2 Continuity, Differentiability and Taylor s theorem

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1 Continuity, Differentiability and Taylor s theorem. Limits of real valued functions Let f(x) bedefinedon(a, b) except possibly at x 0. Definition... We say that lim x x0 f(x) =L if, for every real number ɛ>0, there exists a real number δ>0 such that Equivalently, 0 < x x 0 <δ = f(x) L <ɛ. (.) Remark.. The above definition is equivalent to: for any sequence {x n } with x n x 0, we have f(x n ) L as n. Proof. Suppose lim x x0 f(x) exists. Take ɛ>0 and let {x n } be a sequence converging to x 0. Then there exists N such that x n x 0 <δfor n N. Then by the definition f(x n ) L <ɛ. i.e., f(x n ) L. For the other side, assume that x n c = f(x n ) L. Suppose the limit does not exist. i.e., ɛ 0 > 0 such that for any δ>0, and all x x 0 <δ,wehave f(x) L ɛ 0. Then take δ = n and pick x n in x n x 0 < n,thenx n x 0 but f(x n ) L ɛ 0.Not possible. Theorem... If limit exists, then it is unique. Proof. Proof is easy. Examples: (i) lim( 3x x ) =.Letɛ>0. We have to find δ>0such that (.) holds with L =/. Working backwards, (ii) Prove that lim x f(x) = 4, where f(x) = 3 x <ɛwhenever x <δ:= 3 ɛ. { x x x = Problem: Show that lim sin( x 0 x )doesnotexist. Consider the sequences {x n } = { nπ }, {y n} = { nπ + π }. Then it is easy to see that ( ) ( ) x n,y n 0andsin 0, sin. In fact, for every c [, ], we can find x n y n

2 a sequence z n such that z n 0 and sin( z n ) c as n. By now we are familiar with limits and one can expect the following: Theorem..3. Suppose lim f(x) =L and lim g(x) =M, then x c x c. lim(f(x) ± g(x)) = L ± M. x c. f(x) g(x) for all x in an open interval containing c. Then L M. f 3. lim(fg)(x) =LM and when M 0, lim x c x c g (x) = L M. 4. (Sandwich): Suppose that h(x) satisfies f(x) h(x) g(x) in an interval containing c, and L = M. Then lim h(x) =L. x c Proof. We give the proof of (iii). Proof of other assertions are easy to prove. Let ɛ>0. From the definition of limit, we have δ,δ,δ 3 > 0 such that <δ = f(x) L < = f(x) <Nfor some N>0, <δ = f(x) L < ɛ M, and <δ 3 = g(x) M < ɛ N. Hence for <δ= min{δ,δ,δ 3 },wehave f(x)g(x) LM f(x)g(x) f(x)m + f(x)m LM f(x) g(x) M + M f(x) L <ɛ. To prove the second part, note that there exists an interval (c δ, c + δ) around c such that g(x) 0in(c δ, c + δ). Examples: (i) lim x m =0(m>0). (ii) lim x sin x =0. x 0 x 0 Remark: Suppose f(x) is bounded in an interval containing c and lim g(x) = 0. Then x c lim f(x)g(x) =0. x c Examples: (i) lim x 0 x sin x =0. (ii) lim x 0 x ln x =0.

3 One sided limits: Let f(x) isdefinedon(c, b). The right hand limit of f(x) atc is L, if given ɛ>0, there exists δ>0, such that <δ = f(x) L <ɛ. Notation: lim f(x) =L. Similarly, one can define the left hand limit of f(x) atband is x c + denoted by lim f(x) =L. x b Both theorems above holds for right and left limits. Proof is easy. Problem: Show that lim θ 0 sin θ θ =. Solution: Consider the unit circle centered at O(0, 0) and passing through A(, 0) and B(0, ). Let Q be the projection of P on x axis and the point T is such that A is the projection of T on x axis. Let OT be the ray with AOT = θ, 0 <θ<π/. Let P be the point of intersection of OT and circle. Then ΔOPQ and ΔOTA are similar triangles and hence, Area of ΔOAP < Area of sector OAP < area of ΔOAT. i.e., dividing by sin θ, weget> sin θ θ Now use the fact that sin θ θ sin θ< θ< tan θ is even function. At this stage, it is not difficult to prove the following: Theorem..4. lim x a f(x) =L exists sin θ > cos θ. Now lim cos θ = implies that lim θ 0 + θ 0 + θ lim x a + f(x) = lim x a f(x) =L. =. Limits at infinity and infinite limits Definition..5. f(x) has limit L as x approaches +, if for any given ɛ>0, there exists M>0 such that x>m = f(x) L <ɛ. Similarly, one can define limit as x approaches. Problem: (i) lim x =0, (ii) lim x =0. (iii) lim sin x does not exist. x x Solution: (i) and(ii) are easy. For (iii), Choose x n = nπ and y n = π +nπ. Then x n,y n and sin x n =0,siny n =. Hence the limit does not exist. x Above two theorems on limits hold in this also. 3

