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112 LearnersLast updated on September 27, 2025
Derivative of Delta Function
The derivative of the delta function, often represented as the derivative of δ(x), serves as a tool to understand how a distribution changes in response to a slight shift in its input. Derivatives are crucial in many fields, including signal processing and physics. We will explore the derivative of the delta function in depth.
What is the Derivative of the Delta Function?
Understanding the derivative of the delta function involves analyzing how the delta function, δ(x), behaves under differentiation. The delta function itself is a distribution, not a traditional function, and is defined such that ∫δ(x)f(x)dx = f(0) for any test function f(x). Its derivative, denoted as δ'(x), also acts on test functions and is defined by ∫δ'(x)f(x)dx = -f'(0).
Here are some key concepts:
Delta Function: δ(x) behaves as an identity operator under integration.
Test Function: A smooth function used to probe the behavior of distributions.
Differentiation of Distributions: Techniques to extend the concept of derivatives to distributions.
Derivative of the Delta Function Formula
The derivative of the delta function can be expressed in terms of its action on a test function f(x). The formula is: ∫δ'(x)f(x)dx = -f'(0) This formula is valid for any smooth test function f(x).
It highlights that the derivative of the delta function captures the rate of change at the point of evaluation, effectively negating the derivative of the test function at zero.
Proofs of the Derivative of the Delta Function
To derive the properties of the derivative of the delta function, we use the framework of distributions. The derivative is defined through its action on test functions.
There are several methods to understand this, such as:
- By Integration by Parts
- Using the Sifting Property of the Delta Function
- Using the Definition of Distributions
We will demonstrate that δ'(x) acts on test functions by negating their derivatives at zero using these methods:
By Integration by Parts
The definition of δ'(x) can be shown using integration by parts, where the boundary terms vanish: ∫δ'(x)f(x)dx = -∫δ(x)f'(x)dx = -f'(0) Thus, the derivative of δ(x) acts by negating the test function's derivative at zero.
Using the Sifting Property
The sifting property of δ(x) states that ∫δ(x-a)f(x)dx = f(a). Differentiating both sides with respect to a, we obtain: ∫δ'(x-a)f(x)dx = -f'(a) For a = 0, this simplifies to the previous result.
Using the Definition of Distributions
In distribution theory, δ'(x) is defined such that, for any test function f(x): ∫δ'(x)f(x)dx = -∫δ(x)f'(x)dx Using the properties of δ(x), this holds true, affirming its definition.
Higher-Order Derivatives of the Delta Function
Higher-order derivatives of the delta function involve repeated application of the derivative operation. For distributions, these can be thought of as capturing higher rates of change.
For example, the second derivative δ''(x) acts on test functions as: ∫δ''(x)f(x)dx = f''(0) This concept continues for higher derivatives, providing insight into the behavior of functions or distributions in response to input changes.
Special Cases:
When the delta function is at x = a, δ(x-a) shifts the evaluation point of the test function to a. The derivative of δ(x-a), δ'(x-a), negates the derivative of the test function at a.
At x = 0, δ'(x) specifically targets the rate of change at zero.
Common Mistakes and How to Avoid Them with the Derivative of the Delta Function
When working with the derivative of the delta function, students often make mistakes. These can be avoided by understanding the formalism of distributions. Here are some common mistakes and solutions:
Mistake 1
Misinterpreting the Delta Function
Students often treat δ(x) as a standard function, which can lead to incorrect calculations.
Remember that δ(x) is a distribution, not a traditional function, and must be handled in the context of integration with test functions.
Mistake 2
Ignoring the Role of Test Functions
The derivative of δ(x) must be considered with test functions. Ignoring this leads to misunderstandings.
Always apply δ'(x) to smooth test functions and evaluate the resulting expression.
Mistake 3
Confusing the Sign in Differentiation
When differentiating δ(x), students sometimes forget the negative sign.
The correct operation is ∫δ'(x)f(x)dx = -f'(0). Make sure to include the negative sign when applying this definition.
Mistake 4
Overlooking Higher-Order Derivatives
Students might neglect the possibility of higher-order derivatives of δ(x). These derivatives follow the same logic and apply to test functions.
