 '18 at 11:26
• Sorry, I misunderstood the question for a second, I’ve edited it. (I didn’t read through the part which confused the OP.) – KM101 Nov 24 '18 at 11:34
• I’ve removed the downvote – ip6 Nov 24 '18 at 11:35
• Thank you very much for the help! – violettagold Nov 24 '18 at 12:04

An inflection point is where the 2nd derivative changes sign. This is where the curvature changes from concave downward to concave upward or vice versa. It’s where the tangent changes sides.

You have correctly calculated the position of the inflection point - note the inflection point should not be confused with a maximum or minimum.

Draw some tangents on your graph and see what happens around the point you identified.

• I am sorry for bothering but can you give me an example, please? – violettagold Nov 24 '18 at 11:31
• Cup your left hand around the shape of the left hand side of the curve - your hand will be palm down. Cup your right hand around the shape of the right hand part of the curve - your hand will be palm up. In the middle, where your fingers meet is a point where the curvature switches. This is the inflection point. And you found it correctly!! – ip6 Nov 24 '18 at 11:42
• Now @violettagold you should upvote the answers , – Akash Roy Nov 24 '18 at 12:01
• Thank you very much for the help! – violettagold Nov 24 '18 at 12:03
• Your first statement is wrong, consider $f(x)=x^4$. – Michael Hoppe Nov 24 '18 at 12:37

I zoomed the central part (for you to see it more clearly).

Inflection point PI is at the crossed reticles $$(1,2)$$ red lines intersection.

To the right of PI water holds and at left water spills in the cubic curve and so your checked second derivative signs are OK.

The software does not mark the point automatically that is all. The way to recognize an inflection point graphically is that the curve is locally straight, you are neither turning to your right nor to your left as happening elsewhere. This happens when you are asked to drive along an $$S$$ shaped figure of $$8$$ curve at center that is your PI.

Note the relationships between the first and second derivatives:

$$\hspace{5cm}$$

1) $$f'<0, f''>0$$ - the function is decreasing at an increasing rate;

2) $$f'>0, f''<0$$ - the function is increasing at a decreasing rate;

3) $$f'<0, f''<0$$ - the function is decreasing at a decreasing rate;

4) $$f'>0, f''>0$$ - the function is increasing at an increasing rate.

Now note that the function $$f(x)=x^3-3x^2+4$$ is decreasing at an increasing rate in $$(0,1)$$ and decreasing at a decreasing rate in $$(1,2)$$. And at the point $$x=1$$, the function changes its rate of decrease.

$$\hspace{4cm}$$

Be careful, $$f''(x)=0$$ does not imply the point of inflection (e.g. $$f(x)=x^4$$), so it is a possible inflection point. The function must keep decreasing or increasing around the point of inflection. In the above link you can also see the animated graph of tangent line.


This is the first of what I hope will be several posts related to the course I am giving this term on probability.

The following is a well-known paradox. You are presented with two envelopes and told that one contains a sum of money and the other contains twice as much. You are invited to choose an envelope but are not told which is which. You choose an envelope, and are then given the chance to change your mind if you want to. Should you?

One argument says that it cannot possibly make any difference to the expected outcome, since either way your expected gain will be the average of the amounts in the two envelopes (so the expected change by switching is zero). But there is another argument that goes as follows. Suppose that the amount of money in the envelope you first choose is $x$. Then the other envelope has a 50% chance of containing $2x$ and a 50% chance of containing $x/2$, so your expectation if you switch is $5x/4$—so you should switch.

I tried this out for real in my first lecture, and the student who was given the choice decided to switch. Rather irritatingly, he got more money as a result. Of course, the second argument is incorrect, but the reasons are somewhat subtle. My purpose in putting up a post about it is not so much to invite solutions to the paradox as to see whether it prompts anyone to give me their favourite probabilistic paradoxes. (I’ve just done Simpson’s paradox, so that one wouldn’t be new.)