4 Definition..6. (Horizontal Asymptote:) A line y = b is a horizontal asymptote of y = f(x) if either lim f(x) =b or lim f(x) =b. x x Examples: (i) y = is a horizontal asymptote for + x+ Definition..7. (Infinite Limit): A function f(x) approaches (f(x) )asx x 0 if, for every real B>0, there exists δ>0 such that 0 < x x 0 <δ = f(x) >B. Similarly, one can define for. Also one can define one sided limit of f(x) approaching or. Examples (i) lim x 0 x =, (ii) lim x 0 sin( )doesnotexist. x x For (i) givenb > 0, we can choose δ B. For (ii), choose a sequence {x n } such that sin x n =,say x n = π +nπ and y n = nπ. Then lim f(x n )= and n x n lim f(y n)=0, though x n,y n 0asn. n Definition..8. (Vertical Asymptote:) A line x = a is a vertical asymptote of y = f(x) if either lim f(x) =± or lim f(x) =±. x a + x a Example: f(x) = x+3. x+ x = is a vertical asymptote and y = is a horizontal asymptote.. Continuous functions Definition... A real valued function f(x) is said to be continuous at x = c if (i) c domain(f) (ii) lim x c f(x) exists (iii) The limit in (ii) is equal to f(c). In other words, for every sequence x n c, wemusthavef(x n ) f(c) asn. i.e., for a given ɛ>0, there exists δ>0 such that 0 < <δ = f(x) f(c) <ɛ. { x sin( x Examples: (i) f(x) = ) x 0 0 x =0 is continuous at 0. 4

5 Let ɛ>0. Then f(x) f(0) x. So it is enough to choose δ = ɛ. { x (ii) g(x) = sin x 0 x is not continuous at 0. 0 x =0 Choose x n = π +nπ. Then lim x n =0andf(x n )= x n. The following theorem is an easy consequence of the definition. Theorem... Suppose f and g are continuous at c. Then (i) f ± g is also continuous at c (ii) fg is continuous at c (iii) f is continuous at c if g(c) 0. g Theorem..3. Composition of continuous functions is also continuous i.e., if f is continuous at c and g is continuous at f(c) then g(f(x)) is continuous at c. Corollary: If f(x) is continuous at c, then f is also continuous at c. Theorem..4. If f,g are continuous at c, then max(f,g) is continuous at c. Proof. Proof follows from the relation and the above theorems. Types of discontinuities max(f,g) = (f + g)+ f g Removable discontinuity: f(x) is defined every where in an interval containing a except at x = a and limit exists at x = a OR f(x) is defined also at x = a and limit is NOT equal to function value at x = a. Thenwesaythatf(x) has removable discontinuity at x = a. These functions can be extended as continuous functions by defining the value of f to be the limit value at x = a. Example: f(x) = { sin x x x 0. Here limit as x 0is.Butf(0)isdefinedtobe0. 0 x =0 Jump discontinuity: The left and right limits of f(x) exists but not equal. This type of discontinuities are also called discontinuities of first kind. 5

