Applications to Physics, page 778

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愛因斯坦的狹義相 對論與牛頓力學的 關係也可以用泰勒 級 數 的 方 法 理 解_: 愛因斯坦理論在微 觀尺度下與牛頓力 學相當。

Example 4(page 778). In Einstein’s theory of special relativity the mass of an object moving with velocity v is

m = m₀ q

1 −^vc²²

,

where m₀ is the mass of the object when at rest and c is the speed of light. The kinetic energy of the object is the difference between its total energy and its energy at rest: K = mc²−m0c². (a) Show that when v is very small compared with c, this expression for K agrees with

classical Newtonian physics: K = ¹₂m₀v².

(b) Use Taylor’s Inequality to estimate the difference in these expressions for K when

|v| ≤ 100 m/s.

Solution.

(a) Using the expressions given for K and m, we get

K = mc²− m0c² = m0c² q

1 −^vc²²

− m0c² = m₀c²

 1 − v²

c²

₋¹₂

− 1

! .

With x = −^vc²², the Maclaurin series for (1 + x)⁻¹² is a binomial series with m = −¹₂. Therefore we have

(1 + x)⁻¹² = 1 −1

2x + −¹₂

−³₂

2! x²+ −¹₂

−³₂

−⁵₂

3! x³+ · · ·

= 1 −1 2x +3

8x²− 5

16x³+ · · · , and

K = m₀c²



1 +1 2

v² c² +3

8 v⁴ c⁴ + 5

16 v⁶ c⁶ + · · ·



− 1



= m₀c² 1 2

v² c² +3

8 v⁴ c⁴ + 5

16 v⁶ c⁶ + · · ·

 . If v is much smaller than c, then all terms after the first are very small when compared with the first term. If we omit them, we get

K = m₀c² 1 2

v² c²



= 1 2m₀v².

(b) Let f (x) = m₀c²

(1 + x)⁻¹² − 1

with x = −^vc²². We can use Taylor’s Inequality to write

r₁(x) = f^′′(˜c)

2! x², where −v²

c² ≤ ˜c ≤ 0.

40 11.11 Applications of Taylor Polynomials goo.gl/SbJNXE

Since f^′′(x) = ³₄m₀c²(1 + x)⁻⁵² and we are given that |v| ≤ 100 m/s, so

|f^′′(˜c)| = 3m₀c²

4 (1 + ˜c)⁵² ≤ 3m₀c² 4 1 −¹⁰⁰c²²

⁵₂. Thus, with c = 3 · 10⁸m/s,

|r1(x)| = 1

2 · 3m₀c²

4 1 −¹⁰⁰_c^22
5₂ ·100⁴

c⁴ < (4.17 · 10⁻¹⁰)m₀.

So when |v| ≤ 100 m/s, the magnitude of the error in using the Newtonian expression for kinetic energy is at most (4.17 · 10⁻¹⁰)m₀.

Appendix

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這個例子在數學上 的發展很重要_, 這 個函數的建構告知 其泰勒級數與原函 數除了中心以外都 不相等。 由_(a)到 (d)的討論知道這 個函數的馬克勞林 級數是恆為零的函 數。 既然如此_, 則 不論_n為何_,餘項 和原函數一樣_, 所 以餘項在非中心點 的地方不可能趨近 於零。

Example 5. Consider the function f (x) =

( e^−x21 if x 6= 0 0 if x = 0 . (a) The function f (x) is continuous on R because

x→0lime^−x21 = lim

y→±∞e^−y2 = lim

y→±∞

1

e^y2 = 0 = f (0),

and for x 6= 0, f (x) is a composition of two continuous functions g(x) = e^x and h(x) =

−_x¹², that is, f (x) = (g ◦ h)(x).

(b) We will show that: For x 6= 0, f⁽ⁿ⁾(x) = Pn(y)e^−y2, where y = ¹_x, and Pn(y) is a poly-nomial of y with degree 3n.

(1) When n = 1, we compute f^′(x) = df

dx = df dy

dy

dx = e^−y2(−2y) · (−y²) = 2y³e^−y2 = P₁(y)e^−y2, where P₁(y) = 2y³ is a polynomial of y with degree 3.

(2) Assume that it is true for n = k, that is, f^(k)(x) = ^d_dx^kfk = Pk(y)e^−y2, where Pk(y) is a polynomial with degree 3k.

(3) When n = k + 1, we compute f^(k+1)(x) = d^k+1f

dx^k+1 = d dx

d^kf dx^k = d

dy

 d^kf dx^k

 dy dx = d

dy



Pk(y)e^−y2 (−y²)

= dP_k(y)

dy e^−y2+ P_k(y)e^−y2(−2y)

 (−y²)

=



−y²dP_k(y)

dy + 2y³P_k(y)

 e^−y2.

Let P_k+1(y) = −y^{2 dP}_dy^k(y) + 2y³Pk(y), which is a polynomial of y with degree 3 + 3k = 3(k + 1).

11.11 Applications of Taylor Polynomials goo.gl/SbJNXE 41

(4) By mathematical induction, we know that for x 6= 0, f⁽ⁿ⁾(x) = Pn(y)e^−y2, where y = ¹_x, and Pn(y) is a polynomial of y with degree 3n.

(c) Now, we will show that f⁽ⁿ⁾(0) = 0 for all n ∈ N.

(1) When n = 1, we compute

f^′(0) = lim

x→0

f (x) − f (0) x − 0 = lim

x→0

e^−x21

x = lim

y→±∞

e^−y2

1 y

= lim

y→±∞

y e^y2

(^∞∞),L^′

=== lim

y→±∞

1

2ye^y2 = 0.

