MATH 222 CALCULUS. Second Semester FALL 2004 UW MADISON. Lecture 3 Prof. Sigurd Angenent

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1 MATH Second Semester CALCULUS FALL 4 UW MADISON Lecture 3 Prof. Sigurd Angenent These notes belong to: Name of TA:

2 Math nd Semester Calculus Lecture notes for fall 4. This is a self contained set of lecture notes for Math, Lecture 3 in the fall semester of 4. The notes were written by Sigurd Angenent, starting from an extensive collection of notes and problems compiled by Joel Robbin. PLACE COPYRIGHT/COPYLEFT NOTICE HERE. CONTENTS Part. Methods of Integration 6. The indefinite integral 6. You can always check the answer 6 3. About +C 7 4. Standard Integrals 8 5. Method of substitution 8 6. The double angle trick 7. Integration by Parts 8. Reduction Formulas 9. Partial Fraction Expansion Reduce to a proper rational function Partial Fraction Expansion: The Easy Case Partial Fraction Expansion: The General Case 7 Part. Taylor s Formula and Infinite Series 9. Taylor Polynomials 9. Some special Taylor polynomials 3. The Remainder Term 3 3. Lagrange s Formula for the Remainder Term 4 4. The limit as x, keeping n fixed 6 5. Little-oh 7 6. Computations with Taylor polynomials 9 7. Differentiating Taylor polynomials 3 8. The limit n, keeping x fixed Sequences and their limits Convergence of Taylor Series Leibniz formulas for ln and π/4 38. Some proofs 39.. Proof of Lagrange s formula 39.. Proof of Theorem 4. 4 Part 3. Complex Numbers 4. Complex numbers 4. Argument and Absolute Value 4 3. Geometry of Arithmetic Applications in Trigonometry Unit length complex numbers The Addition Formulas for Sine & Cosine De Moivre s formula 45

3 3 5. Calculus of complex valued functions The Complex Exponential Function Other handy things you can do with complex numbers Partial fractions Certain trigonometric and exponential integrals Complex amplitudes 49 Part 4. Differential Equations 5 8. What is a DiffEq? 5 9. First Order Separable Equations 5 3. First Order Linear Equations The Integrating Factor Variation of constants for st order equations Dynamical Systems and Determinism Higher order equations Constant Coefficient Linear Homogeneous Equations Differential operators The superposition principle The characteristic polynomial Complex roots and repeated roots Inhomogeneous Linear Equations Variation of Constants Undetermined Coefficients Applications of Second Order Linear Equations Spring with a weight The pendulum equation The effect of friction Electric circuits 67 Part 5. Vectors Introduction to vectors Basic arithmetic of vectors Algebraic properties of vector addition and multiplication Geometric description of vectors Geometric interpretation of vector addition and multiplication Parametric equations for lines and planes Parametric equations for planes in space* Vector Bases The Standard Basis Vectors A Basis of Vectors (in general)* Dot Product Algebraic properties of the dot product The diagonals of a parallelogram The dot product and the angle between two vectors Orthogonal projection of one vector onto another Defining equations of lines Distance to a line Defining equation of a plane 8

4 4 4. Cross Product Algebraic definition of the cross product Algebraic properties of the cross product The triple product and determinants Geometric description of the cross product A few applications of the cross product Area of a parallelogram Finding the normal to a plane Volume of a parallelepiped Notation 9 Part 6. Vector Functions and Parametrized Curves Parametric Curves Examples of parametrized curves The derivative of a vector function Higher derivatives and product rules Interpretation of x (t) as the velocity vector Acceleration and Force Tangents and the unit tangent vector 5. Sketching a parametric curve 5. Length of a curve The arclength function Graphs in Cartesian and in Polar Coordinates 7 Part 7. Problems, Problems 9 INTEGRATION Basic Integrals Basic Substitutions Review of the Inverse Trigonometric Functions Integration by Parts and Reduction Formulae 3 Integration of Rational Functions 4 Miscellaneous and Mixed Integrals 6 TAYLOR S FORMULA and POWER SERIES Taylor s formula 9 Little-oh and manipulating Taylor polynomials Lagrange s formula for the remainder Limits of Sequences 3 Convergence of Taylor Series 3 COMPLEX NUMBERS Computing and Drawing Complex Numbers 5 Algebraic properties 6 The Complex Exponential 6 Calculus of Complex Valued Functions 7 DIFFERENTIAL EQUATIONS Separation of Variables 9 Linear Homogeneous 3 Linear Inhomogeneous 3

5 5 Applications 3 VECTORS Computing and drawing vectors 36 Parametric Equations for a Line 38 Orthogonal decomposition of one vector with respect to another 39 The Dot Product 39 The Cross Product 4 PARAMETRIZED CURVES Geometrically described curves. 43 Product rules. 44 Curve sketching, using the tangent vector. 44 Lengths of curves. 45

6 6 Part. Methods of Integration. The indefinite integral We recall some facts about integration from first semester calculus. Definition.. A function y = F(x) is called an antiderivative of another function y = f(x) if F (x) = f(x) for all x.. Example. F (x) = x is an antiderivative of f(x) = x. F (x) = x + 4 is also an antiderivative of f(x) = x. G(t) = sin(t + ) is an antiderivative of g(t) = cos(t + ). The Fundamental Theorem of Calculus states that if a function y = f(x) is continuous on an interval a x b, then there always exists an antiderivative F(x) of f, and one has () b a f(x)dx = F(b) F(a). The best way of computing an integral is often to find an antiderivative F of the given function f, and then to use the Fundamental Theorem (). How you go about finding an antiderivative F for some given function f is the subject of this chapter. The following notation is commonly used for antiderivates: () F(x) = f(x)dx. The integral which appears here does not have the integration bounds a and b. It is called an indefinite integral, as opposed to the integral in () which is called a definite integral. It s important to distinguish between the two kinds of integrals. Here is a list of differences: INDEFINITE INTEGRAL DEFINITE INTEGRAL f(x)dx is a function of x. By definition f(x)dx is any function of x whose derivative is f(x). x is not a dummy variable, for example, xdx = x + C and tdt = t + C are functions of diffferent variables, so they are not equal. b a b a f(x)dx is a number. f(x)dx was defined in terms of Riemann sums and can be interpreted as area under the graph of y = f(x), at least when f(x) >. x is a dummy variable, for example, xdx =, and tdt =, so xdx = tdt.. You can always check the answer Suppose you want to find an antiderivative of a given function f(x) and after a long and messy computation which you don t really trust you get an answer, F(x). You can then throw away the dubious computation and differentiate the F(x) you had found. If