6 { x 0 Example: f(x) =. Easy to see that left and right limits at 0 are different. x 0 Infinite discontinuity: Left or right limit of f(x) is or. Example: f(x) = x has infinite discontinuity at x =0. Discontinuity of second kind: If either lim f(x) or lim f(x) does not exist, then c x c x c + is called discontinuity of second kind. Example: Consider the function { 0 x Q f(x) = x Q Then f does not have left or right limit any point c. Indeed, f(c + n )andf(c + π n ) converges to different values. Properties of continuous functions Definition..5. (Closed set): A subset A of IR is called closed set if A contains all its limit points. (i.e., if{x n } A and x n c, then c A). Theorem..6. Continuous functions on closed, bounded interval is bounded. Proof. Let f(x) be continuous on [a, b] and let {x n } [a, b] be a sequence such that f(x n ) >n. Then {x n } is a bounded sequence and hence there exists a subsequence {x nk } which converges to c. Thenf(x nk ) f(c), a contradiction to f(x nk >n k. Theorem..7. Let f(x) be a continuous function on closed, bounded interval [a, b]. Then maximum and minimum of functions are achieved in [a, b]. Proof. Let {x n } be a sequence such that f(x n ) max f. Then{x n } is bounded and hence by Bolzano-Weierstrass theorem, there exists a subsequence x nk such that x nk x 0 for some x 0. a x n b implies x 0 [a, b]. Since f is continuous, f(x nk ) f(x 0 ). Hence f(x 0 )=maxf. The attainment of minimum can be proved by noting that f is also continuous and min f = max( f). Remark: Closed and boundedness of the interval is important in the above theorem. Consider the examples (i) f(x) = on (0, ) (ii) f(x) =x on IR. x 6

7 Theorem..8. Let f(x) be a continuous function on [a, b] and let f(c) > 0 for some c (a, b), Then there exists δ>0 such that f(x) > 0 in (c δ, c + δ). Proof. Let ɛ = f(c) > 0. Since f(x) is continuous at c, there exists δ>0 such that <δ = f(x) f(c) < f(c) i.e., f(c) <f(x) f(c) < f(c). Hence f(x) > f(c) for all x (c δ, c + δ). Corollary: Suppose a continuous functions f(x) satisfies b f(x)φ(x)dx = 0 for all continuous functions φ(x) on[a, b]. Then f(x) 0on[a, a b]. Proof. Suppose f(c) > 0. Then by above theorem f(x) > 0in(c δ, c + δ). Choose φ(x) sothatφ(x) > 0in(c δ/,c+ δ/) and is 0 otherwise. Then b f(x)φ(x) > 0. A a contradiction. Alternatively, one can choose φ(x) =f(x). Theorem..9. Let f(x) be a continuous function on IR and let f(a)f(b) < 0 for some a, b. Then there exits c (a, b) such that f(c) =0. Proof. Assume that f(a) < 0 <f(b). Let S = {x [a, b] :f(x) < 0}. Then[a, a + δ) S for some δ > 0andS is bounded. Let c = sups. We claim that f(c) = 0. Take x n = c + n,thenx n S, x n c. Therefore, f(c) = lim f(x n ) 0. On the otherhand, taking y n = c n, we see that y n S for n large and y n c, f(c) = lim f(y n ) 0. Hence f(c) =0. Corollary: Intermediate value theorem: Let f(x) be a continuous function on [a, b] and let f(a) <y<f(b). Then there exists c (a, b) such that f(c) =y Remark: A continuous function assumes all values between its maximum and minimum. Problem: (fixed point): Let f(x) be a continuous function from [0, ] into [0, ]. Then show that there is a point c [0, ] such that f(c) =c. Define the function g(x) =f(x) x. Theng(0) 0andg() 0. Now Apply Intermediate value theorem. Application: Root finding: To find the solutions of f(x) = 0, one can think of defining a new function g such that g(x) has a fixed point, which in turn satisfies f(x) =0. Example: () f(x) =x 3 +4x 0 in the interval [, ]. Define g(x) = ( 0 /. 4+x) Wecan check that g maps [, ] into [, ]. So g has fixed point in [, ] which is also solution 7

8 of f(x) = 0. Such fixed points can be obtained as limit of the sequence {x n },where x n+ = g(x n ),x 0 (, ). Note that g (x) = By Mean Value Theorem, z (see next section) 0 (4 + x) 3/ <. x n+ x n = g (z) x n x n x n x n Iterating this, we get x n+ x n < n x x 0. Therefore, {x n } is a Cauchy sequence. (see problem after Theorem.4.4). Uniformly continuous functions Definition: A function f(x) is said to be uniformly continuous on a set S, if for given ɛ>0, there exists δ>0 such that Here δ depends only on ɛ, not on x or y. x, y S, x y <δ = f(x) f(y) <ɛ. Proposition: If f(x) is uniformly continuous function for ANY two sequences {x n }, {y n } such that x n y n 0, we have f(x n ) f(y n ) 0asn. Proof. Suppose not. Then there exists {x n }, {y n } such that x n y n 0and f(x n ) f(y n ) >ηfor some η > 0. Then it is clear that for ɛ = η, there is no δ for which x y <δ = f(x) f(y) <ɛ. Because the above sequence satisfies x y <δ,but its image does not. For converse, assume that for any two sequences {x n }, {y n } such that x n y n 0we have f(x n ) f(y n ) 0. Suppose f is not uniformly continuous. Then by the definition there exists ɛ 0 such that for any δ > 0, x y <δ = f(x) f(y) >ɛ 0. Now take δ = n and choose x n,y n such that x n y n < /n. Then f(x n ) f(y n ) >ɛ 0. A contradiction. Examples: (i) f(x) =x is uniformly continuous on bounded interval [a, b]. Note that x y x + y x y b x y. So one can choose δ< ɛ b. 8

