Calculating pi using the compound formula

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So, what more can we use this formula for? We have previously found that

\displaystyle {{e}^{x}}=\underset{{n\to \infty }}{\mathop{{\lim }}}\,{{\left( {1+\frac{x}{n}} \right)}^{n}}

How about if we explore what this will do if x is an imaginary number. We will look at this, pretending we don’t know how to calculate ez for complex numbers, and that we don’t know the value of π. But we do know how to multiply complex numbers in cartesian form (a+bi).

So, we will explore

\displaystyle {{e}^{x}}=\underset{{n\to \infty }}{\mathop{{\lim }}}\,{{\left( {1+\frac{ib}{n}} \right)}^{n}}

For various values of b.  We can calulate this quite easilly by squaring a number of times, as discussed on the previous page.

To start with, if b=0 we get

\displaystyle {{e}^{{0i}}}={{e}^{0}}=\underset{{n\to \infty }}{\mathop{{\lim }}}\,{{\left( {1+\frac{0}{n}} \right)}^{n}}=\underset{{n\to \infty }}{\mathop{{\lim }}}\,{{\left( 1 \right)}^{n}}=1

I.e., just as expected.  How about other values?  We can test this on this page:

For b=2 we get about  –0.416 + 0.909i and for b=4 we get about – 0.654 – 0.757i so, one might guess that a value somewhere between 2 and 5 would give us an imaginary part of 0.  We may test with b=3, and then we get – 0.990 + 0.141i. This looks rather promising. The real part is close to –1 and the imaginary part somewhat close to 0.

One way to proceed now is to use the interval halving metod. Say we want to solve f(x)=0.  We start with two guesses, one that gives a too big result, and one that gives a too small result. Given that the function is continuous, then we must have a 0 somewhere in between those start values.

I our case we sould try to find a value for b such that the imaginary part of the compound formula is 0.

We pick the mean value of our two guesses and then calculate f(x) of that value.  Say the result is positive, and,  say,  f( of our lower guess) is also positive, then we let the mean value be our new lower guess. We do simmilarly for the upper guess.

We will examine this in the following page.

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