math upper homework-Real analysis due 9 hours later

MATH 105A - Fall 2018 - François Monard - UC Santa Cruz 100

Lecture 24 - 11/30 - L’Hôpital’s rule and the division problem.

A toy problem: find expansions for ratios of functions. Suppose that for x near x0 = 0 18,

a function can be written as

h(x) = a0 + a1x + a2x

2 + . . .

b0 + b1x + b2x2 + . . . , with b0 6= 0,

and where the “. . . ” indicate that the expansion goes up to some finite precision to be determined. A natural question is: how to obtain the expansion h(x) = c0 + c1x + c2x

2 . . . ? Namely, how to obtain the coefficients ck from the coefficients ai and bj ?

We will answer this for the first two terms, noting that the formula become more and more tedious when looking for further terms.

• To compute c0 = limx→0 h(x), the numerator and denominator only need an expansion up to order 1, namely,

h(x) = a0 + o(1)

b0 + o(1) →

a0 b0

as x → 0, so c0 = a0 b0

.

• To compute c1, we study h(x) −c0, and since we need o(x) precision on h to extract the c1x term, then we need o(x) precision on the numerator and denominator, namely:

h(x) = a0 + a1x + xε(x)

b0 + b1x + xη(x) ,

then

h(x) − c0 = a0 + a1x + xε(x)

b0 + b1x + xη(x) − a0 b0

= (b0a1 −a0b1)x + x(b0ε(x) −a0η(x))

b0(b0 + b1x + xη(x))

= (b0a1 −a0b1)

b20 x + xγ(x),

where limx→0 γ(x) = 0. Therefore, c1 = b0a1 −a0b1

b20 .

L’Hôpital’s rule. The previous calculations, together with Taylor expansions, will help us ad- dress the following questions:

1. How to evaluate limx→x0

( f(x) g(x)

) when both f and g vanish at x0 ?

2. What about higher-order derivatives ?

We say that a function f vanishes at order k at x0 if

f(x0) = 0, f ′(x0) = 0, . . . f

(k−1)(x0) = 0, and f (n)(x0) 6= 0.

When f is of class Ck near x0 and vanishes at order k at x0, its Taylor expansion at order k reads

f(x) = f(k)(x0)

k! (x−x0)k + (x−x0)kη(x), lim

x→x0 η(x) = 0.

18If x0 6= 0, one may just replace x by x−x0 in everything that follows.

MATH 105A - Fall 2018 - François Monard - UC Santa Cruz 101

Theorem 89 (L’Hôpital’s rule). Suppose that f and g are of class Cn near x0, that f vanishes of order k at x0 and g vanishes at order m at x0, with both k ≤ n and m ≤ n. Then

lim x→x0

f(x)

g(x) =

{ f(k)(x0)

g(k)(x0) if k ≥ m

does not exist if k < m.

In this sense, as l’Hôpital’s rule suggests, if numerator and denominator both vanish, we can differentiate both and study the ratio of derivatives, and this process terminates on many occasions.

Proof. From the hypothesis of the theorem, we can write

f(x) = f(k)(x0)

k! (x−x0)k + (x−x0)kη(x), lim

x→x0 η(x) = 0,

g(x) = g(m)(x0)

m! (x−x0)m + (x−x0)mη(x), lim

x→x0 η(x) = 0,

where f(k)(x0) 6= 0 and g(m)(x0) 6= 0. Then we immediately have that

f(x)

g(x) =

f(k)(x0) k!

(x−x0)k + (x−x0)kη(x) g(m)(x0)

m! (x−x0)m + (x−x0)mε(x)

= 1

(x−x0)m−k f(k)(x0)

k! + η(x)

g(m)(x0) m!

+ ε(x) .

Observing the form of this ratio, we immediately conclude that the limit as x → x0 does not exists if k < m, and the limit exists and equals

f(k)(x0)

g(k)(x0) if k = m.

Example 55. 1. Find limx→1 lnx x−1 . This can be written as f(x)/g(x) where f(1) = g(1) = 0,

f ′(1) = 1 and g′(1) = 1, so the limit is 1 by l’Hôpital’s rule.

2. Find limx→0 cosx−1 x2

. Differentiating twice numerator and denominator, one arrives at the limit −1/2.

3. Find limx→0 cosx−1+x2/2

(sinx)4 . It may be unwise to differentiate until one gets a finite limit. At the

level of Taylor polynomials, look for the first nonzero term in the expansions of the numerator and denominator. Namely, from the expansion of cosine, we see that cos x−1+x

2

2 = x

4

24 +o(x4),

and since sin x = x + o(x), then one may compute that (sin x)4 = x4 + o(x4). We can then write

cos x− 1 + x2/2 (sin x)4

= x4

24 + o(x4)

x4 + o(x4) =

1/24 + o(1)

1 + o(1) →

1

24 as x → 0.

