Making a continuous equation out of a summation

  • Thread starter Yann
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I have an equation;

[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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  • #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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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
1,425
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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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