Question

Is there a summation rule for continuous functions?

Is there a summation rule for continuous functions? i.e.: does `sum_(n=1)^"∞" f(x)` = `f(sum_(n=1)^"∞" x)` ?

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?

Answer 1

The sum of two continuous functions is continuous.

Let `f(x)` and `g(x)` be two real functions of real variable defined and continuous in `(a,b)`. If `x_0 in (a,b)`:

`lim_(x->x_0) f(x)+g(x) = lim_(x->x_0) f(x) + lim_(x->x_0)g(x) =f(x_0)+g(x_0)`

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

However let `f_n(x)` with `n in NN` be a sequence of real functions of real variable defined and continuous in `(a,b)` and consider the series:

`sum_(n=0)^oo f_n(x)`

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

`f(x) = sum_(n=0)^oo f_n(x)`

that is the sum of the series. But while every partial sum is defined and continuous in `(a,b)` we cannot say the same for `f(x)` 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(x) = cos^n(x)`

that are defined and continuous for every `x in RR`. The sum:

`f(x) = sum_(n=0)^oo cos^n(x) = 1/(1-cosx)`

that we can find based on the sum of the geometric series is not defined for `x = 2kpi`. (and not really the sum of the series for `x=kpi`).

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

`sum_(n=0)^oo sin((2n+1)t)/(2n+1)`

is defined for every `x` but not continuous for `x=kpi`.

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:

`abs (f_n(x)) <= a_n` for `x in (a,b)`

and that the series:

`sum_(n=0)^oo a_n`

is convergent.