. Here is the general result:
Taylor's formula is a more accurate local formula than the "microscope approximation." It has many uses.
Taylor's formula uses the first derivative f'[x] and a number of higher order derivatives: the second derivative f"[x], which is the derivative of f'[x]; the third derivative f(3)[x], which is the derivative of f"[x]; . Here is the general result:
Theorem 22.1Taylor's Small Oh Formula Suppose that f[x] has n ordinary continuous derivatives on the interval (a,b). If x is not near a or b and is small, then
for.
When n=1, Taylor's formula is the "microscope approximation,"
Notice that if
is small, then its square
is small even on a scale of
. This makes the second-order formula a more accurate approximation for
,
, then
is only one one-thousandth of that,
.
Strictly speaking, the "
" in the two formulas cannot be compared.
All we know is that when
is "small enough" both
's are as small as we prescribe.
All we really know is that "eventually" the second-order formula is better than the first-order one.
In the Mathematical Background Chapter 8, we make the approximation more precise.
and Exp[0+dx]=e0+dx.
22.1
It is "clear" that if we view a graph in an powerful microscope and see the graph as indistinguishable from an upward-sloping line at a point x0, then the function must be "increasing" near x0. The previous Project 21 on inverse functions uses this idea in a very computational way.
Certainly, the graph need not be increasing everywhere -- draw y=x2 and consider the point x0=1 with f'[1]=2. Exactly how should we formulate this? Even if you don't care about the symbolic proof of the algebraic formulation, the formulation itself may be useful in cases where you don't have graphs.
Math Background Section 5.6 has a more complete exposition of this topic.
The idea is simple: Compute the change in f[x] using the positive slope straight line and keep track of the error.
Take x1 and x2 so that
The Math Background shows how to make the approximations precise and thus allow x1 and x2 to range in an interval
22.2
The smile and frown icons of text Chapter 9 are based on a simple intuitive mathematical idea: when the slope of the tangent increases, the curve bends up.
We have two questions. (1) How can we formulate bending symbolically? (2) How do we prove that the formulation is true? First things first.
If a curve bends up, it lies above its tangent line.
Draw the picture.
The tangent line at x0 has the formula y=b+m(x-x0) with b=f[x0] and m=f'[x0]. If the graph lies above the tangent, f[x1] should be greater than b+m(x1-x0)=f[x0]+f'[x0](x1-x0) or
We have at least formulated the result as follows.
We want you to use the second-order Taylor formula to show the algebraic form of the smile icon.
If f[x] is twice differentiable on a real interval (a,b), a<x<b, and x is not near a or b, then for any small
Math Background Section 8.1 has a complete exposition of this topic.
In particular, it deals with the question of how far away x1 can be from x0.
Graphically, the approximation of slope given by the symmetric difference is clearly better on a "typical" graph as illustrated below.
A line through the points (x,f[x]) and
0
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A quadratic function q[dx] in the local variable dx that matches the graph y=f[x] at the three x values,
Show that the derivative
It is interesting to compare different numerical approximations to the derivative in a difficult, but known case.
This is done in Project 17 on direct computation of the derivative of an exponential.
The experiments give a concrete form to the error estimates of the previous exercise.
When we only have data (such as in the law of gravity in Chapter 10 of the main text or in the air resistance project in the Scientific Projects), we must use an approximation.
In that case the symmetric formula is best.
22.4
22.5
We know that the first derivative tells us the slope and the second derivative tells us the concavity or convexity (frown or smile), but what do the third, fourth, and higher derivatives tell us?
The symmetric limit interpretation of derivative arose from fitting the curve y=f[x] at the points
To determine a quadradic fit to a curve, we would need three points, say
This approach extends to as many derivatives as we wish.
If we fit to n+1 points, we can determine the n+1 coefficients in the polynomial
Theorem 22.2
, 
, such that f[x] is increasing on
, that is,
, 
, such that f[x] is decreasing on
, that is,
. Since
(see text Section CD 5.5 and Math Background Theorem 5.4), we may write
where m=f'[x0] and
. Let
so
The number m is a real positive number, so
and , since
, 
. This means f[x2]-f[x1]>0 and f[x2]>f[x1].
.
This is the answer to question 1, but now we are faced with question 2. The increment approximation says
so this direct formulation of "bending up" requires that we show that the whole error
stays positive for
. All we have to work with is the increment approximation for f'[x] and the fact that f"[x0]>0. A direct proof is not very easy to give - at least we don't know a direct one.
The second-order Taylor formula will make this easy.
Theorem 22.3
, 
, such that y=f[x] lies above its tangent over
, that is,
, , such that y=f[x] lies below its tangent over
, that is,
with
.
Suppose that f"[x0]>0 at the real value x0. If
, substitute x=x0 and
into Taylor's second-order formula to show the local bending formula.
Use the fact that
.
and an error.
Why is this formula algebraically a better approximation for f'[x] than the one you obtain by solving the ordinary increment approximation
for
? Compare the errors and note the importance of
being small.
is drawn with the tangent at x in one view, while a line through
and
is drawn with the tangent at x in the other.
The second slope is closer to the slope of the tangent, even though the line does not go through the point of tangency.
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to##=0em plus10em&&##=0em plus10em
and
and (x,f[x]) on the first view.
Show that the average of the slopes of the two secant lines on the resulting figure is
, the same as the slope of the symmetric secant line in the second view.
, x, and
, is given by
where y1=f[x], 
, and
.
and
.
, the same as the symmetric secant line slope.
In other words, a quadratic fit gives the same slope approximation as the symmetric one, which is also the same as the average of a left and a right approximation.
All these approximations are "second-order."
Substitute
and
into Taylor's second-order formula to obtain
Add the two to obtain
, x, and
. Two velocities correspond to the difference quotients
Compute the difference of these two differences and divide by the correct x step size.
What formula do you obtain?
for f"[x] and compare this with the answer from part 1 of this exercise.
What does the comparison tell you?
and
and then taking the limit.
A more detailed approach to studying higher order properties of the graph is to fit a polynomial to several points and take a limit.
, x, and
. We would then have three values of the function,
, f[x], and
to use to determine unknown coefficients in the interpolation polynomial p[dx]=a0+a1dx+a2dx2. We could solve for these coefficients in order to make
, f[x]=p[0], and
. This solution can easily be done with computer commands.
The limit of this fit tends to the second-order Taylor polynomial,
so that
for i=0,1,...,n. If the function f[x] is n times continuously differentiable,
Specifically, if
, then
uniformly for x in compact intervals.
The higher derivatives mean no more or less than the coefficients of a local polynomial fit to the function.
In other words, once we understand the geometric meaning of the dx3 coefficient in a cubic polynomial, we can apply that knowledge locally to a thrice differentiable function.
This is explored in detail in Math Background Section 8.5.
.
and
into Taylor's second-order formula to obtain
