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Landscape Urbanism 1(6) Landscape urbanism – large-scale architecture, ecological urban planning or a designerly research policy GUNILLA LINDHOLM Senior lecturer, landscape architect Department of Landscape Architecture, SLU Alnarp, Sweden TEL: +46 (0)40 415429 Abstract Landscape urbanism has been something in between of an ideological urbanistic movement and a buzzword for landscape architects, for about a decade. This concept has an eager and inspiring voice, promising new winds in an urban design and planning caught up with either of a prolonged modernism or the so called “new” urbanism, i.
Voir
- public issues with private issues
- landscape urbanism
- prospective understanding of the urban development
- landscape architecture
- urban design
- public sector
- urban planning
- a.a.
- a. a.
Taylor Polynomials and Taylor Series
Math 126
In many problems in science and engineering we have a functionf(x) which is too
complicated to answer the questions we’d like to ask. In this chapter, we will use local
information near a pointx=bto find a simpler functiong(x), and answer the questions
usingginstead offbe depends upon how closely the function. How useful the answers will
gppastemaxirof, so we also need to estimate, or bound, the error in this approximation:
f−g.
§1. Tangent Line Error Bound.
Ken is at work and his car is located at his home twenty miles north. Fifteen minutes
from now, in the absence of any other information, his best guess is that the car is still at
home. How accurate is this guess? If Ken knows his son will drive the car no faster than
40 miles per hour in the city, how far away can the car be in 15 minutes? A moment’s
reflection should give you the estimate (or bound) that his son can drive the car no further
than 10 miles (4at 40 mph) so that the car will be within 10 miles of home, or nohour more than 30 miles from Ken. We can also see this by using the Fundamental Theorem of
Calculus. Suppose Ken’s son is driving the car in a straight line headed north and suppose
x(tthe distance of the car from Ken at time) is tthen
t)−x(0) =Z0tx0(u)du x( Herex0(u) is the velocity at timeu this problem we know. In|x0(u)| ≤40 for allubetween
0 andt, so that
t Z0x0(u)du≤Z0t|x0(u)|du |x(t)−x(0)|= t ≤Z040du= 40t
1
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There are a couple of steps above that need justification. First we used the inequality
Zx0(u)du≤Z|x0(u)|du The integralRx0(u)durepresents the area which is below the curvey=x0(u) and above
theu-axis minus the area which is above the curve and below theuaxis. Whereas the right-hand side is equal to the total area between the curve and theuaxis, and so the right-hand side is at least as big as the left. Secondly we replaced the function|x0(u)|by
the larger constant 40 and the area under the curveyleast as big as the area= 40 is at under the curvey=|x0(u)|.
This was a rather long-winded way to get the same bound, but it works in general: if |f0(t)| ≤Mfor alltbetweenbandx, then
|f(x)−f(b)| ≤M|x−b|
Test your understanding by writing out the reasoning behind this inequality.
If we have more information, then we can get a better approximation. For example, suppose Ken’s wife called and said that their son left home driving north at 30 miles per
hour. Then we might guess that after 15 minutes he is 304 or 75 miles north of Ken’s
home. Here we’ve used the tangent line approximation
x(t)−x(0)≈x0(0)(t−0)
Recall that the equation of the line which is tangent to the graph ofy=f(x), when x=b, passes through the point (b f(b)) and has slopef0(b equation of the tangent). The
line is then:
or
y−f(b) =f0(b)(x−b)
y=f(b) +f0(b)(x−b)
which we will call thetangent line approximation, or sometimes thefirst Taylor
polynomialforfbased atb(bfor “based”).