9 (ii) f(x) = is not uniformly continuous on (0, ). x Take x n =,y n+ n =,thenfornlarge x n n y n 0 but f(x n ) f(y n ) =. (iii) f(x) =x is not uniformly continuous on IR. Take x n = n + n and y n = n. Then x n y n = n 0, but f(x n) f(y n ) =+ n >. Remarks:. It is easy to see from the definition that if f,g are uniformly continuous, then f ± g is also uniformly continuous.. If f,g are uniformly continuous, then fg need not be uniformly continuous. This can be seen by noting that f(x) =x is uniformly continuous on IR but x is not uniformly continuous on IR. Theorem..0. A continuous function f(x) on a closed, bounded interval [a, b] is uniformly continuous. Proof. Suppose not. Then there exists ɛ>0 and sequences {x n } and {y n } in [a, b] such that x n y n < n and f(x n) f(y n ) >ɛ. But then by Bolzano-Weierstrass theorem, there exists a subsequence {x nk } of {x n } that converges to x 0.Alsoy nk x 0. Now since f is continuous, we have f(x 0 ) = lim f(x nk ) = lim f(y nk ). Hence f(x nk ) f(y nk ) 0, a contradiction. Corollary: Suppose f(x) has only removable discontinuities in [a, b]. Then f, the extension of f, is uniformly continuous. Example: f(x) = sin x for x 0and0forx =0on[0, ]. x Theorem... Let f be a uniformly continuous function and let {x n } be a cauchy sequence. Then {f(x n )} is also a Cauchy sequence. Proof. Let ɛ>0. As f is uniformly continuous, there exists δ>0 such that x y <δ = f(x) f(y) <ɛ. Since {x n } is a Cauchy sequence, there exists N such that Therefore f(x n ) f(x m ) <ɛ. m, n > N = x n x m <δ. 9

10 Example: f(x) = x is not uniformly continuous on (0, ). The sequence x n = n continuous. is cauchy but f(x n)=n is not. Hence f cannot be uniformly.3 Differentiability Definition.3.. A real valued function f(x) is said to be differentiable at x 0 if f(x 0 + h) f(x 0 ) lim h 0 h exists. This limit is called the derivative of f at x 0, denoted by f (x 0 ). Example: f(x) =x f xh + h (x) = lim =x. h 0 h Theorem.3.. If f(x) is differentiable at a, then it is continuous at a. Proof. For x a, we may write, f(x) f(a) f(x) =(x a) + f(a). (x a) Now taking the limit x a and noting that lim(x a) = 0 and lim f(x) f(a) (x a) get the result. = f (a), we Theorem.3.3. Let f,g be differentiable at c (a, b). Then f ± g, fg and f g is also differentiable at c (g(c) 0) Proof. We give the proof for product formula: First note that (fg)(x) (fg)(c) g(x) g(c) f(x) f(c) = f(x) + g(c). Now taking the limit x c, we get the product formula (fg) (c) =f(c)g (c)+f (c)g(c). Since g(c) 0andg is continuous, we get g(x) 0 in a small interval around c. Therefore f g (x) f g(c)f(c)+g(c)f(c) g(x)f(c) (c) =g(c)f(x) g g(x)g(c) 0

