In the last section we have seen how much depends on the proximity to the expansion point of development in the application of the Taylor series, when calculating to calculate function values in the neighbourhood of expansion points for which the function is well known. Therefore, we finally turn to the problem of optimizing the expansion point x0 which till now we always have chosen to be 0.
We obtain thegeneral Taylor series around any pointx0 from our present form very simply by replacing x everywhere by y:=x−x0 and expanding the resulting function of y in the neighborhood of y = 0. Thus we get for infinitely often differentiable functions f(x) the form of the general
Taylor series around the point x0 : f(x) =
∞
P
n=0
f(n)(x0)
n! (x−x0)n,
for the convergence radius R the sufficient quotient criterion:
R= lim
n→∞
an an+1
= lim
n→∞
f(n)(x0) (n+1) fn+1(x0)
and for the error resulting from breaking off after thenm-th term: the Lagrange form of the remainder term
rm(x) = fn+1(x0+θ(x−x0))
(n+ 1)! (x−x0)n+1 again with 0< θ <1.
As an example we expand thesine functionf(x) = sinxnow around the pointx0 = π2 :
f(x) = sinx=
∞
X
n=0
(−1)n(x− π
2)2n/(2n)!
= 1−(x− π
2)2/2! + (x− π
2)4/4!±. . . To the proof:
f0(x) = cosx, f00(x) = −sinx, f000(x) = −cosx, f(4)(x) = sinx, . . . with
f(π
2) = 1, f0(π
2) = 0, f00(π
2) = −1, f000(π
2) = 0, f(4)(π
2) = 1, . . .
altogether: sinx= 1−(x−π2)2/2! + (x−π2)4/4!±. . .=
∞
P
n=0
(−1)n(x− π2)2n/(2n)!
with the convergence radiusR = lim
n→∞|(2(n+ 1))!/(2n)!|= lim
n→∞|(2n+ 1)(2n+ 2)|=∞.
We want to check immediately how the centre of expansion x0 = π2, lying nearer to 100◦, improves our earlier calculation of sin 100◦ = sin(10π/18) which we performed in Section 6.7 with the development pointx0 = 0:
sinx= 1−(x−π/2)2/2! + (x−π/2)4/4!−r4(x) : We get: sin(10π/18) = sin(π/2 +π/18) = 0.984807773−r3(10π/18), where the estimation of the rest term withθ = 0 yields
|r4(10π/18)| ≤(π/18)6/6! = 3.93·10−8.
This result must be compared with the earlier error estimation of 9.38·10−3,i.e. through the better centre of expansion the error can be reduced by more than four orders of magnitude, with comparable effort of calculation.
For other desired values in between we may develop thesine also aroundx =π/4:
f(x) = sinx= (1/√
2)[1 + (x−π/4)−(x−π/4)2/2!−(x−π/4)3/3! + +− −. . .]
since f0(x) = cosx, f00(x) = −sinx, f000(x) = −cosx, f(4)(x) = sinx, . . . withf(π/4) = 1/√
2, f0(π/4) = 1/√
2, f00(π/4) =−1/√
2, f000(π/4) =−1/√
2, f(4)(π/4) = 1/√
2,
altogether: sinx= (1/√
2)[1 + (x−π/4)−(x−π/4)2/2!−(x−π/4)3/3! + +− −. . . with the convergence radius R= lim
n→∞|(n+ 1)!/n!|= lim
n→∞|n+ 1|=∞.
Now we are able to present a series also for the natural logarithm, e.g. aroundx0 = 1 : f(x) = lnx=
∞
P
n=1
(−1)n+1(x−1)n/n= (x−1)−(x−1)2/2 + (x−1)3/3 +. . .
If we replace here x by x+ 1, we arrive again at our earlier Taylor series for ln(x+ 1) around the point 0.
Exercise 6.12 Determine the Taylor series and the convergence regions for:
1) sinx around point x0 =π, 2) ex around point x0 = 1, 3) ex around point x0 = 2 and
4) prove the series for lnx around the point x0 = 1 given above.
Exercise 6.13 Calculate once more √4
e3 up to r3,but now using the Taylor series around x0 = 1 and compare with our earlier calculation around x0 = 0 in Exercise 6.11.
