Find Maclaurin Series of the inverse of a function

Question

I am given a function $f(x)=4x+cos(3x)-1$ and I need to find the Maclaurin series of its inverse up till the 2nd power (assuming the function is only defined over an interval where it is growing, otherwise it wouldn't pass the horizontal line test)

This is what I've done.

$f(x) = 4x+cos(3x)-1$

$f'(x) = 4-3sin(3x)$

$f''(x) = -9cos(3x)$

Now for the function, I guess 0 as a root, which is right, thus since $f(x) = y$ then $x = f^{-1}(y)$ and $0=f^{-1}(0)$

So now by the derivative of an inverse function theorem

$f^{-1'}(0) = \frac{1}{f^{'}(f^{-1}(0))}$

then

$f^{-1}(0) = \frac{1}{4-3sin(3\times0)} = 1/4$

and then for the last one I can find the root easily as cos is 0 when pi/2 is the input so x must be pi/6

so

$0 + y/4 + \pi y^2/12$ is the grade 2 maclaurin series of the inverse of f(x)

When I input my answer, it tells me I am wrong. I've checked my algebra multiple times so I'm almost positive that's not the problem here.

Is it my use of the inverse function theorem that's wrong here?

Answer 1

As you correctly said, you use the identity $f(f^{-1}(y))=y$ to calculate the derivatives of $f^{-1}$.

Differentiating once with respect to $y$ gives: $$(f^{-1})'(y)=\frac{1}{f'(f^{-1}(y))}$$ and substitution $f^{-1}(0)=0$ as you calculated, and $f'(0)=0$, gives you the correct value $(f^{-1})'(0)=\frac{1}{4}$.

Now to find the value of the second derivative at zero, you differentiate again the expression with respect to $y$, to get:

$$(f^{-1})''(y)=-\frac{f''(f^{-1}(y))(f^{-1})'(y)}{(f'(f^{-1}(y)))^2}$$

which you can calculate at $y=0$, since you know all the values that appear on the right hand side.

If you make no mistakes, you should arrive at the value: $(f^{-1})''(0)=\frac{9}{4^3}$.

Note that for the calcualtion, you need to know the value $f''(0)=-9$, rather than finding any roots of $f''$ as I think you mistakenly did in your solution.

Besides, $f''(x)=-9\cos(3x)$ has infinitly many roots, namely $x=\frac{1}{3}(\kappa\pi+\frac{\pi}{2})$, so there is no reason for you to pick $\frac{\pi}{6}$ every any other of them.

Finally as a remark, I'll add that your inverse is defined everywhere, since $f$ is strictly increasing (you have that $f'(x)>0,\forall x\in\mathbb{R}$)

Answer 2

You made some mistake while trying to compute $\left(f^{-1}\right)''(0)$. I have no idea about the meaning of “I can find the root easily as $\cos$ is $0$ when $\pi/2$ is the input so $x$ must be $\pi/6$”.

Since$$f^{-1}(f(x))=x,\label{a}\tag1$$you have $\left(f^{-1}\right)'\bigl(f(0)\bigr)f'(0)=1$. But $f(0)=0$ and $f'(0)=4$. Therefore $\left(f^{-1}\right)'(0)\times4=1$; in other words, $\left(f^{-1}\right)'(0)=\frac14$ (as you already know).

But it also follows from \eqref{a} that $\left(f^{-1}\right)'\bigl(f(x)\bigr)f'(x)=1$ and that therefore$$\left(f^{-1}\right)''(f(x))\times(f'(x))^2+\left(f^{-1}\right)'(f(x))f''(x)=0.$$Putting $x=0$ here, you get that $\left(f^{-1}\right)''(0)\times4^2+\frac14\times(-9)=0$, or $\left(f^{-1}\right)''(0)=\frac9{64}$.

Therefore, the answer is $\dfrac x4+\dfrac{9x^2}{128}$.