Differentiation. Table of contents Definition Arithmetics Composite and inverse functions... 5

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1 Differentiation Table of contents. Derivatives Definition Arithmetics Composite and inverse functions Properties of Differentiable Functions Differentiability and continuity Maximum and minimum Mean value theorem Monotonicity and L Hospital Monotonicity L Hospital s Rule Applications of L Hospital s rule Taylor Expansion Higher order derivative Taylor expansion Applications of Taylor expansion

2 2 Differentiation.. Definition.. Derivatives Definition. Let f be a real function. At a point x 0 inside its domain, if the it f(x) f(x 0 ) () x x 0 x x 0 exists and is finite, we say f is differentiable at x 0, and call the it its derivative at x 0, denoted f (x 0 ). If the it does not exist, we say f is not differentiable at x 0. If f is differentiable at all x E where E R, we say f is differentiable on E. If f is differentiable at every point of its domain, we say f is differentiable. Remark 2. Equivalently, one can define differentiability through the it h 0 That is f is differentiable at x 0 if the above it exists. f(x 0 + h) f(x 0 ). (2) h Exercise. Is the following an equivalent definition of differentiability? Justify your answer. Let f be a real function. At a point x 0 inside its domain, if the it exists, we say f is differentiable at x 0. f(x 0+h) f(x 0 h) h 0 2 h (3) Exercise 2. Let f(x) = a for all x in its domain. Prove that f (x) = 0. Note that you have to establish differentiability of f(x). Remark 3. Recall that in the definition of its, we require 0 < x x 0. This is crucial in the it (3) since at x=x 0 we have 0 0. Example 4. Let f(x) =x. Prove that f(x) is a differentiable function. Solution. We need to prove that f is differentiable at every x 0 R. For every x 0 R, So (x ) =. f(x) f(x 0 ) x x = 0 = =. (4) x x 0 x x 0 x x 0 x x 0 x x 0 Exercise 3. Let f(x) =. Prove it is a differentiable function. Example 5. Let f(x)= Solution. By definition we check the it { x 2 sin x x 0 0 x=0. Prove that f(x) is differentiable at x=0 and find f (0). x 2 sin x x 0 ( = x sin ). (5) 0 x x 0 x

3 3 Since and by Squeeze Theorem we have x xsin x (6) x ( x ) = 0 = x, (7) x 0 x 0 ( x sin ) =0. (8) x 0 x Therefore f(x) is differentiable at x=0 and f (0) =0. Example 6. Let f(x) = Solution. By definition we need to show the it does not exist. Take x n = n π, y n= { x sin x x 0. Prove that f(x) is not differentiable at x =0. 0 x =0 x sin x x 0 ( = sin ) 0 x x 0 x. Then we have (2 n + /2) π (9) As 0, x 0 ( sin x n=0= n; n n n,x n 0, y n 0; (0) ( sin n ( sin n y n x n ) does not exist. ) = sin (n π) = 0 =0, () n n ) = sin((2 n + /2) π)= =. (2) n n.2. Arithmetics. Theorem 7. (Arithmetics of derivatives) Let f, g be differentiable at x 0. Then a) f ± g is differentiable at x 0 with (f ± g) (x 0 ) = f (x 0 ) ± g (x 0 ). Proof. b) (Leibniz rule) f g is differentiable at x 0 with (f g) (x 0 ) = f (x 0 ) g(x 0 ) + f(x 0 ) g (x 0 ). c) If g(x 0 ) 0, then f/g is differentiable at x 0 with ( ) f (x 0 ) = f (x 0 ) g(x 0 ) f(x 0 ) g (x 0 ) g g(x 0 ) 2. (3) a) We have (f + g)(x) (f + g)(x 0 ) x x 0 = f(x) f(x 0) x x 0 + g(x) g(x 0) x x 0. (4)

