Question

Suggest a Taylor series and a method for approximating \(\pi.\)

Short answer

Answer: The approximation of π is given by the sum of the series: \(\pi \approx 4 \cdot \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots\right)\).

Step-by-step solution

Write down the Taylor series expansion formula

To expand a function as a Taylor series around the point \(x = a\), we can use the following formula: \(f(x) = \sum_{n=0}^{\infty}\frac{f^{(n)}(a)}{n!}(x-a)^n,\) where \(f^{(n)}(a)\) denotes the n-th derivative of the function evaluated at the point \(x = a\).

Find the derivatives of the arctan(x) function

We need to find the derivatives of the arctan(x) function: \(f(x) = \arctan(x)\) \(f'(x) = \frac{1}{1+x^2}\) \(f''(x) = -\frac{2x}{(1+x^2)^2}\) \(f'''(x) = \frac{6x^2-2}{(1+x^2)^3}\) Observe that in general, for the n-th derivative of arctan(x) function, we will have a function of the form: \(f^{(n)}(x) = \frac{P_n(x)}{(1+x^2)^n}\) where \(P_n(x)\) is a polynomial in x.

Expand the arctan(x) function at x=0

Now we need to evaluate these derivatives at x=0: \(f(0) = \arctan(0) = 0\) \(f'(0) = \frac{1}{1+0^2} = 1\) \(f''(0) = -\frac{2 \cdot 0}{(1+0^2)^2} = 0\) \(f'''(0) = \frac{6 \cdot 0^2 - 2}{(1+0^2)^3} = -2\) It appears that only odd powers and odd derivatives of x are nonzero when evaluated at x=0.

Expand arctan(x) as a Taylor series

Now we can expand the arctan(x) function using the Taylor series formula: \(\arctan(x) = \sum_{n=0}^{\infty}\frac{f^{(2n+1)}(0)}{(2n+1)!}x^{2n+1} = x - \frac{x^3}{3} + \frac{x^5}{5} - \frac{x^7}{7} + \cdots\)

Use the relation between arctan and π

As \(\pi/4 = \arctan(1)\), we can plug in x=1 into our Taylor series expansion of arctan(x): \(\pi/4 = \arctan(1) = 1 - \frac{1^3}{3} + \frac{1^5}{5} - \frac{1^7}{7} + \cdots\) To approximate π, we can multiply this series by 4: \(\pi \approx 4 \cdot \left(1 - \frac{1}{3} + \frac{1}{5} - \frac{1}{7} + \cdots\right)\)

Convergence and approximation

The series converges to π, although it converges slowly. To approximate π, we can take a certain number of terms from the series. The more terms we take, the better the approximation will be. For instance, taking the first 500 terms would give a relatively accurate approximation of π. However, there are other methods to approximate π that converge faster, but this Taylor series provides a simple way to understand the process.