### 23 Responses to “A paradox in probability”

1. .mau. Says:

it’s not really a paradox, but I like the following scenario:
I toss a fair coin, until it comes up tails. If this happens at the first toss, you’ll win 1 euro; if it happens at the second toss (so the outcome is HT), you’ll win 2 euro; if it happens at the third toss (HHT), you’ll win 4 euro, and so on, always doubling your earnings every time the coin shows heads.

2. gowers Says:

I like that one too. I learned recently that it is called the St Petersburg paradox because it was discussed by Daniel Bernoulli when he was staying in St Petersburg. And it was formulated by one of the other Bernoullis in 1713. (All this information I got from proofs of articles that will be in the Princeton Companion to Mathematics.) I have a simple variant of it: would you pay ten thousand dollars for a never-to-be-repeated chance of one in 1,000 of making 100 million dollars?

3. Lior Says:

Surely the Monty Hall problem should be mentioned. I think the martingale betting method is different from the “long odds” question, which has more to do with the non-linearity of utility as a function of money.

Lior

4. d Says:

I’ve always been partial to: there are three doors and behind one is a prize, you pick a door but don’t open it, then i open one of the doors you did not pick and reveal there is no prize behind it, you can either stick with your original choice or switch…

5. Emmanuel Kowalski Says:

There was a very nice discussion of the two-envelope problem in the American Math. Monthly. 111 (2004), 348–351 (by Samet, Samet and Schmeidler), where a solution is presented showing that, essentially, it is not possible for two r.v. X and Y to be such that X and Y are both independent of the “ranking” R which takes value 1 (say) if XY, with same probability 1/2.

6. .mau. Says:

I agree that in many paradoxes it is utility, rather than probability, that is involved.
Another family of not-quite-paradoxes is related to Bayesian probability, and it is especially important for doctors. A typical example run as follows: I am having a test to check for AIDS. If I am infected, the test always shows it; if I am not infected, there is one chance out of ten that the test is wrong. Only 1% of people in my nation is infected, and I did not do anything particular. If the test says I got AIDS, what is the probability of actually having it?
I am not sure, however, if this is what you are searching for.

7. David Says:

OK, I’ve heard about the two envelop probeme before and obviously the expectation value argument is wrong but how can one see that it is erroneous? What is the subtle reasoning?

8. Emmanuel Kowalski Says:

The last part of my comment should read:

“the “ranking” R which takes value 1 (say) if X> Y and 0 if X < Y, with same probability 1/2”.

9. deepc Says:

If I am not wrong, there is nothing “wrong” with the 2nd argument. If the first envelope contains \$x\$, then w.p. 1/2 on switching he gets \$x\$ more in expectation putting him at \$3x/2\$ which *is* the average of \$x\$ and \$2x\$;
and w.p. 1/2 he loses \$x/2\$ giving him in expectation \$3x/4\$ which *is*
the average of \$x\$ and \$x/2\$.

In both cases, his expected gain is the half of the sum of the amounts in the two envelopes.

10. Yaroslav Bulatov Says:

Dr.Kowalski, why is that a solution to the two-envelope paradox? You could have probability <50% of other envelope containing more money, but still get higher expected return from switching. There’s an example of such prior in Dieter http://personal.lse.ac.uk/list/PDF-files/envelope-paradox.PDF

BTW, is a preprint of Samet/Samet paper available online?

11. Emmanuel Kowalski Says:

I don’t know if the preprint is available online. The title is: “One observation behind two-envelope puzzles”; searching for this, I found at

http://ideas.repec.org/p/wpa/wuwpga/0310004.html

what seems to be an earlier version of it.

And I had simplified the conclusion of their work: there is no assumption that the ranking gives probability 1/2 to the two possibilities, only that the ranking is no constant (i.e. that neither X> Y nor X< Y holds all the time).