7 7 F (x) turns out to be equal to f(x), then your F(x) is indeed an antiderivative and your computation isn t important anymore.. Example. Suppose we want to find lnxdx. My cousin Bruce says it might be F(x) = xln x x. Let s see if he s right: d (xln x x) = x + lnx = lnx. dx x Who knows how Bruce thought of this, but he s right! We now know that lnxdx = xlnx x + C. 3. About +C Let f(x) be a function defined on some interval a x b. If F(x) is an antiderivative of f(x) on this interval, then for any constant C the function F(x) = F(x)+C will also be an antiderivative of f(x). So one given function f(x) has many different antiderivatives, obtained by adding different constants to one given antiderivative. Theorem 3.. If F (x) and F (x) are antiderivatives of the same function f(x) on some interval a x b, then there is a constant C such that F (x) = F (x) + C. Proof. Consider the difference G(x) = F (x) F (x). Then G (x) = F (x) F (x) = f(x) f(x) =, so that G(x) must be constant. Hence F (x) F (x) = C for some constant. It follows that there is some ambiguity in the notation f(x)dx. Two functions F (x) and F (x) can both equal f(x)dx without equaling each other. When this happens, they (F and F ) differ by a constant. This can sometimes lead to confusing situations, e.g. you can check that sinxcosxdx = sin x sinxcosxdx = cos x are both correct. (Just differentiate the two functions sin x and cos x!) These two answers look different until you realize that because of the trig identity sin x+cos x = they really only differ by a constant: sin x = cos x +. To avoid this kind of confusion we will from now on never forget to include the arbitrary constant +C in our answer when we compute an antiderivative. He integrated by parts.

8 8 4. Standard Integrals Here is a list of the standard derivatives and hence the standard integrals everyone should know. f(x)dx = F(x) + C x n dx = xn+ n + + C for all n dx = ln x + C x sinxdx = cosx + C cosxdx = sin x + C tanxdx = lncosx + C dx = arctanx + C + x dx = arcsinx + C (= π arccosx + C) x dx cosx = + sin x ln sin x + C for π < x < π. All of these integrals are familiar from first semester calculus (like Math ), except for the last one. You can check the last one by differentiation (using ln a b = lna lnb simplifies things a bit). 5. Method of substitution The chain rule says that so that df(g(x)) dx = F (G(x)) G (x), F (G(x)) G (x)dx = F(G(x)) + C. 5. Example. Consider the function f(x) = xsin(x +3). It does not appear in the list of standard integrals we know by heart. But we do notice that x = d dx (x + 3). So let s call G(x) = x + 3, and F(u) = cosu, then and so that df(g(x)) dx F(G(x)) = cos(x + 3) = sin(x + 3) }{{} F (G(x)) }{{} x = f(x), G (x) xsin(x + 3)dx = cos(x + 3) + C. You will start noticing things like this after doing several examples.

9 9 The most transparent way of computing an integral by substitution is by introducing new variables. Thus to do the integral f(g(x))g (x)dx where f(u) = F (u), we introduce the substitution u = G(x), and agree to write du = dg(x) = G (x)dx. Then we get f(g(x))g (x)dx = f(u)du = F(u) + C. At the end of the integration we must remember that u really stands for G(x), so that f(g(x))g (x)dx = F(u) + C = F(G(x)) + C. For definite integrals this implies b which you can also write as a f(g(x))g (x)dx = F(G(b)) F(G(a)). (3) b f(g(x))g (x)dx = G(b) a G(a) f(u)du. 5. Substitution in a definite integral. As an example we compute x + x dx, using the substitution u = G(x) = + x. Since du = xdx, the associated indefinite integral is + x }{{ }{{} xdx = u du. } du u To find the definite integral you must compute the new integration bounds G() and G() (see equation (3).) If x runs between x = and x =, then u = G(x) = + x runs between u = + = and u = + =, so the definite integral we must compute is x + x dx = which is in our list of memorable integrals. So we find u du, x + x dx = u du = [ ] lnu = ln.

10 6. The double angle trick If an integral contains sin x or cos x, then you can remove the squares by using the double angle formulas from trigonometry. Recall that cos α sin α = cosα and cos α + sin α =, Adding these two equations gives while substracting them gives cos α = (cosα + ) sin α = ( cosα). 6. Example. The following integral shows up in many contexts, so it is worth knowing: cos xdx = ( + cosx)dx = {x + } sin x + C = x + sinx + C. 4 Since sin x = sin x cos x this result can also be written as cos xdx = x + sin xcosx + C. 7. Integration by Parts If you don t want to memorize the double angle formulas, then you can use Complex Exponentials to do these and many similar integrals. However, you will have to wait until we are in?? where this is explained. The product rule states d df(x) (F(x)G(x)) = G(x) + F(x)dG(x) dx dx dx and therefore, after rearranging terms, F(x) dg(x) = d dx dx This implies the formula for integration by parts F(x) dg(x) dx = F(x)G(x) dx F(x) G (x) F(x) G(x) (F(x)G(x)) df(x) dx G(x). 7. Example Integrating by parts once. }{{} x }{{} e x dx = }{{} x }{{} e x }{{} e x }{{} G(x) F (x) df(x) dx G(x)dx. dx = xe x e x + C. Observe that in this example e x was easy to integrate, while the factor x becomes an easier function when you differentiate it. This is the usual state of affairs when integration by

11 parts works: differentiating one of the factors (F(x)) should simplify the integral, while integrating the other (G (x)) should not complicate things (too much). Another example: sin x = d dx ( cosx) so xsin xdx = x( cosx) ( cosx) dx == xcosx + sin x + C. 7. Example Repeated Integration by Parts. Sometimes one integration by parts is not enough: since e x = d dx ( ex ) one has }{{} x }{{} e x dx = x ex e x xdx F(x) G (x) { = x ex e x } e x 4 x 4 dx { } = x ex e x ex x C Be careful with all the minus signs that appear when you integrate by parts. = x e x xex + 4 ex C The same procedure will work whenever you have to integrate P(x)e ax dx where P(x) is a polynomial, and a is a constant. Each time you integrate by parts, you get this P(x)e ax dx = P(x) eax e ax a a P (x)dx = a P(x)eax P (x)e ax dx. a You have replaced the integral P(x)e ax dx with the integral P (x)e ax dx. This is the same kind of integral, but it is a little easier since the degree of the derivative P (x) is less than the degree of P(x). 7.3 Example My cousin Bruce s computation. Sometimes the factor G (x) is invisible. Here is how you can get the antiderivative of lnx by integrating by parts: lnxdx = lnx }{{} dx }{{} F(x) G (x) = lnx x x xdx = xlnx dx = xlnx x + C. You can do P(x)ln xdx in the same way if P(x) is a polynomial.

