Mathematical Modeling Notes

Taylor Expansions
Fourier Series
Population of the Earth

PROBLEM

Consider a smooth function f(x), and its expansion near x = x₀,

S_i=0^∞ f⁽ⁱ⁾(x₀)(x - x₀)ⁱ / i!

and call E_n(x) the error made by approximating f with its Taylor expansion truncated to order n,

E_n(x) = f(x) - S_i=0ⁿ f⁽ⁱ⁾(x₀)(x - x₀)ⁱ / i!.

(a) Apply this formula to f(x) = cos(x), with x₀ = 0 and n = 5.
(b) Find a condition on |x| which ensures that |E_n(x)| < 0.05.
(c) Use a calculator or Matlab to check your answer to the previous question.

[from Math 485/585, Mathematical Modeling, Spring 2005, University of Arizona]

SOLUTION

(a) f(x) = cos x, x₀ = 0, n = 5

f(x₀) = f(0) = cos 0 = 1

f'(x) = - sin x => f'(x₀) = f'(0) = - sin 0 = 0

f"(x) = - cos x => f"(x₀) = f"(0) = - cos 0 = - 1

f⁽³⁾(x) = sin x => f⁽³⁾(x₀) = f⁽³⁾(0) = sin 0 = 0

f⁽⁴⁾(x) = cos x => f⁽⁴⁾(x₀) = f⁽⁴⁾(0) = cos 0 = 1

f⁽⁵⁾(x) = - sin x => f⁽⁵⁾(x₀) = f⁽⁵⁾(0) = - sin 0 = 0

S_i=0ⁿ f⁽ⁱ⁾(x₀)(x - x₀)ⁱ / i! = S_i=0⁵ f⁽ⁱ⁾(0)xⁱ / i!
= f(0) + f'(0)x + f"(0)x² / 2! + f⁽³⁾(0)x³ / 3! + f⁽⁴⁾(0)x⁴ / 4! + f⁽⁵⁾(0)x⁵ / 5!
= f(0) + f"(0)x² / 2! + f⁽⁴⁾(0)x⁴ / 4!
= 1 - x² / 2! + x⁴ / 4!

E_n(x) = E₅(x) = cos x - 1 + x² / 2! - x⁴ / 4!

(b) The complete Taylor expansion of f(x) = cos x near x = 0 is

cos x = 1 - x² / 2! + x⁴ / 4! - x⁶ / 6! + x⁸ / 8! - ...

Thus,

E₅(x) = - x⁶ / 6! + x⁸ / 8! - ...

or

|E₅(x)| = x⁶ / 6! - x⁸ / 8! + ... < x⁶ / 6!

|E₅(x)| will be < 0.05 if x⁶ / 6! < 0.05 => |x|⁶ / 6! <0.05 => |x|⁶ < (0.05)(6!) = 36 => |x| < 1.817121

(c) To check the result in (b), we first calculate

cos(1.817121) = - 0.24384

Then we calculate

1 - (1.817121)² / 2! + (1.817121)⁴ / 4! = - 0.19668

This differs from the exact value of cos(1.817121) by - 0.19668 + 0.24384 = 0.047157 < 0.05, so as long as |x| < 1.817121, |E₅(x)| will be less than 0.05.

PROBLEM

Show that

p = S_n=1^∞ [4/(2n-1)](-1)^3n-1

[Hint: Find a Fourier series expansion for the function f(x) = x defined for - p < x < p.]

SOLUTION

A function f(x) defined in the region - p < x < p can be written as a Fourier series of the form

f(x) = a₀/2 + S_n=1^∞ a_n cos nx + S_n=1^∞ b_n sin nx

where

a₀ = (1/p) ∫_-p^p f(x) dx
a_n = (1/p) ∫_-p^p f(x) cos nx dx
b_n = (1/p) ∫_-p^p f(x) sin nx dx

Now suppose we let f(x) = x. Then we have

x = a₀/2 + S_n=1^∞ a_n cos nx + S_n=1^∞ b_n sin nx

where

a₀ = (1/p) ∫_-p^p x dx = 0 because x is an odd function of x
a_n = (1/p) ∫_-p^p x cos nx dx = 0 because x cos nx is an odd function of x
b_n = (1/p) ∫_-p^p x sin nx dx

Let u = x and dv = sin nx dx. Then du = dx and v = - (1/n) cos nx, and

∫ x sin nx = - (x/n) cos nx + (1/n) ∫ cos nx dx = - (x/n) cos nx + (1/n²) sin nx
=> ∫_-p^p x sin nx = - (2p/n) cos np = - (2p/n)(-1)ⁿ
=> b_n = - (2/n)(-1)ⁿ + (2/n)(-1)ⁿ⁺¹
=> x = S_n=1^∞ (2/n)(-1)ⁿ⁺¹ sin nx

Let x = p/2. Then

p/2 = S_n=1^∞ (2/n)(-1)ⁿ⁺¹ sin np/2 = S_n=1,3,5^∞ (2/n)(-1)ⁿ⁺¹ sin np/2
= S_n=1,3,5^∞ (2/n)(-1)ⁿ⁺¹(-1)^(n-1)/2 = S_n=1,3,5^∞ (2/n)(-1)^(3n+1)/2

Define m = (n + 1) / 2. Then n = 2m - 1 and

p/2 = S_m=1^∞ [2/(2m-1)](-1)^3m-1 => p = S_n=1^∞[4/(2n-1)](-1)^3n-1

PROBLEM

Show that

S_n=1^∞ 1/n² = p²/6

[Hint: Find a Fourier series expansion for the function f(x) = x² defined for - p < x < p.]