12. JB Says:

Gnedenkko´s book on Probability(chelsea publ) presents what he calls “Bertand Paradoxes” of geometric probabilty: when asked to find the probability that a straight line will intersect a unit disk leaving a chord of lenght 1/2 or higher, he gives 3 “arguments” obtaining the answers 1/2, 1/3, 1/4. Also, S. Ross “Afirst course on probability” or something like that has interesting examples involving infinite sets: given a (huge) bag and balls numbered 1,2, … do the following:
– 1 minute before 12, put balls 1- 10 in the bag and take ball #1 out.
-1/2 minute before 12 add balls 11-20 and take ball 2 out.
……….. and so on …………..

Q1) What happens at 12? bag is empty
Q2) If the ball taken out is selected randomly? prob 1 the bag will be empty at 12.

These are more curious(to me at least) than important but helped me keep some students awake.

13. mmm Says:

Another version of the paradox is:
Suppose you and your friend play a game. In the game, both of you check the amount of money in your wallet, and the one with more money gives all that money to the other. Now you think as follows: Suppose i have x amount
in my wallet. If my friend has >x, i gain more than x. If he has <x, i lose x.
And assume that both are equally likely to win(by symmetry). So your expectation is positive. But the same is true for him, whereas you both can’t have positive expectation as you win as much as he loses. Then what is wrong with the argument?

14. kla Says:

Even though I know this wasn’t supposed to be a discussion on the solution to the paradox, I can’t help it.

I do not know a lot about probability, but I think this paradox can be resolved very simply. It does not matter how (by which distribution) the sums in the envelopes have been determined. All we know is that we are given the information “there are two envelopes, one contains amount A and the other contains amount 2A”. This is something which sounds quite similar to, but is very different from the information “you may open one envelope. If you find amount A in it, you ‘ll be given the possibility to open a second one which will contain amount A/2 or 2A, each with probability 1/2”. Let us call these situations (1) and (2). In situation (1) (the one we are really considering) you open an envelope. It contains amount X and we don’t know if X=A or X=2A. Both are possible with equal probability 1/2. If we change, with probability 1/2 we go from A to 2A, or from 2A to A. No gain or less is to be expected and obviously, the same holds if we do not change.

On the other hand if we are in situation (2), where the amount in the next envelope actually depends on the amount in the first one (that’s the whole point: it does not in situation (1)!!), then naturally one has to go for the second one. If initially one has found amount x, then by iterating this procedure, our obtained amount will do a random walk on x.2^Z (Z={integers}).

15. mabs Says:

Regarding the paradox presented, i believe its solution is, at least in this case, quite straightforward: One envelope contains \$x\$ money, the other \$2x\$. We can initially choose each with probability 1/2, therefore our expected gain is \$x+x/2=3x/2\$. Afterwards, when we consider switching envelopes, the reasoning proceeds thus: I don’t know how much money is in my current envelope, so the other one can either have \$x\$ or \$2x\$, each with probability 1/2. In the first case, i’m bound to lose an amount \$x/2\$ relative to my current expected value (\$3x/2\$); in the second case, i’ll win \$x/2\$. The expected gain, therefore, is \$1/2*(-x/2)+1/2*(x/2)=0\$, as intuition predicts.

Reworking the example to have amounts \$x\$ and \$kx\$ in the envelopes, with \$k\$ any constant other than 2, works as well.

Now, why does the analysis in the problem statement appear paradoxical? The analysis is certainly correct (refer to the calculation in the 3rd paragraph of the post); it’s just that it’s an analysis for a different problem. Consider:

Suppose i have 10 dollars. A friend offers me the following proposition: I give him my 10 dollars, and in return he will flip a coin. If it comes out heads, he’ll give me 5 dollars; tails, and he’ll give me 20 dollars (the \$x/2\$ and \$2x\$ envelopes, respectively). Now, should i accept his offer? Of course! As the calculation in the post shows, my expected gain is 5/4*10 dollars, minus my original 10 dollars, so i come out winning 2.5 dollars on average.

