# Taylor Series Question

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!

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.

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.

FactChecker
Gold Member
Third order in the general case means up to the ##x^3## term.
A tiny bit of "wordsmithing" to be clear: "up to and including"

• DarthMatter
mfb
Mentor
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.
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.
but for sinx it skips the x^4 term so how would this work?
It is 0*x4. There is nothing special about the zero.

vela
Staff Emeritus