Series Expansion of Bessel Function of the First Kind

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Theorem

Let $n \in \Z_{\ge 0}$ be a (strictly) non-negative integer.

Let $\map {J_n} x$ denote the Bessel function of the first kind of order $n$.


Then:

\(\ds \map {J_n} x\) \(=\) \(\ds \dfrac {x^n} {2^n \, \map \Gamma {n + 1} } \paren {1 - \dfrac {x^2} {2 \paren {2 n + 2} } + \dfrac {x^4} {2 \times 4 \paren {2 n + 2} \paren {2 n + 4} } - \cdots}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {k! \, \map \Gamma {n + k + 1} } \paren {\dfrac x 2}^{n + 2 k}\)


Negative Index

\(\ds \map {J_{-n} } x\) \(=\) \(\ds \dfrac {x^{-n} } {2^{-n} \, \map \Gamma {1 - n} } \paren {1 - \dfrac {x^2} {2 \paren {2 - 2 n} } + \dfrac {x^4} {2 \times 4 \paren {2 - 2 n} \paren {4 - 2 n} } - \cdots}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {k! \, \map \Gamma {k + 1 - n} } \paren {\dfrac x 2}^{2 k - n}\)


Proof

We employ Frobenius's method to find the solutions to the Bessel's Equation:

$x^2 \dfrac {\d^2 y} {\d x^2} + x \dfrac {\d y} {\d x} + \paren {x^2 - n^2} y = 0$

for $n \ge 0$, in the form:

$\ds \map y x = \sum_{k \mathop = 0}^\infty A_k x^{k + r}$

defined on $x > 0$, for some constants $r, A_i$, with $A_0 \ne 0$, which are to be determined.


Differentiating the expression with respect to $x$:

\(\ds \map {y'} x\) \(=\) \(\ds \sum_{k \mathop = 0}^\infty A_k \paren {k + r} x^{k + r - 1}\)
\(\ds \map {y' '} x\) \(=\) \(\ds \sum_{k \mathop = 0}^\infty A_k \paren {k + r} \paren {k + r - 1} x^{k + r - 2}\)


Substituting $y, y', y' '$ into Bessel's Equation:

\(\ds 0\) \(=\) \(\ds x^2 \dfrac {\d^2 y} {\d x^2} + x \dfrac {\d y} {\d x} + \paren {x^2 - n^2} y\)
\(\ds \) \(=\) \(\ds x^2 \sum_{k \mathop = 0}^\infty A_k \paren {k + r} \paren {k + r - 1} x^{k + r - 2} + x \sum_{k \mathop = 0}^\infty A_k \paren {k + r} x^{k + r - 1} + \paren {x^2 - n^2} \sum_{k \mathop = 0}^\infty A_k x^{k + r}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty A_k \paren {k + r} \paren {k + r - 1} x^{k + r} + \sum_{k \mathop = 0}^\infty A_k \paren {k + r} x^{k + r} + \sum_{k \mathop = 0}^\infty A_k x^{k + r + 2} - n^2 \sum_{k \mathop = 0}^\infty A_k x^{k + r}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty A_k \paren {\paren {k + r} \paren {k + r - 1} + \paren {k + r} - n^2} x^{k + r} + \sum_{k \mathop = 0}^\infty A_k x^{k + r + 2}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty A_k \paren {\paren {k + r}^2 - n^2} x^{k + r} + \sum_{k \mathop = 0}^\infty A_k x^{k + r + 2}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty A_k \paren {\paren {k + r}^2 - n^2} x^{k + r} + \sum_{k \mathop = 2}^\infty A_{k - 2} x^{k + r}\) Translation of Index Variable of Summation
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty A_k \paren {\paren {k + r}^2 - n^2} x^k + \sum_{k \mathop = 2}^\infty A_{k - 2} x^k\) $x^r \neq 0$

Comparing the constant term on both sides:

\(\ds 0\) \(=\) \(\ds A_0 \paren {r^2 - n^2}\)
\(\ds \leadsto \ \ \) \(\ds r\) \(=\) \(\ds \pm n\)

Take $r = n$.

Comparing the rest of the coefficients:

\(\ds 0\) \(=\) \(\ds A_1 \paren {\paren{n + 1}^2 - n^2}\)
\(\ds 0\) \(=\) \(\ds A_1 \paren {2 n + 1}\)
\(\ds \leadsto \ \ \) \(\ds A_1\) \(=\) \(\ds 0\)
\(\ds 0\) \(=\) \(\ds A_k \paren {\paren{n + k}^2 - n^2} + A_{k - 2}\) for $k \ge 2$
\(\ds 0\) \(=\) \(\ds A_k k \paren {2 n + k} + A_{k - 2}\)
\(\ds \leadsto \ \ \) \(\ds A_k\) \(=\) \(\ds - \dfrac 1 {k \paren {2 n + k} } A_{k - 2}\)

From the recurrence relation above, we see that $A_k = 0$ for odd $k$, and:

\(\ds A_{2 k}\) \(=\) \(\ds - \dfrac 1 {2 k \paren {2 n + 2 k} } A_{2 k - 2}\)
\(\ds \) \(=\) \(\ds \dfrac {-1} {2^2 k \paren {n + k} } A_{2 k - 2}\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {-1}^2} {2^4 k \paren {k - 1} \paren {n + k} \paren {n + k - 1} } A_{2 k - 4}\)
\(\ds \) \(=\) \(\ds \cdots\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {-1}^k} {2^{2 k} k \paren {k - 1} \cdots \paren 1 \paren {n + k} \paren {n + k - 1} \cdots \paren {n + 1} } A_0\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {-1}^k} {2^{2 k} k! \dfrac {\map \Gamma {n + k + 1} } {\map \Gamma {n + 1} } } A_0\)
\(\ds \) \(=\) \(\ds \dfrac {\paren {-1}^k} {2^{n + 2 k} k! \, \map \Gamma {n + k + 1} }\) By choice of $A_0 = \dfrac {2^{- n} } {\map \Gamma {n + 1} }$


Substituting this result to our original equation:

\(\ds \map y x\) \(=\) \(\ds \sum_{k \mathop = 0}^\infty A_k x^{k + n}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty A_{2 k} x^{2 k + n}\) since all even terms vanish
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {2^{n + 2 k} k! \, \map \Gamma {n + k + 1} } x^{2 k + n}\)
\(\ds \) \(=\) \(\ds \sum_{k \mathop = 0}^\infty \dfrac {\paren {-1}^k} {k! \, \map \Gamma {n + k + 1} } \paren {\dfrac x 2}^{2 k + n}\)

$\blacksquare$


Also see


Sources

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