Representing power series as a function - what to do with the constant after integration?












1












$begingroup$


This power series $$f(x)=sum_{n=1}^{infty} {frac{x^{3n}}{3n}}$$ when differentiated, loses $3n$ in the denominator, with one manipulation, one can get $$f'(x)=frac{1}{x(1-x^3)} $$ using the geometric series sum formula. Since this is $f'(x)$, integration is required so we get $$f(x)=log(x)-frac{1}{3}log(1-x^3) + C$$
(the log base is probably $e$). The usual step for finding the constant is to find it's value for $x=0$ in the given power series and the new function. This obviously won't work here or in any other case where a function isn't defined for $x=0$. How do I find the constant then?






power-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    I was talking about log(x) not being defined at x=0.
    $endgroup$
    – user3711671
    Apr 7 at 17:09
















1












$begingroup$


This power series $$f(x)=sum_{n=1}^{infty} {frac{x^{3n}}{3n}}$$ when differentiated, loses $3n$ in the denominator, with one manipulation, one can get $$f'(x)=frac{1}{x(1-x^3)} $$ using the geometric series sum formula. Since this is $f'(x)$, integration is required so we get $$f(x)=log(x)-frac{1}{3}log(1-x^3) + C$$
(the log base is probably $e$). The usual step for finding the constant is to find it's value for $x=0$ in the given power series and the new function. This obviously won't work here or in any other case where a function isn't defined for $x=0$. How do I find the constant then?






power-series







share|cite|improve this question











$endgroup$












  • $begingroup$
    I was talking about log(x) not being defined at x=0.
    $endgroup$
    – user3711671
    Apr 7 at 17:09














1












1








1





$begingroup$


This power series $$f(x)=sum_{n=1}^{infty} {frac{x^{3n}}{3n}}$$ when differentiated, loses $3n$ in the denominator, with one manipulation, one can get $$f'(x)=frac{1}{x(1-x^3)} $$ using the geometric series sum formula. Since this is $f'(x)$, integration is required so we get $$f(x)=log(x)-frac{1}{3}log(1-x^3) + C$$
(the log base is probably $e$). The usual step for finding the constant is to find it's value for $x=0$ in the given power series and the new function. This obviously won't work here or in any other case where a function isn't defined for $x=0$. How do I find the constant then?






power-series







share|cite|improve this question











$endgroup$




This power series $$f(x)=sum_{n=1}^{infty} {frac{x^{3n}}{3n}}$$ when differentiated, loses $3n$ in the denominator, with one manipulation, one can get $$f'(x)=frac{1}{x(1-x^3)} $$ using the geometric series sum formula. Since this is $f'(x)$, integration is required so we get $$f(x)=log(x)-frac{1}{3}log(1-x^3) + C$$
(the log base is probably $e$). The usual step for finding the constant is to find it's value for $x=0$ in the given power series and the new function. This obviously won't work here or in any other case where a function isn't defined for $x=0$. How do I find the constant then?






power-series




power-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Apr 7 at 19:10









Leucippus

19.8k102871




19.8k102871










asked Apr 7 at 16:49









user3711671user3711671

438




438












  • $begingroup$
    I was talking about log(x) not being defined at x=0.
    $endgroup$
    – user3711671
    Apr 7 at 17:09


















  • $begingroup$
    I was talking about log(x) not being defined at x=0.
    $endgroup$
    – user3711671
    Apr 7 at 17:09
















$begingroup$
I was talking about log(x) not being defined at x=0.
$endgroup$
– user3711671
Apr 7 at 17:09




$begingroup$
I was talking about log(x) not being defined at x=0.
$endgroup$
– user3711671
Apr 7 at 17:09










2 Answers
2






active

oldest

votes


















2












$begingroup$

You're not quite on track. One way to know that for sure is that your function isn't defined at $x=0,$ but the power series certainly is!



For $|t|<1,$ we have that $$sum_{n=0}^infty t^n=frac1{1-t},$$ so $$sum_{n=1}^infty t^n=frac1{1-t}-1=frac{t}{1-t}.$$ Thus, we have for $|x|<1$ that $$f'(x)=sum_{n=1}^infty x^{3n-1}=frac1xsum_{n=1}left(x^3right)^n=frac1xcdotfrac{x^3}{1-x^3}=left(1-x^3right)^{-1}cdot-frac13frac{dleft(1-x^3right)}{dx}.$$ This has the antiderivative family $$f(x)=-frac13lnleft(1-x^3right)+C,$$ which is defined at every $x$ in the radius of convergence, allowing you to find your constant in the usual way.




