5 Further Differential Calculus

Transcription

1 MATH1983/1954 Mathematics 0C/1C. General information for the academic year : Lecturer: Dr Theodore Voronov, Room.109, School of Mathematics, Alan Turing Building, Lectures: Tuesdays 11am in Renold/D7 and Wednesdays 9am in Sackville/C14. Tutorials: Thursdays 11am : George Begg/C3 (optometry only) and Fridays 9am : rooms announced on the webpage (foundation). Assessment: coursework 1 (deadline in week 4): 10%, coursework (deadline in week 10): 10%, hour end of semester exam: 80%. Course webpage: 5 Further Differential Calculus Our task is to recall what we have alrea known from Differential Calculus and to develop our knowledge further. 5.1 Differential and derivative The main definition Consider a function y = f(x). We say that x is the argument or independent variable, and y is called the dependent variable. We want to stu the behavior of y when the argument x changes to x + h, for some h. The number h is called the increment of x. Let us examine the difference f(x + h) f(x). It is the increment of y corresponding to an increment of x. It is a function of h (as well as of x) and we denote it y(x; h). ( is the capital Greek letter delta, which is commonly used, among other things, for denoting a difference.) We can also write x(x; h) = h for the increment of x. Example 1. y = kx + n. Then y(x; h) = k(x + h) + n (kx + n) = kh. So y(x; h) = kh is a linear function of h. The coefficient k does not depend on x. In general, y(x; h) is of course not a linear function of the increment h, but y(x; 0) = 0 and we may expect that y(x; h) is approximately linear for small h. 1

2 y Q f(x+h) T f(x) P R 0 x x+h x Definition 1. A function y = f(x) is called differentiable if for each x, y(x; h) = f(x + h) f(x) = k(x) h + α(x, h)h (1) where α(x, h) 0 when h 0, i.e., if the increment y(x; h) can be approximated by a linear function of the increment of the independent variable. The linear function k(x) h of the increment h is called the differential of y = f(x) at x. Notation: df or, so that (x; h) = k(x)h. The number k(x), depending on x, is called the derivative of y = f(x) at x. Notation: y or f, so that f (x) = k(x). Hence the relation between the differential and the derivative: (x; h) = f (x) h, (so they both determine each other). On the diagram above, the points P = (x, f(x)) and Q = (x + h, f(x + h)) on the graph represent the change of y = f(x) when x changes from its initial value to x + h. The horizontal segment P R, where R = (x + h, f(x)), represents the increment h of the independent variable. The vertical segment RQ represents the corresponding increment y(x; h) of y. The straight line P T is tangent to the graph at P. Here the point T is on the intersection of the tangent with the vertical line through R. The segment RT represents the differential (x; h). When h 0, both the increment y(x; h) of y and the differential (x; h) tend to zero. We may

3 say that both are small (or infinitesimal ) when h 0. Their difference, however, which is y(x; h) (x; h) and is represented by the segment QT on the diagram, is smaller than both when h 0 (or, as they say, is an infinitesimal of higher order compared with h). This is exactly what is meant by saying that the line P T is tangent to the graph at P and which is expressed algebraically by the main relation (1). Example. For y = x we see that y(x; h) = h, so dx(x; h) = h. Therefore for each h (and each x), (x; h) = f (x) dx(x; h), and we can simply write = f (x) dx, from where we obtain the familiar expression for the derivative: f (x) = dx, which can be understood now as a genuine fraction. On the diagram above, the derivative y (x) is the slope (or gradient) of the tangent line P T. It is the ratio of the segments RT and P R, and is therefore the limit of the ratio of the segments RQ and P R. We recover the familiar expression for the derivative as the limit of the difference ratio: or, equivalently, f f(x + h) f(x) (x) = lim, h 0 h dx = lim x 0 Note that we have a clear picture of dx and as linear functions of the increment (for fixed x). One should not imagine them as the limit of x and the limit of y for x 0 (since both limits are simply zero), but rather think of as the main linear part of the increment y as a function of x = h when h 0 (in the assumption that this main part is indeed linear, i.e., the assumption of the differentiability). Since in the sequel we deal exclusively with differentiable functions (i.e., for which the differential and derivative make sense), let us look at two examples where there is no differentiability. Geometrically this means that either there is no tangent for the graph y = f(x) or that the slope k(x) of the tangent is not defined, which happens if the tangent is vertical. y x 3

