2

Is my way of solving and my answer correct?

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

And $g(x)=f^{-1}(x)$

Find $g'''(0)$

My attempt:

We know that $g'(x)=\frac{1}{f'(g(x))}$

$f'(x)=1+x+x^2+x^3+x^4=\frac{x^5-1}{x-1}$

$f'(g(x))=\frac{(g(x))^5-1}{(g(x))-1}$

$\Rightarrow g'(x)=\frac{(g(x))-1}{(g(x))^5-1}$

$\Rightarrow g'(x)[(g(x))^5-1]=g(x)-1$

Similarly differentiating again,

$\Rightarrow g'(x)[5(g(x))^4]+[(g(x))^5-1]g''(x)=g'(x)$

Similarly differentiating again,

$\Rightarrow g'(x)[20(g(x))^3g'(x)]+[5(g(x))^4]g''(x)+[(g(x))^5-1]g'''(x)+g''(x)[5(g(x))^4]=g''(x)$

Putting $x=0$,

$g(0)=f^{-1}(0)=0$

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

similarly,$g''(0)=-1$

$\Rightarrow g'''(0)=1$

Obinoscopy
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Vinod Kumar Punia
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1 Answers1

2

As far as I can see, everything looks good. I think there may be a far less messy approach though. First, we find $g(0)$.

$$g(f(x))=x$$ $$g(f(0))=0$$ $$g(0)=0$$

Now, from here on, take derivatives of the equation $f(g(x))=x$. We have

$$f'(g(x))g'(x)=1$$ $$f'(0)g'(0)=1,g'(0)=1$$ $$f''(g(x))[g'(x)]^2+f'(g(x))g''(x)=0$$ $$f''(0)[g'(0)]^2+f'(0)g''(0)=0$$ $$1(1)+g''(0)=0,g''(0)=-1$$

And one last time.

$$f'''(g(x))[g'(x)]^3+2f''(g(x))g'(x)g''(x)+f''(g(x))g'(x)g''(x)+f'(g(x))g'''(x)=0$$ $$f'''(0)[g'(0)]^3+3f''(0)g'(0)g''(0)+f'(0)g'''(0)=0$$ $$2(1)+3(1)(1)(-1)+g'''(0)=0,g'''(0)=1$$

Mike
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