Algorithms for infinite geometric series via long division?

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SUMMARY

This discussion focuses on algorithms for deriving infinite geometric series from functions of the form f(x) = a/(x+b) using long division. The series can be expressed as Σ (a/b)(-x/b)^n, valid under the condition |x| < |b|. Participants clarify that while the series can be derived directly, long division offers an alternative method for obtaining specific terms, such as the 23rd term, which is calculated as -ax^23/b^24. The conversation highlights the confusion surrounding the long division process in this context.

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  • Familiarity with geometric series and their properties
  • Basic knowledge of long division in algebra
  • Experience with calculus concepts, particularly series expansions
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AdkinsJr
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Algorithms for infinite geometric series via long division??

I can't seem to find any algorithms for this on the internet easily.

If I have a function of the form [tex]f(x)=\frac{a}{x+b}[/tex] there should be an algorithm I can use to find some terms of the corresponding series [tex]\sum _{n=0}^{\infty}\frac{a}{b}\left(\frac{-x}{b}\right)^n[/tex]

I can't seem to comprehend how to carry out the division for something like that; obviously it's not absolutely necessary if you know how to find the series using long division, but saw this worked out before and couldn't make sense of it. How does it work?
 
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AdkinsJr said:
I can't seem to find any algorithms for this on the internet easily.

If I have a function of the form [tex]f(x)=\frac{a}{x+b}[/tex] there should be an algorithm I can use to find some terms of the corresponding series [tex]\sum _{n=0}^{\infty}\frac{a}{b}\left(\frac{-x}{b}\right)^n[/tex]

I can't seem to comprehend how to carry out the division for something like that; obviously it's not absolutely necessary if you know how to find the series using long division, but saw this worked out before and couldn't make sense of it. How does it work?

I'm not sure I understand what you want: you already have the infinite series for your function, at least when $$\,\left|\frac{x}{b}\right|<1\Longleftrightarrow |x|<|b|$$
What else do you want?? You want the 23rd term, for ex.? Piece of cake: it is $$\,-\frac{a}{b}\frac{x^{23}}{b^{23}}=-\frac{ax^{23}}{b^{24}}$$

DonAntonio
 


DonAntonio said:
I'm not sure I understand what you want: you already have the infinite series for your function, at least when $$\,\left|\frac{x}{b}\right|<1\Longleftrightarrow |x|<|b|$$
What else do you want?? You want the 23rd term, for ex.? Piece of cake: it is $$\,-\frac{a}{b}\frac{x^{23}}{b^{23}}=-\frac{ax^{23}}{b^{24}}$$

DonAntonio

Yes, I understand how to obtain terms from the infinite series using that theorem/ sum formula. HOwever, there is an alternative procedure using long division.

My question arises from this:



Around 1:00 there's an expression under the integrand containing [tex]\frac{1}{1-\frac{x^2}{4}+\frac{5x^4}{192}-\frac{7x^6}{4608}+...}[/tex]

You can see he inverts it in the next integral but I can't follow the long division that would be involved in that.

I thought I'd ask about a simple case of a geometric series from [tex]\frac{a}{x+b}[/tex] since it's similar in principle with the higher order terms in the denominator. I just find the long division confusing and I can't work it out.

I also thought it was interesting because I read in my calc text that you can find the geometric series using long division, but I don't understand the procedure.
 
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