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












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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







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  • $begingroup$
    I was talking about log(x) not being defined at x=0.
    $endgroup$
    – user3711671
    yesterday
















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
    yesterday














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






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited yesterday









Leucippus

19.8k102871




19.8k102871










asked yesterday









user3711671user3711671

438




438












  • $begingroup$
    I was talking about log(x) not being defined at x=0.
    $endgroup$
    – user3711671
    yesterday


















  • $begingroup$
    I was talking about log(x) not being defined at x=0.
    $endgroup$
    – user3711671
    yesterday
















$begingroup$
I was talking about log(x) not being defined at x=0.
$endgroup$
– user3711671
yesterday




$begingroup$
I was talking about log(x) not being defined at x=0.
$endgroup$
– user3711671
yesterday










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
      yesterday










    • $begingroup$
      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
      yesterday












    • $begingroup$
      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
      yesterday












    • $begingroup$
      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
      yesterday








    • 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
      yesterday














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    2 Answers
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    active

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    2 Answers
    2






    active

    oldest

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    oldest

    votes






    active

    oldest

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    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 10 hours ago

























        answered yesterday









        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
              yesterday










            • $begingroup$
              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
              yesterday












            • $begingroup$
              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
              yesterday












            • $begingroup$
              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
              yesterday








            • 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
              yesterday


















            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
              yesterday










            • $begingroup$
              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
              yesterday












            • $begingroup$
              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
              yesterday












            • $begingroup$
              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
              yesterday








            • 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
              yesterday
















            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 yesterday

























            answered yesterday









            st.mathst.math

            1,113115




            1,113115












            • $begingroup$
              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
              yesterday










            • $begingroup$
              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
              yesterday












            • $begingroup$
              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
              yesterday












            • $begingroup$
              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
              yesterday








            • 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
              yesterday




















            • $begingroup$
              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
              yesterday










            • $begingroup$
              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
              yesterday












            • $begingroup$
              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
              yesterday












            • $begingroup$
              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
              yesterday








            • 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
              yesterday


















            $begingroup$
            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
            yesterday




            $begingroup$
            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
            yesterday












            $begingroup$
            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
            yesterday






            $begingroup$
            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
            yesterday














            $begingroup$
            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
            yesterday






            $begingroup$
            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
            yesterday














            $begingroup$
            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
            yesterday






            $begingroup$
            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
            yesterday






            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
            yesterday






            $begingroup$
            @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
            yesterday




















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