In conclusion, the paradox stems from applying the incorrect analysis to the problem. As we mostly try to find problems in the analysis itself (that is, we neglect to see if the analysis is really meant to solve THIS problem), but the development of the analysis is indeed correct, we get the “What the hell?” feeling.

16. mabs Says:

Well, apparently the LaTeX thingy doesn’t work as i thought. Never tried to use it before, and couldn’t be bothered to read the instructions. So, sorry… In my post above, the \$…\$ things are supposed to be formulae in LaTeX syntax.

Another thing, this time regarding the paradox proposed in the reply above by mmm: It is assumed that, by ‘symmetry’, both participants have the same chance to win, thus the paradox. The assumption is, of course, wrong. You would only have true symmetry if you both had the same amount of money in your respective wallets at the start.

The relative amounts of money each player has is critical to deciding who wins, and it is implicit in the problem statement that, lacking knowledge of how much more (or less) money i have than the other person, i can assume that i occupy a central (‘symmetric’) position in the overall distribution of money contained in people’s wallets in general.

Now, that assumption is obviously at fault. If i have no money at all, i’m guaranteed never to lose. And if only few people carry more than 100 dollars at once in their wallets but i happen to have 150 right now, i’m more than likely to lose.

So, your expected gain depends on how much money you’re carrying in your wallet right now, and you are certainly aware of that amount. And even if you had no way of knowing the general distribution of money inside people’s wallets, you couldn’t claim an expected win based on ‘symmetry’. You’d have to say that your expected gain is indeterminate.

17. m Says:

not sure it would be useful but
there is a nice recent book by Szekely on paradoxes in probability theory

Szekely G.J. Pcaradoxes in probability theory and mathematical statistics,

it is a large listing of paradoxes with references to literature.

18. nicolas Says:

Ok so I know the point of this was not the paradox itself but to hell with it.

The crux of the paradox lies for me in that x itself is a random variable, and we dont recognize it as such when supposedly taking expectancy to compare outcomes : after taking unconditional expectancy, which is supposed to be a number, we end up with x, which is a random variable. This is not possible, and we have been fooled.
The way it is presented, we are considering the case where \$x=2a\$, and say the other envelope has a 1/2 proba of having \$2x=4a\$. this case do not exist and we know so, hence it should not appear in our computation…
So we take an envelope, call its value \$x\$. the other envelope contains either \$2x\$ if x=a, or \$x/2\$ if \$x=2a\$
the expectancy of it value is \$3/2a\$. the expectancy of the other envelope is \$3/2a\$, there is no incentive to switch.

19. M-2: Two Probability Paradoxes « Concrete Nonsense Says:

[…] original post […]

20. Sohbet Says:

Thanks…

Not sure if anyone posted it above, but I think St. Petersburg paradox is a great one to do….

Here’s a lesser known problem; superficially similar to the two-envelope paradox, but is actually not. Two people each pick a number (whether randomly, haphazardly, or otherwise); each writes it down on a piece of paper which is then placed face down. Person A, say, (who only knows their own choice, but not that of the other) has to guess whether the other number, which, to remind you, is hidden from them, is larger (or not) than theirs. The numbers, by the way, can be any two numbers; not necessarily integers, and may be positive or negative etc. The question: is there a strategy whereby the player (A) can guess correctly with probability greater than 0.5 ? In other words, is there a strategy whereby this is a game with positive expectation? Surprisingly, there is such a strategy !

23. bigmoneyoutofpolitics Says:

I hope this thread is still live. Referring to the argument that “your expectation if you switch is 5/4”, you write:

“Of course, the second argument is incorrect, but the reasons are somewhat subtle.”

That is wrong.

You can see this easily by listing the possibilities. The problem is not as simple as you seem to think, but it also has nothing to do with possible prior probability distributions — something that can be easily seen by simply choosing numbers.

