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( 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] .

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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
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