4 Example 3. Consider the function y = x (the absolute value). At x = 0 there is no welldefined tangent to the graph. The graph is not smooth at this point. If we approach zero from the left, the tangent coincides with y = x; if we approach zero from the right, the tangent coincides with y = x. At x = 0 therefore there is no tangent. The function is not differentiable at x = 0. Example 4. Consider the function y = x (the arithmetic square root) for x 0. The tangent at x = 0 is vertical, which means that the derivative becomes infinite when we approach zero from the right. The function is not differentiable at x = Digression: limits In the above diagram, y = f(x), P is a fixed point, Q is a variable point, so x is a fixed number, and h is a variable. Let Q P from either the left (h < 0) or right (h > 0), and consider the values of f(x + h) as h 0. If h 0 and h > 0 (h < 0) write h 0+ (h 0 ). Consider the limits lim h 0 f(x + h), lim h 0+ f(x + h). If a limit gives a definite value, the limit is said to exist. If it does not give a definite value, the limit does not exist. (A precise definition of what is meant by a limit requires some technical language, which is not difficult, but we do not wish to go into it now. An intuitive understanding such as getting closer or as close as possible is entirely sufficient for our purposes.) For the function whose graph is shown in the diagram, the three numbers f(x), lim h 0+ f(x + h), lim h 0 f(x + h) all exist and are equal. In this case y = f(x) is said to be continuous at x. If it happens that the three numbers are not all equal or some of them do not exist, then y = f(x) is discontinuous at x. Many of the functions that we are familiar with are continuous, for example all polynomials, sin x, cos x, exp x. But, many, while being continuous for most values of x also have some points of discontinuity, for example rational functions (at points where the denominator is zero but the numerator is not), tan x, cosec x, sec x, cot x. Example Let f(x) = x and x = then lim ( + h 0 h) = = lim ( + h 0+ h). In this case, both limits and the value of the function exist and all three numbers are equal.. Let f(x) = 1 x and x = 0. Then f(0) is not defined. Also, if x = h and h > 0, then as h 0, takes larger and larger values without bound. We write f(x) as x 0+. Similarly, 1 h if h < 0, 1 h takes negative values with larger and larger magnitude. We write f(x) as x 0. In this case, the value of the function is not defined and either the limits do not exist, or if one allows the possibility of ± as a limit, then the limits are not equal. 3. The function f(x) = x + x x 4

5 is not defined at x = 0 since division by zero is not allowed. It is not the same function as g(x) = x + 1. But for any value of x other than 0, we do have f(x) = g(x) = x + 1 and hence, since g(x) is continuous, lim f(x) = 1, x 0 i.e., as x becomes closer and closer to 0, f(x) becomes closer and closer to 1. In this case the value of the function does not exist, while the limits exist and are equal. Theorem 1 (Properties of limits). Let c be a fixed number, and f(x) and g(x) functions of x. Then, provided the limits exist, ( ) lim f(x) + g(x) = lim f(x) + lim g(x), x c x c x c i.e. the limit of a sum equals the sum of the limits. ( ) ( )( ) lim f(x)g(x) = lim f(x) lim g(x) x c x c x c i.e. the limit of a product equals the product of the limits Examples of differentiation The process of finding the differential of a function is called differentiation. Since finding the differential is equivalent to finding the derivative, the latter is also referred to as differentiation. However, one should not mix the results (the differential the derivative). Consider examples of differentiation directly from definition. Example 6. Let y = x. Then we have ( ) ( ) f(x + h) f(x) (x + h) dx = lim x = lim h 0 h h 0 h ( ) ( ) x + xh + h x xh + h = lim = lim = lim(x + h) = x h 0 h h 0 h h 0 So, the derivative of x is x and the differential is xdx. The next example generalizes this result to higher powers of x. Example 7. Let y = x n, with n a positive integer. Then ( ) ( ) f(x + h) f(x) (x + h) n dx = lim x n = lim h 0 h h 0 h Expand (x + h) n using the binomial theorem only the first few terms will be needed: ( x n + nx n 1 h + 1 = lim n(n 1)xn h + 1n(n 1)(n 6 )xn 3 h h n x n ) h 0 h 5