Finding higher-order derivatives of indeterminate ratios. Suppose now that we would like to find (

f g )(k)(x0) when f and g vanish at x0. From Theorem 89, it is clear that if g vanishes at

a higher order than f, we will not be able to compute anything. Therefore, let us suppose that f vanishes at at least the same order as g. That is to say, there exists some n such that, near x0,

g(x) = g(n)(x0)

n! (x−x0)n + . . . , f(x) =

f(n)(x0)

n! (x−x0)n + . . . ,

MATH 105A - Fall 2018 - François Monard - UC Santa Cruz 102

with g(n)(x0) 6= 0. In that case, a factor (x−x0)n can be factored out and simplified, and the ratio f(x) g(x)

takes the form

f(x)

g(x) = a0 + a1(x−x0) + a2(x−x0)2 . . . b0 + b1(x−x0) + b2(x−x0)2 . . .

,

where for any k = 0, 1, . . . , ak = f(n+k)(x0) (n+k)!

and bk = g(n+k)(x0) (n+k)!

. Then finding the derivatives of f g

is equivalent to obtaining the coefficients c0,c1, . . . in the expansion

f(x)

g(x) = c0 + c1(x−x0) + c2(x−x0)2 . . . ,

which is nothing but the division problem we saw earlier ! Once the coefficients cp are found, the

derivatives will be given by ( f g

)(p) (x0) = p! cp.

The first term c0 is exactly given by l’Hôpital’s rule, while the second term is given by, after simplifications,

c1 = a1b0 −a0b1

b20 = (n + 1)

f(n+1)(x0)g (n)(x0) −f(n)(x0)g(n+1)(x0)

(g(n)(x0))2 .

We can then give this as a second theorem.

Theorem 90. Suppose that f and g are of class Cn+1 near x0, that g vanishes at order n at x0 and f vanishes at least at order n at x0. Then(

f

g

)′ (x0) = (n + 1)

f(n+1)(x0)g (n)(x0) −f(n)(x0)g(n+1)(x0)

(g(n)(x0))2 .

One could write an (increasingly messy) formula for the general derivative, though the formula itself is not as important as understanding how to obtain the coefficients in practice (which is a polynomial division problem, once appropriately set up). In addition, in certain examples, the process can be cut short if one notices certain cancellations.

Example 56. To find any derivative of f(x) = sinx x

at x = 0, from the Taylor expansion sin x =

x− x 3

6 + x

5

120 + . . . , we immediately get, for x 6= 0,

sin x

x = 1 −

x2

6 +

x4

120 + . . .

therefore one may read the limits of any derivative near x = 0 directly from this expansion. This will happen every time the denominator is just a power of x (or of (x−x0) if x0 6= 0).

Example 57. Let f(x) = (ln(x+1))2

cosx−1+2x3 for x 6= 0. Find limx→0 f(x) and f ′(0) if the function is

appropriately defined at x = 0 by f(0) := limx→0 f(x).

Answer: To solve this, we will need the first two nonzero significative terms in the Taylor expansions of the numerator and denominator. By direct computation of the Taylor polynomial of ln(x + 1), we have ln(x + 1) = x− x

2

2 + o(x2) near x = 0, so

(ln(x + 1))2 =

( x−

x2

2 + o(x2)

)2 = x2 −x3 + o(x3).

MATH 105A - Fall 2018 - François Monard - UC Santa Cruz 103

On the other hand, cos x− 1 + 2x3 = −x 2

2 + 2x3 + o(x3). Therefore,

f(x) = x2 −x3 + o(x3) −x2 2

+ 2x3 + o(x3) =

1 −x + o(x) −1/2 + 2x + o(x)

, x 6= 0.

Out of this form, we can use the polynomial division approach as above to find that

lim x→0

f(x) = −2, f ′(0) = −6.

Exercises for Lecture 24:

1. For each of the following functions defined for x 6= 0, find limx→0 f(x) and f ′(0) if the function is appropriately defined at x = 0. You may use the familiar formulas for the derivatives of sine and cosine:

(a) f(x) = x sinx

.

(b) f(x) = 1−cosx x2

.

(c) f(x) = x 2−x sinx

.

(d) f(x) = x 1−cosx−sinx.

(e) f(x) = (ln(1+x))2

x2+x3 .

2. (a) For f(x) = (1 + x) 1 2 , find a general formula for f(k)(0) for k = 0, 1, 2 . . . , and write down

the k-th Taylor polynomial centered at x0 = 0.

(b) Now if x0 > 0, write a Taylor expansion of order 3 for f centered at x0 instead of 0. Instead of differentiating again, write x = x−x0 + x0 and try to extract out of f(x) an expression of the form (1 + h)

1 2 with h a small parameter19, so as to exploit the series

of part (a).

(c) Repeat part (a) for the function f(x) = ln(1 + x), near x0 = 0.

(d) Do as in part (b) to find a Taylor expansion at order 3 of ln x centered at any x0 > 0 (no need to differentiate).

19h is proportional to (x−x0)