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Chapter 1: Problem 32
The function \(J_{1}(x)=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{k !(k+1) ! 2^{2k+1}} \times\) \(x^{2 k+1}\) is called the Bessel function of order 1. Verifythat \(J_{1}(x)\) is a solution of Bessel's equation of order \(1, x^{2}y^{\prime \prime}+x y+\) \(\left(x^{2}-1\right) y=0\).
Short Answer
Expert verified
By substituting the expressions for \(J_{1}(x)\), \(J_{1}'(x)\), and \(J_{1}''(x)\) into the Bessel equation and simplifying, one can verify that they satisfy the equation.
Step by step solution
01
Define Bessel's equation of order 1
The Bessel equation of order 1 is given by: \[ x^{2} y'' + x y' + (x^{2} - 1) y = 0 \]
02
Differentiate the Bessel function
Differentiate the given Bessel function, \(J_{1}(x)\), with respect to \(x\) to find the first and second derivatives.First derivative:\[ J_{1}'(x) = \frac{d}{dx} \bigg( \rightarrow \bigg) \ = \frac{d}{dx} \bigg( \rightarrow \bigg)= \bigg( \rightarrow \bigg) \]Second derivative:\[ J_{1}''(x) = \frac{d^{2}}{dx^{2}} \bigg( \rightarrow \bigg)= \frac{d^{2}}{dx^{2}} \bigg( \rightarrow \bigg)= \bigg( \rightarrow \bigg) \]
03
Substitute the derivatives into Bessel's equation
Substitute the expressions for \(J_{1}(x)\), \(J_{1}'(x)\), and \(J_{1}''(x)\) into the Bessel equation:\[ x^{2} J_{1}''(x) + x J_{1}'(x) + (x^{2} - 1) J_{1}(x) \]
04
Simplify and verify the equality
Simplify the substituted expression to verify if it equals zero. This can involve rearranging and simplifying terms as needed to show\[ x^{2} J_{1}''(x) + x J_{1}'(x) + (x^{2} - 1) J_{1}(x) = 0 \]
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Bessel's equation
Bessel's equation is an important differential equation frequently appearing in various fields like physics and engineering. It is a type of second-order linear differential equation. The general form of Bessel's equation of order \(n\) is:
\[ x^2 y'' + x y' + (x^2 - n^2) y = 0 \]
For this exercise, we are dealing with Bessel's equation of order 1, so the equation becomes:
\[ x^2 y'' + x y' + (x^2 - 1) y = 0 \]
To solve this, we assume a solution and then check if it satisfies the equation by differentiating and substituting back in.
differential equations
Differential equations involve equations that feature derivatives of unknown functions. These are key tools for modeling various natural systems and processes. Here, we are dealing specifically with a second-order differential equation, which means the highest derivative is the second derivative. This requires calculating both the first and second derivatives of a given function. To verify the Bessel function solution, we must compute these derivatives from the original Bessel function and substitute them back into the equation to check if they satisfy it.
mathematical proofs
In mathematics, a proof is a logical argument demonstrating that a specific statement is universally true. Each step logically follows from the last to verify the overall truth. In this context, verifying the Bessel function involves these steps:
- Define Bessel's equation
- Compute the first and second derivatives of the Bessel function
- Substitute these derivatives back into the equation
- Simplify the resulting expression to show it equals zero
This step-by-step process helps in proving that the given function is indeed a solution to the differential equation.
higher-order derivatives
Higher-order derivatives refer to derivatives beyond the first derivative, such as the second, third, and beyond. For this problem, the main focus is on the first and second derivatives of the Bessel function.
The first derivative of the Bessel function \(J_1(x)\) can be found by differentiating the series term-by-term.
The second derivative is then the derivative of this first derivative. Once these derivatives are known, they are substituted back into Bessel's equation to verify that the equation holds true.
Working with higher-order derivatives often involves more complex calculus, and attention to detail is crucial to ensure that each derivative is computed accurately.
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