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Thread: [SOLVED] Equation involving arctan

  1. #1
    Senior Member Spec's Avatar
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    [SOLVED] Equation involving arctan

    Show that $\displaystyle \forall x \in R$, that

    $\displaystyle arctan(x+1) - arctanx = arctan\frac{1}{x^2 + x + 1}$

    tan is a continuous function, so (can anyone tell me why that is important when doing these kinda proofs?)

    $\displaystyle tan(arctan(x+1) - arctanx) = tan\left(arctan\frac{1}{x^2 + x + 1}\right)$

    $\displaystyle u = arctan(x+1), v = arctanx \implies$

    $\displaystyle tan(u-v) = \frac{tanu - tanv}{1 + tanu \cdot tanv} = tan\left(arctan\frac{1}{x^2 + x + 1}\right)$

    Now all I need to prove is that

    $\displaystyle \frac{-\pi}{2} < arctan(x+1) - arctanx < \frac{\pi}{2}, \forall x \in R$

    Suggestions?



    (How do you write the symbol for real numbers in latex by the way?)
    Last edited by Spec; Jan 25th 2008 at 06:50 AM.
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  2. #2
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    Hello, Spec!

    You started off okay . . .


    Show that $\displaystyle \forall x \in R$, that:

    . . $\displaystyle \arctan(x+1) - \arctan(x) \:= \:\arctan\left(\frac{1}{x^2 + x + 1}\right)$
    Let $\displaystyle \theta \:=\:\arctan(x+1) - \arctan(x)$

    Take the tangent:

    . . $\displaystyle \tan\theta \;=\;\tan\left[\arctan(x+1) - \arctan(x)\right]$

    . . . . . .$\displaystyle =\;\frac{\tan[\arctan(x+1)] - \tan[\arctan(x)]}{1 + \tan[\arctan(x+1)]\tan[\arctan(x)]}$

    . . . . . .$\displaystyle = \;\frac{x + 1 - x}{1 + (x+1)x}$

    . . . . . .$\displaystyle =\;\frac{1}{x^2+x+1}$


    We have: .$\displaystyle \tan\theta \;=\;\frac{1}{x^2+x+1}$

    . . Hence: .$\displaystyle \theta \;=\;\arctan\left(\frac{1}{x^2+x+1}\right) $


    Therefore: .$\displaystyle \arctan(x+1) - \arctan(x) \;=\;\arctan\left(\frac{1}{x^2+x+1}\right) $

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  3. #3
    Senior Member Spec's Avatar
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    That much I've already figured out, but according to my math professor, I also need to be able to show that

    $\displaystyle \frac{-\pi}{2} < arctan(x+1) - arctanx < \frac{\pi}{2}, \forall x \in R$
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  4. #4
    Super Member wingless's Avatar
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    Quote Originally Posted by Spec View Post
    $\displaystyle \frac{-\pi}{2} < arctan(x+1) - arctanx < \frac{\pi}{2}, \forall x \in R$
    $\displaystyle \frac{d}{dx}\text{Arctan}(x+1) - \text{Arctan}x = 0$

    $\displaystyle \frac{1}{(x+1)^2+1} - \frac{1}{x^2+1} = 0$

    $\displaystyle \frac{1}{(x+1)^2+1} = \frac{1}{x^2+1}$

    $\displaystyle (\text{ For }(x+1)^2+1 \neq 0, x^2+1 \neq 0)\text{ }$

    $\displaystyle (x+1)^2+1 = x^2+1$
    ... (some easy work)
    $\displaystyle x = -\frac{1}{2}$
    (I leave you the proof of this point is a maxima)

    Now we know that this function has only one extrema. But we must still check it for global maximums/minimums.

    $\displaystyle \lim _{x\to +\infty }\text{Arctan}(x+1)-\text{Arctan}(x)=0$

    $\displaystyle \lim _{x\to -\infty }\text{Arctan}(x+1)-\text{Arctan}(x)=0$

    So, the functions largest value is $\displaystyle -\frac{1}{2}$ and least value is $\displaystyle 0$.
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  5. #5
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    Quote Originally Posted by Spec View Post
    That much I've already figured out, but according to my math professor, I also need to be able to show that

    $\displaystyle \frac{-\pi}{2} < arctan(x+1) - arctanx < \frac{\pi}{2}, \forall x \in R$
    Are you allowed to use $\displaystyle \arctan(x+1) - \arctan(x) \;=\;\arctan\left(\frac{1}{x^2+x+1}\right) \, $ ?

    Note that $\displaystyle \frac{3}{4} \leq x^2 + x + 1 = \left( x + \frac{1}{2} \right)^2 + \frac{3}{4} < \infty$

    Therefore $\displaystyle 0 < \frac{1}{x^2+x+1} \leq \frac{4}{3}$.

    What does that tell you ......?
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