Solutions to Chapter 1 Quiz: Functions

Problem 1 This problem tests your understanding of domains. Remember, the domain of a function is the largest portion of the real numbers where that function is actually defined. √ Question: What is the domain of the function f(x) = ln x? Answer: [1, +∞). We should recognize f as a composition: given an input number x, we first compute its logarithm, and then compute the square-root of that logarithm. The logarithm is only defined for positive x. So, we immediately see that we need x > 0 for ln(x) to make sense. But we are not done! The output of ln x for positive x can be any real number, and in order to make the output of ln x the input of the square root function, we must make sure that we restrict to only those values of x for which ln x 0. Since ln(1) = 0 and natural log is an increasing function, we see that the output of ln x is positive for any x 1. Thus, the domain of f is all x 1, or all x in [1, +∞). Problem 2 This problem tests your ability to compute Taylor series of a composition of functions if you know the Taylor series of each individual function. In short, you have to stuff Taylor series into Taylor series! Moreover, knowledge of big-O is extremely useful in this problem since it helps you figure out where to truncate the series.
1 Question: What is the Taylor series of ln 1−x about x = 0 up to and including terms of order 3? 1 Answer: ln 1−x = x + 1 x2 + 1 x3 + O(x4 ). 2 3 1 Let’s see: we should all at least know the Taylor series of 1−x about x = 0 since it is just the geometric series! That is,

1 = 1 + x + x2 + x3 + O(x4 ). 1−x Here we have stopped at O(x4 ) because we only need to compute up to order 3. So far so good, but this is not the answer we need. In fact we have to compute the natural log of this entire series. So, how do we approximate the expression ln(1 + x + x2 + x3 + O(x4 ))
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up to terms of order 3? Since the expression in parentheses equals 1 for our chosen point x = 0, we are interested in the expansion of ln(1 + y) about y = 0, where y = x + x2 + x3 + O(x4 ). This is also a Taylor series that we have seen before, and so we know that: y2 y 3 + + O(y4 ). 2 3 Again, we stop at O(x4 ) because we only care about terms of order 3 and less! Now, let’s plug in our expression for y in terms of x into this series: ln(1 + y) = y − ln(1 + x + x2 + x3 + O(x4 )) = (x + x2 + x3 + O(x4 )) (x + x2 + x3 + O(x4 ))2 2 (x + x2 + x3 + O(x4 ))3 + + O(x4 ). 3 This looks horrible! But now we can use what we have learned about big-O to simplify things a lot. For instance, that last term on the right side contains a lot of junk: we only need to raise x to the third power, because everything else has a higher degree and is already in O(x4 ). Similarly, when we look at the second term, we only need to square up to the x2 term because again, all the other terms are already in O(x4 ). Let’s use these simplifications and see that (x + x2 )2 x3 ln(1 + x + x2 + x3 + O(x4 )) = (x + x2 + x3 ) − + + O(x4 ). 2 3 That’s much better! Expanding the middle term and collecting the powers of x (be careful about that middle term, it has a negative coefficient and I just got it wrong twice!), we get the final answer: 1 1 1 ln = x + x2 + x3 + O(x4 ), 1−x 2 3 − Problem 3 This problem tests the relation between derivatives and coefficients in Taylor series. Essentially, the only thing you need to know in order to solve the problem is this: the coefficient of the (x − a)n term in the Taylor series of f(x) about x = a is f(n) (a) n!

Question: Use your knowledge of Taylor series, find the sixth derivative f(6) (0) of f(x) = 2 e−x evaluated at x = 0. Answer: -120. Let’s start with the Taylor series of ey : ey = Or, if you are allergic to the yn . n! n=0
∞

symbol we can just use ey = 1 + y + y2 y3 + + O(y4 ). 2! 3!

SOLUTIONS TO CHAPTER 1 QUIZ: FUNCTIONS

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Plugging in y = −x2 , we get the alternating sum e−x = 1 − x2 +
2

x4 x6 − + O(x8 ). 2! 3!

