Krdytkyu

Given that $y=lnBigl(frac{e^{-x}+1}{2}Bigl)$, show $frac{dy}{dx}=frac{1}{2}e^{-y}-1$. Show that the series expansion of $frac{dy}{dx}$ in ascending powers of $x$, up to and including the term in $x^2$ is $-frac{1}{2}+frac{1}{4}x+kx^2+dots$ , where $k$ is to be determined.

I'm able to solve this question but I'm unsure if my $k$ value is correct. My $k$ is $0$. Am I correct? Otherwise, I might have made a mistake somewhere.

My work

1) $f(0)=y=0$

2) $frac{dy}{dx}=frac{1}{2}e^{-y}-1 $
therefore $f'(0)=-frac{1}{2}$

3) $frac{d^2y}{dx^2}=-frac{1}{2}e^{-y}cdotBig(frac{dy}{dx}Bigl) $
therefore $f''(0)=frac{1}{4}$

4) $frac{d^3y}{dx^3}=frac{1}{2}e^{-y}cdotBigl(frac{dy}{dx}Bigl)^2-frac{1}{2}e^{-y}cdotBigl(frac{d^2y}{dx^2}Bigl) $
therefore $f'''(0)=0$

Hence
$$frac{dy}{dx}=f'(0)+f''(0)x+frac{f'''(x)}{2}x^2+dots=-frac{1}{2}+frac{1}{4}x+dots$$

A hint. If you set $$ u=frac{1-e^{-x}}2$$ then, as $tto0$, $u to 0$ and you can use $$ln(1-u)=-u+frac{u^2}2+o(u^3) $$ to get a Maclaurin series expansion of $$y=lnBigl(frac{e^{-x}+1}{2}Bigl).$$

Yes, you are correct $k=0$. This is an alternative solution where we use the expansions of $e^t$ and $(1+t)^{-1}$ at $t=0$:
$$begin{align}
frac{dy}{dx}&=frac{2}{e^{-x}+1}cdot frac{-e^{-x}}{2} =-frac{1}{1+e^x}\
&=-frac{1}{1+1+x+frac{x^2}{2}+o(x^2)}\
&=-frac{1}{2}left(1+frac{x}{2}+frac{x^2}{4}+o(x^2)right)^{-1}\
&=-frac{1}{2}left(1-left(frac{x}{2}+frac{x^2}{4}right)+left(frac{x}{2}+o(x)right)^2+o(x^2)right)\
&=-frac{1}{2}+frac{x}{4}+underbrace{left(-frac{1}{4}+frac{1}{4}right)}_{=0}cdot x^2+ o(x^2).
end{align}$$

Notice that $$2e^y=1+e^{-x}$$therefore by differentiating we have $$2y'e^y=-e^{-x}$$or equivalently$$y'=-({2e^y-1}){1over 2e^y}={1over 2}e^{-y}-1$$also we know that$$k={1over 2}{d^2yover dx^2}|_{x=0}$$since $y(0)=0$ and $y'(0)=-{1over 2}$ we obtain$$y''=-{1over 2}y'e^{-y}to k={1over 8}$$