What is the Inverse Series Method for Calculating Output Fractions?

In summary, the conversation discusses the harmonic series and how it diverges, but also provides a well-known asymptotic approximation for its values. The conversation also presents a method for finding the number of terms needed to approximate a certain output and discusses the relation between the harmonic series and the natural logarithmic function. Lastly, the conversation shares a list of approximations for the difference between the sum of two integers and the concept of a ratio of total fractions approaching e.
  • #1
shintashi
117
1
hi,

im trying to find out how to get the inverse of the series 1/1+1/2+1/3+1/4...
i know the output of 30 is approximately 3.99 and 450 is around 6.6, but I'd like to be able to find the fraction range for outputs of numbers such as 10, 20, and 72. The online series calculators tend to die around output 14. My knowledge of integrals and series is weak at best.
 
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  • #2
Well the series you stated is known as the Harmonic series, and is famously known to diverge because if it did not, it would imply the number of elements in the set of the Prime numbers is finite.

It seems like even though it diverges, you want to approximate its values for a certain amount of terms, and I can help you there. There is a well known asymptotic approximation for the Harmonic series, which diverges as well, but even though it does, its a good approximation because it diverges at a similar rate.

Basically [tex]\lim_{n\to\infty} \frac{ \sum_{k=1}^{n} (1/k) }{\log_e (n) +\gamma} = 1[/tex] where lower case gamma denotes the Euler-Mascheroni constant, which from memory is about 0.577 though there are better numerical approximations.

What that means is that as n becomes larger, [itex]\log_e n + \gamma[/itex] will approximate the n-th partial sum of the harmonic series will less and less error. So say I wanted to know how many terms I would need for an output of 10, an approximate solution can be obtained by setting [itex]\log_e n + \gamma = 10[/itex], so n is approximately 12,370.

As a side note, we can see why the asymptotic expression for the harmonic series is a very plausible thing if we plot y=1/x on the Cartesian plane. Looking at the region for values of x onwards, we know the area bounded by x=1, x=k, y=1/x and y=0 (the area under the curve between x=1 and k) is given by [itex]\log_e k[/itex]. Now draw rectangles, first one between x=1 and 2, with height of 1, then between x=2 and 3, height of 1/2, and general term; rectangle between x=a and a+1, height of 1/a.

We can see that the difference in the area of the rectangles and the area under the curve is only a finite amount, the rectangles approximate the curve better and better. Now it is merely a numerical computation to evaluate the finite amount of difference.
 
  • #3
My interpretation of "inverse" was the other way around: For what n is the series equal to (or larger than) M? That problem only makes sense because the series diverges and so is, for some n, larger than or equal to any M. Unforunately, I also do not see any reasonable way of finding that.
 
  • #4
HallsofIvy said:
My interpretation of "inverse" was the other way around: For what n is the series equal to (or larger than) M? That problem only makes sense because the series diverges and so is, for some n, larger than or equal to any M. Unforunately, I also do not see any reasonable way of finding that.

I believe I did answer that =] In fact I evaluated an example, M=10, then n= 12, 370.
 
  • #5
Here is a simple approximation.

Consider [tex]f(x) = \frac{1}{x}[/tex] on [tex][1,\infty)[/tex] this is a decreasing,continous,positive function. So that means, [tex]\sum_{k=1}^{n-1}\frac{1}{k} \geq \int_1^n \frac{dx}{x} \implies \sum_{k=1}^n \frac{1}{k} \geq \log n + \frac{1}{n}[/tex] and [tex]\sum_{k=2}^n \frac{1}{k} \leq \int_1^n \frac{dx}{x} \implies \sum_{k=1}^n \frac{1}{k} \leq \log n + 1[/tex]. Together these inequalities say [tex]\log n + \frac{1}{n} \leq \sum_{k=1}^n \frac{1}{k} \leq \log n + 1[/tex]
 
  • #6
Update: I fiddled with the series for a while and discovered (probably to no one's surprise) that the difference between the sum of any two integers such as ~11 and ~12, or between ~15 and ~16 is a ratio of total fractions approaching e (2.7182818...)

Here's a small list of the final "k"
(sigma[a,b,1/k] a = first number, b = last number, equation sums all ranges a to b, thus +1/1, +1/2, +1/3… etc. Example: sigma[1,11,1/k] = 3.0198… Rounds down to 3.)

note: I am using ~ for "approximately equals)

1=1
2~4
3~11
4~31
5~83
.
.
.
227
616
1,674
4,550
12,367
33,617
91,380
248,397
675,215
1, 835,426
4,989,205
13,562,065
36,865,516
100,210,865
272,401,372
 

1. What is the Inverse Series method?

The Inverse Series method is a mathematical technique used to find the inverse of a given series or function. It involves finding a new series that yields the original series when multiplied by the original series in a specific way.

2. How is the Inverse Series method different from other methods of finding inverses?

The Inverse Series method is different because it does not require the original series to be a one-to-one function, as long as it is analytic within a certain region. This makes it a more versatile approach compared to other methods such as the inverse function theorem.

3. What are the applications of the Inverse Series method?

The Inverse Series method has various applications in mathematics, physics, and engineering. It is commonly used in solving differential equations, finding the inverse of power series, and in signal processing.

4. Are there any limitations to the Inverse Series method?

Yes, the Inverse Series method may not always yield a convergent inverse series. This can happen when the original series has a radius of convergence that is smaller than the radius of convergence of the inverse series. In such cases, the method may not be applicable.

5. How is the Inverse Series method implemented in practice?

The Inverse Series method involves a series of mathematical calculations and may require the use of computer algorithms. It is often implemented using numerical methods and approximation techniques to find the inverse series with a desired level of accuracy.

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