What form does a Taylor series general term need to be in? Why don't my general term equation give me the first term sometimes?

Question

So this is written in my book for Taylor series:

Most importantly, I notice that the exponent after $(x-a)$, and the factorial, and the n that currently being iterated over are all equal.

To demonstrate my confusion, let's look at the Taylor series for $ln(1+x)$ and $xe^{-2x}$.

so:

$$f'(x) = \frac{1}{x+1},\quad f''(x) = \frac{-1}{(x+1)^2},\quad f'''(x) = \frac{2}{(x+1)^3},\quad f^{(4)}(x) = \frac{-6}{(x+1)^4}$$ and

$$f'(0) = 1, \ f''(0) = -2,\ f'''(0) = 2, \ f^{(4)}(0) = -6 $$

so notice the general term starts at $n = 1$ but the exponent to x is only n. Is this a problem?

so $$ \ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}(n-1)!}{n!} x^n = x - \frac{x^2}{2} + \frac{x^3}{3}$$

So the general term here seems to accurate represent the Taylor series even though the n term doesn't match up with the exponent term for x.

However, $xe^{-2x}$ is different:

so finding the Taylor series form: we know that $e^x$ is: $\sum_{n=0}^{\infty} \frac{x^n}{n!}$. So $$e^{2x} = \sum_{n=0}^{\infty} \frac{(2x)^n}{n!}, \ e^{-2x} = \sum_{n=0}^{\infty} \frac{(-2x)^n}{n!} = \sum_{n=0}^{\infty} \frac{(-2)^nx^n}{n!},\ xe^{-2x} = \sum_{n=0}^{\infty} \frac{(-2)^nx^{n+1}}{n!}$$

So according to the formula, the first term is just f(a) which should equal the general formula when I plug in n = 0 right? But theres a mismatch. When I plug in n = 0, I get $x$. But f(0) = 0. So what gives? What am I doing wrong here?

Answer 1

Let me repeat your question in a concise way.

Let $f(x)=\ln (1+x)$ and $a=0$. On the one hand, formula (6) you quoted form the book says that the $0$-th term of the Taylor series should be $f(a)=f(0)=\ln(1+0)=0$. On the other hand, one has $$ \ln(1+x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}(n-1)!}{n!} x^n = x - \frac{x^2}{2} + \frac{x^3}{3}+\cdots\tag{1} $$ The term $x$ in (1) does not match the $0$-th term in the formula (6), what is going wrong?

It is not that there is mismatch but that you match the formulas in a wrong way. The equality in (1) could be written as

$$ \ln(1+x) = \color{blue}{ 0 +} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}(n-1)!}{n!} x^n = \color{blue} {0+}x - \frac{x^2}{2} + \frac{x^3}{3}+\cdots\tag{2} $$

And (2) "matches" formula (6): $$ f(0)=0\;,\ f'(0)=(-1)^{\color{red}{1}-1}(\color{red}{1}-1)!=1\;,\\ f''(0)=(-1)^{\color{red}{2}-1}(\color{red}{2}-1)!=-1\;,\\ f'''(0)=(-1)^{\color{red}{3}-1}(\color{red}{3}-1)!=2!\;,\\ \vdots $$

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Now, let us look at another Taylor series you mentioned in your post: $$ g(x)=xe^{-2x} = \sum_{n=0}^{\infty} \frac{(-2)^nx^{n+1}}{n!}\tag{3} $$ Again, what confuses you is that $g(0)=0$, which should be the $0$-th term in the Taylor series, but for $n=0$, the term in (3) is $x$.

The problem is that the $n$ in (3) is NOT the same as the $n$ in (6)!!

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In general, if $$ \sum_{n=0}^\infty a_n=\sum_{n=0}^\infty b_n, $$ one can not conclude that $a_n=b_n$.

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Exercise. Find the Taylor series for the function $f(x)=x^3$ at $a=0$ using formula (6) and see how your result "matches" the "general formula".