There are actually two parts to the problem. Smullyan gives a clear statement of one part, which involves only actual amounts (not expected values), and their modes of designation (see, for example, “Satan, Cantor, and Infinity, pp 189-192). He does not resolve the problem, but stating it clearly is the most important step.

The second part of the problem does involve expectation values. This corresponds to your “second argument”. The last part of Smullyan’s account, on p 192 gives a compelling statement similar to yours, but again does not resolve the problem.

I don’t know whether Smullyan ever figured out the solutions but, in my view, his work on the problem is the most important contribution in print because he shows how the problem splits into two parts.

By the way, your system forces me to use the name of an old website of mine, instead of my real name. You use some extremely intrusive software, at the level of Google.


# Is there a fundamental reason to expect $e$ to appear in this probability question?

I came across the following question, whose answer is $e$. I was sort of amazed, since I didn't see a reason why $e$ should be making an appearance. So, phrasing the main question in the title another way: should it have been possible to predict that at least my answer should involve $e$ in some nontrivial way?

Question: Suppose you draw random variables $x_i \in [0,1]$, where the $x_i$ are uniformly distributed. If $x_i > x_{i-1}$, we draw again, and otherwise, we stop. What's the expected number of draws before we stop?

Solution: The probability that we take $n$ draws is given by $\frac{n-1}{n!}$ since there are $n!$ ways to order $x_1,\ldots, x_n$, and exactly $n-1$ choices for the placement of $x_{n-1}$ in the ordering $x_1 < x_1 < \cdots < x_{\widehat{n-1}} < x_n$. That is, for it to take precisely $n$ draws, we need $x_1 < x_2, x_2 < x_3, \ldots, x_{n-2} < x_{n-1}$ but then $x_n < x_{n-1}$. Thus, the expected value is given by $$E(\text{number of draws}) = \sum_{n = 2}^\infty \frac{1}{(n-2)!} = \fbox{e}$$

P.S. It's also possible I simply made a mistake, and if that's the case please point it out and I can edit or delete the question accordingly.

• See also this question. – John Bentin Jul 25 '17 at 20:14
• The number of months in a year is very close to $12\dfrac1e~.~$ Coincidence ? – Lucian Jul 25 '17 at 20:16
• Another combinatorial problem where $e$ pops up is the number of derangements in a card deck, en.wikipedia.org/wiki/Inclusion%E2%80%93exclusion_principle – An aedonist Jul 25 '17 at 20:18
• Does it surprise you that $\sum \frac1{n!}$ comes up in a probability question? If not then it shouldn't surprise you that $e$ comes up. What should surprise you is that $e=\sum \frac1{n!}$ in the first place. – Gregory Grant Jul 25 '17 at 20:25
• @DerekAllums Yeah if you take $\sum \frac1{n!}$ to be the definition of $e$ then it shouldn't surprise you that $e=\sum \frac1{n!}$ – Gregory Grant Jul 25 '17 at 20:40

First: I would take a slightly different route, but that's a matter of taste: Use that $$E[X]=\sum_{n=0}^\infty nP(X=n)=\sum_{n=0}^\infty P(X> n).$$ Similar to your reasoning, $X> n$ occurs iff (up to almost impossible equalities) $x_1<x_2<\ldots <x_n$, which happens with probability $\frac 1{n!}$. Hence, (again) $$\tag1 E[X]=\sum_{n=0}^\infty\frac1{n!}=e.$$

But regarding your main question: Could we expect $e$ to raise its head? Well, perhaps. The problem is about random order, hence about permutations, and (hand-wave) $e$ very often occurs in the context of permutations - which is of course owed to it being the sum if reciprocals of factorials. Then again, before one follows this thought further, one has already written down $(1)$ ...

I'll give a different way of approaching the problem that shows another way that $e$ can arise.

As with Hagen von Eitzen's answer, let $X$ be the random variable that represents the number of draws.