4 4 Differentiation Since The it [ f(x) f(x0 ) + g(x) g(x ] 0) = f x x 0 x x 0 x x (x 0 )+ g (x 0 ) (5) 0 (f + g)(x) (f + g)(x 0 ) x x 0 (6) also exists and equals f (x 0 ) + g (x 0 ). The case f g can be proved similarly. b) We have Since (f g)(x) (f g)(x 0 ) x x 0 = f(x) g(x) g(x 0) x x 0 + g(x 0 ) f(x) f(x 0) x x 0. (7) f(x) = f(x 0 ); x x 0 g(x) g(x 0 ) = g f(x) f(x (x 0 ); 0 ) = f (x 0 ) (8) x x 0 x x 0 x x 0 x x 0 we reach (f g)(x) (f g)(x 0 ) = f(x 0 ) g x x 0 x x (x 0 ) + f (x 0 ) g(x 0 ). (9) 0 c) We only prove the last one. In light of b), it suffices to prove Write ( ) = g (x 0 ) g g 2 (x 0 ). (20) g(x) g(x 0 ) x x 0 g(x) g(x 0 ) x x 0 = g(x) g(x 0 ). (2) Note that both the denominator and the numerator have its, and furthermore the it of the denominator is not 0. So we have the it of the ratio exists and g(x) g(x 0 ) x x x 0 = x g(x) g(x 0 ) x 0 x x 0 x 0 g(x) g(x 0 ) x x 0 g(x) g(x 0 ) = g (x 0 ) g(x 0 ) 2. (22) Thus ends the proof. Example 8. Let n N. Prove that x n is differentiable everywhere for n > 0 and differentiable at x 0 for all n < 0. Solution. Let P(n) be x n is differentiable everywhere. We prove P(n) is true for all n N through induction. Base: P() = x is differentiable everywhere has already been proved above. P(n) P(n + ). Assume x n is differentiable everywhere. Let x 0 R be arbitrary. Then since both x n and x are differentiable at x 0, we have is differentiable at x 0. x n+ =(x n ) x (23)

5 5 Let P(n) be x n is differentiable at every x 0 0. We prove P(n) is true for all n N through induction. Base: P()= /x is differentiable at every x 0 0. We have shown that x is differentiable everywhere. Let x 0 0. We check = 0 (24) x x 0 x x 0 therefore is differentiable at x 0. By the above theorem we have /x differentiable at x 0. P(n) P(n + ): Left as exercise. Theorem 9. The following functions are differentiable everywhere..3. Composite and inverse functions. sinx, cosx, e x, Polynomials. (25) Theorem 0. (Chain rule) If f is differentiable at x 0 and g is differentiable at f(x 0 ), then the composite function g f is differentiable at x 0 and satisfy (g f) (x 0 ) = g (f(x 0 )) f (x 0 ). (26) Proof. Set g(y) g(f(x 0 )) y f(x h(y)8 0 ) y f(x 0 ). (27) g (f(x 0 )) y = f(x 0 ) Then we have h(y) satisfying y f(x 0 )h(y)= h(f(x 0 )). Now write (g f)(x) (g f)(x 0 ) x x 0 = h(f(x)) f(x) f(x 0) x x 0. (28) By Lemma 3 we have x x 0 f(x) = f(x 0 ). Thus taking it of both sides of (28) we reach ) =h(f(x 0 )) f (x 0 ) (29) (g f)(x) (g f)(x 0 ) ( = x x 0 x x 0 and the proof ends. ) ( h(f(x)) x x 0 Remark. Naturally one may want to prove through and try to show f(x) f(x 0 ) x x 0 x x 0 (g f)(x) (g f)(x 0 ) = g(f(x)) g(f(x 0)) f(x) f(x 0 ) (30) x x 0 f(x) f(x 0 ) x x 0 g(f(x)) g(f(x 0 )) = g x (f(x 0 )). (3) x 0 f(x) f(x 0 ) However this does not work because it may happen that f(x) f(x 0 )=0. The above trick overcomes this difficulty.

