Power series of arctan'x

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  • #1
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how could i expand something such as arctan'x into a power series. also how would you be able to find the power series for it?


so far i have managed to work out that:

arctan'x = [itex] \frac{1}{1 + x^2} [/itex]

[itex]\frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 +...+ (- 1)^n x^{2n}[/itex]

how do you work out the radius of convergence though: i know it is : |x|< 1.. but how do you work it out please?
 
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Answers and Replies

  • #2
dextercioby
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Integrate the series you wrote term by term. Watch out for the first line you wrote. You're missing a derivative operator acting on the "arctan" function.
 
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  • #3
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just to clear confusion.. i did mean the derivate of arctanx .. i.e. d/dx arctan x , hence arctan'x....

how would i show the radius of convergence as |x|<1 though please?

to work it out i tried it on
[itex](-1)^n x^{2n}[/itex]
i ended up with

[itex]a_{n+1} / a_{n} = \frac{|x|^{2n + 2}}{|x|^{2n}} = |x|^2/1 [/itex] as n tends to infinity... ...

so radius of convergence is |x|< 1...

is this working out correct?
 
  • #5
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the way my book (Stewart) does it is they say that since it's a geometric series the series will be convergent when |-x^2n|<1 = x^2<1=|x|<1
 
  • #6
D H
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That's the ratio test at work. The alternating series test also works here.
 
  • #7
Gib Z
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Another way to check would have been to see where the expression [itex]\frac{1}{1+x^2} = 1 - x^2 + x^4 - x^6 +...+ (- 1)^n x^{2n}[/itex] is valid, since that is the basis for the new power series. We can see that the expression fails for values of x larger than 1. Really, its just a tiny variation of what DH and dex said :(
 
  • #8
HallsofIvy
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In general, a power series will converge as long as has no reason not too!

[tex]\frac{1}{1+x^2}[/tex] is defined for all complex x except i or -i. The radius of the "disk" of convergence in the complex numbers is 1 so, restricting to the real numbers, the radius of the interval of convergence is also 1.

Of course, you can look at it as a geometric series: it is of the form arn with a= 1, r= -x2: its sum is [tex]\frac{1}{1+x^2}[/tex] and it converges as long as |-x2|< 1 or |x|< 1.

Similarly, the ratio test gives the same result: |x|< 1.

Oh, and the root test: [itex]^n\sqrt{a_n}= |x|< 1[/itex] as well.

I think we have determined that the radius of convergence is 1!
 
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