So this is written in my book for Taylor series:

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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 $xe^{-2x}$.


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

  • 2
    $\begingroup$ You shouldn't be concerned about matching. Simplifying by multiplying out terms will affect the index of summation and how it's expressed in summation notation, but how a Taylor Series is derived stays the same. You're getting too worked up by the notation. $\endgroup$ – Andrew Li Apr 24 at 4:19
  • $\begingroup$ do my answers look right? $\endgroup$ – Jwan622 Apr 24 at 5:26
  • $\begingroup$ What do you mean by "plug in $n=0$? You should "plug in $a=0$". $\endgroup$ – Jack Apr 24 at 12:24
  • $\begingroup$ Note that $f''(0) = -1$ not $-2$. Also, your question is hard to read really. Could you please make it simpler and very focused? You know the general term...so what exactly is the question? $\endgroup$ – NoChance Apr 24 at 14:51
  • $\begingroup$ @Jack I plug in n = 0 because that's what $t_0$ is which should = f(a) right? $\endgroup$ – Jwan622 Apr 25 at 4:48

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

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)!!

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

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".

  • $\begingroup$ Very neat presentation indeed. Thank you. $\endgroup$ – NoChance Apr 25 at 18:03

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