# Is there a summation rule for continuous functions?

## Is there a summation rule for continuous functions? i.e.: does ${\sum}_{n = 1}^{\text{∞}} f \left(x\right)$ = $f \left({\sum}_{n = 1}^{\text{∞}} x\right)$ ? Obviously, I know that isn't true, but is there some other transformation which applies to all continuous functions, or is there not? If there isn't a rule for all functions f(x), is there any for specifically any of the three primary trig functions?

##### 1 Answer
Jun 10, 2018

The sum of two continuous functions is continuous.

Let $f \left(x\right)$ and $g \left(x\right)$ be two real functions of real variable defined and continuous in $\left(a , b\right)$. If ${x}_{0} \in \left(a , b\right)$:

${\lim}_{x \to {x}_{0}} f \left(x\right) + g \left(x\right) = {\lim}_{x \to {x}_{0}} f \left(x\right) + {\lim}_{x \to {x}_{0}} g \left(x\right) = f \left({x}_{0}\right) + g \left({x}_{0}\right)$

and clearly the result extends to the sum of any finite sum of continuous functions.

However let ${f}_{n} \left(x\right)$ with $n \in \mathbb{N}$ be a sequence of real functions of real variable defined and continuous in $\left(a , b\right)$ and consider the series:

${\sum}_{n = 0}^{\infty} {f}_{n} \left(x\right)$

For the values of $x$ for which the series is convergent we have a new function:

$f \left(x\right) = {\sum}_{n = 0}^{\infty} {f}_{n} \left(x\right)$

that is the sum of the series. But while every partial sum is defined and continuous in $\left(a , b\right)$ we cannot say the same for $f \left(x\right)$ because there can be values of $x$ for which the function is either not defined or not continuous.

For instance consider the functions:

${f}_{n} \left(x\right) = {\cos}^{n} \left(x\right)$

that are defined and continuous for every $x \in \mathbb{R}$. The sum:

$f \left(x\right) = {\sum}_{n = 0}^{\infty} {\cos}^{n} \left(x\right) = \frac{1}{1 - \cos x}$

that we can find based on the sum of the geometric series is not defined for $x = 2 k \pi$. (and not really the sum of the series for $x = k \pi$).

Similarly we can demonstrate based on Fourier analysis that the series:

${\sum}_{n = 0}^{\infty} \sin \frac{\left(2 n + 1\right) t}{2 n + 1}$

is defined for every $x$ but not continuous for $x = k \pi$.

A sufficient condition for the sum of the series to be continuous, is that the series is totally convergent, that is for every $n$ we can find ${a}_{n}$ such that:

$\left\mid {f}_{n} \left(x\right) \right\mid \le {a}_{n}$ for $x \in \left(a , b\right)$

and that the series:

${\sum}_{n = 0}^{\infty} {a}_{n}$

is convergent.