You will get to know and carry out many more series expansions during your studies.
There are expansion series also for fields, for instance the famous multipole series. You will expand the frequently used periodic functions “after Fourier in the quadratic mean”
with cosine and sine as basis functions, and later on also transform non-periodic functions
“after Fourier” with help of the exponential functions. Complex functions can be developed
“after Laurent” even in the neighbourhood of certain singularities. Finally in quantum mechanics, you will perform several perturbation theoretical expansions around the few solvable systems like the harmonic oscillator. Theoretical physics is in large part the high art of dealing with series expansions.
Chapter 7
INTEGRATION
The second main pillar in mathematical methods for all natural sciences is the integral calculus. In some sense integration is the inverse of differentiation. While differentiation assigns to a function f(x) its gradient f0(x) the integral calculus deals with problems in which something is known about the gradient of a function and other functions having this gradient are sought. This assignment is much more difficult than differentiation, it has however a central meaning for natural science. It turns out that the basic laws of mechanics, electrodynamics and quantum mechanics can be formulated as differential equationswhich make statements about the derivatives of functions. For example, New- ton’s second law makes a statement about the path or trajectoryx(t) of a particle under the influence of a force K, and makes this statement using the second derivative of x(t), i.e. the acceleration:
m(dtd)2x(t) =K.
For a given force, e.g. gravity or the electric or magnetic force, a particle trajectory is to be determined. To integrate in this context means to solve the basic equations, applying the theory to the various cases which are encountered in practice.
In previous chapters you have learned and practiced the technique of differentiation. It is actually not that difficult, if you follow some rules. Integration however is an “art”, as you will soon see. However, since no artist ever appeared out of nowhere, but rather had to develop his talent through learning, gaining experience and practice, the same goes for integration. The following chapters will give you plenty of opportunity to do this.
7.1 Work
First we take another look at the uniform motion of a mass point on a straight line, for example the centre of mass of a car on a highway. After we have answered our first
question for the velocity with help of the differential calculus, we look as physicists to one level deeper for the reason: Why does the car move on this straight plane highway section with the observed uniform speed? Obviously, the reason for the uniform straight motion according to Newton is a force, namely the force K, with which the engine pushes the car against inertia, wind and other frictional loss. Latest at the gas station thequestion ariseshow much work A has the engine done over the travelled distance ∆x. The work done is proportional to the needed force was, and proportional to the travelled distance
∆x. In fact it is exactly equal to the product of both quantities: A=K∆x,geometrically equal to therectangular area K times ∆x,if we draw in a Cartesian (i.e. right angled) coordinate system with the travelled distance x in the direction of the 1-axis and the operating forceK in direction of the 2-axis.
Figure 7.1: Constant force K as a function of the travelled distance from a to b =a+ ∆x. The yellow rectangular area is proportional to the workA =K∆x.
Following this idealized case of a constant operating force we turn our attention to a more realistic case, where the force over the distance is increased linearly from an initial value K0 by giving more gas: K(x) = sx+K0 :
From the figure we immediately see how we can help ourselves, if the question arises (while filling up the gas tank) how much work was done over the entire distance. We take the mean value ξ= (a+b)/2 between the starting point a and the end point b, read out the corresponding function value of the force K(ξ) = K((a+b)/2), and multiply this with the travelled distance ∆x=b−a:
A=K(ξ)∆x=K(a+b2 )(b−a).
Figure 7.2: Force rising linearly with distance from the initial valueK0 to K(x) =sx+K0 between the points a and b
In this way we reduce the physical question for the work to the geometrical question for the “area under the force line over the interval ∆x”, more precisely for the content of the area which is bordered on the top by a known function, to the left by the straight line x = a, to the right by the line x = b, and on the bottom by the 1-axis K(x) = 0.
The advanced question for the work done by a force varying according to an arbitrary function is herewith traced back to the mathematical problem of the determination of area of a rectangle, in which one side (in this case the upper one) is replaced by a curve.
In the following we would like to explore this more general mathematical question and to give an answer which will fulfill all wishes of physicists concerning work, and many other wishes beyond.