Given some \(f(x)\), what’s the taylor series of \(f(x)^p\) for some real \(p\) around \(x=0\)? One can always calculate derivatives by hand, but we’d present a more intuitive method.

For the sake of simplicity, we will assume that \(f(x)>0\) on a small ball around \(x=0\). The idea is that we express \(f(x)^p = f(0)(1+h(x))^p\) for some \(h(x)\) derived from the taylor series of \(f(x)\). Then we use the taylor series of \((1+x)^p\) around \(x=0\) to deduce the taylor series of \(f(x)^p\). Even though this sounds abstract, this might be what you are used to doing unconciously anyways.

Example: What’s the taylor series of \(\cos^2 x\)? We know that \(\cos x = 1 - \frac{x^2}{2} + O(x^4).\) From binomial expansion we also know \((1+x)^2 = 1+2x+x^2.\) So

\[\begin{align*} \cos^2 x &= \left(1-\tfrac{x^2}{2} + O(x^4)\right)^2 \\ &= \left(1-\tfrac{x^2}{2}\right)^2 + O(x^4) \\ &= 1 - x^2 + O(x^4). \\ \end{align*}\]

We were able to simplify \((1-\tfrac{x^2}{2} + O(x^4))^2\) into \((1-\tfrac{x^2}{2})^2 + O(x^4)\) because any cross term that involves \(O(x^4)\) is going to be \(O(x^4)\).

This is evidently quicker than finding the first and second derivative of \(\cos^2 x\) at \(x=0\).

Example: What’s the taylor series of \(\frac{1}{\cos x}\)? From geometric series, we know that \(\frac{1}{1-x} = 1 + x + x^2 + O(x^3)\). So

\[\begin{align*} 1/{\cos x} &= 1/(1 - (\tfrac{x^2}{2} + O(x^4))) \\ &= 1/(1-\tfrac{x^2}{2}) + O(x^4) \\ &= 1 + \tfrac{x^2}{2} + O(x^4). \end{align*}\]

We were able to simplify \(1/(1 - \tfrac{x^2}{2} + O(x^4))\) into \(1/(1-\tfrac{x^2}{2}) + O(x^4)\) for a similar reason to the previous example. Any cross term involving \(O(x^4)\) in the geometric series is going to be of \(O(x^4)\) and can be ignored.

We could’ve reasonbly guessed this expression from the previous example as \(\cos x \tfrac{1}{\cos x} = 1\) and \((1+\tfrac{x^2}{2})(1-\tfrac{x^2}{2}) = 1 + O(x^4)\).

Non-integer exponents

We will make use of the following. Let \(p\) be a real number. Then for \(x \in (-1, 1)\), then the taylor series of \((1+x)^p\)

\[(1+x)^p = 1 + px + \tfrac{p(p-1)}{2!}x^2 + \tfrac{p(p-1)(p-2)}{3!}x^3 + \dots\]

Intuitively we can deduce the above by repeatedly taking derivatives of \((1+x)^p\) and evaluating it at \(x=0\) (Similar to calculating expectations from probability generating functions).

If \(p\) is an integer, the above reduces to the binomial theorem. For example,

\[\begin{align*}(1+x)^3 &= 1 + 3x + \tfrac{3(3-1)}{2!} x^2 + \tfrac{3(3-1)(3-2)}{3!}x^3 + \dots \\ &= 1 + 3x + 3x^2 + x^3. \end{align*}\]

With this at hand, we can deduce even more taylor series.

Example: What’s the taylor series of \(\sqrt{\cos x}\)? We now know that \(\sqrt{1+x} = 1 + \tfrac{1}{2}x + O(x^2)\). So

\[\begin{align*} \sqrt{\cos x} &= \sqrt{1-\tfrac{x^2}{2} + O(x^4)} \\ &= \sqrt{1-\tfrac{x^2}{2}} + O(x^4) \\ &= 1 - \tfrac{x^2}{4} + O(x^4). \\ \end{align*}\]

Example: What’s the taylor series of \(1/\sqrt{\cos x}\)? We now know that \(1/\sqrt{1+x} = 1 - \tfrac{1}{2}x + O(x^2)\). So

\[\begin{align*} 1/\sqrt{\cos x} &= 1/\sqrt{1-\tfrac{x^2}{2} + O(x^4)} \\ &= 1/\sqrt{1-\tfrac{x^2}{2}} + O(x^4) \\ &= 1 + \tfrac{x^2}{4} + O(x^4). \\ \end{align*}\]

We could’ve reasonbly guessed this expression from the previous example as \(\sqrt{\cos x} \frac{1}{\sqrt{\cos x}} = 1\) and \((1+x^2)(1-x^2) = 1 + O(x^4)\).

You may find the general Leinbiz rule interesting. We can also formulate the taylor series of \((1+x)^p\) using Gamma functions.

Going further

More generally one can ask what the taylor series of \(f(g(x))\) around \(x=0\). Again we can get an explicit formula by calculating derivatives using the product rule, but for most cases better methods apply. Again we will assume \(f(x), g(x) > 0\) on a ball around \(x=0\).

Example: What’s the taylor series of \(\cos\left(\cos x\right)\) around \(x=0\)? We have

\[\begin{align*} \cos\left(\cos x\right)&= \cos\left(1 - \tfrac{x^2}{2} + O(x^4)\right). \\ \end{align*}\]

Evidently we need the taylor series of \(\cos x\) around \(x=1\). We can compute that \(\cos(1+x) = \cos(1) - \sin(1) x + O(x^2)\). As such

\[\begin{align*} \cos\left(\cos x\right)&= \cos(1) + \sin(1) \tfrac{x^2}{2} + O(x^4). \\ \end{align*}\]

In essence, you can deduce the chain rule using taylor series. Given \(f(g(x))\), we can taylor expand it around some \(x=a\). We will make use of two taylor series—The taylor series of \(g(x)\) around \(x=a\) and the taylor series of \(f(x)\) around \(x=g(a)\).

\[\begin{align*} g(a+x) &= g(a) + g'(a)x + O(x^2), \\ f\left(g(a)+x\right) &= f(g(a)) + f'(g(a))x + O(x^2). \end{align*}\]

Hence

\[\begin{align*} f(g(x+a)) &= f\left(g(a) + g'(a)x + O(x^2)\right) \\ &= f\left(g(a) + g'(a)x\right) + O(x^2) \\ &= f(g(a)) + f'(g(a))g'(a)x + O(x^2). \end{align*}\]