6 6 Differentiation Theorem 2. (Derivative of inverse function) Let f be differentiable at x 0 with f (x 0 ) 0. Then if f has an inverse function g, then g is differentiable at y 0 = f(x 0 ) and satisfies g (f(x 0 ))=/f (x 0 ) or equivalently g (y 0 ) =/f (g(y 0 )). Proof. Since f has an inverse function, f is either strictly increasing or strictly decreasing. Furthermore g is continuous, and also strictly increasing or decreasing. Let y 0 = f(x 0 ). We compute g(y) g(y 0 ) y y 0 = g(y) g(y 0 ) f(g(y)) f(g(y 0 )) = ( ) f(g(y)) f(g(y0 )). (32) g(y) g(y 0 ) Note that as f, g are both strictly increasing/decreasing, all the denominators in the above formula are nonzero. To show that the it exists, we recall that F(x) exists at x 0 if for all x n x 0 the it of F(x n ) exists. Take y n y 0. By continuity of g we have g(y n ) g(y 0 ). The differentiability of f at g(y 0 ), that is the existence of the it x g(y 0 ) f(x) f(g(y 0)), then gives x g(y 0 ) Thus ends the proof. f(g(y n )) f(g(y 0 )) = f n (g(y 0 )) = f (x 0 ) 0. (33) g(y n ) g(y 0 ) Example 3. Assume that we are given tan (x)=, find cos 2 x arctan. Solution. We have arctan (y)= tan (x) = cos2 (x). (34) What we need now is to represent cos 2 (x) by y = tanx. It is clear that cos 2 x= so + y 2 arctan (y)= + y 2. Example 4. Assume that we are given (e x ) = e x. Find (ln x). Solution. We have (ln ) (y) = (e x ) = e x = y since y =e x. (35) Example 5. (f (x 0 ) = 0) Consider f(x) = x 3. Then g(y) = y /3. We see that at x 0 = 0, g is not differentiable. Exercise 4. Prove the following Toy L Hospital s Rule. Let f, g be differentiable at x 0, and furthermore f(x 0 ) = g(x 0 )=0. Then if g (x 0 ) 0, we have Then apply it to evaluate the following its: f(x) x x0 g(x) = f (x 0) g (x 0 ). (36) Do you encounter any difficulty? x x 2 x ; x sin x 0 x ; x cos x. (37) 0 x 2

7 7 2.. Differentiability and continuity. 2. Properties of Differentiable Functions Lemma 6. (Differentiable functions are continuous) If f(x) is differentiable at x 0, then f(x) is continuous at x 0. Proof. Since f(x) is differentiable at x 0, we have by definition Now write f(x) f(x 0 ) =L R (38) x x 0 x x 0 f(x) = f(x 0 ) +(x x 0 ) f(x) f(x 0) (39) x x 0 and take it x x 0, we have [ ][ ] f(x) f(x f(x) = f(x 0 ) + x x 0 0 ) = f(x 0 ) +0 L= f(x 0 ) (40) x x 0 x x 0 x x 0 x x 0 Therefore f(x) is continuous at x 0. Remark 7. One can also prove using definition as follows. Since f(x) is differentiable at x 0, we have by definition f(x) f(x 0 ) =L R (4) x x 0 x x 0 Take δ >0 such that for all 0 < x x 0 <δ, f(x) f(x 0) < L x x 0 f(x) f(x 0) <( L +) x x 0. (42) { } ε Now for any ε > 0, take δ = min δ,. We have, for all 0 < x x 0 <δ, L + f(x) f(x 0 ) <( L +) x x 0 <( L +)δ ε. (43) Exercise 5. Prove or disprove the converse claim: If f(x) is continuous at x 0, then it is differentiable at x Maximum and minimum. Definition 8. (Local maximum/minimum) Let f:[a,b] R be a real function. We say f has a local maximum at x 0 (a, b) if there exists some δ > 0 such that f(x) f(x 0 ) for all x (x 0 δ, x 0 +δ). This x 0 is said to be a local maximizer. We say f has a local minimum at x 0 if there exists some δ >0 such that f(x) f(x 0 ) for all x (x 0 δ,x 0 +δ). This x 0 is said to be a local minimizer. Exercise 6. Find all local maxima/minima for the following functions. a) f(x) =. b) f(x) = x 2 ; c) f(x) = sin x; d) f(x) = x sin x; e) f(x) = sin (/x). Theorem 9. If f is differentiable at its local maximizer or minimizer, then the derivative is 0 there. Proof. Assume x 0 is a local maximizer. Take x n (x 0, x 0 + δ) with n x n = x 0. differentiable at x 0, we have Since f is f f(x) f(x (x 0 ) = 0 ) f(x = n ) f(x 0 ). (44) x x 0 x x 0 n x x 0