Keeping the giant box above in mind, we examine the coefficient of the x6 term: it equals (6) 1 − 3! . We know that this coefficient equals f 6!(0) , so in fact f(6) (0) = − 6! = −(6 × 5 × 4) = −120. 3! Problem 4 This problem tests the application of Taylor series to compute complicated-looking limits. Again, understanding the big-O helps to make an otherwise nasty computation very easy. Question: Recall that the Taylor series for arctan is
∞

arctan(x) = k=0 (−1)k

x2k+1 2k + 1

for |x| < 1. Using this, compute arctan x x→0 x3 + 7x lim
1 Answer: 7 . Let’s write out the first few terms of the arctan series:

arctan(x) = x − So, our answer is given by

x3 x5 x7 + − + O(x9 ). 3 5 7 x3 3
7 x5 −x 5 7 x3 + 7x

x− arctan x lim 3 = lim x→0 x + 7x x→0 Now let’s simplify using big-O:

+

+ O(x9 )

.

arctan x x + O(x3 ) lim = lim . x→0 x3 + 7x x→0 7x + O(x3 ) Note that there is an x that is a common factor of both numerator and denominator. Cancel it to get arctan x 1 + O(x2 ) 1 = lim = . lim 3 x→0 7 + O(x2 ) x→0 x + 7x 7 Another way of attacking this problem is by using l’Hopital’s rule (in Mathematics, there is usually more then one way of solving a problem!). Remember that the derivative of 1 arctan x is 1+x2 . Then, arctan x 1 1 2 lim = lim 1+x 2 = lim = x→0 x3 + 7x x→0 7 + 3x x→0 (1 + x2 )(7 + 3x2 ) 7
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Problem 5 This problem tests the same concepts as the last one. You need to use the Taylor expansion for sines/cosines near 0. Question: Compute limx→0 cos(3x)−cos(5x) . x2 Answer: 8. The limit is taken as x → 0 so we can expand the cosines in the numerator at x = 0. 2 Remember cos(y) = 1 − x + O(x4 ) near y = 0. Plugging y = 3x gives 2! 9x2 cos(3x) = 1 − + O(x4 ), 2! and plugging y = 5x gives 25x2 + O(x4 ), cos(3x) = 1 − 2! So, the difference is 9 − 25 2 x + O(x4 ) = 8x2 + O(x4 ). 2 2 Now just divide by x and take the limit as x → 0 to get the answer! cos(3x) − cos(5x) = − Problem 6 This problem only tests your knowledge of standard Taylor series when the input variables have been changed slightly. Question: Determine which value is approximated by √ √ √ √ ( 2π)3 ( 2π)4 ( 2π)5 2 1 + 2π + π + + + + H.O.T. 3! 4! 5! √ Answer: e 2π . √ n Look at the terms of order 3 and above: they are all of the form y where y = 2π. n! Looking at the smaller terms, they also follow the same pattern: for example, consider the order 2 term: √ ( 2π)2 2π2 = = π2 . 2! 2 √ n Okay, so the series looks a lot like ∞ y = ey where y = 2π. Thus, this series is simply k=0 n! √ approximating e 2π . Problem 7 This problem only tests your understanding of the summation notation Question: Which of the following expressions describes the sum √ √ 2 2 3 3 2 −x + x − x + x4 + H.O.T. 4 9 16 Choose all that apply. Answer. The two correct answers were √ √ ∞ ∞ n n n + 1 n+1 (−1)n 2 x and (−1)n+1 x n (n + 1)2 n=1 n=0 .

SOLUTIONS TO CHAPTER 1 QUIZ: FUNCTIONS

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These are equal: the index n of the first series corresponds to the index n − 1 of the second series. You can easily plug in low values of n to confirm that these give us the expression from the question. Any other series can be ruled out by plugging in values of n and comparing coefficients. √ 2n For example, when n = 2 the series ∞ (−1)n n2 xn gives us a coefficient of 1 for the x2 n=1 2 term, which is clearly wrong. Problem 8 This problem only tests your ability to recognize and manipulate the geometric series. All you had to know was ∞ 1 yk for |y| < 1. = 1 − y k=0 Question: Use the geometric series to evaluate the sum
∞

