Math for Non-Geeks/ Monotonic functions

oben Monotonicity criterion The monotonicity criterion is quite intuitive: if the derivative of a function (i.e. the slope) is positive, it goes up, if the derivative is negative, it goes down. Mathematically, if the derivative of a differentiable function is non-negative or (non-positive) on an interval , then is monotonically increasing (or decreasing) on . If is even strictly positive (or negative) , then is strictly monotonically increasing (or decreasing). In the first case, even inversion of the statement is true: If a differentiable function is monotonically increasing on , then and if the function is monotonically decreasing on , then . However, the inversion does not hold true in the strict case, monotone functions do not always have or . For instance, is strictly monotone, but . Monotonicity criterion for differentiable functions Let be continuous and differentiable on . Then, there is on monotonically increasing on on monotonically decreasing on on strictly monotonically increasing on on strictly monotonically decreasing on Proof The four directions "" follow from the mean value theorem. The two directions "" follow by differentiability of the function: monotonicity criterion for differentiable functions We first show the four directions "" and then the two "". 1. From on we get that in monotonically increasing on . Let for all and let with . We need to show . By assumption, is continuous on and differentiable on . By the mean-value theorem, there is a with By assumption, , and hence . Since we have in the enumerator . This is equivalent to , i.e. is monotonically increasing. 2. From on we get that in monotonically decreasing on . Let for all and let with . We need to show . By assumption, is continuous on and differentiable on . By the mean-value theorem, there is a with Now, , and hence . Since we have . This is equivalent to , i.e. is monotonically decreasing. 3. on implies that is strictly monotonically increasing on We prove this by contradiction: Let be not strictly monotonically increasing. That means, we have some with and . We need to find a with . Now, is continuous on and differentiable on . So by the mean value theorem, we can find a with Since , the enumerator of the quotient is non-positive, and because of the denominator is positive. Thus, the whole fraction is non-positive, and therefore . 4. on implies that is strictly monotonically increasing on Another proof by contradiction: Let be not strictly monotonically decreasing. That means, we have some with and . We need to find a with . Now, is continuous on and differentiable on . So by the mean value theorem, we can find a with Since , the enumerator of the quotient is non-positive, and because of the denominator is positive. Thus, the whole fraction is non-positive, and therefore . Now, the two directions "" follow: 1. being monotonically increasing on implies on Let with . By monotonicity, . Further, let with . Then we have for the difference quotient If , then . The enumerator and denominator of the difference quotient are thus non-negative, and so is the total quotient. Similarly, in the case of and enumerator and denominator are non-positive. Thus, the whole fraction is again non-negative. Now we form the differential quotient by taking the limit . This limit exists because is differentiable on . Furthermore, the inequality remains valid because of the monotonicity rule for limit values. Thus, we have Since and have been arbitrary, we get on all of . 2. being monotonically decreasing on implies on Let again with . By monotonicity, . Further, let with . Then we have for the difference quotient If , then and thus the total quotient is non-positive. An analogous statement holds in the case and . By forming the differential quotient we now obtain Since and have been arbitrary, we get on all of . Examples: monotonicity criterion Quadratic and cubic functions Monotonicity of quadratic and cubic functions thumb|Graphs of the functions and thumb|Graphs of the functions and For the quadratic power function there is So is strictly monotonically decreasing by the monotonicity criterion on and strictly monotonically increasing on . For the cubic power function there is So by the monotonicity criterion, is monotonically increasing on and strictly monotonically increasing on and . The cubic power function is even strictly monotonically increasing on all of . The fact that with is strictly monotonically increasing, although only and not , stems from its derivative being zero at only a single point (namely 0). In the end of this article, we will treat a criterion, which tells us when a function is strictly monotone, even if there is not everywhere . Question Why is strictly monotonically increasing on ? We must show: From with we get . For the cases and we have already shown this with the monotonicity criterion. So we only have to look at the case . Here there is with the arrangement axioms (missing): So is strictly monotonically increasing on all of . In the example we have seen that the statement " implies strict monotonicity" does not hold true! This means that from the fact that increases strictly monotone, we can in general not conclude that . In the example of the function one can also see that the statement " implies strictly monotone falling" does not hold true in general. Exponential and logarithm function Monotonicity of the exponential and logarithm function For the exponential function there is for all : Therefore, according to the monotonicity criterion, is strictly monotonically increasing on all of . For the (natural) logarithm function there is for all : So is strictly monotonically increasing on (not including the 0). Question What is the monotonicity behaviour of the logarithm function extended to , i.e.? There is Above we have shown that for . So is strictly monotonically increasing on , as well . For on the other hand there is . So is strictly monotonically decreasing on . Trigonometric functions Monotonicity of the sine function For the sine function there is So for all , the is strictly monotonically increasing on the intervals and strictly monotonically decreasing on the intervals . Question Where does the cosine function show monotone behaviour? Here, . So for all , the is strictly monotonically increasing on the intervals and strictly monotonically decreasing on the intervals . Monotonicity of the tangent function For the tangent function there is for all : Hence, for all , the is strictly monotonically increasing on the intervals . Question Where does the cotangent function show monotone behaviour? For all , there is So for all , the is strictly monotonically decreasing on the intervals . Exercise Monotonicity intervals and existence of a zero Monotonicity intervals and existence of a zero Where is the following polynomial function monotone? Prove that has exactly one zero. thumb|Graph of the function Monotonicity intervals The function is differentiable on all of , with So According to the monotonicity criterion, is strictly monotonically increasing on and on . Further, According to the monotonicity criterion, is strictly monotonically decreasing on . has exactly one zero For , we have the following table of values: Based on the monotonicity properties and the continuity of that we have previously investigated, we can read off that: On is strictly monotonically increasing. Because of there is for all . On is then strictly monotonically decreasing. So there is also for all . Subsequently increases on again strictly monotonically. Because of and , there must be an with by the mean value theorem. Because of the strict monotonicity of on , there cannot be any further zeros. Necessary and sufficient criterion for strict monotonicity Necessary and sufficient criterion for strict monotonicity Prove that: a continuous function which is differentiable to is strictly monotonically increasing exactly when there is for all The zero set of contains no open interval. As an application: Show that the function is strictly monotonically increasing on all of . From the monotonicity criterion we already know that is monotonically increasing exactly when . So we only have to show that is strictly monotonically increasing exactly when the second condition is additionally fulfilled. strictly monotonically increasing the set of zeros of does not contain an open interval. We perform a proof by contradiction. In other words, we show: If the set of zeros of contains an open interval, is not strictly monotonically increasing. Assume there is with for all . Then, by the mean value theorem there is a with So . If now , then since is monotonically increasing, there is So there is for all . Hence, is not strictly monotonically increasing. the set of zeros of does not contain an open interval strictly monotonically increasing We perform a proof by contradiction. In other words, we show: if is monotonically, but not strictly monotonically increasing, then the zero set of contains an open interval. Assume there is with with . Because of the monotonicity of there is So for all . That means is constant on . Hence, there is for all : so the set of zeros of does contain an open interval, and we get a contradiction. Exercise is strictly monotonically increasing is differentiable for all where as for all . Hence is monotonically increasing. Further there is So the set of zeros of contains only isolated points, and thus no open interval. Therefore is strictly monotonically increasing on . unten