Find $frac{partial^{200}f}{partial x^{200}}(0)$ if $log(1+f(x)+f^4(x))= x^{50}$, $f(0)=0$ and f is...












0












$begingroup$


I understand that I need to somehow use Taylor series, but I have no idea what to do next. Can you give me a hint?






taylor-expansion







share|cite|improve this question









$endgroup$












  • $begingroup$
    In general, if $p in mathbb{R}[X]$ is a fixed polynomial, $I$ a (nondegenerate) interval and $f in mathrm{C}^{infty}(I, mathbb{R})$ is infinitely derivable such that $f(I)$ does not contain any roots of $p$, can you see how the derivatives of $mathrm{ln}|tilde{p}circ f|$ would behave (by $widetilde{p}$ I mean the polynomial function attached to $p$)? Especially those of an order that exceeds the degree of $p$?
    $endgroup$
    – ΑΘΩ
    Dec 8 '18 at 12:46












  • $begingroup$
    Just to make sure, when you write $f^4(x)$ do you mean $[f(x)]^4$ or $frac{d^4f}{dx^4}$?
    $endgroup$
    – glowstonetrees
    Dec 8 '18 at 12:57










  • $begingroup$
    I mean $(f(x))^4$.
    $endgroup$
    – TsarN
    Dec 8 '18 at 13:00
















0












$begingroup$


I understand that I need to somehow use Taylor series, but I have no idea what to do next. Can you give me a hint?






taylor-expansion







share|cite|improve this question









$endgroup$












  • $begingroup$
    In general, if $p in mathbb{R}[X]$ is a fixed polynomial, $I$ a (nondegenerate) interval and $f in mathrm{C}^{infty}(I, mathbb{R})$ is infinitely derivable such that $f(I)$ does not contain any roots of $p$, can you see how the derivatives of $mathrm{ln}|tilde{p}circ f|$ would behave (by $widetilde{p}$ I mean the polynomial function attached to $p$)? Especially those of an order that exceeds the degree of $p$?
    $endgroup$
    – ΑΘΩ
    Dec 8 '18 at 12:46












  • $begingroup$
    Just to make sure, when you write $f^4(x)$ do you mean $[f(x)]^4$ or $frac{d^4f}{dx^4}$?
    $endgroup$
    – glowstonetrees
    Dec 8 '18 at 12:57










  • $begingroup$
    I mean $(f(x))^4$.
    $endgroup$
    – TsarN
    Dec 8 '18 at 13:00














0












0








0





$begingroup$


I understand that I need to somehow use Taylor series, but I have no idea what to do next. Can you give me a hint?






taylor-expansion







share|cite|improve this question









$endgroup$




I understand that I need to somehow use Taylor series, but I have no idea what to do next. Can you give me a hint?






taylor-expansion




taylor-expansion






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 8 '18 at 12:34









TsarNTsarN

465




465












  • $begingroup$
    In general, if $p in mathbb{R}[X]$ is a fixed polynomial, $I$ a (nondegenerate) interval and $f in mathrm{C}^{infty}(I, mathbb{R})$ is infinitely derivable such that $f(I)$ does not contain any roots of $p$, can you see how the derivatives of $mathrm{ln}|tilde{p}circ f|$ would behave (by $widetilde{p}$ I mean the polynomial function attached to $p$)? Especially those of an order that exceeds the degree of $p$?
    $endgroup$
    – ΑΘΩ
    Dec 8 '18 at 12:46












  • $begingroup$
    Just to make sure, when you write $f^4(x)$ do you mean $[f(x)]^4$ or $frac{d^4f}{dx^4}$?
    $endgroup$
    – glowstonetrees
    Dec 8 '18 at 12:57










  • $begingroup$
    I mean $(f(x))^4$.
    $endgroup$
    – TsarN
    Dec 8 '18 at 13:00


















