- #1

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[tex]f(x) = \sum_{i=1}^{x-1} s^i[/tex]

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

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- Thread starter Yann
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- #1

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[tex]f(x) = \sum_{i=1}^{x-1} s^i[/tex]

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

- #2

JasonRox

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

- #3

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Maybe something along the lines of the zeta function?

- #4

JasonRox

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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:

[tex]f(x) = \sum_{i=1}^{n} 0x[/tex]

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.

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- #5

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

Well, it IS possible, I found the equation;

[tex]h(x) = \frac{s^x-1}{s-1}-1[/tex]

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

HallsofIvy

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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

(It's missing the first i= 0 term which is why you need to subtract 1)

- #7

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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[/tex]

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.

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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