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
1






active

oldest

votes


















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









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3031056%2ffind-frac-partial200f-partial-x2000-if-log1fxf4x-x5%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    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




        2,336418






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3031056%2ffind-frac-partial200f-partial-x2000-if-log1fxf4x-x5%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Probability when a professor distributes a quiz and homework assignment to a class of n students.

            Image
            0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
            Read more

            Aardman Animations

            Image
            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 ...
            Read more

            Are they similar matrix

            Image
            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 ...
            Read more