How to Derive Bessel Function Recurrences from the Generating Function

The generating function for the Bessel functions of the first kind packages all integer orders into one expression. Differentiating it in two different ways produces the standard recurrence relations: one connects neighboring orders, and another connects neighboring orders to the derivative with respect to the argument.

Here, $n$ is any integer, $x$ is the function argument, and $J_n(x)$ denotes the Bessel function of the first kind. The quotient forms below assume $x\ne 0$; equivalent forms without division are also given.

Start with the generating function

The generating function is

$$\exp\!\left(\frac{x}{2}\left(t-t^{-1}\right)\right) =\sum_{n=-\infty}^{\infty}J_n(x)t^n,$$

where $t\ne 0$. Denote the left-hand side by $G(t,x)$. The series is a Laurent series in $t$, so its coefficients can be compared after differentiating with respect to either variable.

The key idea is to differentiate the same function in two ways. Differentiation with respect to $t$ brings down the order index $n$ on the series side. Differentiation with respect to $x$ brings down a factor involving $t-t^{-1}$, which shifts the Bessel order.

Differentiate with respect to t

From the exponential expression,

$$t\frac{\partial G}{\partial t} =\frac{x}{2}(t+t^{-1})G.$$

On the series side,

$$t\frac{\partial G}{\partial t} =\sum_{n=-\infty}^{\infty}nJ_n(x)t^n.$$

For the right-hand side, multiply the series for $G$ by $t+t^{-1}$. The coefficient of $t^n$ is $J_{n-1}(x)+J_{n+1}(x)$. Equating coefficients gives

$$nJ_n(x)=\frac{x}{2}\bigl(J_{n-1}(x)+J_{n+1}(x)\bigr),$$

or, for $x\ne 0$,

$$J_{n-1}(x)+J_{n+1}(x)=\frac{2n}{x}J_n(x).$$

This is the order recurrence. It relates three consecutive orders at the same argument.

Differentiate with respect to x

Differentiating the exponential expression with respect to $x$ gives

$$\frac{\partial G}{\partial x}=\frac{1}{2}(t-t^{-1})G.$$

The derivative of the series is

$$\frac{\partial G}{\partial x} =\sum_{n=-\infty}^{\infty}J_n'(x)t^n.$$

The coefficient of $t^n$ in $(t-t^{-1})G$ is $J_{n-1}(x)-J_{n+1}(x)$. Thus

$$J_n'(x)=\frac{1}{2}\bigl(J_{n-1}(x)-J_{n+1}(x)\bigr).$$

This is the derivative recurrence. It is valid without division by $x$.

Combine the two identities

Adding and subtracting the order and derivative recurrences isolates each neighboring order. The results are

$$J_n'(x)=J_{n-1}(x)-\frac{n}{x}J_n(x)$$

and

$$J_n'(x)=\frac{n}{x}J_n(x)-J_{n+1}(x),$$

for $x\ne 0$. Equivalently, these can be written without quotients as

$$xJ_n'(x)=xJ_{n-1}(x)-nJ_n(x)$$

and

$$xJ_n'(x)=nJ_n(x)-xJ_{n+1}(x).$$

The same formulas can be rearranged into useful product-derivative identities, valid for $x\ne 0$ whenever a displayed power has a negative exponent:

$$\frac{d}{dx}\bigl(x^nJ_n(x)\bigr)=x^nJ_{n-1}(x), \qquad \frac{d}{dx}\bigl(x^{-n}J_n(x)\bigr)=-x^{-n}J_{n+1}(x).$$

When a displayed power has a negative exponent, the product has a removable extension at $x=0$, and the identity extends there by continuity. The second formula follows directly from the corresponding derivative recurrence.

Worked example: the derivative of J₂(3)

Set $n=2$ and $x=3$. The two derivative formulas give

$$J_2'(3)=J_1(3)-\frac{2}{3}J_2(3)$$

and

$$J_2'(3)=\frac{2}{3}J_2(3)-J_3(3).$$

Using $J_1(3)\approx0.33905896$, $J_2(3)\approx0.48609126$, and $J_3(3)\approx0.30906272$, the first expression is approximately

$$0.33905896-\frac{2}{3}(0.48609126)\approx0.01499812.$$

The second gives $\frac{2}{3}(0.48609126)-0.30906272\approx0.01499812$ as well. Agreement is a useful check that the signs and order indices have been handled correctly.

Common mistakes

• Losing an index shift: The coefficient of $t^n$ in $tG$ is $J_{n-1}(x)$, while the coefficient in $t^{-1}G$ is $J_{n+1}(x)$.
• Changing the sign in the x-derivative: The generating function contains $t-t^{-1}$, so the coefficient difference is $J_{n-1}-J_{n+1}$, not the reverse.
• Dividing by zero: The quotient recurrences require $x\ne0$. Use their multiplied forms at $x=0$.
• Mixing up Bessel conventions: These derivations use $J_n$, the Bessel function of the first kind. For modified Bessel functions, the generating function has $t+t^{-1}$ in the exponent, and the corresponding sign patterns differ.

Modified Bessel function variant

If the intended functions are the modified Bessel functions $I_n$, start instead from

$$\exp\!\left(\frac{x}{2}(t+t^{-1})\right) =\sum_{n=-\infty}^{\infty}I_n(x)t^n.$$

The same coefficient-matching method gives $I_n'(x)=\tfrac12(I_{n-1}(x)+I_{n+1}(x))$ and $I_{n-1}(x)-I_{n+1}(x)=\tfrac{2n}{x}I_n(x)$. The plus sign in the generating function changes which neighboring-order combination appears in each identity.