  • $begingroup$
    In general, if $p in mathbb{R}[X]$ is a fixed polynomial, $I$ a (nondegenerate) interval and $f in mathrm{C}^{infty}(I, mathbb{R})$ is infinitely derivable such that $f(I)$ does not contain any roots of $p$, can you see how the derivatives of $mathrm{ln}|tilde{p}circ f|$ would behave (by $widetilde{p}$ I mean the polynomial function attached to $p$)? Especially those of an order that exceeds the degree of $p$?
    $endgroup$
    – ΑΘΩ
    Dec 8 '18 at 12:46












  • $begingroup$
    Just to make sure, when you write $f^4(x)$ do you mean $[f(x)]^4$ or $frac{d^4f}{dx^4}$?
    $endgroup$
    – glowstonetrees
    Dec 8 '18 at 12:57










  • $begingroup$
    I mean $(f(x))^4$.
    $endgroup$
    – TsarN
    Dec 8 '18 at 13:00
















$begingroup$
In general, if $p in mathbb{R}[X]$ is a fixed polynomial, $I$ a (nondegenerate) interval and $f in mathrm{C}^{infty}(I, mathbb{R})$ is infinitely derivable such that $f(I)$ does not contain any roots of $p$, can you see how the derivatives of $mathrm{ln}|tilde{p}circ f|$ would behave (by $widetilde{p}$ I mean the polynomial function attached to $p$)? Especially those of an order that exceeds the degree of $p$?
$endgroup$
– ΑΘΩ
Dec 8 '18 at 12:46






$begingroup$
In general, if $p in mathbb{R}[X]$ is a fixed polynomial, $I$ a (nondegenerate) interval and $f in mathrm{C}^{infty}(I, mathbb{R})$ is infinitely derivable such that $f(I)$ does not contain any roots of $p$, can you see how the derivatives of $mathrm{ln}|tilde{p}circ f|$ would behave (by $widetilde{p}$ I mean the polynomial function attached to $p$)? Especially those of an order that exceeds the degree of $p$?
$endgroup$
– ΑΘΩ
Dec 8 '18 at 12:46














$begingroup$
Just to make sure, when you write $f^4(x)$ do you mean $[f(x)]^4$ or $frac{d^4f}{dx^4}$?
$endgroup$
– glowstonetrees
Dec 8 '18 at 12:57




$begingroup$
Just to make sure, when you write $f^4(x)$ do you mean $[f(x)]^4$ or $frac{d^4f}{dx^4}$?
$endgroup$
– glowstonetrees
Dec 8 '18 at 12:57












$begingroup$
I mean $(f(x))^4$.
$endgroup$
– TsarN
Dec 8 '18 at 13:00




$begingroup$
I mean $(f(x))^4$.
$endgroup$
– TsarN
Dec 8 '18 at 13:00










1 Answer
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$begingroup$

Firstly, I would write it as



$$1+f(x)+[f(x)]^4 = e^{x^{50}}$$



Next, express the RHS as a Taylor series centered at $0$:



$$e^{x^{50}} = 1+x^{50}+frac 12 x^{100}+frac 16 x^{150}+frac{1}{24}x^{200}+cdots$$



The Taylor series of $f$ centered at $0$ is



$$f(x) = frac{f(0)}{0!}x^0+frac {f'(0)}{1!}x^1 + frac {f''(0)}{2!}x^2+frac{f'''(0)}{3!}x^3+cdots = sum_{n=0}^inftyfrac{f^{(n)}(0)}{n!}x^n$$



We are given that $f(0)=0$, so



$$f(x) = frac {f'(0)}{1!}x^1 + frac {f''(0)}{2!}x^2+frac{f'''(0)}{3!}x^3+cdots$$



We take the $4$-th power on both sides to obtain the Taylor series for $f^4$. So



$$[f(x)]^4 = [f'(0)]^4x^4 + 2[f'(0)]^3f''(0)x^5+cdots$$



Now put it altogether and compare the coefficients of the powers of $x$:



begin{align}
1+f(x)+[f(x)]^4 & = e^{x^{50}} \
implies 1+bigg(frac {f'(0)}{1!}x^1 + frac {f''(0)}{2!}x^2+frac{f'''(0)}{3!}x^3+cdotsbigg) + big([f'(0)]^4x^4+cdotsbig)& = big(1+ cdotsbig)
end{align}