11 Hence (f/g)(x) (f/g)(c) = { f(x) f(c) g(c) f(c) } g(x) g(c) g(x)g(c) Now taking the limit x c, weget ( f g ) (c) = g(c)f (c) f(c)g (c). g (c) Theorem.3.4. (Chain Rule): Suppose f(x) is differentiable at c and g is differentiable at f(c), then h(x) :=g(f(x)) is differentiable at c and Proof. Define the function h as h (c) =g (f(c))f (c) h(y) = { g(y) g(f(c)) y f(c) g (f(c)) y f(c) y = f(c) Then the function h is continuous at y = f(c) andg(y) g(f(c)) = h(y)(y f(c)), so g(f(x)) g(f(c)) = h(f(x)) f(x) f(c). Now taking limit x c, we get the required formula. Local extremum: Apointx = c is called local maximum of f(x), if there exists δ>0 such that 0 < <δ = f(c) f(x). Similarly, one can define local minimum: x = b is a local minimum of f(x) if there exists δ>0 such that 0 < x b <δ = f(b) f(x). Theorem.3.5. Let f(x) be a differentiable function on (a, b) and let c (a, b) isalocal maximum of f. Then f (c) =0. Proof. Let δ be as in the above definition. Then x (c, c + δ) = x (c δ, c) = f(x) f(c) f(x) f(c) 0 0.

12 Now taking the limit x c, wegetf (c) =0. Theorem.3.6. Rolle s Theorem: Let f(x) be a continuous function on [a, b] and differentiable on (a, b) such that f(a) =f(b). Then there exists c (a, b) such that f (c) =0. Proof. If f(x) is constant, then it is trivial. Suppose f(x 0 ) >f(a) forsomex 0 (a, b), then f attains maximum at some c (a, b). Other possibilities can be worked out similarly. Theorem.3.7. Mean-Value Theorem (MVT): Let f be a continuous function on [a, b] and differentiable on (a, b). Then there exists c (a, b) such that f(b) f(a) =f (c)(b a). Proof. Let l(x) be a straight line joining (a, f(a)) and (b, f(b)). Consider the function g(x) =f(x) l(x). Then g(a) =g(b) = 0. Hence by Rolle s theorem 0=g (c) =f (c) f(b) f(a) b a Corollary: If f is a differentiable function on (a, b) andf =0,thenf is constant. Proof. By mean value theorem f(x) f(y) = 0 for all x, y (a, b). Example: Show that cos os y x y. Use Mean-Value theorem and the fact that sin x. Problem: If f(x) is differentiable and sup f (x) <Cfor some C. Then, f is uniformly continuous. Apply mean value theorem to get f(x) f(y) C x y for all x, y. Definition: A function f(x) is strictly increasing on an interval I, ifforx, y I with x < y we have f(x) < f(y). We say f is strictly decreasing if x < y in I implies f(x) >f(y). Theorem.3.8. A differentiable function f is (i) strictly increasing in (a, b) if f (x) > 0 for all x (a, b). (ii) strictly decreasing in (a, b) if f (x) < 0. Proof. Choose x, y in (a, b) suchx<y.thenbymvt,forsomec (x, y) f(x) f(y) x y = f (c) > 0. Hence f(x) <f(y).

13 .4 Taylor s theorem and Taylor Series Let f be a k times differentiable function on an interval I of IR. We want to approximate this function by a polynomial P n (x) such that P n (a) =f(a) atapointa. Moreover, if the derivatives of f and P n also equal at a then we see that this approximation becomes more accurate in a neighbourhood of a. So the best coefficients of the polynomial can be calculated using the relation f (k) (a) =P n (k) (a),k =0,,,..., n. The best is in the sense that if f(x) itself is a polynomial of degree less than or equal to n, thenbothf n f (k) (a) and P n are equal. This implies that the polynomial is (x a) k. Then we write k! k=0 f(x) =P n (x)+r n (x) in a neighbourhood of a. From this, we also expect the R n (x) 0 as x a. In fact, we have the following theorem known as Taylor s theorem: 4 Plot of sin(x) and its Taylor polynomials 3 x sin(x) 0 x x 3 /3!+x 5 /5! sin(x) x x 3 /6 3 4 pi pi/ 0 pi/ pi π x π Figure : Approximation of sin(x) by Taylor s polynomials Theorem.4.. Let f(x) and its derivatives of order m are continuous and f (m+) (x) exists in a neighbourhood of x = a. Then there exists c (a, x)( or c (x, a)) such that f(x) =f(a)+f (a)(x a) f (m) (x a)m (a) + R m (x) m! where R m (x) = f (m+) (c) (m +)! (x a)m+. 3