8 8 Differentiation But as f(x n ) f(x 0 ) 0 for all n, by comparison theorem we reach f (x 0 ) 0. Now take x n (x 0 δ, x 0 ) with n x n = x 0. Similar argument as above gives f (x 0 ) 0. Therefore f (x 0 )= 0. The proof for the local minimizer case is similar and left as exercise. Remark 20. It may happen that f is not differentiable at its maximizer or minimizer. For example f(x) = x. Example 2. Consider f(x) = x sin (/x). Then its local maximizers and minimizers can be obtained by solving 0 = f (x)= sin(/x) x cos(/x) tan(/x) =/x. (45) x2 The solutions have to be obtained numerically as it is not possible to represent them using elementary functions. Note. It is important to note that the local maximizers and local minimizers of f(x) are not x n = (2 n + /2) π and y n= 2.3. Mean value theorem. (2 n /2) π! Theorem 22. (Rolle s Theorem) Let f be continuous on [a, b] and differentiable on (a, b). If f(a) = f(b) then there is ξ (a,b) such that f (ξ) =0. Remark 23. Before proving the theorem, we illustrate the necessity of the assumptions. { x 0 x < f is continuous on [a,b]. If not, f(x) =. 0 x= f is differentiable on (a, b). If not, f(x)= x over [,]. Proof. Since f is continuous on [a,b], there are x min,x max [a,b] such that f(x min ) is the minimum and f(x max ) is the maximum. If one of them is different from a,b, then f =0 there due to Theorem 6. Otherwise we have f(a)= f(b)= f(x min )= f(x max ) f(x) is constant on [a,b], consequently f (x) =0 for all x (a, b). Exercise 7. (Rolle over R) Let f be continuous and differentiable on R. If x + f(x)= x f(x), then there is ξ R such that f (ξ) =0. Theorem 24. (Mean Value Theorem) Let f be continuous on [a,b] and differentiable on (a,b). Then there is a point ξ (a, b) such that Proof. Set g(x) = f(x) f(b) f(a) b a f (ξ) = f(b) f(a). (46) b a (x a) and apply Rolle s Theorem. Remark 25. When the interval has infinite size, the Mean Value Theorem may not hold (even if we accept (f(b) f(a))/ =0). An example is f(x) = arctanx. Theorem 26. (Cauchy s extended mean value theorem) Let f, g be continuous over [a, b] and differentiable over (a, b). Further assume g(a) g(b). Then there is ξ (a, b) such that f(b) f(a) g(b) g(a) = f (ξ) g (ξ). (47) f(b) f(a) Exercise 8. Prove Cauchy s MVT. (Hint: take h(x) = f(x) g(x) and apply MVT). g(b) g(a)

9 9 3.. Monotonicity. 3. Monotonicity and L Hospital Theorem 27. Let f be defined over [a, b] R. Here a, b can be extended real numbers. Suppose f is continuous on [a, b] and differentiable on (a, b). Then a) f is increasing if and only if f (x) 0 for all x (a,b); f is decreasing if and only if f (x) 0 for all x (a,b). Proof. b) f is strictly increasing if f (x) > 0 for all x (a, b); f is strictly decreasing if f (x) < 0 for all x (a, b). c) f is a constant if and only if f (x) =0 for all x (a, b). a) We prove the increasing case here. Let f be increasing, we show f (x) 0. Take any x 0 (a, b). Since f is increasing, f(x) f(x 0 ) when x > x 0 and f(x) f(x 0 ) when x < x 0, thus f(x) f(x 0 ) x x 0 0 (48) for all x x 0. As f is differentiable at x 0, taking it of both sides leads to f (x 0 ) 0. Let f (x) 0 for all x (a, b). Assume f is not increasing. Then there are x < x 2 such that f(x )> f(x 2 ). Apply Mean Value Theorem we have there must exist ξ (x,x 2 ) (a,b) such that f (ξ) = f(x ) f(x 2 ) x x 2 <0. (49) Contradiction. b) The proof is similar to the corresponding part of a). c) The proof is left as exercise. Remark 28. Note that f(x) strictly increasing f (x)>0 everywhere. An examples is f(x)=x 3. Example 29. Prove that e x > +x for all x >0. Proof. Let f(x)=e x x. We see that f(0)=0. To show f(x)>0 it suffices to show f is strictly increasing. Calculate f (x) =e x >0 (50) for all x >0. Therefore f is strictly increasing and consequently f(x) > 0 for all x >0. Example 30. Prove for all x >. x ln( +x) x (5) + x Proof. For the first inequality let f(x)=ln(+x) is f(x) f(0). Calculate f (x) = x. We have f(0)=0 so all we need to show + x x ( + x) 2. (52)

