How does this Taylor Polynomial work? The Taylor Polynomial is defined as following:
$$P_n(x) = 1 + \dfrac{1}{2}x - \dfrac{1}{8}x^2 + \cdots + (-1)^n \dfrac{1.3.5 \cdots (2n - 3)}{2.4.6 \cdots 2n}x^n$$
If $n = 4$, then the last term in the numerator expansion would be $2 \cdot 4 - 3 = 5$.
So we will have 5 terms totally, running from $x^0$ to $x^4$.
But then how was the numerator generated?
First term = ?
Second term = 1
Third term = 1.3
Fourth term = 1.3.5
Fifth term = ?
I don't understand how they have $1$ for the first term as well as the fifth term. It did not make any sense to me. Any idea?
Edit
The basic function for this polynomial is $f(x) = \sqrt{x + 1}$ at $x = 0$.
Thanks,  
 A: I think you have some signs wrong.
$n=0$ term: $(-1)^0 = 1$ (note that an empty product is considered to be 1).
$n=1$ term: $(-1)^1 (\frac{1}{2}) x^1 = -\frac{1}{2} x $ (again in the numerator, an empty product)
$n=2$ term: $(-1)^2 \frac{1}{2\cdot 4} x^2 = \frac{1}{8} x^2$  
$n=3$ term: $(-1)^3 \frac{1 \cdot 3}{2 \cdot 4 \cdot 6} x^3 = - \frac{1}{16} x^3$
$n=4$ term: $(-1)^4 \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6 \cdot 8} x^4 = \frac{5}{128} x^4$
etc
A: It looks like there is an offset of 1 somewhere.  $-\frac{1}{8}x^2$ looks like $(-1)^1\frac{1}{2\cdot 4}x^2$.  Then the next term would be $(-1)^2\frac{1\cdot 3}{2 \cdot 4 \cdot 6}x^3=+\frac{3}{48}x^3$ followed by $(-1)^3\frac{1\cdot 3\cdot 5}{2 \cdot 4 \cdot 6\cdot 8}x^4=-\frac{15}{384}x^4$.  What basic function is this the Taylor Polynomial for?  Also please refer to the terms by the power of $x$ in them-is the first term the constant $1$ or the $\frac{1}{2}x$ term?
Added:  Wolfram Alpha agrees with these signsfor $\sqrt{1+x}$, so the exponent of $-1$ must be $n+1$ or $n-1$.
