Making a continuous equation out of a summation

In summary, there is a discussion about transforming the equation f(x) = \sum_{i=1}^{x-1} s^i into a continuous function. It is mentioned that using x as a bound to the summation can make things more complicated and there are easier ways to construct the continuous function. One suggested example is the function f(x) = \sum_{i=1}^{n} 0x. Another suggestion is to use Power Series and Taylor Series before delving into Zeta Functions. However, Yann mentions finding an equation h(x) = \frac{s^x-1}{s-1}-1, which is not exactly the same as the summation but is equal to it at integer
  • #1
Yann
48
0
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 ?
 
Mathematics news on Phys.org
  • #2
Nope, how would you evaluate the function when x = 2.3?
 
  • #3
Maybe something along the lines of the zeta function?
 
  • #4
Werg22 said:
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.
 
Last edited:
  • #5
JasonRox said:
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
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
JasonRox said:
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?
 

1. How do you convert a summation into a continuous equation?

To convert a summation into a continuous equation, you can use the formula for the summation of a series, which is ∑(aᵢ) = (1/2)n(a₁+aₙ), where aᵢ is the i-th term in the series, n is the number of terms, and a₁ and aₙ are the first and last terms respectively. This formula allows you to replace the summation symbol with the continuous sigma (∑) symbol in the equation.

2. What is the purpose of making a continuous equation out of a summation?

The purpose of converting a summation into a continuous equation is to make it easier to work with and to represent a larger number of terms in a more compact form. This can be especially useful in mathematical and scientific calculations where large numbers of terms are involved.

3. Can you apply this method to any type of summation?

Yes, this method can be applied to any type of summation as long as the series is finite and the terms follow a pattern or can be expressed algebraically. It may not be applicable to infinite series or series with non-algebraic terms.

4. How do you handle the limits of the summation when converting to a continuous equation?

The limits of the summation become the limits of integration in the continuous equation. The lower limit of the summation becomes the lower limit of integration, and the upper limit of the summation becomes the upper limit of integration. This allows you to evaluate the continuous equation over the same range of values as the original summation.

5. Are there any drawbacks to using a continuous equation instead of a summation?

One potential drawback is that a continuous equation may not accurately represent the behavior of the individual terms in the series. It is also important to note that a continuous equation is an approximation and may not give an exact result, especially in cases where the series has a large number of terms or the terms are not well-defined. In these cases, it may be necessary to use the actual summation to get a more precise result.

Similar threads

  • General Math
Replies
6
Views
839
  • General Math
Replies
1
Views
761
  • General Math
Replies
7
Views
1K
  • General Math
Replies
5
Views
947
Replies
3
Views
234
  • General Math
Replies
11
Views
1K
Replies
7
Views
821
  • General Math
Replies
5
Views
930
  • General Math
Replies
9
Views
1K
  • General Math
Replies
5
Views
1K
Back
Top