Smoothness of power series

## Smoothness of power series
            ---
            ## Are power series continuous?
            Polynomials are continuous functions, and power series look like "infinite polynomials", so maybe they are?
            But they can only be continuous within their open interval of convergence.

Let's try: Suppose $y\in(-R,R)$.

$\displaystyle \lim\_{x\to y} \sum\_{n=1}^\infty c_n x^n \overset{\text{?}}= \sum\_{n=1}^\infty c_n \lim\_{x\to y}(x^n) = \sum\_{n=1}^\infty c_n y^n$, using continuity of the $n$<sup>th</sup> power function.
            
            ---
            ## Are power series continuous?
            Polynomials are continuous functions, and power series look like "infinite polynomials", so maybe they are?
            But they can only be continuous within their open interval of convergence.

Let's try: Suppose $y\in(-R,R)$.

But really $\displaystyle \sum\_{n=1}^\infty c_n x^n = \lim\_{k\to\infty} \sum\_{n=1}^k c_n x^n$,
            so we have done

$$
            \lim\_{x\to y} \sum\_{n=1}^\infty c_n x^n = \lim\_{x\to y} \lim\_{k\to\infty} \sum\_{n=1}^k c_n x^n \overset{\text{?}}= \lim\_{k\to\infty} \lim\_{x\to y} \sum\_{n=1}^k c_n x^n = \lim\_{k\to\infty} \sum\_{n=1}^k c_n \lim\_{x\to y} x^n = \sum\_{n=1}^\infty c_n \lim\_{x\to y}(x^n)
            $$

an exchange of limits.
            
            ---
            ## Are power series continuous?
            Polynomials are continuous functions, and power series look like "infinite polynomials", so maybe they are?
            But they can only be continuous within their open interval of convergence.

Let's try: Suppose $y\in(-R,R)$.

But really $\displaystyle \sum\_{n=1}^\infty c_n x^n = \lim\_{k\to\infty} \sum\_{n=1}^k c_n x^n$,
            so we have done

an exchange of limits.

And we know we can't always exchange limits.
            
            ---
            ## Are power series continuous?
            Polynomials are continuous functions, and power series look like "infinite polynomials", so maybe they are?
            But they can only be continuous within their open interval of convergence.

#### Theorem
            Let $\displaystyle f(x) = \sum\_{n=0}^\infty c_n(x-a)^n$ a power series with radius of convergence $R$.

Then $f$ is differentiable on $(a-R,a+R)$ and
            $$
              f'(x) = \sum\_{n=1}^\infty nc_n(x-a)\^{n-1} = \sum\_{n=0}^\infty (n+1)c\_{n+1}(x-a)^n.
            $$

Proof:
            Uses uniform continuity; see MATH2350.

#### Conclusion
            So yes, power series $f$ is continuous, and even differentiable!
            
            ---
            ## Term by term differentiation of a power series
            #### Theorem
            Let $\displaystyle f(x) = \sum\_{n=0}^\infty c_n(x-a)^n$ a power series with radius of convergence $R$.

Then $f$ is differentiable on $(a-R,a+R)$ and
            $$
              f'(x) = \sum\_{n=1}^\infty nc_n(x-a)\^{n-1} = \sum\_{n=0}^\infty (n+1)c\_{n+1}(x-a)^n.
            $$

#### Example
            If $\displaystyle f(x) = \sum\_{n=2}^\infty \frac{(x-5)^n}{3^nn^2}$ on $[2,8]$, then $\displaystyle f'(x) = \sum\_{n=2}^\infty \frac{(x-5)^{n-1}}{3^nn}$ on $(2,8)$.

The theorem does not tell us whether the power series converges or diverges at $x=2$ or $x=8$.
            But certainly it does not converge to $f'(2)$ or $f'(8)$ because they do not exist.
            
            ---
            ## Term by term differentiation of a power series
            #### Theorem
            Let $\displaystyle f(x) = \sum\_{n=0}^\infty c_n(x-a)^n$ a power series with radius of convergence $R$.

Then $f$ is differentiable on $(a-R,a+R)$ and
            $$
              f'(x) = \sum\_{n=1}^\infty nc_n(x-a)\^{n-1} = \sum\_{n=0}^\infty (n+1)c\_{n+1}(x-a)^n.
            $$

#### Example
            Find the radius of convergence and derivative of $\displaystyle f(x) = \sum\_{n=0}^\infty \frac{x^n}{n!}$.

Recall: $0!=1$ by definition.
            
            ---
            ## Term by term differentiation of a power series
            #### Example
            Find the radius of convergence and derivative of $\displaystyle f(x) = \sum\_{n=0}^\infty \frac{x^n}{n!}$.

Recall: $0!=1$ by definition.
            #### Solution
            $\displaystyle \left\lvert\frac{x\^{n+1}/(n+1)!}{x^n/n!}\right\rvert = \frac{\lvert x \rvert}{n+1} \underset{n\to\infty}\longrightarrow 0$, for all $x\in\mathbb R$.
            So, by the ratio test, $f(x)$ converges for all $x\in(-\infty,\infty)$.
            In other words, the radius of convergence is $R=\infty$.
            
            ---
            ## Term by term differentiation of a power series
            #### Example
            Find the radius of convergence and derivative of $\displaystyle f(x) = \sum\_{n=0}^\infty \frac{x^n}{n!}$.

Therefore, by the differentiation of power series theorem,

$$
              f'(x) = \sum\_{n=1}^\infty \frac{x\^{n-1}n}{n!} = \sum\_{n=1}^\infty \frac{x\^{n-1}}{(n-1)!} = \sum\_{n=0}^\infty \frac{x^n}{n!} = f(x) \qquad \text{on }(-\infty,\infty).
            $$

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            ## Term by term differentiation of a power series
            #### Example
            Find the radius of convergence and derivative of $\displaystyle f(x) = \sum\_{n=0}^\infty \frac{x^n}{n!}$.

Recall: $0!=1$ by definition.
            #### Solution

We can calculate $f(0)= \frac1{0!} = 1$.
            
            ---
            ## Term by term differentiation of a power series
            #### Example
            Find the radius of convergence and derivative of $\displaystyle f(x) = \sum\_{n=0}^\infty \frac{x^n}{n!}$.

Recall: $0!=1$ by definition.
            #### Solution

We can calculate $f(0)= \frac1{0!} = 1$.

Therefore $f(x) = \mathrm{e}^x$.
            
            ---
            ## Term by term differentiation of a power series
            #### Example
            Find the radius of convergence and derivative of $\displaystyle f(x) = \sum\_{n=0}^\infty \frac{x^n}{n!}$.

Recall: $0!=1$ by definition.
            #### Solution

We can calculate $f(0)= \frac1{0!} = 1$.

Therefore $f(x) = \mathrm{e}^x$.

In fact this is actually a good *definition* of $\mathrm{e}^x$, because otherwise what does $\mathrm{e}^\pi$ mean?