# 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?

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?

• 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. – Andrew Li Apr 24 at 4:19
• do my answers look right? – Jwan622 Apr 24 at 5:26
• What do you mean by "plug in $n=0$? You should "plug in $a=0$". – Jack Apr 24 at 12:24
• 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? – NoChance Apr 24 at 14:51
• @Jack I plug in n = 0 because that's what $t_0$ is which should = f(a) right? – 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".

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