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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 ...view middle of the document...

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

3

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...

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