6 = lim (nx n n(n 1)xn h + 16 ) n(n 1)(n )xn 3 h h n 1 h 0 canceling first the x n s and then an h from the numerator and denominator. Now, let h 0 then all but the first term becomes zero, so dx = nxn 1 Note: Cancelling the h s, eliminated the problem of division by 0 and effectively enabled us to put h = 0 in the remaining formula. For the next example we need the following remarkable limit. Lemma 1. For θ measured in radians, as θ 0, we have Alternatively, sin θ θ sin θ lim θ 0 θ 1. = 1. Proof. We use a geometric argument. Let the circle be of radius r. D θ C A B Since the line CD is a tangent to the circle, angle ADC is a right angle. Now consider the three areas: ADB < sector ADB < ADC. The area of ADB is 1 r sin θ, the area of the sector is θ π πr = 1 r θ and the area of ADC is 1 r tan θ. Hence, dividing by 1 r gives 1 r sin θ 1 r θ 1 r tan θ sin θ θ tan θ 6

7 The first inequality gives while the second gives sin θ θ 1 cos θ sin θ θ. Now let θ 0 then cos θ 1 and so sin θ is trapped between 1 and a number tending to 1. θ Hence the result. Example 8. Let y = sin x. Then ( ) f(x + h) f(x) dx = lim h 0 h = lim h 0 ( ) sin(x + h) sin x Now, in the identity sin X sin Y = cos( X+Y ) sin( X Y ), let X = x + h, Y = x then dx = lim h 0 = lim h 0 ( ( cos( x+h ) sin( h ) h cos(x + h )sin ( h ) h = lim (cos(x + h ) ) lim h 0 h 0 ) ( sin( h h ) ) ) h. By Lemma 1, From here, sin( h ) h sin θ lim θ 0 θ = 1. 1 as h 0, which leaves us with (cos(x dx = lim + h ) ) = cos x. h 0 Similarly, we may differentiate x n for positive integral n, and cos x (see example sheets) and also the functions e x and ln x. We can also deduce the derivative of x α for any rational or real power α. We shall come back to these examples after deducing the fundamental properties of differentiation. 7

8 5. Properties of differentiation and further examples 5..1 Fundamental properties We shall consider the main properties of differential and derivative. Since, for a function y = f(x), the differential and the derivative and can be expressed from each other as = f (x) dx and we mainly speak about differentials. f (x) = dx, Theorem (Linearity). Let u(x) and v(x) be functions of x and let k be a constant number. Then d(ku) = k du d(u + v) = du + dv, and similarly for the derivatives. Proof. Let f(x) = u(x) + v(x). Consider f(x + h) f(x). We have, from the definition, u(x + h) u(x) = u (x)h + α(x, h)h where α 0 when h 0, and v(x + h) v(x) = v (x)h + β(x, h)h where β 0 when h 0. By adding, we obtain f(x + h) f(x) = ( u (x) + v (x) ) h + ( α(x, h) + β(x, h) ) h. Now, α(x, h) + β(x, h) 0 when h 0, so we see that df(x)(h) = ( u (x) + v (x) ) h or df = du + dv as claimed. Similarly for ku. A function f(x) may be the product of two simpler functions, e.g. f(x) = (x + 1)e x is a product of x + 1 and e x. Then f(x) may be differentiated using the product rule. 8

9 Theorem 3 (Product Rule). For arbitrary differentiable functions u(x) and v(x), d(uv) = du v + u dv and (uv) = u v + uv. Proof. By the definition, we have u(x + h) u(x) = u (x)h + α(x, h)h where α 0 when h 0, and v(x + h) v(x) = v (x)h + β(x, h)h where β 0 when h 0. Consider the increment of f(x) = u(x)v(x) : u(x + h)v(x + h) = ( u(x) + u (x)h + α(x, h)h )( v(x) + v (x)h + β(x, h)h ) = u(x)v(x) + u (x)v(x)h + u(x)v (x)h+ ( ) u(x)β(x, h) + u (x)β(x, h)h + α(x, h)v(x) + α(x, h)v (x)h + α(x, h)β(x, h)h h. Note that each term inside the large bracket tends to zero when h 0. Therefore we may write the obtained equality as u(x + h)v(x + h) = u(x)v(x) + ( u (x)v(x) + u(x)v (x) ) h + γ(x, h)h where γ 0 when h 0. Hence df(x)(h) = ( u (x)v(x) + u(x)v (x) ) h, i.e., df = du v + u dv as claimed. The product rule is often referred also to as the Leibniz rule. Example 9. Let y = x sin x. Let u = x and v = sin x. Then = d(uv) = du v+u du = d(x ) sin x+x d sin x = x dx sin x+x cos x dx = (x sin x+x cos x) dx. 9