12 8. Reduction Formulas Consider the integral I n = x n e ax dx. Integration by parts gives you I n = x n a eax nx n a eax dx = a xn e ax n x n e ax dx. a We haven t computed the integral, and in fact the integral that we still have to do is of the same kind as the one we started with (integral of x n e ax instead of x n e ax ). What we have derived is the following reduction formula I n = a xn e ax n a I n (R) which holds for all n. For n = the reduction formula says I = a eax, i.e. e ax dx = a eax + C. When n the reduction formula tells us that we have to compute I n if we want to find I n. The point of a reduction formula is that the same formula also applies to I n, and I n, etc., so that after repeated application of the formula we end up with I, i.e., an integral we know. 8. Example. To compute x 3 e ax dx we use the reduction formula three times: I 3 = a x3 e ax 3 a I = a x3 e ax 3 { a a x e ax } a I = a x3 e ax 3 { a a x e ax ( a a xeax )} a I Insert the known integral I = a eax + C and simplify the other terms and you get x 3 e ax dx = a x3 e ax 3 a x e ax + 6 a 3 xeax 6 a 4 eax + C. 8. Reduction formula requiring two partial integrations. Consider S n = x n sin xdx. Then for n one has S n = x n cosx + n x n cosxdx = x n cosx + nx n sinx n(n ) x n sinxdx. Thus we find the reduction formula S n = x n cosx + nx n sin x n(n )S n.

13 3 Each time you use this reduction, the exponent n drops by, so in the end you get either S or S, depending on whether you started with an odd or even n. 8.3 A reduction formula where you have to solve for I n. We try to compute I n = (sin x) n dx by a reduction formula. Integrating by parts twice we get I n = (sin x) n sin xdx = (sin x) n cosx ( cosx)(n )(sinx) n cosxdx = (sin x) n cosx + (n ) (sin x) n cos xdx. We now use cos x = sin x, which gives I n = (sin x) n cosx + (n ) {sin n x sin n x } dx = (sin x) n cosx + (n )I n (n )I n. You can think of this as an equation for I n, which, when you solve it tells you ni n = (sin x) n cosx + (n )I n and thus implies I n = n sinn xcos x + n n I n. Since we know the integrals I = (sin x) dx = dx = x + C and I = sin xdx = cosx + C (S) the reduction formula (S) allows us to calculate I n for any n. 8.4 A reduction formula which will be handy later. In the next section you will see how the integral of any rational function can be transformed into integrals of easier functions, the hardest of which turns out to be dx I n = ( + x ) n. When n = this is a standard integral, namely dx I = = arctanx + C. + x When n > integration by parts gives you a reduction formula. Here s the computation: I n = ( + x ) n dx x = ( + x ) n x( n) ( + x ) n xdx x = ( + x ) n + n x ( + x dx ) n+

14 4 Apply to get x ( + x ) n+ = ( + x ) ( + x ) n+ = x ( + x ) n x ( + x ) n+ x { x ( + x dx = ) n+ ( + x ) n x } ( + x ) n+ Our integration by parts therefore told us that x I n = ( + x ) n + n( ) I n I n+, which you can solve for I n+. You find the reduction formula I n+ = x n ( + x ) n + n n I n. dx = I n I n+. As an example of how you can use it, we start with I = arctanx + C, and conclude that dx ( + x ) = I = I + = x ( + x ) + I = x + x + arctanx + C. Apply the reduction formula again, now with n =, and you get dx ( + x ) 3 = I 3 = I + = x ( + x ) + I { } = x 4 ( + x ) + 3 x 4 + x + arctanx = x 4 ( + x ) + 3 x 8 + x arctanx + C. 9. Partial Fraction Expansion A rational function is one which is a ratio of polynomials, f(x) = P(x) Q(x) = p nx n + p n x n + + p x + p q d x d + q d x d + + q x + q. Such rational functions can always be integrated, and the trick which allows you to do this is called a partial fraction expansion. The whole procedure consists of several steps which are explained in this section. The procedure itself has nothing to do with integration: it s just a way of rewriting rational functions. It is in fact useful in other situations, such as finding Taylor series (see Part of these notes) and computing inverse Laplace transforms (see MATH 39.)

15 5 9.. Reduce to a proper rational function A proper rational function is a rational function P(x)/Q(x) where the degree of P(x) is strictly less than the degree of Q(x). the method of partial fractions only applies to proper rational functions. Fortunately there s an additional trick for dealing with rational functions that are not proper. If P/Q isn t proper, i.e. if degree(p) degree(q), then you divide P by Q, with result P(x) R(x) = S(x) + Q(x) Q(x) where S(x) is the quotient, and R(x) is the remainder after division. In practice you would do a long division to find S(x) and R(x). 9. Example. Consider the rational function f(x) = x3 x + x. Here the numerator has degree 3 which is more than the degree of the denominator (which is ). To apply the method of partial fractions we must first do a division with remainder. One has x + = S(x) x x x 3 x + x 3 x x x x x x + = R(x) so that f(x) = x3 x + x = x + + x + x When we integrate we get x 3 { x + x dx = x + + x + } x dx = x + x + x + x dx. The rational function which still have to integrate, namely x+ x, is proper, i.e. its numerator has lower degree than its denominator. 9.. Partial Fraction Expansion: The Easy Case To compute the partial fraction expansion of a proper rational function P(x)/Q(x) you must factor the denominator Q(x). Factoring the denominator is a problem as difficult as finding all of its roots; in Math we shall only do problems where the denominator is already factored into linear and quadratic factors, or where this factorization is easy to find. In the easiest partial fractions problems, all the roots of Q(x) are reall and distinct, so the denominator is factored into distinct linear factors, say P(x) Q(x) = P(x) (x a )(x a ) (x a n ). To integrate this function we find constants A, A,..., A n so that P(x) Q(x) = A + A + + A n. (#) x a x a x a n

