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1 Mean Value Theorem 1 1.1 Applications of the Mean Value Theorem Seunghee Ye Ma 8: Week 5 Oct 28 Week 5 Summary In Section 1, we go over the Mean Value Theorem and its applications. In Section 2, we will recap what we have covered so far this term. Topics Page 1 Mean Value Theorem 1 1.1 Applications of the Mean Value Theorem . .1 2 Midterm Review 5 2.1 Proof Techniques . .5 2.2 Sequences . .6 2.3 Series . .7 2.4 Continuity and Differentiability of Functions . .9 1 Mean Value Theorem The Mean Value Theorem is the following result: Theorem 1.1 (Mean Value Theorem). Let f be a continuous function on [a; b], which is differentiable on (a; b). Then, there exists some value c 2 (a; b) such that f(b) − f(a) f 0(c) = b − a Intuitively, the Mean Value Theorem is quite trivial. Say we want to drive to San Francisco, which is 380 miles from Caltech according to Google Map. If we start driving at 8am and arrive at 12pm, we know that we were driving over the speed limit at least once during the drive. This is exactly what the Mean Value Theorem tells us. Since the distance travelled is a continuous function of time, we know that there is a point in time when our speed was ≥ 380=4 >>> speed limit. As we can see from this example, the Mean Value Theorem is usually not a tough theorem to understand. The tricky thing is realizing when you should try to use it. Roughly speaking, we use the Mean Value Theorem when we want to turn the information about a function into information about its derivative, or vice-versa. 1.1 Applications of the Mean Value Theorem Example 1.1. Consider the equation (x + y)n = xn + yn If either x or y is zero, the equation holds. Also, if x = −y and n is odd, the equation holds. Are there other values of x; y 2 R and n 2 N for which we can find solutions for above equation? Solution. We will show that the answer is no. First, we note the following lemma we can prove using the Mean Value Theorem: 0 Lemma 1.1. Let f be a differentiable function with k distinct roots a1 < a2 < ··· < ak. Then, f has at least k − 1 distinct roots b1 < b2 < ··· < bk−1 such that a1 < b1 < a2 < b2 < ··· < ak−1 < bk−1 < ak Page 1 of 10 Seunghee Ye Ma 8: Week 5 Oct 28 Proof. Since f is differentiable on [a1; a2], by the Mean Value Theorem, we can find b1 2 (a1; a2) such that 0 f(a2) − f(a1) 0 − 0 f (b1) = = = 0 a2 − a1 a2 − a1 In other words, b1 is a root of f. Repeating this argument for [ai; ai+1] for i = 1; : : : ; k − 1, we find k − 1 0 distinct roots b1; : : : ; bk−1 of f with the desired property. How can we use this lemma to solve our problem? Let's fix a nonzero value y 2 R and n 2 N, and look at the equation n n n fn(x) = (x + y) − x − y Note that we are trying to find out which values of x satisfy f(x) = 0. Currently we know that the following values of x are roots of f(x): 1. If n is even, the only root we know is x = 0. 2. If n is odd, we know that x = 0 and x = −y are roots 0 Can there be more distinct roots? By Lemma 1.1 we know that if fn(x) has k distinct roots, fn(x) must 0 have at least k − 1 distinct roots. So let's see how many roots fn(x) can have. Differentiating we obtain 0 n−1 n−1 fn(x) = n(x + y) − nx 0 Now, if x is a root of fn(x), we have 0 = n(x + y)n−1 − nxn−1 (1) , nxn−1 = n(x + y)n−1 (2) , xn−1 = (x + y)n−1 (3) Now, we have two cases depending on whether n is even or odd. If n is even, n − 1 is odd. Hence, (3) is equivalent to x = x + y , y = 0 0 which contradicts our nonzero choice of y. Hence, when n is even, fn(x) does not have a root. Therefore, fn(x) can have at most 1 distinct root, namely x = 0. If n is odd, then n − 1 is even. Now (3) is equivalent to jxj = jx + yj , ±x = x + y , y = 0 or y = −2x 0 y Again, since we chose y to be nonzero, fn(x) has a unique root x = − 2 . By Lemma 1.1, we conlcude that