10.5 Accuracy of Approximations, and the Mean Value Theorem

]> 10.5 Accuracy of Approximations, and the Mean Value Theorem

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10.5 Accuracy of Approximations, and the Mean Value Theorem

We now ask, how accurate are any of the approximations here, from the trivial one, the constant approximation $f (x) = f (x_{0})$ , the linear approximation, etc.

Suppose $x > x_{0}, m$ is the minimum value of the k-th derivative of $f$ between these two arguments, and $M$ is the maximum value of that derivative there.

We will invoke a principle, which in its simplest form is the statement: the faster you move, the further you go, other things being equal. Here we claim that if we invent a new function $f_{M}$ by replacing the actual value of the k-th derivative of $f$ throughout the interval $(x_{0}, z)$ by its maximum value over that interval, then $f_{M}$ and all its first $k = 1$ derivatives will obey $f_{M} (x') \geq f^{(j)} (x')$ for all $x'$ in that interval.

Think of it this way: if you increase your speed $f'$ to the value $M$ you increase the distance traveled. If alternately you increase acceleration $f "$ to $M$ , by the same argument, that will increase speed, and hence will increase distance traveled. And so on. If you increase a higher derivative, that increase will trickle down to increase all the lower derivatives, and ultimately $f$ itself.

The nice thing about doing this is that the degree $k$ approximation to $f_{M}$ at $x_{0}$ is exact at argument $x$ , because $f_{M}$ 's k-th derivative is constant in the interval between $x_{0}$ and $x$ . Now the degree $k$ approximation to $f_{M}$ is the degree $k - 1$ approximation to $f$ plus $\frac{M {(x - x_{0})}^{k}}{k!}$ .

Our inequality above applied with $j = 0$ therefore tells us that the (k - 1)-th approximation to f, plus $\frac{M {(x - x_{0})}^{k}}{k!}$ is at least $f (x)$ , while by the same argument applied in the opposite order with $M$ replaced by $m$ , we can deduce that the same approximation plus $\frac{m {(x - x_{0})}^{k}}{k!}$ is at most $f (x)$ .

The upshot of all this is have bounds on how far off the degree $k - 1$ approximation to $f$ at $x_{0}$ is from $f$ at argument $x$ : their difference lies between $\frac{m {(x - x_{0})}^{k}}{k!}$ and $\frac{M {(x - x_{0})}^{k}}{k!}$ .

We can go one step further and notice that this tells us that the error in the degree $k - 1$ approximation can be written as $\frac{q {(x - x_{0})}^{k}}{k!}$ where $q$ lies between $m$ and $M$ .

Since $m$ and $M$ are the minimum and maximum values of $f^{(k)}$ between $x_{0}$ and $x$ , if $f^{(k)}$ takes on all values in between its maximum and minimum (which it must if it is differentiable in that interval), it will take on the value $q$ . We can therefore write $q = f^{(k)} (x')$ for some $x'$ in that interval.

This allows us to translate our conclusion here into the following statement.

Theorem:

The error in the degree $k - 1$ approximation to $f$ at $x_{0}$ evaluated at argument $x$ is

f^{(k)} (x') \frac{{(x - x_{0})}^{k}}{k!}

for some $x'$ in the interval $(x_{0}, x)$ , if $f^{(k)}$ is continuous in that interval.

Exercises:

10.5 State this theorem for $k = 1$ . This result is called "the mean value theorem".

10.6 Repeat the argument above for the situation that occurs when $x < x_{0}$ . How does the conclusion change? What is different in the argument?