16 6 Then the integral is P(x) Q(x) dx = A ln x a + A ln x a + + A n ln x a n + C. One way to find the coefficients A i in (#) is called the method of equating coefficients. In this method we multiply both sides of (#) with Q(x) = (x a ) (x a n ). The result is a polynomial of degree n on both sides. Equating the coefficients of these polynomial gives a system of n linear equations for A,..., A n. You get the A i by solving that system of equations. Another much faster way to find the coefficients A i is the Heaviside Trick 3. Multiply equation (#) by x a i and then plug in 4 x = a i. On the right you are left with A i so A i = P(x)(x a i) Q(x) = x=ai 9. Example continued. To integrate x + x P(a i ) (a i a ) (a i a i )(a i a i+ ) (a i a n ). x = (x )(x + ). we factor the denominator, The partial fraction expansion of x + x then is x + x = x + (x )(x + ) = A x + B x +. ( ) Multiply with (x )(x + ) to get x + = A(x + ) + B(x ) = (A + B)x + (A B). The functions of x on the left and right are equal only if the coefficient of x and the constant term are equal. In other words we must have A + B = and A B =. These are two linear equations for two unknowns A and B, which we now proceed to solve. Adding both equations gives A =, so that A = ; from the first equation one then finds B = A = 3. So x + x = / x 3/ x +. Instead, we could also use the Heaviside trick: multiply ( ) with x to get x + x + = A + B x x + Take the limit x and you find + + = A, i.e. A =. Similarly, after multiplying ( ) with x + one gets x + x = Ax + x + B, 3 Named after OLIVER HEAVISIDE, an electrical engineer in the late 9th and early ieth century. 4 More properly, you should take the limit x ai. If you multiply both sides with x a i and then set x = a i, then you have effectively multiplied both sides of the equation with zero!

17 7 and letting x you find B = ( ) + ( ) = 3, as before. Either way, the integral is now easily found, namely, x 3 x + x dx = x x + + x + x dx = x + x + { / x 3/ x + } dx = x + x + ln x 3 ln x + + C Partial Fraction Expansion: The General Case Buckle up. When the denominator Q(x) contains repeated factors or quadratic factors (or both) the partial fraction decomposition is more complicated. In the most general case the denominator Q(x) can be factored in the form (4) Q(x) = (x a ) k (x a n ) kn (x + b x + c ) l (x + b m x + c m ) lm Here we assume that the factors x a,..., x a n are all different, and we also assume that the factors x + b x + c,..., x + b m x + c m are all different. It is a theorem from advanced algebra that you can always write the rational function P(x)/Q(x) as a sum of terms like this (5) P(x) Q(x) = + A (x a i ) k + + Bx + C (x + b j x + c j ) l + How did this sum come about? For each linear factor (x a) k in the denominator (4) you get terms A x a + A (x a) + + A k (x a) k in the decomposition. There are as many terms as the exponent of the linear factor that generated them. For each quadratic factor (x + bx + c) l you get terms B x + C x + bx + c + B x + C (x + bx + c) + + B mx + C m (x + bx + c) l. Again, there are as many terms as the exponent l with which the quadratic factor appears in the denominator (4). In general, you find the constants A..., B... and C... by the method of equating coefficients. 9.3 Example. To do the integral x + 3 x (x + )(x + ) dx

18 8 apply the method of equating coefficients to the form x + 3 x (x + )(x + ) = A x + A x + A 3 x + + B x + C x + B x + C + (x + ). (EX) Solving this last problem will require solving a system of seven linear equations in the seven unknowns A, A, A 3, B, C, B, C. A computer program like Maple can do this easily, but it is a lot of work to do it by hand. In general, the method of equating coefficients requires solving n linear equations in n unknowns where n is the degree of the denominator Q(x). See Problem 7 for a worked example where the coefficients are found.!! Unfortunately, in the presence of quadratic factors or repeated linear factors the Heaviside trick does not give the whole answer; you must use the method of equating coefficients. Once you have found the partial fraction decomposition (EX) you still have to integrate the terms which appeared. The first three terms are of the form A(x a) p dx and they are easy to integrate: Adx = Aln x a + C x a and Adx (x a) p = A ( p)(x a) p + C if p >. The next, fourth term in (EX) can be written as B x + C x + dx = B x x + dx + C dx x + = B ln(x + ) + C arctanx + C integration const. While these integrals are already not very simple, the integrals Bx + C (x dx with p > + bx + c) p which can appear are particularly unpleasant. If you really must compute one of these, then complete the square in the denominator so that the integral takes the form Ax + B ((x + b) + a ) p dx. After the change of variables u = x + b and factoring out constants you have to do the integrals du u du (u + a ) p and (u + a ) p. Use the reduction formula we found in example 8.4 to compute this integral. An alternative approach is to use complex numbers (which are on the menu for this semester.) If you allow complex numbers then the quadratic factors x + bx + c can be factored, and your partial fraction expansion only contains terms of the form A/(x a) p, although A and a can now be complex numbers. The integrals are then easy, but the answer has complex numbers in it, and rewriting the answer in terms of real numbers again can be quite involved.!!

19 Part. Taylor s Formula and Infinite Series 9. Taylor Polynomials All continuous functions which vanish at x = a are approximately equal at x = a but some are more approximately equal than others. Definition.. The Taylor polynomial of a function y = f(x) of degree n at a point a is the polynomial Recall n! = 3 n, and by definition! = (6) T n f(x; a) = f(a) + f (a)(x a) + f (a)! Note that the zeroth order Taylor polynomial is just a constant, while the first order Taylor polynomial is T f(x; a) = f(a), T f(x; a) = f(a) + f (a)(x a). (x a) + + f(n) (a) (x a) n. n! This is exactly the linear approximation of f(x) for x close to a which was derived in st semester calculus. The Taylor polynomial generalizes this first order approximation by providing higher order approximations to f. Most of the time we will take a = in which case we write T n f(x) instead of T n f(x; a), and we get a slightly simpler formula (7) T n f(x) = f() + f ()x + f ()! x + + f(n) () x n. n! You will see below that for many functions f(x) the Taylor polynomials T n f(x) give better and better approximations as you add more terms (i.e. as you increase n). For this reason the limit when n is often considered, which leads to the infinite sum T f(x) = f() + f ()x + f () x + f () x 3 +! 3! At this point we will not try to make sense of the sum of infinitely many numbers. Theorem.. The Taylor polynomial has the following property: it is the only polynomial P(x) of degree n whose value and whose derivatives of orders,,..., and n are the same as those of f, i.e. it s the only polynomial of degree n for which holds. P(a) = f(a), P (a) = f (a), P (a) = f (a),..., P (n) (a) = f (n) (a)