10 Example 10. Let y = x ln x. We can let u = x and v = ln x. Then = d ( x ln x ) = dx ln x + x d ln x = ln x dx + x 1 dx = (ln x + 1) dx. x Remark 1. We have used the (possibly familiar) formula d ln x = dx x, which we have not deduced yet from the definition. This is done below, see Example 18. Suppose we wish to differentiate the function y = sin (x ). This is a function of a function in the sense that the output from the function x is the input for the sine function. Let u = x, then y = sin u, and we now have two simpler functions, each of which we can differentiate. The following theorem enables us to differentiate the composite function y = sin (x ). A function like this is called a function of a function or the composition of functions. In the example, y is the composition of the function sin and the function x. Theorem 4 (Chain Rule). Let y = f(u) and u = g(x) then the differential of the function y = f(g(x)) is given by and the derivative is given by = du dx = du du dx dx, du dx. Remark. The formula for the differential of y as a function of x can be rewritten as = du du, where in the RHS, du is the differential of u as a function of x. Note however that this formula has the same form as the formula for where y is regarded as a function of u as an independent variable. This demonstrates the fact that the general formula for the differential = f (u)du holds regardless of whether u is an independent variable or a function of some other variable. This invariance of form is a particular advantage of working with differentials. Proof. Example 11. Differentiate y = sin (x ). Let u = x then y = sin u and Hence, using the chain rule = du du = cos u, du dx = x. du dx dx = (cos u) x dx = x cos x dx. 10

11 With practice it is possible to use the chain rule without explicitly making the substitution. Example 1. Let y = e sin x. Let u = sin x, then y = e u. Now, applying the chain rule: dx = du du dx = eu cos x = cos x e sin x Example 13. Let y = ln (x + 3). Let u = x + 3, then y = ln u. Now, applying the chain rule: dx = du du dx = 1 u 1 = 1 x + 3 Example 14. Let y = 1 w = w 1 for any function w of x. Now, applying the chain rule: dx = dw dw dx = w dw dx = 1 w There is one corollary from the product rule and the chain rule, which concerns differentiation of a quotient. Theorem 5 (Quotient Rule). Consider the function dw dx y = u v for differentiable functions u and v of x. Then rule: or = du v u dv v y = u v u v v Proof. Apply the product rule to uv 1 and then the chain rule to v 1. Example 15. Consider y = tan x = sin x. By the quotient rule: cos x = d sin x cos x sin x d cos x cos x So d tan x = dx cos x, and this is the formula for the differential of tangent. = (cos x dx) cos x sin x ( sin x) dx = cos x (cos x + sin x) dx cos x = dx cos x. This was an example of an application of the quotient rule. However, we can give an advice: try to avoid using the quotient rule! Better work directly with the product rule and chain rule. Example 16. Let y = ex x. We have: = d(e x ) 1 x + ex d ( ) 1 = e x dx 1x ( x + ex dx ) = ex (x 1) dx. x x 11

12 5.. Implicit Differentiation Let y = f(x), then y may be called an explicit function of x, e.g., y = x. Substituting a value of x gives a value of y. However, consider the example x + y = 1, which is the equation of a circle. In this case substituting a value of x such that 1 x +1 gives a value for y. To find y further operations are required, such as taking the square root. Hence, y, for all values of x, is not a function of x, since rearranging the expression to get y = ± 1 x, we may have two values for y, or 1 value or even none (a function must give exactly one value). Nevertheless, in a neighborhood of a point of the circle (except for the points ( 1, 0) and (1, 0)), we can find a unique value of y continuously depending on x. A situation such as above is described by the term implicit function : if there is a relation between x and y written as an equation f(x, y) = 0, () we may think that it defines y as a function of x implicitly, so that y becomes a genuine function of x near each point of the curve on the xy plane specified by equation (). (It becomes an explicit function if we somehow manage to solve the equation for y.) For a function y = y(x) defined implicitly, by an equation such as (), it is possible to find the derivative /dx without explicitly solving the equation. Example 17. We can differentiate the equation of the circle to obtain: Hence and rearranging this gives: x dx + y = 0. x + y dx = 0 dx = x y. Note that the right-hand side contains y, and it may be eliminated from the formula by solving the equation for y, if required. (From the answer obtained for the slope = x of the tangent dx y line to the circle at a point (x, y), we can immediately deduce that this tangent is perpendicular to the line through the origin and the point (x, y), whose slope is y x.) Implicit differentiation is very useful in finding the derivatives of inverse functions. 1