Comparing terms at order $x^1$, $x^2$ and $x^3$, we see that $f'(0)=f''(0)=f'''(0)=0$. Thus, we now know that the Taylor series of $f$ is of the form



$$f(x) = frac {f^{(4)}(0)}{4!}x^4 + frac {f^{(5)}(0)}{5!}x^5+frac{f^{(6)}(0)}{6!}x^6+cdots$$



where $f^{(n)}(0)$ denotes the $n$-th derivative of $f$ evaluated at $x=0$.



We can now take $4$-th powers again:



$$[f(x)]^4 = bigg(frac{f^{(4)}(0)}{4!}bigg)^4x^{16} +cdots$$



and, upon substituting into the equation $$1+f(x)+[f(x)]^4 = e^{x^{50}}$$ we find that $$f^{(4)}(0)=cdots=f^{(15)}(0)=0$$



Repeat again and we find further that $$f^{(16)}(0)=cdots=f^{(49)}(0)=0=f^{(51)}(0)=cdots=f^{(63)}(0)$$



whereas $frac{f^{50}(0)}{50!}=1$ due to the $x^{50}$ term from $exp(x^{50})$. So now we have



$$f(x) = x^{50} + frac {f^{(64)}(0)}{64!}x^{64}+frac{f^{(65)}(0)}{65!}x^{65}+cdots$$



and



$$[f(x)]^4 = x^{200}+cdots$$



Finally, substitute into the equation and compare the coefficients of the $x^{200}$ term:



begin{align}
frac{f^{(200)}(0)}{200!} + 1 & = frac{1}{24} \
implies f^{(200)}(0) & = -200! cdot frac{23}{24}
end{align}






share|cite|improve this answer









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    0












    $begingroup$

    Firstly, I would write it as



    $$1+f(x)+[f(x)]^4 = e^{x^{50}}$$



    Next, express the RHS as a Taylor series centered at $0$:



    $$e^{x^{50}} = 1+x^{50}+frac 12 x^{100}+frac 16 x^{150}+frac{1}{24}x^{200}+cdots$$



    The Taylor series of $f$ centered at $0$ is



    $$f(x) = frac{f(0)}{0!}x^0+frac {f'(0)}{1!}x^1 + frac {f''(0)}{2!}x^2+frac{f'''(0)}{3!}x^3+cdots = sum_{n=0}^inftyfrac{f^{(n)}(0)}{n!}x^n$$



    We are given that $f(0)=0$, so



    $$f(x) = frac {f'(0)}{1!}x^1 + frac {f''(0)}{2!}x^2+frac{f'''(0)}{3!}x^3+cdots$$



    We take the $4$-th power on both sides to obtain the Taylor series for $f^4$. So



    $$[f(x)]^4 = [f'(0)]^4x^4 + 2[f'(0)]^3f''(0)x^5+cdots$$



    Now put it altogether and compare the coefficients of the powers of $x$:



    begin{align}
    1+f(x)+[f(x)]^4 & = e^{x^{50}} \
    implies 1+bigg(frac {f'(0)}{1!}x^1 + frac {f''(0)}{2!}x^2+frac{f'''(0)}{3!}x^3+cdotsbigg) + big([f'(0)]^4x^4+cdotsbig)& = big(1+ cdotsbig)
    end{align}



    Comparing terms at order $x^1$, $x^2$ and $x^3$, we see that $f'(0)=f''(0)=f'''(0)=0$. Thus, we now know that the Taylor series of $f$ is of the form



    $$f(x) = frac {f^{(4)}(0)}{4!}x^4 + frac {f^{(5)}(0)}{5!}x^5+frac{f^{(6)}(0)}{6!}x^6+cdots$$



    where $f^{(n)}(0)$ denotes the $n$-th derivative of $f$ evaluated at $x=0$.