20 Proof. We do the case a =, for simplicity. Let n be given, and let s take a polynomial P(x) of degree n, say, P(x) = a + a x + a x + a 3 x a n x n, and let s see what its derivatives look like. They are: P(x) = a + a x + a x + a 3 x 3 + a 4 x 4 + P (x) = a + a x + 3a 3 x + 4a 4 x 3 + P () (x) = a + 3a 3 x + 3 4a 4 x + P (3) (x) = 3a a 4 x + P (4) (x) = 3 4a 4 + When you set x = all the terms which have a positive power of x vanish, and you are left with the first entry on each line, i.e. and in general P() = a, P () = a, P () () = a, P (3) () = 3a 3, etc. P (k) () = k!a k for k n. For k n + the derivatives p (k) (x) all vanish of course, since P(x) is a polynomial of degree n. Therefore, if we want P to have the same values and derivatives at x = of orders,,..., n as the function f, then we must have k!a k = P (k) () = f (k) () for all k n. Thus a k = f(k) () for k n. k!.3 Example: Compute the Taylor polynomials of degree, and of f(x) = e x at a =, and plot them One has so that f(x) = e x = f (x) = e x = f (x) = e x, f() =, f () =, f () =. Therefore the first three Taylor polynomials of e x at a = are T f(x) = T f(x) = + x T f(x) = + x + x. The graphs are found in Figure. As you can see from the graphs, the Taylor polynomial T f(x) of degree is close to e x for small x, by virtue of the continuity of e x The Taylor polynomial of degree, i.e. T f(x) = captures the fact that e x by virtue of its continuity does not change very much if x stays close to x =. The Taylor polynomial of degree, i.e. T f(x) = + x corresponds to the tangent line to the graph of f(x) = e x, and so it also captures the fact that the function f(x) is increasing near x =. Clearly T f(x) is a better approximation to e x than T f(x). The graphs of both y = T f(x) and y = T f(x) are straight lines, while the graph of y = e x is curved (in fact, convex). The second order Taylor polynomial captures this convexity. In fact, the graph of y = T f(x) is a parabola, and since it has the same first

21 y=e x y=+x+x / y=+x y= y=e x.5 FIGURE. Polynomials approximating e x near x = and second derivative at x =, its curvature is the same as the curvature of the graph of y = e x at x =. So it seems that y = T f(x) = + x + x / is an approximation to y = e x which beats both T f(x) and T f(x)..4 Example: Find the Taylor polynomials of f(x) = sin x When you start computing the derivatives of sin x you find f(x) = sin x, f (x) = cosx, f (x) = sinx, f (3) (x) = cosx, and thus f (4) (x) = sinx. So after four derivatives you re back to where you started, and the sequence of derivatives of sin x cycles through the pattern sin x, cosx, sinx, cosx, sin x, sinx, cosx, sinx, cosx, sinx,... on and on. At x = you then get the following values for the derivatives f (j) (), j f (j) ()

22 This gives the following Taylor polynomials T f(x) = T f(x) = x T f(x) = x T 3 f(x) = x x3 3! T 4 f(x) = x x3 3! T 5 f(x) = x x3 3! + x5 5! Note that since f () () = the Taylor polynomials T f(x) and T f(x) are the same! The second order Taylor polynomial in this example is really only a polynomial of degree. In general the Taylor polynomial T n f(x) of any function is a polynomial of degree at most n, and this example shows that the degree can sometimes be strictly less. T T 5 π π π π T 3 T 7 FIGURE. Taylor polynomials of f(x) = sin x

23 3. Some special Taylor polynomials Here is a list of functions whose Taylor polynomials are suffiently regular that you can write a formula for the nth term. T n e x = + x + x! + x3 3! + + xn n! T n+ {sin x} = x x3 3! + x5 5! x7 7! + + x n+ ( )n+ (n + )! T n {cosx} = x! + x4 4! x6 xn + + ( )n 6! (n)! { } T n = + x + x + x 3 + x x n (Geometric Series) x T n {ln( + x)} = x x + x3 3 x4 xn + + ( )n+ 4 n All of these Taylor polynomials can be computed directly from the definition, by repeatedly differentiating f(x).. The Remainder Term The Taylor polynomial T n f(x) is almost never exactly equal to f(x), but often it is a good approximation, especially if x is small. To see how good the approximation is we define the error term or, remainder term. Definition.. If f is an n times differentiable function on some interval containing a, then R n f(x; a) = f(x) T n f(x; a) is called the n th order remainder (or error) term in the Taylor polynomial of f. If a =, as will be the case in most examples we do, then we write R n f(x) = f(x) T n f(x).. Example. If f(x) = sin x then we have found that T 3 f(x) = x 6 x3, so that R 3 {sin x} = sin x x + 6 x3. This is a completely correct formula for the remainder term, but it s rather useless: there s nothing about this expression that suggests that x 6 x3 is a much better approximation to sinx than, say, x + 6 x3. The usual situation is that there is no simple formula for the remainder term..3 An unusual example, in which there is a simple formula for R n f(x) Consider f(x) = x + 3 x 5 x 3. Then you find T f(x) = x + 3 x, so that R f(x) = f(x) T f(x) = 5 x 3. The moral of this example is this: Given a polynomial f(x) you find its n th degree Taylor polynomial by taking all terms of degree n in f(x); the remainder R n f(x) then consists of the remaining terms.

