Is there a summation rule for continuous functions?

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

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(x)f(x) and g(x)g(x) be two real functions of real variable defined and continuous in (a,b)(a,b). If x_0 in (a,b)x0(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.