    We can now take $4$-th powers again:



    $$[f(x)]^4 = bigg(frac{f^{(4)}(0)}{4!}bigg)^4x^{16} +cdots$$



    and, upon substituting into the equation $$1+f(x)+[f(x)]^4 = e^{x^{50}}$$ we find that $$f^{(4)}(0)=cdots=f^{(15)}(0)=0$$



    Repeat again and we find further that $$f^{(16)}(0)=cdots=f^{(49)}(0)=0=f^{(51)}(0)=cdots=f^{(63)}(0)$$



    whereas $frac{f^{50}(0)}{50!}=1$ due to the $x^{50}$ term from $exp(x^{50})$. So now we have



    $$f(x) = x^{50} + frac {f^{(64)}(0)}{64!}x^{64}+frac{f^{(65)}(0)}{65!}x^{65}+cdots$$



    and



    $$[f(x)]^4 = x^{200}+cdots$$



    Finally, substitute into the equation and compare the coefficients of the $x^{200}$ term:



    begin{align}
    frac{f^{(200)}(0)}{200!} + 1 & = frac{1}{24} \
    implies f^{(200)}(0) & = -200! cdot frac{23}{24}
    end{align}






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Firstly, I would write it as



      $$1+f(x)+[f(x)]^4 = e^{x^{50}}$$



      Next, express the RHS as a Taylor series centered at $0$:



      $$e^{x^{50}} = 1+x^{50}+frac 12 x^{100}+frac 16 x^{150}+frac{1}{24}x^{200}+cdots$$



      The Taylor series of $f$ centered at $0$ is



      $$f(x) = frac{f(0)}{0!}x^0+frac {f'(0)}{1!}x^1 + frac {f''(0)}{2!}x^2+frac{f'''(0)}{3!}x^3+cdots = sum_{n=0}^inftyfrac{f^{(n)}(0)}{n!}x^n$$



      We are given that $f(0)=0$, so



      $$f(x) = frac {f'(0)}{1!}x^1 + frac {f''(0)}{2!}x^2+frac{f'''(0)}{3!}x^3+cdots$$



      We take the $4$-th power on both sides to obtain the Taylor series for $f^4$. So



      $$[f(x)]^4 = [f'(0)]^4x^4 + 2[f'(0)]^3f''(0)x^5+cdots$$



      Now put it altogether and compare the coefficients of the powers of $x$:



      begin{align}
      1+f(x)+[f(x)]^4 & = e^{x^{50}} \
      implies 1+bigg(frac {f'(0)}{1!}x^1 + frac {f''(0)}{2!}x^2+frac{f'''(0)}{3!}x^3+cdotsbigg) + big([f'(0)]^4x^4+cdotsbig)& = big(1+ cdotsbig)
      end{align}



      Comparing terms at order $x^1$, $x^2$ and $x^3$, we see that $f'(0)=f''(0)=f'''(0)=0$. Thus, we now know that the Taylor series of $f$ is of the form



      $$f(x) = frac {f^{(4)}(0)}{4!}x^4 + frac {f^{(5)}(0)}{5!}x^5+frac{f^{(6)}(0)}{6!}x^6+cdots$$



      where $f^{(n)}(0)$ denotes the $n$-th derivative of $f$ evaluated at $x=0$.



      We can now take $4$-th powers again:



      $$[f(x)]^4 = bigg(frac{f^{(4)}(0)}{4!}bigg)^4x^{16} +cdots$$



      and, upon substituting into the equation $$1+f(x)+[f(x)]^4 = e^{x^{50}}$$ we find that $$f^{(4)}(0)=cdots=f^{(15)}(0)=0$$



      Repeat again and we find further that $$f^{(16)}(0)=cdots=f^{(49)}(0)=0=f^{(51)}(0)=cdots=f^{(63)}(0)$$



      whereas $frac{f^{50}(0)}{50!}=1$ due to the $x^{50}$ term from $exp(x^{50})$. So now we have



      $$f(x) = x^{50} + frac {f^{(64)}(0)}{64!}x^{64}+frac{f^{(65)}(0)}{65!}x^{65}+cdots$$



      and



      $$[f(x)]^4 = x^{200}+cdots$$



      Finally, substitute into the equation and compare the coefficients of the $x^{200}$ term:



      begin{align}
      frac{f^{(200)}(0)}{200!} + 1 & = frac{1}{24} \
      implies f^{(200)}(0) & = -200! cdot frac{23}{24}
      end{align}






