Re: A89: Re: help me too please!
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Re: A89: Re: help me too please!
> >4(1-1/3+1/5-1/7+1/9-1/11+1/13....)
>
> I did it to the 10000 and it came within .0001 (or something like that).
> Does anyone know why this works? Anyone know how many decimals of PI the
> calculator has stored? (I know, it's probably at the back of my manual)
(Note: 'arctan' is called 'tan^-1' on the calculator.)
Try 'arctan(1)' on your calculator, it should answer 'pi/4' (in "exact"
mode!). This means that if we could calculate arctan(1) in some way,
"manually", and then multiply the answer by 4, we would get the exact value
of pi.
So, how do we calculate arctan(1)?
Well, one approach is to use taylor series: Try 'taylor(arctan(x),x,9)', and
the calc answers 'x-x^3/3+x^5/5-x^7/7+x^9/9'. (Compare this to your sum
above, with x=1 !). However, as this sum really has an infinite number of
terms, we cannot get an exact answer to arctan(1), only approximations...
Try this: '(pi*1E9)-3141592653'. The answer shows the last digits of pi
known by the calculator.
//Johan
References: