Math for Non-Geeks/ Constant functions

oben "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 Let be an interval and a differentiable function with for all . Then is constant. 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 Identitätssatz 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. Identity theorem Let be two differentiable functions with . Then, there is for all . Where is a constant number. We define the auxiliary function This function is differentiable since and are differentiable and Hence, there must be for all with a constant number . Or equivalently, Application: characterization of the exponential function Characterization of the exponential function Let be differentiable. Further let and for all we have Then, there is for all with a constant . If and in addition , then precisely . 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 . 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 Interval assumption for constant functions The condition that the function is defined on an interval is necessary for the criterion for constancy! This is illustrated by the following task: 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 thumb|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 Using the criterion for constancy, identities of functions can also be proven very well: Trigonometric Pythagorean theorem Show that for all there is Here, and . 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 Function equation for Show: 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 Logarithm representation of the arsinh Show that for all there is 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 Characterization of sin and cos Let be two differentiable functions with Prove that: There is for all There is exactly one pair of functions which meets the above conditions, namely and . Hint: Consider the help function in the second sub-task . 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 . 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 zero, both summands and must be zero. That means So and , which is what we wanted to show. unten