1 Mean Value Theorem 1 1.1 Applications of the Mean Value Theorem

Home , Alternating series test, 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 ∈ (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 ∈ R and n ∈ 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

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Proof. Since f is diﬀerentiable on [a1, a2], by the Mean Value Theorem, we can ﬁnd b1 ∈ (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 ∈ R and n ∈ 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

|x| = |x + y| ⇔ ±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 |p00(t)| ≥ 4.

Solution. We will prove this by contradiction.

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Suppose for every t ∈ [0, 1], we have |p00(t)| < 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. |p00(t)| < 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 ∈ (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 = |p (c)| < 4 b − a where c ∈ (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 |p (t)| = = < 4 t − 0 t t

Hence, |p0(t)| < 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) |p (1 − t)| = = < 4 1 − (1 − t) t t

Hence, |p0(1 − t)| < 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

•| p0(t)| < 4t for all t ∈ (0, 1)

•| p0(1 − t)| < 4t for all t ∈ (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 ∀t ∈ (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 ∀t ∈ (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

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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 ∈ [0, 1], we can find c ∈ (0, t) such that

p(t) − p(0) p(t) = = p0(c) t − 0 t

However, we know that |p0(c)| < 4c < 4t for all c ∈ (0, t) ⊂ (0, 1). Hence, we get

p(t) 0 2 = |p (c)| < 4c < 4t ⇒ |p(t)| < 4t t

This isn’t exactly the bound |p(t)| < 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 ∈ (t, 2t) ⊂ (0, 1) such that

p(2t) − p(t) = p0(c) 2t − t Hence, we get

|p(2t) − p(t)| = t|p0(c)| < t · 4c < t · 4 · 2t = 8t2 ⇒ |p(2t)| < |p(t)| + 8t2 < 4t2 + 8t2

Hence, for all t such that 2t ∈ [0, 1], we have

|p(2t)| < 4t2 + 8t2 = 4t2(1 + 2)

Let’s do one more step. If we look at an interval of the form [2t, 3t] ⊂ [0, 1], we can ﬁnd c ∈ (2t, 3t) ⊂ (0, 1) such that p(3t) − p(2t) = p0(c) 3t − 2t Hence, we get

|p(3t) − p(2t)| = t|p0(c)| < t · 4c < t · 4 · 3t = 12t2 ⇒ |p(3t)| < |p(2t)| + 12t2 < 4t2 + 8t2 + 12t2

Hence, for all t such that 3t ∈ (0, 1), we conclude that

|p(3t)| < 4t2 + 8t2 + 12t2 = 4t2(1 + 2 + 3)

By a simple inductive argument, we can conclude that for whenever nt ∈ (0, 1) for n ∈ N, we have

|p(nt)| < 4(1 + 2 + ··· + n)t2 = 2n(n + 1)t2

t Now, let t ∈ (0, 1). Then, we can write t = n · n for all n ∈ N. By above inequality, we have that for all n, 2 t t 2 n + 1 |p(t)| = p n · < 2n(n + 1) = 2t n n n

Since this is true for all n ∈ N, we can take the limit as n goes to inﬁnity. Then, we obtain

|p(t)| < 2t2 ∀t ∈ (0, 1)

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(Technically, what we get is |p(t) ≤ 2t2 but we know that above inequality is true from integrating) Similarly, 2 1 we can show that p(1 − t) ≥ −2t + 1. Hence, by letting t = 2 , we get

1 1 1 < p < 2 2 2

which is a contradiction.

2 Midterm Review

2.1 Proof Techniques

Example 2.1. Suppose that a1, . . . , an ∈ [0, 1]. Show that

n n Y X (1 − ai) ≥ 1 − ai i=1 i=1

Proof. We proceed by induction. The base case is trivial. For n = 1, we have

1 1 Y X (1 − ai) = 1 − a1 ≥ 1 − a1 = 1 − ai i=1 i=1

Now, assume that the statement is true for all k ≤ n. Pick a1, . . . , an+1 ∈ [0, 1]. Note that

n+1 n Y Y (1 − ai) = (1 − an+1) (1 − ai) i=1 i=1

By induction hypothesis we see that

n+1 n Y Y (1 − ai) = (1 − an+1) (1 − ai) (4) i=1 i=1 n ! X ≥ (1 − an+1) 1 − ai (5) i=1 n ! n ! X X = 1 − an+1 − ai + an+1 ai (6) i=1 i=1 n X ≥ 1 − an+1 − ai (7) i=1 n+1 X = 1 − ai (8) i=1

√ Example 2.2. Show that 5 39 is irrational. √ 5 Proof. We proceed by√ contradiction. Assume that 39 is rational. Then, we can ﬁnd relatively prime 5 p integers p, q such that 39 = q . By raising both sides to the ﬁfth power we get

p5 39 = ⇒ 39q5 = p5 q5

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Note that 3 divides 39q5. Hence, 3 must also divide p5. However, since 3 is a prime number, it means that 3 divides p: one way to deﬁne prime numbers is to deﬁne them as numbers p with the property that if p | xy, then p | x or p | y. This is also not very hard to prove by contradiction. Now that we know 3 | p, we can write p = 3r. Hence, we have

