Taylor approximations

[clairez93]

1. The problem statement, all variables and given/known data

1. Use Taylor's Theorem to determine the accuracy of the approximation.

arcsin(0.4) = 0.4 + \frac{(0.4)^{3}}{2*3}}

2. Determine the degree of the Maclaurin polynomial required for the error in the approximation of the function at the indicated value fo x to be less than 0.0001. Use a symbolic differentiation utility to obtain and evaluate the required derivatives.

f(x) = ln(x+1) approximate f(1.5)

2. Relevant equations



3. The attempt at a solution

1.
f(x) = arcsin (0.4)
x = 0.4
a = 0.5
N = 3

R_{n}(x) = \frac{f^{N+1}(c)}{(N+1)!}(x-a)^{N+1}
R_{3}(0.4) = \frac{f^{4}(c)}{4!}(0.4-0.5)^{4}
= \frac{\frac{-3c(2c^{2}+3)}{(c^{2}-1)^{3}\sqrt{1-c^{2}}}}{4!}(0.4-0.5)^{4} \leq \frac{(0.4-0.5)^{4}}{4!} = 4.166666667 * 10^{-6}

Book Answer: R_{3} \leq 7.82 * 10^{-3}


No idea what I did wrong here.

2.
f(x) = ln(x+1)
a = 0
x = 1.5
N = ?

R_{n}(x) = \frac{f^{N+1}(c)}{(N+1)!}(x-a)^{N+1}
R_{n}(x) = \frac{f^{N+1}(c)}{(N+1)!}(1.5)^{N+1}
|R_{n}(1.5)| \leq 0.0001

To make a long sheet of work short, I got all the way up to N=12 before I finally got to 0.0001. Here is the work for N=12:

N=12: |R_{12}(1.5)| = |\frac{f^{13}(c)}{13!}(1.5)^{13}| = |\frac{(479001600)}{(c+1)^{13}} * \frac{(1.5)^{13}}{13!}| = |\frac{(479001600)(3.1254*10^{-8})}{(c+1)^{13}}| = \frac{14.9707}{(c+1)^{13}}

\frac{14.9707}{(1.5+1)^{13}} = 0.0001

The book answer says N=9, for which when I tested I got this:

N=9: |R_{9}(1.5)| = |\frac{f^{10}(c)}{10!}(1.5)^{10}| = |\frac{(-362880)}{(c+1)^{10}} * \frac{(1.5)^{10}}{10!}| = |\frac{(-362880)(0.000106)}{(c+1)^{10}}| = \frac{5.7665}{(c+1)^{10}}

\frac{5.77665}{(1.5+1)^{10}} = 0.000605 which isn't exactly less than 0.001.


When I went back through my notes to see if I did something wrong, I realized that since the function was decreasing, (c+1), on the bottom of the fraction, maybe I should have plugged in 0, since c should be greater than or equal to 0 and less than or equal to 1.5, thus the biggest R could be would be whatever I get when I plug 0 in, not 1.5.

However, if I plug in 0 for these instead of 1.5, it seems to make the problem worse, since as you can see, N=9, with 0 instead of 1.5 for C would get 1 on the bottom and thus R_9 would be 5.77665, which is very very far from 0.0001, much farther than when I used 1.5.

I'm thoroughly frustrated by this one. Please help!

[g_edgar]

You chose a=0.5. Probably that is a good choice. But then what would be the Taylor series? Not the one shown. So the writers of the problem did not choose a=0.5. What did they choose? The point is that the Taylor series for their choice is much simpler to write down than for your choice a=0.5.

[clairez93]

I think they chose a=0? Is that right?

[g_edgar]

a = 0 is the way to get that Taylor series, yes. So do the rest of the problem on that basis.