Taylor expansion of composite functions

Question

Suppose one has a function of the form: $$f(x)=\sqrt{1+a\cos^{2}(x)}$$ where $a$ is some parameter.

Firstly, what is the correct method for Taylor expanding such a composite function? Should one just repeatedly use the chain rule $(f\circ g)'(x)=f'(g(x))g'(x)$ to determine the coefficients of the Taylor expansion, such that it has the form: $$(f\circ g)(x)=(f\circ g)(x_{0})+(f\circ g)'(x_{0})(x-x_{0})+\frac{1}{2}(f\circ g)''(x_{0})(x-x_{0})^{2}+\cdots$$ or is there a different method?

Secondly, although not technically a Taylor expansion, if the parameter $a<<1$ is it reasonable to approximate the function as $$\sqrt{1+a\cos^{2}(x)}\approx 1+\frac{a}{2}\cos^{2}(x)$$ i.e. naively plugging in $a\cos^{2}(x)$ to the Taylor expansion for $\sqrt{1+x}=1+\frac{x}{2}-\frac{x^{2}}{8}+\mathcal{O}(x^{3})$?

Answer 1

Your method would work, but it'd involve computing a lot of unpleasant derivatives. A better method would be to expand $cos^2(x)$, then expand the square root: $$\sqrt{1+a \cos^2(x)} = \sqrt{1 + a \left(1 - \frac{1}{2} x^2 + \mathcal{O}(x^4)\right)^2} \\ = \sqrt{1 + a \left(1 - x^2 + \mathcal{O}(x^4)\right)} \\ = \sqrt{1 + a - a x^2 + \mathcal{O}(x^4)} \\ = \sqrt{1+a} \sqrt{1 - \frac{a}{1+a} x^2 + \mathcal{O}(x^4)} \\ = \sqrt{1+a} \left(1 - \frac{a}{2(1+a)} x^2 + \mathcal{O}(x^4)\right) \\ = \sqrt{1+a} - \frac{a}{2\sqrt{1+a}} x^2 + \mathcal{O}(x^4)$$

Answer 2

Hint

If $f (X)=P_n(X)+X^n\epsilon_1 (X) $ and $g (y)=Q_m (y)+y^m\epsilon_2 (y) $ then

$$f (g (y))=P_n (g (y))+g^n (y)\epsilon_1(Q_m(y)+y^m\epsilon_2 (y)) $$

when $y\to 0$, you need $Q_m (0)=0$ to satisfy $\epsilon_1\to 0$.

Starting from $$\sqrt {1+X}=1+\frac X2-\frac {X^2}{8}+... $$

you can replace $X $ by $g (X) $ only if $\lim_{X\to 0}g (X)=0$. which is not the case for $a\cos^2 (x) $ since it goes to $a $.

so you write your function as

$$\sqrt {1+a+a (\cos^2 (x)-1)}=$$ $$\sqrt {1+a}\sqrt {1+\frac {a(\cos^2 (x )-1)}{1+a} }$$ and now you can replace $X $ by $$g (x)=\frac {a (\cos^2 (x)-1)}{1+a} .$$

Answer 3

$$f(x)=\sqrt{1+a - a\sin^{2}(x)} = \sqrt{1+a}\sqrt{1 - {a\over a+1}\sin^2x}$$ $$ = \sqrt{1+a}\left(1 - {1\over2}{a\over a+1}\sin^2x + {1\over2!}{1\over2}\left({1\over2}-1\right)\left(-{a\over a+1}\right)^2\sin^4x\right) +O(x^6)$$ $$ = \sqrt{1+a}\left(1 - {1\over4}{a\over a+1}(1-\cos2x) - {1\over32}{a^2\over (a+1)^2}(1-2\cos 2x+\cos^22x)\right)+O(x^6)$$ $$ = \sqrt{1+a}\left(1 - {1\over4}{a\over a+1}(1-\cos2x) - {1\over64}{a^2\over (a+1)^2}(2-4\cos 2x+ 1+\cos4x)\right)+O(x^6)$$ $$ = \sqrt{1+a} - \sqrt{1+a}\left({1\over4}{a\over a+1} + {1\over16}{a^2\over (a+1)^2}\right)(1-\cos2x) +\sqrt{1+a} + {1\over64}{a^2\over (a+1)^2}(1-\cos4x)+O(x^6)$$ $$ = \sqrt{1+a} - \sqrt{1+a}\left({1\over4}{a\over a+1} + {1\over16}{a^2\over (a+1)^2}\right)\left({4x^2\over2!}-{16x^4\over4!}\right)$$ $$ + {\sqrt{1+a}\over64}{a^2\over (a+1)^2}\left({16x^2\over2!}-{256x^4\over4!}\right)+O(x^6)$$ $$ = \sqrt{1+a} - {\sqrt{1+a}\over4}{a\over a+1}x^2 + {\sqrt{1+a}\over4!}\left({4a\over a+1} - {3a^2\over (a+1)^2}\right)x^4+O(x^6).$$