<< Chapter < Page Chapter >> Page >

( cos 2 + sin 2 = 1 and exp ( i π = - 1 . )

  1. Prove that cos 2 ( z ) + sin 2 ( z ) = 1 for all complex numbers z .
  2. Prove that cos ( π ) = - 1 . HINT: We know from part (a) that cos ( π ) = ± 1 . Using the Mean Value Theorem for the cosine function on the interval [ 0 , π ] , derive a contradiction from the assumption that cos ( π ) = 1 .
  3. Prove that exp ( i π ) = - 1 . HINT: Recall that exp ( i z ) = cos ( z ) + i sin ( z ) for all complex z . (Note that this does not yet tell us that e i π = - 1 . We do not yet know that exp ( z ) = e z . )
  4. Prove that cosh 2 z - sinh 2 z = 1 for all complex numbers z .
  5. Compute the derivatives of the tangent and hyperbolic tangent functions tan = sin / cos and tanh = sinh / cosh . Show in fact that
    tan ' = 1 cos 2 and tanh ' = 1 cosh 2 .

Here are two more elementary consequences of the Mean Value Theorem.

  1. Suppose f and g are two complex-valued functions of a real (or complex) variable,and suppose that f ' ( x ) = g ' ( x ) for all x ( a , b ) (or x B r ( c ) . ) Prove that there exists a constant k such that f ( x ) = g ( x ) + k for all x ( a , b ) (or x B r ( c ) . )
  2. Suppose f ' ( z ) = c exp ( a z ) for all z , where c and a are complex constants with a 0 . Prove that there exists a constant c ' such that f ( z ) = c a exp ( a z ) + c ' . What if a = 0 ?
  3. (A generalization of part (a)) Suppose f and g are continuous real-valued functions on the closed interval [ a , b ] , and suppose there exists a partition { x 0 < x 1 < ... < x n } of [ a , b ] such that both f and g are differentiable on each subinterval ( x i - 1 , x i ) . (That is, we do not assume that f and g are differentiable at the endpoints.) Suppose that f ' ( x ) = g ' ( x ) for every x in each open subinterval ( x i - 1 , x i ) . Prove that there exists a constant k such that f ( x ) = g ( x ) + k for all x [ a , b ] . HINT: Use part (a) to conclude that f = g + h where h is a step function, and then observe that h must be continuous and hence a constant.
  4. Suppose f is a differentiable real-valued function on ( a , b ) and assume that f ' ( x ) 0 for all x ( a , b ) . Prove that f is 1-1 on ( a , b ) .

Let f : [ a , b ] R be a function that is continuous on its domain [ a , b ] and differentiable on ( a , b ) . (We do not suppose that f ' is continuous on ( a , b ) . )

  1. Prove that f is nondecreasing on [ a , b ] if and only if f ' ( x ) 0 for all x ( a , b ) . Show also that f is nonincreasing on [ a , b ] if and only if f ' ( x ) 0 for all x ( a , b ) .
  2. Conclude that, if f ' takes on both positive and negative values on ( a , b ) , then f is not 1-1. (See the proof of [link] .)
  3. Show that, if f ' takes on both positive and negative values on ( a , b ) , then there must exist a point c ( a , b ) for which f ' ( c ) = 0 . (If f ' were continuous, this would follow from the Intermediate Value Theorem. But, we are not assuming here that f ' is continuous.)
  4. Prove the Intermediate Value Theorem for Derivatives: Suppose f is continuous on the closed bounded interval [ a , b ] and differentiable on the open interval ( a , b ) . If f ' attains two distinct values v 1 = f ' ( x 1 ) < v 2 = f ' ( x 2 ) , then f ' attains every value v between v 1 and v 2 . HINT: Suppose v is a value between v 1 and v 2 . Define a function g on [ a , b ] by g ( x ) = f ( x ) - v x . Now apply part (c) to g .

Here is another perfectly reasonable and expected theorem, but one whose proof is tough.

Inverse function theorem

Suppose f : ( a , b ) R is a function that is continuous and 1-1 from ( a , b ) onto the interval ( a ' , b ' ) . Assume that f is differentiable at a point c ( a , b ) and that f ' ( c ) 0 . Then f - 1 is differentiable at the point f ( c ) , and

f - 1 ' ( f ( c ) ) = 1 f ' ( c ) .