SOLUTION

A function f(x) defined in the region - p < x < p can be written as a Fourier series of the form

f(x) = a₀/2 + S_n=1^∞ a_n cos nx + S_n=1^∞ b_n sin nx

where

a₀ = (1/p) ∫_-p^p f(x) dx
a_n = (1/p) ∫_-p^p f(x) cos nx dx
b_n = (1/p) ∫_-p^p f(x) sin nx dx

Now suppose we let f(x) = x². Then we have

x² = a₀/2 + S_n=1^∞ a_n cos nx + S_n=1^∞ b_n sin nx

where

a₀ = (1/p) ∫_-p^p x² dx = 2p²/3
a_n = (1/p) ∫_-p^p x² cos nx dx
b_n = (1/p) ∫_-p^p x² sin nx dx = 0 because x² sin nx is an odd function of x

Let u = x² and dv = cos nx dx. Then du = 2x dx, v = (1/n) sin nx, and

∫ x² cos nx dx = (x²/n) sin nx - (2/n) ∫ x sin nx dx

Let u = x and dv = sin nx dx. Then du = dx, v = - (1/n) cos nx, and

∫ x sin nx = - (x/n) cos nx + (1/n) ∫ cos nx dx = - (x/n) cos nx + (1/n²) sin nx
=> ∫ x² cos nx dx = (x²/n) sin nx + (2x/n²) cos nx - (2/n³) sin nx
=> ∫_-p^p x² cos nx dx = (4p/n²) cos np = (4p/n²)(-1)ⁿ
=> a_n = (4/n²)(-1)ⁿ
=> x² = p²/3 + S_n=1^∞ (4/n²)(-1)ⁿ cos nx

Let x = p. Then

p² = p²/3 + S_n=1^∞ (4/n²)(-1)ⁿ cos np = p²/3 + S_n=1^∞ (4/n²)(-1)ⁿ(-1)ⁿ = p²/3 + S_n=1^∞ (4/n²)
=> S_n=1^∞ (1/n²) = p²/6

PROBLEM

Show that S_n=1^∞ 1/n⁴ = p⁴/90.

[Hint: Find a Fourier series expansion for the function f(x) = x⁴ defined for - p < x < p.]

SOLUTION

A function f(x) defined in the interval - p < x < p. It can be written as a Fourier series of the form

f(x) = a₀/2 + S_n=1^∞ a_n cos nx + S_n=1^∞ b_n sin nx

where

a₀ = (1/p) ∫_-p^p f(x) dx
a_n = (1/p) ∫_-p^p f(x) cos nx dx, n > 1
b_n = (1/p) ∫_-p^p f(x) sin nx dx, n > 1

Now suppose we let f(x) = x⁴. Then we have

a₀ = (1/p) ∫_-p^p x⁴ dx = (1/p)(x⁵/5)|_-p^p = 2p⁴/5
a_n = (1/p) ∫_-p^p x⁴ cos nx dx
b_n = (1/p) ∫_-p^p x⁴ sin nx dx = 0 because x⁴ sin nx is an odd function of x

Let u = x⁴ and dv = cos nx dx. Then du = 4x³ dx, v = (1/n) sin nx

and

∫ x⁴ cos nx dx = (1/n) x⁴ sin nx - (4/n) ∫ x³ sin nx dx

Let u = x³ and dv = sin nx dx. Then du = 3x² dx, v = - (1/n) cos nx, and

∫ x³ sin nx dx = - (1/n) x³ cos nx + (3/n) ∫ x² cos nx dx

Let u = x² and dv = cos nx dx. Then du = 2x dx, v = (1/n) sin nx, and

∫ x² cos nx dx = (1/n) x² sin nx - (2/n) ∫ x sin nx dx

Let u = x and dv = sin nx dx. Then du = dx, v = - (1/n) cos nx, and

∫ x sin nx dx = - (1/n) x cos nx + (1/n) ∫ cos nx dx = - (1/n) x cos nx + (1/n²) sin nx
=> ∫ x² cos nx dx = (1/n) x² sin nx + (2/n²) x cos nx - (2/n³) sin nx
=> ∫ x³ sin nx dx = - (1/n) x³ cos nx + (3/n²) x² sin nx + (6/n³) x cos nx - (6/n⁴) sin nx
=> ∫ x⁴ cos nx dx = (1/n) x⁴ sin nx + (4/n²) x³ cos nx - (12/n³) x² sin nx - (24/n⁴) x cos nx +
(24/n⁵) sin nx
=> ∫_-p^p x⁴ cos nx dx = (8p³/n²) cos np - (48p/n⁴) cos np = (8p³/n²)(-1)ⁿ - (48p/n⁴)(-1)ⁿ
=> a_n = (8p²/n²)(-1)ⁿ - (48/n⁴)(-1)ⁿ
=> x⁴ = p⁴/5 + 8p² S_n=1^∞ (1/n²)(-1)ⁿ cos nx - 48 S_n=1^∞ (1/n⁴)(-1)ⁿ cos nx

