Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: John Machin's Formula

  1. Dec 28, 2004 #1
    I'm a bit puzzled by this problem, which involves the John Machin's formula. Here it goes:

    Show that for [tex] xy \neq -1 [/tex],

    [tex] \arctan x - \arctan y = \arctan \frac{x-y}{1+xy} [/tex]

    ïf the left side lies between [tex] -\pi /2 [/tex] and [tex] \pi /2 [/tex].

    By the way, John Machin's formula is:

    [tex] 4 \arctan \frac{1}{5} - \arctan \frac{1}{239} = \arctan \frac{\pi}{4} [/tex]

    He used this to find [tex] \pi [/tex] correct to 100 decimal places.

    Anyhow, here is what I've done so far:

    If

    [tex] \arctan x = \sum _{n=0} ^{\infty} \left( -1 \right) ^n \frac{x^{2n+1}}{2n+1} [/tex]

    [tex] \arctan y = \sum _{n=0} ^{\infty} \left( -1 \right) ^n \frac{y^{2n+1}}{2n+1} [/tex]

    [tex] \arctan \frac{x-y}{1+xy} = \sum _{n=0} ^{\infty} \left( -1 \right) ^n \frac{\left( \frac{x-y}{1+xy} \right)^{2n+1}}{2n+1} [/tex]

    Then

    [tex] \sum _{n=0} ^{\infty} \left( -1 \right) ^n \frac{\left( x^{2n+1} - y^{2n+1}\right)}{2n+1} = \sum _{n=0} ^{\infty} \left( -1 \right) ^n \frac{\left( \frac{x-y}{1+xy} \right)^{2n+1}}{2n+1} [/tex]

    Therefore

    [tex] x^{2n+1} - y^{2n+1} = \left( \frac{x-y}{1+xy} \right)^{2n+1} [/tex]

    I'm not sure I am on the right path. Maybe, I simply need to pick a couple of values to demonstrate that the given relationship is true, but it feels like it might not be enough. In other words, "show" up there looks a bit ambiguous. Any help is highly appreciated.

    Thanks
     
  2. jcsd
  3. Dec 28, 2004 #2

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    My friend,why use calculus to get nowhere,when u can use (circular) trigonometry to prove in an elegant way?????????
    Take "tan"out both sides of your identity and use what i'll write below:
    [tex] \tan(a-b)=:\frac{\sin(a-b)}{\cos(a-b)}=\frac{\sin a\cos b-\sin b\cos a}{\cos a\cos b+\sin a\sin b} [/tex]

    Simplify the last fraction through the product of 'cosines' to find the celebrated formula
    [tex] \tan(a-b)=\frac{\tan a-\tan b}{1+\tan a\tan b} [/tex]
    Substitute in the last identity:
    [tex] a\rightarrow \arctan x;b\rightarrow \arctan y [/tex]
    ,make use of the fact that "tan",on the interval give in the problem is a uniform function and get exactly the formula u would get if taking "tan" from the identity u need to prove.

    Daniel.

    PS.Trigonometry is beauty... :approve:
     
  4. Dec 29, 2004 #3
    Thanks for your input, Daniel.

    Let me see if I understand it...

    [tex] \arctan x - \arctan y = \arctan \frac{x-y}{1+xy} \qquad (1) [/tex]​

    [tex] x - y = \frac{x-y}{1+xy} \qquad (2) [/tex]​

    [tex] \tan \left( x - y \right) = \tan \left( \frac{x-y}{1+xy} \right) \qquad (3) [/tex]​

    [tex] \tan \left( x - y \right) = \frac{\sin \left( x - y \right)}{\cos \left( x - y \right)} \qquad (4) [/tex]​

    [tex] \tan \left( x - y \right) = \frac{\sin x \cos y - \cos x \sin y}{\cos x \cos y + \sin x \sin y} \qquad (5) [/tex]​

    [tex] \tan \left( x - y \right) = \left( \frac{\frac{1}{\cos x \cos y}}{\frac{1}{\cos x \cos y}} \right) \frac{\sin x \cos y - \cos x \sin y}{\cos x \cos y + \sin x \sin y} \qquad (6) [/tex]​

    [tex] \tan \left( x - y \right) = \frac{\tan x - \tan y}{1 + \tan x \tan y} \qquad (7) [/tex]​

    [tex] \arctan x - \arctan y = \arctan \frac{x-y}{1+xy} \qquad (8) [/tex]​
     
  5. Dec 29, 2004 #4

    dextercioby

    User Avatar
    Science Advisor
    Homework Helper

    U didn't. :tongue2: I can't explain in another way the bunch of c*** at numbers (2) and (3)...

    Formulas (4) pp.(7) give a proof for the trigonometrical identity that can be used to prove your identity.I stated that proof just to let u know i didn't invent it nor rediscovered it hundreds of years later.

    I said to apply tangent on both sides of your identity:
    [tex] \tan(\arctan x-\arctan y) =\tan[\arctan(\frac{x-y}{1+xy})] [/tex]

    `Work the left hand side using the formula i've given proof:
    [tex] \tan(\arctan x-\arctan y) =\frac{\tan(\arctan x)-\tan(\arctan y)}{1+\tan(\arctan x)\tan(\arctan y)}=\frac{x-y}{1+xy} [/tex] (1)

    Work the right hand side:
    [tex] \tan[\arctan(\frac{x-y}{1+xy})] =\frac{x-y}{1+xy} [/tex] (2)

    Interpretation of the relations (1) and (2) is that you have shown that:
    [tex] \tan A=C (1') \tan B=C (2') [/tex](3)
    From (3) it follows immediately
    [tex] \tan A=\tan B [/tex] (4)
    From (4) and from tha fact that on the interval [itex] (-\frac{\pi}{2},\frac{\pi}{2}) [/itex] the function "tan" is a uniform/surjective function,u get that
    [tex] A=B [/tex] (5)
    ,which is nothing but your identity in symbolic form.

    Daniel.
     
  6. Dec 29, 2004 #5
    Oh... I now see what you mean. Thank you.
     
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook