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Math Help - inverse tan x power series

  1. #1
    Member jacs's Avatar
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    inverse tan x power series

    For the power series of inverse tan (x) the geometric series is:
    1 - t ^2 + t^ 4 - t^ 6 +...+ t^( 4n)

    find the sum and hence prove that for 0 < t < x:
    1/(1+t2) < 1 - t^2 + t^ 4 - t^ 6 +...+ t^(4n)


    I havent got a clue how to approach this.
    thanks
    jacs
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by jacs
    For the power series of inverse tan (x) the geometric series is:
    1 - t ^2 + t^ 4 - t^ 6 +...+ t^( 4n)
    I'm a bit puzzled by the reference to the series for the inverse tan
    function here. It seems unrelated to the problem.

    Just a mo - it seems that this series is the derivative (term-by-term)
    of the Gregory series:

    <br />
\arctan(x)=x-x^3/3+x^5/5- .. +(-1)^n x^{2n+1}/(2n+1) ..<br />

    which may be what it wants you to use.

    RonL
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  3. #3
    Member jacs's Avatar
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    I am more than a bit puzzled myself, in fact, have no idea. But that is what the question said. I have got only as far as:
    a = 1 and n = 2n + 1
    <br />
r = -t^2<br />
    <br />
Sn = ((-t^2)^{2n+1} - 1)/(-t^2 - 1)<br />

    but i am not sure about my n = 2n + 1 and even if that is right, i have no idea how to proceed from there to get the proof.
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by CaptainBlack
    I'm a bit puzzled by the reference to the series for the inverse tan
    function here. It seems unrelated to the problem.

    Just a mo - it seems that this series is the derivative (term-by-term)
    of the Gregory series:

    <br />
\arctan(x)=x-x^3/3+x^5/5- .. +(-1)^n x^{2n+1}/(2n+1) ..<br />

    which may be what it wants you to use.

    RonL
    Start with:

    <br />
\arctan(x)=x-x^3/3+x^5/5- .. +(-1)^n x^{2n+1}/(2n+1) ..<br />

    then differentiate to get:

    <br />
\frac{d}{dx}\arctan(x)=\frac{1}{1+x^2}=-x^2+x^4- .. +(-1)^n x^{2n} ..<br />
.

    Now if we truncate the series after a positive term the remainder is again
    a geometric series like what we had originally but multipled by -1 times the
    last term included in the sum, and the sum of the series comprising the
    remainder is negative. Which will prove the result (after a bit of jiggery-pokery).


    (or use the remainder form for taylor series, that should also give the
    result)

    We also need the restriction that |x|<1.


    RonL
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  5. #5
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    There is a much simpler way. If you used a finite geometric expansion and then take the limit it will hold for x=\pm 1 but you will have a remainder term. It can be shown very easily that the remainder term approaches zero because,
    \int_0^t \frac{x^n}{1+x^2} dx<\int_0^t \frac{x}{1+x^2}
    Since the dominating series converges to zero so too the remainder term.
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by ThePerfectHacker
    There is a much simpler way. If you used a finite geometric expansion and then take the limit it will hold for x=\pm 1 but you will have a remainder term. It can be shown very easily that the remainder term approaches zero because,
    \int_0^t \frac{x^n}{1+x^2} dx<\int_0^t \frac{x}{1+x^2}
    Since the dominating series converges to zero so too the remainder term.
    Except the question apparently wants the student to use the series for
    arctan.

    RonL
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