MHB Solution for POTW #230: Calculate g^(3)(0) for f(x)

[anemone]

Here is this week's POTW:

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Let $$f(x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}$$.

Set $$g(x)=f^{-1}(x)$$, compute $$g^{(3)}(0)$$.

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[anemone]

Congratulations to the following members for their correct solution::)

1. kaliprasad
2. lfdahl

Solution from lfdahl:

Spoiler

$f$ and its derivatives evaluated in $x=0$:

\[f(x) = x + \frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5} \\\\ f'(x) = 1 + x + x^2+ x^3+x^4 \\\\ f''(x)= 1+2x + 3x^2+4x^3 \\\\ f^{(3)}(x)=2+6x+12x^2 \\\\ f(0)=0, \: \: \: f'(0)=1,\: \: \: f''(0)=1,\: \: \: f^{(3)}(0)=2\]$g$ evaluated in $x=0$:

\[g(f(x)) = x \Rightarrow g(f(0)) = 0 \Rightarrow g(0)=0 \\\\ \]

In order to evaluate $g^{(3)}(0)$, I need the first three derivatives of $f(g(x))$:

1st differentiation:

\[\frac{\mathrm{d} }{\mathrm{d} x}f(g(x)) = 1\Rightarrow f'(g(x))g'(x)=1 \\\\ f'(g(0))g'(0)=1 \Rightarrow f'(0)g'(0) = 1 \Rightarrow g'(0)=1\]

2nd differentiation:

\[\frac{\mathrm{d} }{\mathrm{d} x}f'(g(x))g'(x)=0 \\\\ f''(g(x))(g'(x))^2+f'(g(x))g''(x)=0 \\\\ f''(0)(g'(0))^2+f'(0)g''(0)=0 \Rightarrow 1 \cdot 1^2 + g''(0)=0 \\\\ \Rightarrow g''(0) = -1\]

3rd differentiation:

\[\frac{\mathrm{d} }{\mathrm{d} x}\left [ f''(g(x))(g'(x))^2+f'(g(x))g''(x) \right ]=0 \\\\ f^{(3)}(g(x))(g'(x))^3+3f''(g(x))g'(x)g''(x)+f'(g(x))g^{(3)}(x)=0 \]

\[ f^{(3)}(0)(g'(0))^3+3f''(0)g'(0)g''(0)+f'(0)g^{(3)}(0)=0 \\\\ 2\cdot 1^3+3 \cdot 1 \cdot 1 \cdot (-1)+1\cdot g^{(3)}(0)=0 \\\\ \Rightarrow g^{(3)}(0) = 1\]