Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

The basic function for this polynomial is $f(x) = \sqrt{x + 1}$ at $x = 0$.


share|cite|improve this question
up vote 3 down vote accepted

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: $(-1)^4 \frac{1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6 \cdot 8} x^4 = \frac{5}{128} x^4$


share|cite|improve this answer
Many thanks, now I know what's empty product ;) – Chan Apr 11 '11 at 23:50

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 $n-1$.

share|cite|improve this answer
If my interpretation is correct, it is the Taylor series of $2 - \sqrt{1+x}$. – Robert Israel Apr 11 '11 at 23:51
Thanks. Basic function was added. About the power of $x$, I really don't know. I copied it exactly from my notes, and I guess that's what confused me. – Chan Apr 11 '11 at 23:53

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.