10
$\begingroup$

I was told by a fellow student that sometimes one cannot represent certain functions by a taylor series. I was also told that sometimes using a taylor series in a proof is invalid. Is any of this true? When is it invalid to use taylor series expansion?

Edit: By certain functions, I mean well behaved functions with nice properties, entire, countinuous, etc.

$\endgroup$
  • $\begingroup$ I meant continuous functions sorry. $\endgroup$ – mtheorylord Nov 7 '16 at 1:57
  • $\begingroup$ Continuous functions can not have a Taylor series as well. $\endgroup$ – user223391 Nov 7 '16 at 1:57
  • 4
    $\begingroup$ No, actually, he's quite right. Functions that can be represented as Taylor Series are called analytic. Note $\text{Functions} \supset \text{Real Functions} \supset \text{Continuous Functions} \supset \text{Differentiable Functions} \supset \text{Infinitely Differentiable Functions} \supset \text{Analytic Functions}$ $\endgroup$ – MathematicsStudent1122 Nov 7 '16 at 1:58
  • 3
    $\begingroup$ Do you really mean entire? An entire function. is by definition representable by Taylor series in the whole complex plane. $\endgroup$ – Robert Israel Nov 7 '16 at 2:00
  • 11
    $\begingroup$ A more moderate approach would be to assume that he/she is mistaken. $\endgroup$ – user259242 Nov 7 '16 at 2:14
11
$\begingroup$

Take a look at $$f(x)=\begin{cases}e^{-1/x^2}&x\ne0\\0&x=0\end{cases}$$ and take the Taylor expansion around $x=0$

$$f(x)=f(0)+f'(0)x+\frac{f''(0)}2x^2+\dots$$

Now, you will find out that

$$f(0)=0$$

$$f'(0)=0$$

$$f''(0)=0$$

$$etc.$$

So be applying Taylor's theorem here, one has

$$e^{-1/x^2}=0$$

which is nonsense.


Then, there is a second case. As Ethan Alwaise mentions, any series expansion makes no sense if it doesn't converge. Take, for example, the expansion of $\frac1{1-r}$ at $r=0$. Then consider that expansion for $r=2$. You should get something along the following lines:

$$\frac1{1-2}=1+2+4+8+\dots+2^n+\dots$$

which is also nonsense.

$\endgroup$
  • 3
    $\begingroup$ A function can have a Taylor Series, even if the Taylor Series does not converge to the function. So, technically, $0$ is the Taylor Series for $f(x)=e^{-1/x^2}$ for $x\ne 0$ and $f(0)=0$, although the series does not converge to $f(x)$ for $x\ne 0$.. $\endgroup$ – Mark Viola Nov 7 '16 at 2:22
  • 7
    $\begingroup$ Interestingly, $1+2+4+\ldots$ does converge to $-1$ in the 2-adic numbers. $\endgroup$ – user76284 Nov 7 '16 at 5:06
  • 4
    $\begingroup$ "Applying Taylor's theorem here, one has $e^{-1/x^2}=0$ which is nonsense." -- Taylor's theorem doesn't state that. After all, if it "returned nonsense" in some cases, it wouldn't be a theorem... $\endgroup$ – Federico Poloni Nov 7 '16 at 10:36
  • $\begingroup$ @Dr.MV Yes, but you wouldn't say that $e^{-1/x^2}$ is representable by the Taylor series as the OP's question asks. $\endgroup$ – Simply Beautiful Art Nov 7 '16 at 14:25
  • $\begingroup$ Correct. $f(x)$ has a Taylor series, but the series does not converge to $f(x)$. In fact, the remainder of the Taylor expansion of $f(x)$, in this case, is $f(x)$. $\endgroup$ – Mark Viola Nov 7 '16 at 15:38
9
$\begingroup$

If by represents a function by its Taylor series you mean equal to its Taylor series, then your friend is right.

Indeed - the Taylor series at $0$ of $f(x)=\left\{\begin{array}{ll}e^{-1/x}\textrm{, if }x>0\\0\textrm{, otherwise}\end{array}\right.$ is zero but $f$ is nonzero. Howeover, $f$ is $\mathcal{C}^{\infty}$.

You are looking for analytic functions.

$\endgroup$
6
$\begingroup$

He is right, many functions don't have a Taylor series. For example $f(x)=\begin{cases}1 \text{ for } x\in \Bbb{Q}\\0\text{ else}\end{cases}$ is discontinuous everywhere so not differentiable everywhere and thus doesn't have a Taylor series.

$\endgroup$
5
$\begingroup$

To add to Zachary's answer, a function may have a Taylor series that is valid only within a restricted domain. For example, the function $f(x) = 1/(1 - x)$ is defined on $\mathbb{R} \setminus \{1\}$. It has a Taylor series $$f(x) = \sum_{n=0}^{\infty}x^n$$ on the interval $x \in (-1,1)$, but for $\vert x \vert \geq 1$, the above series diverges.

$\endgroup$
  • 2
    $\begingroup$ There are now four answers and they won't always appear in the same order. Thus, "the above answer" is unclear. Perhaps, you could refer to it with a link - like this. $\endgroup$ – Mark McClure Nov 7 '16 at 2:20
2
$\begingroup$

Let $N$ be an integer greater than zero and let $f$ be a $C^{N+1}$ function defined on an interval $[a,b]$ which contains zero. By the fundamental theorem of calculus, $$ f(x) = f(0) + \int_0^x f'(t) dt $$ Integrating by parts (choose $u = f'(t)$, $v = (t-x)$) we obtain $$ \begin{align} \int_0^x f'(t)dt &= \left. (t-x) f'(t) \right|_0^x - \int_0^x (t-x) f''(t) dt \\ &= x f'(0) - \int_0^x (t-x) f''(t) dt. \end{align} $$ Therefore $$ f(x) = f(0) + xf'(0) - \int_0^x (t-x) f''(t) dt. $$ We can continue inductively integrating the last term by parts to obtain $$ f(x) = \sum_{n=0}^N f^{(n)}(0) \frac{x^n}{n!} + R_N(x), $$ where the remainder term $R_N(x)$ is defined by $$ R_N(x) = \frac{(-1)^N}{N!} \int_0^x (t-x)^N f^{(N+1)}(t) dt. $$ This gives an exact computation of the error in the $N$th order Taylor approximation.

For a counter-example such as $f(x) = e^{-1/x}$ for $x > 0$ and $0$ otherwise, it is easy to see that for any $x > 0$ the remainder terms $R_N(x)$ diverge to infinity, and therefore the Taylor series does not converge to $f(x)$, even though the series itself converges.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.