Ensure you understand how δ''(x) and beyond operate.
Mistake 5
Forgetting the Zero at Boundaries
When integrating by parts to find δ'(x), students may overlook that boundary terms vanish. This can lead to incorrect results.
Always verify boundary terms are zero in integration by parts.
Examples Using the Derivative of the Delta Function
Problem 1
Calculate the effect of δ'(x) on the test function f(x) = e^(-x^2).
Consider f(x) = e^(-x^2). The effect of δ'(x) on this test function is given by: ∫δ'(x)e^(-x^2)dx = -f'(0) Differentiating f(x), f'(x) = -2xe^(-x^2) At x = 0, f'(0) = 0 Thus, the effect of δ'(x) on e^(-x^2) is 0.
Explanation
We determine the effect of δ'(x) by evaluating the derivative of the test function at zero.
Since the derivative is zero at that point, δ'(x) has no effect on this test function.
Problem 2
A system is modeled by the impulse response δ'(t-1). How does this affect the input signal g(t) = sin(t) at t = 1?
The derivative δ'(t-1) acts on the input signal g(t) = sin(t) as follows: ∫δ'(t-1)sin(t)dt = -g'(1) The derivative of g(t) is g'(t) = cos(t) At t = 1, g'(1) = cos(1) Thus, δ'(t-1) affects g(t) by -cos(1) at t = 1.
Explanation
The derivative δ'(t-1) shifts the evaluation point of the derivative of the input signal to t = 1 and negates it.
The result is -cos(1), which shows how the impulse response affects g(t).
Problem 3
Derive the second derivative effect of δ''(x) on the function h(x) = ln(1+x^2).
First, find the first derivative of h(x): h'(x) = 2x/(1+x^2) Second, find the second derivative: h''(x) = (2(1+x^2) - 4x^2)/(1+x^2)^2 At x = 0, h''(0) = 2 So the effect of δ''(x) on h(x) is 2.
Explanation
To find the effect of δ''(x), we calculate the second derivative of the test function at zero.
This gives the effect of δ''(x) as 2, showing how the second derivative influences the function.
Problem 4
Prove: ∫δ'(x-a)g(x)dx = -g'(a).
Using the sifting property of the delta function, ∫δ(x-a)g'(x)dx = g'(a). Differentiating under the integral sign, we have: ∫δ'(x-a)g(x)dx = -g'(a) This confirms the definition of δ'(x-a) by negating the derivative of g(x) at a.
Explanation
In this proof, we apply the sifting property and show how the derivative of the delta function affects the test function, confirming the expected result by negating g'(a).
Problem 5
Solve: ∫δ'(x)cos(x)dx.
The effect of δ'(x) on cos(x) is given by: ∫δ'(x)cos(x)dx = -d/dx[cos(x)] at x=0 The derivative of cos(x) is -sin(x) Thus, at x=0, the result is -(-sin(0)) = 0.
Explanation
We apply the definition of δ'(x) by finding the derivative of the test function cos(x) at zero.
Since sin(0) is zero, the effect of δ'(x) on cos(x) is zero.
FAQs on the Derivative of the Delta Function
1.Find the derivative of the delta function δ(x).
2.Can the delta function derivative be used in real-life applications?
3.Is it possible to differentiate the delta function at a specific point, like x = 1?
4.What rule is used to derive the action of δ'(x) on a test function?
5.Are δ(x) and its derivative δ'(x) treated as functions?
Important Glossaries for the Derivative of the Delta Function
- Delta Function: A distribution that acts as an identity operator under integration.
- Derivative of a Distribution: Extension of traditional derivatives to distributions.
- Test Function: A smooth function used to probe the properties of distributions.
- Integration by Parts: A technique used for deriving properties of distribution derivatives.
- Sifting Property: A unique property of the delta function that evaluates at a specific point.
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Jaskaran Singh Saluja
About the Author
Jaskaran Singh Saluja is a math wizard with nearly three years of experience as a math teacher. His expertise is in algebra, so he can make algebra classes interesting by turning tricky equations into simple puzzles.
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