3k+1 xk k=0 Don’t forget to indicate what restrictions there are on x. 3 Answer: 2−3x on |x| < 1 . 3 This is easy! The series looks like ∞ yk for y = 3x (but there is an extra factor of 3), k=0 3 so immediately we know that the answer equals 1−y , and plugging in y gives us 3 . 1 − 3x And what about the restriction on x? Well, the series only converges for |y| < 1, which for us means 3|x| < 1 or |x| < 1 . That’s all! 3 Problem 9 This problem tests your ability to compute the Taylor series of a polynomial. A polynomial written in the usual form a0 + a1 x + a2 x2 + . . . an xn is its own Taylor series about x = 0. But as we have seen in the lectures, things look very different when we expand about points that are not 0... Question: What is the Taylor series expansion about x = 2 of x3 − 2x2 + 3x − 4? Answer: 2 + 7(x − 2) + 4(x − 2)2 + (x − 2)3 + O((x − 2)4 ). Let’s write p(x) = x3 − 2x2 + 3x − 4. The formula for the first few terms of the Taylor series T (x) of p about the point x = a is given by p(2) (a) p(3) (a) (x − a)2 + (x − a)3 + O((x − a)4 ) 2! 3! The constant term is just p(2) = 23 − 23 + 6 − 4 = 2 , but now we should start computing the derivatives of p(x) at x = 2 to figure out the higher terms. Here they are: T (x) = p(a) + p (a) + p (x) = 3x2 − 4x + 3, so p (2) = 12 − 8 + 3 = 7 . Next, p (x) = 6x − 4, so p (2) = 12 − 4 = 8 .

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Finally, p(3) (x) = 6 for all possible x, including x = 2 Of course, all higher derivatives are zero. So, plugging in the boxed numbers for the derivatives of p(x) at x = 2 into the Taylor series formula, we get T (x) = 2 + 7(x − 2) + 8 6 (x − 2)2 + (x − 2)3 . 2! 3!

Reduce the fractions to get the answer. Note: the Taylor series of any function about a point x = a must be expressed in terms of powers of (x − a). So, even if there were choices that were algebraically equivalent to this answer, you would not have received credit for selecting them. Problem 10 This problem tests your understanding of big-O. Remember, f(x) belongs to O(g(x)) as x → a if and only if we have some constant C so that |f(x)| C · |g(x)| when x is near a .

Question: Exactly two of the following statements are correct. Select the two correct statements. √ (1) 16x4 − 2 is in O(x2 ) as x → +∞. (2) ln(1 + x + x2 ) is in O(xn ) for all n 1 as x → +∞. 2 (3) ex is in O(x2 ) as x → +∞. (4) cosh 2x is in O(xn ) for all n 0 as x → +∞. √ (5) 7 x is in O(x4 ) as x → 0. (6) 3x4 − 14 is in O(x2 ) as x → +∞ (7) ex belongs to O(ln x) as x → +∞. (8) 7x3 is in O(x4 ) as x → 0. Answer: Only the first two statements are true. Note: the choices on Quizzes are randomized, so you may have seen these statements in some different order on your Quiz. I’ve shown why the first statement is true and why the fourth and fifth ones are false. The methods are very similar for the other statements, so try them out on your own. 1. To see why this statement is true, we will try to compute the limit √ 16x4 − 2 lim x→+∞ x2 Let’s slide the denominator into the square root of the numerator: the only difference this makes is that we have to square it, so we get: 16x4 − 2 2 = lim 16 − 4 4 x→+∞ x→+∞ x x √ and clearly, as x → +∞ this equals 16 = 4 since the limit exists and the statement is true. lim

2 x4

term vanishes. So, the

SOLUTIONS TO CHAPTER 1 QUIZ: FUNCTIONS

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4. To see that this statement is false, remember the definition of cosh: e2x + e−2x cosh(2x) = 2 It is easy to see – by using the Taylor series of e2x and e−2x – that the Taylor series of cosh(2x) has strictly positive coefficients for all even powers of n, so in fact the series must grow faster than any xn . Thus, the statement is false. 5. To see why this statement is false, examine the limit √ 7 x lim , x→0 x4 √ 1 and note that since x = x 2 we may re-write this limit as 7 7 lim 4− 1 = lim 7 . x→0 x 2 x→0 x 2 For x positive but small, this goes to +∞. And for negative but small x it does not even exist!