Question

Prove that 


\frac{d}{dx}(j_{n}^{2}(x))=\frac{x}{2n}[j_{n+1}^{2}(x)-j_{n+1}^{2}(x)]

22 Mar 2023
Answer :
Word Count : 924

To prove the given identity, we start with the definition of the Bessel function of order n, j_n(x), which is given by:

j_n(x) = sum_{k=0}^{infty}frac{ (-1)^k}{(k!(k+n)!)^{2k+n}}

Differentiating this with respect to x, we get:

j'n(x) = \sum (k=0)^\infty (-1)^k/(k!(k+n)!)(x/2)^{2k+n-1}

Multiplying both sides by x and rearranging, we get:

xj'n(x) = \sum (k=0)^\infty (-1)^k/(k!(k+n)!)x^{2k+n}

Now, we can use the recurrence relation for the Bessel function to express j_{n+1}(x) in terms of j_n(x):

j_{n+1}(x) = (2n/x)j_n(x) - j_{n-1}(x)

Squaring both sides and rearranging, we get:

j_{n+1}^2(x) - j_n(x)j_{n+2}(x) = (j_n(x) - j_{n-1}(x))(j_{n+1}(x) - j_{n}(x))/xMultiplying both sides by x/2n and summing over n, we get:

 

Expanding the first term on the left-hand side, we get:

j_1^2(x)/2 + \sum _(n=1)^\infty [j_{n+1}^2(x) - j_n(x)j_{n+2}(x)]/2n = \sum _(n=0)^\infty [j_n(x) - j_{n-1}(x)][j_{n+1}(x) - j_n(x)]/(2n+2)(2n)Using the identity for the product of differences of consecutive terms of a series, we can simplify the right-hand side as:

\sum _(n=0)^\infty [j_{n+1} - j_n(x)]^2/(2n+2)

Differentiating both sides with respect to x, we get:

xj_1^2(x)/2 + x\sum _(n=1)^\infty [j_{n+1}^2(x) - j_n(x)j_{n+2}(x)]/2n = x\sum _(n=0)^\infty [j_{n+1} - j_n(x)][j_{n+1} - j_n(x)]/(2n+2)

Simplifying the terms on the left-hand side using the recurrence relation for the Bessel function, we get:

xj_1^2(x)/2 + x\sum _(n=1)^\infty [j_{n}(x) - j_{n-1}(x)]^2/(2n) = x\sum _(n=0)^\infty [j_{n+1} - j_n(x)][j_{n+1} - j_n(x)]/(2n+2) ______ _____ ____ ______ ____ ____ _______ ____.
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