13 5..3 Differentiation of inverse functions Example 18. Differentiate ln x. We have e ln x = x. By differentiating both sides and applying the chain rule we obtain: d ( e ln x) = dx e ln x d (ln x) = dx d (ln x) = dx e ln x d (ln x) = dx x. Therefore the formula for differentiating natural logarithm is d ln x = dx x or d ln x dx = 1 x. Example 19. Consider arcsin x, the inverse function for sin. (Note that π arcsin x π.) We have sin(arcsin x) = x. By differentiating both sides and applying the chain rule we obtain: d (sin(arcsin x)) = dx cos(arcsin x) d arcsin x = dx 1 sin (arcsin x) d arcsin x = dx 1 x d arcsin x = dx d arcsin x = dx 1 x. (We have legitimately expressed cos(arcsin x) as 1 sin (arcsin x), with the positive sign, because in the range π θ π, cos θ 0, hence cos θ = + 1 sin θ.) Therefore: d arcsin x = dx 1 x and d dx arcsin x = 1 1 x. Example 0. Let y = arctan x, the inverse of the tangent. We have tan(arctan x) = x and can differentiate both sides: ( ) d tan(arctan x) = dx. 13

14 By the chain rule we obtain 1 cos (arctan x) d arctan x = dx, so d arctan x = cos (arctan x) dx, To simplify the RHS, we need to express cos (rather, its square) in terms in tan. We have tan θ = sin θ cos θ tan θ = sin θ cos θ = 1 cos θ cos θ = 1 cos θ 1, from which we get tan θ + 1 = 1 cos θ or cos θ = tan θ. Therefore d arctan x = which is the formula for the differential of arctan x. dx 1 + tan arctan x = dx 1 + x, 5..4 More examples and summary Example 1. Consider y = x α = e α ln x for an arbitrary real α. (Unlike the case of integral powers x n, this is well-defined for x > 0.) We have, by the chain rule, = e α ln x d(α ln x) = e α ln x α d ln x = α x α dx x = α xα 1 dx. We note that the formula is the same as for the integral powers such as x n. We produce a summary of results. Some of these have been proved, or solved in examples. Others are on exercise sheets. 14

15 Function y = x n Differential = nx n 1 dx (integral power) y = x α = α x α 1 dx (arbitrary power, x > 0) y = e x = e x dx y = a x = a x ln a dx (base a > 0, a 1) y = sin x = cos x dx y = cos x = sin x dx y = tan x = dx cos x y = cot x = dx sin x y = ln x = dx x dx y = arcsin x = 1 x y = arccos x = dx 1 x y = arctan x = dx 1 + x Using these basic functions, and the fundamental properties of differentiation such as the product rule and the chain rule, it is possible to differentiate a wide range of functions. Two more examples may suffice: ex Example. Let y =. This is a quotient, so we use that rule or the product rule, but sin x along the way we also need apply the chain rule. dx = v du u dv dx dx v = (sin x)(xex) (e x )(cos x) sin x e x = (x sin x cos x) sin x Example 3. Let y = (1 + cos x ) 6. This is a function of a function of a function: let u = 1+cos x then y = u 6 may be differentiated with respect to u. To differentiate u = 1+cos x we may use the chain rule again to differentiate u w.r.t x. So let v = x then u = 1 + cos v. The formula for the derivative becomes dx = du dv du dv dx Note: this is the product of all the functions, differentiated with respect to their argument and multiplied together (justifying the term chain rule). Hence, dx = 6u5 ( sin v) x = 6(1 + cos x ) 5 ( x sin x ) = 1x sin x (1 + cos x ) 5 15