2
y
(b f(b))
x
y=f(x)
Figure 1.Tangent line through (b f(b))
In Chapter 3 of Stewart, we found that the tangent line was useful for approximating complicated functions: the graph of the linear functiony=f(b) +f0(b)(x−b) is close to
the graph ofy=f(x) ifxis nearb. In other words f(x) =f(b) +f0(b)(x−b) + error
How big is the error?
y
b
y=f(b) +f0(b)(x−b)
x
error
y=f(x)
Figure 2.Error in tangent line approximation. Tangent Line Error Bound.If|f00(t)| ≤Mfor alltbetweenxandbthen |error|=f(x)−[f(b) +f0(b)(x−b)]≤M2|x−b|2
In terms of the previous example, if we know that Ken’s son starts with a speed of 30 mph and accelerates no more that 20mihr2while driving away, then at timetwe can
estimate his location as
x(t)≈x(0) +x0(0)t
and the error in this approximation can be bounded by the Tangent Line Error Bound: x(t)−[x(0) +x0(0)t]≤20t (13) 2 2
3
In particular, after 15 minutes, the error
x(4)−[20 + 30∙4] is at most2(4)2=8 the details to be sure you understand.miles. Check Another question we could ask is: how long will it take Ken’s son to be 25 miles from
Ken? Replace the complicated (and unknown) functionx(t) with the linear approximation y=x(0) +x0(0)tto answer the question:
25 =|x(0) +x0(0)t|=|20 + 30t|
sot= 16, or 10 minutes would be the approximate answer.
A more difficult question is: when can Ken be sure his son is at least 25 miles from
Ken? To answer this question, we write the error bound (13) in a different form:
−10t2≤x(t)−(20 + 30t)≤10t2
Adding 20 + 30tto the left hand inequality we obtain
20 + 30t−10t2≤x(t)
So to guarantee 25≤x(t), it is sufficient to have
25≤20 + 30t−10t2
or at least (3−7) might take this long if he is2 hours, which is about 11 minutes. (It
“decelerating” because we really assumed that theabsolute valueof the acceleration was
at most 20.)
Why is theTangent Line Error Boundtrue? Supposex > bthen by the Funda-
mental Theorem of Calculus
f(x)−f(b) =Zbxf0(t)dt
4
Treatxas a constant for the moment and setU=f0(t),V
to obtain
=t−xand integrate by parts
f(x)−f(b) =f0(t)(t−x)bx−Zx(t−x)f00(t)dt b
=f0(b)(x−b) +Zxbf00(t)(x−t)dt
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Subtractingf0(b)(x−b) from both sides we obtain f(x)−[f(b) +f0(b)(x−b)]≤Zbx|f00(t)|(x−t)dt≤ZxbM(x−t)dt=M(2x−b)2 Here we used the same ideas as the inequalites in (1.1) and the fact thatx−t≥0.
The case whenx < b might test your understanding of the Youis proved similarly.
above argument by writing out a proof for that case.
Example 1.1. Find a bound for the error in approximating the functionf(x) = tan−1(x)
by the first Taylor polynomial (tangent line approximation) based atb= 1 on the interval
I= [911].
The first step is to find the tangent line approximation based at 1:
y=4+2(x−1)
Then calculate the second derivative:
f00(x) = (1−2+xx2)2
and forxin the intervalI, |f00(x)|(1+2=xx2)2 By computing a derivative, you can show that|f00(x)|is decreasing on the intervalIso
that its maximum value onIis equal to its value atx= 09, and
|f00(x)| ≤20((+1099)2)2≤055
5
SettingM= 055, the Tangent Line Error Bound gives
tan−1(x)−[4+2(x−4)]≤2(x−1)2≤00028 In other words, if we use the simple functiony=4+2(x−1) instead of the more complicated tan−1(x), we will make an error in the values of the function of no more than
00028 whenxis in the intervalI.
Example 1.2.For the same functionf(x) = tan−1(x), find an intervalJso that the
error is at most0001onJ.
The intervalJwill be smaller than the intervalI, since001< 0028, so the same
bound holds forxinJ:
or
≤(− tan−1(x)−[4+2(x−4)]2x1)2
The error will be at most 0001 if
2(x−1)2≤0001
|x−1|< 0603
Thus if we setJ= [94106] then the error is at most 0001 whenxis inJ.
