# Thread: [SOLVED] Little doubt : Taylor's polynomial

1. ## [SOLVED] Little doubt : Taylor's polynomial

I must find $\displaystyle P_{5,0}(x)$, in other words the Taylor's polynomial of order $\displaystyle 5$ evaluated in $\displaystyle 0$ of the function $\displaystyle f(x)=\frac{\sin (x)}{1+x^2}$. But it says to consider the Taylor's polynomials of the functions $\displaystyle \sin (x)$ and $\displaystyle \frac{1}{1+x^2}$. I found them, so my bet would be to multiply them... but as you can imagine, I got a polynomial of degree $\displaystyle 10$. So my little doubt is : I only take on consideration the terms of degree $\displaystyle \leq 5$ and I'm done? In other words, I throw up the terms of greater degree than $\displaystyle 5$ and I get the Taylor's polynomial of $\displaystyle f$ of order $\displaystyle 5$. Am I right?

2. Originally Posted by arbolis
I must find $\displaystyle P_{5,0}(x)$, in other words the Taylor's polynomial of order $\displaystyle 5$ evaluated in $\displaystyle 0$ of the function $\displaystyle f(x)=\frac{\sin (x)}{1+x^2}$. But it says to consider the Taylor's polynomials of the functions $\displaystyle \sin (x)$ and $\displaystyle \frac{1}{1+x^2}$. I found them, so my bet would be to multiply them... but as you can imagine, I got a polynomial of degree $\displaystyle 10$. So my little doubt is : I only take on consideration the terms of degree $\displaystyle \leq 5$ and I'm done? In other words, I throw up the terms of greater degree than $\displaystyle 5$ and I get the Taylor's polynomial of $\displaystyle f$ of order $\displaystyle 5$. Am I right?
That is based upon the conventions that your book uses. I have seen "order" used to denote number of terms and degree of a polynomial. But offhand, I would say that is the correct assumption.