Let x = p. Then

p⁴ = p⁴/5 + 8p² S_n=1^∞ (1/n²)(-1)ⁿ(-1)ⁿ - 48 S_n=1^∞ (1/n⁴)(-1)ⁿ(-1)ⁿ
= p⁴/5 + 8p² S_n=1^∞ 1/n² - 48 S_n=1^∞ 1/n⁴

It can be shown that

S_n=1^∞ 1/n² = p²/6

Thus,

p⁴ = p⁴/5 + 8p⁴/6 - 48 S_n=1^∞ 1/n⁴ = p⁴/5 + 4p⁴/3 - 48 S_n=1^∞ 1/n⁴
=> S_n=1^∞ 1/n⁴ = (p⁴/5 + 4p⁴/3 - p⁴)/48 = (1/5 + 4/3 - 1)p⁴/48 = (1/5 + 1/3)p⁴/48 = p⁴/90
=> S_n=1^∞ 1/n⁴ = p⁴/90

PROBLEM

Using the data in the following table, which shows the estimated population of the Earth at various points in time (Haub 1995, Curtin 2007), estimate the total number of people who have ever lived.

Year	Population
47994 BC1	2
8000 BC	5,000,000
1	300,000,000
1650	500,000,000
1850	1,265,000,000
2002	6,215,000,000
2007	6,500,000,000

¹50,000 years ago as of the year 2007

Curtin, Ciara 2007, "Fact or Fiction? Do Living People Outnumber the Dead?" Scientific American, September 2007, 50.
Haub, Carl 1995, "How Many People Have Ever Lived on Earth?" Population Today, February 1995.

SOLUTION

Let t be the average lifespan of a human being and assume that all human beings live this long and that the population is always evenly distributed in age from 0 to t. Then the fraction of the population which dies each year is 1/t. For example, if t = 70 years, the fraction of the population which dies each year is 1/t = 1/70 = 1.429%.

The above table contains seven data points and represents six time intervals. Let Y_i be the ith year listed in the table. Let E_i be the number of years which have elapsed between Y_i and Y_i+1, during the ith time interval. Then P_i and P_i+1 are related by

P_i+1 = P_i(1 + R_i)^E_i

where R_i is the average annual population growth rate during the ith time interval. Solving for R_i, we get

R_i = [(P_i+1/P_i)^(1/E_i)] - 1

The relationship between the annual population growth rate R_i, the annual birth rate B_i, and the annual death rate D_i is

R_i = B_i - D_i

From above, if t = 70 years, D_i = 1/t = 1/70 = 0.01429. For given values of R_i and D_i, the annual birth rate is

B_i = R_i + D_i

Thus, we can calculate the annual birth rates that are listed in the following table. The annual (birth, death) rate is the number which, when multiplied by the current population at the beginning of a particular year, gives the total number of (births, deaths) that occur during that year.

Data Point (i)	Year (Yi)	Population (Pi)	Elapsed Years (Ei)	Annual Growth Rate (Ri)	Annual Death Rate (Di)	Annual Birth Rate (Bi)
1	47994 BC	2	39994	0.000368	0.014286	0.014654
2	8000 BC	5,000,000	8000	0.000512	0.014286	0.014798
3	1	300,000,000	1649	0.000310	0.014286	0.014596
4	1650	500,000,000	200	0.004652	0.014286	0.018938
5	1850	1,265,000,000	152	0.010528	0.014286	0.024814
6	2002	6,215,000,000	5	0.009008	0.014286	0.023293
7	2007	6,500,000,000	-	-	-	-

Using the values of B_i for each of the time intervals, the number of births in each year can be calculated. Example: In the year 1850, the population of the world is estimated to be 1.265 x 10⁹. The birth rate during this time interval is 0.024814, so the number of births in 1850 is estimated to be (0.024814)(1.265 x 10⁹) = 31,389,371.

If this is done for all the years from 47994 BC to 2007, the total number of births is 34 billion. If we assume different values for the average lifespan t, we get the total number of births shown in the following table.

Average Lifespan t	Total Births
10	197,369,926,000
20	101,964,238,000
30	70,162,341,800
40	54,261,393,700
50	44,720,824,900
60	38,360,445,600
70	33,817,317,600

If we assume t = 20 years, we get a result which is close to the estimate given by Curtin (2007). The results in the table were generated by the program SciAm070824.f.