Arctangent

Discussion in 'Math' started by Alberto, Aug 19, 2009.

  1. Alberto

    Thread Starter Active Member

    Nov 7, 2008
    169
    36
    DELETED! End of my partecipation to AAC forum.
     
    Last edited: Jan 10, 2011
  2. rspuzio

    Active Member

    Jan 19, 2009
    77
    0
    To make life simpler, we can first restrict the range
    to the case where the value lies between 0 and 1 by
    using the identities

      \arctan  (-x) =    \arctan  (-x)

    and

     \arctan (x) = {\pi \over 2} - <br />
 \arctan \left( {1 \over x} \right) .

    Note that these formulae, and the ones to follow all
    give the arc tangent in radians so, if you want degrees,
    you will have to convert the answer.

    If we only need limited accuracy, a rational approximation
    is the way to go. The following formula is accurate to
    two decimal places:

     \arctan x \approx {x \over 1 + 0.28 x^2}

    Converting to degrees, we have the following:

     \arctan x ({\rm degrees}) \approx {57 x \over 1 <br />
+ 0.28 x^2} ,

    which gives the answer to within a degree.

    If we want five-place accuracy, there is the following
    polynomial approximation:

     \arctan x \approx 0.999,866,0 x - 0.330,299,5 x^3 +<br />
0.180,141,0 x^5 - 0.085,133,0 x^7 + 0.020,835,1 x^9

    This should be good enough for all but the most demanding
    applications but, if we want arbitrary precision, we could use
    the following series expansion which has nice convergence
    properties throughout the range of x:

     \arctan x = {x \over 1 + x^2} \.<br />
\left( 1 + {2 \over 3} \cdot {x^2 \over 1 + x^2} <br />
 + {2 \cdot 4 \over 3 \cdot 5} \cdot<br />
     \left( {x^2 \over 1 + x^2} \right)^2<br />
 + {2 \cdot 4 \cdot 6 \over 3 \cdot 5 \cdot 7} \cdot<br />
     \left( {x^2 \over 1 + x^2} \right)^3 <br />
+ \cdots \right)

    Also, for the purpose of numerical computation, this
    series folds up quite nicely:

     \arctan x = {x \over 1 + x^2}<br />
\left( 1 + {2 \over 3} \cdot {x^2 \over 1 + x^2}<br />
 \left( 1 + {4 \over 5} \cdot {x^2 \over 1 + x^2}<br />
  \left( 1 + {6 \over 7} \cdot {x^2 \over 1 + x^2}<br />
   \left( 1 + \cdots \right) \right) \right) \right)

    Because the terms in this series are all positive, it
    has the "Price is Right" property --- it will estimate
    the value as closely as you want without ever going over.

    Let's illustrate this by computing the arc tangent of -3.
    Using our identities, we see that

     \arctan (-3) = - \arctan (3) =<br />
\arctan \left( {1 \over 3} \right)- {\pi \over 2}

    so we'll first compute the arc tangent of a third, then
    go back and turn that into the arctangent of minus three.
    (See the other thread for a debate over whether I should
    have said "negative three" instead. :eek:)

    By our rough-and-dirty approximation:

      \arctan \left( {1 \over 3} \right) \approx<br />
{0.33 \over 1 + 0.28 (0.33)^2} \approx 0.32

    or, in degrees,

      \arctan \left( {1 \over 3} \right) \approx<br />
{57 \cdot 0.33 \over 1 + 0.28 (0.33)^2} <br />
\approx 19^{\circ} .

    Using the polynomial approximation,

     \begin{align}<br />
\arctan \left( {1 \over 3} \right) &\approx <br />
0.999,866,0 (0.333,333,3) <br />
- 0.330,299,5 (0.333,333,3)^3 <br />
+ 0.180,141,0 (0.333,333,3)^5 \\<br />
&- 0.085,133,0 (0.333,333,3)^7 <br />
+ 0.020,835,1 (0.333,333,3)^9 \\<br />
&\approx 0.321,758,8 \end{align} .

    Finally, the series becomes

      \arctan \left( {1 \over 3} \right) =<br />
{3 \over 10} \left(<br />
  1<br />
  + {2 \over 3} \cdot {1 \over 10}<br />
  + {8 \over 15} \cdot \left( {1 \over 10} \right)^2<br />
  + {48 \over 105} \cdot \left( {1 \over 10} \right)^3<br />
   + {384 \over 945} \cdot \left( {1 \over 10} \right)^4<br />
  + {3840 \over 10395} \cdot \left( {1 \over 10} \right)^5<br />
  + {46080 \over 135135} \cdot \left( {1 \over 10} \right)^6<br />
  + \cdots \right)

    As we see from the powers of a tenth, we will gain at least a
    decimal place of accuracy for each extra term we include.
    Computing the successive terms numerically, we get the
    following approximations.

    0.300,000,000
    0.320,000,000
    0.321,600,000
    0.321,737,142
    0.321,749,333
    0.321,750,441
    0.321,750,543
    . . . . . . .

    So we're good to seven decimal places so far and have our
    best value so far,  \arctan (1/3) =  0.321,750,5 \cdots .
    In particular, comparing with this, we see that the polynomial
    approximation is good to five decimal places, as advertised.
    Subtracting the pi, we get

     \arctan (-3) = \arctan (1/3) - \pi/2 = -1.249,045,7 \cdots

    Converting into degrees,

     \arctan (1/3) = 18.434,948^{\circ}
     \arctan (-3) = -71.565,051^{\circ}

    or, in the good old Babylonian notation,
     \arctan (1/3) = 18^{\circ} 26' 6''
     \arctan (-3) = -71^{\circ} 33' 54'' .

    Alright, lesson over, now go compute some angles :)
     
    Last edited: Aug 19, 2009
  3. KL7AJ

    AAC Fanatic!

    Nov 4, 2008
    2,040
    287
    Smart@$$ !! :D


    Eric