MacLaurin series, inverse tan of x

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SUMMARY

The discussion focuses on finding the first four terms of the Maclaurin series for the inverse tangent function, arctan(x), using its derivative, which is 1/(1+x^2). Participants clarify that instead of calculating derivatives, one can utilize the geometric series expansion for 1/(1+x^2) and integrate term by term to derive the series for arctan(x). The series expansion for 1/(1+x^2) is identified as 1 - x^2 + x^4 - x^6, leading to the conclusion that integrating this series provides the desired Maclaurin series for arctan(x).

PREREQUISITES
  • Understanding of Maclaurin series and their general form
  • Familiarity with derivatives and their applications in series expansions
  • Knowledge of geometric series and polynomial long division
  • Basic integration techniques for series
NEXT STEPS
  • Study the derivation of Maclaurin series for common functions
  • Learn about geometric series and their convergence properties
  • Practice polynomial long division with rational functions
  • Explore integration techniques for power series
USEFUL FOR

Students and educators in mathematics, particularly those studying calculus and series expansions, as well as anyone looking to deepen their understanding of the inverse tangent function and its applications.

Joza
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I was asked to find dy/dx of inverse tan of x , which is 1/(1+x^2)


Then its says, using that dy/dx ^^^^^ equal to a particular series, find the first 4 terms of inverse tan of x.

I'm confused? What is it asking here??
 
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It says find the 4 first terms of the maclaurin series of the inverse of tan(x). But it wants you to find those terms based only on the knowledge that the derivative of the inverse of tan(x) is 1/(1+x^2) (this means without any further calculations/differentiabtions)

Hint: the nth term of a Maclaurin series is of the form \frac{f^{(n)}(0)}{n!}x^n and the series expansion of f(z)=\frac{1}{1-z} you have somewhere in your notes.
 
But it has given me a series there already, so what do I do with that?
 
the point is that the maclaurin presription for finding the series is impractical since the derivatives are tedious to calculate. the obvious geometric series for 1/(1+x^2) can be integrated term by etrm far easier.
 
Sorry I don't understand...I only have high school maths and I'm struggling a bit with it.
 
Ok Well the series they gave you for 1/(1+x^2) can be done by polynomial long division, and is 1-x^2+x^4-x^6.... And we also know that dy/dx \arctan x = 1/(1+x^2)

So It basically wanted you to substitute in the series for the derivative, ie \frac{dy}{dx} \arctan x = 1 - x^2 + x^4 - x^6 + x^8...

Then it wanted you to realize that the series for arctan x could be achieved by integrating both sides...high school math is all you needed :)
 

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