# Making a continuous equation out of a summation

1. Mar 25, 2007

### Yann

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 ?

2. Mar 25, 2007

### JasonRox

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

3. Mar 25, 2007

### Werg22

Maybe something along the lines of the zeta function?

4. Mar 25, 2007

### JasonRox

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: Mar 25, 2007
5. Mar 25, 2007

### Yann

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...

6. Mar 25, 2007

### HallsofIvy

Staff Emeritus
No, it is NOT "exactly the same as the summation". It is the same when x is an integer but different when x is not an integer.

If by
[tex]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)

7. Mar 25, 2007

### Werg22

That's not what I meant. I was talking about a function that is equal to this sum at integers x. I see, that's what Yann found. The function assumes the values of the summation at integers values. Is that what you meant to find Yann?

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