16 5.3 Taylor formula There is a very useful notation called the little o and big O notation. Consider a function f(x). We say that some other function g(x) is an infinitesimal of higher order with respect to f(x) when x a or is a little o of f(x), notation: if g(x) = o(f(x)) for x a, g(x) f(x) 0 for x a. (Here a can be finite or infinite, and it is also possible that we have a infinite sequence f(n) where n = 1,, 3,... instead of a function f(x) of a continuous argument x, and the limit is considered for n. We write f(x) and x a for concreteness.) x a. Note that in this notation the function f(x) itself is not assumed to be tending to zero for Example 4. Suppose g(x) = o(1) for x a. According to the above definition, this means that g(x) 1 = g(x) 0 for x a. Functions g(x) tending to zero (for x a) are called infinitesimals. Example 5. Suppose h 0. Then h itself, h, h 3, etc., are all infinitesimals. Now, h is an infinitesimal of higher order compared to h (when h 0) or for h 0 because h = o(h) h h = h 0. This little o notation can be used for expressing the definition of a derivative. Recall that the number f (x), the derivative of f(x) at x, is defined by the identity This can be written as f(x + h) = f(x) + f (x) h + α(h) h where α(h) 0 for h 0. f(x + h) = f(x) + f (x) h + o(h) for h 0. 16

17 An advantage of using such a notation is that it allows to avoid introducing explicitly (irrelevant) functions such as α(h). One should keep in mind that o(h) stands not for a particular function, but for an arbitrary function of this class. From here follow simple rules of manipulation: o(h) + o(h) = o(h), o(h)o(h) = o(h); also o(h) f(h) = o(h) for an arbitrary function f(x) which is bounded near h = 0. Now there is the big O notation. We say that a function g(x) is of the same order as a function f(x) for x a or that g(x) is a big O of f(x) for x a, if there is a constant C such that g(x) = O(f(x)), g(x) f(x) C for x close to a. Big O carries more information than little o as the following example shows. Example 6. In the setup of Example 5, we have f(h) = O(h) f(h) = o(1). Indeed, if f(h) = O(h), that means that there is a constant C such that f(h) C h for all h near zero. Clearly, if h 0, then it follows that f(h) 0 or that f(h) = o(1) for h 0 (see Example 4). Also, f(h) = O(h n+1 ) f(h) = o(h n ) for h 0. Indeed, for f(h) being O(h n+1 ) (for h 0) means that there is a constant C such that f(h) C h n+1. Obviously, from here f(h) h C h, so f(h) 0 when h 0, and this n h n is f(x) being o(h n ). Example 7. What does it mean that f(x) = O(1) for x a? We have, by the definition, f(x) 1 = f(x) C for all x close to a. That simply means that f(x) is bounded in a neighborhood of x = a. Example 8. Suppose f(x) = O(x n ) (where n = 1,, 3,...) for x 0. Then Indeed, we are given that f(x) x = O(xn 1 ). f(x) x n C 17

18 for a constant C (for all x close to 0). This is the same as f(x)/x x n 1 C, which means that f(x)/x = O(x n 1 ) as claimed. Symbolically we have the equality for x 0. O(x n ) x = O(x n 1 ) The main statement of this subsection is as follows. Theorem 6 (Taylor formula). Suppose a function f(x) can be differentiated infinitely many times. Then for an arbitrary n = 1,, 3,..., f(x) = f(x 0 ) + f (x 0 ) (x x 0 ) + 1 f (x 0 ) (x x 0 ) + 1 3! f (x 0 ) (x x 0 ) for x x 0. 1 n! f (n) (x 0 ) (x x 0 ) n + O((x x 0 ) n+1 ), (3) Before proving this theorem, we can make some remarks explaining its meaning. Remark 3. When n = 1, equation (3) gives f(x) = f(x 0 ) + f (x 0 )(x x 0 ) + O((x x 0 ) ), for x x 0. Writing x = x 0 + h, we recognize the defining of the derivative at x 0, in a slightly stronger form. Indeed, O((x x 0 ) ) is also o((x x 0 )). Therefore the Taylor formula for n 1 should be regarded as the approximation of a function by a polynomial of degree n as compared to the approximation by a linear function, which is the idea of (first) derivative. Remark 4. The appearance of the higher derivatives such as f (n) (x 0 ) and the factorials in (3) is not by chance. If we suppose that there is a general formula such as f(x) = a 0 + a 1 (x x 0 ) + a (x x 0 ) + a 3 (x x 0 ) a n (x x 0 ) n + O((x x 0 ) n+1 ), (4) (for x x 0 ) with some coefficients a 0, a 1, a,..., then the the coefficients a k are uniquely defined and have to be exactly as in (3). Indeed, substitute x = x 0 into (4). We get f(x 0 ) = a 0. 18

