# Thread: [SOLVED] Taylor inequality to estimate accuracy of approximation

1. ## [SOLVED] Taylor inequality to estimate accuracy of approximation

$\displaystyle f(x) = sec(x)$ $\displaystyle \rightarrow$ $\displaystyle a=0$, $\displaystyle n=2$, $\displaystyle -0.3 \leq x \leq 0.3$

Here is my Taylor polynomial approximation:

$\displaystyle T_2(x) = 1 + \frac{1}{2}x^2$

So, then I found my (n+1) derivative:

$\displaystyle f^3(x) = secxtanx(tan^2x + 5sec^2x)$

So,
$\displaystyle \left|R_n\right| \leq \frac{M}{(n+1)!} \left|x-a\right|^{n+1}$

CAn someone show me how to do this part?
I know that I need to find a suitable M. How do I do that with these trig fuctions? Can I just choose zero for my x, since it is within my interval?

Thanks!!

2. no--- a is 0

Rn depends on x -- the max value of x on your interval is .3 So if you're trying to find an upper bound for the error for all x in the interval use .3

To find M I'd do it graphically making sure to use |f^3(x)| on your graph

3. Originally Posted by Calculus26
no--- a is 0

Rn depends on x -- the max value of x on your interval is .3 So if you're trying to find an upper bound for the error for all x in the interval use .3

To find M I'd do it graphically making sure to use |f^3(x)| on your graph

Right, forgot that my a was = 0. Thanks!
The instructions say to find M and then check my answer by graphing. So to find M, do I just plug in .3 into my third derivative and solve? Is that what I use for my M???

4. Generally no

but since sec(x) and tan(x) are increasing 0n (0,.3)

and from symmetry concerns |sec(x)tan(x)| is the same on (-.3,.3)

you can just use f^3 (.3) for M in this case