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Firstly, I would write it as



        $$1+f(x)+[f(x)]^4 = e^{x^{50}}$$



        Next, express the RHS as a Taylor series centered at $0$:



        $$e^{x^{50}} = 1+x^{50}+frac 12 x^{100}+frac 16 x^{150}+frac{1}{24}x^{200}+cdots$$



        The Taylor series of $f$ centered at $0$ is



        $$f(x) = frac{f(0)}{0!}x^0+frac {f'(0)}{1!}x^1 + frac {f''(0)}{2!}x^2+frac{f'''(0)}{3!}x^3+cdots = sum_{n=0}^inftyfrac{f^{(n)}(0)}{n!}x^n$$



        We are given that $f(0)=0$, so



        $$f(x) = frac {f'(0)}{1!}x^1 + frac {f''(0)}{2!}x^2+frac{f'''(0)}{3!}x^3+cdots$$



        We take the $4$-th power on both sides to obtain the Taylor series for $f^4$. So



        $$[f(x)]^4 = [f'(0)]^4x^4 + 2[f'(0)]^3f''(0)x^5+cdots$$



        Now put it altogether and compare the coefficients of the powers of $x$:



        begin{align}
        1+f(x)+[f(x)]^4 & = e^{x^{50}} \
        implies 1+bigg(frac {f'(0)}{1!}x^1 + frac {f''(0)}{2!}x^2+frac{f'''(0)}{3!}x^3+cdotsbigg) + big([f'(0)]^4x^4+cdotsbig)& = big(1+ cdotsbig)
        end{align}



        Comparing terms at order $x^1$, $x^2$ and $x^3$, we see that $f'(0)=f''(0)=f'''(0)=0$. Thus, we now know that the Taylor series of $f$ is of the form



        $$f(x) = frac {f^{(4)}(0)}{4!}x^4 + frac {f^{(5)}(0)}{5!}x^5+frac{f^{(6)}(0)}{6!}x^6+cdots$$



        where $f^{(n)}(0)$ denotes the $n$-th derivative of $f$ evaluated at $x=0$.



        We can now take $4$-th powers again:



        $$[f(x)]^4 = bigg(frac{f^{(4)}(0)}{4!}bigg)^4x^{16} +cdots$$



        and, upon substituting into the equation $$1+f(x)+[f(x)]^4 = e^{x^{50}}$$ we find that $$f^{(4)}(0)=cdots=f^{(15)}(0)=0$$



        Repeat again and we find further that $$f^{(16)}(0)=cdots=f^{(49)}(0)=0=f^{(51)}(0)=cdots=f^{(63)}(0)$$



        whereas $frac{f^{50}(0)}{50!}=1$ due to the $x^{50}$ term from $exp(x^{50})$. So now we have



        $$f(x) = x^{50} + frac {f^{(64)}(0)}{64!}x^{64}+frac{f^{(65)}(0)}{65!}x^{65}+cdots$$



        and



        $$[f(x)]^4 = x^{200}+cdots$$



        Finally, substitute into the equation and compare the coefficients of the $x^{200}$ term:



        begin{align}
        frac{f^{(200)}(0)}{200!} + 1 & = frac{1}{24} \
        implies f^{(200)}(0) & = -200! cdot frac{23}{24}
        end{align}






        share|cite|improve this answer









        $endgroup$



        Firstly, I would write it as



        $$1+f(x)+[f(x)]^4 = e^{x^{50}}$$



        Next, express the RHS as a Taylor series centered at $0$:



        $$e^{x^{50}} = 1+x^{50}+frac 12 x^{100}+frac 16 x^{150}+frac{1}{24}x^{200}+cdots$$



