Summing x(1/2)^x: An Explanation

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In summary: Sorry, I didn't mean to go beyond the scope of your question!)In summary, the conversation discusses the summation of the infinite series x(1/2)^x for x=1,2,3,4,... A possible method to find the sum is by multiplying the series by (1/2) and subtracting it from the original series, resulting in the summation of a geometric series in powers of 1/2. The final value of the sum is 2.
  • #1
kuahji
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I have the summation x(1/2)^x for (x=1,2,3,4,...)

So I set it up as s=1(1/2)+2(1/4)+3(1/8)+4(1/16)...

This is however where I'm lost, I'm not exactly sure how to sum an infinite sequence, it hasn't really been introduced in any of my math courses, it just popped up in a statistics problem though.

The book shows the next step as 1/2s=1/4+2(1/8)+3(1/16)...
then
1/2s=1/2+1/4+1/8+1/6+...=1
Thus s=2

But I'm not sure exactly what they are doing in the first & second step there. Any explanations would be helpful.
 
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  • #2
In the second step they just multiplied s by (1/2). In the next step they subtract the second series for (1/2)s from the first series for s. E.g. the 1/8 in the last series comes from 3(1/8) in the first series minus 2(1/8) in the second series. FInally, the summed the geometric series in powers of 1/2.
 
  • #3
kuahji said:
I have the summation x(1/2)^x for (x=1,2,3,4,...)

So I set it up as s=1(1/2)+2(1/4)+3(1/8)+4(1/16)...

This is however where I'm lost, I'm not exactly sure how to sum an infinite sequence, it hasn't really been introduced in any of my math courses, it just popped up in a statistics problem though.

The book shows the next step as 1/2s=1/4+2(1/8)+3(1/16)...
then
1/2s=1/2+1/4+1/8+1/6+...=1
Thus s=2

But I'm not sure exactly what they are doing in the first & second step there. Any explanations would be helpful.
In the first step, assuming you mean (1/2)s and not 1/(2s), they just multiplied everything by 1/2, obviously: (1/2)s= (1/2)(1/2)+(1/2)(2/4)+ (1/2)(3/8)+ (1/2)(4/16)+ (1/2)(5/32)+ (1/2)(6/64)+ ... Now, how they got the next equation is beyond me! I thought at first that they were just adding "pairs" of fractions: 1/4+ 2(1/8)= 1/4+ 1/4= 1/2 obviously, but then 3/16+ 4/32= 3/16+ 2/32= 5/32, NOT 1/4! Perhaps they are adding additional terms but I can't find them.

The result, x= 2 is, however, correct. Here's how I would do it:

(If you don't mind I am going to use "n" for your "x" so I can use x as a variable.)

I notice that the derivative of xn is n xn-1 so the derivative of
(1/2)xn, evaluated at x= 1/2, is (1/2)n (1/2)n-1= n(1/2)n.

The point of that is that [itex]d[(1/2) \sum x^n]/dx][/itex] is [itex]\sum n x^{n-1}[/itex] which, at x= 1/2 is the sum you want.

That helps because [itex]\sum x^n[/itex] is a geometric series and its sum is 1/(1-x) so [itex](1/2) \sum x^n= 1/(2(1-x))[/itex]. The derivative of (1/2)(1-x)-1 is (1/2)(1-x)-2 and so the original sum is (1/2)(1- 1/2)2= 2.
 
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Related to Summing x(1/2)^x: An Explanation

1. What is the formula for summing x(1/2)^x?

The formula for summing x(1/2)^x is ∑ x(1/2)^x = 1/4(3x + 2).

2. How is this formula derived?

This formula is derived using the geometric series formula, ∑ ar^n = a / (1 - r), where a is the first term, r is the common ratio, and n is the number of terms. In this case, a = 1/2, r = 1/2, and n = x. Plugging in these values gives us ∑ (1/2)(1/2)^x = 1/4(1 + 1/2 + 1/4 + ... + (1/2)^x).

3. What is the significance of this sum?

This sum is significant because it represents the area under the curve of the function f(x) = x(1/2)^x. This function is commonly used in mathematics and economics to model various phenomena, and understanding its sum can provide insight into these phenomena.

4. Can this formula be used to find the sum for any value of x?

Yes, this formula can be used to find the sum for any positive integer value of x. However, it is not applicable for non-integer values of x.

5. Are there any real-world applications for this formula?

Yes, this formula has applications in fields such as finance, biology, and computer science. In finance, it can be used to calculate the present value of a series of cash flows. In biology, it can be used to model population growth. In computer science, it can be used to analyze the complexity of algorithms.

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