Derivative Function
In Mathematics, the Darboux theorem is a theorem named after Jean Gaston Darboux. It states that every function that is derived from the differentiation of another function has intermediate value property: the image of an interval will also be an interval.
When the function $f$ is continuous differentiable $(f$ in $C^1[(x, y])$, this is the result of the intermediate value theorem. But even when the function $f$ is not continuous, Darboux theorem lays down every limitation on what it can be.
Darboux Theorem Statement
If $f$ is differentiable on $[a,b]$ and if is a number between $f^\prime (a)$ and $f^\prime(b)$, then there is minimum one point $c \in (a,b)$ such that $f^\prime(c)$ is .
This states that the converse of the intermediate theorem is not true because the derivative function of a differentiation function is not certainly continuous. For example,
$ f(t) = \left\{\begin{matrix} t^2 sin\left(\dfrac{1}{t}\right) & \text{for } t \neq 0, \\ 0 & \text{for } t = 0 \end{matrix}\right.$
is continuous and has discontinuous derivatives which satisfy the intermediate value property given by the Darboux theorem.
Darboux Theorem Proof 1
Suppose that $f^\prime(a) < \lambda f^\prime(b)$. Let $F : [a,b] \to$ be defined by $f(x) = f(x)- \lambda x$. Then, $f$ is differentiable on $[a,b]$, because so is the function $f$ by hypothesis. We find $F^\prime(a)= f^\prime(a) - \lambda < 0$ and $F^\prime(b) = f^\prime(b) - \lambda > 0$. Note that $F^\prime(a) < 0$ means $F^\prime(t_1) < F (a)$ for some $t_1 \in (a, b)$. Also for, $F^\prime(b) > 0 $, we can find the value of $t_2(a, b)$ such that $F^\prime(t_2) < F(b)$. Hence, neither $a$ nor $b$ are the points where $F$ can attain absolute minimum. As $F$ is continuous on $[a,b]$, it must attains its relative minimum at some point $c \in (a,b)$ by extreme value theorem for some continuous functions. This implies $F^\prime(c)$ by Fermat's theorem and hence $f^\prime(c) = \lambda$ as desired. This proof follows the similar argument of $f^\prime(b) < \lambda < f^\prime(a)$.
Darboux Theorem Proof 2
Let us assume that $f^\prime(a) < \lambda < f^\prime(b)$ and consider a function $F: [a,b] \to R$ defined by $F(x) = f^\prime(x)- \lambda x$. Then $F^\prime(x) = f^\prime(x)- \lambda$ and determine that $F^\prime(a) = f^\prime(a)- \lambda < 0$ and $F^\prime(b) = f^\prime(b)- \lambda > 0$.This implies that function $F$ is monotonic on $[a,b]$ in the way there exist $x,y,z \in [a,b]$ such that $x < y < z$ satisfying only one the following condition $(1)$ or $(2)$.
$F(x) < F(y)$ and $F(y) >F(z)$. The theorem will be proved for this case.
$F(x) > F(y)$ and $F(y) < F(z)$. Proof in the case will follow a similar argument.
Suppose $(a )$ holds, then we can observe following three cases
$F(x) < F(z)$
$F(z) < F(x)$
$F(z) = F(x)$
Case 1: Suppose that $F(x) > F(z)$. Then $F(x) < F(z) < F(y)$. $F$ is differentiable on $(a,b)$ so is $f$ by hypothesis. Hence $F$ is also continuous on $(a,b)$. Hence, the intermediate value theorem applies to $F$ on $[x,y] \subseteq (a,b)$ and we get $d\subseteq (x,y) \subseteq (a,b)$ such that $F(d) = F(z)$. Note that $a < x < d< y < z < b$.
Now, we can use the Rolle’s theorem to $F$ on the closed interval $(d,z)$ and gets $c \in (d,z)\subseteq (x,z)\subseteq (a,b) $ such that $F^\prime(c) = 0$, which results $f^\prime (c) = \lambda$ as desired.
Case 2: Suppose that $F(z) < F(x)$. Then $F(z) < F(x) < F(y)$. This theorem further follows by argument similar to that of case 1.
Case 3: Suppose that $F(z) = F(x)$. Then, we can finally apply rolle’s theorem for $F$ directly on $[x,z]$ and obtain $c(x,z) \subseteq (a,b)$ such that $F^\prime(c) = 0$ leading to $f^\prime(c) = \lambda$ as desired.
