Taylor Series Question

  • #1
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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!
 

Answers and Replies

  • #2
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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
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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
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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"
 
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  • #5
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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.
 
  • #6
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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.
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.
 

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