Consider the closely related problem where $x_i \in \{1, 2, \ldots, N\}$ instead. At most $N + 1$ draws can happen in this case. We can calculate the expectation using the binomial theorem.

$$E[X] = \sum_{k=0}^N{P(X > k)} =\sum_{k=0}^{N}\frac{N\choose k}{N^k} = \left(1 + \frac1{N}\right)^N$$

Here, $N \choose k$ is the number of strictly increasing ways of choosing $x_1, \ldots, x_k$ which is then divided by the total number of ways of choosing them to get the probability.

Of course then, the expectation approaches $e$ in the limit as $N \to \infty$. (I don't think this is a particularly good way to solve the problem, as one still has to justify how this limit is applicable, but hopefully it provides some insight into the appearance of $e$ regardless.)

As for the fundamental question, I don't really see a way of knowing that $e$ will appear until it has more or less appeared, but $e$ does seem to arise here quite naturally, so maybe it should not be so surprising in general when $e$ arises in a combinatorial context.

• Very cool - and I suppose this further illustrates the point that I shouldn't be surprised by the appearance of $e$, if my definition of $e$ is, for example, $\lim_{n \to \infty} \left(1+\frac{1}{n}\right)^n$. – Derek Allums Jul 26 '17 at 13:04


# Moderate Percentages Solved QuestionAptitude Discussion

 Q. When processing flower-nectar into honeybees' extract, a considerable amount of water gets reduced. How much flower-nectar must be processed to yield 1kg of honey, if nectar contains $50%$ water, and the honey obtained from this nectar contains $15%$ water?
 ✖ A. 1.5 kgs ✔ B. 1.7 kgs ✖ C. 3.33 kgs ✖ D. None of these

Solution:
Option(B) is correct

Flower-nectar contains $50\%$ of non-water part.

In honey this non-water part constitutes $85\% (100-15)$.

Therefore $0.5X$ Amount of flower-nectar = $0.85X$ Amount of honey = $0.85\times 1$ kg

Therefore amount of flower-nectar needed

$=\left(\dfrac{0.85}{0.5}\right)\times 1$

$=1.7$ kg

## (7) Comment(s)

Babar
()

50% of 100g is 50 g. if we are to get 1 kg honey we have to have 2 kg raw material because there is 50% water. there are also 15% additional water contents so we have to have more than 2 kg raw material. the given answers are wrong.

Azex
()

Doubt.

Extraction process--> Nector(+50%water) --> Honey(+15%water)--> Honey (pure).

End result is 1 Kg honey.

That means if we go one step up, before its extraction it will contain 15% water as well. So now total quantity:

X-15% of X = 1kg honey.

==> .85X = 1

X= 1/.85 ~ 1.176 Kg (Honey plus water)

Now this mixture was extracted from Nector-water mixture.

let us assume it is extracted from X kg of flower nector.

Therefore,

X- 50% of X = 1.176

.5X = 1.176

=> X = 2.352 Kg

Amit
()

i also got this answer but it is correct or not?

PPatel
()

Are there any other different method to solve this question ?

Param
()

As of now no other substitute becoz it is already simple. Let's make it simpler with some verbal concepts.

1. Honey in Nectar = Honey in honeybee = Honey we got. i.e no loss of honey anywhere only the proportion of water kept changing at different stage.

2. The value of honey at all stages is equal . so calculate the value at all three stages at keep them equal.

3. First stage - Honey is 50% of something.say, Value= 0.5X

4. Second stage - Honey is 85% of something. say, Value =0.85X

5. Last stage - Honey is 1 Kg. (Given)

so, 0.5X=0.85X=1kg. X=1.7 Kg (Pretty Obvious will me more than the honey)

Hope it helps.

Salman
()

this is a good place for the preparation of aptitude tests.

Khirodsaikia
()

Please, include all the subjects which are important for ssc exam