Constant functions – Serlo

"A function is constant, if its derivative vanishes", i.e. . This is the main statement which we want to make concrete in this article.

Criterion for constant functions Bearbeiten

Theorem

Let   be an interval and   a differentiable function with   for all  . Then   is constant.

Proof

Let   with   be given. Let further   be differentiable on   such that for all   we have  . By the mean value theorem there is a   with

 

We know that   must hold. So:

 

Since   there is  . Now we multiply both sides with   :

 

Or in other words,  . Since this holds for all   and   in   , the function   must be constant.

Identity theorem of differential calculus Bearbeiten

The first conclusion of   implying that a function is constant is that functions with identical derivatives are identical except for one constant. This result will prove very useful later on in the fundamental theorem of calculus.

Theorem (Identity theorem)

Let   be two differentiable functions with  . Then, there is   for all  . Where   is a constant number.

Proof (Identity theorem)

We define the auxiliary function

 

This function is differentiables ince   and   are differentiable and

 

Hence, there must be   for all   with a constant number  . Or equivalently,

 

Application: characterization of the exponential function Bearbeiten

Theorem (Characterization of the exponential function)

Let   be differentiable. Further let   and for all   we have

 

Then, there is   for all   with a constanten  . If   and in addition  , then precisely  .

Proof (Characterization of the exponential function)

We define the auxiliary function

 

By the product and the chain rule, it is differentiable. There is

 

According to the criterion for a function being constant there is a   with   for all  . But this is now equivalent to

 

If now   and additionally  , then

 

So  .

Hint

Alternatively we can write   as   and apply the quotient rule to determine the derivative  . Furthermore, the function   satisfies the differential equation  , since there is:

 

Exercises Bearbeiten

Interval assumption for constant functions Bearbeiten

The condition that the function   is defined on an interval is necessary for the criterion for constancy! This is illustrated by the following task:

Exercise

Find a differentiable function   with   and   for all   which is not constant.

  must be chosen here such that it is not an interval. Otherwise it would follow from the previous sentence that   is constant. So, let us take two intervals

Solution

 
The function  

We define   and set  

 

The function   is obviously not constant. But for all   there is  .

For proving this we first consider a  . Let   be a sequence in   which converges towards  . Then there is an   such that for all   the inequality   is fulfilled. From this we get  . There is hence  for all   and

 

So:

 

The proof that also for all   the equation   is fulfilled goes completely analogous.

Trigonometric Pythagorean theorem Bearbeiten

Using the criterion for constancy, identities of functions can also be proven very well:

Exercise (Trigonometric Pythagorean theorem)

Show that for all   there is

 

Here,   and  .

Solution (Trigonometric Pythagorean theorem)

We define the auxiliary function

 

According to the chain and sum rule for derivatives, this is differentiable on all of  , and there is

 

Thus   is a constant number  . We can determine it by calculating  :

 

So   is constant   and there is:

 

Function equation for arctan Bearbeiten

Exercise (Function equation for  )

Show:   for  

Solution (Function equation for  )

We define   and  . The function   is differentiable on   according to the sum and chain rule for derivatives. Now,

 

According to the criterion for constancy,   is constant. To determine the exact value it is sufficient to use a concrete value. We choose   and get

 

Here, we used that   and hence  . This implies our claim.

Exercise: identity theorem Bearbeiten

Exercise (Logarithm representation of the arsinh)

Show that for all   there is

 

Proof (Logarithm representation of the arsinh)

The function   is differentiable on all of  , see also the by the examples for derivatives. Its derivative is

 

by the chain and sum rule, the function   is differentiable on all of   . There is:

 

We have   for all   so by the identity theorem,   with a constant  . But now, since  :

 

In addition

 

So   which implies the claim.

Characterization of sin and cos Bearbeiten

Exercise (Characterization of sin and cos)

Let   be two differentiable functions with

 

Prove that:

  1. There is   for all  
  2. There is exactly one pair of functions which meets the above conditions, namely   and  .

Hint: Consider the help function in the second sub-task  .

Solution (Characterization of sin and cos)

Solution sub-exercise 1:

We consider the auxiliary function

 

where   and   fulfil the conditions from above. Then   is differentiable by the sum and chain rule, and

 

According to the criterion for constancy, there is   for some  . We have

 

So   and we get the claim  .

Solution sub-exercise 2:

We consider the differential auxiliary function

 

For this function

 

According to the criterion for constancy,there is   with  . We have

 

So  . Now both   and   for all  . In order for the sum   to be be zero, both summands   and   must be zero. That means

 

So   and  , which is what we wanted to show.