Discover the Power of Maclaurin's Series: Solving 1/(1+x^2) for -1 > x > 1

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Discussion Overview

The discussion revolves around deriving the Maclaurin series for the function 1/(1+x^2) within the interval -1 > x > 1. Participants explore different methods to arrive at the series expansion and the underlying principles involved.

Discussion Character

  • Exploratory
  • Technical explanation
  • Mathematical reasoning

Main Points Raised

  • One participant presents the series expansion for 1/(1+x^2) as a sum of terms involving powers of x, questioning how to derive this representation.
  • Another participant identifies the series as a Maclaurin series and outlines the general form of the series expansion, referencing the binomial series and conditions for convergence.
  • A different approach is suggested, where a participant proposes using the geometric series formula and substituting y with -x^2 to derive the same result without delving into the general theory of power series.
  • One participant confirms the validity of the geometric series approach by discussing the relationship between the series and its abbreviated form.

Areas of Agreement / Disagreement

Participants present multiple methods for deriving the series, indicating that there is no single agreed-upon approach. The discussion remains open with various perspectives on how to achieve the series expansion.

Contextual Notes

Some assumptions regarding the convergence of the series and the conditions under which the geometric series applies are not fully explored, leaving potential gaps in understanding.

newton1
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1/(1+x^2)=1-x^2 + x^4-x^6+...+(-1)^n(x^2n)+... -1 > x > 1
how to get this??...
 
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It's basically Maclaurin's series.

f(x) = f(0) + x*f'(0) + (x^2)/2! * f''(0) +...

Which derives the binomial series:
(1+a)^n = 1+ na + n(n-1)/2! * a^2 + n(n-1)(n-2)/3! * a^3 ... As long as |a|<1

Substitute a = x^2 and n = -1 et viola!
 


Originally posted by Newton1
1/(1+x^2)=1-x^2 + x^4-x^6+...+(-1)^n(x^2n)+... -1 > x > 1
how to get this??...

If you do not want to go all the way to the general theory of series of powers (Taylor and MacLaurin series), you can simply use the result of the geometric series

1/(1-y)=1+y+y^2+...+y^n+...

which can be obtained with basic arithmetic arguments and substistute y with -x^2.
 
Yes. Because, if you abbreviate the right hand side as S, then obviously
y*S = S-1
 

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