The formula f - 1 ' ( f ( c ) ) = 1 / f ' ( c ) is no surprise. This follows directly from the Chain Rule. For, if f - 1 ( f ( x ) ) = x , and f and f - 1 are both differentiable, then f - 1 ' ( f ( c ) ) f ' ( c ) = 1 , which gives the formula. The difficulty with this theorem is in proving that the inverse function f - 1 of f is differentiable at f ( c ) . In fact, the first thing to check is that the point f ( c ) belongs to the interior of the domain of f - 1 , for that is essential if f - 1 is to be differentiable there, and here is where the hypothesis that f is a real-valued function of a real variable is important. According to [link] , the 1-1 continuous function f maps [ a , b ] onto an interval [ a ' , b ' ] , and f ( c ) is in the open interval ( a ' , b ' ) , i.e., is in the interior of the domain of f - 1 .

According to part (2) of [link] , we can prove that f - 1 is differentiable at f ( c ) by showing that

lim x f ( c ) f - 1 ( x ) - f - 1 ( f ( c ) ) x - f ( c ) = 1 f ' ( c ) .

That is, we need to show that, given an ϵ > 0 , there exists a δ > 0 such that if 0 < | x - f ( c ) | < δ then

| f - 1 ( x ) - f - 1 ( f ( c ) ) x - f ( c ) - 1 f ' ( c ) | < ϵ .

First of all, because the function 1 / q is continuous at the point f ' ( c ) , there exists an ϵ ' > 0 such that if | q - f ' ( c ) | < ϵ ' , then

| 1 q - 1 f ' ( c ) | < ϵ

Next, because f is differentiable at c , there exists a δ ' > 0 such that if 0 < | y - c | < δ ' then

| f ( y ) - f ( c ) y - c - f ' ( c ) | < ϵ ' .

Now, by [link] , f - 1 is continuous at the point f ( c ) , and therefore there exists a δ > 0 such that if | x - f ( c ) | < δ then

| f - 1 ( x ) - f - 1 ( f ( c ) | < δ '

So, if | x - f ( c ) | < δ , then

| f - 1 ( x ) - c | = | f - 1 ( x ) - f - 1 ( f ( c ) ) | < δ ' .

But then, by [link] ,

| f ( f - 1 ( x ) ) - f ( c ) f - 1 ( x ) - c - f ' ( c ) | < ϵ ' ,

from which it follows, using [link] , that

| f - 1 ( x ) - f - 1 ( f ( c ) ) x - f ( c ) - 1 f ' ( c ) | < ϵ ,

as desired.

REMARK A result very like [link] is actually true for complex-valued functions of a complex variable. We will have to show that if c is in the interior of the domain S of a one-to-one, continuously differentiable, complex-valued function f of a complex variable, then f ( c ) is in the interior of the domain f ( S ) of f - 1 . But, in the complex variable case, this requires a somewhat more difficult argument. Once that fact is established, the proof that f - 1 is differentiable at f ( c ) will be the same for complex-valued functions of complex variables as it is here for real-valued functions of a real variable.Though the proof of [link] is reasonably complicated for real-valued functions of a real variable, the corresponding result for complex functions is much more deep, and that proofwill have to be postponed to a later chapter. See [link] .

Questions & Answers

what are components of cells
ofosola Reply
twugzfisfjxxkvdsifgfuy7 it
Sami
58214993
Sami
what is a salt
John
the difference between male and female reproduction
John
what is computed
IBRAHIM Reply
what is biology
IBRAHIM
what is the full meaning of biology
IBRAHIM
what is biology
Jeneba
what is cell
Kuot
425844168
Sami
what is cytoplasm
Emmanuel Reply
structure of an animal cell
Arrey Reply
what happens when the eustachian tube is blocked
Puseletso Reply
what's atoms
Achol Reply
discuss how the following factors such as predation risk, competition and habitat structure influence animal's foraging behavior in essay form
Burnet Reply
cell?
Kuot
location of cervical vertebra
KENNEDY Reply
What are acid
Sheriff Reply
define biology infour way
Happiness Reply
What are types of cell
Nansoh Reply
how can I get this book
Gatyin Reply
what is lump
Chineye Reply
what is cell
Maluak Reply
what is biology
Maluak
what is vertibrate
Jeneba
what's cornea?
Majak Reply
what are cell
Achol
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, Analysis of functions of a single variable. OpenStax CNX. Dec 11, 2010 Download for free at http://cnx.org/content/col11249/1.1
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Analysis of functions of a single variable' conversation and receive update notifications?

Ask