(2) Assume that it is true for n = k, that is, f^(k)(0) = 0.

(3) When n = k + 1, we compute f^(k+1)(0) = lim

x→0

f^(k)(x) − f^(k)(0)

x − 0 = lim

x→0

f^(k)(x)

x = lim

y→±∞

P_k(y)e^−y2

1 y

= lim

y→±∞

yP_k(y) e^y2 = 0.

Remark that we can apply l’ Hospital Rule_3n−1

2  times to get the limit is 0.

(4) By mathematical induction, we know that f⁽ⁿ⁾(0) = 0 for all n ∈ N.

(d) Since f (0) = 0 and f⁽ⁿ⁾(0) = 0 for all n ∈ N, the Maclaurin series of f (x) is

M (x) =

∞

X

n=0

f⁽ⁿ⁾(0)

n! xⁿ= f (0) + f^′(0)

1! x +f^′′(0)

2! x²+ · · · +f⁽ⁿ⁾(0)

n! xⁿ+ · · · = 0.

This is a zero function, so the interval of convergence of M (x) is R. We compute the remainder

r_n(x) = f (x) − Tn(x) = f (x).

We get for any x 6= 0, lim

n→∞r_n(x) = e^−x21 6= 0. Therefore, f (x) is not equal to its Maclaurin series.

這裡引進集合符號 C^k(R),表示k次 求導之後仍連續的 函 數 所 成之集 合。

而 C^∞(R) 的元 素是不論微分幾次 函數都連續。 至於 能夠用泰勒級數重 新表示的函數稱為 解析函數_, 集合以 C^ω(R)表示。

(e) For any integer k ≥ 0, let C^k(R) be the set (in fact, it is a vector space) consisting of all functions f (x) that the derivatives f^′(x), f^′′(x), . . . , f^(k)(x) exist and are continuous on R. So C⁰(R), which is also denoted by C(R), consists of all continuous functions on R, and C^∞(R) = ∩^∞k=0C^k(Ω) consists of all smooth functions (continuous derivatives of all orders) on R (光滑函數_).

Denote C^ω(R) be the set consisting of all smooth functions f (x) that for all x ∈ R, there exists R > 0 such that f (x) equals its Taylor series expansion on (x − R, x + R).

We say a function f (x) ∈ C^ω(R) is analytic (解析函數).

42 11.11 Applications of Taylor Polynomials goo.gl/SbJNXE

(f) The above discussion shows that the function f (x) is a smooth function, but not an analytic function because f (x) is not analytic at x = 0. So the conclusion is C^ω(R) ( C^∞(R).

這個例子告知: 存 在光滑函數並非解 析函數。

Remark that we have the following relations:

C^ω(R) ( C^∞(R) · · · ( C²(R) ( C¹(R) ( C⁰(R).

Example 6. Recall that the binomial series is

∞

We will check the convergence of the binomial series at the endpoints.

(a) If m ≤ −1, then

n! is an alternating series. We compute

|Cn^m| = so it is a decreasing sequence. Next, we calculate

|Cn^m| =

11.11 Applications of Taylor Polynomials goo.gl/SbJNXE 43

By the Alternating Series Test, P^∞

n=0

(c) Before we check the case m > 0, we introduce the Raabe’s Test:

The Raabe’s Test. Suppose a series P^∞

n=1

a_n satisfies

n→∞lim then the series is absolutely convergent.

Remark that the p-series P^∞

n=1 1

n^p satisfies the condition, so the Raabe’s Test is a Com-parison Test with p-series.

If m > 0, then

44 11.11 Applications of Taylor Polynomials goo.gl/SbJNXE

Example 7. We will prove (1 + x)^m= P^∞

n=0

C_n^mxⁿ on |x| < 1.

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這裡提供二項式函 數及其泰勒級數的 相等之證明。 注意 到這邊不是透過驗 證餘項趨近於零的 方式, 而是用微分 方程的方法處理。

(a) Let g(x) = P^∞

n=0

C_n^mxⁿ on the interval of convergence (−1, 1). We will show that (1 + x)g^′(x) = mg(x) on the interval of convergence (−1, 1).

We compute g^′(x) = P^∞

n=1

C_n^mnxⁿ⁻¹ on the interval of convergence (−1, 1), and

(1 + x)g^′(x) = (1 + x) X∞ n=1

C_n^mnxⁿ⁻¹= X∞ n=1

C_n^mnxⁿ⁻¹+ X∞ n=1

C_n^mnxⁿ

=

∞

X

n=0

C_n+1^m (n + 1)xⁿ+

∞

X

n=0

C_n^mnxⁿ

= X∞ n=0

m(m − 1)(m − 2) · · · (m − n + 1)(m − n)(n + 1)

(n + 1)! xⁿ

+

∞

X

n=0

m(m − 1)(m − 2) · · · (m − n + 1)n

n! xⁿ

= X∞ n=0

m(m − 1)(m − 2) · · · (m − n + 1)((m − n) + n)

n! xⁿ

= m

∞

X

n=0

C_n^mxⁿ= mg(x).

(b) Solve the differential equation (1 + x)g^′(x) = mg(x), g(0) = 1, |x| < 1. It is separable equation, so we have

g^′(x)

g(x) = m

1 + x ⇒ d

dx(ln g(x)) = m

1 + x ⇒ ln g(x) = m ln(1 + x) + C.

Since g(0) = 1, we know that C = 0. Hence ln g(x) = m ln(1 + x) = ln(1 + x)^m and it implies g(x) = P^∞

n=0

C_n^mxⁿ= (1 + x)^m on |x| < 1.

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