# Taylor Series Question

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1. Jan 7, 2015

### Austin

I am just trying to clarify this point which I am unsure about:

If I am asked to write out (for example) a third order taylor polynomial for sin(x), does that mean I would write out 3 terms of the series OR to the x^3 term.

x-x^3/3!+x^5/5!

or just

x-x^3/3!

Also, I have a question for the remainder theorem for taylor series (or Lagrange). If I am doing the remainder for a a polynomial for sinx written out as x-x^3/3!, the remainder theorem says use f^n+1(c)(x-a)^n+1/(n+1)!...but for sinx it skips the x^4 term so how would this work? I would assume you would just use the next term: (x^5/5!) but it does not say this anywhere I can find so I am unsure.

Thanks!

2. Jan 7, 2015

### jbstemp

The 3rd order taylor series for sin(x) would only be up to the x^3 term.

In the remainder formula, n refers to the number of terms, not necessarily the powers. In this case x is the first term, -x^3/3! is the 2nd term, and x^5/5! is the 3rd term.

3. Jan 7, 2015

### DarthMatter

Third order in the general case means up to the $x^3$ term. When you remember that a lot of terms are zero in your expression because they involve the sine of zero, maybe it becomes more clear.

4. Jan 7, 2015

### FactChecker

A tiny bit of "wordsmithing" to be clear: "up to and including"

5. Jan 7, 2015

### Staff: Mentor

The n has the same meaning as in "expansion to order n", but the formula is true for every n so it does not really matter as long as you use the same n for both expansion and error.
It is 0*x4. There is nothing special about the zero.

6. Jan 8, 2015

### vela

Staff Emeritus
By definition, the remainders are given by
\begin{align*}
R_3(x) &= \sin x - T_3(x) \\
R_4(x) &= \sin x - T_4(x)
\end{align*} where $T_n(x)$ is the n-th order Taylor polynomial. For $\sin x$ about x=0, you have $T_3(x)=T_4(x)$ because the $x^4$ term vanishes; therefore, the remainders $R_3(x)$ and $R_4(x)$ have to be equal. Typically, however, you don't calculate the actual remainder. Instead, you find an upper bound. If you find an upper bound for $R_4(x)$, it would give you an idea for how well $T_4(x)$ approximates $\sin x$, and since $T_3(x)=T_4(x)$, it also tells you how well $T_3(x)$ approximates $\sin x$. So why find an upper bound for $R_4(x)$ instead of $R_3(x)$? It's because you generally get a better, more stringent estimate for the upper bound by using the formula for higher-order remainder.