10 0 Differentiation Thus f(x) 0 when x > 0 and f(x) 0 when x <0. Consequently f(x) f(0). For the second inequality let g(x)=x ln(+x). We have g(0)=0 and need to show g(x) g(0) for all x. Calculate g (x) = x + x. (53) For x > we have g (x) >0 if x >0 and <0 if x < 0. Example 3. Prove for < x <. arctan +x x = arctan x+ π 4 (54) Proof. Set x=0 we have Therefore all we need to show is arctan +0 0 = arctan 0 + π 4. (55) h(x)8 arctan +x arctanx (56) x is a constant for < x <. Once this is shown, we have h(x) =h(0) = π 4. Taking derivative, we have h (x) = + ( ) +x x ( +x x ) 2 +x 2 = ( x) ( ) (+x) ( x) 2 ( x) 2 + (+x) 2 ( x) 2 =0. (57) +x2 Thus ends the proof L Hospital s Rule. Theorem 32. (L Hospital s Rule) Let x 0 (a,b) and f(x), g(x) be differentiable on (a,b) {x 0 }. Assume that x x 0 f(x) = x x 0 g(x) = 0. Then if x x 0 f (x) g (x) exists and g (x) 0 for x (a,b) {x 0 }, the following holds. f(x) x x 0 g(x) = x f (x) x 0 g (x). (58) Remark 33. Note that there are four conditions for the application of L Hospital s rule:. f(x), g(x) are differentiable on (a,b) {x 0 }; 2. x x 0 f(x) = x x 0 g(x) =0; 3. x x 0 f (x) g (x) exists; 4. g (x) 0 for x (a,b) {x 0 }. Proof. (of L Hospital s Rule) Since x x 0 f(x) = x x 0 g(x) = 0 we can define f(x 0 ) = g(x 0 )=0. After such definition f, g becomes continuous over (a,b). Now for any x (a,b), we have f(x) g(x) = f(x) f(x 0) g(x) g(x 0 ) = f (ξ) g (ξ) (59)

11 for some ξ between x, x 0, thanks to Cauchy s MVT. Now taking it x x 0, we have ξ x 0 and the conclusion follows Applications of L Hospital s rule. Example 34. Find x 0 x sinx x 2. We see that the conditions for L Hospital s rule is satisfied. Therefore x sinx sin x+xcosx x 0 x 2 = x 0 2 x if the latter exists. Now this second it still satisfies the conditions for L Hospital and consequently we have 2 cosx x sin x. (6) 2 sinx +xcosx = x 0 2 x x 0 But this last it exists and equals. Therefore (60) x sinx x 0 x 2 =. (62) Remark 35. L Hospital s rule still holds when x 0 = ±, x x0 f (x) g (x) = ±, or x x 0 f, x x 0 g = ±. We will prove for the following situation and leave other cases as exercise. x 0 =+ ; x f(x)= x g(x) =+ ; x x 0 f (x) g (x) = L R. Proof. Let ε >0 be arbitrary. We try to find M >0 such that f(x) g(x) L < ε x > M. (63) { Set any positive number δ < min Take M >0 such that 2, ε +4( L + ) Take x 0 =M +. Fix x 0 now. Then take M >0 such that Now for all x > M, we have }. f (x) g (x) L <δ x >M. (64) f(x 0 ) <δf(x); g(x 0 ) <δg(x) x > M. (65) which gives δ f(x) +δ g(x) < f(x) f(x 0) g(x) g(x 0 ) < +δ f(x) δ g(x) δ f(x) f(x 0 ) +δ g(x) g(x 0 ) < f(x) g(x) < +δ f(x) f(x 0 ) δ g(x) g(x 0 ) (66) (67) Now we apply Cauchy s MVT to obtain f(x) f(x 0 ) g(x) g(x 0 ) = f (ξ) g (ξ) (68)

