Finding the Root of an Infinite Series (for example, Sigma(i=0 to n) 1/x^n)

[ted_gress]

Hi,
I was wondering if there is a way to find the root of an infinite series.

For example:

Sigma(i=0 to n) 1/x^n.

Not specifically that equation but in general.

 

[ksdhart2]

Erm... sorry, but I don't understand what you're asking. You mention "infinite series," but the example you show is definitely not infinite. The summand also doesn't involve the index variable, so for any x and n:

\(\displaystyle \displaystyle \sum_{i=1}^n \frac{1}{x^n} = nx^{-n}\)

That aside, it's also unclear what is meant by root in this context. Do you mean in the sense of the root (aka zero) of a polynomial? So you want to solve, for instance:

\(\displaystyle \displaystyle \sum_{i=1}^n \frac{1}{x^{in}} = 0\)

Or do you mean in the sense of a square/cube/whatever root? So you want to find the numeral value of, for instance:

\(\displaystyle \displaystyle \sqrt{\sum_{i=1}^n sin(ix)}\)

 

[ted_gress]

Erm... sorry, but I don't understand what you're asking. You mention "infinite series," but the example you show is definitely not infinite. The summand also doesn't involve the index variable, so for any x and n:

\(\displaystyle \displaystyle \sum_{i=1}^n \frac{1}{x^n} = nx^{-n}\)

That aside, it's also unclear what is meant by root in this context. Do you mean in the sense of the root (aka zero) of a polynomial? So you want to solve, for instance:

\(\displaystyle \displaystyle \sum_{i=1}^n \frac{1}{x^{in}} = 0\)

Or do you mean in the sense of a square/cube/whatever root? So you want to find the numeral value of, for instance:

\(\displaystyle \displaystyle \sqrt{\sum_{i=1}^n sin(ix)}\)

Sorry, bad example. I mean an infinite series and by roots I mean like roots of a polynomial (zeroes)

So:

Sigma (n=0 to infinity) 1/n^k. where k is some constant.
And is there a way I can insert math operators into these posts so its not so cryptic?

 

[HallsofIvy]

Hi,
I was wondering if there is a way to find the root of an infinite series.

For example:

Sigma(i=0 to n) 1/x^n.

Not specifically that equation but in general.

What? the way that is written, since the index is i, not n, it is simply (1/x^n) n_1 times. That is equal to (n+1)/x^n.

IF you intended \(\displaystyle \sum_{n=0}^\infty \frac{1}{x^n}\), that is a "geometric series" and its sum is \(\displaystyle \frac{1}{1- \frac{1}{x}}= \frac{x}{x- 1}\).

 

[ksdhart2]

Sorry, bad example. I mean an infinite series and by roots I mean like roots of a polynomial (zeroes)

So:

Sigma (n=0 to infinity) 1/n^k. where k is some constant.
And is there a way I can insert math operators into these posts so its not so cryptic?

Ah, okay. Well, in that case, in general, finding the zeroes of an infinite series is something that can be fairly easily or extremely difficult. Some series lend themselves easier to the task than others. First, determine if the series converges or not. Obviously if the series diverges for every value of the non-indexed variable, it cannot have any zeroes. Some series may conditionally converge for some values of the non-indexed variable, but not others. And some series may converge for all values of the non-indexed variable. The example you gave only conditionally converges. For n = 1, the series diverges, for n = 2 the series converges to \(\displaystyle \dfrac{\pi^2}{6}\), for n = 3, the series converges to \(\displaystyle \zeta(3) \approx 1.2021\), etc. Unless I'm mistaken, I believe it converges for all real n > 1. An example of a series that converges for all values is \(\displaystyle \displaystyle \sum_{n=1}^{\infty} \frac{(x-n)^6}{n^n}\). A (trivial) example of a series that always diverges is \(\displaystyle \displaystyle \sum_{n=1}^{\infty} 1^x\).

Finding if a series converges or diverges can sometimes be difficult in and of itself. Here is a page from Paul's Online Calculus Notes explaining some techniques for determining the radius of convergence of a series. And here is another page that has some other techniques you might find useful for determining if a series converges or diverges.

Now once you know if a series converges or diverges, you can then begin to find any zeroes. This is easiest if the series has a nice closed form, which you can just use standard techniques to find the zeroes of. For example, the series \(\displaystyle \displaystyle \sum_{n=1}^{\infty} \dfrac{52^n}{n} \cdot (38x - 21)^n\) converges to \(\displaystyle -ln(1093-1976x)\) when \(\displaystyle |38x - 21| < \dfrac{1}{52}\). This function, then, has only one zero, namely when \(\displaystyle 1093 - 1976x = 1 \implies x = \dfrac{21}{38}\). The example you gave, \(\displaystyle \displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n^k} = \zeta(k)\) for all real \(\displaystyle k > 1\). But finding the zeroes of the Riemann Zeta function is... not an easy task.

As for "inserting math," the forum has a built-in LaTeX parser, which makes writing math stuff quite easy. For instance, if you write

[tex]\displaystyle \sum_{n=1}^{\infty} (summand)[/tex]

, it will render as \(\displaystyle \displaystyle \sum_{n=1}^{\infty} (summand)\). Some other basic functions you'll find handy: To write a fraction, use

[tex]\dfrac{numerator}{denominator}[/tex]

to display \(\displaystyle \dfrac{numerator}{denominator}\). Most Greek letters can be represented by simply typing a slash and then the name of the letter. So, \alpha gives \(\displaystyle \alpha\), \omega gives \(\displaystyle \omega\), \Omega (note the capital O, this definitely matters) gives \(\displaystyle \Omega\), and so on.

 

[ted_gress]

Actually what I'm looking for is how to find solutions for functions that are defined by series. So just finding the radius or interval isn't enough. And yes, the Riemann zeta function would be an example of a harder one. I'm looking for something a little bit more simpler.