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

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

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

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