19 If we differentiate both sides of (4) once, we obtain f (x) = 0 + a 1 + a (x x 0 ) + a 3 3(x x 0 ) a n n(x x 0 ) n 1 + O((x x 0 ) n ). (5) (One can easily see that differentiating O((x x 0 ) k ) gives O((x x 0 ) k 1 ) for any k =, 3,....) By substituting x = x 0 to (5), we get f (x 0 ) = a 1. If we differentiate the second time, i.e., differentiate (5), we get f (x) = a + a 3 3 (x x 0 ) a n n(n 1)(x x 0 ) n + O((x x 0 ) n 1 ), (6) from where f (x 0 ) = a. Similarly, by differentiating formula (4) k times and substituting x = x 0, we obtain f (k) (x 0 ) = k a k = k! a k for all k. So we necessarily have a k = 1 k! f (k) (x 0 ). To establish Taylor s formula, we first prove an auxiliary statement known as Hadamard s Lemma. Lemma (Hadamard s Lemma). For an arbitrary infinitely differentiable function f(x), there is an expansion where g(x) is also infinitely differentiable. f(x) = f(x 0 ) + g(x)(x x 0 ) (7) (Note that (7) is Taylor s formula for n = 0. Indeed, g(x)(x x 0 ) = O((x x 0 )) because g(x)(x x 0 ) x x 0 = g(x) is differentiable and in particular bounded near x 0.) Proof of Hadamard s Lemma. To obtain (7), consider h(t) = f ( x 0 + t(x x 0 ) ). h(0) = f(x 0 ) and h(1) = f(x). By the chain rue, We have dh dt = f ( x 0 + t(x x 0 ) ) (x x 0 ). 19

20 Therefore gives f(x) f(x 0 ) = 1 0 h(1) = h(0) = 1 0 dh dt dt f ( x 0 + t(x x 0 ) ) ( 1 (x x 0 ) dt = Denoting the function in brackets by g(x) we arrive at 0 f ( x 0 + t(x x 0 ) ) ) dt (x x 0 ). f(x) f(x 0 ) = g(x) (x x 0 ), as desired. Remark 5. Hadamard s Lemma should be compared with Theorem 3 from (the Remainder Theorem ) and Corollary 3 there. In the context of differentiable functions, it tells the same as Theorem.3 tells for polynomials: the remainder of the division of a function f(x) by x x 0 is the value f(x 0 ) at x 0 ; and if f(x 0 ) = 0, the function f(x) is divisible (within the class of infinitely differentiable functions) by x x 0. Now we can prove the main Theorem 6. Proof of Taylor s formula. Consider a function f(x) and fix a point x 0. From Hadamard s lemma, we can write f(x) = a 0 + g 0 (x) (x x 0 ) (8) where a 0 is a constant and g 0 (x) is a function of the same class as f(x) (possessing infinitely many derivatives). We can apply Hadamard s lemma to g 0 (x): g 0 (x) = a 1 + g 1 (x)(x x 0 ). (9) By substituting (9) into (8) and multiplying through, we arrive at f(x) = a 0 + a 1 (x x 0 ) + g 1 (x) (x x 0 ) (10) with a new function g 1 (x). We apply Hadamard s lemma to g 1 (x), etc. Repeating this many times, we can arrive at Taylor s formula for arbitrary n. More formally, if we alrea have a formula f(x) = a 0 + a 1 (x x 0 ) + a (x x 0 ) + a 3 (x x 0 ) a n 1 (x x 0 ) n 1 + g n 1 (x) (x x 0 ) n, (11) 0