12 2 Differentiation for some ξ (x 0,x). Recalling our choice of x 0, we conclude f(x) f(x 0) g(x) g(x 0 ) L < δ. (69) This gives +δ f(x) f(x 0 ) δ g(x) g(x 0 ) L f(x) f(x 0 ) g(x) g(x 0 ) L + 2 δ f(x) f(x 0 ) δ g(x) g(x 0 ) <δ + 2 δ ( L +δ) (70) δ and δ +δ f(x) f(x 0 ) g(x) g(x 0 ) L <δ + 2 δ ( L +δ). (7) +δ Recall that 0< δ < 2 and δ < Therefore ε, we have +4( L + ) δ + 2 δ 2 δ ( L +δ) <δ +4δ ( L +) < ε, δ + ( L +δ) < δ +2δ ( L +)< ε. (72) δ +δ Thus ends the proof. f(x) g(x) L < ε x > M. (73) Exercise 9. Let a < b <c be real numbers. Let L R. Prove that b L < max { a L, c L }. (74) Exercise 0. Is it possible to prove the above situation through the following: Set F(z)8 (f(x(z))), G(z)8 (g(x(z))) where x(z) = /z. Exercise. List all possible situations for L Hospital s rule and write down detailed proof for each. (Hint: Some proofs can be obtained from others by replacing f(x) f(x) etc. ) Example 36. Find x 0x lnx. We have ln x x ln x= x 0 x 0 /x = x 0 /x /x 2 = x ( x) =0. (75) 0 Remark 37. L Hospital s rule only applies to the situations 0/0, (± )/(± ). Exercise 2. Show that applying L Hospital s rule to x sin x 0 leads to wrong result. Then explain which + x step in the proof breaks down for this it. { Example 38. Let f(x) = e x 2 x 0 0 x=0. Prove that f is differentiable at 0 and find f (0). Solution. We need to study the it e (/x2) 0 = x 0 x 0 e (/x2) x 0 x. (76)

13 3 4.. Higher order derivative. 4. Taylor Expansion Definition 39. (Second and higher order derivatives) Let f(x) be differentiable on (a, b). Let f (x) be its derivative. If f (x) is differentiable at x 0, then we denote f (x 0 )8 (f ) (x 0 ) (77) and say f(x) is twice differentiable at x 0. We say f(x) is twice differentiable on (a,b) if f (x) exists for all x (a,b). Similarly, for any n > 2, if f (n ) (x) is differentiable at x 0, then we denote f (n) (x 0 )8 ( f (n )) (x 0 ) (78) and say f(x) is n-th differentiable at x 0. We say f(x) is n-th differentiable on (a,b) if f (n) (x) exists for all x (a,b). Exercise 3. Let n, m N. Let f(x) be n+m-th differentiable at x 0. Prove that f (n) (x) is m-th differentiable at x 0 and furthermore (Hint: Induction.) (f (n) ) (m) (x 0 ) = f (n+m) (x 0 ). (79) Example 40. Let f(x) =e 3x. Compute f (n) (x) for n N. We claim f (n) (x) = 3 n e 3x and prove this by induction. Denote by P(n) the claim: f (n) (x) = 3 n e 3x. Then P(): f (x)= 3 e 3x thanks to chain rule. Therefore P() holds. P(n) P(n + ). Assume P(n): f (n) (x)= 3 n e 3x. Then by definition Thus P(n +) holds and the proof ends. f (n+) (x) = (3 n e 3x ) =3 n 3e 3x = 3 (n+) e 3x. (80) Exercise 4. Let f(x) = sin x. Calculate f (n) (x) for all n N. Justify your answer. Remark 4. Note that for f (n) (x 0 ) to exist, f (n ) (x) must exist over (x 0 δ,x 0 +δ) for some δ >0. Exercise 5. Explain why existence of f (x 0 ) is not enough to define f (x 0 ). Remark 42. (Notation) Usually we use f, f, f to denote first, second, third order derivatives, while switch to f (4), f (5), for higher order derivatives Taylor expansion. Theorem 43. (Lagrange form of the remainder) Let f be such that f (k) (x) exists on (a, b) for k =0,,n+. Then for every x, x 0 (a, b) the following holds: where P n (x)8 f(x 0 )+ f (x 0 ) (x x 0 ) + f (x 0 ) 2 f(x) =P n (x) +R n (x) (8) (x x 0 ) f(n) (x 0 ) n! (x x 0 ) n. (82)

