# What is the justification for taylor series for functions with one or no critical points?

Some(but not all) smooth functions can be represented by taylor series. And the common justification people give why this is possible(like in this question, and that) is something along this line:

They say consider a function like the sine curve. It looks like a collection of parabolas with infinitely many turning points. A polynomial of degree $n$ can have up to $n-1$ turning points. So that as we increase the degree of the polynomial approximating the sine curve to infinity, its turning points increase and becomes a sine curve.

So after this justification one can say that a sin curve is a polynomial of infinite degree:

$sin(x)=a+bx+cx^2+dx^3+...$ and it follows that by differentiating succesively and solving for the coefficients of the polynomial one gets:

$sin(0)=a,sin'(0)=b,\dfrac{sin"(0) }{2}=c,\dfrac{sin'''(0) }{2*3}=d, ...$

So that after figuring out the coefficients, we can rewrite the sine as:

$sin(x)=sin(0)+sin'(0)x+\dfrac{sin"(0) }{2!}x^2+\dfrac{sin'''(0) }{3!}x^3+...$

Which is the taylor expansion around $x=0$ for the sine curve.

The fact that functions like sine and cosine are representable as infinite series of polynomial is convincing and intuitive, given the fact that they look like infinitely many parabolas next to each other. However I fail to grasp why this should work at all with functions that don't even have critical points, like $e^x$, or a function that has only one critical point like $cosh(x)$.

Why should it be possible that $e^x$ or $cosh(x)$ $=a+bx+cx^2+dx^3+...$ ?

What is the justification that a lot of analytic functions like $e^x$ and $cosh(x)$ are representable as an infinite series of polynomials?

I believe that you're asking the wrong question. The real question is "Why did I actually believe that stuff about sines and cosines, when it's merely intuitive hand-waving?"

Suppose I said "Look at $\cosh x$; it has one "bowl", so maybe a sum of $\cosh$ functions, suitably transformed, should be able to represent the sine function." Would you believe that? It might be true, it might not. Seriously...I have no idea whether it is or isn't. But the notion that "having a certain number of bumps" is all that you need to persuade you that two functions might match up, not just at a countable number of points, but at every point of the (uncountable) real numbers ... well, that makes you seem credulous to me.

(To be fair: I've done exactly this hand-waving myself in front of a class of calculus students, and it's part of what I was told when I learned this stuff. But sometimes age makes you think a little more deeply. )

A more persuasive argument (to me, anyhow) is that with $f(x) = \sin x$, you can get a linear function (namely $y = x$) that looks like the sine function near $x = 0$; you can put a tolerance on that, say "within 10 percent", and find that the fit is good over some interval $[-a, a]$.

Then you can ask "What about higher orders?" and find that $y = x - \frac{x^3}{36}$ fits within 10 percent over an interval $[-b, b]$, where $b > a$, and that the interval of 1% fit grows too, etc. And pretty soon, you can make a good case that you can fit, within any precision you want, over as large an interval as you like (this takes some real work!), by using higher and higher degree polynomials.

What's nice about that argument is that it's based on concrete experience -- you can write down the coeffs and check stuff -- and that it applies equally well to $f(x) = e^x$.

Indeed, a slightly crazy teacher might introduce the function $$f(x) = \begin{cases} 0 & x = 0\\ exp(-x^{-2}) & \text{otherwise} \end{cases}$$ at this point and show that this function cannot be approximated by polynomials, leading the brightest students to ask "Where's the dividing line between "good" functions and bad ones?"

Mostly that doesn't happen, of course, more's the pity.

Let me try to recover something from the sines-and-polynomials argument: you could reasonably ask "if I hoped to represent sine as a polynomial, what degree would the polynomial have to have?" Given the infinite number of local maxima for sine, the answer is "not any finite degree", so your only hope is that sine might equal some infinite-degree polynomial (once you've clearly defined what such a think might mean, using clear definitions of convergence of series, etc.)

That's more or less the opposite of the argument you posed, but it's still solid. And it also applies to $f(x) = e^x$. Suppose $f$ were equal to some $n$th-degree polynomial $p$. Then the first, second, third, etc., derivatives of $f$, at $x = 0$, would have to match those of $p$. But at most $n$ of those derivatives at 0, for $p$, are nonzero, while ALL the derivatives of $f$ at $0$ are equal to $1$. Hence $p$ cannot have any finite degree. So your only hope is that there might be an infinite-degree polynomial matching $f(x) = e^x$.

• Thanks for your very enlightening answer, both for its informative mathematical content and for making me more critical of any proposition that I'm presented with. I'll think more carefully next time. – Omar Nagib Nov 16 '15 at 18:30

As you mention yourself, not every smooth function is representable by a power series (is real analytic), so one has to be very careful with justifications why every function you run across is real analytic, and some of those that you mention are probably bogus. On the other hand the set of real analytic functions is large and useful. For those functions that you mention explicitly, there is indeed good reason, namely that the differential equations that they are defined. As you probably know, the exponential function is characterised by $$\exp'=\exp,\qquad\exp(0)=1.$$ Now if we consider a power series $$f(x)=\sum_{k=0}^\infty a_kx^k$$ then $f'=f$, $f(0)=0$ translates to $$a_k=(k+1)a_{k+1},\qquad a_0=1,$$ which has the unique solution $a_k=\frac1{k!}$. So there you have the exponential function as a power series. (Of course some of these steps have to be justified.) Similarly for $\sin$ and $\cos$, which are characterised by $$\sin'=\cos,\quad\cos'=\sin,\quad\sin(0)=0,\quad\cos(0)=1.$$