More generally, let's suppose that you've used a power series $$sum_{n=0}^infty a_n(x-a)^n$$ for some real $a,$ and have (by differentiation and then integration) determined that $$f(x)=C+sum_{n=0}^infty a_n(x-a)^n$$ for some constant $C.$ Even without determining the radius of convergence, if you haven't made any errors, then the function is necessarily defined at $x=a,$ so you can use $x=a$ to determine what your value of $C$ will be. Namely, we will have $C=f(a)-a_0.$






share|cite|improve this answer











$endgroup$





















    4












    $begingroup$

    Differentiating $f(x)$:



    $$f(x)=frac{x^3}{3}+frac{x^6}{6}+cdotsimplies f'(x)=x^2+x^5+cdots=sum_{n=1}^{infty}x^{3n-1}=frac{x^2}{1-x^3}$$ for $|x|<1$, which results in $$f(x)=-frac13ln(1-x^3)+C$$ after integration. Your function now is well-defined at $x=0$, and this implies $C=0$.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Right, I've made a mistake at the geometric series, n starts at 1, not 0.My question still remains, what should one do if plugging in x=0 can't be done (because of functions like log(x), 1/x etc)?The interesting thing is that if log(x) remains untouched, C will equal -log(x).
      $endgroup$
      – user3711671
      Apr 7 at 17:09










    • $begingroup$
      Then plug in another number, e.g. $x=1$. The constant after integration can be "fixed" by moving the graph of the function up or down the $y$-axis. This can be done by using any point; not only $x=0$.
      $endgroup$
      – st.math
      Apr 7 at 17:10












    • $begingroup$
      But what about the infinite sum of 1/3n then?Shouldn't I find out what the constant is by inserting the value for which the sum and it's function representation are equal (when x=0 both are equal to zero)?
      $endgroup$
      – user3711671
      Apr 7 at 17:13












    • $begingroup$
      You can plug in any point where the function is defined. After plugging in $x=0$ into the sum, you get $f(0)=0$. The closed form you obtained after integration ($f(x)=ln(1-x^3)/3+C$) is undetermined by the constant $C$ only, which you can find after plugging in $0$ there, which directly implies $f(0)=0+C$ or $C=0$ (because we know that $f(0)$ has to be $0$ by the sum). You will get the same result $C=0$ after plugging in $x=0.5$, for example.
      $endgroup$
      – st.math
      Apr 7 at 17:15








    • 1




      $begingroup$
      @user3711671: The only kicker is that your value of $x$ must be within the interval of convergence for the power series. The center point is typically the simplest to plug in. I explain a bit further in my answer.
      $endgroup$
      – Cameron Buie
      Apr 7 at 17:28














    Your Answer








    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%2f3178460%2frepresenting-power-series-as-a-function-what-to-do-with-the-constant-after-int%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    You're not quite on track. One way to know that for sure is that your function isn't defined at $x=0,$ but the power series certainly is!



    For $|t|<1,$ we have that $$sum_{n=0}^infty t^n=frac1{1-t},$$ so $$sum_{n=1}^infty t^n=frac1{1-t}-1=frac{t}{1-t}.$$ Thus, we have for $|x|<1$ that $$f'(x)=sum_{n=1}^infty x^{3n-1}=frac1xsum_{n=1}left(x^3right)^n=frac1xcdotfrac{x^3}{1-x^3}=left(1-x^3right)^{-1}cdot-frac13frac{dleft(1-x^3right)}{dx}.$$ This has the antiderivative family $$f(x)=-frac13lnleft(1-x^3right)+C,$$ which is defined at every $x$ in the radius of convergence, allowing you to find your constant in the usual way.




    More generally, let's suppose that you've used a power series $$sum_{n=0}^infty a_n(x-a)^n$$ for some real $a,$ and have (by differentiation and then integration) determined that $$f(x)=C+sum_{n=0}^infty a_n(x-a)^n$$ for some constant $C.$ Even without determining the radius of convergence, if you haven't made any errors, then the function is necessarily defined at $x=a,$ so you can use $x=a$ to determine what your value of $C$ will be. Namely, we will have $C=f(a)-a_0.$






    share|cite|improve this answer











    $endgroup$


















      2












      $begingroup$

      You're not quite on track. One way to know that for sure is that your function isn't defined at $x=0,$ but the power series certainly is!