39q5 = (3r)5 = 35r5 ⇒ 13q5 = 34r5

2.2 Sequences Example 2.3. Let x, y be a pair of positive real numbers such that x < y. Show that

lim (xn + yn)1/n = y n→∞

Proof. We will use the squeeze theorem to solve the problem. Note that

y ≤ (xn + yn)1/n ≤ (yn + yn)1/n = 21/ny

But we know that lim y = y n→∞ and lim 21/ny = y lim 21/n = y n→∞ n→∞ Hence, by the squeeze theorem, we conclude that

lim (xn + yn)1/n = y n→∞

Example 2.4. Consider the following sequence:

∞ Pn 1 k X 2 a = `=1 ` n k! k=0

Determine whether {an} converges and ﬁnd the limit if it exists.

Proof. So this is a pretty crazy looking sequence. But there’s nothing to be afraid of. Recall that if f(x) is

continuous at a, and we have a sequence {an} which converges to a, then f(an) converges to f(a). Let’s try to use this. First, we formulate the problem in the right way so we can use the theorem. What should be our f(x)? x x Pn 1 Note that an looks suspiciously like the function e . In fact, if f(x) = e , an = f( `=1 `2 ). Great. We know that f(x) = ex is continuous on R. Pn 1 P∞ 1 Now, note that `=1 `2 is the n-th partial sum of a familiar sequence n=1 n2 . We already know that this sequence of partial sums converge:

n ∞ X 1 X 1 π2 lim = = n→∞ `2 n2 6 `=1 n=1

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Therefore, we conclude that ∞ Pn 1 k 2 X 2 π 2 lim `=1 ` = f = eπ /6 n→∞ k! 6 k=0

√ p √ q p √ Example 2.5. Show that the sequence 1, 2, 2 2, 2 2 2,... converges.

Proof. We will show that this sequence is monotonically increasing and bounded. First, let’s show that the sequence is monotonically increasing. Note that the sequence can be recursively deﬁned as follows: √ a1 = 1, an+1 = 2an √ Let’s proceed by induction. For n = 1, a2 = 2 ≥ 1 = a1. Now suppose ak ≤ ak+1 for all k = 1, . . . , n − 1. We want to show that an+1 ≥ an. We have:

2 p 2 √ 2 2 an = ( 2an−1) = 2an−1 ≤ 2an = ( 2an) = an+1

Hence, the sequence is monotonically increasing. Now, let’s show that the sequence is bounded from above. How are we going to do this? Induction!

We claim that the sequence is bounded above by 2. The base case is trivial: a1 = 1 ≤ 2. Now suppose for all k = 1, . . . , n, we have ak ≤ 2. Then, note that √ √ √ √ √ an+1 = 2an = 2 · an ≤ 2 · 2 = 2

Therefore, an+1 ≤ 2. Hence, we conclude that an ≤ 2 for all n. Since {an} is a monotonically increasing sequence which is bounded from above, we conclude that the sequence converges. (In fact, lim an = 2) n→∞

2.3 Series Example 2.6. Show that the following series converges conditionally

∞ X (−1)n n log n n=2 2

Proof. First, let’s show that the series converges. To do this, we can use the Alternating Series Test. By the Alternating Series Test, it suﬃces to show that

1 lim = 0 n→∞ n log2 n Note that for all n ≥ 2 we have the following inequalities:

1 1 0 ≤ ≤ n log2 n n 1 However, we have lim 0 = 0 and lim = 0. Therefore, by the squeeze theorem we conclude n→∞ n→∞ n 1 lim = 0 n→∞ n log2 n Hence, the Alternating Series Test tells us that above series converges.

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Now, we show that the series does not converge absolutely. In other words, we want to show that the series ∞ X 1 n log n n=2 2 diverges. First, let’s write out the ﬁrst few terms of the series.

∞ X 1 1 1 1 1 1 = + + + + + ··· n log n 2 log 2 3 log 3 4 log 4 5 log 5 6 log 6 n=2 2 2 2 2 2 2

Now, we observe the following set of inequalities:

1 1 1 1 ≥ 1 · = 1 · = · (9) 2 log2 2 2 log2 2 2 · 1 2 1 1 1 1 1 1 1 + ≥ 2 · = 2 · 2 = · (10) 3 log2 3 4 log2 4 4 log2 4 2 · 2 2 2 1 1 1 1 1 2 1 1 1 + + + ≥ 4 · = 2 · 3 = · (11) 5 log2 5 6 log2 6 7 log2 7 8 log2 8 8 log2 8 2 3 2 3 We start to see a pattern. More precisely, we see that