fn(x) can have at most 2 distinct roots, namely x = 0 and x = −y. Therefore, there can be no other type of roots to the original equation than the ones listed. Example 1.1 showed how we could turn information about our function (specifically, knowledge of where its roots are) into information about the derivative. The Mean Value Theorem can also be used to turn information about the derivative into information about the function as we illustrate here: Example 1.2. Let p(t) denote the current location of a particle moving in a one-dimensional space. Suppose that p(0) = 0; p(1) = 1 and p0(0) = p0(1) = 0. Show that there must be some point in time in [0; 1] where jp00(t)j ≥ 4. Solution. We will prove this by contradiction. Page 2 of 10 Seunghee Ye Ma 8: Week 5 Oct 28 Suppose for every t 2 [0; 1], we have jp00(t)j < 4. Where can we go from here? We have some boundary conditions, i.e. p(0) = 0; p(1) = 1; p0(0) = p0(1) = 0, and one piece of global information, i.e. jp00(t)j < 4. How can we turn this knowledge of the second derivative to information about the rest of the function? What does the Mean Value Theorem say about p0(t)? It tells us that on any interval [a; b], we can find c 2 (a; b) such that p0(b) − p0(a) = (p0)0(c) = p00(c) b − 1 In particular, we know that for all [a; b] ⊂ [0; 1] the Mean Value Theorem tells us that 0 0 p (b) − p (a) 00 = jp (c)j < 4 b − a where c 2 (a; b) ⊂ [0; 1]. In particular, if we set a = 0; b = t and remember that p0(0) = 0, we see that 0 0 0 0 p (t) − p (0) p (t) − 0 jp (t)j = = < 4 t − 0 t t Hence, jp0(t)j < 4t. Similarly, if we let a = 1 − t; b = 1 and remember p0(1) = 0, we get 0 0 0 0 p (1) − p (1 − t) 0 − p (1 − t) jp (1 − t)j = = < 4 1 − (1 − t) t t Hence, jp0(1 − t)j < 4t. Great. So we managed to turn information about the second derivative to information about the first derivative. Let's pretend for the moment that you are back in your high school calculus course and you know how to find antiderivatives. In this situation, we have a function p(t) with the following properties: • p(0) = 0; p(1) = 1 •j p0(t)j < 4t for all t 2 (0; 1) •j p0(1 − t)j < 4t for all t 2 (0; 1) Well, if p0(t) < 4t, we can integrate to get p(t) < 2t2 + C. Since p(0) = 0, we see that p(t) < 2t2 8t 2 (0; 1) Now, since p0(1 − t) < 4t we know that −p0(1 − t) > −4t. Integrating, we get p(1 − t) > −2t2 + C and using p(1) = 1, we see that p(1 − t) > −2t2 + 1 8t 2 (0; 1) 1 But what happens when we take a look at t = 2 ? In our first inequality, we have 1 12 1 p < 2 = 2 2 2 But in our second inequality we have 1 12 1 p 1 − > −2 + 1 = 2 2 2 Hence, we get 1 1 1 < p < 2 2 2 Page 3 of 10 Seunghee Ye Ma 8: Week 5 Oct 28 which is a contradiction. Of course, this was assuming we knew how to find antiderivatives. Let's solve this problem using just the tools we have available to us. In fact, let's use the Mean Value Theorem one more time to solve the problem. Earlier, we turned information about the second derivative into information about the first derivative. Can we try the same trick to get information about the original function, p(x)? By the Mean Value Theorem, we know that for all t 2 [0; 1], we can find c 2 (0; t) such that p(t) − p(0) p(t) = = p0(c) t − 0 t However, we know that jp0(c)j < 4c < 4t for all c 2 (0; t) ⊂ (0; 1). Hence, we get p(t) 0 2 = jp (c)j < 4c < 4t ) jp(t)j < 4t t This isn't exactly the bound jp(t)j < 2t2 we obtained by integrating. But we can take this bound and look at the interval [t; 2t] ⊂ [0; 1]. By the Mean Value Theoren, we can find c 2 (t; 2t) ⊂ (0; 1) such that p(2t) − p(t) = p0(c) 2t − t Hence, we get jp(2t) − p(t)j = tjp0(c)j < t · 4c < t · 4 · 2t = 8t2 ) jp(2t)j < jp(t)j + 8t2 < 4t2 + 8t2 Hence, for all t such that 2t 2 [0; 1], we have jp(2t)j < 4t2 + 8t2 = 4t2(1 + 2) Let's do one more step.
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