14 Proof. Define the functions F and g as F (y) =f(x) f(y) f (y)(x y)... f (m) (y) (x y) m, m! ( ) m+ x y g(y) =F (y) F (a). x a Then it is easy to check that g(a) =0. Alsog(x) =F (x) =f(x) f(x) = 0. Therefore, by Rolle s theorem, there exists some c (a, x) such that g (c) =0=F (c)+ On the other hand, from the definition of F, (m +)()m (x a) m+ F (a). F (c) = f (m+) (c) () m. m! Hence F (a) = (x a)m+ f (m+) (c) and the result follows. (m +)! Examples: (i) e x =+x + x! + x3 xn ! n! ec,c (0,x)or(x, 0) depending on the sign of x. (ii) sinx = x x3 3! + x5 xn ! n! sin(c + nπ ),c (0,x)or(x, 0). (iii) cosx = x! + x4 xn ! n! cos(c + nπ ),c (0,x)or(x, 0). Problem: Find the order n of Taylor Polynomial P n,aboutx = 0 to approximate e x in (, ) so that the error is not more than Solution: We know that p n (x) =+x xn. The maximum error in [-,] is n! So n is such that e (n+)! R n (x) or n 5. (n +)! max [,] x n+ e x e (n +)!. Problem: Find the interval of validity when we approximate cos x with nd order polynomial with error tolerance 0 4. Solution: Taylor polynomial of degree for cos x is x x3. So the remainder is (sin c). 3! 4

15 Since sin c, the error will be atmost 0 4 if x3 3! 0 4. Solving this gives x < Taylor s Series Suppose f is infinitely differentiable at a and if the remainder term in the Taylor s formula, R n (x) 0asn.Thenwewrite f(x) = f (n) (a) (x a) n. n! This series is called Taylor series of f(x) aboutthepointa. Suppose there exists C = C(x) > 0, independent of n, such that f (n) (x) C(x). Then x a n+ R n (x) 0 if lim = 0. For any fixed x and a, we can always find N such n (n +)! that x a <N.Letq := x a N (x a) n+ (n +)! 0asn thanks to q<. <. Then = x a x a... x a N < x a N (N )! qn N+ x a N... x a n + In case of a = 0, the formula obtained in Taylor s theorem is known as Maclaurin s formula and the corresponding series that one obtains is known as Maclaurin s series. Example: (i) f(x) =e x. In this case R n (x) = xn+ (n +)! f (n+) (c) = xn+ (n +)! ec = xn+ ( x n+ Therefore for any given x fixed, lim n R n (x) = lim n Example: (ii) f(x) =sinx. In this case it is easy to see that R n (x) sin x and follow as in example (i). Maxima and Minima: Derivative test (n +)! eθx, for some θ (0, ). ) e θx =0. (n +)! x n+ (n +)! sin(c + nπ ). Now use the fact that Definition.4.. A point x = a is called critical point of the function f(x) if f (a) =0. 5

16 Second derivative test: Apointx = a is a local maxima if f (a) =0,f (a) < 0. Suppose f(x) is continuously differentiable in an interval around x = a and let x = a be a critical point of f. Then f (a) = 0. By Taylor s theorem around x = a, there exists, c (a, x) (orc (x, a)), f(x) f(a) = f (c) (x a). If f (a) < 0. Then by Theorem..8, f (c) < 0in x a <δ. Hence f(x) <f(a) in x a <δ,which implies that x = a is a local maximum. Similarly, one can show the following for local minima: f (a) =0,f (a) > 0. x = a is a local minima if Also the above observations show that if f (a) =0,f (a) =0andf (3) (a) 0, then the sign of f(x) f(a) depends on (x a) 3. i.e., it has no constant sign in any interval containing a. Such point is called point of inflection or saddle point. We can also derive that if f (a) =f (a) =f (3) (a) = 0, then we again have x = a is a local minima if f (4) (a) > 0 and is a local maxima if f (4) (a) < 0. Summarizing the above, we have: Theorem.4.3. Let f be a real valued function that is differentiable n times and f (n) is continuous at x = a. Then. If f (k) (a) =0for k =,,...n and f (n) (a) > 0 then a is a point of local minimum of f(x). If f (k) (a) =0for k =,,...n and f (n) (a) < 0 then a is a point local maximum of f(x). 3. If f (k) =0for k =,,...n and f (n ) (a) 0, then a is point of inflection. i.e., f has neither local maxima nor local minima at x = a. L Hospitals Rule: Suppose f(x) and g(x) are differentiable n times, f (n),g (n) are continuous at a and f (k) (a) =g (k) (a) =0fork =0,,,..., n. Also if g (n) (a) 0. Then by Taylor s 6

17 theorem, f(x) lim x a g(x) = lim f (n) (c) x a g (n) (c) = f (n) (a) g (n) (a) In the above, we used the fact that g (n) (x) 0 nearx = a andg (n) (c) g (n) (a) as x a. Similarly, we can derive a formula for limits as x approaches infinity by taking x = y. f(x) lim x g(x) = lim f(/y) y 0 g(/y) ( /y )f (/y) = lim y 0 ( /y )g (/y) f (x) = lim x g (x).5 Power series & Taylor Series Given a sequence of real numbers {a n }, the series a n () n is called power series with center c. It is easy to see that a power series converges for x = c. Power series is a function of x provided it converges for x. If a power series converges, then the domain of convergence is either a bounded interval or the whole of IR. So it is natural to study the largest interval where the power series converges. Remark: If a n x n converges at x = r, then a n x n converges for x < r. Proof: We can find C>0such that a n x n C for all n. Then a n x n a n r n x r n C x r n. Conclusion follows from comparison theorem. Theorem.5.. Consider the power series R = (We define R =0if β = and R = if β =0). Then β. a n x n converges for x <R. a n x n diverges for x >R. 3. No conclusion if x = R. 7 a n x n. Suppose β = lim sup n a n and

18 Proof. Proof of (i) follows from the root test. For a proof, take α n (x) = a n x n and α = lim sup n α n. For (ii), one can show that if x >R,then there exists a subsequence {a n } such that a n 0. Notice that α = β x. For(iii), observe as earlier that the series with a n = and b n n = will have R =. n Similarly, we can prove: Theorem.5.. Consider the power series a n x n. Suppose β = lim a n+ a n and R = β (We define R =0if β = and R = if β =0). Then. a n x n converges for x <R. a n x n diverges for x >R. 3. No conclusion if x = R. Definition.5.3. The real number R in the above theorems is called the Radius of convergence of power series. Examples: Find the interval of convergence of (i) x n n (ii) x n n! (iii) n x 3n. β = lim a n+ a n =, and we know that the series does not converge for x =,. Sotheintervalofconvegenceis(, ).. β = lim a n+ a n = 0. Hence the series converges everywhere. 3. To see the subsequent non-zero terms, we write the series as n (x 3 ) n = n y n. For this series β y = lim sup n a n =. Therefore, β x = /3 and R = /3. { n n is even Example.5.4. Let a n = n n is odd. Then lim sup a n+ a n =4and the limit of a n+ a n does not exist. But lim sup a n /n =/. So the radius of convergence of the series an x n is /. The following theorem is very useful in identifying the domain of convergence of some Taylor series. Theorem.5.5. (Term by term differentiation and integration): Suppose f(x) = a n x n converges for x <R. Then. na n x n converges in x <Rand is equal to f (x).. a n+ nx n+ converges in x <Rand is equal to f(x)dx. 8

19 From this theorem one concludes that a power series is infinitely differentiable with in its radius of convergence. Now it is natural to ask weather this series coincides with the Taylor series of the resultant function. The answer is yes and it is simple to prove that if f(x) = a n x n,thena n = f (n). The above theorem is useful to find the domain of n! convergence of Taylor series of some functions. Example: The Taylor series of f(x) =tan x and a domain of its convergence. tan x = Taking x = we get interesting sum dx +x = =x x3 3 + x x + x =tan () = π 4. Though the function tan x is defined on all of IR, we see that the power series converges on (, ). We can apply Abel s theorem on alternating series to show that the series converges at x =,. For approximation, we can use the error approximation of alternating series discussed in the previous section. The total error if we approximate tan x by s n (x), then the maximum error is x n+. /// n+ We note that the power series may converge to a function on small interval, even though the function is defined on a much bigger interval. For example the function log( + x) has power series that converges on (, ), but log( + x) isdefinedon(, ). The domain of convergence of power series is symmetric about the center but the domain of definition of function. For instance, for a function defined on (, 3) the radius of convergence of its power series(about 0) cannot be more than. Another interesting application is to integrate the functions for which we have no clue. For example, () erf(x) = x 0 e t dt = x 0 ( t! + t4! +... = x x3 3 + x () x 0 sin t t = x 0 t 3! + t4 5! = x x3 3! 3 + x5 5!