1 1 n−1 1 1 1 n−1 n−1 + ··· n n ≥ 2 n n = · (2 + 1) log2(2 + 1) 2 log2 2 2 log2 2 2 n

Now, we are ready to show that the series diverges absolutely. Consider the partial sums S2n . We have

2n X 1 S n = (12) 2 n log n n=2 2 1 1 1 1 1 = + + + ··· + n−1 n−1 + ··· + n n (13) 2 log2 2 3 log2 3 4 log2 4 (2 + 1) log2(2 + 1) 2 log2 2 1 1 1 1 1 ≥ · 1 + · + ··· + · (14) 2 2 2 2 n 1 1 1 = 1 + + ··· + (15) 2 2 n

But note that we have now bounded the partial sum S2n from below by half of the n-th partial sum of the harmonic series. However, we already know that the harmonic series diverges, which implies that the partial

sums of the harmonic series is a divergent sequence. Therefore, by Comparison Test, we conclude that S2n ∞ X 1 also diverges. Thus, we conclude that the series diverges. n log n n=2 2 ∞ X (−1)n Combining the two results, we conclude that the series converges conditionally. n log n n=2 2

Example 2.7. For n ∈ N, let α(n) denote the number of prime factors of n. For example, α(2) = 1, α(4) = 2, α(6) = 2, α(120) = 5, etc. Determine whether the following series converges:

∞ X α(n) n3 n=1

Proof. Whenever we have a series and we want to get a sense of whether it converges, it is a good idea to P 1 compare it to series that we know already. In this particular example, we will compare it to n2 . First, note that α(n) ≤ n for all n. Why? Suppose you have a number with k prime factors. The smallest

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such number is 2k > k. Using this, we see that

α(n) n 1 ≤ = n3 n3 n2 Now, by the comparison test, we conclude that our series converges.

2.4 Continuity and Differentiability of Functions Example 2.8. Consider ( x2 x ∈ f(x) = Q 0 x 6∈ Q Is f(x) continuous and/or differentiable at 0? Proof. We will show that f(x) is indeed differentiable at 0. Since differentiable functions are always continuous, it suffices to show that f(x) is differentiable at 0. We will, however, also show that f(x) is continuous at 0 to get more practice with ε − δ proofs. √ Let ε > 0. Then, we choose δ = ε. Now, let x be any real number such that |x − 0| < δ. Then, ( |x2| x ∈ |f(x) − f(0)| = |f(x) − 0| = |f(x)| = Q 0 x 6∈ Q

If x 6∈ Q, then |f(x) − 0| = 0 < ε so we are done. If x ∈ Q, we have

|f(x)| = |x2| < δ2 = ε

Therefore, lim f(x) = 0 = f(0) and thus, f(x) is continuous at 0. x→0 Now, we show that f(x) is differentiable at 0. Let’s look at the definition of the derivative. We say that f(x) is differentiable at 0 if the limit f(h) − f(0) lim h→0 h exists. Since f(0) = 0, this is just f(h) lim h→0 h But note that because of the way f(x) is defined, for all h, we have

2 2 2 2 h f(h) h −h ≤ |f(h)| ≤ h ⇒ − ≤ ≤ h h h

But note that 2 2 h h lim = lim − = 0 h→0 h h→0 h f(h) Therefore, by the squeeze theorem, we conclude that lim = 0. Hence, f(x) is diﬀerentiable at 0 and h→0 h f 0(0) = 0.

Example 2.9. Show that the function ( x sin 1 x 6= 0 f(x) = x 0 x = 0

is continuous but not diﬀerentiable at 0.

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1 Proof. Recall that the function g(x) = sin x does not have a limit as x → 0. Let’s see why f(x) is continuous at 0. To show that f(x) is continuous at 0, we must show that for all ε > 0, we can ﬁnd δ > 0 such that if |x| < δ, then |f(x) − f(0)| = |f(x)| < ε. As always, we start by examining the quantity that we want to make small i.e. |f(x)|. For nonzero x, we have that 1 |f(x)| = x sin x Since | sin y| ≤ 1, we can bound this as follows 1 |f(x)| = x sin ≤ |x| x

But by choosing δ, we get to control |x|. Let’s start the proof. Let ε > 0 and let δ = ε. Suppose |x − 0| = |x| < δ. Then,

|f(x) − 0| = |f(x)| ≤ |x| < δ = ε

Therefore, we conclude that lim f(x) = 0 = f(0). Hence, f(x) is continuous at 0. x→0 Now, we show that f(x) is not diﬀerentiable at 0. In other words, we what to show that the following limit does not exist:

f(h) − f(0) f(h) h sin 1 1 lim = lim = lim h = lim sin h→0 h h→0 h h→0 h h→0 h

However, we have already seen that this limit does not exist. Thus, by the definition of differentiability, we conclude that f(x) is not differentiable at 0.

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