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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 ' ) ≥ 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 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 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 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 m ( x − x 0 ) k k ! and 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 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 ' ) ( 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?

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