21 with constants a 0, a 1,..., a n 1 and a function g n 1 (x), we apply Hadamard s lemma to g n 1 (x) : and substitute (1) into (11). This gives, after simplification, g n 1 (x) = a n + g n (x)(x x 0 ), (1) f(x) = a 0 + a 1 (x x 0 ) + a (x x 0 ) + a 3 (x x 0 ) a n 1 (x x 0 ) n 1 + a n (x x 0 ) n ) + g n (x) (x x 0 ) n+1. (13) The remainder term g n (x) (x x 0 ) n+1 is obviously O ( (x x 0 ) n+1) because g(x) is differentiable and hence bounded near x = x 0. From Remark 4 we know that the coefficients a k are necessarily f (k) /k! as claimed. This completes the proof. The Taylor formula (3) for f(x) is referred to as the Taylor expansion of a function f(x) at x 0 to order n (or the nth order expansion). Particularly useful are the Taylor expansions at x 0 = 0, sometimes called the Maclaurin expansions. Example 9 (Generalized binomial formula). Let α be a real number. Find the Taylor expansion of f(x) = (1 + x) α at zero. Solution. Clearly f(0) = 1. We have f (x) = α(1 + x) α 1, f (x) = α(α 1)(1 + x) α, etc. For an arbitrary n, f (n) (x) = α(α 1)... (α n + 1)(1 + x) α n. It follows that f (n) (0) = α(α 1)... (α n + 1). Hence the Taylor expansion takes the form (1+x) α α(α 1) = 1 +α x + x α(α 1)(α ) + x 3 α(α 1)... (α n + 1) x n +O(x n+1 ), 3! n! for x 0. This is a generalization of the binomial formula for (1 + x) n for a positive integer n. (Note that (1 + x) n is a polynomial of degree n, so the Taylor expansion of it of order greater than n terminates at the nth term. Unlike that, the Taylor expansion of (1 + x) α for a real power α does not terminate.) Example 30. Find the generalized binomial formula for α = 1. Solution. From the general formula above we find the nth coefficient as ( 1)( )... ( 1 n + 1)/n! = ( 1)( )... ( n)/n! = ( 1) n. Therefore (1 + x) 1 = 1 x + x x ( 1) n x n + O(x n+1 ) for x 0. Example 31. Find the Taylor expansion of 1 + x (at zero) to the third power of x. This is a particular case of the generalized binomial formula, for α = 1. We obtain: 1

22 1 1 + x = 1 + x + 1 ( ) 1 1! 1 x ! ( ) ( ) x 3 + O(x 4 ) = x 1 8 x x3 + O(x 4 ). Example 3. Find the Taylor expansion at zero for the function f(x) = e x. Solution: we have f (x) = e x, f (x) = e x, etc. At zero, f(0) = f (0) = f (0) =... = 1. Therefore e x = 1 + x + 1! x + 1 3! x ! x n! xn + O(x n+1 ). Example 33. Find the Taylor expansion at zero for the function f(x) = sin x. Solution: we have f (x) = cos x, f (x) = sin x, f (x) = cos x, etc. At zero, f(0) = 0, f (0) = 1, f (0) = 0, f (0) = 1, etc. alternating +1 and 1 for the derivatives of odd orders.) Therefore (We get 0 for all derivatives of even orders and sin x = x 1 3! x ! x5 1 7! x ( 1) n 1 (n + 1)! xn+1 + O(x n+3 ). Example 34. Find the Taylor expansion at zero for the function f(x) = cos x. Solution: similar to the previous example. We have f (x) = sin x, f (x) = cos x, f (x) = sin x, etc. At zero, we have f(0) = 1 and we get 0 for all derivatives of odd orders and alternating +1 and 1 for the derivatives of even orders.) Therefore cos x = 1 1! x + 1 4! x4 1 6! x ( 1) n 1 (n)! xn + O(x n+ ). Example 35. Find the Taylor expansion at zero for the function f(x) = ln(1 + x). Solution: f(0) = ln 1 = 0, and we have f (x) = 1 1+x = (1+x) 1, f (x) = ( 1)(1+x), f (x) = ( 1)( )(1+x) 3, and f (n) (x) = ( 1)()... ( n+1)(1+x) n = ( 1) n 1 (n 1)! (1+x) n, from where f (n) (0) = ( 1) n 1 (n 1)!. Therefore 1 n! f (n) (0) = ( 1) n 1 1 n has the form ln(1 + x) = x 1 x x3 1 4 x ( 1) n 1 1 n xn + O(x n+1 ). and the Taylor expansion Taylor formula has many applications. One such application is to finding limits. Example 36. Calculate the limit: lim x x 1. x

23 It is not possible to simply substitute x = 0 into the top and bottom of the fraction, because the result will be the indeterminacy 0. To find the actual value of the limit we can use the Taylor 0 expansion of the square root to the first order: 1 + x = x + O(x ) = 1 + x + O(x ). Hence lim x x 1 x = lim x x + O(x ) 1 x = lim x 0 x + O(x ) x ( ) = lim 1 + O(x) = 1. x 0 More examples of calculating limits by this method the reader can find in the problem sheet and the worked solutions. 3