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

1. Aug 24, 2008

### viciousp

1. The problem statement, all variables and given/known data
The approximation $$e^{x}=1+x+(x^{2}/2)$$ is used when $$X$$ is small estimate the error when $$\left|x \right|<0.1$$

2. Relevant equations
$$\left|R_{n} \right|<\frac{M(x-a)^{n+1}}{(n+1)!}$$

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

My book on the other hand used $$3^{0.1}$$ instead of $$e^{0.1}$$ and as a result the answer in the book was larger than my answer. So which answer is the right answer?

2. Aug 24, 2008

### HallsofIvy

Staff Emeritus
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 text book 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. Aug 25, 2008

### cellotim

The only reason I can imagine is that $$3^{0.1}$$ is easier to calculate because it's the solution of $$x^{10} = 3$$ and a Newton's method iteration would provide an accurate estimation quickly. The assumption, I suppose, is that if you are approximating $$e^x$$, you don't have the luxury of calculating it explicitly.