      For $|t|<1,$ we have that $$sum_{n=0}^infty t^n=frac1{1-t},$$ so $$sum_{n=1}^infty t^n=frac1{1-t}-1=frac{t}{1-t}.$$ Thus, we have for $|x|<1$ that $$f'(x)=sum_{n=1}^infty x^{3n-1}=frac1xsum_{n=1}left(x^3right)^n=frac1xcdotfrac{x^3}{1-x^3}=left(1-x^3right)^{-1}cdot-frac13frac{dleft(1-x^3right)}{dx}.$$ This has the antiderivative family $$f(x)=-frac13lnleft(1-x^3right)+C,$$ which is defined at every $x$ in the radius of convergence, allowing you to find your constant in the usual way.




      More generally, let's suppose that you've used a power series $$sum_{n=0}^infty a_n(x-a)^n$$ for some real $a,$ and have (by differentiation and then integration) determined that $$f(x)=C+sum_{n=0}^infty a_n(x-a)^n$$ for some constant $C.$ Even without determining the radius of convergence, if you haven't made any errors, then the function is necessarily defined at $x=a,$ so you can use $x=a$ to determine what your value of $C$ will be. Namely, we will have $C=f(a)-a_0.$






      share|cite|improve this answer











      $endgroup$
















        2












        2








        2





        $begingroup$

        You're not quite on track. One way to know that for sure is that your function isn't defined at $x=0,$ but the power series certainly is!



        For $|t|<1,$ we have that $$sum_{n=0}^infty t^n=frac1{1-t},$$ so $$sum_{n=1}^infty t^n=frac1{1-t}-1=frac{t}{1-t}.$$ Thus, we have for $|x|<1$ that $$f'(x)=sum_{n=1}^infty x^{3n-1}=frac1xsum_{n=1}left(x^3right)^n=frac1xcdotfrac{x^3}{1-x^3}=left(1-x^3right)^{-1}cdot-frac13frac{dleft(1-x^3right)}{dx}.$$ This has the antiderivative family $$f(x)=-frac13lnleft(1-x^3right)+C,$$ which is defined at every $x$ in the radius of convergence, allowing you to find your constant in the usual way.




        More generally, let's suppose that you've used a power series $$sum_{n=0}^infty a_n(x-a)^n$$ for some real $a,$ and have (by differentiation and then integration) determined that $$f(x)=C+sum_{n=0}^infty a_n(x-a)^n$$ for some constant $C.$ Even without determining the radius of convergence, if you haven't made any errors, then the function is necessarily defined at $x=a,$ so you can use $x=a$ to determine what your value of $C$ will be. Namely, we will have $C=f(a)-a_0.$






        share|cite|improve this answer











        $endgroup$



        You're not quite on track. One way to know that for sure is that your function isn't defined at $x=0,$ but the power series certainly is!



        For $|t|<1,$ we have that $$sum_{n=0}^infty t^n=frac1{1-t},$$ so $$sum_{n=1}^infty t^n=frac1{1-t}-1=frac{t}{1-t}.$$ Thus, we have for $|x|<1$ that $$f'(x)=sum_{n=1}^infty x^{3n-1}=frac1xsum_{n=1}left(x^3right)^n=frac1xcdotfrac{x^3}{1-x^3}=left(1-x^3right)^{-1}cdot-frac13frac{dleft(1-x^3right)}{dx}.$$ This has the antiderivative family $$f(x)=-frac13lnleft(1-x^3right)+C,$$ which is defined at every $x$ in the radius of convergence, allowing you to find your constant in the usual way.




        More generally, let's suppose that you've used a power series $$sum_{n=0}^infty a_n(x-a)^n$$ for some real $a,$ and have (by differentiation and then integration) determined that $$f(x)=C+sum_{n=0}^infty a_n(x-a)^n$$ for some constant $C.$ Even without determining the radius of convergence, if you haven't made any errors, then the function is necessarily defined at $x=a,$ so you can use $x=a$ to determine what your value of $C$ will be. Namely, we will have $C=f(a)-a_0.$







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Apr 8 at 17:58

























        answered Apr 7 at 17:07









        Cameron BuieCameron Buie

        86.8k773161




        86.8k773161























            4












            $begingroup$

            Differentiating $f(x)$:



            $$f(x)=frac{x^3}{3}+frac{x^6}{6}+cdotsimplies f'(x)=x^2+x^5+cdots=sum_{n=1}^{infty}x^{3n-1}=frac{x^2}{1-x^3}$$ for $|x|<1$, which results in $$f(x)=-frac13ln(1-x^3)+C$$ after integration. Your function now is well-defined at $x=0$, and this implies $C=0$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Right, I've made a mistake at the geometric series, n starts at 1, not 0.My question still remains, what should one do if plugging in x=0 can't be done (because of functions like log(x), 1/x etc)?The interesting thing is that if log(x) remains untouched, C will equal -log(x).
              $endgroup$
              – user3711671
              Apr 7 at 17:09










