Constant functions – Serlo

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"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.