Compact form of the Lagrange inversion formula

Posted on by John

The Lagrange inversion formula can be used to find the power series for the inverse of a function. I wrote about a different approach this problem a couple years ago, that time using Bell polynomials. This time I’ll give a formula that is more direct and easier to remember.

Suppose we have a function A(x) and can compute its derivatives. We want to find a power series for B(x) where B(A(x)) = x. We assume A(0) = 0 and A‘(0) ≠ 0.

The kth coefficient in the power series for B(x) is given by

b_k = \frac{1}{k} [k-1] \left(\frac{x}{A(x)}\right)^k

where [k − 1] in front of a function means to take the (k − 1)st coefficient in its power series.

Let’s apply this to get the first few terms of the series for tangent. Since inverse tangent has a simpler power series than tangent, we’ll set A(x) = arctan(x) so that B(x) is tangent, i.e. the inverse of the inverse tangent.

Of course we could just find the power series for tangent directly, and this is just a demonstration. Power series inversion is more useful when you can’t simply find the series for the inverse function directly.

We will compute the coefficients b1, b3, and b5 to get a 5th order series for tangent. Why don’t we need to compute b2 and b4? Because tangent is an odd function, we know that its power series coefficients with even indices are zero.

(You can see that this happens in general by looking at the equation above. If A(x) is an odd function, then x / A(x) is even, and so are its kth powers. The coefficients with odd index in the power series for an even function are zero.)

\begin{align*} \left(\frac{x}{\arctan(x)}\right)^1 &= 1 + \frac{1}{3}x^2 - \frac{4}{45}x^4 + \cdots \\ \left(\frac{x}{\arctan(x)}\right)^3 &= 1 + \phantom{\frac{1}{3}}x^2 + \frac{1}{15}x^4 + \cdots \\ \left(\frac{x}{\arctan(x)}\right)^5 &= 1 + \frac{5}{3}x^2 + \frac{2}{3}x^4 + \cdots \end{align*}

So b1 equals the 0th coefficient in the power series for x/arctan(x), which is 1.

Next b3 equals 1/3 times the 2nd coefficient in the power series for (x/arctan(x))3, and so b3 = 1/3.

Finally, b5 equals 1/5 times the 4th coefficient in the power series for (x/arctan(x))5, and so b5 = 2/15.

This tells us the power series for tangent is given by

\tan(x) = x + \frac{1}{3} x^3 + \frac{2}{15}x^5 + \cdots

and we could check by computing the power series directly that these terms are correct.

Incidentally, we can extend the formula at the top of this post to include powers of the inverse function. That is, the coefficients in the power series for B(x)n are given by

[k] ((B(x)^n) = \frac{n}{k} [k-n] \left(\frac{x}{A(x)}\right)^k [k] ((B(x)^n) = \frac{n}{k} [k-n] \left(\frac{x}{A(x)}\right)^k

which reduces to the formula up top when n = 1.

3 thoughts on “Compact form of the Lagrange inversion formula”

  1. This is neat to see.

    I believe there are two typos in the series expansions for (x/arctan(x))^n. The 4th order term of x/arctan(x) should be -4/45 x^4 and the 4th order term of (x/arctan(x))^5 should be 2/3 x^4.

  2. GlennF

    Happened to be looking at this old post.. I can confirm the typos that Jason Merrill pointed out back in August. There is an additional typo where
    it’s stated that b_3 = 2/15, but you meant to say b_5 = 2/15.

    The 4th order term for x/arctan(x) doesn’t effect the overall discussion, but the 4th order term for (x/arctan(x))^5 does since it is used to compute b_5. The value of 2/15 is correct, but the value of the coefficient of (x/arctan(x))^5 used to compute it should be 2/3.

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