A word of caution: we rarely can tell exactly how big the the Tangent Line error is,
that is, the exact difference between the function and its first Taylor polynomial. The point
of the Tangent Line Error Bound is to give some control or bound on how big the error can be. Sometimes we cannot tell exactly how big|f00(t)|is, but many times we can say it
is no more than some numberM numbers. SmallerMof course give better (or smaller)
bounds for the error. In Example 1.2, we found the maximum of the second derivative on
the larger intervalIis actually a better (smaller) bound on the smaller interval . ThereJ,
but that bound would have been hard to find before we even knew whatJis!
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§ Approximation2. Quadratic
In many situations, the tangent line approximation is not good enough. For example,
ifh(t) is the height of baseball thrown into the air, thenh(t) is influenced by its initial positionh(0), initial vertical velocityh0(0) and vertical accelerationh00=−gdue to gravity. A simple model for theh(t) can be found by integration: h(t)≈h(0) +h0(0)t−g2t2(21) However, there are other forces on the baseball, for example air resistance is important.
The right-hand side of (2.1) is called a quadratic approximation to the functionh. How
do we find a quadratic approximation to a functiony=f(x) and how accurate is this
approximation? The secret to solving these problems is to notice that the equation of the
tangent line showed up in our integration by parts in (1 integrate (14). Let’s4) by parts
again.
Treatxas a constant again and setU=f00(t),V=−2(x−t)2and integrate (14) by parts to obtain
f(x)−f(b) =f0(b)(x−b) +Zxbf00(t)(x−t)dt
x =f0(b)(x−b)−f00(t)2(x−t)2bx+2Zbf000(t)(x−t)2dt Moving everything except the integral to the left-hand side, f(x)−[f(b) +f0(b)(x−b) +2f00(b)(x−b)2] =2Rxbf000(t)(x−t)2dt
Definition.We call T2(x) =f(b) +f0(b)(x−b) +2f00(b)(x−b)2 the quadratic approximation orsecond Taylor polynomialforfbased atb.
By the same argument used to prove the Tangent Line Error Bound, we obtain:
7
(22)
Quadratic Approximation Error Bound.If|f000(t)| ≤Mfor alltbetweenxandb
then
|f(x)−T2(x)|=f(x)−[f(b) +f0(b)(x−b) +2f00(b)(x−b)2]≤6|x−b|3
The differencef(x)−T2(x) is the error in the approximation offby the second Taylor polynomial based atb third derivative measures how rapidly the second derivative. The
is changing. In the baseball example above, if we can bound how rapidly the acceleration can change:|h000(t)| ≤M, then we can bound how closely the second Taylor polynomial h(0) +h0(0)t+2h00(0)t2approximates the true heighth(t). Another important property of the second Taylor polynomial, which you can verify
by differentiation, is thatT2has the same value, the same derivative, and the same second derivative asfatb:
T2(b) =f(b) T20(b) =f0(b) T200(b) =f00(b) The second Taylor polynomialT2is the only quadratic polynomial with this property. Example 2.1.Find a bound for the error in approximating the functionf(x) = cos(x)
using the second Taylor polynomial (quadratic approximation) based atb= 0on the
intervalI= [−2 2].
The first step is to find the second Taylor polynomial based atb= 0 forf(x) = cos(x). We findf0(x) =−sin(x),f00(x) =−cos(x) and T2(x) = cos(0) + (−sin(0))(x−0) +2(−cos(0))(x−0)2= 1−2x2 y
− 2
y= 1−2x2
y= cos(x) x 2
Figure 3.cos(x) and its second Taylor polynomial based atb= 0.