            • $begingroup$
              Then plug in another number, e.g. $x=1$. The constant after integration can be "fixed" by moving the graph of the function up or down the $y$-axis. This can be done by using any point; not only $x=0$.
              $endgroup$
              – st.math
              Apr 7 at 17:10












            • $begingroup$
              But what about the infinite sum of 1/3n then?Shouldn't I find out what the constant is by inserting the value for which the sum and it's function representation are equal (when x=0 both are equal to zero)?
              $endgroup$
              – user3711671
              Apr 7 at 17:13












            • $begingroup$
              You can plug in any point where the function is defined. After plugging in $x=0$ into the sum, you get $f(0)=0$. The closed form you obtained after integration ($f(x)=ln(1-x^3)/3+C$) is undetermined by the constant $C$ only, which you can find after plugging in $0$ there, which directly implies $f(0)=0+C$ or $C=0$ (because we know that $f(0)$ has to be $0$ by the sum). You will get the same result $C=0$ after plugging in $x=0.5$, for example.
              $endgroup$
              – st.math
              Apr 7 at 17:15








            • 1




              $begingroup$
              @user3711671: The only kicker is that your value of $x$ must be within the interval of convergence for the power series. The center point is typically the simplest to plug in. I explain a bit further in my answer.
              $endgroup$
              – Cameron Buie
              Apr 7 at 17:28


















            4












            $begingroup$

            Differentiating $f(x)$:



            $$f(x)=frac{x^3}{3}+frac{x^6}{6}+cdotsimplies f'(x)=x^2+x^5+cdots=sum_{n=1}^{infty}x^{3n-1}=frac{x^2}{1-x^3}$$ for $|x|<1$, which results in $$f(x)=-frac13ln(1-x^3)+C$$ after integration. Your function now is well-defined at $x=0$, and this implies $C=0$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Right, I've made a mistake at the geometric series, n starts at 1, not 0.My question still remains, what should one do if plugging in x=0 can't be done (because of functions like log(x), 1/x etc)?The interesting thing is that if log(x) remains untouched, C will equal -log(x).
              $endgroup$
              – user3711671
              Apr 7 at 17:09










            • $begingroup$
              Then plug in another number, e.g. $x=1$. The constant after integration can be "fixed" by moving the graph of the function up or down the $y$-axis. This can be done by using any point; not only $x=0$.
              $endgroup$
              – st.math
              Apr 7 at 17:10












            • $begingroup$
              But what about the infinite sum of 1/3n then?Shouldn't I find out what the constant is by inserting the value for which the sum and it's function representation are equal (when x=0 both are equal to zero)?
              $endgroup$
              – user3711671
              Apr 7 at 17:13












            • $begingroup$
              You can plug in any point where the function is defined. After plugging in $x=0$ into the sum, you get $f(0)=0$. The closed form you obtained after integration ($f(x)=ln(1-x^3)/3+C$) is undetermined by the constant $C$ only, which you can find after plugging in $0$ there, which directly implies $f(0)=0+C$ or $C=0$ (because we know that $f(0)$ has to be $0$ by the sum). You will get the same result $C=0$ after plugging in $x=0.5$, for example.
              $endgroup$
              – st.math
              Apr 7 at 17:15








            • 1




              $begingroup$
              @user3711671: The only kicker is that your value of $x$ must be within the interval of convergence for the power series. The center point is typically the simplest to plug in. I explain a bit further in my answer.
              $endgroup$
              – Cameron Buie
              Apr 7 at 17:28
















            4












            4








            4





            $begingroup$

            Differentiating $f(x)$:



            $$f(x)=frac{x^3}{3}+frac{x^6}{6}+cdotsimplies f'(x)=x^2+x^5+cdots=sum_{n=1}^{infty}x^{3n-1}=frac{x^2}{1-x^3}$$ for $|x|<1$, which results in $$f(x)=-frac13ln(1-x^3)+C$$ after integration. Your function now is well-defined at $x=0$, and this implies $C=0$.






