Maclaurin Series

The subject of Maclaurin Series is one that will be carefully justified in Calculus. However, the idea is so wonderful that it is difficult to keep it a secret. All of the functions that one develops in advanced algebra and precalculus can be approximated to any degree of accuracy by polynomial functions and thereby evaluated by just adding, subtracting and multiplying. It is this very technique that allows your calculators to evaluate those functions without having any of the values stored away wasting memory. The Maclaurin Series will converge rapidly for numbers near zero. The series can be shifted to any other number if the required input is not near zero. In this case, the series are known as Taylor Series. Some examples of familiar functions are the following.

I.

$$e^x = \frac{1}{0!}+\frac{x}{1!}+\frac{x^2}{2!}+\frac{x^3}{3!}+\cdots$$ $x\in \Re$

II.

$$\sin x = \frac{x}{1!}-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots$$ $x\in \Re$

III.

$$\cos x = \frac{1}{0!}-\frac{x^2}{2!}+\frac{x^4}{4!}-\frac{x^6}{6!}+\cdots$$ $x\in \Re$

IV.

$$\ln \vert x+1 \vert= \frac{x}{1}-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+\cdots$$ $-1<x\le1$

V.

$$\arctan x = \frac{x}{1}-\frac{x^3}{3}+\frac{x^5}{5}-\frac{x^7}{7}+\cdots$$ $-1\le x\le1$

VI.

$$(x+1)^k = x^k +kx^{k-1} +\frac{k(k-1)}{1\cdot2}x^{k-2}+\frac{k(k-1)(k-2)} {1\cdot2\cdot3}x^{k-3}+\cdots$$ $k\in\Re$

To get some feel for these series make the following substitutions for $x$.

1. In [I], let $x = 1$ to find the sum of $$\sum_{n=0}^{\infty}\frac{1}{n!} =\frac{1}{0!}+\frac{1}{1!}+\frac{1}{2!}+\frac{1}{3!}+\cdots$$
2. In [IV], let $x = 1$ to find the sum of $$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n}= 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\cdots$$.
3. In [V], let $x = 1$ to find the sum of $$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{2n-1}= 1-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots$$
4. In [I] let $x = i\theta$ where $i^2=-1$. Separate the real and imaginary parts and use [II] and [III] to show that $e^{\pi i}= -1$ or $e^{\pi i}+1=0$. This is considered by many to be one of the most beautiful of all mathematical formulae with its collecting the five most important constants in mathematics.
5. 1. Show $$\arctan x -\arctan y = \arctan\frac{x-y}{1+xy}.$$
   2. Show $$\arctan \frac{120}{119} -\arctan\frac{1}{239} = \frac{\pi}{4}.$$
   3. Use the fact that $\arctan x$ is odd and replace $y$ with $-x$ in part (a). Use this formula twice to show that $4\arctan \frac{1}{5} = \arctan \frac{120}{119}$, so that
      $4\arctan \frac{1}{5} -\arctan\frac{1}{239} = \frac{\pi}{4}.$
   4. Use the first seven terms of $\arctan \frac{1}{5}$ and the first three terms of $\arctan\frac{1}{239}$ to compute $\pi$ to 10 decimal places by adding only 10 terms.
6. (BAMM 2001) Find $\sqrt[3]{1729}$ to 4 decimal places in 40 seconds. (Hint: recall that 1729 is the famous taxicab number that Ramanujan stated was the smallest integer that can be written as a sum of two cubes in two ways; $9^3+10^3$ and $1^3+12^3$.) Now use [VI] with $x=12^3$ and $k=1/3$.