24 4.4 Another unusual, but important example where you can compute R n f(x) Consider the function f(x) = x. Then repeated differentiation gives and thus f (x) = Consequently, ( x), f() (x) = ( x) 3, f(3) (x) = 3 ( x) 4,... f (n) (x) = 3 n ( x) n+. f (n) () = n! = n! f(n) () =, and you see that the Taylor polynomials of this function are really simple, namely T n f(x) = + x + x + x 3 + x x n. But this sum should be really familiar: it is just the Geometric Sum (each term is x times the previous term). Its sum is given by 5 which we can rewrite as T n f(x) = + x + x + x 3 + x x n = xn+ x, The remainder term therefore is T n f(x) = x xn+ xn+ = f(x) x x. R n f(x) = f(x) T n f(x) = xn+ x. 3. Lagrange s Formula for the Remainder Term Theorem 3.. Let f be an n+ times differentiable function on some interval I containing x =. Then for every x in the interval I there is a ξ between and x such that R n f(x) = f(n+) (ξ) (n + )! xn+. (ξ between and x means either < ξ < x or x < ξ <, depending on the sign of x.) This theorem (including the proof) is similar to the Mean Value Theorem. The proof is a bit involved, and I ve put it at the end of this chapter. There are calculus textbooks which, after presenting this remainder formula, give a whole bunch of problems which ask you to find ξ for given f and x. Such problems completely miss the point of Lagrange s formula. The point is that even though you usually can t compute the mystery point ξ precisely, Lagrange s formula for the remainder term allows you to estimate it. Here is the most common way to estimate the remainder: 5 Multiply both sides with x to verify this, in case you had forgotten the formula!

25 5 Theorem 3. (Estimate of remainder term). If f is an n + times differentiable function on an interval containing x =, and if you have a constant M such that f (n+) (t) M for all t between and x, then R n f(x) M x n+ (n + )!. Proof. We don t know what ξ is in Lagrange s formula, but it doesn t matter, for wherever it is, it must lie between and x so that our assumption implies f (n+) (ξ) M. Put that in Lagrange s formula and you get the stated inequality. 3.3 How to compute e in a few decimal places Consider f(x) = e x. We computed the Taylor polynomials before. If you set x =, then you get e = f() = T n f() + R n f(), and thus, taking n = 8, e = +! +! + 3! + 4! + 5! + 6! + 7! + 8! + R 8(). By Lagrange s formula there is a ξ between and such that R 8 () = f(9) (ξ) 9 = eξ 9! 9!. (remember: f(x) = e x, so all its derivatives are also e x.) We don t really know where ξ is, but since it lies between and we know that < e ξ < e. So the remainder term R 8 () is positive and no more than e/9!. Estimating e < 3, we find Thus we see that 9! < R 8() < 3 9!. +! +! + 3! + + 7! + 8! + 9! < e < +! +! + 3! + + 7! + 8! + 3 9! or, in decimals,.78, 8... < e <.78, Error in the approximation sin x x In many calculations involving sin x for small values of x one makes the simplifying approximation sin x x, justified by the known limit sin x lim x x =. Question: How big is the error in this approximation? To answer this question, we use Lagrange s formula for the remainder term again. Let f(x) = sin x. Then the first degree Taylor polynomial of f is T f(x) = x. The approximation sin x x is therefore exactly what you get if you approximate f(x) = sin x by its first degree Taylor polynomial. Lagrange tells us that where, since f (x) = sinx, f(x) = T f(x) + R f(x), i.e. sin x = x + R f(x), R f(x) = f (ξ) x =! sin ξ x

26 6 for some ξ between and x. As always with Lagrange s remainder term, we don t know where ξ is precisely, so we have to estimate the remainder term. The easiest (but not the best: see below) way to do this is to say that no matter what ξ is, sin ξ will always be between and. Hence the remainder term is bounded by ( ) R f(x) x, and we find that x x sin x x + x. Question: How small must we choose x to be sure that the approximation sin x x isn t off by more than %? If we want the error to be less than % of the estimate, then we should require x to be less than % of x, i.e. x <. x x <. So we have shown that, if you choose x <., then the error you make in approximating sin x by just x is no more than %. A final comment about this example: the estimate for the error we got here can be improved quite a bit in two different ways: () You could notice that one has sin x x for all x, so if ξ is between and x, then sinξ ξ x, which gives you the estimate R f(x) x 3 instead of x as in ( ). () For this particular function the two Taylor polynomials T f(x) and T f(x) are the same (because f () = ). So T f(x) = x, and we can write sin x = f(x) = x + R f(x), In other words, the error in the approximation sin x x is also given by the second order remainder term, which according to Lagrange is given by R f(x) = cosξ x 3 cos ξ = R f(x) 3! 6 x 3, which is the best estimate for the error in sin x x we have so far. 4. The limit as x, keeping n fixed Lagrange s formula for the remainder term let s us write a function y = f(x) which is defined on some interval containing x = in the following way (8) f(x) = f() + f ()x + f() ()! x + + f(n) () x n + f(n+) (ξ) n! (n + )! xn+ Note that the last term contains the ξ from Lagrange s theorem, which depends on x, and of which you only know that it lies between and x. Here is an IMPORTANT THEOREM which is very helpful when you want to compute Taylor polynomials. Theorem 4.. If f(x) and g(x) are n times differentiable functions and if T n f(x) = T n g(x), then (9) lim x f(x) g(x) x n =.

27 7 5. Little-oh Conversely, if the functions f(x) and g(x) satisfy (9) then they have the same nth degree Taylor polynomials. In other words, if two functions have the same nth degree Taylor polynomial, then their difference is much smaller than x n for small enough n. Landau introduced the following notation which many people find useful. By definition o(x n ) is an abbreviation for any function h(x) which satisfies So you can rewrite (9) as h(x) lim x x n =. f(x) g(x) = o(x n ) (x ), or f(x) = g(x) + o(x n ), (x ) The last equation is pronounced as f(x) is equal to g(x) plus little-oh of x n as x goes to zero. The intuitive interpretation is that f and g(x) are almost the same, and that they differ by something which is negligible compared to x n at least for very small values of x. With Landau s notation Theorem 4. says that two n times differentiable functions f(x) and g(x) have the same nth degree Taylor polynomial if and only if f(x) = g(x) + o(x n ) (x ) The nice thing about Landau s little-oh is that you can compute with it, as long as you obey the following (at first sight rather strange) rules which will be proved in class x n o(x m ) = o(x n+m ) (x ) o(x n ) o(x m ) = o(x n+m ) (x ) x m = o(x n ) (x ) o(x n ) + o(x m ) = o(x n ) (x ) o(cx n ) = o(x n ) (x ) if n < m if n < m for any constant C 5. Example: prove one of these little-oh rules Let s do the first one, i.e. let s show that x n o(x m ) is o(x n+m ) as x. Remember, if someone writes x n o(x m ), then the o(x m ) is an abbreviation for some function h(x) which satisfies lim x h(x)/x m =. So the x n o(x m ) we are given here really is an abbreviation for x n h(x). We then have x n h(x) lim x x n+m = lim h(x) x x m =, since h(x) = o(xm ). 5. Can you see that x 3 = o(x ) by looking at the graphs of these functions? A picture is of course never a proof, but have a look at figure 3 which shows you the graphs of y = x, x, x 3, x 4, x 5 and x. As you see, when x approaches, the graphs of higher powers of x approach the x-axis (much?) faster than do the graphs of lower powers. You should also have a look at figure 4 which exhibits the graphs of y = x, as well as several linear functions y = Cx (with C =,, 5 and.) For each of these linear functions one has x < Cx if x is small enough how small is small enough depends on C. The smaller the constant C, the closer you have to keep x to to be sure that x is smaller than Cx. Nevertheless, no matter how small C is, the parabola will eventually always reach the region below the line y = Cx.