            share|cite|improve this answer











            $endgroup$



            Differentiating $f(x)$:



            $$f(x)=frac{x^3}{3}+frac{x^6}{6}+cdotsimplies f'(x)=x^2+x^5+cdots=sum_{n=1}^{infty}x^{3n-1}=frac{x^2}{1-x^3}$$ for $|x|<1$, which results in $$f(x)=-frac13ln(1-x^3)+C$$ after integration. Your function now is well-defined at $x=0$, and this implies $C=0$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Apr 7 at 17:04

























            answered Apr 7 at 16:56









            st.mathst.math

            1,228115




            1,228115












            • $begingroup$
              Right, I've made a mistake at the geometric series, n starts at 1, not 0.My question still remains, what should one do if plugging in x=0 can't be done (because of functions like log(x), 1/x etc)?The interesting thing is that if log(x) remains untouched, C will equal -log(x).
              $endgroup$
              – user3711671
              Apr 7 at 17:09










            • $begingroup$
              Then plug in another number, e.g. $x=1$. The constant after integration can be "fixed" by moving the graph of the function up or down the $y$-axis. This can be done by using any point; not only $x=0$.
              $endgroup$
              – st.math
              Apr 7 at 17:10












            • $begingroup$
              But what about the infinite sum of 1/3n then?Shouldn't I find out what the constant is by inserting the value for which the sum and it's function representation are equal (when x=0 both are equal to zero)?
              $endgroup$
              – user3711671
              Apr 7 at 17:13












            • $begingroup$
              You can plug in any point where the function is defined. After plugging in $x=0$ into the sum, you get $f(0)=0$. The closed form you obtained after integration ($f(x)=ln(1-x^3)/3+C$) is undetermined by the constant $C$ only, which you can find after plugging in $0$ there, which directly implies $f(0)=0+C$ or $C=0$ (because we know that $f(0)$ has to be $0$ by the sum). You will get the same result $C=0$ after plugging in $x=0.5$, for example.
              $endgroup$
              – st.math
              Apr 7 at 17:15








            • 1




              $begingroup$
              @user3711671: The only kicker is that your value of $x$ must be within the interval of convergence for the power series. The center point is typically the simplest to plug in. I explain a bit further in my answer.
              $endgroup$
              – Cameron Buie
              Apr 7 at 17:28




















            • $begingroup$
              Right, I've made a mistake at the geometric series, n starts at 1, not 0.My question still remains, what should one do if plugging in x=0 can't be done (because of functions like log(x), 1/x etc)?The interesting thing is that if log(x) remains untouched, C will equal -log(x).
              $endgroup$
              – user3711671
              Apr 7 at 17:09










            • $begingroup$
              Then plug in another number, e.g. $x=1$. The constant after integration can be "fixed" by moving the graph of the function up or down the $y$-axis. This can be done by using any point; not only $x=0$.
              $endgroup$
              – st.math
              Apr 7 at 17:10












            • $begingroup$
              But what about the infinite sum of 1/3n then?Shouldn't I find out what the constant is by inserting the value for which the sum and it's function representation are equal (when x=0 both are equal to zero)?
              $endgroup$
              – user3711671
              Apr 7 at 17:13












            • $begingroup$
              You can plug in any point where the function is defined. After plugging in $x=0$ into the sum, you get $f(0)=0$. The closed form you obtained after integration ($f(x)=ln(1-x^3)/3+C$) is undetermined by the constant $C$ only, which you can find after plugging in $0$ there, which directly implies $f(0)=0+C$ or $C=0$ (because we know that $f(0)$ has to be $0$ by the sum). You will get the same result $C=0$ after plugging in $x=0.5$, for example.
              $endgroup$
              – st.math
              Apr 7 at 17:15








            • 1




              $begingroup$
              @user3711671: The only kicker is that your value of $x$ must be within the interval of convergence for the power series. The center point is typically the simplest to plug in. I explain a bit further in my answer.
              $endgroup$
              – Cameron Buie
              Apr 7 at 17:28


















            $begingroup$
            Right, I've made a mistake at the geometric series, n starts at 1, not 0.My question still remains, what should one do if plugging in x=0 can't be done (because of functions like log(x), 1/x etc)?The interesting thing is that if log(x) remains untouched, C will equal -log(x).
            $endgroup$
            – user3711671
            Apr 7 at 17:09




            $begingroup$
            Right, I've made a mistake at the geometric series, n starts at 1, not 0.My question still remains, what should one do if plugging in x=0 can't be done (because of functions like log(x), 1/x etc)?The interesting thing is that if log(x) remains untouched, C will equal -log(x).
            $endgroup$
            – user3711671
            Apr 7 at 17:09












