Am I right or is my book right (Taylor remainder)

In summary, when using the approximation e^{x}=1+x+(x^{2}/2) for small values of X, the maximum error can be estimated by using the third derivative of e^{x}, which can be either e^{0.1} or 3^{0.1}. Both values give slightly different estimates, with e^{0.1} being more accurate but 3^{0.1} being easier to calculate. Therefore, both answers are correct and which one to use depends on the context and purpose of the calculation.
  • #1
viciousp
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0

Homework Statement


The approximation [tex] e^{x}=1+x+(x^{2}/2) [/tex] is used when [tex] X [/tex] is small estimate the error when [tex]\left|x \right|<0.1[/tex]

Homework Equations


[tex]\left|R_{n} \right|<\frac{M(x-a)^{n+1}}{(n+1)!}[/tex]

The Attempt at a Solution


Since the Taylor expansion goes to the second power I used the third derivative of [tex]e^{x}[/tex] which is just itself and found the maximum value that it can be between the domain [0,0.1] which is at [tex]e^{0.1}[/tex] then continuing the formula [tex](0.1-0)^{3}[/tex] then i divided it by 3! which gave me an answer of [tex]1.84*10^{-4}[/tex].

My book on the other hand used [tex]3^{0.1}[/tex] instead of [tex]e^{0.1}[/tex] and as a result the answer in the book was larger than my answer. So which answer is the right answer?
 
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  • #2
Depends upon what you mean by "estimate". Remember that neither of those answers will be the error- you can't determine the exact error this way. What you are saying is that the error is less than 1.98x10-4 while the textbook is saying it is less than a larger value (because e< 3) so both are correct. I don't know if your book has some reason for using "3" rather than "e". e gives a slightly more accurate estimate but they are both estimates.
 
  • #3
The only reason I can imagine is that [tex] 3^{0.1} [/tex] is easier to calculate because it's the solution of [tex] x^{10} = 3 [/tex] and a Newton's method iteration would provide an accurate estimation quickly. The assumption, I suppose, is that if you are approximating [tex] e^x [/tex], you don't have the luxury of calculating it explicitly.
 

1. What is Taylor remainder and why is it important in mathematics?

Taylor remainder is a concept in mathematics that is used to measure the accuracy of approximating a function with its Taylor polynomial. It is important because it helps us understand the limitations of polynomial approximations and allows us to determine how closely a polynomial can approximate a given function.

2. How is Taylor remainder calculated?

The Taylor remainder is calculated using the remainder term of the Taylor polynomial, which is given by the formula: R_n(x) = f^{(n+1)}(c)(x-a)^{n+1} / (n+1)!, where f^{(n+1)}(c) is the (n+1)th derivative of the function evaluated at a point c between x and a.

3. Can Taylor remainder be negative?

Yes, Taylor remainder can be negative. The remainder term in the Taylor polynomial formula can be positive or negative depending on the values of the function's derivatives at the given point. A negative remainder indicates that the Taylor polynomial is an overestimate of the function, while a positive remainder indicates an underestimate.

4. How does Taylor remainder relate to Taylor series?

Taylor series is a way of representing a function as an infinite sum of terms, where each term is a derivative of the function evaluated at a point. Taylor remainder is the error term in the Taylor series, which measures the difference between the function and its Taylor polynomial approximation.

5. What are some applications of Taylor remainder?

Taylor remainder has many applications in mathematics, physics, and engineering. It is used to approximate functions in numerical analysis, to determine the convergence of series, and to find upper and lower bounds for integrals. It is also used in error analysis to quantify the accuracy of a numerical solution or measurement.

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