Power Series Representation for Arctan(x)

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SUMMARY

The discussion focuses on deriving the power series representation for the integral of the arctangent function, specifically F(x) = ∫(0 to x) arctan(t) dt. The series expansion for arctan(x) is established as ∑(n=0 to ∞) (-1)^n * x^(2n+1)/(2n+1). The user identifies en as n-1, but struggles with determining the correct values for an and pn. The correct expressions are an = 1/(2n(2n-1)) and pn = 2n, although the user doubts their accuracy despite confirmation from Mathematica.

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  • Understanding of power series and convergence
  • Familiarity with integration techniques for series
  • Knowledge of the arctangent function and its properties
  • Experience with symbolic computation tools like Mathematica
NEXT STEPS
  • Study the derivation of power series for trigonometric functions
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  • Explore the use of Mathematica for symbolic integration and series expansion
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Mathematics students, educators, and anyone interested in advanced calculus, particularly those working with series expansions and integrals of trigonometric functions.

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Homework Statement


F(x)=∫(0 to x) tan^(-1)t dt. f(x)= infinite series ∑n=1 (-1)^(en)(an)x^(pn)?
en=?
an=?
pn=?
I know en = n-1


Homework Equations





The Attempt at a Solution


Start with the geometric series
1/(1 - t) = ∑(n=0 to ∞) t^n.

Let t = -x^2:
1/(1 + x^2) = ∑(n=0 to ∞) (-1)^n * x^(2n).

Integrate both sides from 0 to x:
arctan x = ∑(n=0 to ∞) (-1)^n * x^(2n+1)/(2n+1).

Now that we have a series for arctan x...
f(x) = ∫(0 to x) arctan t dt
= ∫(0 to x) [∑(n=0 to ∞) (-1)^n * t^(2n+1)/(2n+1)] dt
= ∑(n=0 to ∞) (-1)^n * [∫(0 to x) t^(2n+1)/(2n+1) dt]
= ∑(n=0 to ∞) (-1)^n * x^(2n+2)/[(2n+1)(2n+2)].

Shifting indices up one unit, we have
f(x) = ∑(n=1 to ∞) {(-1)^(n-1)/[(2n-1)(2n)]} x^(2n).


This gives me
en = n-1
an=1/(2n(2n-1))
pn = 2n

however an and pn are wrong. where am i going wrong?
 
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Your work looks okay to me, and your answer matches what Mathematica spits out. Why do you think it's wrong?
 

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