I'm a bit puzzled by this problem, which involves the John Machin's formula. Here it goes:(adsbygoogle = window.adsbygoogle || []).push({});

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

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# Homework Help: John Machin's Formula

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