14 4 Differentiation is called the Taylor polynomial of f at x 0 to degree n, and R n (x)8 f(x) P n (x) (83) is called the Remainder. Under the assumptions of the theorem, we have the following formula for the remainder: R n (x) = f(n+) (ξ) (n + )! (x x 0 ) n+ (84) where ξ is between x, x 0. This formula is called the Lagrange form of the remainder. Remark 44. It is important to understand P n (x) depends on x 0, that is it changes when x 0 changes; For every fixed x 0, P n (x) is a polynomial. However the dependence of P n (x) on x 0 may be more complicated. The ξ in R n (x): = f(n+) (ξ) (x x (n + )! 0 ) n+ depends on both x and x 0, that is when x or x 0 changes, so does ξ. Furthermore the there is no convenient formula yielding ξ from x, x 0. One way to understand the role of ξ is as follows. For any fixed x, it is clear that there is r R such that f(x) = f(x 0 )+ f (x 0 ) (x x 0 ) + + f(n) (x 0 ) n! (x x 0 ) n + r (x x 0 ) n+. (85) Thus what the theorem actually says is: ξ between x,x 0 such that r = f(n+) (ξ). (n + )! Proof. In the following x is fixed. Take r R such that f(x) = f(x 0 ) + f (x 0 )(x x 0 ) + f (x 0 ) 2 holds for this particular x. Now set [ g(t) = f(t) (x x 0 ) f(n) (x 0 ) n! f(x 0 )+ f (x 0 ) (t x 0 ) + + f(n) (x 0 ) n! (x x 0 ) n +r (x x 0 ) n+. (86) (t x 0 ) n +r (t x 0 ) n+ ] then we have g(x 0 ) = g(x) = 0. Applying Rolle s theorem, we obtain ξ between x 0, x such that g (ξ ) = 0. On the other hand clearly g (x 0 ) = 0. Thus we have ξ 2 between ξ and x 0 (thus also between x, x 0 ) such that g (ξ 2 ) = 0. Apply this n times we conclude that there is ξ such that g (n+) (ξ) = 0, which gives r = f(n+) (ξ) (n +)!. (88) Thus ends the proof. Remark 45. Note that the case n=0 is exactly Rolle s theorem. Also note that one cannot prove the above theorem through induction. Remark 46. A more clever proof is as follows. Set [ F(t) = f(x) f(t) + f (t) (x t)+ + f(n) (t) (x t) ], n n! G(t) =(x t) n+ (89) (87)

