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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?

Leucippus
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user3711671
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2 Answers2

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Differentiating $f(x)$:

$$f(x)=\frac{x^3}{3}+\frac{x^6}{6}+\cdots\implies f'(x)=x^2+x^5+\cdots=\sum_{n=1}^{\infty}x^{3n-1}=\frac{x^2}{1-x^3}$$ for $|x|<1$, which results in $$f(x)=-\frac13\ln(1-x^3)+C$$ after integration. Your function now is well-defined at $x=0$, and this implies $C=0$.

log_math
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  • 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). – user3711671 Apr 07 '19 at 17:09
  • 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 $x=0$. – log_math Apr 07 '19 at 17:10
  • 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)? – user3711671 Apr 07 '19 at 17:13
  • 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. – log_math Apr 07 '19 at 17:15
  • How will I find the series value for any $x$ that isn't equal to zero?I will just get an infinite series with unknown sum.When $x=0$ the sum will simply be equal to 0 so there is nothing to do with it. – user3711671 Apr 07 '19 at 17:24
  • Yes, this is why $x=0$ would be the easiest way to determine the constant. This does not change the fact that any other value of $x$ would work as well. – log_math Apr 07 '19 at 17:25
  • That's what I'm saying, but how do I find the series value for any other value of $x$?What is $\sum_{n=1}^{\infty} {\frac{(1/2)^{3n}}{3n}}$ equal to? – user3711671 Apr 07 '19 at 17:28
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    @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. – Cameron Buie Apr 07 '19 at 17:28
  • @user3711671 It is equal to $\frac13\ln(1-0.5^3)$ ;) But why make it so complicated when you can work with $x=0$? – log_math Apr 07 '19 at 17:34
  • How did you find that result? – user3711671 Apr 07 '19 at 17:52
  • I did not actually calculate it, I just plugged it into the already-determined closed form of $f(x)$. (There is a minus sign missing, though; it should be $-\frac13\ln(1-0.5^3)$.) – log_math Apr 07 '19 at 17:54
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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 $$f'(x)=\sum_{n=1}^\infty x^{3n-1}=\frac1x\sum_{n=1}\left(x^3\right)^n=\frac1x\cdot\frac{x^3}{1-x^3}=\left(1-x^3\right)^{-1}\cdot-\frac13\frac{d\left(1-x^3\right)}{dx}.$$ This has the antiderivative family $$f(x)=-\frac13\ln\left(1-x^3\right)+C,$$ 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 $C=f(a)-a_0.$

Cameron Buie
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