            $begingroup$
            Then plug in another number, e.g. $x=1$. The constant after integration can be "fixed" by moving the graph of the function up or down the $y$-axis. This can be done by using any point; not only $x=0$.
            $endgroup$
            – st.math
            Apr 7 at 17:10






            $begingroup$
            Then plug in another number, e.g. $x=1$. The constant after integration can be "fixed" by moving the graph of the function up or down the $y$-axis. This can be done by using any point; not only $x=0$.
            $endgroup$
            – st.math
            Apr 7 at 17:10














            $begingroup$
            But what about the infinite sum of 1/3n then?Shouldn't I find out what the constant is by inserting the value for which the sum and it's function representation are equal (when x=0 both are equal to zero)?
            $endgroup$
            – user3711671
            Apr 7 at 17:13






            $begingroup$
            But what about the infinite sum of 1/3n then?Shouldn't I find out what the constant is by inserting the value for which the sum and it's function representation are equal (when x=0 both are equal to zero)?
            $endgroup$
            – user3711671
            Apr 7 at 17:13














            $begingroup$
            You can plug in any point where the function is defined. After plugging in $x=0$ into the sum, you get $f(0)=0$. The closed form you obtained after integration ($f(x)=ln(1-x^3)/3+C$) is undetermined by the constant $C$ only, which you can find after plugging in $0$ there, which directly implies $f(0)=0+C$ or $C=0$ (because we know that $f(0)$ has to be $0$ by the sum). You will get the same result $C=0$ after plugging in $x=0.5$, for example.
            $endgroup$
            – st.math
            Apr 7 at 17:15






            $begingroup$
            You can plug in any point where the function is defined. After plugging in $x=0$ into the sum, you get $f(0)=0$. The closed form you obtained after integration ($f(x)=ln(1-x^3)/3+C$) is undetermined by the constant $C$ only, which you can find after plugging in $0$ there, which directly implies $f(0)=0+C$ or $C=0$ (because we know that $f(0)$ has to be $0$ by the sum). You will get the same result $C=0$ after plugging in $x=0.5$, for example.
            $endgroup$
            – st.math
            Apr 7 at 17:15






            1




            1




            $begingroup$
            @user3711671: The only kicker is that your value of $x$ must be within the interval of convergence for the power series. The center point is typically the simplest to plug in. I explain a bit further in my answer.
            $endgroup$
            – Cameron Buie
            Apr 7 at 17:28






            $begingroup$
            @user3711671: The only kicker is that your value of $x$ must be within the interval of convergence for the power series. The center point is typically the simplest to plug in. I explain a bit further in my answer.
            $endgroup$
            – Cameron Buie
            Apr 7 at 17:28




















            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%2f3178460%2frepresenting-power-series-as-a-function-what-to-do-with-the-constant-after-int%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

            Plaza Victoria

            Image
            Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l
            Read more

            In PowerPoint, is there a keyboard shortcut for bulleted / numbered list?

            Image
            .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty,.everyoneloves__bot-mid-leaderboard:empty{ height:90px;width:728px;box-sizing:border-box; } 2 OneNote has two useful keyboard shortcuts for bulleted and numbered lists, namely Ctrl + . and Ctrl + - . Is there something similar for PowerPoint where the work with lists is also very common? Haven't found anything here: Use keyboard shortcuts in PowerPoint. keyboard-shortcuts microsoft-powerpoint share | improve this question edited Aug 14 '14 at 14:49 Linger 2,826 10 28 40
            Read more

            How to put 3 figures in Latex with 2 figures side by side and 1 below these side by side images but in...

            Image
            1 0 I want to put 3 images in LateX such as 2 figures are side by side horizotally and 3rd figure below these 2 side by side figures but in middle. I have following code: begin{figure}[H] centering begin{minipage}{.5textwidth} centering includegraphics[width=.50linewidth]{Text/Images/Genelec_8010_AP.jpg} captionof{figure}{Genelec 8010 AP} label{fig:Genelec 8010} end{minipage}% begin{minipage}{.5textwidth} centering includegraphics[width=.69linewidth]{Text/Images/Genelec_8020_CPM.jpg} captionof{figure}{Genelec 8020 CPM} label{fig:Genelec 8020} end{minipage}% end{figure} %%Here is my 3rd picture code begin{figure} centering includegraphics[width=5cm]{Text/Images/Genelec_8030_BPM.jpg} caption{Genelec 8030 BPM} label{fig:Genelec 8030} end{figure} I want to ask ho
            Read more