Power series proof without Taylor

Question

After some research I have found out that in order to find the power series expansion of a function I can apply Taylor's formula. In order to prove Taylor or Maclaurin formulas I need to assume power series as valid. Eventually I cannot find a proof to understand why some functions such as $exp(x)$ or $sin x$ can be expanded using power series, one that will not assume Taylor as valid, while Taylor uses power series. My mind is struggling in a circular argument. I've also checked similar questions here and they use the same technique. Any help?

PS: I found out in Tom Apostol- Calculus, Vol I, that the proof is linked to a circle of absolute convergence, but it is not detailed

PS2: To clarify, I am looking for a proof for power series. Why can some functions be written as power series, and if possible, one that does not make use of Taylor's expansion.

Answer 1

Your concern is a good one.

First, a word about terminology: If a function $f(x)$ is infinitely differentiable at some value $a$ (that is, if the $n$-th derivative of $f(x)$ at $x=a$ exists and is finite for all natural numbers $n$) then the Taylor series for $f(x)$ at $x=a$ is always formally valid. A real question is whether the Taylor series converges at various values of $(x-a)$, and this may be what you meant to ask about.

For example, the Taylor series about $x=0$ for $f(x)=\frac1{1-x}$ does not converge at $x=2$.

Another way for the Taylor series to be "invalid" is if the series converges, but to a value other than $f(x-a)$. For example, the Taylor series for $e^{-1/x^2}$ about $x=0$ converges for all values of $x$, but is identically the zero polynomial, which does not match $f(x)$ at any non-zero value of $x$.

Now for your practical problem: To show the series for $e^x$ converges at any arbitrary value of $x$.

Take $n = 1 + \lfloor |2x| \rfloor$. Then the series terms up to the $n$-th term obviously converge; they are just a finite polynomial. The $n$-th term is some real number $r = x^n/n!$, and because of the factorial in the denominator, each subsequent term is (in absolute value) at most half of its predecessor. Therefore, the sum of the absolute values of the terms is less than or equal to $r(1+\frac12+\frac14\ldots)=2r$. So the series converges.

Answer 2

I am not sure how detailed or how rigorous you are looking for.

Taylor series don't always converge, or may only converge withing finite intervals. But if the Taylor series converges, you want to prove that Taylor polynomial converges to your function.

Suppose $T_n(x)$ is the $n$ degree Taylor approximation of your function.

Then $R_n(x) = f(x) - T_n(x)$ is the residual (or the error in the approximation.)

We want to show that as $n$ gets large $R_n(x)$ and ultimately where we need to show $\forall \epsilon, \forall x\in I, \exists N : n>N\implies |R_n(x) |<\epsilon$

So we need a formula for $R_n$

$R_n(x) = \frac {1}{n!} \int_a^x (x-t)^n f^{(n+1)}(t) dt$

If you want to prove that this is indeed the the formula for the residual, I am going to leave that to you. You can prove it by induction.

But, if $f^{(n+1)}(x)$ exists and is bounded.

$|f^{(n+1)}(x)| <M\\ |R_n(x)| < |\frac {(x-a)^{n+1}}{n!} |M$

There is an $N$ such that $n>N\implies |R_n(x)| < \epsilon$

Answer 3

Take the function $f$ defined by

$f(x)=e^{\frac{-1}{x^2}}$ if $x\neq 0$ and $f(0)=0$.

for each integer $n\geq 0 ,\; f$ has a Taylor expansion of order $n$ but it has not a power series expansion.

we can write $$f(x)=0+x^n\epsilon(x)$$ but not $$f(x)=0+0+0+....$$