# polynomial approximations put into sigma notation (just for fun)

We're doing polynomial approximations in my Calc 2 class and because I've been told by virtually every person who has taken calc 2 before me that Taylor series is the most difficult part of the class, I've been trying to put as many of the Taylor polynomials I get into sigma notation. It's pretty fun actually! It blows my mind that you can put these long series into a short compact form that gives you any term you want so I've been doing quite a few today.

I'm wondering how you'd do $f(x) = \ln(2+x)$, however.

I calculated the following derivatives: \begin{align*} f^{(1)}(x) &= \frac{1}{(2+x)}\\ f^{(2)}(x) &= \frac{-1}{(2+x)^2}\\ f^{(3)}(x) &= \frac{4+2x}{(2+x)^3} \end{align*} and then solved for $f^*(0)$ for each derivative, which gave me the values: \begin{align*} f(0) &= \ln(2)\\ f^{(1)}(0) &= \frac{1}{2}\\ f^{(2)}(0) &= -\frac{1}{4}\\ f^{(3)}(0) &= \frac{1}{4} \end{align*} So with these Taylor coefficients, I can finally get the Taylor polynomial approximation (for $n=3$ in this case): $$y = \ln(2) + \frac{x}{2} - \frac{x^2}{8} + \frac{x^3}{24}$$

Now, how do I put this into sigma notation? I can see a pattern with the numbers and the alternating signs, but I don't see how I'd possibly get a $\ln(2)$ to pop out as the first term.

• Did you notice that $$f^{(3)}(x) = \frac{4+2x}{(2+x)^4} = \frac{2(2+x)}{(2+x)^4} = 2(2+x)^{-3}\quad ?$$ Oct 16, 2011 at 2:35
You'll just have to handle the constant term specially. Since we know that $\log(x+1)$ has a nice Taylor series, we can write $$\log(2+x)=\log\left(2(\frac{x}{2} + 1)\right)=\log 2 + \log\left(\frac{x}{2}+1\right) =\log2+\sum_{k=1}^\infty \frac{(-1)^{k+1}}{k2^k}x^k$$ So one shouldn't expect the Taylor series to match up nicely going from the constant term to the rest.
You seem to be using the quotient rule to get the successive derivatives; this is obscuring what is going on here. Note that if $g(x) = f'(x) = \frac{1}{x+2} = (x+2)^{-1}$, after that you can just use the Chain Rule and you have: \begin{align*} g(x) &= (x+2)^{-1}\\ g'(x) &= -(x+2)^{-2}\\ g''(x) &= (-1)(-2)(x+2)^{-3}\\ g^{(3)}(x) &= (-1)(-2)(-3)(x+2)^{-4}\\ g^{(4)}(x) &= (-1)(-2)(-3)(-4)(x+2)^{-5}\\ &\vdots\\ g^{(n)}(x) &= (-1)^{n}(n!)(x+2)^{-n-1}\\ &\vdots \end{align*} and after a few steps one can see the general formula for $g^{(n)}(x)$, hence for $f^{(n+1)}(x)$, hence for $f^{(n+1)}(0)$.
That said, you can't really make the first term fit (that is, the constant term $f(0)=\ln(2)$). This is often the case: the first (or first few) terms don't quite follow an obvious rule, but after a while the terms "settle down" into a nice pattern (not always the case: sometimes we can't spot a pattern in any case). So we just write the first few terms separately, and then the rest. So here $$y = \ln(2) + \sum_{n=0}^{\infty}\left(\frac{(-1)^nn!}{2^{n+1}}\cdot\frac{x^{n+1}}{(n+1)!}\right) = \ln(2) + \sum_{n=1}^{\infty}\frac{(-1)^{n-1}}{n2^n}x^n.$$