28 8 y=x n.5 n= n= n= FIGURE 3. How the powers stack up 5.3 Example: Little-oh arithmetic is a little funny Both x and x 3 are functions which are o(x) as x, i.e. x = o(x) (x ) and x 3 = o(x) (x ) Nevertheless x x 3. So in working with little-oh we are giving up on the principle that says that two things which both equal a third object must themselves be equal; i.e. a = b and b = c implies a = c, but not when you re using little-ohs! You can also put it like this: just because two quantities both are much smaller than x as x, they don t have to be equal. In particular, you can never cancel little-ohs!!! In other words, the following is pretty wrong o(x ) o(x ) = (x ). Why? The two o(x ) s both refer to functions h(x) which satisfy lim x h(x)/x =, but there are many such functions, and the two o(x ) s could be abbreviations for different functions h(x). x x x/ x/5 x/ FIGURE 4. In the end (as x ) the parabola always wins

29 9 Contrast this with the following computation, which at first sight looks wrong even though it is actually right: o(x ) o(x ) = o(x ) (x ). In words: if you subtract two quantities both of which are negligible compared to x for small x then the result will also be negligible compared to x for small x. 6. Computations with Taylor polynomials In principle the definition of T n f(x) lets you compute as many terms of the Taylor polynomial as you want, but in many (most) examples the computations quickly get out of hand. To see what can happen go though the following example: 6. How not to compute the Taylor polynomial of degree of f(x) = /(+x ) Diligently computing derivatives one by one you find f (x) = x ( + x ) so f () = f (x) = 6x ( + x ) 3 so f () = f (3) (x) = 4 x x3 ( + x ) 4 so f (3) () = f (4) (x) = 4 x + 5x 4 ( + x ) 5 so f (4) () = 4 = 4! f (5) (x) = 4 3x + x3 3x 5 ( + x ) 6 so f (4) () = f (6) (x) = 7 + x 35x 4 + 7x 6 ( + x ) 7 so f (4) () = 7 = 6!. Can you find f () (x)? After a lot of work we give up at the sixth derivative, and all we have found is { } T 4 + x = x + x 4 x 6. By the way, f () (x) = x + 75 x 4 76 x x 8 86 x + 3 x ( + x ) 3 and 4796 =!. 6. The right approach to finding the Taylor polynomial of any degree of f(x) = /( + x ) Start with the Geometric Series: if g(t) = /( t) then g(t) = + t + t + t 3 + t t n + o(t n ). Now substitute t = x in this limit, ( ( x g( x ) = x + x 4 x ( ) n x n + o ) ) n

30 3 Since o (( x ) n) = o(x n ) and g( x ) = ( x ) = + x, we have found + x = x + x 4 x ( ) n x n + o(x n ) By Theorem (4.) this implies T n { + x } = x + x 4 x ( ) n x n. 6.3 Example of multiplication of Taylor series Finding the Taylor series of e x /(+x) directly from the definition is another recipe for headaches. Instead, you should exploit your knowledge of the Taylor series of both factors e x and /( + x): Then multiply these two e x + x = e x = + x + x! + 3 x 3 3! + 4 x 4 4! + o(x 4 ) = + x + x x3 + 3 x4 + o(x 4 ) + x = x + x x 3 + x 4 + o(x 4 ). ( + x + x x3 + 3 x4 + o(x 4 ) ) ( x + x x 3 + x 4 + o(x 4 ) ) = x + x x 3 + x 4 + o(x 4 ) + x x + x 3 x 4 + o(x 4 ) + x x 3 + x 4 + o(x 4 ) x3 4 3 x4 + o(x 4 ) + 3 x4 + o(x 4 ) = + x + x + 3 x3 + 3 x4 + o(x 4 ) (x ) 6.4 Taylor s formula and Fibonacci numbers The Fibonacci numbers are defined as follows: the first two are f = and f =, and the others are defined by the equation (Fib) So f n = f n + f n f = f + f = + =, f 3 = f + f = + = 3, f 4 = f 3 + f = 3 + = 5, etc. The equation (Fib) lets you compute the whole sequence of numbers, one by one, when you are given only the first few numbers of the sequence (f and f in this case). Such an equation for the elements of a sequence is called a recursion relation.

31 3 Let Now consider the function f(x) = x x. T f(x) = c + c x + c x + c 3 x 3 + be its Taylor series. Due to Lagrange s remainder theorem you have, for any n, x x = c + c x + c x + c 3 x c n x n + o(x n ) (x ). Multiply both sides with x x and you get = ( x x ) (c + c x + c x + + c n + o(x n )) (x ) = c + c x + c x + + c n x n + o(x n ) c x c x c n x n + o(x n ) c x c n x n o(x n ) (x ) = c + (c c )x + (c c c )x + (c 3 c c )x (c n c n c n )x n + o(x n ) (x ) Compare the coefficients of powers x k on both sides for k =,,..., n and you find c =, c c = = c = c =, c c c = = c = c + c = and in general c n c n c n = = c n = c n + c n Therefore the coefficients of the Taylor series T f(x) are exactly the Fibonacci numbers: c n = f n for n =,,, 3,... Since it is much easier to compute the Fibonacci numbers one by one than it is to compute the derivatives of f(x) = /( x x ), this is a better way to compute the Taylor series of f(x) than just directly from the definition. 6.5 More about the Fibonacci numbers In this example you ll see a trick that lets you compute the Taylor series of any rational function. You already know the trick: find the partial fraction decomposition of the given rational function. Ignoring the case that you have quadratic expressions in the denominator, this lets you represent your rational function as a sum of terms of the form A (x a) p. These are easy to differentiate any number of times, and thus they allow you to write their Taylor series. Let s apply this to the function f(x) = /( x x ) from the example 6.4. First we factor the denominator. The number x x = x + x = x = ± 5. φ = , 33, 988, 749, 89...