        The Taylor series of $f$ centered at $0$ is



        $$f(x) = frac{f(0)}{0!}x^0+frac {f'(0)}{1!}x^1 + frac {f''(0)}{2!}x^2+frac{f'''(0)}{3!}x^3+cdots = sum_{n=0}^inftyfrac{f^{(n)}(0)}{n!}x^n$$



        We are given that $f(0)=0$, so



        $$f(x) = frac {f'(0)}{1!}x^1 + frac {f''(0)}{2!}x^2+frac{f'''(0)}{3!}x^3+cdots$$



        We take the $4$-th power on both sides to obtain the Taylor series for $f^4$. So



        $$[f(x)]^4 = [f'(0)]^4x^4 + 2[f'(0)]^3f''(0)x^5+cdots$$



        Now put it altogether and compare the coefficients of the powers of $x$:



        begin{align}
        1+f(x)+[f(x)]^4 & = e^{x^{50}} \
        implies 1+bigg(frac {f'(0)}{1!}x^1 + frac {f''(0)}{2!}x^2+frac{f'''(0)}{3!}x^3+cdotsbigg) + big([f'(0)]^4x^4+cdotsbig)& = big(1+ cdotsbig)
        end{align}



        Comparing terms at order $x^1$, $x^2$ and $x^3$, we see that $f'(0)=f''(0)=f'''(0)=0$. Thus, we now know that the Taylor series of $f$ is of the form



        $$f(x) = frac {f^{(4)}(0)}{4!}x^4 + frac {f^{(5)}(0)}{5!}x^5+frac{f^{(6)}(0)}{6!}x^6+cdots$$



        where $f^{(n)}(0)$ denotes the $n$-th derivative of $f$ evaluated at $x=0$.



        We can now take $4$-th powers again:



        $$[f(x)]^4 = bigg(frac{f^{(4)}(0)}{4!}bigg)^4x^{16} +cdots$$



        and, upon substituting into the equation $$1+f(x)+[f(x)]^4 = e^{x^{50}}$$ we find that $$f^{(4)}(0)=cdots=f^{(15)}(0)=0$$



        Repeat again and we find further that $$f^{(16)}(0)=cdots=f^{(49)}(0)=0=f^{(51)}(0)=cdots=f^{(63)}(0)$$



        whereas $frac{f^{50}(0)}{50!}=1$ due to the $x^{50}$ term from $exp(x^{50})$. So now we have



        $$f(x) = x^{50} + frac {f^{(64)}(0)}{64!}x^{64}+frac{f^{(65)}(0)}{65!}x^{65}+cdots$$



        and



        $$[f(x)]^4 = x^{200}+cdots$$



        Finally, substitute into the equation and compare the coefficients of the $x^{200}$ term:



        begin{align}
        frac{f^{(200)}(0)}{200!} + 1 & = frac{1}{24} \
        implies f^{(200)}(0) & = -200! cdot frac{23}{24}
        end{align}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 8 '18 at 13:28









        glowstonetreesglowstonetrees

        2,336418




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            Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com
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            Are they similar matrix

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            0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19
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            “minimization” problem in Euclidean space related to orthonormal basis

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            3 $begingroup$ I have stumbled upon an exercice for second year undegraduate student majoring in economics which I find quite demanding. I have an idea for the solution, but it seems awfully complicated, and I am wondering if there was a simpler way to solve it. This goes as follows: Let $(E, langle, rangle)$ be a euclidean space of dimension $n geq 2$ . 1) For any $x in E$ , show that $||x|| = sqrt{n}$ if and only if there exists $(e_1, dots, e_n)$ an orthonormal basis of $E$ such that $x =e_1+ ldots + e_n$ . 2) For any $(x,y) in E$ , show that $||x|| = sqrt{n}$ , $||y|| = sqrt{frac{n(n+1)(2n+1)}{6}}$ and $langle x,y rangle = frac{n(n+1)}{2}$ if and only if there eixsts $(e_1, ldots, e_n)$ an orthonormal basis of $E$ such that $x = e_1 + ldots + e_n$ and $y = e_1 + 2e_2 + ldots + ne_n$ The first q
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