8
To bound the error, we find|f000(x)|=|sin(x)| ≤1. SettingM= 1 in the Quadratic
Approximation Error Bound
|cos(x)−[1−2x2]| ≤6|x3| ≤0014
since|x| ≤02 onI we were . Ifa bit cleverer, we might have noticed that|f000(x)|=
|sin(x)| ≤ |x| ≤02 (see page 212 in Stewart) so that we could have takenM= 02 instead
ofM= 1, which gives the better error bound of 000027.
Using local information to make an estimate is something that you do every day. If
you are driving down an hill and see a pedestrian in the crosswalk, you feel the speed of
your car, the acceleration due to the hill, and the speed of the pedestrian to decide whether
or not to apply the brakes. You mentally approximate how long it will take to get to the
intersection, where the pedestrian will be when you get there, and add a margin of safety
to protect against an error in your approximation.
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§ Order Approximation and Taylor’s Inequality3. Higher
In this section we will extend the ideas of the two preceeding sections to approxima-
tions by higher degree polynomials. The ideas are the same. But first we introduce some
notation to make it easier to describe the results.
n! =n∙(n−1)∙(n−2)∙ ∙ ∙2∙1
read “nfactorial”, is the product of the firstnintegers, so that 1! = 1, 2! = 2, 3! = 6,
4! = 24 and so forth. Also
f(k)(x)
denotes thekthderivative offatx. If we integrate equation (22) by parts again, we
obtain
f(x) =f(b) +f0(b)(x−b) +2f00(b)(x−b)2+3∙2f(3)(b)(x−b)3+3∙2Rxbf(4)(t)(x−t)3dt The pattern continues (and can be proved by mathematical induction) by integrating by
parts: f(x) =f(b) +f0(b)(x−b) + +n1!f(n)(b)(x−b)n+ +n!1Zbxf(n+1)(t)(x−t)ndt(31) The dots “ the pattern continues until the term after” in the above formula mean that
the dots is reached.
Definition.ThenthTaylor polynomial forfbased atbis Tn(x) =f(b) +f0(b)(x−b)1+21f(2)(b)(x−b)2+ +n!1f(n)(b)(x−b)n ∙ There are a lot of symbols in the above formula. Keep in mind thatxis the variable. The baseb Theis a fixed number.nthTaylor polynomial has terms which are numbers
(calledcoefficients) times powers of (x−b).
Equation (3.1) gives a formula for the errorf(x)−Tn(x can bound this error in). We the same way we bounded the error whenn= 1 andn= 2 in the preceeding sections.
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Taylor’s Inequality.SupposeIis an interval containingb. If|f(n+1)(t)| ≤Mfor allt
inIthen |f(x)−Tn(x)| ≤(Mn+ 1)!|x−b|n+1 for allxinI, whereTnis thenthTaylor polynomial forfbased atb. By equation (3.1) and the definition ofTn, ifx > bthen |f(x)−Tn(x)| ≤n!1Zxb|f(n+1)(t)(x−t)n|dt ≤n1!ZxbM(x−t)ndt=(n+M(1!)x−b)n+1 The casex < bcan be proved similarly.
A few observations might be useful.
•The Tangent Line Error Bound is just Taylor’s Inequality withn= 1 and
•the Quadratic Approximation Error Bound is just Taylor’s Inequality withn= 2. •ThenthTaylor polynomialTnbased atbhas the same value asfatband the same firstnderivatives asfatb fact. InTnis the only polynomial of degreenwith this
property.
•The right-hand side in Taylor’s Inequality is similar to the last term inTn+1 has. It
the same power ofx−band the same (n but otherwise it is different.+ 1)!
Check the third observation for yourself by finding the first few derivatives ofTnand
evaluating them atb rigorous proof uses mathematical induction.. The
Example 3.1.Supposef(x) =1−x the. FindnthTaylor poly atb= 0.
We first find the derivatives off(x) = (1−x)−1:
f0(x) = (1−x)−2
f00(x) = 2(1−x)−3
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nomialTn(x)forfbased