15 5 We have F(x)= G(x) =0, F(x 0 ) = R n (x), G(x 0 ) =(x x 0 ) n+, and F (t) = f(n+) (t) n! (x t) n, G (t) = (n +) (x t) n. (90) Now apply Cauchy s MVT: There is ξ between x 0, x and ξ x 0, x (this can be written as 0 < ξ x 0 < x x 0 ) such that f (n+) (ξ) (n +)! = F (ξ) G (ξ) = F(x) F(x 0) G(x) G(x 0 ) = R n(x) (x x 0 ) n+ R n (x) = f(n+) (ξ) (x x 0 ) n+. (9) (n +)! Exercise 6. Try to prove the theorem through induction and explain the difficulty you encounter. Exercise 7. Check out the youtube video ( proving Taylor expansion formula and explain where the mistake is. Compare it with the Khan Academy video on the same topic. Exercise 8. Prove the theorem as follows. Fix x, x 0. Define [ F(t) = f(t) f(x 0 ) + f (x 0 ) (t x 0 ) + + f(n) (x 0) (t x 0 ) ]; n n! G(t) = (t x 0 ) n+. (92) Apply Cauchy s MVT n times to F(t) F(x0) G(t) G(x0). Example 47. Calculate Taylor expansion with Lagrange form of remainder (to degree 2 that is n =2) of the following functions at x 0 = 0. Solution. f(x)= sin(sin x). We calculate: f(x) = sin(sinx); f(x)= x 4 +x+ (93) f (x)= [cos(sinx)] cosx f (0) =; (94) f (x) = [ sin (sinx) cosx] cosx [cos(sinx)] sin x f (0) = 0; (95) f (x) = {[ sin (sin x)] cos 2 x} {[cos(sin x)] sin x} = cos(sin x) cos 3 x+2sin(sinx) cosxsinx +sin(sin x) sinx cosx cos(sinx) cosx = cosx[(cos 2 x +) cos(sinx) 3sinx (sin(sinx))]. (96) Thus the Taylor polynomial at x 0 =0 to degree 2 reads: which simplifies to 0 + (x 0) (x 0)2 + f (ξ) 6 sin(sinx)= x+ cos ξ [(cos2 ξ +) cos(sin ξ) 3sin ξ (sin (sin ξ))] 6 Here ξ lies between 0 and x. f(x)= x 4 +x+. We calculate: and (x 0) 3 (97) x 3. (98) f(0)=, f (x)= 4 x 3 + f (0) =, f (x) = 2x 2 f (0) =0 (99) f (x)= 24 x. (00)

16 6 Differentiation Therefore the Taylor polynomial at x 0 =0 to degree 2 reads where ξ lies between 0 and x. x 4 +x+=+x+(4 ξ) x 3 (0) Example 48. Calculate Taylor expansion (to degree 2) of the following functions at the specified x 0 s. f(x)= sinx, x 0 = π 2 ; f(x) =x4 + x+, x 0 = ; f(x)= e x, x 0 =2. (02) Solution. f(x)= sinx, x 0 = π 2. We have ( ) π f(x 0 ) = sin = ; (03) 2 Therefore the answer is f (x) = cosx f (x 0 )= 0; (04) f (x) = sinx f (x 0 ) = ; (05) sinx = 2 where ξ is between x and π/2. f(x)= x 4 +x+, x 0 =. We have So the answer is where ξ is between x and. f(x)= e x, x 0 =2. We have So the answer is f (x)= cosx. (06) ( x π 2 ) 2 ( cos ξ x π ) 3 (07) 6 2 f(x 0 ) =3; f (x) =4x 3 + f (x 0 ) = 5 (08) f (x) = 2 x 2 f (x 0 ) = 2; f (x) = 24x. (09) x 4 +x+=3+5(x ) +6(x ) 2 +4 ξ (x ) 3 (0) f(x 0 )= f (x 0 ) = f (x 0 ) =e 2, f (x)= e x. () e x = e 2 + e 2 (x 2) + e2 2 (x 2)2 + eξ 6 (x 2)3 (2) where ξ is between x and. Exercise 9. Find a computer system (the most convenient one would be that can plot functions. Plot the Taylor polynomial to degrees 2, 3, 4, 5 of f(x) = sin x. Observe how they approximates f(x) near x 0. What happens far away from x 0? 4.3. Applications of Taylor expansion. Example 49. Prove the following. a) e x >+x+ x2 2 + x3 6 for all x >0.

17 7 b) Proof. ) cosx ( x2 2 < for all x (, ). 24 a) The Taylor polynomial with Lagrange remainder for e x at 0 is e x = +x+ x2 2 + x3 6 + eξ 24 x4. (3) Since x > 0, ξ (note that it depends on x, that is ξ = ξ(x) is in fact a function of x) is also positive. Consequently eξ 24 x4 > 0 for all x. So e x >+x+ x2 2 + x3 6 holds for all x > 0. b) The Taylor polynomial with Lagrange remainder for cosx at 0 is (up to degree 2): with ξ between 0 and x. Thus we have ( ) cosx x2 = 2 cosx= x2 2 + sin ξ 6 x3 (4) sin ξ 6 for all x (,). This is not enough so we expand one more term: x 3 < 6. (5) which gives Thus ends the proof. cosx= x2 2 + cos ξ 24 x4 (6) ) = x 4 < 24. (7) cosx ( x2 2 cos ξ 24