Is the series expansion of $e^x$, $\sin x$ & $\cos x$ valid for any real $x$?

Question

$$e^x=1+\frac{x}{1!}+\frac{x^2}{2!}+\cdots$$

$$\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots$$

$$\cos x=1-\frac{x^2}{2!}+\frac{x^4}{4!}-\cdots$$

Are the above valid for any real $x$?

Answer 1

Yes, that property locally holds for all analytic function. You can show that this property holds for the functions in question globally as well.

Proof: The radius of convergence of a power series defined by $an$ $$ \sum_{n=0}^{\infty}a_{n}z^{n} $$ is $$ R = \frac{1}{\limsup |a_{n}|^{1/n}} $$ by the ratio test. For the above cases: $$\frac{1}{\limsup|a_n|^{1/n}} = \lim \sqrt[n]{n!} =\infty$$ Note: To see the last equality one can use the Stirling-formula

Above I showed that the power series, which is in our case the Taylor series of the functions in question converges for all real numbers. Now we only have to show that $\lim_{n\to\infty} f(x) - T_n(x) = 0 \forall x\in\mathbb{R}$. To prove this one can use Taylor's theorem: $$\lim_{n\to\infty} f(x) - T_n(x) = \lim_{n\to\infty} \frac{f^{(n+1)}(\xi)}{(n+1)!} x^{n+1}$$

For $f(x) = e^x$:$~~~~~~~~~~~~$ $\lim_{n\to\infty} \frac{f^{(n+1)}(\xi)}{(n+1)!} x^{n+1} = \lim_{n\to\infty} \frac{e^\xi x^{n+1}}{(n+1)!} = 0 (\forall x \in \mathbb{R})$
For $f(x) = sin(x), cos(x)$: $~~~~$ $\lim_{n\to\infty} \left|\frac{f^{(n+1)}(\xi)}{(n+1)!} x^{n+1}\right| \le \lim_{n\to\infty} \frac{x^{n+1}}{(n+1)!} = 0 (\forall x \in \mathbb{R})$

So we can conclude that the Taylor series of the functions in queston are not only convergent but also are convergin to the corresponding function.

Answer 2

As others have pointed out, you can use convergence tests to show that all three series have an infinite radius of convergence; that is, all three series converge for all real $x$. However, that is not enough to conclude that they converge to the functions specified. That requires further work.

The approach taken in most calculus books is to use a formula for the Taylor series remainder. But there is another approach. Let me illustrate the alternative approach in the case of $e^x$. Let $$ f(x) = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots, $$ which is defined for all $x$ because the series has an infinite radius of convergence. By term-by-term differentiation, we have $f'(x) = f(x)$, and therefore $$ \frac{d}{dx}\left(\frac{f(x)}{e^x}\right) = \frac{e^x f(x) - e^x f(x)}{(e^x)^2} = 0. $$ Thus $f(x)/e^x = C$ for some constant $C$, so $f(x) = Ce^x$. To find $C$ we evaluate at $x=0$: $C = Ce^0 = f(0) = 1 + 0 + 0 + \cdots = 1$, $f(x) = e^x$.

Similar arguments can be given for $\sin x$ and $\cos x$ (although it's a little trickier). For details, see my book Calculus: A Rigorous First Course.