32 3 is called the Golden Ratio. It satisfies 6 φ + φ = 5. The roots of our polynomial x + x are therefore x = 5 and we can factor x x as follows = φ, x + = + 5 = φ. x x = (x + x ) = (x x )(x x + ) = (x )(x + φ). φ So f(x) can be written as f(x) = x x = A (x = φ )(x + φ) x φ The Heaviside trick will tell you what A and B are, namely, A = The nth derivative of f(x) is φ + φ =, B = 5 φ + φ = 5 f (n) (x) = A( )n n! ( ) n+ + B( )n n! x (x + φ) n+ φ + B x + φ Setting x = and dividing by n! finally gives you the coefficient of x n in the Taylor series of f(x). The result is the following formula for the nth Fibonacci number c n = f(n) () = A( ) n n! ( ) n! n! n+ + B( ) n ( ) n+ n! n! (φ) n+ = Aφ n+ B φ φ Using the values for A and B you find f n = c n = { φ n+ } 5 φ n+ 7. Differentiating Taylor polynomials If T n f(x) = a + a x + a x + + a n x n is the Taylor polynomial of a function y = f(x), then what is the Taylor polynomial of its derivative f (x)? Theorem 7.. The Taylor polynomial of degree n of f (x) is given by T n {f (x)} = a + a x + + na n x n. In other words, the Taylor polynomial of the derivative is the derivative of the Taylor polynomial. 6 To prove this, use φ = + 5 = = + 5.

33 33 Proof. Let g(x) = f (x). Then g (k) () = f (k+) (), so that ($) T n g(x) = g() + g ()x + g () () x! + + g(n ) () xn (n )! = f () + f () ()x + f (3) () x! + + f(n) () xn (n )! On the other hand, if T n f(x) = a + a x + + a n x n, then a k = f (k) ()/k!, so that In other words, ka k = k (k )! f(k) () = f(k) () (k )!. So, continuing from ($) you find that a = f (), a = f () (), 3a 3 = f(3) (), etc.! T n {f (x)} = T n g(x) = a + a x + + na n x n as claimed. 7. Example. We compute the Taylor polynomial of f(x) = /( x) by noting that Since f(x) = F (x), where F(x) = x. T n+ F(x) = + x + x + x x n+, theorem 7. implies that { } T n ( x) = + x + 3x + 4x (n + )x n 7.3 Example: Taylor polynomials of arctan x Let f(x) = arctan x. Then know that f (x) = + x. By substitution of t = x in the Taylor polynomial of /( t) we had found { } T n {f (x)} = T n + x = x + x 4 x ( ) n x n + o ( x n). This Taylor polynomial must be the derivative of T n+ f(x), so we have T n+ {arctanx} = x x3 3 + x5 xn+ + + ( )n 5 n +.

34 34 8. The limit n, keeping x fixed 8.. Sequences and their limits We shall call a sequence any ordered sequence of numbers a, a, a 3,...: for each positive integer n we have to specify a number a n. 8. Examples of sequences definition first few number in the sequence a n = n,, 3, 4,... b n =,,,,... c n = n,, 3, 4,... d n = ( 3) n 3, 9, 7, 8,... E n = +! +! + 3! + + n!,,, 3, 7 4, 43 6,... S n = T n+ {sin x} = x x3 3! + + ( )n xn+ (n + )! x, x x3 3!, x x3 3! + x5 5!,... The last two sequences are derived from the Taylor polynomials of e x (at x = ) and sinx (at any x). The last example S n really is a sequence of functions, i.e. for every choice of x you get a different sequence. Definition 8.. A sequence of numbers (a n ) n= converges to a limit L, if for every ǫ > there is a number N ǫ such that for all n > N ǫ one has One writes a n L < ǫ. lim a n = L n 8.3 Example: lim n n = The sequence c n = /n converges to. To prove this let ǫ > be given. We have to find an N ǫ such that c n < ǫ for all n > N ǫ. The c n are all positive, so c n = c n, and hence c n < ǫ n < ǫ n > ǫ, which prompts us to choose N ǫ = /ǫ. The calculation we just did shows that if n > ǫ = N ǫ, then c n < ǫ. That means that lim n c n =. 8.4 Example: lim n an = if a < As in the previous example one can show that lim n n =, and more generally, that for any constant a with < a < one has Indeed, lim n an =. a n = a n = e nln a < ǫ

35 35 holds if and only if n ln a < lnǫ. Since a < we have ln a < so that dividing by ln a reverses the inequality, with result a n < ǫ n > lnǫ ln a The choice N ǫ = (lnǫ)/(ln a ) therefore guarantees that a n < ǫ whenever n > N ǫ. One can show that the operation of taking limits of sequences obeys the same rules as taking limits of functions. Theorem 8.5. If then one has lim a n = A and n lim a n ± b n = A ± B n lim a nb n = AB n lim n a n b n = A B lim b n = B, n (assuming B ). The so-called sandwich 7 theorem for ordinary limits also applies to limits of sequences. Namely, one has Theorem 8.6 ( Sandwich theorem ). If a n is a sequence which satisfies b n < a n < c N for all n, and if lim n b n = lim n c n =, then lim n a n =. Finally, one can show this: Theorem 8.7. If f(x) is a function which is continuous at x = A, and a n is a sequence which converges to A, then ( ) lim f(a n) = f lim a n = f(a). n n 8.8 Example. Since lim n /n = and since f(x) = cosx is continuous at x = we have lim cos = cos =. n n 8.9 Example. Limits of rational functions of n can all be computed by dividing numerator and denominator by the highest occurring power of n. Here is an example: n lim n n + 3n = lim n ( n ) + 3 n = + 3 = 7 Earl of Sandwich, John Montague, was born Nov. 3, 78 and died April 3, 79. In John Montague s life time, he was a Politician, Naval Administrator, member of the Bedford Gang, Duke of Bedford, Lord of the Admiralty, Secretary of state, leader of the Monks, invented the sandwich, discovered the Hawaiian Islands, and devoted his life to gambling. When the Earl of Sandwich was gambling, he would sit at the gaming table for hours without meals. One day he sent his menservants to get him: slices of bread, meat, and cheese. Thus made the sandwich. He did this so he wouldn t get his cards greasy. The sandwich was named after him in 76. (from site/academic/kimd site/earls.html)