Series de Maclaurin. calculo de Pi con Serie

http://www.intmath.com/series-expansion/2-maclaurin-series.php

2. Maclaurin Series

By M. Bourne

In the last section, we learned about Taylor Series, where we found an approximating polynomial for a particular function in the region near some value x=a.
We now take a particular case of Taylor Series, in the region near x=0. Such a polynomial is called the Maclaurin Series.
The infinite series expansion for f(x) about x=0 becomes:

f(x)≈f(0)+f’(0)x +f’’(0)2!x2 +f’’’(0)3!x3 +fiv(0)4!x4 +...

f’(0) is the first derivative evaluated at x=0, f’’(0) is the second derivative evaluated at x=0, and so on.
[Note: Some textbooks call the series on this page "Taylor Series" (which they are, too), or "series expansion" or "power series".]

Example: Expanding sin x

Find the Maclaurin Series expansion for f(x)=sin x.

Answer

We need to find the first, second, third, etc derivatives and evaluate them at x = 0. Starting with:

f(x) = sin x

f(0) = 0

First dervative:

f '(x) = cos x

f '(0) = cos 0 = 1

Second dervative:

f ''(x) = −sin x

f ''(0) = −sin 0 = 0

Third dervative:

f '''(x) = −cos x

f '''(0) = −cos 0 = −1

Fourth dervative:

f ^iv(x) = sin x

f ^iv(0) = sin 0 = 0

We observe that this pattern will continue forever.
Now to substitute the values of these derivatives into the Maclaurin Series:

f(x) ≈f(0)+f’(0)x +f’’(0)2!x2 +f’’’(0)3!x3 +fiv(0)4!x4+...

We have:

sin x=0+(1)(x) +0 +−13!x3 +0 +15!x5 +0 +−17!x7+...

This gives us:

sin x=x−16x3+1120x5 −15040x7+...

We plot our answer

sin x=x−16x3 +1120x5 −15040x7+...
to see if the polynomial is a good approximation to f(x)=sin x.

We observe that our polynomial (in red) is a good approximation to f(x)=sin x (in blue) near x=0. In fact, it is quite good between -3 ≤ x ≤ 3.

Example: Expanding ex

Find the Maclaurin Series expansion of f(x)=ex.

Answer

Exercise

Find the Maclaurin Series expansion of cos x.

Answer

Finding Pi Using Infinite Series

In the 17th century, Leibniz used the series expansion of arctan x to find an approximation of π.
We start with the first derivative:

ddxarctan x=11+x2

The value of this derivative when x=0 is 1. That is, f’(0)=1.
Similarly for the subsequent derivatives:
d2dx2arctan x=−2x(1+x2)2

f’’(0)=0

d3dx3arctan x=23x2−1(1+x2)3

f’’’(0)=−2

d4dx4arctan x=−24xx2−1(1+x2)4

fiv(0)=0

d5dx5arctan x=245x4−10x2+1(1+x2)5

fv(0)=24

Now we can substitute into the Maclaurin Series formula:
arctan x≈0+(1)x+02x2−23!x3+04!x4+245!x5=x−x33+x55−x77+...
Considering that (see the 45-45 triangle)

arctan1=π4

we can substitute x=1 into the above expression and get the following expansion for π

π=4(1−13+15−17+...)

All very well, but it was not a good way to find the value of π because this expansion converges very slowly.
Even after adding 1000 terms, we don't have 3 decimal place accuracy.

4∑n=01000(−1)n2n+1=3.142591654...

(We know now that π = 3.141 592 653 5..., and we know many other more efficient ways to find π.)
Here's the graph of y = arctan x (in blue) compared to the polynomial we just found (in red).