Problem:
Question 1:
Consider the function $f(t)=\cos(t)$. What is the error bound associated with expanding the function $f(t+\Delta t)$ about $ t $using $\Delta t=0.1$. In particular, compute the error bound when truncated at the second-derivative (keep only the first two significant-digits).
Question 2:
Consider the function $f(t)=\cos(t)$. What is the error bound associated with expanding the function $f(t+\Delta t)$ about $t $ using $\Delta t=0.1$. In particular, compute the error bound when truncated at the third-derivative (keep only the first two significant-digits).
Solution:
Consider the Taylor expansions:
For question 1:
$ \cos(t + \Delta t) = \cos(t) - \Delta t \sin(t) - \frac{\Delta t^2}{2}\cos(c) $
So the error bound is the maximum value of the last term, which is $ \frac{0.1^2}{2} = 0.005 $
For question 2:
$ \cos(t + \Delta t) = \cos(t) - \Delta t \sin(t) - \frac{\Delta t^2}{2}\cos(t) + \frac{\Delta t^3}{6}\sin(c) $
So the error bound is the maximum value of the last term, which is $ \frac{0.1^3}{6} = 0.00017 $
Question 1:
Consider the function $f(t)=\cos(t)$. What is the error bound associated with expanding the function $f(t+\Delta t)$ about $ t $using $\Delta t=0.1$. In particular, compute the error bound when truncated at the second-derivative (keep only the first two significant-digits).
Question 2:
Consider the function $f(t)=\cos(t)$. What is the error bound associated with expanding the function $f(t+\Delta t)$ about $t $ using $\Delta t=0.1$. In particular, compute the error bound when truncated at the third-derivative (keep only the first two significant-digits).
Solution:
Consider the Taylor expansions:
For question 1:
$ \cos(t + \Delta t) = \cos(t) - \Delta t \sin(t) - \frac{\Delta t^2}{2}\cos(c) $
So the error bound is the maximum value of the last term, which is $ \frac{0.1^2}{2} = 0.005 $
For question 2:
$ \cos(t + \Delta t) = \cos(t) - \Delta t \sin(t) - \frac{\Delta t^2}{2}\cos(t) + \frac{\Delta t^3}{6}\sin(c) $
So the error bound is the maximum value of the last term, which is $ \frac{0.1^3}{6} = 0.00017 $
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