# Making a continuous equation out of a summation

I have an equation;

$$f(x) = \sum_{i=1}^{x-1} s^i$$

Where s is a constant. Is it possible to transform f(x) into continuous functions ? If so, how ?

JasonRox
Homework Helper
Gold Member
Nope, how would you evaluate the function when x = 2.3?

Maybe something along the lines of the zeta function?

JasonRox
Homework Helper
Gold Member
Maybe something along the lines of the zeta function?

He's using x as a bound to the summation. Not actually in the summation.

If you want to change the question to putting x in the equation, there are much much much easier ways to construct the continuous function. I tend to avoid making things harder than they really are.

Here is the easiest example:

$$f(x) = \sum_{i=1}^{n} 0x$$

If you want to get more complicated than that, then I would go into Power Series and Taylor Series before heaving off go into the Zeta Functions.

Last edited:
Nope, how would you evaluate the function when x = 2.3?

Well, it IS possible, I found the equation;

$$h(x) = \frac{s^x-1}{s-1}-1$$

It's exacly the same as the summation, but it's continuous. However, how could you get h(x) from f(x), I'm sure trial and error isn't the only way to do this...

HallsofIvy
$$f(x)= \sum_{1}^{x-1} s^i[/itex] you mean "sum over integers i as long as i<= x-1", then that is (almost) a geometric and so gives the formula you give as long as x is an integer. (It's missing the first i= 0 term which is why you need to subtract 1) He's using x as a bound to the summation. Not actually in the summation. If you want to change the question to putting x in the equation, there are much much much easier ways to construct the continuous function. I tend to avoid making things harder than they really are. Here is the easiest example: [tex]f(x) = \sum_{i=1}^{n} 0x$$