Let $f(x)=x+\frac{x^2}{2} + \frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}$ and let $g(x)=f^{-1} (x)$. Find $g’’’(0)$

Question

My method is extremely inefficient, but here it is

$$g(f(x))=x$$ $$g’(f(x)).f’(x)=1$$Differentiating wrt x multiple times

$$g’’’(f(x))(f’(x))^3 + (g’’(f(x)))(2f’(x))(f’’(x)) + g’’(f(x)).f’(x).f’’(x) + g’(f(x)).f’’’(x)=0$$

I may have made some computation errors in this, but I can’t seem to find any (I apologize if there are)

$f(x)=0$ at $x=0$ $$g’’’(0) +2 g’’(0) + g’’(0) +2g’(0)=0$$

How do I proceed from here? Is there a better way to approach this ?

if instead of $f(x)$ you take $-\log(1-x)$, you'll get the same result — user8268, Aug 26 '20 at 17:40

DatBoi · Accepted Answer · 2020-08-27T11:01:45.487

Take $$g(f(x))=x$$ $$\Rightarrow g'(f(x)).f'(x)=1$$ $$\Rightarrow g'(f(x))=\frac{1}{f'(x)}$$ $$g''(f(x)).f'(x)=-\frac{f''(x)}{(f'(x))^2}$$ $$\Rightarrow g''(f(x))=-\frac{f''(x)}{(f'(x))^3}$$ I'll let you differentiate once again.

Note that I have not combined any other function with $g'(f(x))$ It is done so that it restricts $g'(f(x))$ from popping up in the final $g'''(x)$ expression. Same goes with $g''(x)$.

Spoilers

$$g'''(f(x)).f'(x)=-\frac{f'''(x).(f'(x))^3-3(f''(x))^2.(f'(x))^2}{(f'(x))^6}$$ $$g'''(f(x))=-\frac{f'''(x).(f'(x))^3-3(f''(x))^2.(f'(x))^2}{(f'(x))^7}$$

$$f(x)=x+\frac{x^2}{2}+\frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}$$ $$f'(x)=1+x^2+x^3+x^4+x^5$$ $$f''(x)=1+2x+3x^2+4x^3+5x^4$$ $$f'''(x)=2+6x+...$$

To conclude with, substitute $0$ in the above four(as$ f(0)=0$) $$f(0)=0$$ $$f'(0)=1$$ $$f''(0)=1$$ $$f'''(0)=2$$ $$\therefore g'''(0)=-\frac{f'''(0).(f'(0))^3-3(f''(0))^2.(f'(0))^2}{(f'(0))^7}$$ $$g'''(0)=-\frac{2.1-3.1.1}{1}$$ $$\therefore g'''(0)=1$$ :)

Z Ahmed · Answer 2 · 2023-02-03T04:44:47.987

2

If $f(x)$ is an infinite series, then

$$y=f(x)=x+x^2/2+x^3/3+x^4/4+.....=-\ln(1-x) \implies y=-\ln(1-x) $$ $$ \implies x=1-e^{-y} \implies x=f^{-1}(y)=1-e^{-y} \implies f^{-1}(x)= 1-e^{-x}=g(x).$$ Then $$g'(x)=e^{-x}, ~g''(x)=-e^{-x} \implies g'''(x)=e^{-x} \implies g'''(0)=1$$

edited Feb 03 '23 at 04:44

answered Aug 26 '20 at 17:58

Z Ahmed

43,235

i think we are asked to find $g'''(0$ – Albus Dumbledore Aug 26 '20 at 18:00
and how can we say f(x)=ln(1-x) – Albus Dumbledore Aug 26 '20 at 18:01
@Quantum I think it’s the expansion of $\ln (1-x)$ – Aditya Aug 26 '20 at 18:02
1

agreed but in the original question it was not an infinirte series – Albus Dumbledore Aug 26 '20 at 18:03
@Quantum Oh yes I have now found $g'''(0)=1$, your $ f(x)$ is an infinite series which is well known as $x+x^2/2+x^3/3+x^4/4+x^5/5+......$ – Z Ahmed Aug 26 '20 at 18:04
yes i know that , i didnt see that the question was edited initially it was not so, sorry – Albus Dumbledore Aug 26 '20 at 18:06
In the question, it is not meant to be an infinite series though. However, your answer is correct so there might have been a typo in the source question. But can you please confirm it again, since it wasn’t a typo on my part – Aditya Aug 26 '20 at 18:06
Then why dots were there ! – Z Ahmed Aug 26 '20 at 18:06
I just added the dots in my question because I didn’t want to type the whole thing out. In the source problem, the whole expression was typed out, clearly terminating at $\frac{x^5}{5}$ – Aditya Aug 26 '20 at 18:08
@Aditya did you get the problem from vikas guptas book if so it was not a typo .I have the same book and the question given was not an infinite series – Albus Dumbledore Aug 26 '20 at 18:08
@Quantum I am not aware of that book, and I didn’t pick it up from there. I agree though, my question didn’t have an infinite series either – Aditya Aug 26 '20 at 18:09
@Aditya i have found the value of g'''(0) if f(x) was terminating and it also is 1 so it may not be a typo – Albus Dumbledore Aug 26 '20 at 18:14
@Quantum how did you do it? – Aditya Aug 27 '20 at 05:52
@Aditya by brute force – Albus Dumbledore Aug 27 '20 at 06:00

score 1 · Answer 3 · answered Aug 27 '20 at 08:28

If $f(x)$ is a polynomial, OP's approach is correct. The complete solution is here:

$$f(x)=x+x^2/2+x^3/3+x^4/4+x^5/5 \implies f(0)=0, f'(0)=1,f''(0)=1,=f'''(0)=2,$$

$$y=f(x) \implies x=f^{-1}{y} \implies x=g(y), gf(x)=1 \implies g'(f(x))f'(x)=1~~~~(1)$$ $$\implies g'(0)=1/f'(0)=1$$ D. (1) w.r.t. $x$, we get $$(2)~~~ g''(f(x)(f'(x))^2+g'(f(x)) f''(x)=0 \implies g''(0)+g'(0)f''(0)=0 \implies g''(0)=-1.$$ D. (2) w.r.t. $x$, we get $$g'''(f(x))(f'(x))^3+3g''(f(x)) f'(x) f''(x)+g'(f(x)) f'(x) f'''(x)=0 ```(3)$$ Using values of : $g(0),g'(0),g''(0),f'(0),f''(0), f'''(0)$ in above, we get $$g'''(0) +3g''(0)+2g'(0)=0 \implies g'''(0)=3-2=1$$

@Aditya if $f(x)$ is the polynomial and not the series, you may see my new answer, Your approach itself is correct. — Z Ahmed, Aug 27 '20 at 08:32

score 0 · Answer 4 · answered Aug 26 '20 at 17:44

0

Hint : $g(0)=0$ and $$g' = \frac{1}{f' \circ g}$$

You deduce $g'(0)$ and $g''$, then $g''(0)$ and $g'''$, then finally $g'''(0)$.

answered Aug 26 '20 at 17:44

TheSilverDoe

29,720

Is there any easier way to solve this, because it is getting quite heavy at $g’’’(x)$ – Aditya Aug 26 '20 at 17:50

Let $f(x)=x+\frac{x^2}{2} + \frac{x^3}{3}+\frac{x^4}{4}+\frac{x^5}{5}$ and let $g(x)=f^{-1} (x)$. Find $g’’’(0)$

4 Answers4