We arrive at the same conclusion if $(2)$ holds by the same argument.
Darboux Problems and Solutions
Consider the function $f(x) = sin (x)$
1. What Values of the Derivatives of $f$ are assured by Darboux Theorem on the Interval $[1,3]$?
Solution:
As $f^\prime (x)= cos x , f^\prime (1)= 0.54030$ and $f^\prime (3)= - 0.98999$. Hence, the assured values are from $- 0.98999$ to $0.54030$.
2. Does Darbous Property Guarantee any Value on the Interval $[0, 2 \pi]$.
No, $f^\prime(x)= 1$ at both the ending points. Hence, there are no values between $1$ and $1$.
FAQs on Darboux's Theorem
1. What is the statement of Darboux's Theorem in calculus?
Darboux's Theorem states that if a function f is differentiable on a closed interval [a, b], then its derivative, f'(x), must take on every value between f'(a) and f'(b). In simpler terms, the derivative function has the Intermediate Value Property, meaning it cannot skip any values within its range on that interval, even if the derivative itself is not a continuous function.
2. How is Darboux's Theorem different from the Intermediate Value Theorem (IVT)?
The key difference lies in the function to which each theorem applies.
- The Intermediate Value Theorem (IVT) applies to functions that are continuous. It guarantees that if a function f is continuous on [a, b], it will achieve every value between f(a) and f(b).
- Darboux's Theorem applies specifically to derivative functions. It states that the derivative f' possesses the intermediate value property, regardless of whether f' is continuous or not.
Essentially, IVT is about continuous functions, while Darboux's Theorem is about the special properties of functions that are derivatives.
3. What is the core idea behind the proof of Darboux's Theorem?
The proof of Darboux's Theorem hinges on creating a helper function and applying two fundamental calculus theorems: the Extreme Value Theorem and Fermat's Theorem on local extrema. The general steps are:
- An auxiliary function, like g(x) = f(x) - kx, is defined, where 'k' is the intermediate value we want the derivative to equal.
- The derivative of this helper function, g'(x), is shown to be positive at one endpoint and negative at the other.
- Since f(x) is differentiable, g(x) is continuous. The Extreme Value Theorem ensures that g(x) must have a minimum (or maximum) value within the interval.
- This minimum cannot occur at the endpoints, so it must be at an interior point 'c'.
- By Fermat's Theorem, the derivative at this interior minimum must be zero. Therefore, g'(c) = 0, which implies f'(c) - k = 0, proving that f'(c) = k.
4. Can you provide a simple example of Darboux's Theorem in action?
Certainly. Let's consider the function f(x) = x³ on the interval [0, 2]. The derivative is f'(x) = 3x². At the endpoints of the interval:
- At x = 0, the derivative is f'(0) = 3(0)² = 0.
- At x = 2, the derivative is f'(2) = 3(2)² = 12.
Darboux's Theorem guarantees that for any value 'k' between 0 and 12, there is a point 'c' in (0, 2) such that f'(c) = k. For example, if we choose k = 6, we solve 3c² = 6, which gives c = √2. Since √2 is approximately 1.414, it lies within the interval (0, 2), confirming the theorem.
5. Does Darboux's Theorem mean the derivative of a function must be continuous?
No, and this is a crucial point of the theorem. A function's derivative does not have to be continuous for Darboux's Theorem to hold. The theorem's power is that it establishes that all derivative functions have the intermediate value property, a property often associated with continuous functions. There are classic examples of functions that are differentiable everywhere but whose derivatives have discontinuities; even in these cases, the derivative never 'skips' a value.
6. What is the main significance or application of Darboux's Theorem?
The primary significance of Darboux's Theorem is theoretical. It acts as a classification tool, helping to define which types of functions can be considered the derivative of another function. Its main application is in real analysis to understand the behaviour of derivatives. For instance, it proves that a function with a simple 'jump' discontinuity (where it jumps from one value to another without covering the points in between) cannot be the derivative of any function on that interval.
7. Is Darboux's Theorem related to 'Darboux Sums' in integration?
No, they are distinct concepts named after the same French mathematician, Jean-Gaston Darboux.
- Darboux's Theorem is about the intermediate value property of derivatives, a topic in differential calculus.
- Darboux Sums (upper and lower sums) are used to formally define the Riemann integral, which is a topic in integral calculus.
While both are fundamental concepts in calculus, they apply to different branches: one for derivatives and the other for integration.
