Find $k$ in Maclaurin series expansion of $frac{dy}{dx}=-frac{1}{2}+frac{1}{4}x+kx^2+…$ where...











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






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    Looks OK, but the title mismatches with the main text, where you calculate the Maclaurin expansion of $y'$.
    – gammatester
    Nov 21 at 15:55

















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






sequences-and-series taylor-expansion







share|cite|improve this question




















  • 1




    Looks OK, but the title mismatches with the main text, where you calculate the Maclaurin expansion of $y'$.
    – gammatester
    Nov 21 at 15:55















up vote
2
down vote

favorite
2









up vote
2
down vote

favorite
2






2






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






sequences-and-series taylor-expansion







share|cite|improve this question
















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






sequences-and-series taylor-expansion




sequences-and-series taylor-expansion






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edited Nov 21 at 17:14









Robert Z

92.7k1060130




92.7k1060130










asked Nov 21 at 15:46









Henias

615




615








  • 1




    Looks OK, but the title mismatches with the main text, where you calculate the Maclaurin expansion of $y'$.
    – gammatester
    Nov 21 at 15:55
















  • 1




    Looks OK, but the title mismatches with the main text, where you calculate the Maclaurin expansion of $y'$.
    – gammatester
    Nov 21 at 15:55










1




1




Looks OK, but the title mismatches with the main text, where you calculate the Maclaurin expansion of $y'$.
– gammatester
Nov 21 at 15:55






Looks OK, but the title mismatches with the main text, where you calculate the Maclaurin expansion of $y'$.
– gammatester
Nov 21 at 15:55












3 Answers
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1
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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 $$y=lnBigl(frac{e^{-x}+1}{2}Bigl).$$






share|cite|improve this answer




























    up vote
    1
    down vote













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






    share|cite|improve this answer






























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      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$$y''=-{1over 2}y'e^{-y}to k={1over 8}$$






      share|cite|improve this answer





















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        3 Answers
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        up vote
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        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 $$y=lnBigl(frac{e^{-x}+1}{2}Bigl).$$






        share|cite|improve this answer

























          up vote
          1
          down vote













          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 $$y=lnBigl(frac{e^{-x}+1}{2}Bigl).$$






          share|cite|improve this answer























            up vote
            1
            down vote










            up vote
            1
            down vote









            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 $$y=lnBigl(frac{e^{-x}+1}{2}Bigl).$$






            share|cite|improve this answer












            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 $$y=lnBigl(frac{e^{-x}+1}{2}Bigl).$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 21 at 15:54









            Dan Kent

            237




            237






















                up vote
                1
                down vote













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






                share|cite|improve this answer



























                  up vote
                  1
                  down vote













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






                  share|cite|improve this answer

























                    up vote
                    1
                    down vote










                    up vote
                    1
                    down vote









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






                    share|cite|improve this answer














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







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Nov 21 at 16:08

























                    answered Nov 21 at 15:56









                    Robert Z

                    92.7k1060130




                    92.7k1060130






















                        up vote
                        1
                        down vote













                        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$$y''=-{1over 2}y'e^{-y}to k={1over 8}$$






                        share|cite|improve this answer

























                          up vote
                          1
                          down vote













                          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$$y''=-{1over 2}y'e^{-y}to k={1over 8}$$






                          share|cite|improve this answer























                            up vote
                            1
                            down vote










                            up vote
                            1
                            down vote









                            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$$y''=-{1over 2}y'e^{-y}to k={1over 8}$$






                            share|cite|improve this answer












                            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$$y''=-{1over 2}y'e^{-y}to k={1over 8}$$







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 21 at 19:24









                            Mostafa Ayaz

                            13.6k3836




                            13.6k3836






























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